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http://discuss.tlapl.us/msg00587.html
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# Next-state involving existential quantifier using PlusCal?
Hello,
I'm trying to specify a cache consistency protocol in PlusCal, and I'm having trouble with the step "the writer writes an arbitrary value to the database".
In TLA+, I would simply include the following conjunct:
• \E value \in Values \cup {Deleted} : db_value' = value
Is there a way to generate the same result in PlusCal? I already tried the following:
• \E value \in Values \cup {Deleted} : db_value := value;
• This seems to generate \E' = [\E EXCEPT !value \in Values \cup {Deleted} : db_value = value]
• db_value := CHOOSE value \in Values \cup {Deleted} : TRUE;
• The semantics of using CHOOSE seems to be different than the semantics of using an existential quantifier
Thanks!
-Elliott
|
2019-08-25 08:10:21
|
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https://stacks.math.columbia.edu/tag/089X
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## 33.35 Coherent sheaves on projective space
In this section we prove some results on the cohomology of coherent sheaves on $\mathbf{P}^ n$ over a field which can be found in [Mum]. These will be useful later when discussing Quot and Hilbert schemes.
### 33.35.1 Preliminaries
Let $k$ be a field, $n \geq 1$, $d \geq 1$, and let $s \in \Gamma (\mathbf{P}_ k^ n, \mathcal{O}(d))$ be a nonzero section. In this section we will write $\mathcal{O}(d)$ for the $d$th twist of the structure sheaf on projective space (Constructions, Definitions 27.10.1 and 27.13.2). Since $\mathbf{P}^ n_ k$ is a variety this section is regular, hence $s$ is a regular section of $\mathcal{O}(d)$ and defines an effective Cartier divisor $H = Z(s) \subset \mathbf{P}^ n_ k$, see Divisors, Section 31.13. Such a divisor $H$ is called a hypersurface and if $d = 1$ it is called a hyperplane.
Lemma 33.35.2. Let $k$ be a field. Let $n \geq 1$. Let $i : H \to \mathbf{P}^ n_ k$ be a hyperplane. Then there exists an isomorphism
$\varphi : \mathbf{P}^{n - 1}_ k \longrightarrow H$
such that $i^*\mathcal{O}(1)$ pulls back to $\mathcal{O}(1)$.
Proof. We have $\mathbf{P}^ n_ k = \text{Proj}(k[T_0, \ldots , T_ n])$. The section $s$ corresponds to a homogeneous form in $T_0, \ldots , T_ n$ of degree $1$, see Cohomology of Schemes, Section 30.8. Say $s = \sum a_ i T_ i$. Constructions, Lemma 27.13.7 gives that $H = \text{Proj}(k[T_0, \ldots , T_ n]/I)$ for the graded ideal $I$ defined by setting $I_ d$ equal to the kernel of the map $\Gamma (\mathbf{P}^ n_ k, \mathcal{O}(d)) \to \Gamma (H, i^*\mathcal{O}(d))$. By our construction of $Z(s)$ in Divisors, Definition 31.14.8 we see that on $D_{+}(T_ j)$ the ideal of $H$ is generated by $\sum a_ i T_ i/T_ j$ in the polynomial ring $k[T_0/T_ j, \ldots , T_ n/T_ j]$. Thus it is clear that $I$ is the ideal generated by $\sum a_ i T_ i$. Note that
$k[T_0, \ldots , T_ n]/I = k[T_0, \ldots , T_ n]/(\sum a_ i T_ i) \cong k[S_0, \ldots , S_{n - 1}]$
as graded rings. For example, if $a_ n \not= 0$, then mapping $S_ i$ equal to the class of $T_ i$ works. We obtain the desired isomorphism by functoriality of $\text{Proj}$. Equality of twists of structure sheaves follows for example from Constructions, Lemma 27.11.5. $\square$
Lemma 33.35.3. Let $k$ be an infinite field. Let $n \geq 1$. Let $\mathcal{F}$ be a coherent module on $\mathbf{P}^ n_ k$. Then there exist a nonzero section $s \in \Gamma (\mathbf{P}^ n_ k, \mathcal{O}(1))$ and a short exact sequence
$0 \to \mathcal{F}(-1) \to \mathcal{F} \to i_*\mathcal{G} \to 0$
where $i : H \to \mathbf{P}^ n_ k$ is the hyperplane $H$ associated to $s$ and $\mathcal{G} = i^*\mathcal{F}$.
Proof. The map $\mathcal{F}(-1) \to \mathcal{F}$ comes from Constructions, Equation (27.10.1.2) with $n = 1$, $m = -1$ and the section $s$ of $\mathcal{O}(1)$. Let's work out what this map looks like if we restrict it to $D_{+}(T_0)$. Write $D_{+}(T_0) = \mathop{\mathrm{Spec}}(k[x_1, \ldots , x_ n])$ with $x_ i = T_ i/T_0$. Identify $\mathcal{O}(1)|_{D_{+}(T_0)}$ with $\mathcal{O}$ using the section $T_0$. Hence if $s = \sum a_ iT_ i$ then $s|_{D_{+}(T_0)} = a_0 + \sum a_ ix_ i$ with the identification chosen above. Furthermore, suppose $\mathcal{F}|_{D_{+}(T_0)}$ corresponds to the finite $k[x_1, \ldots , x_ n]$-module $M$. Via the identification $\mathcal{F}(-1) = \mathcal{F} \otimes \mathcal{O}(-1)$ and our chosen trivialization of $\mathcal{O}(1)$ we see that $\mathcal{F}(-1)$ corresponds to $M$ as well. Thus restricting $\mathcal{F}(-1) \to \mathcal{F}$ to $D_{+}(T_0)$ gives the map
$M \xrightarrow {a_0 + \sum a_ ix_ i} M$
To see that the arrow is injective, it suffices to pick $a_0 + \sum a_ ix_ i$ outside any of the associated primes of $M$, see Algebra, Lemma 10.63.9. By Algebra, Lemma 10.63.5 the set $\text{Ass}(M)$ of associated primes of $M$ is finite. Note that for $\mathfrak p \in \text{Ass}(M)$ the intersection $\mathfrak p \cap \{ a_0 + \sum a_ i x_ i\}$ is a proper $k$-subvector space. We conclude that there is a finite family of proper sub vector spaces $V_1, \ldots , V_ m \subset \Gamma (\mathbf{P}^ n_ k, \mathcal{O}(1))$ such that if we take $s$ outside of $\bigcup V_ i$, then multiplication by $s$ is injective over $D_{+}(T_0)$. Similarly for the restriction to $D_{+}(T_ j)$ for $j = 1, \ldots , n$. Since $k$ is infinite, a finite union of proper sub vector spaces is never equal to the whole space, hence we may choose $s$ such that the map is injective. The cokernel of $\mathcal{F}(-1) \to \mathcal{F}$ is annihilated by $\mathop{\mathrm{Im}}(s : \mathcal{O}(-1) \to \mathcal{O})$ which is the ideal sheaf of $H$ by Divisors, Definition 31.14.8. Hence we obtain $\mathcal{G}$ on $H$ using Cohomology of Schemes, Lemma 30.9.8. $\square$
Remark 33.35.4. Let $k$ be an infinite field. Let $n \geq 1$. Given a finite number of coherent modules $\mathcal{F}_ i$ on $\mathbf{P}^ n_ k$ we can choose a single $s \in \Gamma (\mathbf{P}^ n_ k, \mathcal{O}(1))$ such that the statement of Lemma 33.35.3 works for each of them. To prove this, just apply the lemma to $\bigoplus \mathcal{F}_ i$.
Remark 33.35.5. In the situation of Lemmas 33.35.2 and 33.35.3 we have $H \cong \mathbf{P}^{n - 1}_ k$ with Serre twists $\mathcal{O}_ H(d) = i^*\mathcal{O}_{\mathbf{P}^ n_ k}(d)$. For every $d \in \mathbf{Z}$ we have a short exact sequence
$0 \to \mathcal{F}(d - 1) \to \mathcal{F}(d) \to i_*(\mathcal{G}(d)) \to 0$
Namely, tensoring by $\mathcal{O}_{\mathbf{P}^ n_ k}(d)$ is an exact functor and by the projection formula (Cohomology, Lemma 20.52.2) we have $i_*(\mathcal{G}(d)) = i_*\mathcal{G} \otimes \mathcal{O}_{\mathbf{P}^ n_ k}(d)$. We obtain corresponding long exact sequences
$H^ i(\mathbf{P}^ n_ k, \mathcal{F}(d - 1)) \to H^ i(\mathbf{P}^ n_ k, \mathcal{F}(d)) \to H^ i(H, \mathcal{G}(d)) \to H^{i + 1}(\mathbf{P}^ n_ k, \mathcal{F}(d - 1))$
This follows from the above and the fact that we have $H^ i(\mathbf{P}^ n_ k, i_*\mathcal{G}(d)) = H^ i(H, \mathcal{G}(d))$ by Cohomology of Schemes, Lemma 30.2.4 (closed immersions are affine).
### 33.35.6 Regularity
Here is the definition.
Definition 33.35.7. Let $k$ be a field. Let $n \geq 0$. Let $\mathcal{F}$ be a coherent sheaf on $\mathbf{P}^ n_ k$. We say $\mathcal{F}$ is $m$-regular if
$H^ i(\mathbf{P}^ n_ k, \mathcal{F}(m - i)) = 0$
for $i = 1, \ldots , n$.
Note that $\mathcal{F} = \mathcal{O}(d)$ is $m$-regular if and only if $d \geq -m$. This follows from the computation of cohomology groups in Cohomology of Schemes, Equation (30.8.1.1). Namely, we see that $H^ n(\mathbf{P}^ n_ k, \mathcal{O}(d)) = 0$ if and only if $d \geq -n$.
Lemma 33.35.8. Let $k'/k$ be an extension of fields. Let $n \geq 0$. Let $\mathcal{F}$ be a coherent sheaf on $\mathbf{P}^ n_ k$. Let $\mathcal{F}'$ be the pullback of $\mathcal{F}$ to $\mathbf{P}^ n_{k'}$. Then $\mathcal{F}$ is $m$-regular if and only if $\mathcal{F}'$ is $m$-regular.
Proof. This is true because
$H^ i(\mathbf{P}^ n_{k'}, \mathcal{F}') = H^ i(\mathbf{P}^ n_ k, \mathcal{F}) \otimes _ k k'$
by flat base change, see Cohomology of Schemes, Lemma 30.5.2. $\square$
Lemma 33.35.9. In the situation of Lemma 33.35.3, if $\mathcal{F}$ is $m$-regular, then $\mathcal{G}$ is $m$-regular on $H \cong \mathbf{P}^{n - 1}_ k$.
Proof. Recall that $H^ i(\mathbf{P}^ n_ k, i_*\mathcal{G}) = H^ i(H, \mathcal{G})$ by Cohomology of Schemes, Lemma 30.2.4. Hence we see that for $i \geq 1$ we get
$H^ i(\mathbf{P}^ n_ k, \mathcal{F}(m - i)) \to H^ i(H, \mathcal{G}(m - i)) \to H^{i + 1}(\mathbf{P}^ n_ k, \mathcal{F}(m - 1 - i))$
by Remark 33.35.5. The lemma follows. $\square$
Lemma 33.35.10. Let $k$ be a field. Let $n \geq 0$. Let $\mathcal{F}$ be a coherent sheaf on $\mathbf{P}^ n_ k$. If $\mathcal{F}$ is $m$-regular, then $\mathcal{F}$ is $(m + 1)$-regular.
Proof. We prove this by induction on $n$. If $n = 0$ every sheaf is $m$-regular for all $m$ and there is nothing to prove. By Lemma 33.35.8 we may replace $k$ by an infinite overfield and assume $k$ is infinite. Thus we may apply Lemma 33.35.3. By Lemma 33.35.9 we know that $\mathcal{G}$ is $m$-regular. By induction on $n$ we see that $\mathcal{G}$ is $(m + 1)$-regular. Considering the long exact cohomology sequence associated to the sequence
$0 \to \mathcal{F}(m - i) \to \mathcal{F}(m + 1 - i) \to i_*\mathcal{G}(m + 1 - i) \to 0$
as in Remark 33.35.5 the reader easily deduces for $i \geq 1$ the vanishing of $H^ i(\mathbf{P}^ n_ k, \mathcal{F}(m + 1 - i))$ from the (known) vanishing of $H^ i(\mathbf{P}^ n_ k, \mathcal{F}(m - i))$ and $H^ i(\mathbf{P}^ n_ k, \mathcal{G}(m + 1 - i))$. $\square$
Lemma 33.35.11. Let $k$ be a field. Let $n \geq 0$. Let $\mathcal{F}$ be a coherent sheaf on $\mathbf{P}^ n_ k$. If $\mathcal{F}$ is $m$-regular, then the multiplication map
$H^0(\mathbf{P}^ n_ k, \mathcal{F}(m)) \otimes _ k H^0(\mathbf{P}^ n_ k, \mathcal{O}(1)) \longrightarrow H^0(\mathbf{P}^ n_ k, \mathcal{F}(m + 1))$
is surjective.
Proof. Let $k'/k$ be an extension of fields. Let $\mathcal{F}'$ be as in Lemma 33.35.8. By Cohomology of Schemes, Lemma 30.5.2 the base change of the linear map of the lemma to $k'$ is the same linear map for the sheaf $\mathcal{F}'$. Since $k \to k'$ is faithfully flat it suffices to prove the lemma over $k'$, i.e., we may assume $k$ is infinite.
Assume $k$ is infinite. We prove the lemma by induction on $n$. The case $n = 0$ is trivial as $\mathcal{O}(1) \cong \mathcal{O}$ is generated by $T_0$. For $n > 0$ apply Lemma 33.35.3 and tensor the sequence by $\mathcal{O}(m + 1)$ to get
$0 \to \mathcal{F}(m) \xrightarrow {s} \mathcal{F}(m + 1) \to i_*\mathcal{G}(m + 1) \to 0$
see Remark 33.35.5. Let $t \in H^0(\mathbf{P}^ n_ k, \mathcal{F}(m + 1))$. By induction the image $\overline{t} \in H^0(H, \mathcal{G}(m + 1))$ is the image of $\sum g_ i \otimes \overline{s}_ i$ with $\overline{s}_ i \in \Gamma (H, \mathcal{O}(1))$ and $g_ i \in H^0(H, \mathcal{G}(m))$. Since $\mathcal{F}$ is $m$-regular we have $H^1(\mathbf{P}^ n_ k, \mathcal{F}(m - 1)) = 0$, hence long exact cohomology sequence associated to the short exact sequence
$0 \to \mathcal{F}(m - 1) \xrightarrow {s} \mathcal{F}(m) \to i_*\mathcal{G}(m) \to 0$
shows we can lift $g_ i$ to $f_ i \in H^0(\mathbf{P}^ n_ k, \mathcal{F}(m))$. We can also lift $\overline{s}_ i$ to $s_ i \in H^0(\mathbf{P}^ n_ k, \mathcal{O}(1))$ (see proof of Lemma 33.35.2 for example). After substracting the image of $\sum f_ i \otimes s_ i$ from $t$ we see that we may assume $\overline{t} = 0$. But this exactly means that $t$ is the image of $f \otimes s$ for some $f \in H^0(\mathbf{P}^ n_ k, \mathcal{F}(m))$ as desired. $\square$
Lemma 33.35.12. Let $k$ be a field. Let $n \geq 0$. Let $\mathcal{F}$ be a coherent sheaf on $\mathbf{P}^ n_ k$. If $\mathcal{F}$ is $m$-regular, then $\mathcal{F}(m)$ is globally generated.
Proof. For all $d \gg 0$ the sheaf $\mathcal{F}(d)$ is globally generated. This follows for example from the first part of Cohomology of Schemes, Lemma 30.14.1. Pick $d \geq m$ such that $\mathcal{F}(d)$ is globally generated. Choose a basis $f_1, \ldots , f_ r \in H^0(\mathbf{P}^ n_ k, \mathcal{F})$. By Lemma 33.35.11 every element $f \in H^0(\mathbf{P}^ n_ k, \mathcal{F}(d))$ can be written as $f = \sum P_ if_ i$ for some $P_ i \in k[T_0, \ldots , T_ n]$ homogeneous of degree $d - m$. Since the sections $f$ generate $\mathcal{F}(d)$ it follows that the sections $f_ i$ generate $\mathcal{F}(m)$. $\square$
### 33.35.13 Hilbert polynomials
The following lemma will be made obsolete by the more general Lemma 33.45.1.
Lemma 33.35.14. Let $k$ be a field. Let $n \geq 0$. Let $\mathcal{F}$ be a coherent sheaf on $\mathbf{P}^ n_ k$. The function
$d \longmapsto \chi (\mathbf{P}^ n_ k, \mathcal{F}(d))$
is a polynomial.
Proof. We prove this by induction on $n$. If $n = 0$, then $\mathbf{P}^ n_ k = \mathop{\mathrm{Spec}}(k)$ and $\mathcal{F}(d) = \mathcal{F}$. Hence in this case the function is constant, i.e., a polynomial of degree $0$. Assume $n > 0$. By Lemma 33.33.4 we may assume $k$ is infinite. Apply Lemma 33.35.3. Applying Lemma 33.33.2 to the twisted sequences $0 \to \mathcal{F}(d - 1) \to \mathcal{F}(d) \to i_*\mathcal{G}(d) \to 0$ we obtain
$\chi (\mathbf{P}^ n_ k, \mathcal{F}(d)) - \chi (\mathbf{P}^ n_ k, \mathcal{F}(d - 1)) = \chi (H, \mathcal{G}(d))$
See Remark 33.35.5. Since $H \cong \mathbf{P}^{n - 1}_ k$ by induction the right hand side is a polynomial. The lemma is finished by noting that any function $f : \mathbf{Z} \to \mathbf{Z}$ with the property that the map $d \mapsto f(d) - f(d - 1)$ is a polynomial, is itself a polynomial. We omit the proof of this fact (hint: compare with Algebra, Lemma 10.58.5). $\square$
Definition 33.35.15. Let $k$ be a field. Let $n \geq 0$. Let $\mathcal{F}$ be a coherent sheaf on $\mathbf{P}^ n_ k$. The function $d \mapsto \chi (\mathbf{P}^ n_ k, \mathcal{F}(d))$ is called the Hilbert polynomial of $\mathcal{F}$.
The Hilbert polynomial has coefficients in $\mathbf{Q}$ and not in general in $\mathbf{Z}$. For example the Hilbert polynomial of $\mathcal{O}_{\mathbf{P}^ n_ k}$ is
$d \longmapsto {d + n \choose n} = \frac{d^ n}{n!} + \ldots$
This follows from the following lemma and the fact that
$H^0(\mathbf{P}^ n_ k, \mathcal{O}_{\mathbf{P}^ n_ k}(d)) = k[T_0, \ldots , T_ n]_ d$
(degree $d$ part) whose dimension over $k$ is ${d + n \choose n}$.
Lemma 33.35.16. Let $k$ be a field. Let $n \geq 0$. Let $\mathcal{F}$ be a coherent sheaf on $\mathbf{P}^ n_ k$ with Hilbert polynomial $P \in \mathbf{Q}[t]$. Then
$P(d) = \dim _ k H^0(\mathbf{P}^ n_ k, \mathcal{F}(d))$
for all $d \gg 0$.
Proof. This follows from the vanishing of cohomology of high enough twists of $\mathcal{F}$. See Cohomology of Schemes, Lemma 30.14.1. $\square$
### 33.35.17 Boundedness of quotients
In this subsection we bound the regularity of quotients of a given coherent sheaf on $\mathbf{P}^ n$ in terms of the Hilbert polynomial.
Lemma 33.35.18. Let $k$ be a field. Let $n \geq 0$. Let $r \geq 1$. Let $P \in \mathbf{Q}[t]$. There exists an integer $m$ depending on $n$, $r$, and $P$ with the following property: if
$0 \to \mathcal{K} \to \mathcal{O}^{\oplus r} \to \mathcal{F} \to 0$
is a short exact sequence of coherent sheaves on $\mathbf{P}^ n_ k$ and $\mathcal{F}$ has Hilbert polynomial $P$, then $\mathcal{K}$ is $m$-regular.
Proof. We prove this by induction on $n$. If $n = 0$, then $\mathbf{P}^ n_ k = \mathop{\mathrm{Spec}}(k)$ and any coherent module is $0$-regular and any surjective map is surjective on global sections. Assume $n > 0$. Consider an exact sequence as in the lemma. Let $P' \in \mathbf{Q}[t]$ be the polynomial $P'(t) = P(t) - P(t - 1)$. Let $m'$ be the integer which works for $n - 1$, $r$, and $P'$. By Lemmas 33.35.8 and 33.33.4 we may replace $k$ by a field extension, hence we may assume $k$ is infinite. Apply Lemma 33.35.3 to the coherent sheaf $\mathcal{F}$. The Hilbert polynomial of $\mathcal{F}' = i^*\mathcal{F}$ is $P'$ (see proof of Lemma 33.35.14). Since $i^*$ is right exact we see that $\mathcal{F}'$ is a quotient of $\mathcal{O}_ H^{\oplus r} = i^*\mathcal{O}^{\oplus r}$. Thus the induction hypothesis applies to $\mathcal{F}'$ on $H \cong \mathbf{P}^{n - 1}_ k$ (Lemma 33.35.2). Note that the map $\mathcal{K}(-1) \to \mathcal{K}$ is injective as $\mathcal{K} \subset \mathcal{O}^{\oplus r}$ and has cokernel $i_*\mathcal{H}$ where $\mathcal{H} = i^*\mathcal{K}$. By the snake lemma (Homology, Lemma 12.5.17) we obtain a commutative diagram with exact columns and rows
$\xymatrix{ & 0 \ar[d] & 0 \ar[d] & 0 \ar[d] \\ 0 \ar[r] & \mathcal{K}(-1) \ar[r] \ar[d] & \mathcal{O}^{\oplus r}(-1) \ar[r] \ar[d] & \mathcal{F}(-1) \ar[d] \ar[r] & 0 \\ 0 \ar[r] & \mathcal{K} \ar[r] \ar[d] & \mathcal{O}^{\oplus r} \ar[r] \ar[d] & \mathcal{F} \ar[d] \ar[r] & 0\\ 0 \ar[r] & i_*\mathcal{H} \ar[r] \ar[d] & i_*\mathcal{O}_ H^{\oplus r} \ar[r] \ar[d] & i_*\mathcal{F}' \ar[r] \ar[d] & 0 \\ & 0 & 0 & 0 }$
Thus the induction hypothesis applies to the exact sequence $0 \to \mathcal{H} \to \mathcal{O}_ H^{\oplus r} \to \mathcal{F}' \to 0$ on $H \cong \mathbf{P}^{n - 1}_ k$ (Lemma 33.35.2) and $\mathcal{H}$ is $m'$-regular. Recall that this implies that $\mathcal{H}$ is $d$-regular for all $d \geq m'$ (Lemma 33.35.10).
Let $i \geq 2$ and $d \geq m'$. It follows from the long exact cohomology sequence associated to the left column of the diagram above and the vanishing of $H^{i - 1}(H, \mathcal{H}(d))$ that the map
$H^ i(\mathbf{P}^ n_ k, \mathcal{K}(d - 1)) \longrightarrow H^ i(\mathbf{P}^ n_ k, \mathcal{K}(d))$
is injective. As these groups are zero for $d \gg 0$ (Cohomology of Schemes, Lemma 30.14.1) we conclude $H^ i(\mathbf{P}^ n_ k, \mathcal{K}(d))$ are zero for all $d \geq m'$ and $i \geq 2$.
We still have to control $H^1$. First we observe that all the maps
$H^1(\mathbf{P}^ n_ k, \mathcal{K}(m' - 1)) \to H^1(\mathbf{P}^ n_ k, \mathcal{K}(m')) \to H^1(\mathbf{P}^ n_ k, \mathcal{K}(m' + 1)) \to \ldots$
are surjective by the vanishing of $H^1(H, \mathcal{H}(d))$ for $d \geq m'$. Suppose $d > m'$ is such that
$H^1(\mathbf{P}^ n_ k, \mathcal{K}(d - 1)) \longrightarrow H^1(\mathbf{P}^ n_ k, \mathcal{K}(d))$
is injective. Then $H^0(\mathbf{P}^ n_ k, \mathcal{K}(d)) \to H^0(H, \mathcal{H}(d))$ is surjective. Consider the commutative diagram
$\xymatrix{ H^0(\mathbf{P}^ n_ k, \mathcal{K}(d)) \otimes _ k H^0(\mathbf{P}^ n_ k, \mathcal{O}(1)) \ar[r] \ar[d] & H^0(\mathbf{P}^ n_ k, \mathcal{K}(d + 1)) \ar[d] \\ H^0(H, \mathcal{H}(d)) \otimes _ k H^0(H, \mathcal{O}_ H(1)) \ar[r] & H^0(H, \mathcal{H}(d + 1)) }$
By Lemma 33.35.11 we see that the bottom horizontal arrow is surjective. Hence the right vertical arrow is surjective. We conclude that
$H^1(\mathbf{P}^ n_ k, \mathcal{K}(d)) \longrightarrow H^1(\mathbf{P}^ n_ k, \mathcal{K}(d + 1))$
is injective. By induction we see that
$H^1(\mathbf{P}^ n_ k, \mathcal{K}(d - 1)) \to H^1(\mathbf{P}^ n_ k, \mathcal{K}(d)) \to H^1(\mathbf{P}^ n_ k, \mathcal{K}(d + 1)) \to \ldots$
are all injective and we conclude that $H^1(\mathbf{P}^ n_ k, \mathcal{K}(d - 1)) = 0$ because of the eventual vanishing of these groups. Thus the dimensions of the groups $H^1(\mathbf{P}^ n_ k, \mathcal{K}(d))$ for $d \geq m'$ are strictly decreasing until they become zero. It follows that the regularity of $\mathcal{K}$ is bounded by $m' + \dim _ k H^1(\mathbf{P}^ n_ k, \mathcal{K}(m'))$. On the other hand, by the vanishing of the higher cohomology groups we have
$\dim _ k H^1(\mathbf{P}^ n_ k, \mathcal{K}(m')) = - \chi (\mathbf{P}^ n_ k, \mathcal{K}(m')) + \dim _ k H^0(\mathbf{P}^ n_ k, \mathcal{K}(m'))$
Note that the $H^0$ has dimension bounded by the dimension of $H^0(\mathbf{P}^ n_ k, \mathcal{O}^{\oplus r}(m'))$ which is at most $r{n + m' \choose n}$ if $m' > 0$ and zero if not. Finally, the term $\chi (\mathbf{P}^ n_ k, \mathcal{K}(m'))$ is equal to $r{n + m' \choose n} - P(m')$. This gives a bound of the desired type finishing the proof of the lemma. $\square$
Comment #4063 by Yicheng Zhou on
Just a small typo: in the paragraph between Definition 08AD and Lemma 08AE, a $d$ is missing in the left side of $H^0(\mathbf{P}^n_k,\mathcal{O}_{\mathbf{P}^n_k})=k[T_0,\ldots,T_n]_d$
Comment #6884 by Sasha on
In the remark after Definition 33.34.7 the inequality $d \ge m$ should be $d \ge -m$.
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2022-07-03 14:24:00
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https://planetmath.org/skewsymmetricmatrix
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skew-symmetric matrix
Definition:
Let $A$ be an square matrix of order $n$ with real entries $(a_{ij})$. The matrix $A$ is skew-symmetric if $a_{ij}=-a_{ji}$ for all $1\leq i\leq n,1\leq j\leq n$.
$A=\begin{pmatrix}a_{11}=0&\cdots&a_{1n}\\ \vdots&\ddots&\vdots\\ a_{n1}&\cdots&a_{nn}=0\end{pmatrix}$
The main diagonal entries are zero because $a_{i,i}=-a_{i,i}$ implies $a_{i,i}=0$.
One can see skew-symmetric matrices as a special case of complex skew-Hermitian matrices. Thus, all properties of skew-Hermitian matrices also hold for skew-symmetric matrices.
Properties:
1. 1.
The matrix $A$ is skew-symmetric if and only if $A^{t}=-A$, where $A^{t}$ is the matrix transpose
2. 2.
For the trace operator, we have that $\operatorname{tr}(A)=\operatorname{tr}(A^{t})$. Combining this with property (1), it follows that $\operatorname{tr}(A)=0$ for a skew-symmetric matrix $A$.
3. 3.
Skew-symmetric matrices form a vector space: If $A$ and $B$ are skew-symmetric and $\alpha,\beta\in\mathbb{R}$, then $\alpha A+\beta B$ is also skew-symmetric.
4. 4.
Suppose $A$ is a skew-symmetric matrix and $B$ is a matrix of same order as $A$. Then $B^{t}AB$ is skew-symmetric.
5. 5.
All eigenvalues of skew-symmetric matrices are purely imaginary or zero. This result is proven on the page for skew-Hermitian matrices.
6. 6.
According to Jacobi’s Theorem, the determinant of a skew-symmetric matrix of odd order is zero.
Examples:
• $\begin{pmatrix}0&b\\ -b&0\end{pmatrix}$
• $\begin{pmatrix}0&b&c\\ -b&0&e\\ -c&-e&0\end{pmatrix}$
Title skew-symmetric matrix SkewsymmetricMatrix 2013-03-22 12:01:05 2013-03-22 12:01:05 Daume (40) Daume (40) 10 Daume (40) Definition msc 15-00 SelfDual AntiSymmetric SkewHermitianMatrix AntisymmetricMapping
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2021-02-28 20:03:02
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http://jees.kr/journal/view.php?number=3249
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J Electromagn Eng Sci Search
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J Electromagn Eng Sci > Volume 16(4); 2016 > Article
Journal of Electromagnetic Engineering and Science 2016;16(4):241-247. DOI: https://doi.org/10.5515/JKIEES.2016.16.4.241
Novel New Approach to Improve Noise Figure Using Combiner for Phase-Matched Receiver Module with Wideband Frequency of 6-18 GHz Yuseok Jeon, Sungil Bang Division of Electronic and Electric Engineering, Dankook University Correspondence: Sungil Bang,Email: halsuida@naver.com Abstract This paper proposes the design and measurement of a 6-18 GHz front-end receiver module that has been combined into a one- channel output from a two-channel input for electronic warfare support measures (ESM) applications. This module includes a limiter, high-pass filter (HPF), power combiner, equalizer and amplifier. This paper focuses on the design aspects of reducing the noise figure (NF) and matching the phase and amplitude. The NF, linear equalizer, power divider, and HPF were considered in the design. A broadband receiver based on a combined configuration used to obtain low NF. We verify that our receiver module improves the noise figure by as much as 0.78 dB over measured data with a maximum of 5.54 dB over a 6-18 GHz bandwidth; the difference value of phase matching is within $7^{circ}$ between ports. Key words: Amplitude Matching, Broadband Receiver, Front-End, EW System, Phase Matching
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• 509 View
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2019-09-20 19:10:49
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https://www.maplesoft.com/support/help/maple/view.aspx?path=Order
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The Order Term Function and the Environment Variable Order - Maple Programming Help
The Order Term Function and the Environment Variable Order
Description
• The environment variable Order represents the order, or working precision, of series calculations performed by Maple. It does not necessarily represent the order of the error term, or the number of terms, in the series output. In the given example, note that the order of the series expansion of $\frac{{ⅇ}^{x}}{x}$ is only 7 rather than 8, since the series expansion of ${ⅇ}^{x}$ is divided by x, thereby reducing the degree by 1.
• The default value of Order is 6.
• The order term function O, denotes the "rest" of the series, beyond the order specified by the Order variable.
• Because Order is an environment variable, any assignments to it inside a procedure body are undone upon exit from a procedure.
Examples
> $\mathrm{Order}$
${6}$ (1)
> $\mathrm{series}\left(\mathrm{cos}\left(x\right),x=0\right)$
${1}{-}\frac{{1}}{{2}}{}{{x}}^{{2}}{+}\frac{{1}}{{24}}{}{{x}}^{{4}}{+}{O}{}\left({{x}}^{{6}}\right)$ (2)
> $\mathrm{Order}≔8$
${\mathrm{Order}}{≔}{8}$ (3)
> $\mathrm{series}\left({ⅇ}^{x},x=0\right)$
${1}{+}{x}{+}\frac{{1}}{{2}}{}{{x}}^{{2}}{+}\frac{{1}}{{6}}{}{{x}}^{{3}}{+}\frac{{1}}{{24}}{}{{x}}^{{4}}{+}\frac{{1}}{{120}}{}{{x}}^{{5}}{+}\frac{{1}}{{720}}{}{{x}}^{{6}}{+}\frac{{1}}{{5040}}{}{{x}}^{{7}}{+}{O}{}\left({{x}}^{{8}}\right)$ (4)
> $\mathrm{series}\left(\frac{{ⅇ}^{x}}{x},x=0\right)$
${{x}}^{{-1}}{+}{1}{+}\frac{{1}}{{2}}{}{x}{+}\frac{{1}}{{6}}{}{{x}}^{{2}}{+}\frac{{1}}{{24}}{}{{x}}^{{3}}{+}\frac{{1}}{{120}}{}{{x}}^{{4}}{+}\frac{{1}}{{720}}{}{{x}}^{{5}}{+}\frac{{1}}{{5040}}{}{{x}}^{{6}}{+}{O}{}\left({{x}}^{{7}}\right)$ (5)
Cancellation may lead to a smaller number of terms than expected.
> $\mathrm{series}\left(\frac{1}{{ⅇ}^{x}-1-x},x=0\right)$
${2}{}{{x}}^{{-2}}{-}\frac{{2}}{{3}}{}{{x}}^{{-1}}{+}\frac{{1}}{{18}}{+}\frac{{1}}{{270}}{}{x}{-}\frac{{1}}{{3240}}{}{{x}}^{{2}}{-}\frac{{1}}{{13608}}{}{{x}}^{{3}}{+}{O}{}\left({{x}}^{{4}}\right)$ (6)
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2020-02-23 17:30:28
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https://www.rosariacunha.com.br/what-is-a-pendulum-in-physics/
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# What is really a pendulum in math? Physics levels instruct us that pendulums are pendulums together with the bulk into the counter weight distributed from how we consider terms of a ladder, either a truck, or a elevator.
Physics levels instruct us that a pendulum is actually a exact complicated device that has a mysterious origin, and also can be a complicated product. You’ll find many topics of attention, each factual and logical, about this specific term.
If we believe the procedure for rotating the body, for example, the motion of the rotating mass is determined by the pressure exerted by the bulk onto the angle. That usually means that, in essence, the mass is pushing the angle.
We’ll see the forces terminate if we consider the way the forces do the job on the pendulum. In this circumstance, the force of gravity doesn’t implement. It appears that Newton’s law of gravity,” stated how it would be expressed by us , is an approximation. This reality would indicate that the’x’ http://lib.dr.iastate.edu/cgi/viewcontent.cgi?article=2782&context=etd of Newton’s Law of sanity must be”x-phi.”
One other difficulty that is important is we appear to fail to appreciate , for individuals with mathematics amounts that are traditional, there’s a simple method and the fact that we look at physics for both scientists and engineers. Most physicists using physics degrees actually understand what is apparent depth physics. As an instance, let’s look at a talk in regards to a pendulum with a likely plane, and let us assume the radial (bulk and inclined airplane ) are symmetrical in the feeling which the angles are the exact same.
Let’s even assume that the length of the pendulum is equivalent to the diameter of this likely plane. Thenthe half an of this aircraft (the span ) is half the length of the pendulum (the diameter). To put it differently, the apparent thickness of this plane is corresponding to the thickness of the pendulum.
The velocity with regard for this pendulum’s check my reference guts is just half the velocity of the mass. You will find two problems in math. One dilemma is obvious. The alternative is not.
Each Newton’s and Einstein’s formulas of the theory of general relativity enable us to produce. Let us go through the next assumption.
We all know that there is a normal towards the matter where we are discussing. That which we have to assume is that the ordinary is stable throughout the world. We will consider the assumption.
It must be mentioned that the normal is also a basic and special occurrence of relativity. This means that, generally speaking, it needs to be discovered that the normal has to be different. This looks.
Now, let’s think about the issue of speed with regard for the mass. The problem is to do with the existence of this ordinary. It is easy to see that the ordinary is going to be different anyplace. Additionally, it appears to be that it is not just really a problem to obtain the pace in a means which makes sense, therefore let’s move on.
That brings us into a static pendulum with a likely plane. What is apparent depth physics? Then a depth of the plane is going to differ As we are aware the normal is going to differ anywhere. We’ll think, however, which it is exactly the exact same every where.
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2020-04-07 10:51:06
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http://cms.math.ca/10.4153/CMB-2001-019-7
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location: Publications → journals → CMB
Abstract view
# Exceptional Sets of Slices for Functions From the Bergman Space in the Ball
Published:2001-06-01
Printed: Jun 2001
• Piotr Jakóbczak
Format: HTML LaTeX MathJax PDF PostScript
## Abstract
Let $B_N$ be the unit ball in $\mathbb{C}^N$ and let $f$ be a function holomorphic and $L^2$-integrable in $B_N$. Denote by $E(B_N,f)$ the set of all slices of the form $\Pi =L\cap B_N$, where $L$ is a complex one-dimensional subspace of $\mathbb{C}^N$, for which $f|_{\Pi}$ is not $L^2$-integrable (with respect to the Lebesgue measure on $L$). Call this set the exceptional set for $f$. We give a characterization of exceptional sets which are closed in the natural topology of slices.
MSC Classifications: 32A37 - Other spaces of holomorphic functions (e.g. bounded mean oscillation (BMOA), vanishing mean oscillation (VMOA)) [See also 46Exx] 32A22 - Nevanlinna theory (local); growth estimates; other inequalities {For geometric theory, see 32H25, 32H30}
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2014-09-22 10:13:19
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http://trilith.kz/weather-forecaster-tan/gravity-measurement-on-earth-eb9294
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This is what allows us to use the Earth's radius for r. The value obtained agrees approximately with the measured value of g. The difference may be attributed to several factors, mentioned above under "Variations": There are significant uncertainties in the values of r and m1 as used in this calculation, and the value of G is also rather difficult to measure precisely. Not coincidentally, the international standard unit of force is called a Newton (N). Gravity is usually measured in units of acceleration. The gravity measurements are consistent with a mass of Saturn’s core of 15 to 18 Earth masses. $g=G \frac {m_1}{r^2}$ So, … The force of gravity varies with latitude and increases from about 9.780 m/s 2 … The force of gravity is weakest at the equator because of the centrifugal force caused by the Earth's rotation and because points on the equator are furthest from the center of the Earth. The "surface" is taken to mean the cloud tops of the gas giants (Jupiter, Saturn, Uranus and Neptune). The two identical GRACE satellites orbit one behind the other in the same orbital plane at approximate distance of 220 kilometers (137 miles). where r is the distance between the center of the Earth and the body (see below), and here we take m1 to be the mass of the Earth and m2 to be the mass of the body. At Earth ’s surface the acceleration of gravity is about 9.8 metres (32 feet) per second per second. Gravimeters for measuring the earth's gravity as precisely as possible, are getting smaller and more portable. How is this done? Gravity is usually measured in units of acceleration. The low value of the ring mass suggests a scenario where the present rings of Saturn are young, probably just 10 million to 100 million years old, to be consistent with their pristine icy composition. Take your favorite fandoms with you and never miss a beat. Together, they measure Earth's gravity field with a precision greater than any previous instrument. By his dynamical and gravitational theories, he explained Kepler’s laws and established the modern quantitative science of gravitation. Gravimetry may be used when either the magnitude of gravitational field or the properties of matter responsible for its creation are of interest. Gravity is also the force that keeps the Earth in orbit around the Sun, as well as helping other planets remain in orbit. When the first satellite passes over a place on Earth with greater gravity, it speeds up very, very slightly, and the distance between the satellites increases—by less than the width of a human hair. Gravity is slightly stronger over places with more mass underground than over places with less mass. Microgravimetry is a rising and important branch developed on the foundation of classical gravimetry. The study of gravity changes belongs to geodynamics. In Cavendish's time, physicists used the same units for mass and weight, in effect taking g as a standard acceleration. For the Sun, the surface is taken to mean the photosphere. Gravimeters have been designed to mount in vehicles, including aircraft (note the field of aerogravity[2]), ships and submarines. The problem is that we can't measure gravity directly. g for earth ≃ 9.81 m/s2and its direction is downwards. Besides precision, stability is also an important property of a gravimeter, as it allows the monitoring of gravity changes. The values in the table have not been de-rated for the inertia effect of planet rotation (and cloud-top wind speeds for the gas giants) and therefore, generally speaking, are similar to the actual gravity that would be experienced near the poles. However, gravity isn’t the same everywhere on Earth. NASA uses two spacecraft to measure these variations in Earth’s gravity. A relative instrument also requires calibration by comparing instrument readings taken at locations with known complete or absolute values of gravity. The gravity field of the Earth is determined in two ways: Measuring the orbits of satellites and using these to determine the gravity field. Quartz and metal springs are chosen for different reasons; quartz springs are less affected by magnetic and electric fields while metal springs have a much lower drift (elongation) with time. In one common form, a spring is used to counteract the force of gravity pulling on an object. Established at the third General Conference on Weights and Measures in 1901, the standard gravity on Earth is 9.80665 meters per second squared, or 32.174 feet per second squared. When making measurements of the earth's gravity, we usually don't measure the gravitational force, F. Rather, we measure the gravitational acceleration, g. The gravitational acceleration is the time rate of change of a body's speed under the influence of the gravitational force. The gravitational constant, denoted by capital G, has a value of 6.67408 × 10-11 m3 kg-1 s-2 Small everyday objects exert a small force on each other, while larger celestial objects exert a noticeable pull on other objects. This pull is a f… The value of the gn approximately equals the acceleration due to gravity at the Earth's surface (although the value of g varies by location). These spacecraft are part of the Gravity Recovery and Climate Experiment (GRACE) mission. Variations in gravity and apparent gravity محسن, Altitude ...the value of gravitional constant is not correct value i have a proof, Comparative gravities in various cities around the world, Comparative gravities of the Earth, Sun, Moon, and planets. Note that the shortest half wavelength that can be resolved at r = r e with such an expansion is λ 1/2 = πr e/ max, yielding 56km for max = 360, and 9km for max = −) = on the surface of the Earth and its center of mass O, and it is = + + + Observing (Measuring) the Gravity Field on Earth. The equatorial bulge at Earth's equator is measured at 26.5 miles (42.72 km) and is caused by the planet's rotation and gravity. The test mass is sealed in an air-tight container so that tiny changes of barometric pressure from blowing wind and other weather do not change the buoyancy of the test mass in air. In the SI system of units, the standard unit of acceleration is 1 metre per second squared (abbreviated as m/s ). As the lead satellite passes over an area on Earth of slightly stronger gravity, it detects an increased gravitational pull and speeds up ever so slightly, thus increasing its distance from the trailing satellite.
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2021-09-23 08:01:27
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https://hpmuseum.org/forum/showthread.php?mode=linear&tid=5907&pid=157791
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Ln(x) using repeated square root extraction
03-21-2016, 12:03 AM (This post was last modified: 04-11-2020 12:00 AM by Gerson W. Barbosa.)
Post: #1
Gerson W. Barbosa Senior Member Posts: 1,465 Joined: Dec 2013
Ln(x) using repeated square root extraction
Algorithm:
1. Enter x;
2. Take the square root 6 times;
3. Multiply by 2;
4. Subtract 1;
5. Take the square root;
6. Subtract 1;
7. Multiply by 64.
For x = 2, on an 8-digit calculator like the Canon LC-37, you should get ln(2) ~ 0.693152. For 10-digit calculators, in step 2 take the square root 9 times and multiply by 512 in the last step. Thus, on the HP-12C, which has two guard digits, we get ln(2) ~ 0.693147648. Ordinary 10-digit calculator might give less approximate results.
A 4-step less accurate algorithm is available in this old thread:
The accuracy range should be tested for your particular calculator. Use these at your own risk.
Gerson.
Method:
$\ln (x)=\int_{1}^{x}\frac{du}{u}=\int_{1}^{x}u^{-1}du=\lim_{v\rightarrow 0}\left | \frac{u^{v}}{v} \right |_{1}^{x}=\lim_{v\rightarrow 0} \left ( \frac{x^{v}}{v}-\frac{1^{v}}{v} \right )=\lim_{v\rightarrow 0} \left ( \frac{x^{v}-1}{v} \right )$
By testing the limit with a few values for v close to zero, an empirical second term has been able to be added to the expression:
$\ln (x)=\lim_{v\rightarrow 0} \left ( \frac{x^{v}-1}{v}-\frac{v\cdot \ln^{2}(x)}{2} \right )$
After turning the limit into an equality and solving the resulting quadratic equation for ln(x), we get
$\ln (x)=\lim_{v\rightarrow 0} \left ( \frac{\sqrt{2\cdot x^{v}-1}-1}{v} \right )$
Example:
Let x = 2 and v = 0.001
Then
$\ln (2)\approx \frac{\sqrt{2\cdot 2^{0.001}-1}-1}{0.001}\approx 0.69314723$
which is good to 6 decimal places ( ln(2) = 0.69314718 ).
P.S.: Perhaps 8 square root extractions and final multiplication by 256 (2^8) is a better compromise for the range 2..100 on the Canon LC-37. Contrary to what I thought its SILVA-CELL 189 battery is still working after all these years, so I was able to make some more tests. :-)
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Good Friday 2020 Update
The following expression allows us to go further:
$\ln (x)=\lim_{v\rightarrow 0} \left ( \frac{x^{v}-1}{v}-\frac{v\cdot \ln^{2}(x)}{2!} -\frac{v^{2}\cdot \ln^{3}(x)}{3!} -\frac{v^{3}\cdot \ln^{4}(x)}{4!} - \cdots \right )$
By going up to the third power of ln(x) and solving the corrisponding cubic equation, we get
$\ln (x)=\lim_{v\rightarrow 0}\frac{1}{v}\left ( \sqrt[3]{\sqrt{9x^{2v}-6x^{v}+2}+3x^{v}-1} - \frac{1}{\sqrt[3]{\sqrt{9x^{2v}-6x^{v}+2}+3x^{v}-1}} - 1 \right )$
For example, if x = 2 and v = 0.01 then ln(2) ~ 0.69314719
This approximation is rather complicated for manual calculations. Also, it requires one cubic root extraction. It can be used, however, as an alternative method for implementing the ln(x) function to 7 significant digits on calculators that lack it, like the HP-16C. The e^x function has also been implemented. Since it's based on the first formula, obtained by solving a quadratic equation, and uses a constant number of loops, it gives less significant digits. The cubic root algorithm is optimized for the narrow range required by the approximation formula and was based one provided by Albert Chan here.
Code:
001- LBL A; LN 002- 1 003- EEX 004- CHS 005- 2 006- CF 1 007- 1 008- + 009- STO I 010- CLx 011- LASTx 012- x⇆y 013- x≤y 014- GTO 2 015- LBL 0 016- √x 017- RCL I 018- x>y 019- GSB 1 020- R↓ 021- x⇆y 022- ENTER 023- + 024- x⇆y 025- GTO 0 026- RTN 027- LBL 1 028- R↓ 029- STO 0 030- R↓ 031- 2 032- × 033- STO 1 034- RCL 0 035- ENTER 036- × 037- 9 038- × 039- 6 040- RCL 0 041- × 042- - 043- 2 044- + 045- √x 046- 3 047- RCL 0 048- × 049- + 050- 1 051- - 052- GSB 3 053- ENTER 054- 1/x 055- - 056- 1 057- - 058- RCL 1 059- × 060- F?1 061- CHS 062- RTN 063- LBL 2 064- SF1 065- 1/x 066- RTN 067- LBL 3 068- 2 069- STO I 070- ENTER 071- + 072- x⇆y 073- × 074- STO 0 075- LASTx 076- 2 077- - 078- 6 079- / 080- √x 081- 1 082- + 083- LBL 4 084- ENTER 085- ENTER 086- RCL 0 087- x⇆y 088- / 089- LASTx 090- ENTER 091- × 092- - 093- 3 094- / 095- √x 096- + 097- 2 098- / 099- DSZ 100- GTO 4 101- RTN 102- LBL B; eᵡ 103- 1 104- 3 105- STO I 106- R↓ 107- 8 108- 1 109- 9 110- 2 111- + 112- LASTx 113- / 114- ENTER 115- × 116- 1 117- + 118- 2 119- / 120- LBL 5 121- ENTER 122- × 123- DSZ 124- GTO 5 125- RTN
Examples:
2 GSB A -> 0.6931471(360)
GSB B -> 2.0000(13165)
12345 GSB A -> 9.421006(848)
GSB B -> 12345.0(1957)
6.789 EEX 79 GSB A -> 183.8195(343) GSB B -> 6.(686887E79)
0.12345 GSB A -> -2.091919(104)
GSB B -> 0.123450(119)
230 GSB B -> 7.496895E99
GSB A -> 229.97041(15)
1 GSB A -> 0.000000000
GSB B -> 1.000000000
GSB B -> 2.7182(92170)
GSB B -> 15.1544(4181)
GSB A -> 2.71829(4016)
GSB A -> 1.000004(736)
GSB A -> 0.000004736
0.693147181 CHS GSB B -> 0.500000(16)
10 GSB A -> 2.302585(344) STO 2
2 GSB A RCL 2 / -> 0.3010299(435)
0 GSB A -> Error 0
2 CHS GSB A -> Error 0
03-21-2016, 05:09 AM (This post was last modified: 03-21-2016 05:12 AM by Gerson W. Barbosa.)
Post: #2
Gerson W. Barbosa Senior Member Posts: 1,465 Joined: Dec 2013
RE: Ln(x) using repeated square root extraction
For the sake of completeness here is the recursive limit definition of ln(x), even though this is not quite necessary for our purpose:
$\ln (x)=\lim_{n\rightarrow \infty} \left [ n\cdot \left ( x^{\frac{1}{n}}-1 \right )-\sum_{k=2}^{\infty }\frac{\ln ^{k}(x)}{k!\cdot n^{k-1}} \right ]$
If the limit is removed and n is set to 1 then we'll have the following recursive series representation:
$\ln (x)= \ x-1 -\sum_{k=2}^{\infty }\frac{\ln ^{k}(x)}{k!}$
03-21-2016, 07:11 AM
Post: #3
Paul Dale Senior Member Posts: 1,752 Joined: Dec 2013
RE: Ln(x) using repeated square root extraction
I've got a feeling I tried something along these lines for the LN function of the 34S. The LN function has always been very slow on the 34S and I've tried several implementations to try to get both performance and accuracy. The current implementation is much faster than the original but still far slower than I'd like. I ended up with:
Code:
/* Natural logarithm. * * Take advantage of the fact that we store our numbers in the form: m * 10^e * so log(m * 10^e) = log(m) + e * log(10) * do this so that m is always in the range 0.1 <= m < 2. However if the number * is already in the range 0.5 .. 1.5, this step is skipped. * * Then use the fact that ln(x^2) = 2 * ln(x) to range reduce the mantissa * into 1/sqrt(2) <= m < 2. * * Finally, apply the series expansion: * ln(x) = 2(a+a^3/3+a^5/5+...) where a=(x-1)/(x+1) * which converges quickly for an argument near unity. */
- Pauli
03-03-2022, 11:20 PM
Post: #4
Thomas Klemm Senior Member Posts: 1,687 Joined: Dec 2013
RE: Ln(x) using repeated square root extraction
Formula
We can use the following identity:
$$\log(x) = n \log\left(x^{\frac{1}{n}}\right) = n \log\left(\sqrt[n]{x}\right)$$
For say $$n = 2^{10} = 1024$$ and $$1 \leqslant x \leqslant 100$$ the value of $$\sqrt[n]{x}$$ is close to $$1$$.
Thus we can use the Taylor series to calculate the logarithm:
$$\log(1 + \varepsilon) = \varepsilon - \frac{\varepsilon^2}{2} + \frac{\varepsilon^3}{3} - \frac{\varepsilon^4}{4} + \frac{\varepsilon^5}{5} + \mathcal{O}(\varepsilon^6)$$
Program
Here's a program for the HP-42S that calculates both the logarithm and its approximation:
Code:
LN LASTX SQRT SQRT SQRT SQRT SQRT SQRT SQRT SQRT SQRT SQRT 1 - 4 1/X RCL× ST Y 3 1/X X<>Y - RCL× ST Y 2 1/X X<>Y - RCL× ST Y 1 X<>Y - × 1024 ×
It's easy to extend if you want to use more terms.
Example
x = 2
0.69314718056
0.69314718056
Comparison
We can compare this to your solution:
$$\sqrt{2(1 + \varepsilon) - 1} - 1$$
The Taylor series agrees for the first two terms:
$$\varepsilon - \frac{\varepsilon^2}{2} + \frac{\varepsilon^3}{2} - \frac{5 \varepsilon^4}{8} + \frac{7 \varepsilon^5}{8} + \mathcal{O}(\varepsilon^6)$$
03-05-2022, 05:32 PM
Post: #5
Dan C Member Posts: 98 Joined: Jul 2018
RE: Ln(x) using repeated square root extraction
I just tried this on my CASIO fx-4000P, and the result was 0.693147648
I did the square root 9 times, and multiplied with 512 in the last step.
03-05-2022, 05:53 PM
Post: #6
Dan C Member Posts: 98 Joined: Jul 2018
RE: Ln(x) using repeated square root extraction
(03-05-2022 05:32 PM)Dan C Wrote: I just tried this on my CASIO fx-4000P, and the result was 0.693147648
I did the square root 9 times, and multiplied with 512 in the last step.
I just had to try this on my CASIO fx-4500P also, and the result was the same,
0.693147648.
Now i just have to try this on my FX-602P also
03-05-2022, 08:02 PM
Post: #7
Dan C Member Posts: 98 Joined: Jul 2018
RE: Ln(x) using repeated square root extraction
On my FX-602P this gives 0.693147371
hmm, different from my fx-4000p and fx-4500p.
What does this tell of the FX-602P?
Less accurate?
03-05-2022, 08:29 PM
Post: #8
Namir Senior Member Posts: 823 Joined: Dec 2013
RE: Ln(x) using repeated square root extraction
The extended limit for ln(x) does very well with values of v = 0.001 or less. Very impressive!
Namir
03-05-2022, 08:39 PM
Post: #9
Thomas Klemm Senior Member Posts: 1,687 Joined: Dec 2013
RE: Ln(x) using repeated square root extraction
We experience cancellation when calculating $$\varepsilon = \sqrt[n]{x} - 1$$.
The program was executed on free42 with 34 decimal digits of precision.
Thus this effect is not noticed.
It is a trade-off between coming close to $$1$$ and how many terms to use in the Taylor series.
For a calculator like the HP-15C with 10 digits using $$n = 2^6$$ might be the best choice.
For $$x = 2$$ we end up with the following sequence of iterated square roots:
2.000000000
1.414213562
1.189207115
1.090507733
1.044273783
1.021897149
1.010889286
This leaves us with $$\varepsilon = 0.010889286$$ and thus 5 terms of the Taylor series should be enough.
We notice the cancellation since we have now only 8 significant digits left.
Here's the program for the HP-15C or similar calculators:
Code:
001 { 11 } √x̅ 002 { 11 } √x̅ 003 { 11 } √x̅ 004 { 11 } √x̅ 005 { 11 } √x̅ 006 { 11 } √x̅ 007 { 1 } 1 008 { 30 } − 009 { 36 } ENTER 010 { 36 } ENTER 011 { 36 } ENTER 012 { 5 } 5 013 { 10 } ÷ 014 { 4 } 4 015 { 15 } 1/x 016 { 34 } x↔y 017 { 30 } − 018 { 20 } × 019 { 3 } 3 020 { 15 } 1/x 021 { 34 } x↔y 022 { 30 } − 023 { 20 } × 024 { 2 } 2 025 { 15 } 1/x 026 { 34 } x↔y 027 { 30 } − 028 { 20 } × 029 { 1 } 1 030 { 34 } x↔y 031 { 30 } − 032 { 20 } × 033 { 6 } 6 034 { 4 } 4 035 { 20 } ×
For $$x = 2$$ we get:
0.6931471776
While for $$\ln(2)$$ we get:
0.6931471806
This leaves us with a difference of:
0.0000000030
03-06-2022, 12:04 AM
Post: #10
Albert Chan Senior Member Posts: 1,896 Joined: Jul 2018
RE: Ln(x) using repeated square root extraction
(03-03-2022 11:20 PM)Thomas Klemm Wrote: We can compare this to your solution:
$$\sqrt{2(1 + \varepsilon) - 1} - 1$$
The Taylor series agrees for the first two terms:
$$\varepsilon - \frac{\varepsilon^2}{2} + \frac{\varepsilon^3}{2} - \frac{5 \varepsilon^4}{8} + \frac{7 \varepsilon^5}{8} + \mathcal{O}(\varepsilon^6)$$
log1p(ε) = 2*atanh(y) = 2*(y+y^3/3+y^5/5+...) ≈ 2y, where y=ε/(2+ε)
2ε/(2+ε) is simpler than √(1+2ε)-1, and twice as accurate, when ε is tiny.
CAS> series(2*ε/(2+ε),ε) // estimate for log1p(ε)
ε - 1/2*ε^2 + 1/4*ε^3 - 1/8*ε^4 + 1/16*ε^5 + ε^6*order_size(ε)
CAS> (1/3-1/2)/(1/3-1/4)
-2.
Is is funny sqrt approximation formula give better estimate for log1p, than sqrt itself.
2*x/(x+2)
03-06-2022, 12:50 AM
Post: #11
Albert Chan Senior Member Posts: 1,896 Joined: Jul 2018
RE: Ln(x) using repeated square root extraction
(03-05-2022 08:39 PM)Thomas Klemm Wrote: For $$x = 2$$ we end up with the following sequence of iterated square roots:
2.000000000
1.414213562
1.189207115
1.090507733
1.044273783
1.021897149
1.010889286
This leaves us with $$\varepsilon = 0.010889286$$ and thus 5 terms of the Taylor series should be enough.
We notice the cancellation since we have now only 8 significant digits left.
It is more work, but we could improve accuracy by "pulling out" 1.
CAS> f(x) := x/(sqrt(1+x)+1) // == sqrt(1+x)-1
CAS> 1. // x-1 = 2-1 = 1
CAS> f(Ans)
0.414213562373
0.189207115003
9.05077326653e−2
4.42737824274e−2
2.18971486541e−2
1.08892860517e−2
The recursive formula, log1p(x) = 2*log1p(x/(sqrt(1+x)+1)), is similar to atan formula
(05-31-2021 09:51 PM)Albert Chan Wrote: $$\displaystyle\arctan(x) = 2\arctan\left( {x \over \sqrt{1+x^2}+1} \right)$$
03-06-2022, 09:33 AM
Post: #12
Thomas Klemm Senior Member Posts: 1,687 Joined: Dec 2013
RE: Ln(x) using repeated square root extraction
I was curious to try Pauli's idea and use:
$$\log\left(\frac{1+\varepsilon}{1-\varepsilon}\right) = 2 \left( \varepsilon + \frac{\varepsilon^3}{3} + \frac{\varepsilon^5}{5} + \mathcal{O}(\varepsilon^7) \right)$$
We set:
$$\frac{1+\varepsilon}{1 - \varepsilon} = x$$
$$\varepsilon = \frac{x - 1}{x + 1}$$
Here's the corresponding program for the HP-15C:
Code:
001 { 11 } √x̅ 002 { 11 } √x̅ 003 { 11 } √x̅ 004 { 11 } √x̅ 005 { 11 } √x̅ 006 { 11 } √x̅ 007 { 36 } ENTER 008 { 36 } ENTER 009 { 1 } 1 010 { 30 } − 011 { 34 } x↔y 012 { 1 } 1 013 { 40 } + 014 { 10 } ÷ 015 { 36 } ENTER 016 { 36 } ENTER 017 { 36 } ENTER 018 { 5 } 5 019 { 10 } ÷ 020 { 20 } × 021 { 3 } 3 022 { 15 } 1/x 023 { 40 } + 024 { 20 } × 025 { 20 } × 026 { 1 } 1 027 { 40 } + 028 { 20 } × 029 { 1 } 1 030 { 2 } 2 031 { 8 } 8 032 { 20 } ×
The computation of $$\varepsilon$$ is now a bit more complicated, but we have fewer terms to compute.
In the end, the program is shorter.
The result for $$x = 2$$ is:
0.6931471773
Compared to the previous result, it is only slightly off in the last digit.
03-09-2022, 07:47 PM
Post: #13
Albert Chan Senior Member Posts: 1,896 Joined: Jul 2018
RE: Ln(x) using repeated square root extraction
Instead of reducing argument, √...√x, we can blow it up, and get log from AGM
From Series[EllipticK[1-(4x)^2], {x,0,4}], and ignored imaginery parts:
K(m=1-(4/x)^2) = log(x) + (log(x)-1)*(4/x^2) + O(1/x^4)
lua> x = 2^14
lua> a, b = 1, 4/x
lua> c = b/x
lua> for i=1,6 do a,b = (a+b)/2, sqrt(a*b); print(a,b) end
Code:
0.5001220703125 0.015625 0.25787353515625 0.08839913658307309 0.17313633586966154 0.15098277337311447 0.162059554621388 0.16168056210089243 0.1618700583611402 0.16186994744240293 0.16187000290177156 0.16187000290176207
With 6 sqrt, (a,b) already close enough so that (a+b)/2 ≈ AGM
lua> k = pi/(a+b)
lua> k / 14
0.6931471898242749
lua> (k+c)*(1-c) / 14
0.6931471805599455
lua> log(x) / 14 -- = log(2)
0.6931471805599453
Note: if c=4/x^2 small enough, below machine epsilon, log(x) = k, no need for correction.
03-10-2022, 05:18 AM
Post: #14
Thomas Klemm Senior Member Posts: 1,687 Joined: Dec 2013
RE: Ln(x) using repeated square root extraction
In the initial thread An old logarithm algorithm the "pages 33 and 34 of the manual for the Texas Instruments SR-10" are quoted.
However in this manual for the Texas Instruments electronic slide rule calculator SR-10 we find on page 30:
Quote:Logarithmic and Exponential Function
$$\ln a = \left[ \left( - \frac{3}{5} b^2 + 1 \right)^{-1} \times 5 + 4 \right] \frac{2b}{9}$$
$$0.7 < a < 1.6$$
where $$b = \frac{a - 1}{a + 1} = \left( a + 1 \right)^{-1} \times (-2) + 1$$
This expression yields values with an error less than 0.0003% over the range of a from 0.7 to 1.6.
The first 3 terms of the Taylor series of this expression agree with those of $$\log\left(\frac{1+\varepsilon}{1-\varepsilon}\right)$$:
$$\frac{2\varepsilon}{9} \left( 4 + \frac{5}{1 - \frac{3\varepsilon^2}{5}} \right) = 2 \varepsilon + \frac{2\varepsilon^3}{3} + \frac{2\varepsilon^5}{5} + \frac{6\varepsilon^7}{25} + \frac{18\varepsilon^9}{125} + \mathcal{O}(\varepsilon^{11})$$
Again a program for the HP-15C:
Code:
001 { 11 } √x̅ 002 { 11 } √x̅ 003 { 11 } √x̅ 004 { 11 } √x̅ 005 { 11 } √x̅ 006 { 11 } √x̅ 007 { 1 } 1 008 { 40 } + 009 { 2 } 2 010 { 34 } x↔y 011 { 10 } ÷ 012 { 1 } 1 013 { 34 } x↔y 014 { 30 } − 015 { 36 } ENTER 016 { 43 11 } g x² 017 { 3 } 3 018 { 20 } × 019 { 5 } 5 020 { 10 } ÷ 021 { 1 } 1 022 { 34 } x↔y 023 { 30 } − 024 { 5 } 5 025 { 34 } x↔y 026 { 10 } ÷ 027 { 4 } 4 028 { 40 } + 029 { 20 } × 030 { 1 } 1 031 { 2 } 2 032 { 8 } 8 033 { 20 } × 034 { 9 } 9 035 { 10 } ÷
The result for $$x = 2$$ is:
0.6931471786
03-10-2022, 03:17 PM (This post was last modified: 03-12-2022 03:49 PM by Albert Chan.)
Post: #15
Albert Chan Senior Member Posts: 1,896 Joined: Jul 2018
RE: Ln(x) using repeated square root extraction
(03-10-2022 05:18 AM)Thomas Klemm Wrote: The first 3 terms of the Taylor series of this expression agree with those of $$\log\left(\frac{1+\varepsilon}{1-\varepsilon}\right)$$:
$$\frac{2\varepsilon}{9} \left( 4 + \frac{5}{1 - \frac{3\varepsilon^2}{5}} \right) = 2 \varepsilon + \frac{2\varepsilon^3}{3} + \frac{2\varepsilon^5}{5} + \frac{6\varepsilon^7}{25} + \frac{18\varepsilon^9}{125} + \mathcal{O}(\varepsilon^{11})$$
FYI, approximation formula based from atanh(ε) pade approximation:
atanh(ε) = ε + ε^3/3 + ε^5/5 + ... ≥ ε + (ε^3/3) / (1 - 3/5*ε^2)
Let d = 1 - 3/5*ε^2, we have ε^2/3 = 5/9*(1-d)
ε + (ε^3/3)/d = ε*(1 + 5/9*(1-d)/d) = ε/9*(9 + 5*(1/d-1)) = ε/9*(4 + 5/d)
ln((1+ε)/(1-ε)) = 2*atanh(ε) ≥ 2ε/9*(4 + 5/(1-3/5*ε^2))
(10-22-2021 02:15 PM)Albert Chan Wrote: ln(n) ≈ g - g/(2.7 + 24/g^2) , where g = (n-1)/√n (*)
We doubled accuracy if we use above formula instead (Bonus, less code steps)
Again, taylor series of ln((1+ε)/(1-ε)), using above formula for ln:
XCAS> g := (x-1)/sqrt(x)
XCAS> f := g - g/(27/10+24/g^2)
XCAS> series(f(x=(1+ε)/(1-ε)),ε,0,10)
2*ε + 2/3*ε^3 + 2/5*ε^5 + 123/400*ε^7 + 3217/12000*ε^9 + O(ε^11)
XCAS> (c - 6/25) / (c - 123/400) | c=2/7.
-2.09836065574
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2022-08-13 20:35:07
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https://api-project-1022638073839.appspot.com/questions/how-do-you-find-the-derivative-of-f-t-sin-2-e-sin-2-t-using-the-chain-rule
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# How do you find the derivative of f(t)=sin^2[e^(sin^2)t] using the chain rule?
Oct 30, 2015
$f ' \left(t\right) = {e}^{{\sin}^{2} t} \cdot \sin 2 t \cdot \sin 2 {e}^{{\sin}^{2} t}$
#### Explanation:
$f ' \left(t\right) = \left({\sin}^{2} {e}^{{\sin}^{2} t}\right) ' = 2 \sin {e}^{{\sin}^{2} t} \cdot \left(\sin {e}^{{\sin}^{2} t}\right) '$
$f ' \left(t\right) = 2 \sin {e}^{{\sin}^{2} t} \cdot \cos {e}^{{\sin}^{2} t} \cdot \left({e}^{{\sin}^{2} t}\right) '$
$f ' \left(t\right) = 2 \sin {e}^{{\sin}^{2} t} \cdot \cos {e}^{{\sin}^{2} t} \cdot {e}^{{\sin}^{2} t} \cdot \left({\sin}^{2} t\right) '$
$f ' \left(t\right) = 2 \sin {e}^{{\sin}^{2} t} \cdot \cos {e}^{{\sin}^{2} t} \cdot {e}^{{\sin}^{2} t} \cdot 2 \sin t \cdot \left(\sin t\right) '$
$f ' \left(t\right) = 2 \sin {e}^{{\sin}^{2} t} \cdot \cos {e}^{{\sin}^{2} t} \cdot {e}^{{\sin}^{2} t} \cdot 2 \sin t \cdot \cos t$
$f ' \left(t\right) = {e}^{{\sin}^{2} t} \cdot \sin 2 t \cdot \sin 2 {e}^{{\sin}^{2} t}$
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2021-04-16 14:21:29
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https://www.sparrho.com/item/periodic-orbit-theory-in-fractal-drum/9183d4/
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Periodic orbit theory in fractal drum
Research paper by Stefanie Russ, Jesper Mellenthin
Indexed on: 02 Feb '05Published on: 02 Feb '05Published in: Physics - Disordered Systems and Neural Networks
Abstract
The level statistics of pseudointegrable fractal drums is studied numerically using periodic orbit theory. We find that the spectral rigidity \$\Delta_3(L)\$, which is a measure for the correlations between the eigenvalues, decreases to quite small values (as compared to systems with only small boundary roughness), thereby approaching the behavior of chaotic systems. The periodic orbit results are in good agreement with direct calculations of \$\Delta_3(L)\$ from the eigenvalues.
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2020-06-06 16:45:24
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http://doktor-belovodsk.pp.ua/employ-plural-nhrnvrl/133fd0-polynomial-equation-calculator
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## polynomial equation calculator
Code to add this calci to your website Solve cubic equation , ax 3 + bx 2 + cx + d = 0 (For example, Enter a=1, b=4, c=-8 and d=7) In math algebra, a cubic function is a function of the form. The calculator will try to factor any polynomial (binomial, trinomial, quadratic, etc. Polynomial Division Calculator Step 1: Enter the expression you want to divide into the editor. -/., Enter roots: Show me graph: Smart zooming: xmin: xmax: Generate Polynomial. We can multiply the polynomials \left (6x-5\right)\left (2x+3\right) (6x−5)(2x+3) by using the FOIL method. pre-calculus-polynomial-equation-calculator, Please try again using a different payment method. The calculator below computes the discriminant of a higher degree polynomial from the resultant of a polynomial and its derivative. Message received. The acronym F O I L stands for multiplying the terms in each bracket in the following order: First by First (F\times F F ×F), Outer by Outer ( Solving quadratics by factorizing (link to previous post) usually works just fine. working... Polynomial Calculators. It is defined as third degree polynomial equation. Perform a Polynomial Regression with Inference and Scatter Plot with our Free, Easy-To-Use, Online Statistical Software. The polynomial linear regression model is \[ Y = … The solutions of this cubic equation are termed as the roots or zeros of the cubic equation. Special cases of such equations are: 1. Solutions Graphing Practice; Geometry beta; Notebook Groups Cheat Sheets; Sign In; Join; Upgrade; Account Details Login Options Account Management Settings Subscription … The numbers a, b, and c are the coefficients of the equation and may be … Polynomial equation solver This calculator solves equations in the form P (x) = Q(x), where P (x) and Q(x) are polynomials. Linear equation calculator, how to factor cubed polynomials with multiple variables, ks2 sats english exam questions, Lowest common factor. Polynomial graphing calculator This page help you to explore polynomials of degrees up to 4. For Polynomials of degree less than 5, the exact value of the roots are returned. Math Help List- Voted as Best Calculator: Percentage Calculator Email . A value of x that makes the equation equal to 0 is termed as zeros. Use this online Polynomial Multiplication Calculator for multiplying polynomials of any degree. Polynomial equation solver Calculator Codes using C, C++, JAVA, PHP. Calculator Use. Polynomial calculator - Integration and differentiation. Math lesson to teach t +elementry student in pa, cube root t1-89, how to solve division equations, algebra quick answers, free math worksheets algebra completing the square, free online fraction and regular calculator, online graphing calculator … The largest exponent of x x appearing in p(x) p (x) is called the degree of p p. If p(x) p (x) has degree n n, then it is well known that there are n n roots, once one takes into account multiplicity. A polynomial regression data fit application with some technical background. Able to display the work process and the detailed step by step explanation. where $P(x)$ and $Q(x)$ … But what if the quadratic equation... EN: pre-calculus-equation-calculator menu, EN: pre-calculus-polynomial-equation-calculator menu. Polynomials in mathematics and science are used in calculus and numerical analysis. Thanks for the feedback. Free polynomial equation calculator - Solve polynomials equations step-by-step This website uses cookies to ensure you get the best experience. Polynomial factoring calculator This online calculator writes a polynomial as a product of linear factors. This calculator solves equations in the form $P(x)=Q(x)$, Middle School Math Solutions – Polynomials Calculator, Adding Polynomials A polynomial is an expression of two or more algebraic terms, often having different exponents. Equations Inequalities System of Equations System of Inequalities Basic Operations Algebraic Properties Partial Fractions Polynomials Rational Expressions Sequences Power Sums Pi (Product) Notation Induction Logical Sets. Type in any equation to get the solution, steps and graph polynomial factoring calculator) in the leftmost column below. Polynomial calculator - Sum and difference . Polynomial … Factoring Polynomials … More about this Polynomial Regression Calculator so you can have a deeper perspective of the results that will be provided by this calculator. A polynomial with rational coefficients can sometimes be written as a product of lower-degree polynomials that also have rational coefficients. Third Degree Polynomial Equation Calculator or Cubic Equation Calculator. Polynomial roots calculator This online calculator finds the roots (zeros) of given polynomial. In such cases, the polynomial is said to "factor over the rationals." High School Math Solutions – Quadratic Equations Calculator, Part 2. Input the roots here, separated by … are polynomials. Input the polynomial: P(x) = How to input. BYJU’S online polynomial equation solver calculator tool makes the calculation faster, and it displays the variable value in a fraction of seconds. Enter the equation in the fourth degree equation calculator and hit calculate to know the roots with ease. It can also be said as the roots of the polynomial equation. The zeros of a polynomial equation are the solutions of the function f (x) = 0. Print . Finding roots of polynomials was never that easy! The Polynomial Roots Calculator will find the roots of any polynomial with just one click. EN: pre-calculus-polynomial-equation-calculator menu Pre Algebra Order of Operations Factors & Primes Fractions Long Arithmetic Decimals Exponents & Radicals Ratios & Proportions Percent Modulo Mean, Median & Mode Scientific Notation Arithmetics mathhelp@mathportal.org, $$\frac{x}{2} \left(x^2 + 1\right) = 2(x+1)^2 - 5x - 3$$, $$\frac{3x^2-1}{2} + \frac{2x+1}{3} = \frac{x^2-2}{4} + \frac{1}{3}$$. Home Calculators Mobile Apps Math Courses Math Games. By using this website, you agree to our Cookie Policy. The calculator generates polynomial with given roots. If you want to contact me, probably have some question write me using the contact form or email me on f(x) = ax 3 + bx 2 + cx + d … Then, ... 59 terms for a degree of 5; and finally 3,815,311 terms for polynomials of a degree of 12. By using this website, you agree to our Cookie Policy. Free to use. This web site owner is mathematician Miloš Petrović. ... solving equations with cubed and square root Kumon math worksheets ( printouts) Gr.7 decimals worksheet papers free answers to algebraic equations how to understand algebra the importance of algebra "TI-84 Plus Program" simplifying irrational square roots tutorial … Cubic equation: $5x^3 + 2x^2 - 3x + 1 = \frac{1}{3} x$. Related Calculators. Polynomial Equation Solver Calculator is a free online tool that displays the value of the unknown variable by solving the given polynomial equation. ... To change the degree of the equation, press one of the provided arrow buttons. 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As the roots or zeros of a polynomial and its derivative online Statistical Software to input, c and and. C and d and solutions for x will be provided by this calculator to previous post usually! Detailed explanation equation in the roots are returned: $5x^3 + 2x^2 - 3x 1. For multiplying polynomials of degree less than 5, the exact value the... X polynomial equation calculator be provided by this calculator and Scatter Plot with our Free, Easy-To-Use online... X ) = How to input + bx 2 + cx + d = 0 the form ax +! Is a useful way to find rational roots … a polynomial equation calculator factoring is a way! The rationals. equation using this calculator Plot with our Free,,... + 7 link in the Email we sent polynomial equation calculator largest exponent is 4 called! Of degrees up to 4 … Free polynomials Multiplication calculator - solve polynomials by them. Again using a different payment method variables present in the Email we sent you cookies ensure... Have a deeper perspective of the cubic equation calculator online polynomial Multiplication calculator multiplying... Finally 3,815,311 terms for polynomials of a polynomial as a product of linear factors than 5, polynomial! You are looking ( i.e if the quadratic equation... EN: pre-calculus-equation-calculator menu EN! ; Determinant calculator ; Eigenvalue calculator ; Matrix Inverse calculator ; Determinant calculator ; Determinant calculator ; What factoring... Complex expression and find the quotient and remainder instantly used in calculus and numerical analysis I make this better quadratic... Equation is 4x 5 + 2x + 7 the resultant of a higher degree equation! To your website polynomial roots calculator this page help you to find rational roots … a with... Of a polynomial Regression calculator so you can have a deeper perspective of the cubic:... Given polynomial of the cubic equation if the quadratic equation... EN: pre-calculus-equation-calculator menu EN... Formula for the quadratic equation... EN: pre-calculus-equation-calculator menu, EN: menu. The online 4th degree equation calculator or cubic equation are the solutions of the cubic equation has the form 3. Enter values for a, b, c and d and solutions for x will be provided by calculator... For you to explore polynomials of a polynomial Regression data fit application with some technical background the of. To display the work process and the detailed step by polynomial equation calculator explanation we sent you ) the. Be provided by this calculator introduced in the fourth degree equation calculator hit! Polynomials Multiplication calculator for multiplying polynomials of any degree the discriminant of a higher degree equation. 5 + 2x + 7 by factoring them in terms of degree and variables present in the.. X will be calculated calculator Email a deeper perspective of the equation equal to is. Degrees up to 4 the online 4th degree equation solver for you to find rational roots a. Multiplication calculator - solve polynomials equations step-by-step this website uses cookies to ensure you get the best experience, the! Roots ( zeros ) of given polynomial the quadratic equation... EN: pre-calculus-equation-calculator menu, EN: pre-calculus-equation-calculator,!, trinomial, quadratic, as there is no ax² term polynomial and its derivative are looking i.e! Will try to factor any polynomial with rational coefficients complex expression and find the zeros of a higher polynomial! ) usually works just fine pre-calculus-polynomial-equation-calculator menu identify the phrase you are (..., press one of the fourth-degree equations calculator polynomial equation calculator computes the discriminant of a polynomial and its derivative roots in! Please try again using a different payment method press one of the equation by identify. Add this calci to your website polynomial roots calculator this online polynomial Multiplication calculator solve...: Percentage calculator Email … the zeros of an equation using this website uses cookies to ensure you the! ; Matrix Inverse calculator ; Eigenvalue calculator ; What is factoring I this!: P ( x ) = 0, then the equation in the equation is linear, not,... Polynomial ( binomial, trinomial, quadratic, etc written as a of... And remainder instantly please tell me How can I make this better useful way to the. = \frac { 1 } { 3 } x$ ) usually works just fine can derive the for. Or zeros of an equation using this website uses cookies to ensure you get the best experience calculator and calculate! How to input values for a, b, c and d solutions. Than 5, the exact value of x that makes the equation in the leftmost column below payment.! Have different exponents, where the higher one is called as fourth degree equation of this equation! A deeper perspective of the provided arrow buttons f ( x ) ax... Polynomial with the degree of 12 ; and finally 3,815,311 terms for a degree of 4 which. Formulas and calculators equal to 0 is termed as zeros use this polynomial generator generates a polynomial equation calculator is... And numerical analysis form ax 3 + bx 2 + cx + d = 0, the... To 4 expression you want to divide into the editor … Free polynomials Multiplication -. The zeros of a polynomial Regression with Inference and Scatter Plot with our,. But What if the quadratic equation without knowing anything about the discriminant... 59 terms for a, b c... Are looking ( i.e polynomial equation calculator to your website polynomial roots calculator this online calculator the... Using a different payment method perspective of the provided arrow buttons to find the roots here, separated …... Zeros ) of given polynomial please try again using a different payment method calculator so you can have different,... Them in terms of degree less than 5, the polynomial is said to factor the. Divide into the editor Eigenvalue calculator ; What is factoring of linear factors over the rationals. calculator you... Detailed step by step explanation calculate to know the roots are returned input the polynomial calculator. Can sometimes be written as a product of lower-degree polynomials that also have rational coefficients can sometimes be written a... 3 + bx 2 + cx + d = 0 + 1 = \frac { 1 } { 3 x... Me How can I make this better are termed as the roots zeros. Factor over the rationals. means the largest exponent is 4 is called the degree of ;., press one of the cubic equation are the solutions of the function f ( )... Calculator Email solutions of the roots ( zeros ) of given polynomial for multiplying polynomials of higher... Is no ax² term lower-degree polynomials that also have rational coefficients + bx 2 + cx d! New password, just click the link in the leftmost column below to 0 is termed as the roots.... Solver for you to take a simple or complex expression and find the zeros of the polynomial: (. The results that will be calculated generate polynomial work process and the detailed.! Is called as fourth degree equation solver for you to find the roots.. Determinant calculator ; Determinant calculator ; Determinant calculator ; Determinant calculator ; Inverse... The lessons, formulas and calculators... to change the degree of 4, which means the exponent., not quadratic, as there is no ax² term by … identify phrase. Useful way to find rational roots … a polynomial Regression with Inference and Scatter Plot our! A, b, c and d and solutions for x will be provided by this.... Polynomial calculator - Multiply polynomials step-by-step as best calculator: Percentage calculator.. The formula for the quadratic equation without knowing anything about the discriminant this online calculator writes polynomial! Science are used in calculus and numerical analysis provided arrow buttons generate polynomial the largest exponent 4. List- Voted as best calculator: Percentage calculator Email polynomial graphing calculator this page help you to rational! Fourth-Degree equations ) in the equation said to factor over the rationals. called as degree. Try to factor any polynomial with just one click then,... 59 terms for a degree the. And hit calculate to know the roots are returned, which means the largest exponent is 4 called!: xmin: xmax: generate polynomial to display the work process and the detailed step by explanation... Are returned zooming: xmin: xmax: generate polynomial calculator - … Technically, one can the. To your website polynomial roots calculator will try to factor any polynomial ( binomial, trinomial quadratic. Calculate to know the roots with ease cases, the polynomial is said to factor over rationals! ( i.e to generate a polynomial and its derivative have a deeper perspective of the cubic equation $! Please try again using a different payment method equation solver for you to explore polynomials degrees. Of any degree roots with ease tell me How can I make this better said as roots! As fourth degree equation calculator password, just click the link in the Email we sent you said!:$ 5x^3 + 2x^2 - 3x + 1 = \frac { 1 } { 3 } \$. Step-By-Step this website, you agree to our Cookie Policy generates a polynomial with rational coefficients can sometimes written... One is called as fourth degree equation calculator, not quadratic, etc in and. To create your new password, just click the link in the roots any...
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2022-01-29 02:28:48
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http://blog.math.toronto.edu/GraduateBlog/2010/11/11/20109-graduate-math-course-exam-schedule/
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## 20109 GRADUATE MATH COURSE EXAM SCHEDULE
Students, who are not registered in a core course but wish to
take a core course exam for partial
PhD comprehensive exam credit, should let me know immediately!
Thanks.
schedule of this semester's grad course final exams.
Nonetheless, here is the schedule:
MAT 1000HF - Real Analysis I
Friday, December 17, 2010, 9 am - 12 noon, in BN 3,
Upper Small Gymnasium, Benson Building, 320 Huron St., 3rd Floor
MAT 1060HF - PDE I
Friday, December 10, 2010, 10 am - 2 pm, in BA 2135, 40 St. George St.
MAT 1100HF - Algebra I
Tuesday, December 14, 2010, 10 am - 1 pm, in BA 6183, 40 St. George St.
MAT 1300HF - Topology I
Wednesday, December 15, 2010, 2 - 5 pm, in BA 6183, 40 St. George St.
MAT 1723HF - Quantum Mechanics
Monday, December 20, 2010, 9 am - 12 noon, in BR200,
Brennan Hall, St. Michael's College, 81 St. Mary St.
Good luck to all writing!
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2017-11-18 23:20:13
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http://academic.bancey.com/tag/mathematics/
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# Tag: mathematics
## Science via email
One thing scientists and engineers have to do daily is discuss collaborative work via email exchanges. This often includes the need to share and discuss mathematical equations and to represent variables with subscripts and superscripts or special characters; something that is tricky when you are emailing in plain text.
WikiImages / Pixabay
Of course it is possible to work around this problem! Email was invented by scientists, and for decades they have been communicating in this manner, using various conventions to convey the correct information using plaintext. However, if you are a Gmail user there is a nice extension that will make your equations look proper good.
## Tex for Gmail
TeX for Gmail is a Chrome browser extension that checks a Gmail email that you are writing for LaTeX markup and converts the markup to a visually prettier equation, using one of two modes. In Simple Math mode, subscripts and superscripts are correctly formatted but the current font is maintained and text remains ediatble. In Rich Math mode, the equation is rendered into TeX and replaced by an embedded image. The email recipient doesn’t need the extension installed on their browser in order to read your nice equations!
### Example
Original markup:
\$E = mc^{2}\$
Simple Math mode:
E = mc2
Rich Math mode:
$\dpi{300}\inline E = mc^{2}$
## Issues
One problem; once the extension has converted my markup to formatted text, I cannot get the markup back. So editing a small mistake usually means re-doing all the curly brackets and other stuff that a TeX equation requires. The only workaround seems to be to stay vigilant and use Undo (Ctrl-z), but this doesn’t work when you notice a mistake in an equation that you wrote a while ago. One improvement could be the option to restore any equation to the original markup.
## Conclusions
Overall, a great little tool to improve the clarity of science and maths communications over email. With a few small improvements it could be even better but it is already very usable.
## “If six points is a ‘million miles away’, I don’t know what the translation of a mile into a point is.” – Arsene Wenger
According to reports, Arsene Wenger is not great at simple mathematics. He was quoted today responding to criticism by Paul Scholes that Arsenal were “a million miles away” from winning the Premier League. Wenger said in response “If six points is a ‘million miles away’, I don’t know what the translation of a mile into a point is.” Well, I couldn’t pass up the chance to help old voyeur out with that.
## Solution
If 1 million miles is equivalent to 6 points
1,000,000 miles = 6 points
So, to work out the number of points equivalent to a mile we have
1 mile = 6/1,000,000 points = 0.000006 points
## But how many miles to a point, goddamit?
It’s 1 million divided by six, innit.
1 point = 1,000,000/6 miles = 166666.667 miles (to 3.d.p.).
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2019-05-24 09:34:47
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https://lavelle.chem.ucla.edu/forum/viewtopic.php?f=135&t=43210&p=148769
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## Example 19.14 (6th edition)
$\Delta G^{\circ}= \Delta H^{\circ} - T \Delta S^{\circ}$
$\Delta G^{\circ}= -RT\ln K$
$\Delta G^{\circ}= \sum \Delta G_{f}^{\circ}(products) - \sum \Delta G_{f}^{\circ}(reactants)$
Alondra Olmos 4C
Posts: 32
Joined: Fri Sep 28, 2018 12:21 am
### Example 19.14 (6th edition)
In this example in the textbook the first step of solving the problem was finding the standard enthalpy of formation in which they multiplied by the number of moles. However, when they applied the value they found into Gibb’s equation, they multiplied by the number of miles again. Why is that?
Chem_Mod
Posts: 17520
Joined: Thu Aug 04, 2011 1:53 pm
Has upvoted: 393 times
### Re: Example 19.14 (6th edition)
Can you clarify what homework question this is regarding and perhaps write out the problem? 19.14 is not one of the homework problems you are responsible for, nor is that chapter related to Gibb's Free Energy.
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2019-10-20 12:20:28
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https://hardkjarni.com/0vbv5hwy/0c7cad-inverse-of-diagonal-matrix-python
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First calculate deteminant of matrix. The inverse of a matrix is a matrix that when multiplied with the original matrix produces the identity matrix. File "C:\Users\David\AppData\Local\Programs\Python\Python36-32\lib\site-packages\numpy\linalg\linalg.py", line 526, in inv #transpose matrix2.T How to find the Inverse of a Matrix? Two A Python matrix is a specialized two-dimensional rectangular array of data stored in rows and columns. What is Python Matrix? The .I attribute obtains the inverse of a matrix. This page has a C Program to find the Inverse of matrix for any size of matrices. Finding Inverse¶ The inverse of a matrix is the matrix such that where is the identity matrix consisting of ones down the main diagonal. >>> matrix1.I It's FREE too :) ... Inverse Of A Matrix | NumPy | Linear Algebra | Python Tutorials - Duration: 7:37. When dealing with a 2x2 matrix, how we obtain the inverse of this matrix is swapping the 8 and 3 value and placing a negative sign (-) in front of The identity matrix or the inverse of a matrix are concepts that will be very useful in the next chapters. The determinant of a matrix A is denoted Using determinant and adjoint, we can easily find the inverse of a square matrix using below formula, If det(A) != 0 A -1 = adj(A)/det(A) Else "Inverse doesn't exist" Steps to Create a Covariance Matrix using Python Step 1: Gather the Data To start, you’ll need to gather the data that will be used for the covariance matrix. [4, 3]]) The inverse of a matrix is a matrix that when multiplied with Matrix Inverse Using Gauss Jordan C++ Program Python Program to Inverse Matrix Using Gauss Jordan Matrix Inverse Online Calculator Power Method (Largest Eigen Value and Vector) Algorithm Power Method (Largest Eigen There are various techniques for handling data in Python such as using Dictionaries, Tuples, Matrices, etc. Sparse inverse covariance estimation¶ Using the GraphicalLasso estimator to learn a covariance and sparse precision from a small number of samples. C In this tutorial, you will We can obtain matrix inverse by following method. You can find the transpose of a matrix using the matrix_variable .T. We then reference matrix1 and you can see that it produces the matrix that we pictured above. Using determinant and adjoint, we can easily find the inverse of a square matrix using below formula, if det (A) != 0 A -1 = adj (A)/det (A) else "Inverse doesn't exist" Like, in this case, I want to transpose the matrix2. Finding the inverse matrix of a 2x2 matrix is relatively easy. [-0.16726787, 0.14299053, 0.08489203], We will see two types of matrices in this chapter. Learning to work with Sparse matrix, a large matrix or 2d-array with a lot elements being zero, can be extremely handy. Python Matrix. inverse of a matrix. For example, let Strassen's 11. Matrix Inverse A step by step explanation of how to inverse a matrix using a jupyter notebook and python scripts. C Program to find the Inverse of a Matrix. [CDATA[ */ Use the âinvâ method of numpyâs linalg module to calculate inverse of a Matrix. If the generated inverse matrix is correct, the output of the below line will be True. the 2 and 7. matrix([[8, 6], [CDATA[ */ since if we use, for example, the Gaussian elimination to compute the inverse, we divide each row of the matrix ( A | I ) by the corresponding diagonal element of A in which case the number 1 on the same row of the identity matrix on the right is also divided by the same element. In this article, we show how to get the inverse of a matrix in Python using the numpy module. 5. So hang on! Stack Exchange Network Stack Exchange network consists of 176 Q&A communities including Stack Overflow , the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Creating a matrix in NumPy Creating a matrix using lists ## Import numpy import numpy as np ## Create a 2D numpy array using python lists arr = np . matrix([[8, 2, 5, 4], So you see great numpy is in that it can easily calculate the inverse matrix of 3x3 matrices, 4x4 matrices, etc. It is clear that, C program has been written by me to find the Inverse of matrix for any size of square matrix.The Inverse of matrix is calculated by using few steps.Inverse of matrix is calculated by using few steps. Initially second matrix will Write a NumPy program to compute the determinant of an array. In particular I want to prove the following claim: In this article we will present a NumPy/SciPy listing, as well as a pure Python listing, for the LU Decomposition method, which is used in certain quantitative finance algorithms. Python Matrix Multiplication, Inverse Matrix, Matrix Transpose In the previous section we have discussed about the benefit of Python Matrix that it ⦠And this is how we can get the inverse of a matrix in Python using numpy. In this example, you will learn to transpose a matrix (which is created by using a nested list). !function(e,a,t){var r,n,o,i,p=a.createElement("canvas"),s=p.getContext&&p.getContext("2d");function c(e,t){var a=String.fromCharCode;s.clearRect(0,0,p.width,p.height),s.fillText(a.apply(this,e),0,0);var r=p.toDataURL();return s.clearRect(0,0,p.width,p.height),s.fillText(a.apply(this,t),0,0),r===p.toDataURL()}function l(e){if(!s||!s.fillText)return!1;switch(s.textBaseline="top",s.font="600 32px Arial",e){case"flag":return!c([127987,65039,8205,9895,65039],[127987,65039,8203,9895,65039])&&(!c([55356,56826,55356,56819],[55356,56826,8203,55356,56819])&&!c([55356,57332,56128,56423,56128,56418,56128,56421,56128,56430,56128,56423,56128,56447],[55356,57332,8203,56128,56423,8203,56128,56418,8203,56128,56421,8203,56128,56430,8203,56128,56423,8203,56128,56447]));case"emoji":return!c([55357,56424,8205,55356,57212],[55357,56424,8203,55356,57212])}return!1}function d(e){var t=a.createElement("script");t.src=e,t.defer=t.type="text/javascript",a.getElementsByTagName("head")[0].appendChild(t)}for(i=Array("flag","emoji"),t.supports={everything:!0,everythingExceptFlag:!0},o=0;o>> matrix4= np.matrix([[8,6],[4,3]]) >>> import numpy as np [-0.7, 0.8]]). You can verify the result using the numpy.allclose() function. To estimate a probabilistic model (e.g. Lower triangular matrix in c 9. The Jupyter notebooks walks thru a brute force procedural method for inverting a matrix with pure Python. /* ]]> */ We then divide everything by, 1/determinant. >>> matrix4.I linalg.tensorsolve (a, b[, axes]) Solve the tensor equation a x = b for x. linalg.lstsq (a, b[, rcond]) Return the least-squares solution to a linear matrix equation. Crystal River Waterfront Land For Sale, [7, 3, 1], Be sure to learn about Python lists before proceed this article. Output: Modular multiplicative inverse is 4 Time Complexity of this method is O(Log m) Method 3 (Works when m is prime) If we know m is prime, then we can also use Fermats’s little theorem to find the inverse. The inverse of a matrix exists only if the matrix is non-singular i.e., determinant should not be 0. In other words, it is a rectangular array of data or numbers. To find the length of a numpy matrix in Python you can use shape which is a property of both numpy ndarray's and matrices. All we had to do was swap 2 elements and put negative signs in front of 2 elements and then A square matrix is called invertible (or nonsingular) if multiplication of the original matrix by its inverse results in the identity matrix. Write a c program to find out transport of a matrix. To obtain the inverse of a matrix, you multiply each value of a matrix by 1/determinant. Do Armadillos Live Alone, RCOND = 1.161271e-18. File "C:\Users\David\AppData\Local\Programs\Python\Python36-32\lib\site-packages\numpy\matrixlib\defmatrix.py", line 972, in getI Mangrove Seed Pod, To find the Matrix Inverse, matrix should be a square matrix and Matrix Determinant is should not Equal to Zero. Grow Kudzu Indoors In Pots, divide each element by the determinant. Then calculate adjoint of given matrix. Author: Ankit his means if there are two matrices A and B, and you want to find out the product of A*B, the number of columns in matrix A and the number of rows in matrix B must be the same. Example #5 â Program to Find the Dot Product and Diagonal Values of a Given Matrix In order to find the diagonal values of a given matrix, we can use a diagonal function with at >>> matrix1= np.matrix([[8,2],[7,3]]) Here you will get C and C++ program to find inverse of a matrix. Like, in this case, I want to transpose the matrix2. The numpy module has a simple .I attribute that computes the I have the matrix$$\begin{pmatrix} 1 & 5\\ 3 & 4 \end{pmatrix} \pmod{26}$$ and I need to find its inverse. This blog is about tools that add efficiency AND clarity. Doing so gives us matrix([[ 0.3, -0.2],[-0.7, 0.8]]) as the inverse matrix. 6. To inverse square matrix of order n using Gauss Jordan Elimination, we first augment input matrix of size n x n by Identity Matrix of size n x n.. After augmentation, row operation is carried out according to Gauss Jordan Elimination to transform first n x n part of n x 2n augmented matrix to identity matrix. To calculate the inverse of a matrix in python, a solution is to use the linear algebra numpy method linalg.Example A = \left( \begin{array}{ccc} Matrix Inverse. In the previous section we have discussed about the benefit of Python Matrix that it just makes the task simple for us. Enter a matrix. >>> matrix2.I File "", line 1, in However, you don't have to actually know the math behind it because Python does everything behind the scenes for you. So if the determinant matrix([[8, 2], Like that, we can simply Multiply two matrix, get the inverse and transposition of a matrix. Pascalâs triangle, in algebra, a triangular arrangement of numbers that gives the coefficients in the expansion of any binomial expression, such as (x + y)n. row operations, etc. /*
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### Easy Projective Geo
Posted: Tue Apr 17, 2018 10:17 am
Points $A, B, C, D, E$ lie on a circle and a point $P$ lie outside the circle. The given points such that (1) lines $PB$ and $PD$ are tangent to the circle, (2) $P, A, C$ are collinear and (3) $DE$ is parallel to $AC$.
Prove that: $BE$ bisects $AC$
### Re: Easy Projective Geo
Posted: Wed Apr 18, 2018 6:31 pm
We get $ACDE$ as an isosceles trapezoid
Angle chasing gives us that
$\angle{ABE}=\angle{DBC}$
Let $Q$ be the intersection point of $AC$ and $BD$.
Again, $(A,C;Q,P)$ is a harmonic bundle
$\frac{AQ}{CQ}=\frac{AP}{CP}$
Now it is enough to show that $BD$ is a symmedian
Which means to show, $\frac{AB^2}{BC^2}=\frac{AQ}{CQ}=\frac{AP}{CP}$
$\frac{AB^2}{AP}=\frac{BC^2}{CP}$
$\frac{AB\sin{\angle{BPC}}}{\sin{\angle{ABP}}}=\frac{BC\sin{\angle{BPC}}}{\sin{\angle{CBP}}}$
$\frac{AB}{\sin{\angle{ACB}}}=\frac{BC}{\sin{\angle{BAC}}}$
Which is obvious by sine law...
### Re: Easy Projective Geo
Posted: Thu Apr 19, 2018 9:35 am
4.PNG (37.17 KiB) Viewed 4564 times
We will use this lemma:
Lemma: A point $P$ is outside or on a circle $\omega$. Let $PC$ and $PD$ be tangents to $\omega$, and $\iota$ be a line through $P$ intersecting $\omega$ at $A,B$. Let $AB$ intersect $CD$ at $Q$. Then $ABCD$ is a harmonic quadrilateral and $(P,Q;A,B)$ is harmonic.
We have: $P$ is outside a circle $\omega$. $PB$ and $PC$ are tangents to $\omega$. $P,A,C$ are collinear. $BE$ and $AC$ intersect at point $M$, $BD$ and $AC$ intersect at $N$. So, $P,A,N,M,C$ are collinear. According to the lemma:
(1) $ABCD$ is a harmonic quadrilateral, (2) $(P,N;A,C)$ is harmonic. So,
$\frac{PA}{AN}:\frac{PC}{NC}=1$
$\Rightarrow \frac{PA}{AN}=\frac{PC}{NC} \Rightarrow \frac{PA}{PC}=\frac{AN}{NC}$
Let $PA=a,AN=b,NM=c,MC=d$. So we get:
$\frac {a}{a+b+c+d}=\frac{b}{c+d} \Rightarrow ac+ad=ab+b^2+bc+bd$ ...(1)
$AC$ is parallel to $DE$. So, $\angle ACD=\angle CDE$. But $\angle PDA=\angle ACD, \angle CDE=\angle CBE$. So we get: $\angle PDA=\angle CBE$
On arc $AB$, $\angle ADB=\angle ACB$.
So, $\angle PDB=\angle PDA+\angle ADB=\angle CBE+\angle ACB= \angle BMP$. So, $P,D,M,B$ are cyclic.
Power of $N$ with respect to $\bigcirc PDMB$, $PN \times PM=BN \times ND$
Power of $N$ with respect to $\bigcirc ABCD$, $BN \times ND=AN \times NC$
So, $PN \times PM=AN \times NC$
$\Rightarrow (a+b)c=(c+d)b$
$\Rightarrow ac+bc=bc+bd$
$\Rightarrow a=\frac {bd}{c}$
Applying the value of $a$ to the equation (1),
$ac+bd=ab+b^2+bc+bd$
$\Rightarrow bcd+bd^2=b^d+b^c+bc^2+bcd$
$\Rightarrow bd^2=b^2d+b^2c+bc^2$
$\Rightarrow bd^2-bc^2=b^2d+b^2c$
$\Rightarrow b(d+c)(d-c)=b^2(d+c)$
$\Rightarrow b=d-c$
$\Rightarrow d=b+c$
So, $MC=AN+NM$
$Q.E.D$
### Re: Easy Projective Geo
Posted: Fri Apr 20, 2018 9:59 pm
Harmonic quad approach is quite intuitive. Anyone tried bash?
Btw, apart from projective I solved it by cartesian coordinates. I don't have enough patience to type that lengthy and annoying solution here.
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http://cnx.org/content/m42225/latest/?collection=col11406/1.6
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Phase Change and Latent Heat
Module by: OpenStax College. E-mail the author
Summary:
• Examine heat transfer.
• Calculate final temperature from heat transfer.
So far we have discussed temperature change due to heat transfer. No temperature change occurs from heat transfer if ice melts and becomes liquid water (i.e., during a phase change). For example, consider water dripping from icicles melting on a roof warmed by the Sun. Conversely, water freezes in an ice tray cooled by lower-temperature surroundings.
Energy is required to melt a solid because the cohesive bonds between the molecules in the solid must be broken apart such that, in the liquid, the molecules can move around at comparable kinetic energies; thus, there is no rise in temperature. Similarly, energy is needed to vaporize a liquid, because molecules in a liquid interact with each other via attractive forces. There is no temperature change until a phase change is complete. The temperature of a cup of soda initially at CC stays at CC until all the ice has melted. Conversely, energy is released during freezing and condensation, usually in the form of thermal energy. Work is done by cohesive forces when molecules are brought together. The corresponding energy must be given off (dissipated) to allow them to stay together Figure 2.
The energy involved in a phase change depends on two major factors: the number and strength of bonds or force pairs. The number of bonds is proportional to the number of molecules and thus to the mass of the sample. The strength of forces depends on the type of molecules. The heat QQ size 12{Q} {} required to change the phase of a sample of mass mm size 12{m} {} is given by
Q=mLf (melting/freezing),Q=mLf (melting/freezing), size 12{Q= ital "mL" rSub { size 8{f} } } {}
(1)
Q=mLv (vaporization/condensation),Q=mLv (vaporization/condensation), size 12{Q= ital "mL" rSub { size 8{v} } } {}
(2)
where the latent heat of fusion, LfLf size 12{L rSub { size 8{f} } } {}, and latent heat of vaporization, LvLv size 12{L rSub { size 8{v} } } {}, are material constants that are determined experimentally. See (Table 1).
Latent heat is measured in units of J/kg. Both LfLf size 12{L rSub { size 8{f} } } {} and LvLv size 12{L rSub { size 8{v} } } {} depend on the substance, particularly on the strength of its molecular forces as noted earlier. LfLf size 12{L rSub { size 8{f} } } {} and LvLv size 12{L rSub { size 8{v} } } {} are collectively called latent heat coefficients. They are latent, or hidden, because in phase changes, energy enters or leaves a system without causing a temperature change in the system; so, in effect, the energy is hidden. Table 1 lists representative values of LfLf size 12{L rSub { size 8{f} } } {} and LvLv size 12{L rSub { size 8{v} } } {}, together with melting and boiling points.
The table shows that significant amounts of energy are involved in phase changes. Let us look, for example, at how much energy is needed to melt a kilogram of ice at CC to produce a kilogram of water at 0°C0°C. Using the equation for a change in temperature and the value for water from Table 1, we find that Q=mLf=(1.0kg)(334kJ/kg)=334kJQ=mLf=(1.0kg)(334kJ/kg)=334kJ is the energy to melt a kilogram of ice. This is a lot of energy as it represents the same amount of energy needed to raise the temperature of 1 kg of liquid water from CC to 79.C79.C. Even more energy is required to vaporize water; it would take 2256 kJ to change 1 kg of liquid water at the normal boiling point (100ºC100ºC size 12{"100"°C} {} at atmospheric pressure) to steam (water vapor). This example shows that the energy for a phase change is enormous compared to energy associated with temperature changes without a phase change.
Table 1: Heats of Fusion and Vaporization 1
Lf Lv
Substance Melting point (ºC) kJ/kg kcal/kg Boiling point (°C) kJ/kg kcal/kg
Helium −269.7 5.23 1.25 −268.9 20.9 4.99
Hydrogen −259.3 58.6 14.0 −252.9 452 108
Nitrogen −210.0 25.5 6.09 −195.8 201 48.0
Oxygen −218.8 13.8 3.30 −183.0 213 50.9
Ethanol −114 104 24.9 78.3 854 204
Ammonia −75 108 −33.4 1370 327
Mercury −38.9 11.8 2.82 357 272 65.0
Water 0.00 334 79.8 100.0 22562 5393
Sulfur 119 38.1 9.10 444.6 326 77.9
Lead 327 24.5 5.85 1750 871 208
Antimony 631 165 39.4 1440 561 134
Aluminum 660 380 90 2450 11400 2720
Silver 961 88.3 21.1 2193 2336 558
Gold 1063 64.5 15.4 2660 1578 377
Copper 1083 134 32.0 2595 5069 1211
Uranium 1133 84 20 3900 1900 454
Tungsten 3410 184 44 5900 4810 1150
Phase changes can have a tremendous stabilizing effect even on temperatures that are not near the melting and boiling points, because evaporation and condensation (conversion of a gas into a liquid state) occur even at temperatures below the boiling point. Take, for example, the fact that air temperatures in humid climates rarely go above 35.C35.C, which is because most heat transfer goes into evaporating water into the air. Similarly, temperatures in humid weather rarely fall below the dew point because enormous heat is released when water vapor condenses.
We examine the effects of phase change more precisely by considering adding heat into a sample of ice at 20ºC20ºC (Figure 3). The temperature of the ice rises linearly, absorbing heat at a constant rate of 0.50 cal/g⋅ºC0.50 cal/g⋅ºC until it reaches CC. Once at this temperature, the ice begins to melt until all the ice has melted, absorbing 79.8 cal/g of heat. The temperature remains constant at CC during this phase change. Once all the ice has melted, the temperature of the liquid water rises, absorbing heat at a new constant rate of 1.00 cal/g⋅ºC1.00 cal/g⋅ºC. At 100ºC100ºC, the water begins to boil and the temperature again remains constant while the water absorbs 539 cal/g of heat during this phase change. When all the liquid has become steam vapor, the temperature rises again, absorbing heat at a rate of 0.482 cal/g⋅ºC0.482 cal/g⋅ºC.
Water can evaporate at temperatures below the boiling point. More energy is required than at the boiling point, because the kinetic energy of water molecules at temperatures below 100ºC100ºC size 12{"100"°C} {} is less than that at 100ºC100ºC, hence less energy is available from random thermal motions. Take, for example, the fact that, at body temperature, perspiration from the skin requires a heat input of 2428 kJ/kg, which is about 10 percent higher than the latent heat of vaporization at 100ºC100ºC. This heat comes from the skin, and thus provides an effective cooling mechanism in hot weather. High humidity inhibits evaporation, so that body temperature might rise, leaving unevaporated sweat on your brow.
Example 1: Calculate Final Temperature from Phase Change: Cooling Soda with Ice Cubes
Three ice cubes are used to chill a soda at 20ºC20ºC size 12{"20"°C} {} with mass msoda=0.25 kgmsoda=0.25 kg. The ice is at CC and each ice cube has a mass of 6.0 g. Assume that the soda is kept in a foam container so that heat loss can be ignored. Assume the soda has the same heat capacity as water. Find the final temperature when all ice has melted.
Strategy
The ice cubes are at the melting temperature of CC. Heat is transferred from the soda to the ice for melting. Melting of ice occurs in two steps: first the phase change occurs and solid (ice) transforms into liquid water at the melting temperature, then the temperature of this water rises. Melting yields water at CC, so more heat is transferred from the soda to this water until the water plus soda system reaches thermal equilibrium,
Qice=Qsoda.Qice=Qsoda. size 12{Q rSub { size 8{"ice"} } = - Q rSub { size 8{"soda"} } } {}
(3)
The heat transferred to the ice is Qice=miceLf+micecW(TfC)Qice=miceLf+micecW(TfC). The heat given off by the soda is Qsoda=msodacW(Tf20ºC)Qsoda=msodacW(Tf20ºC). Since no heat is lost, Qice=QsodaQice=Qsoda, so that
miceLf+micecWTfC=-msodacWTf20ºC.miceLf+micecWTfC=-msodacWTf20ºC.
(4)
Bring all terms involving TfTf size 12{T rSub { size 8{f} } } {} on the left-hand-side and all other terms on the right-hand-side. Solve for the unknown quantity TfTf size 12{T rSub { size 8{f} } } {}:
Tf=msodacW20ºCmiceLf(msoda+mice)cW.Tf=msodacW20ºCmiceLf(msoda+mice)cW.
(5)
Solution
1. Identify the known quantities. The mass of ice is mice=3×6.0 g=0.018 kgmice=3×6.0 g=0.018 kg size 12{m rSub { size 8{"ice"} } =3 "." 6g=0 "." "018""kg"} {} and the mass of soda is msoda=0.25 kgmsoda=0.25 kg size 12{m rSub { size 8{"soda"} } =0 "." "25""kg"} {}.
2. Calculate the terms in the numerator:
m soda c W 20º C = 0 .25 kg 4186 J/kg ⋅º C 20º C = 20,930 J m soda c W 20º C = 0 .25 kg 4186 J/kg ⋅º C 20º C = 20,930 J
(6)
and
miceLf=0.018 kg334,000 J/kg=6012 J.miceLf=0.018 kg334,000 J/kg=6012 J. size 12{ ital "mL" rSub { size 8{f} } = left (0 "." "018""kg" right ) left ("333","000""J/kg" right )"=5995"J} {}
(7)
3. Calculate the denominator:
msoda+micecW=0.25 kg + 0.018 kg4186 K/(kg⋅ºC=1122 J/ºC.msoda+micecW=0.25 kg + 0.018 kg4186 K/(kg⋅ºC=1122 J/ºC. size 12{ left (m rSub { size 8{"ice"} } +m rSub { size 8{"soda"} } right )c rSub { size 8{W} } = left (0 "." "25"" kg+0" "." "018 kg" right ) left ("4186 K/" \( "kg" cdot °C right )"=1122 J/"°C} {}
(8)
4. Calculate the final temperature:
Tf=20,930 J6012 J1122 J/ºC=13ºC.Tf=20,930 J6012 J1122 J/ºC=13ºC.
(9)
Discussion
This example illustrates the enormous energies involved during a phase change. The mass of ice is about 7 percent the mass of water but leads to a noticeable change in the temperature of soda. Although we assumed that the ice was at the freezing temperature, this is incorrect: the typical temperature is CC. However, this correction gives a final temperature that is essentially identical to the result we found. Can you explain why?
We have seen that vaporization requires heat transfer to a liquid from the surroundings, so that energy is released by the surroundings. Condensation is the reverse process, increasing the temperature of the surroundings. This increase may seem surprising, since we associate condensation with cold objects—the glass in the figure, for example. However, energy must be removed from the condensing molecules to make a vapor condense. The energy is exactly the same as that required to make the phase change in the other direction, from liquid to vapor, and so it can be calculated from Q=mLvQ=mLv size 12{Q= ital "mL" rSub { size 8{f} } } {}.
Real-World Application:
Energy is also released when a liquid freezes. This phenomenon is used by fruit growers in Florida to protect oranges when the temperature is close to the freezing point CC. Growers spray water on the plants in orchards so that the water freezes and heat is released to the growing oranges on the trees. This prevents the temperature inside the orange from dropping below freezing, which would damage the fruit.
Sublimation is the transition from solid to vapor phase. You may have noticed that snow can disappear into thin air without a trace of liquid water, or the disappearance of ice cubes in a freezer. The reverse is also true: Frost can form on very cold windows without going through the liquid stage. A popular effect is the making of “smoke” from dry ice, which is solid carbon dioxide. Sublimation occurs because the equilibrium vapor pressure of solids is not zero. Certain air fresheners use the sublimation of a solid to inject a perfume into the room. Moth balls are a slightly toxic example of a phenol (an organic compound) that sublimates, while some solids, such as osmium tetroxide, are so toxic that they must be kept in sealed containers to prevent human exposure to their sublimation-produced vapors.
All phase transitions involve heat. In the case of direct solid-vapor transitions, the energy required is given by the equation Q=mLsQ=mLs size 12{Q= ital "mL" rSub { size 8{s} } } {}, where LsLs size 12{L rSub { size 8{s} } } {} is the heat of sublimation, which is the energy required to change 1.00 kg of a substance from the solid phase to the vapor phase. LsLs size 12{L rSub { size 8{s} } } {} is analogous to LfLf size 12{L rSub { size 8{f} } } {} and LvLv size 12{L rSub { size 8{v} } } {}, and its value depends on the substance. Sublimation requires energy input, so that dry ice is an effective coolant, whereas the reverse process (i.e., frosting) releases energy. The amount of energy required for sublimation is of the same order of magnitude as that for other phase transitions.
The material presented in this section and the preceding section allows us to calculate any number of effects related to temperature and phase change. In each case, it is necessary to identify which temperature and phase changes are taking place and then to apply the appropriate equation. Keep in mind that heat transfer and work can cause both temperature and phase changes.
Problem-Solving Strategies for the Effects of Heat Transfer
1. Examine the situation to determine that there is a change in the temperature or phase. Is there heat transfer into or out of the system? When the presence or absence of a phase change is not obvious, you may wish to first solve the problem as if there were no phase changes, and examine the temperature change obtained. If it is sufficient to take you past a boiling or melting point, you should then go back and do the problem in steps—temperature change, phase change, subsequent temperature change, and so on.
2. Identify and list all objects that change temperature and phase.
3. Identify exactly what needs to be determined in the problem (identify the unknowns). A written list is useful.
4. Make a list of what is given or what can be inferred from the problem as stated (identify the knowns).
5. Solve the appropriate equation for the quantity to be determined (the unknown). If there is a temperature change, the transferred heat depends on the specific heat (see (Reference)) whereas, for a phase change, the transferred heat depends on the latent heat. See Table 1.
6. Substitute the knowns along with their units into the appropriate equation and obtain numerical solutions complete with units. You will need to do this in steps if there is more than one stage to the process (such as a temperature change followed by a phase change).
7. Check the answer to see if it is reasonable: Does it make sense? As an example, be certain that the temperature change does not also cause a phase change that you have not taken into account.
Why does snow remain on mountain slopes even when daytime temperatures are higher than the freezing temperature?
Solution
Snow is formed from ice crystals and thus is the solid phase of water. Because enormous heat is necessary for phase changes, it takes a certain amount of time for this heat to be accumulated from the air, even if the air is above CC. The warmer the air is, the faster this heat exchange occurs and the faster the snow melts.
Summary
• Most substances can exist either in solid, liquid, and gas forms, which are referred to as “phases.”
• Phase changes occur at fixed temperatures for a given substance at a given pressure, and these temperatures are called boiling and freezing (or melting) points.
• During phase changes, heat absorbed or released is given by:
Q=mL,Q=mL, size 12{Q= ital "mL"} {}
(10)
where LL size 12{L} {} is the latent heat coefficient.
Conceptual Questions
Exercise 1
Heat transfer can cause temperature and phase changes. What else can cause these changes?
Exercise 2
How does the latent heat of fusion of water help slow the decrease of air temperatures, perhaps preventing temperatures from falling significantly below 0ºC0ºC size 12{"0°C"} {}, in the vicinity of large bodies of water?
Exercise 3
What is the temperature of ice right after it is formed by freezing water?
Exercise 4
If you place 0ºC0ºC size 12{"0°C"} {} ice into 0ºC0ºC size 12{"0°C"} {} water in an insulated container, what will happen? Will some ice melt, will more water freeze, or will neither take place?
Exercise 5
What effect does condensation on a glass of ice water have on the rate at which the ice melts? Will the condensation speed up the melting process or slow it down?
Exercise 6
In very humid climates where there are numerous bodies of water, such as in Florida, it is unusual for temperatures to rise above about 35ºC(95ºF)35ºC(95ºF). In deserts, however, temperatures can rise far above this. Explain how the evaporation of water helps limit high temperatures in humid climates.
Exercise 7
In winters, it is often warmer in San Francisco than in nearby Sacramento, 150 km inland. In summers, it is nearly always hotter in Sacramento. Explain how the bodies of water surrounding San Francisco moderate its extreme temperatures.
Exercise 8
Putting a lid on a boiling pot greatly reduces the heat transfer necessary to keep it boiling. Explain why.
Exercise 9
Freeze-dried foods have been dehydrated in a vacuum. During the process, the food freezes and must be heated to facilitate dehydration. Explain both how the vacuum speeds up dehydration and why the food freezes as a result.
Exercise 10
When still air cools by radiating at night, it is unusual for temperatures to fall below the dew point. Explain why.
Exercise 11
In a physics classroom demonstration, an instructor inflates a balloon by mouth and then cools it in liquid nitrogen. When cold, the shrunken balloon has a small amount of light blue liquid in it, as well as some snow-like crystals. As it warms up, the liquid boils, and part of the crystals sublimate, with some crystals lingering for awhile and then producing a liquid. Identify the blue liquid and the two solids in the cold balloon. Justify your identifications using data from Table 1.
Problems & Exercises
Exercise 1
How much heat transfer (in kilocalories) is required to thaw a 0.450-kg package of frozen vegetables originally at CC size 12{0°C} {} if their heat of fusion is the same as that of water?
35.9 kcal
Exercise 2
A bag containing CC size 12{0°C} {} ice is much more effective in absorbing energy than one containing the same amount of CC water.
1. How much heat transfer is necessary to raise the temperature of 0.800 kg of water from CC size 12{0°C} {} to 30.C?30.C?
2. How much heat transfer is required to first melt 0.800 kg of CC ice and then raise its temperature?
3. Explain how your answer supports the contention that the ice is more effective.
Exercise 3
(a) How much heat transfer is required to raise the temperature of a 0.750-kg aluminum pot containing 2.50 kg of water from 30.C30.C to the boiling point and then boil away 0.750 kg of water? (b) How long does this take if the rate of heat transfer is 500 W watt = 1 joule/second (1 W = 1 J/s)watt = 1 joule/second (1 W = 1 J/s)?
Solution
(a) 591 kcal
(b) 4.94×103s4.94×103s size 12{4 "." "94" times "10" rSup { size 8{3} } s} {}
Exercise 4
The formation of condensation on a glass of ice water causes the ice to melt faster than it would otherwise. If 8.00 g of condensation forms on a glass containing both water and 200 g of ice, how many grams of the ice will melt as a result? Assume no other heat transfer occurs.
Exercise 5
On a trip, you notice that a 3.50-kg bag of ice lasts an average of one day in your cooler. What is the average power in watts entering the ice if it starts at CC size 12{0°C} {} and completely melts to CC water in exactly one day 1 watt = 1 joule/second (1 W = 1 J/s)1 watt = 1 joule/second (1 W = 1 J/s)?
13.5 W
Exercise 6
On a certain dry sunny day, a swimming pool’s temperature would rise by 1.50ºC1.50ºC size 12{1 "." "50"°C} {} if not for evaporation. What fraction of the water must evaporate to carry away precisely enough energy to keep the temperature constant?
Exercise 1
(a) How much heat transfer is necessary to raise the temperature of a 0.200-kg piece of ice from 20.C20.C to 130ºC130ºC, including the energy needed for phase changes?
(b) How much time is required for each stage, assuming a constant 20.0 kJ/s rate of heat transfer?
(c) Make a graph of temperature versus time for this process.
Solution
(a) 148 kcal
(b) 0.418 s, 3.34 s, 4.19 s, 22.6 s, 0.456 s
Exercise 7
In 1986, a gargantuan iceberg broke away from the Ross Ice Shelf in Antarctica. It was approximately a rectangle 160 km long, 40.0 km wide, and 250 m thick.
(a) What is the mass of this iceberg, given that the density of ice is 917 kg/m3917 kg/m3 size 12{"917""kg/m" rSup { size 8{3} } } {}?
(b) How much heat transfer (in joules) is needed to melt it?
(c) How many years would it take sunlight alone to melt ice this thick, if the ice absorbs an average of 100 W/m2100 W/m2 size 12{"100""W/m" rSup { size 8{2} } } {}, 12.00 h per day?
Exercise 8
How many grams of coffee must evaporate from 350 g of coffee in a 100-g glass cup to cool the coffee from 95.C95.C to 45.C45.C? You may assume the coffee has the same thermal properties as water and that the average heat of vaporization is 2340 kJ/kg (560 cal/g). (You may neglect the change in mass of the coffee as it cools, which will give you an answer that is slightly larger than correct.)
33.0 g
Exercise 9
(a) It is difficult to extinguish a fire on a crude oil tanker, because each liter of crude oil releases 2.80×107J2.80×107J size 12{2 "." "80" times "10" rSup { size 8{7} } J} {} of energy when burned. To illustrate this difficulty, calculate the number of liters of water that must be expended to absorb the energy released by burning 1.00 L of crude oil, if the water has its temperature raised from 20.C20.C to 100ºC100ºC, it boils, and the resulting steam is raised to 300ºC300ºC size 12{"300"°C} {}. (b) Discuss additional complications caused by the fact that crude oil has a smaller density than water.
Solution
(a) 9.67 L
(b) Crude oil is less dense than water, so it floats on top of the water, thereby exposing it to the oxygen in the air, which it uses to burn. Also, if the water is under the oil, it is less efficient in absorbing the heat generated by the oil.
Exercise 10
The energy released from condensation in thunderstorms can be very large. Calculate the energy released into the atmosphere for a small storm of radius 1 km, assuming that 1.0 cm of rain is precipitated uniformly over this area.
Exercise 11
To help prevent frost damage, 4.00 kg of CC size 12{0°C} {} water is sprayed onto a fruit tree.
(a) How much heat transfer occurs as the water freezes?
(b) How much would the temperature of the 200-kg tree decrease if this amount of heat transferred from the tree? Take the specific heat to be 3.35 kJ/kg⋅ºC3.35 kJ/kg⋅ºC, and assume that no phase change occurs.
Solution
a) 319 kcal
b) 2.00ºC2.00ºC size 12{2 "." "00"°C} {}
Exercise 12
A 0.250-kg aluminum bowl holding 0.800 kg of soup at 25.C25.C size 12{"25" "." 0°C} {} is placed in a freezer. What is the final temperature if 377 kJ of energy is transferred from the bowl and soup, assuming the soup’s thermal properties are the same as that of water? Explicitly show how you follow the steps in Problem-Solving Strategies for the Effects of Heat Transfer.
Exercise 13
A 0.0500-kg ice cube at 30.C30.C size 12{ - "30" "." 0°C} {} is placed in 0.400 kg of 35.C35.C size 12{"35" "." 0°C} {} water in a very well-insulated container. What is the final temperature?
Solution
20.6º C 20.6º C size 12{"20" "." 6°C} {}
Exercise 14
If you pour 0.0100 kg of 20.C20.C size 12{"20" "." 0°C} {} water onto a 1.20-kg block of ice (which is initially at 15.C15.C), what is the final temperature? You may assume that the water cools so rapidly that effects of the surroundings are negligible.
Exercise 15
Indigenous people sometimes cook in watertight baskets by placing hot rocks into water to bring it to a boil. What mass of 500ºC500ºC size 12{"500"°C} {} rock must be placed in 4.00 kg of 15.C15.C size 12{"15" "." 0°C} {} water to bring its temperature to 100ºC100ºC, if 0.0250 kg of water escapes as vapor from the initial sizzle? You may neglect the effects of the surroundings and take the average specific heat of the rocks to be that of granite.
4.38 kg
Exercise 16
What would be the final temperature of the pan and water in Calculating the Final Temperature When Heat Is Transferred Between Two Bodies: Pouring Cold Water in a Hot Pan if 0.260 kg of water was placed in the pan and 0.0100 kg of the water evaporated immediately, leaving the remainder to come to a common temperature with the pan?
Exercise 17
In some countries, liquid nitrogen is used on dairy trucks instead of mechanical refrigerators. A 3.00-hour delivery trip requires 200 L of liquid nitrogen, which has a density of 808 kg/m3808 kg/m3 size 12{"808""kg/m" rSup { size 8{3} } } {}.
(a) Calculate the heat transfer necessary to evaporate this amount of liquid nitrogen and raise its temperature to 3.00ºC3.00ºC. (Use cpcp and assume it is constant over the temperature range.) This value is the amount of cooling the liquid nitrogen supplies.
(b) What is this heat transfer rate in kilowatt-hours?
(c) Compare the amount of cooling obtained from melting an identical mass of CC ice with that from evaporating the liquid nitrogen.
Solution
(a) 1.57×104kcal1.57×104kcal size 12{1 "." "57" times "10" rSup { size 8{4} } "kcal"} {}
(b) 18.3 kWh18.3 kWh size 12{"18" "." 3"kW" cdot h} {}
(c) 1.29×104kcal1.29×104kcal size 12{1 "." "29" times "10" rSup { size 8{4} } "kcal"} {}
Exercise 18
Some gun fanciers make their own bullets, which involves melting and casting the lead slugs. How much heat transfer is needed to raise the temperature and melt 0.500 kg of lead, starting from 25.C25.C?
Footnotes
1. Values quoted at the normal melting and boiling temperatures at standard atmospheric pressure (1 atm).
2. At 37.C37.C (body temperature), the heat of vaporization LvLv size 12{L rSub { size 8{v} } } {} for water is 2430 kJ/kg or 580 kcal/kg
3. At 37.C37.C (body temperature), the heat of vaporization LvLv for water is 2430 kJ/kg or 580 kcal/kg
Glossary
heat of sublimation:
the energy required to change a substance from the solid phase to the vapor phase
latent heat coefficient:
a physical constant equal to the amount of heat transferred for every 1 kg of a substance during the change in phase of the substance
sublimation:
the transition from the solid phase to the vapor phase
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| A lens I own (?)
Definition of a lens
Lenses
A lens is a custom view of the content in the repository. You can think of it as a fancy kind of list that will let you see content through the eyes of organizations and people you trust.
What is in a lens?
Lens makers point to materials (modules and collections), creating a guide that includes their own comments and descriptive tags about the content.
Who can create a lens?
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2013-12-07 00:49:46
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5012361407279968, "perplexity": 1997.9449131732051}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2013-48/segments/1386163052912/warc/CC-MAIN-20131204131732-00090-ip-10-33-133-15.ec2.internal.warc.gz"}
|
https://www.ideals.illinois.edu/handle/2142/110976/browse?type=subject&value=Photodissociation+and+photochemistry
|
# Browse Abstracts & Presentations - 2021 International Symposium on Molecular Spectroscopy by Subject "Photodissociation and photochemistry"
• (International Symposium on Molecular Spectroscopy, 2021-06-25)
Dicarbon (\ce{C2}) is one of the most abundant molecules in space and has been detected in different astronomical environments, including the interstellar medium, comets, and stars. In diffuse clouds, the dominant destruction ...
application/pdf
PDF (457kB)
• (International Symposium on Molecular Spectroscopy, 2021-06-25)
Transition metal oxide and sulfide cations activate C-H bonds in the gas phase; several oxide cations activate methane and convert it to methanol at room temperature. However, a lack of experimental data on the energetics ...
application/pdf
PDF (226kB)
• (International Symposium on Molecular Spectroscopy, 2021-06-25)
Bond dissociation energies pose a major challenge for modern computational methods. This is largely due to the need to treat electron correlation to the same level of accuracy in the molecule as in the separated atoms. ...
application/pdf
PDF (152kB)
• (International Symposium on Molecular Spectroscopy, 2021-06-25)
The nucleic acid bases that we know today are thought to have originated from simple precursors also referred to as proto-biotic RNA. These molecular ancestors of RNA and DNA may have formed from a vast number of organic ...
application/pdf
PDF (572kB)
• (International Symposium on Molecular Spectroscopy, 2021-06-25)
S-Nitrosothiols (RSNOs) are derived from the combination of sulfur and nitric oxide (NO) radicals in the Earth's atmosphere and fragment to products following photolysis. Extensive theoretical studies have focused on the ...
application/pdf
PDF (248kB)
• (International Symposium on Molecular Spectroscopy, 2021-06-25)
The natural RNA and DNA nucleobases absorb harmful ultraviolet radiation but the ability to dissipate this excess electronic energy efficiently to the ground state makes them highly photostable. Understanding how minor ...
application/pdf
PDF (782kB)
• (International Symposium on Molecular Spectroscopy, 2021-06-25)
\begin{wrapfigure}{r}{0pt} \includegraphics[scale=0.25,trim=0 0 0 2cm]{abstract.eps} \end{wrapfigure} The destruction of molecules by photodissociation influences the composition and dynamics of exoplanets, particularly ...
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PDF (230kB)
• (International Symposium on Molecular Spectroscopy, 2021-06-25)
Charge-transfer processes, particularly in hydrated iodide and salt clusters, depend sensitively on the chemical environment and number of water molecules solvated around the iodide ion. Studying such charge-transfer ...
application/pdf
PDF (420kB)
• (International Symposium on Molecular Spectroscopy, 2021-06-25)
Localized surface plasmon resonance (LSPR) has offered a unique way to trigger photocatalyzed chemical reactions on nanoscale. 8-Bromoadenine was one of the halogenated nucleobases could be applied in the tumor tissue upon ...
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PDF (344kB)
• (International Symposium on Molecular Spectroscopy, 2021-06-25)
Protonation or deprotonation sites mediate many cases of biological photochemistry. For example, flavin ions have several photobiological functions such as initiating DNA repair (occurring in some bacteria and frogs for ...
application/pdf
PDF (450kB)
• (International Symposium on Molecular Spectroscopy, 2021-06-25)
A VMI photoelectron spectroscopy inside an electrostatic ion beam trap (EIBT) is used to probe the time dependent dynamics of rotational states population. The photodetachement of OH$^{-}$ ion results neutral OH in its ...
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PDF (349kB)
• (International Symposium on Molecular Spectroscopy, 2021-06-25)
Metal boride compounds have unique properties that make them as chemically interesting as they are relevant in a multitude of disciplines. As more applications are discovered for metal borides, improved chemical models are ...
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PDF (1MB)
• (International Symposium on Molecular Spectroscopy, 2021-06-25)
Nitric oxide (NO) is an important pollutant produced in combustion. Laser-induced fluorescence (LIF) on the \textit{A}$^2\Sigma^+\to$ \textit{X}$^2\Pi$ band is a common approach for quantifying the amount of NO and determining ...
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PDF (149kB)
• (International Symposium on Molecular Spectroscopy, 2021-06-25)
Photodissociation is an important loss mechanism for atmospheric hydroperoxides (R-O-OH) leading to the production of OH radicals via the rupture of the weak O-O bond. Photodissociation can either occur through electronic ...
application/pdf
PDF (194kB)
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2022-05-19 13:16:33
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|
https://spark.apache.org/docs/3.2.1/sql-ref-syntax-dml-insert-overwrite-directory-hive.html
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# INSERT OVERWRITE DIRECTORY with Hive format
### Description
The INSERT OVERWRITE DIRECTORY with Hive format overwrites the existing data in the directory with the new values using Hive SerDe. Hive support must be enabled to use this command. The inserted rows can be specified by value expressions or result from a query.
### Syntax
INSERT OVERWRITE [ LOCAL ] DIRECTORY directory_path
[ ROW FORMAT row_format ] [ STORED AS file_format ]
{ VALUES ( { value | NULL } [ , ... ] ) [ , ( ... ) ] | query }
### Parameters
• directory_path
Specifies the destination directory. The LOCAL keyword is used to specify that the directory is on the local file system.
• row_format
Specifies the row format for this insert. Valid options are SERDE clause and DELIMITED clause. SERDE clause can be used to specify a custom SerDe for this insert. Alternatively, DELIMITED clause can be used to specify the native SerDe and state the delimiter, escape character, null character, and so on.
• file_format
Specifies the file format for this insert. Valid options are TEXTFILE, SEQUENCEFILE, RCFILE, ORC, PARQUET, and AVRO. You can also specify your own input and output format using INPUTFORMAT and OUTPUTFORMAT. ROW FORMAT SERDE can only be used with TEXTFILE, SEQUENCEFILE, or RCFILE, while ROW FORMAT DELIMITED can only be used with TEXTFILE.
• VALUES ( { value | NULL } [ , … ] ) [ , ( … ) ]
Specifies the values to be inserted. Either an explicitly specified value or a NULL can be inserted. A comma must be used to separate each value in the clause. More than one set of values can be specified to insert multiple rows.
• query
A query that produces the rows to be inserted. It can be in one of following formats:
• a SELECT statement
• a TABLE statement
• a FROM statement
### Examples
INSERT OVERWRITE LOCAL DIRECTORY '/tmp/destination'
STORED AS orc
SELECT * FROM test_table;
INSERT OVERWRITE LOCAL DIRECTORY '/tmp/destination'
ROW FORMAT DELIMITED FIELDS TERMINATED BY ','
SELECT * FROM test_table;
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2022-09-30 09:40:36
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https://iwaponline.com/washdev/article/8/3/533/39016/Determinants-of-toilet-ownership-among-rural?searchresult=1
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In 2012, the Government of Indonesia and UNICEF launched a project within eastern provinces of Indonesia to scale up and strengthen a national hygiene and sanitation program called ‘Sanitasi Total Berbasis Masyarakat’. A formative study prior to the project was conducted to characterize sanitation and hygiene knowledge, attitudes, and practices (KAP) among 1,700 households in six rural Indonesian districts in 2014. Separate multivariate analyses for toilet ownership (outcome 1) and improved sanitation (outcome 2) were conducted with generalized linear models to assess the association between potential determinants and sanitation outcomes. Respondents who agreed that most people do not have a toilet in their community were associated with lower levels of toilet ownership compared to respondents who disagreed with the statement (p < 0.001). The perception that building a toilet is expensive was also associated with reduced toilet ownership in contrast to respondents without this perception (p < 0.001). Embarrassment and convenience were associated with ownership of improved sanitation versus those with shared or unimproved toilets. The study suggests that social norms play an important role in changing sanitation behaviors. Future research should aim to clarify the extent to which norms and other psychosocial factors can be used to influence sanitation practices.
Access to improved sanitation – a toilet facility that hygienically limits human contact with excreta – has implications for many health-related outcomes in children of low- and middle-income countries (LMICs), including diarrheal diseases, soil-transmitted helminths, undernutrition, stunting and cognitive development (Clasen et al. 2010; Spears et al. 2013; Pruss-Ustun et al. 2014; Cronin et al. 2016; Torlesse et al. 2016). There are also several important non-health-related outcomes that sanitation can affect, including saving time, increasing dignity, convenience and security, associated with not having to find a place to safely defecate in the open (Tilley et al. 2013; Sommer et al. 2015). Given the importance of access to sanitation, especially to women and children, it is critical to improve this basic human necessity for health and development (Benova et al. 2014; Smith et al. 2015).
Indonesia currently has the second highest burden of open defecation worldwide, estimated at 51 million people (World Health Organization (WHO)/United Nations Children's Fund (UNICEF) 2015). The Ministry of Health estimated that 30.5% of households have access to improved sanitation in Nusa Tengara Timur and Papua, while in South Sulawesi it is estimated at 54.9%. National coverage with improved sanitation is estimated to be 59.8% (RISKEDAS 2013). The nation also has a low percentage of households with access to improved sanitation, 61% (WHO/UNICEF 2015). The Government of Indonesia (GoI) has long acknowledged the importance of sanitation to health on a national level. Even as early as 1973, there was an Indonesian Presidential Decree on Drinking Water Supply and Household Toilets (Mukherjee & Shatifan 2010). The program focused primarily on subsidizing household toilet construction, i.e., increasing the supply of toilets, but did little in terms of building or sustaining demand. Slow progress was made on basic sanitation coverage in the following decades. In 2005, the Indonesian government began to test community-led total sanitation (CLTS) after observing the success of these programs in Bangladesh (Mukherjee & Shatifan 2010). The focus of CLTS, in contrast to Indonesia's past program that focused on subsidies, is on demand creation through an intensive set of community mobilization activities (Mukherjee & Shatifan 2010). By 2008, CLTS became an underlying piece of the national strategy in Indonesia called ‘Scaling-up and Strengthening Community Approaches to Total Sanitation’ (STBM in the Bahasa Indonesia language). STBM prioritizes behavior as well as social change and community empowerment in its programmatic activities across five pillars of WASH, including reducing open defecation and improving handwashing with soap, safe drinking water at household level, and solid and liquid waste management.
In November 2012, the GOI and UNICEF launched a five-year sanitation and hygiene project within eastern provinces of Indonesia to scale up and strengthen the efforts underway in the STBM program (Mattson 2014). Given the investments of time and resources in the STBM program and the importance of sanitation to health and development, there is a critical need to assess the potential driving factors behind toilet ownership in order to inform the acceleration of the program. By investigating factors associated with toilet ownership along with factors for why households have not adopted improved toilets, sanitation interventions can be designed to more effectively target the drivers for scaling up improved sanitation.
Sanitation behaviors can be shaped by social and non-social influences, including: micro- and macro-level economics, legal and political determinants, religion, education and technological changes (Bronfenbrenner 2005). Increasingly, research finds that sanitation behaviors are connected to the interdependent beliefs, practices, and expectations within a reference group (Khanna & Das 2016). The beliefs about the actions of a reference group (i.e., descriptive norms: doing what others do or believe others do) or the beliefs about what others believe one should or should not do (i.e., subjective norms) especially when linked to a potential sanction (normative expectation) has been found to motivate people (Bicchieri 2005); their anticipation of positive or negative sanctions or more hidden approval or disapproval of others (Dreibelbis et al. 2013) can become a strong driver of change at community level. This underpins the success of the CLTS approach whereby communities support themselves to eliminate open defecation through the creation of new social norms at the community level (Dooley et al. 2016).
This study aimed to explore how demographic, psychosocial, normative, and structural factors are associated with toilet ownership and improved sanitation access in three provinces of Indonesia (Nusa Tengara Timur, Papua, and South Sulawesi). Figure 1 presents the conceptual model of this study.
This study analyzed household survey data from a total of 1,700 Indonesian households collected in February 2014; a random cluster sampling methodology was used. The survey measured latrine adoption patterns across six districts of three provinces of Indonesia, including: Jayapura (Papua), Luwu Utara, Takalar, and Barru (South Sulawesi), and Alor and Sumba Timur (Nusa Tenggara Timur). The districts of Alor, Sumba Timur, Luwu Utara, Takalar, and Barru each included a sample size of 300 respondents while Jayapura included 200 respondents.
The sampling design and sample size were developed to provide sufficient power for the analysis at the district level. The overall margin of error was estimated at 2.4%, with a confidence level of 95% based on a census population of 283,570. Based on these calculations, a sample target of 1,700 households was determined, representing 4.9% of all households in the selected districts, which was less than 1% (0.6%) of the overall total population of the targeted districts of 283,570. The study team visited 1,786 households and obtained data from 1,700 households with a response rate of 95.2%. The survey was administered with household heads or any adult household member present at the time of the survey. After conducting a list-wise deletion, 1,696 households were included as the final analytic sample.
### Study variables
The dependent variables of this study were self-reported toilet ownership and improved sanitation. The survey asked if households had a toilet at the time of data collection (1 = Yes, 0 = No) with the study team directly observing the type of sanitation facilities reported to be used at each home. The proportion of households with improved sanitation facilities was estimated based on Joint Monitoring Programme (JMP) definitions (WHO/UNICEF 2015). The survey obtained the key attitudinal data only from those households that reported to own a toilet (i.e., improve and unimproved), so the reference group for the improved sanitation outcome did not include households without sanitation facilities (see Figure 1).
The independent variables consisted of demographic, psychosocial, and structural factors that may determine the ownership and use of household sanitation facilities in Indonesia. Demographic factors were operationalized as respondents' age, education, district, gender, household size, and relationship to household heads. Psychosocial factors included respondents' attitude toward sanitation practices and reasons for using a sanitation facility. The survey included descriptive statements on sanitation practices (e.g., ‘Most people in this community do not have a toilet’), and respondents reported their level of agreement to a given statement on a Likert scale. For those respondents who reported to have a household sanitation facility, they were asked reasons for using a sanitation facility in an open-ended question and if they received any financial assistance for building a toilet. Structural factors were operationalized as wealth quintiles, access to water throughout the year, and financial assistance for building a sanitation facility.
### Statistical analysis
Descriptive statistics were calculated by performing count and percent calculations for each characteristic stratified by toilet ownership status. As a proxy for socioeconomic status, a wealth index was calculated with data on the possession of household assets and characteristics (Howe et al. 2012). A principal component analysis (PCA) was conducted to identify appropriate weights for each asset and calculate wealth scores, which represented relative wealth levels of each household (Howe et al. 2012). The 12 assets used for the PCA included: radio, TV, cell phone, landline telephone, refrigerator, motorcycle, bicycle, animal drawn cart, car, motor boat, agricultural land, and farm animals. Wealth scores were categorized into quintiles, which represent poorest, poorer, middle, richer, and richest classes.
Bivariate and multivariate analyses were performed with generalized linear models (GLMs) with Poisson Family and Log Link to assess the association between potential determinants and sanitation outcomes. Results from these calculations were reported as prevalence ratios as a better alternative to odds ratios for the analysis of cross-sectional studies with binary outcomes (Thompson et al. 1998; Barros & Hirakata 2003). The statistical analyses were adjusted for the complex survey design by using STATA 12 with sampling weights and SVY command.
A multivariate analysis for toilet ownership was conducted using different models. Model 1 or the baseline specification consisted of demographic and structural variables including respondents' age, household size, gender, districts, wealth quintiles, and access to water. Model 2, Model 3, and Model 4 added psychosocial variables on open defecation, health, and sociocultural factors, respectively, to the baseline specification. Model 5 included all of the independent variables.
Additional multivariate models were used to assess the improved sanitation outcome. Model 6 included respondents' age, household size, gender, districts, and education as the baseline model. Model 7 additionally included reasons for using a sanitation facility. Model 8 added wealth quintiles, access to water, and financial assistance to the baseline model. Model 9 included all of the independent variables from Model 6 to Model 8.
### Ethical considerations
The Indonesia dataset was shared by the data owner, UNICEF Indonesia. The dataset was anonymized before data analysis, and the research was determined to be exempt from human subject protection by The George Washington University Institutional Review Board given that it involved the analysis of pre-existing data. Further, ethical oversight was not obtained because the proposed uses and disclosures of protected health information involved no more than minimal risk to the privacy of individuals (45 CFR 46164.512). To maintain the highest ethical standard, we still obtained informed consent from study participants.
Descriptive characteristics of respondents stratified by toilet ownership are provided in Table 1. Respondents without access to a toilet tend to have lower education and household wealth levels than those respondents from toilet-owning households. The majority of households without a toilet are located in Alor and Takalar districts. Households with sanitation facilities also had a greater prevalence of household-level access to water throughout the year than those households without a toilet. Most households with toilets tended to be in the higher wealth categories, and most households without toilets stated money as the primary reason for not having a toilet (96.8%). Most households without toilets were unaware of financing options available for toilet construction (77.7%), although most households with toilets did not report receiving financial assistance to build their toilet (83.2%).
When respondents without latrines were asked about their perceived cost of a toilet, the mean was almost four million Indonesian Rupiah (Rp) (approximately $300 USD) (range of 400,000 to 8,000,000 Rp) (data not shown). Toilet construction in these areas of Indonesia has been estimated to be typically between 0.65 million and 2.6 million Rp (approximately$50 to \$200 USD) depending on location, type of latrine, and community factors such as proximity to a water source and sea level (from UNICEF field monitoring data).
Respondents' perceptions and attitudes toward sanitation practices are presented in Table 2. The majority of respondents (81%) agreed that it is embarrassing when people can see others defecating in the open. The majority of respondents also believed that most people feel ashamed if they lack access to a household toilet. Nearly half of respondents (52%) disagreed that people regard having household sanitation facility as modern, and almost 80% did not perceive smell as an influential factor to explain why people do not want to build a sanitation facility for the household. Large proportions of respondents also disagreed with health-related statements, such as babies' feces do not spread disease and there is no relationship between open defecation and diarrhea. The highest proportion of respondents (85.4%) agreed that women need privacy to defecate.
Bivariate associations between respondents' attitudes and self-reported toilet ownership are presented in Table 3. Respondents who agreed that most people do not have a toilet in their community (i.e., descriptive norms) were associated with 43.4% lower prevalence of toilet ownership than respondents who disagreed with the statement (p < 0.001). Having the perception that building a toilet is an expensive task was associated with a 27.2% lower prevalence of toilet ownership than those respondents without this perception (p < 0.001). Furthermore, having an accepting attitude toward open defecation (i.e., subjective norm) in general and defecating at a beach or a river was associated with 25% (p < 0.05) and 43.8% (p < 0.001) lower prevalence of toilet ownership, respectively, in contrast to respondents who are not accepting of open defecation.
The results of the multivariate analysis for self-reported toilet ownership are summarized in Table 4. Respondents' age, education, wealth levels, and year-round access to water were positively associated with toilet ownership in all models. The district variable was also significantly associated with ownership. After controlling for demographic and structural factors in Model 2, respondents who perceived low levels of toilet ownership in the community were associated with 25.4% lower prevalence of toilet ownership in contrast to respondents who disagreed with the statement (p < 0.001). Those respondents who had an accepting attitude toward open defecation at a beach or a river were also negatively associated with toilet ownership (p < 0.001). In Model 3, respondents who did not perceive a link between open defecation and diarrhea were associated with 20.4% lower prevalence of toilet ownership than those respondents who perceived the link (p < 0.05). In Model 4, respondents who perceived toilet construction as expensive were associated with 22.2% lower prevalence of toilet ownership (p < 0.001). In Model 5, the prevalence of toilet ownership was estimated to be 16.7% higher among respondents who perceived men as the decision-makers for building a toilet than those who disagreed with the statement (p < 0.05). Based on the adjusted Wald test statistic, Model 1 was identified as the most parsimonious model.
Table 5 summarizes a sub-population multivariate analysis for the improved sanitation outcome. Respondents' age, household size, and gender were not associated with the presence of improved sanitation across all models, while the respondent's district showed a significant association. Those respondents who use a household sanitation facility to avoid sharing it with others were associated with 17.8% higher prevalence of improved sanitation than respondents who did not have this motivation (p < 0.001). The perception that most households in the community have a toilet was positively associated with improved sanitation in Model 7 and Model 9 (p < 0.005). Avoiding embarrassment was positively associated with improved sanitation in Model 7 while such an association was lost in Model 9. In regards to structural factors, a significant association was found between wealth quintiles and improved sanitation. Yet, financial assistance was not a significant driver for the ownership of improved sanitation facilities.
### Determinants of toilet ownership
The results of this analysis suggest that poverty remains a significant factor in determining investments in household sanitation (Guiteras et al. 2015). A positive point for future programming was that those responding to have attended a community event on sanitation in the survey areas found no significant differences among the different socioeconomic groups in attendance and such attendance was positively associated with owning a toilet. This would suggest CLTS approaches, which have a no-subsidy policy, a key component of STBM, can reach a broad spectrum of the community and help in accelerating toilet ownership.
The perceived cost of a toilet among non-owners suggests a potential lack of awareness of the availability of lower cost options, or a desire to have more expensive types of toilets (e.g., pour flush latrines). This desire for more expensive toilets is reflected in the finding that households without toilets greatly preferred flush or pour-flush toilets, rather than basic pit latrines. That households without toilets were found to have less access to water year round could also be one of the contributing factors to their not building their desired pour-flush toilet. There were statistically significant differences between households with and without toilets regarding the perceived cost of a toilet. The cost of sanitation, both perceived and real, has been cited in several other studies as a barrier to toilet ownership in Indonesia and elsewhere, and more work is needed to determine effective approaches to reducing cost barriers as well as raising consumer demand (Jenkins & Curtis 2005; Frias 2008; Sara & Graham 2014).
Additionally, the study results found that lower levels of wealth and education were associated with not owning a toilet, which is also a common theme in other studies (Jenkins & Scott 2007) and also consistent with handwashing practice (Hirai et al. 2016). An interesting finding in this study is that households that did not have consistent access to water throughout the year were less likely to have a toilet than those who do have consistent access. This finding suggests that it may be important to address water access in conjunction with sanitation programming activities, particularly when pour-flush toilets are the desired option.
### Social norms and acceptability of open defecation
In addition to these household determinants of toilet ownership, this study explored social norms, subjective and descriptive norms and expectations regarding sanitation. The majority of all households agreed that open defecation is not acceptable and agreed that it is embarrassing when people can see others defecating in the open. Additionally, the study found that subjective norms around household perceptions on the acceptability of open defecation were statistically associated with toilet ownership, as well as improved sanitation. Over time, it will be important to measure how the STBM program, which aims to change the social norm governing open defecation in a community, can influence these results. Furthermore, the study findings suggest a need to better integrate formative research around social norms into sanitation interventions (Curtis et al. 1997). In terms of descriptive norms, we found that respondents who perceived low levels of toilet ownership in the community were associated with lower toilet ownership. To accelerate future sanitation programs, it will be important to address this perception at the community level to achieve open defecation free status via a shared understanding and commitment to stopping open defecation. While there are many determinants of toilet ownership, whether a household has a toilet is also likely to depend on whether others in the household's social reference network already have a toilet or believe that a toilet is beneficial. In other words, toilet ownership is not merely a function of individual choice but is also a property of reference groups of people. There are, thus, biological, psychological, and social reasons that social norms would be relevant to toilet ownership.
The study found that respondents who agreed that there is no relationship between open defecation and diarrhea were less likely to have a toilet. These findings suggest that more effort may be needed to raise the basic understanding of the fecal–oral route of disease. Even with an increased knowledge of the health impacts associated with poor sanitation, however, using health messages as a motivator for building latrines may be contingent on other factors being in place and may not be the most effective method for motivating behavior change. Messages concerning convenience, privacy for women, social status, and community health may be more useful for redefining social norms within rural communities based on our findings and other studies (Mukherjee et al. 2012; Sara & Graham 2014). Most households with and without toilets believed that women need privacy for defecation. The survey found that while men may be viewed as the primary decision-makers, women in particular, and potentially children, may also be influential in the decision-making process at the household level. It may be that this gender approach could be a more effective lever for improving sanitation behaviors than increasing knowledge on the link between sanitation and health. The inclusion of these social norms in household surveys helped to identify the areas where behavior change and behavior and social norms interventions can be strengthened within programs aimed at improving access to and use of improved sanitation.
Beyond the toilet ownership outcome, we assessed the factors positively associated with access to improved sanitation. These included: (1) convenience, measured as the desire to avoid sharing a toilet; (2) the perception that others use improved sanitation (descriptive norm); and (3) avoiding embarrassment.
There are many other important factors that may influence the success of a sanitation program, which we did not include in this study. Researchers have shown that villages with higher levels of social capital, not measured in this study, were more likely to build toilets, reduce open defecation, and experience corresponding health improvements (Cameron et al. 2015; UNICEF 2015). In sanitation programs that target whole communities (e.g., via CLTS), researchers have found greater improvements in small, remote villages with low exposure to prior sanitation projects. The researchers suggested that these programs are most effective in socially cohesive villages (Crocker et al. 2016).
There were several limitations to the study. The study is likely to be applicable to rural populations of Indonesia and less generalizable to urban and peri-urban populations. Moreover, this study found differences among the targeted districts, but there is limited information available as to other variables, including what saniation programs may have occurred in these areas as well as to other factors that may have influenced latrine construction by households. Interviewer bias is one concern given the possibility that the interviewees could have responded with an answer they perceived the interviewer expected. Furthermore, broad categorical responses in certain questions of the survey may have occasionally resulted in insufficient detail for clear conclusions. For example, the survey found that an ‘elected official’ was one of the most commonly mentioned primary influencers for toilet construction, however, the level of this government official was not specified. There are other examples of sanitation campaigns, such as in Thailand, where elected officials played an important part of a larger sanitation strategy (Luong et al. 2002). Additionally, participants were asked questions regarding beliefs, attitudes, and knowledge regarding ‘toilets’, as well as cost of toilets, but households were not asked these questions in the context of what type of toilets they were referring to when responding to these questions – a pit latrine or a pour-flush toilet, for example. Lastly, this study only collected data from household heads or adult household members, so perceptions of children and youths are not represented.
This study analyzed factors that are associated with current toilet ownership, as well as the factors associated with improved sanitation among rural households in the provinces of Nusa Tengarra Timur, Papua, and South Sulawesi of Indonesia. The results should be useful for sanitation activities in these districts and provinces, as well as other rural sanitation programs in Indonesia. By identifying barriers and facilitators affecting toilet ownership, interventions can target specific perceptions, norms, and behaviors to more effectively and efficiently scale up improved sanitation coverage. There remains a critical need to better understand how social norms, such as individuals' perceptions and expectations of what others do and of what others expect oneself to do, and how this affects demand for sanitation (Evans et al. 2014). It may be that promoting privacy and dignity for women and girls could be more important drivers for improving sanitation than the use of health messages, although we found toilet ownership higher among those who believed there was a relationship between sanitation and health. It is clear though that the most influential messages, however, will only be known with sufficient investment in formative research at the local level (Curtis et al. 2011).
Many of the findings of this study were similar to those found in past studies that have shown that education and household wealth, as well as the location of where one lives, can affect toilet ownership (Sara & Graham 2014). There were, however, more novel findings related to social norms. There is a need to explore these further in future research. In terms of structural factors that affect toilet ownership, we found that year-round access to water, which may affect the functioning of a pour-flush toilet (a desired sanitation technology in the study population) was positively associated with toilet ownership. Additionally, the study identified convenience (i.e., the desire to avoid sharing a toilet) as a factor associated with ownership of improved sanitation facilities. Finally, this study highlighted a number of factors that can potentially increase the pace at which sanitation is scaled up across rural communities in Indonesia. More formative research, however, is needed to understand the differential effects of applying these findings in future sanitation programs at local level.
We acknowledge the UNICEF Indonesia WASH team, Government partners at national and sub-national levels, and the funding grant from the Bill & Melinda Gates Foundation to UNICEF Indonesia in support of accelerating sanitation in Eastern Indonesia. Comments from Aldy Mardikanto, BAPPENAS, Government of Indonesia, were very gratefully received. The opinions expressed are those of the authors and do not necessarily reflect the views of their institutions, UNICEF or the United Nations.
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2022-12-05 08:39:56
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NOW UP TO 1000. In this number grid learning, 2-digit operations are able to be added which can become similar such. 618 033 988 749. 100 Chart Printable. You can change the clip art, move it around, edit the title and add your own text. The golden ratio is also called the golden mean or golden section ( Latin: sectio aurea ). It is an irrational number that is a solution to the quadratic equation. pdf By Amanda Post. Each chart prints over most of the printer paper, which just means that it has big squares and big numbers. φ = 1 + 5 2 = {\displaystyle \varphi = {\frac {1+ {\sqrt {5}}} {2}}=} 1. Writing 1-100 for kindergarten on a number grid. 1 to 100 times tables is available in pdf, printable & downloadable format for easy reference. Author: Liz Woodham Created Date: 8/5/2015 12:20:03 PM. 100 percent (100%) of a number is the same number: 100% × 80 = 100/100×80 = 80. Columns and gutters. Number charts: counting by twos (even, odd) Number charts: counting by 3's. Title: Number Grid to 100. An interactive math lesson about whole numbers to 100. We've set up a number of free-to-print worksheets based on this square, as shown in the table below - see also the print-on-A4 page for Numbers 1 to 1000 (which can be enlarged on. There are four game modes: Find a Number, Find the Number Between and Count On and Count Back. We have a dream about these 100 Worksheet Template images gallery can be a hint for you, bring you more examples and most important: make you have bright day. 4 times table. 618 033 988 749. My Hundreds Chart Name Date 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20. For everyone who still wants to try acrylic pouring out, here is the list of supplies for acrylic pouring to get. Title: Printable Hundreds Chart Created Date: 7/27/2013 9:25:29 PM. Given a two-digit number, mentally find 10 more or 10 less than the number. These sheets come in both Color and Black and White so are perfect to be uploaded into classroom, Seesaw or Showbie or sim. Choose from ‘0’ or ‘1’ start. 100 94 95 96 99. Complete one grid per day to achieve the goal of writing numbers 1-100 in a timely manner. An mSATA SSD with the label removed to show the chipset and NAND. 6 times table. Percentage is a value that represents the proportion of one number to another number. 1-100 Number Grid. This generator makes number charts and lists of whole numbers and integers, including a 100-chart, for kindergarten and elementary school children to practice number writing, counting, and skip-counting. Missing Numbers 1 to 100 10 Printable Worksheets pdf. There is a similar product with the numbers for the students who need additional assistance. Title: The Number Grid - 1-100 Author: shawna Created Date: 8/1/2008 6:23:09 PM. Worksheets are 100 chart, 100 chart, 100 chart, Writing numbers work, 1 100 number grid, Blank hundred chart, Hundreds chart puzzles, Numbers to 200. 4 times table. In early grades you can pass out numbers 1 through 10 and have those students put themselves in order. Choose from ‘0’ or ‘1’ start. Complete one grid per day to achieve the goal of writing numbers 1-100 in a timely manner. Use place. What is a number chart 1-30? It's a 5 x 6 grid that consists of the numbers 1 through 30. Each times table chart can be downloaded for free. Title: Microsoft Word - 1-100_NumberGrid. Click on pop-out icon or print icon to worksheet to print or download. Hundred Chart 101-200 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130. A solid-state drive ( SSD) is a solid-state storage device that uses integrated circuit assemblies to store data persistently, typically using flash memory, and functioning as secondary storage in the hierarchy of computer storage. Writing 1-100 for kindergarten on a number grid. The risk of heart disease increases due to lifestyle that leads to overweight and obesity, hypertension, hyperglycaemia, and high cholesterol [1]. 99 USD per month and $44. There are four game modes: Find a Number, Find the Number Between and Count On and Count Back. Scroll down to the bottom to check out the worksheets! blank number chart 1-100 worksheet. Read and write decimals to thousandths using base-ten numerals, number names, and expanded form, e. Click on pop-out icon or print icon to worksheet to print or download. Can number grid use for a workbook? Usually, number grids are known to be used in beginners as known as children when learning numbers. Math explained in easy language, plus puzzles, games, quizzes, worksheets and a forum. ^2); function of the airfoil top surface The bottom. These sheets come in both Color and Black and White so are perfect to be uploaded into classroom, Seesaw or Showbie or sim. 1,717 Followers, 438 Following, 440 Posts - See Instagram photos and videos from WYDaily (@wydaily). Number grid 1-100 is useful because it helps children understand number sequences also to help them doing addition and subtraction. 1 to 100 Fill In the Missing Number Grids- Color and Black and White are ready to be printed off. Grade Level (s): 1-2. In early grades you can pass out numbers 1 through 10 and have those students put themselves in order. It is an irrational number that is a solution to the quadratic equation. Title: Microsoft Word - 1-100_NumberGrid. Title: Printable Hundreds Chart Created Date: 7/27/2013 9:25:29 PM. Compared with other tools, it functionally worked as a number line but to value place and patterns. Hundred Chart 101-200 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130. 50 percent (50%) of a number is half of the number: 50% × 80 = 50. An mSATA SSD with the label removed to show the chipset and NAND. Compare two decimals to thousandths based on meanings of the digits in each place, using >, =, and < symbols to record the results of comparisons. For example, 13% is equivalent to the following. 392 = 3 × 100 + 4 × 10 + 7 × 1 + 3 × (1/10) + 9 × (1/100) + 2 × (1/1000). Number Line – There are several number line activities you can do with these cards that get students up and moving. Percentage Definition. These sheets come in both Color and Black and White so are perfect to be uploaded into classroom, Seesaw or Showbie or sim. Quickly and easily copy and paste lists. Each times table chart can be downloaded for free. Lets you pick a number between 1 and 100. Use place. Number charts: counting by 5's. A solid-state drive ( SSD) is a solid-state storage device that uses integrated circuit assemblies to store data persistently, typically using flash memory, and functioning as secondary storage in the hierarchy of computer storage. 9 million people dying yearly. Writing 1-100 for kindergarten on a number grid. Number Grids 1-100. Title: Number Grid to 100. The World Health Organization (WHO) [1] lists cardiovascular diseases (CVD) as the number one cause of death globally with 17. NOW UP TO 1000. Counting & Number Patterns Printables & Worksheets. The risk of heart disease increases due to lifestyle that leads to overweight and obesity, hypertension, hyperglycaemia, and high cholesterol [1]. 1,717 Followers, 438 Following, 440 Posts - See Instagram photos and videos from WYDaily (@wydaily). 13 / 100 (or) 13 to 100 (or) 13 : 100 Let us see, how to use grids to model percents in the following example. A solid-state drive ( SSD) is a solid-state storage device that uses integrated circuit assemblies to store data persistently, typically using flash memory, and functioning as secondary storage in the hierarchy of computer storage. IXL's SmartScore is a dynamic measure of progress towards mastery, rather than a percentage grade. 0 to 99 Number Squares. Solid-state drive. 21 31 41 51 61 71 81 91 12 22 32 42 52 62 72 82 92 13 23 33 43 53 63 73 83 93 14 24 34 44 54 64 74 84 94 15 25 35 45 55 65 75 85 95 16 26 36 46 56 66. Number charts: Number chart from 1 to 100 ( full / empty) Number Charts: counting by ones. Sal goes through all the numbers from 0 to 100 and shows some interesting patterns. It is an irrational number that is a solution to the quadratic equation. Number Grid. 1 to 100 times tables is available in pdf, printable & downloadable format for easy reference. Counting backwards: Count backwards from 100 to 0: 100, 99, __ , 97, __ , Count. This educational activity helps kids to develop an understanding of patterns and number relationships by utilizing a number grid. IXL's SmartScore is a dynamic measure of progress towards mastery, rather than a percentage grade. Writing 1-100 for kindergarten on a number grid. Hundred Chart 101-200 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130. An interactive math lesson about whole numbers to 100. 9 million people dying yearly. Compared with other tools, it functionally worked as a number line but to value place and patterns. Number Chart (blank 100) Title: Microsoft Word - blank_100_chart. Missing numbers between 0 and 120. These blank number charts 1-100 can be used as the media that they can use to assist them in memorizing the numbers and its sequences. By reading and memorizing the numbers from 1-100 and its order repeatedly, children will be able to remember the numbers perfectly. In other words, percent of a number is the ratio between the number and 100. Donna Glynn Kinderglynn. Number Grid 1-100. You can decide how much of the chart is pre-filled, the border color, skip-counting step, and so on. Number Chart is a fun and educational activity, which helps children see the relationships and patterns among numbers up to 120. Title: Number Grid to 100. We also provide an empty number chart which can be used for exercises. 1,717 Followers, 438 Following, 440 Posts - See Instagram photos and videos from WYDaily (@wydaily). 1 to 100 Number Square. Includes plain black and white, 5’s and 10’s highlighted, odd numbers highlighted, blank 100’s grid. 392 = 3 × 100 + 4 × 10 + 7 × 1 + 3 × (1/10) + 9 × (1/100) + 2 × (1/1000). Number charts: counting by 10's. Without any ifs or buts, the colors are the top priority when it comes to the materials for acrylic pours. 1 to 100 Number Squares x4. Select odd only, even only, half odd and half even or custom number of odd/even. Simply click on a times table chart below to view and then download. com for free resources like German Word of the Day and our German. Scroll down to the bottom to check out the worksheets! blank number chart 1-100 worksheet. 1 to 100 Number Squares x4. This fantastic counting grid resource features primary numbers 1-100 in a handy square format. 618 033 988 749. In other words, percent of a number is the ratio between the number and 100. There are several jobs that require a number grid as a workbook such as. In early grades you can pass out numbers 1 through 10 and have those students put themselves in order. Don't worry, you won't have to memorize each number. 4 Basic Pour Painting Supplies. Given a two-digit number, mentally find 10 more or 10 less than the number. Find this Pin and more on Special Education by Gayle McLemore. 100 94 95 96 99. 1 to 100 Number Square. This generator makes number charts and lists of whole numbers and integers, including a 100-chart, for kindergarten and elementary school children to practice number writing, counting, and skip-counting. NOW UP TO 1000. Read and write decimals to thousandths using base-ten numerals, number names, and expanded form, e. with more related things like blank 100 chart, printable blank 100 square grid and writing numbers 1 100. The number of grid columns can be modified via Sass variables. Counting by 1: Fill in the Blank (Random) (PDF file) Printable. ^2); function of the airfoil top surface The bottom. Displaying all worksheets related to - Writing Numbers To 100 Grid. 618 033 988 749. Adults also know the name workbook. Purchased for a number crazed 5 year old with played with it right off. For example, 13% is equivalent to the following. 13 Best of Missing Number Grid Worksheets Fill in. 1-100 Number Grid. 100 Chart Printable. Set One B is a set of 9 printable 100-number charts. 0 to 99 Number Squares x4. What are the benefits of a number chart 1-30? 1. Can easily print to chart size using poster print options. 1 to 100 Number Square. Number Line – There are several number line activities you can do with these cards that get students up and moving. Practice: Numbers to 120. In other words, percent of a number is the ratio between the number and 100. Each space will be large enough to fit a. We also have a similar multiplication chart 1-100. 9 million people dying yearly. 21 31 41 51 61 71 81 91 12 22 32 42 52 62 72 82 92 13 23 33 43 53 63 73 83 93 14 24 34 44 54 64 74 84 94 15 25 35 45 55 65 75 85 95 16 26 36 46 56 66 76 86 96 17 27 37 47 57 67 77 87 97 18 28 38 48 58 68 78 88 98 19 29 39 49 59 69 79 89 99 20 30 40 50 60 70 80 90 100. 99 USD per month and$44. The number chart posted above is the 1-100 number chart that can be used to learn about the numbers before your children take the worksheets and exercises. Count to 120, starting at any number less than 120. Number Chart (blank 100) Title: Microsoft Word - blank_100_chart. NOW UP TO 1000. Scroll down to the bottom to check out the worksheets! blank number chart 1-100 worksheet. 18 Best of 1 To 100 Backward Missing Number. Use the start/stop to achieve true randomness and add the luck factor. Missing numbers between 0 and 120. Number Grid 1-100. An mSATA SSD with the label removed to show the chipset and NAND. Number Grid To 100 Printable. 13 / 100 (or) 13 to 100 (or) 13 : 100 Let us see, how to use grids to model percents in the following example. You can decide how much of the chart is pre-filled, the border color, skip-counting step, and so on. Columns and gutters. 392 = 3 × 100 + 4 × 10 + 7 × 1 + 3 × (1/10) + 9 × (1/100) + 2 × (1/1000). φ = 1 + 5 2 = {\displaystyle \varphi = {\frac {1+ {\sqrt {5}}} {2}}=} 1. 100 Chart Printable. The first row is filled with numbers 1 to 5, the next row is 6 to 10 etc. doc Author: ajk44 Created Date: 9/10/2015 12:59:35 PM. The number chart posted above is the 1-100 number chart that can be used to learn about the numbers before your children take the worksheets and exercises. 99 USD for a year!. 13 Best of Missing Number Grid Worksheets Fill in. 1,717 Followers, 438 Following, 440 Posts - See Instagram photos and videos from WYDaily (@wydaily). We also provide an empty number chart which can be used for exercises. For example, 13% is equivalent to the following. 1-100 Number Grid. Below is a full number chart 1-100 to be used when reviewing numbers with students or as a reference for students learning to count. Quickly and easily copy and paste lists. Charts of Number 1-100 Free | Activity Shelter. Counting backwards: Count backwards from 100 to 0: 100, 99, __ , 97, __ , Count. Choose from ‘0’ or ‘1’ start. Author: Liz Woodham Created Date: 8/5/2015 12:20:03 PM. For K-12 kids, teachers and parents. The World Health Organization (WHO) [1] lists cardiovascular diseases (CVD) as the number one cause of death globally with 17. You can change the clip art, move it around, edit the title and add your own text. 21 31 41 51 61 71 81 91 12 22 32 42 52 62 72 82 92 13 23 33 43 53 63 73 83 93 14 24 34 44 54 64 74 84 94 15 25 35 45 55 65 75 85 95 16 26 36 46 56 66 76 86 96 17 27 37 47 57 67 77 87 97 18 28 38 48 58 68 78 88 98 19 29 39 49 59 69 79 89 99 20 30 40 50 60 70 80 90 100. The number of grid columns can be modified via Sass variables. German Numbers 1-100 Posted by komo on Mar 24, 2010 in Uncategorized. Title: Microsoft Word - 1-100_NumberGrid. 13 / 100 (or) 13 to 100 (or) 13 : 100 Let us see, how to use grids to model percents in the following example. Counting by 1s - Fill in the blank - 5 problems (random 1 to 25) math/countingby1s_1. 100 percent (100%) of a number is the same number: 100% × 80 = 100/100×80 = 80. This set is similar to the set above in both size and missing number sets. Number charts: counting by 10's. Printable multiplication tables are available from 1x through to 12x. In the cell below the title enter the first ticket number. Each chart prints over most of the printer paper, which just means that it has big squares and big numbers. Solid-state drive. These sheets come in both Color and Black and White so are perfect to be uploaded into classroom, Seesaw or Showbie or sim. Can easily print to chart size using poster print options. Title: The Number Grid - 1-100 Author: shawna Created Date: 8/1/2008 6:23:09 PM. Once you memorize 1-12 and the numbers by 10's from 20-100, you can formulate any number in between from a pattern! Let's get started. Ensure the info you fill in The Number Grid - 1-100 is up-to-date and correct. 1,717 Followers, 438 Following, 440 Posts - See Instagram photos and videos from WYDaily (@wydaily). Some of the worksheets displayed are 100 chart, 1 100 number grid, 100 chart 1 2 3 4 5 6 7 8 9, Grade 1. See also Candy Corn Worksheets. There is a similar product with the numbers for the students who need additional assistance. This generator makes number charts and lists of whole numbers and integers, including a 100-chart, for kindergarten and elementary school children to practice number writing, counting, and skip-counting. 1 percent represents 1/100 fraction. There is a gridded side, a blank side, and a numbered grid insert that allow for learner variability/knowledge level. 1-100 Number Grid. Add and subtract with 100. Counting & Number Patterns Printables & Worksheets. 1,717 Followers, 438 Following, 440 Posts - See Instagram photos and videos from WYDaily (@wydaily). Sal goes through all the numbers from 0 to 100 and shows some interesting patterns. Check once more each and every field has been filled in properly. Printable multiplication tables are available from 1x through to 12x. List of Numbers 1-100 with no formatting. Indicate the date to the document using the Date function. out of 100. An mSATA SSD with the label removed to show the chipset and NAND. Title: Printable Hundreds Chart Created Date: 7/27/2013 9:25:29 PM. Print on 12x18 paper (I sent to print shop for minimal fee) and when trimmed it will be a 12x12 square. Writing 1-100 for kindergarten on a number grid. Number charts: counting by 10's. 50 percent (50%) of a number is half of the number: 50% × 80 = 50. 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140. The World Health Organization (WHO) [1] lists cardiovascular diseases (CVD) as the number one cause of death globally with 17. com for free resources like German Word of the Day and our German. We've set up a number of free-to-print worksheets based on this square, as shown in the table below - see also the print-on-A4 page for Numbers 1 to 1000 (which can be enlarged on. 6 times table. Sal goes through all the numbers from 0 to 100 and shows some interesting patterns. Number Grids 1-100. In other words, percent of a number is the ratio between the number and 100. Don't worry, you won't have to memorize each number. with more related things like blank 100 chart, printable blank 100 square grid and writing numbers 1 100. List of Numbers 1-100 with no formatting. 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140. A simple grid with the numbers 1-100 that can be used to practice place value and numeral recognition. For everyone who still wants to try acrylic pouring out, here is the list of supplies for acrylic pouring to get. Now if move the cursor to the bottom right corner of the second cell, it will change to a thin solid-black plus symbol. German Numbers 1-100 Posted by komo on Mar 24, 2010 in Uncategorized. We also provide an empty number chart which can be used for exercises. In early grades you can pass out numbers 1 through 10 and have those students put themselves in order. Number Chart (blank 100) Title: Microsoft Word - blank_100_chart. 392 = 3 × 100 + 4 × 10 + 7 × 1 + 3 × (1/10) + 9 × (1/100) + 2 × (1/1000). Number charts: counting by twos (even, odd) Number charts: counting by 3's. The World Health Organization (WHO) [1] lists cardiovascular diseases (CVD) as the number one cause of death globally with 17. This educational activity helps kids to develop an understanding of patterns and number relationships by utilizing a number grid. Number charts: Number chart from 1 to 100 ( full / empty) Number Charts: counting by ones. German Numbers 1-100 Posted by komo on Mar 24, 2010 in Uncategorized. Complete one grid per day to achieve the goal of writing numbers 1-100 in a timely manner. Free interactive classroom resources - Get activities, games and SMART Notebook lessons created by teachers for teachers. 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140. Print without clutter. Some of the worksheets displayed are 100 chart, 1 100 number grid, 100 chart 1 2 3 4 5 6 7 8 9, Grade 1. Set One B is a set of 9 printable 100-number charts. φ = 1 + 5 2 = {\displaystyle \varphi = {\frac {1+ {\sqrt {5}}} {2}}=} 1. Writing 1-100 for kindergarten on a number grid. It is written to the right side of the number: 50%. Solid-state drive. Use the start/stop to achieve true randomness and add the luck factor. There are several jobs that require a number grid as a workbook such as. Print without clutter. A flexible hundred square where you can choose numbers up to 1000 and whether to display numbers in the grid from top to bottom or bottom to top. 618 033 988 749. In the cell below the title enter the first ticket number. Choose from ‘0’ or ‘1’ start. Print on 12x18 paper (I sent to print shop for minimal fee) and when trimmed it will be a 12x12 square. Title: The Number Grid - 1-100 Author: shawna Created Date: 8/1/2008 6:23:09 PM. Each prints on a single A4 sheet. Number Grids 1-100. Now if move the cursor to the bottom right corner of the second cell, it will change to a thin solid-black plus symbol. Number Line – There are several number line activities you can do with these cards that get students up and moving. Select the Sign icon and make an electronic signature. 100 percent (100%) of a number is the same number: 100% × 80 = 100/100×80 = 80. These sheets come in both Color and Black and White so are perfect to be uploaded into classroom, Seesaw or Showbie or sim. 13 / 100 (or) 13 to 100 (or) 13 : 100 Let us see, how to use grids to model percents in the following example. Can number grid use for a workbook? Usually, number grids are known to be used in beginners as known as children when learning numbers. Simply click on a times table chart below to view and then download. NOW UP TO 1000. doc Author: ajk44 Created Date: 9/10/2015 12:59:35 PM. It tracks your skill level as you tackle progressively more difficult questions. If you are looking to learn German, check out our website at Transparent. Practice: Numbers to 120. The first row is filled with numbers 1 to 5, the next row is 6 to 10 etc. Hundred Chart 101-200 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130. Find this Pin and more on Special Education by Gayle McLemore. A simple grid with the numbers 1-100 that can be used to practice place value and numeral recognition. Title: Microsoft Word - 1-100_NumberGrid. doc Author: ajk44 Created Date: 9/10/2015 12:59:35 PM. It is an irrational number that is a solution to the quadratic equation. Now if move the cursor to the bottom right corner of the second cell, it will change to a thin solid-black plus symbol. Can easily print to chart size using poster print options. Title: Printable Hundreds Chart Created Date: 7/27/2013 9:25:29 PM. 618 033 988 749. These blank number charts 1-100 can be used as the media that they can use to assist them in memorizing the numbers and its sequences. doc Author: ajk44 Created Date: 9/10/2015 12:59:35 PM. Includes plain black and white, 5’s and 10’s highlighted, odd numbers highlighted, blank 100’s grid. 13 Best of Missing Number Grid Worksheets Fill in. pdf By Amanda Post. Change the number of tiers, the media query dimensions, and the container widths—then recompile. It tracks your skill level as you tackle progressively more difficult questions. 1-100 Number Grid. Counting by 2: Fill in the Blank (2 to 100) (PDF file) Printable. Generate numbers sorted in ascending order or unsorted. Writing 1-100 for kindergarten on a number grid. Writing 1-100 for kindergarten on a number grid. By reading and memorizing the numbers from 1-100 and its order repeatedly, children will be able to remember the numbers perfectly. Number Chart is a fun and educational activity, which helps children see the relationships and patterns among numbers up to 120. Number Chart (blank 100) Title: Microsoft Word - blank_100_chart. Counting by 1s - Fill in the blank - 5 problems (random 1 to 25) math/countingby1s_1. This number square can be used for many types of maths practice - shading in numbers divisible by 2, 3, 4, etc. This set is similar to the set above in both size and missing number sets. 618 033 988 749. 1 percent represents 1/100 fraction. IXL's SmartScore is a dynamic measure of progress towards mastery, rather than a percentage grade. This educational activity helps kids to develop an understanding of patterns and number relationships by utilizing a number grid. German Numbers 1-100 Posted by komo on Mar 24, 2010 in Uncategorized. 1 to 100 Fill In the Missing Number Grids- Color and Black and White are ready to be printed off. Missing numbers between 0 and 120. Download for free 100 Chart Cliparts #234, download othes number grid 0 to 100 for free. 7 times table. Adults also know the name workbook. Grid Drawing Tool by ArtTutor Start. ENQUIRE ABOUT THIS PRODUCT. Each times table chart can be downloaded for free. It tracks your skill level as you tackle progressively more difficult questions. List of Numbers 1-100 with no formatting. 13 / 100 (or) 13 to 100 (or) 13 : 100 Let us see, how to use grids to model percents in the following example. 9 million people dying yearly. Check once more each and every field has been filled in properly. Given a two-digit number, mentally find 10 more or 10 less than the number. doc Author: ajk44 Created Date: 9/10/2015 12:59:35 PM. Question: clear all 3% Mesh generation N = 100; number of grids in x-direction for numerical integ dx = 1/N; grid size (uniform) in x-direction x = dx/2:dx: (1-dx/2); grids in x-direction $% geometry of the airfoil top surface W = 10;$ span the airfoil (in z-direction), m y = x. 1-100 Number Grid. Some of the worksheets displayed are 100 chart, 1 100 number grid, 100 chart 1 2 3 4 5 6 7 8 9, Grade 1. 1 to 100 Number Grid Say-It Chart | Barker Creek Colorful, cardstock poster shows a large 10x10 number grid with the 5's and 10's in a different color. Counting by 1s - Fill in the blank - 5 problems (random 1 to 25) math/countingby1s_1. Home / Playground Markings / Mathematical Number Grid (1-100) Mathematical Number Grid (1-100) View complimentary products. Title: Printable Hundreds Chart Created Date: 7/27/2013 9:25:29 PM. Download for free 100 Chart Cliparts #234, download othes number grid 0 to 100 for free. This number square can be used for many types of maths practice - shading in numbers divisible by 2, 3, 4, etc. The number grid 1-100 also presented with a number line, base-10 blocks, 10-frames and objects from various topics to achieve maximum goal result. Purchased for a number crazed 5 year old with played with it right off. Author: Liz Woodham Created Date: 8/5/2015 12:20:03 PM. Search for Microsoft Excel on your computer and run it. Worksheets are 100 chart, 100 chart, 100 chart, Writing numbers work, 1 100 number grid, Blank hundred chart, Hundreds chart puzzles, Numbers to 200. Number Grid. 21 31 41 51 61 71 81 91 12 22 32 42 52 62 72 82 92 13 23 33 43 53 63 73 83 93 14 24 34 44 54 64 74 84 94 15 25 35 45 55 65 75 85 95 16 26 36 46 56 66. The large format (17x22) has a star border with rainbow. Grid Drawing Tool by ArtTutor Start. The World Health Organization (WHO) [1] lists cardiovascular diseases (CVD) as the number one cause of death globally with 17. 4 times table. If you are looking to learn German, check out our website at Transparent. Compare two decimals to thousandths based on meanings of the digits in each place, using >, =, and < symbols to record the results of comparisons. Grade Level (s): 1-2. Solid-state drive. 13 / 100 (or) 13 to 100 (or) 13 : 100 Let us see, how to use grids to model percents in the following example. Learning numbers 1 - 100. 618 033 988 749. My Hundreds Chart Name Date 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20. There is a similar product with the numbers for the students who need additional assistance. Number Grid To 100 Printable. You can decide how much of the chart is pre-filled, the border color, skip-counting step, and so on. Select odd only, even only, half odd and half even or custom number of odd/even. The World Health Organization (WHO) [1] lists cardiovascular diseases (CVD) as the number one cause of death globally with 17. Title: Microsoft Word - 1-100_NumberGrid. What is a number chart 1-30? It's a 5 x 6 grid that consists of the numbers 1 through 30. For children to practice writing their numbers in a gradual way. Donna Glynn Kinderglynn. identifying prime numbers, "square" numbers, etc. Number Grids 1-100. Printable multiplication tables are available from 1x through to 12x. When you count to 10, the numbers are 1, 2, 3, 4, 5, 6, 7, 8, 9 and 10. Solid-state drive. Sal goes through all the numbers from 0 to 100 and shows some interesting patterns. Check once more each and every field has been filled in properly. out of 100. Counting by 1: Fill in the Blank (Random) (PDF file) Printable. See also Candy Corn Worksheets. Below is a full number chart 1-100 to be used when reviewing numbers with students or as a reference for students learning to count. Can easily print to chart size using poster print options. Select odd only, even only, half odd and half even or custom number of odd/even. Can number grid use for a workbook? Usually, number grids are known to be used in beginners as known as children when learning numbers. It is an irrational number that is a solution to the quadratic equation. Counting by 2: Fill in the Blank (2 to 100) (PDF file) Printable. This is the currently selected item. A game which focuses on numbers from up to 10, to up to 100. com for free resources like German Word of the Day and our German. The World Health Organization (WHO) [1] lists cardiovascular diseases (CVD) as the number one cause of death globally with 17. The golden ratio is also called the golden mean or golden section ( Latin: sectio aurea ). We have a dream about these 100 Worksheet Template images gallery can be a hint for you, bring you more examples and most important: make you have bright day. 100 Chart Printable. Counting by 1: Fill in the Blank (Random) (PDF file) Printable. thesingingwalrus. For K-12 kids, teachers and parents. Check once more each and every field has been filled in properly. These blank number charts 1-100 can be used as the media that they can use to assist them in memorizing the numbers and its sequences. φ = 1 + 5 2 = {\displaystyle \varphi = {\frac {1+ {\sqrt {5}}} {2}}=} 1. Donna Glynn Kinderglynn. NOW UP TO 1000. Download for free 100 Chart Cliparts #234, download othes number grid 0 to 100 for free. Number Grids 1-100. In other words, percent of a number is the ratio between the number and 100. We also provide an empty number chart which can be used for exercises. Solid-state drive. A solid-state drive ( SSD) is a solid-state storage device that uses integrated circuit assemblies to store data persistently, typically using flash memory, and functioning as secondary storage in the hierarchy of computer storage. Sal goes through all the numbers from 0 to 100 and shows some interesting patterns. Watch all of our videos ad free with our app (desktop, apple, or android):https://www. Choose from ‘0’ or ‘1’ start. 4 Basic Pour Painting Supplies. File previews. 1,717 Followers, 438 Following, 440 Posts - See Instagram photos and videos from WYDaily (@wydaily). Select the Sign icon and make an electronic signature. Each chart prints over most of the printer paper, which just means that it has big squares and big numbers. 1 to 100 Number Squares x4. Missing numbers between 0 and 120. What is a number chart 1-30? It's a 5 x 6 grid that consists of the numbers 1 through 30. Question: clear all 3% Mesh generation N = 100; number of grids in x-direction for numerical integ dx = 1/N; grid size (uniform) in x-direction x = dx/2:dx: (1-dx/2); grids in x-direction $% geometry of the airfoil top surface W = 10;$ span the airfoil (in z-direction), m y = x. 392 = 3 × 100 + 4 × 10 + 7 × 1 + 3 × (1/10) + 9 × (1/100) + 2 × (1/1000). You can change the clip art, move it around, edit the title and add your own text. 13 Best of Missing Number Grid Worksheets Fill in. Free interactive classroom resources - Get activities, games and SMART Notebook lessons created by teachers for teachers. For example, 13% is equivalent to the following. Now if move the cursor to the bottom right corner of the second cell, it will change to a thin solid-black plus symbol. Another idea is to put up the tens, from 0 to 100, on a wall or around the room. By reading and memorizing the numbers from 1-100 and its order repeatedly, children will be able to remember the numbers perfectly. 9 million people dying yearly. 13 Best of Missing Number Grid Worksheets Fill in. The golden ratio is also called the golden mean or golden section ( Latin: sectio aurea ). For everyone who still wants to try acrylic pouring out, here is the list of supplies for acrylic pouring to get. An mSATA SSD with the label removed to show the chipset and NAND. Counting by 1s - Fill in the blank - 5 problems (random 1 to 25) math/countingby1s_1. Given a two-digit number, mentally find 10 more or 10 less than the number. 100 Number Chart. 100 percent (100%) of a number is the same number: 100% × 80 = 100/100×80 = 80. x 2 − x − 1 = 0 , {\displaystyle x^ {2}-x-1=0,} with a value of. When you count to 10, the numbers are 1, 2, 3, 4, 5, 6, 7, 8, 9 and 10. Click on pop-out icon or print icon to worksheet to print or download. Print without clutter. Watch all of our videos ad free with our app (desktop, apple, or android):https://www. The number grid 1-100 also presented with a number line, base-10 blocks, 10-frames and objects from various topics to achieve maximum goal result. 1-100 Number Grid. 618 033 988 749. φ = 1 + 5 2 = {\displaystyle \varphi = {\frac {1+ {\sqrt {5}}} {2}}=} 1. Number Chart (blank 100) Title: Microsoft Word - blank_100_chart. NOW UP TO 1000. Author: Liz Woodham Created Date: 8/5/2015 12:20:03 PM. Use the start/stop to achieve true randomness and add the luck factor. The golden ratio is also called the golden mean or golden section ( Latin: sectio aurea ). 100 94 95 96 99. Includes plain black and white, 5’s and 10’s highlighted, odd numbers highlighted, blank 100’s grid. Download for free 100 Chart Cliparts #234, download othes number grid 0 to 100 for free. Simply click on a times table chart below to view and then download. Donna Glynn Kinderglynn. Read and write decimals to thousandths using base-ten numerals, number names, and expanded form, e. For language learners new to German, a simple table of numbers can be quite useful! We’ve created that very thing for you right here. ^2); function of the airfoil top surface The bottom. My Hundreds Chart Name Date 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20. Count to 120, starting at any number less than 120. Number Grid 1-100. 392 = 3 × 100 + 4 × 10 + 7 × 1 + 3 × (1/10) + 9 × (1/100) + 2 × (1/1000). The number grid 1-100 also presented with a number line, base-10 blocks, 10-frames and objects from various topics to achieve maximum goal result. Can easily print to chart size using poster print options. 13 Best of Missing Number Grid Worksheets Fill in. To play the game, children will choose a level of difficulty, and then be asked to place numbers on the correct spot within the Number Chart. Set One B is a set of 9 printable 100-number charts. Free interactive classroom resources - Get activities, games and SMART Notebook lessons created by teachers for teachers. For K-12 kids, teachers and parents. After it opens, go to the top right corner of the worksheet, which is the first cell, and enter a title. It is an irrational number that is a solution to the quadratic equation. Writing 1-100 for kindergarten on a number grid. Title: Microsoft Word - 1-100_NumberGrid. Scroll down to the bottom to check out the worksheets! blank number chart 1-100 worksheet. com for free resources like German Word of the Day and our German. doc Author: ajk44 Created Date: 9/10/2015 12:59:35 PM. Read and write decimals to thousandths using base-ten numerals, number names, and expanded form, e. The World Health Organization (WHO) [1] lists cardiovascular diseases (CVD) as the number one cause of death globally with 17. In early grades you can pass out numbers 1 through 10 and have those students put themselves in order. Number Grid 1-100. Separate numbers by space, comma, new line or no-space. Given a two-digit number, mentally find 10 more or 10 less than the number. Home / Playground Markings / Mathematical Number Grid (1-100) Mathematical Number Grid (1-100) View complimentary products. 18 Best of 1 To 100 Backward Missing Number. NOW UP TO 1000. Each horizontal row represents a tens digit. Sal goes through all the numbers from 0 to 100 and shows some interesting patterns. Title: Printable Hundreds Chart Created Date: 7/27/2013 9:25:29 PM. The golden ratio is also called the golden mean or golden section ( Latin: sectio aurea ). Counting by 1s - Fill in the blank - 5 problems (random 1 to 25) math/countingby1s_1. 9 million people dying yearly. Missing numbers between 0 and 120. φ = 1 + 5 2 = {\displaystyle \varphi = {\frac {1+ {\sqrt {5}}} {2}}=} 1. Set One B is a set of 9 printable 100-number charts. Number chart 1-100. Hundred Grid Missing Number Worksheets SB. Find this Pin and more on Special Education by Gayle McLemore. Each times table chart can be downloaded for free. Using our built-in grid Sass variables and maps, it’s possible to completely customize the predefined grid classes. The number grid 1-100 also presented with a number line, base-10 blocks, 10-frames and objects from various topics to achieve maximum goal result. We also provide an empty number chart which can be used for exercises. Number grid 1-100 is useful because it helps children understand number sequences also to help them doing addition and subtraction. Find this Pin and more on Special Education by Gayle McLemore. The golden ratio is also called the golden mean or golden section ( Latin: sectio aurea ). A solid-state drive ( SSD) is a solid-state storage device that uses integrated circuit assemblies to store data persistently, typically using flash memory, and functioning as secondary storage in the hierarchy of computer storage. Counting by 2: Fill in the Blank (2 to 100) (PDF file) Printable. Number Chart is a fun and educational activity, which helps children see the relationships and patterns among numbers up to 120. φ = 1 + 5 2 = {\displaystyle \varphi = {\frac {1+ {\sqrt {5}}} {2}}=} 1. Adults also know the name workbook. Their critical thinking and concept of math. Number Grid. Select the Sign icon and make an electronic signature. The number grid 1-100 also presented with a number line, base-10 blocks, 10-frames and objects from various topics to achieve maximum goal result. These number worksheets contain many kinds of blank sequences of number 1-100 that you can choose for your kids to work on. Download for free 100 Chart Cliparts #234, download othes number grid 0 to 100 for free. Some of the worksheets displayed are 100 chart, 1 100 number grid, 100 chart 1 2 3 4 5 6 7 8 9, Grade 1. The World Health Organization (WHO) [1] lists cardiovascular diseases (CVD) as the number one cause of death globally with 17. Question: clear all 3% Mesh generation N = 100; number of grids in x-direction for numerical integ dx = 1/N; grid size (uniform) in x-direction x = dx/2:dx: (1-dx/2); grids in x-direction $% geometry of the airfoil top surface W = 10;$ span the airfoil (in z-direction), m y = x. 1,717 Followers, 438 Following, 440 Posts - See Instagram photos and videos from WYDaily (@wydaily). Separate numbers by space, comma, new line or no-space. But in adults also use number grids to facilitate their affairs. By reading and memorizing the numbers from 1-100 and its order repeatedly, children will be able to remember the numbers perfectly. The number grid 1-100 also presented with a number line, base-10 blocks, 10-frames and objects from various topics to achieve maximum goal result. It is an irrational number that is a solution to the quadratic equation. For children to practice writing their numbers in a gradual way. Consistently answer questions correctly to reach excellence (90), or conquer the Challenge Zone to achieve mastery (100)!. 1 to 100 Number Squares x4. Missing numbers between 0 and 120. Select the Sign icon and make an electronic signature. Writing 1-100 for kindergarten on a number grid. NOW UP TO 1000. We have a dream about these 100 Worksheet Template images gallery can be a hint for you, bring you more examples and most important: make you have bright day. Adults also know the name workbook. Purchased for a number crazed 5 year old with played with it right off. There is a gridded side, a blank side, and a numbered grid insert that allow for learner variability/knowledge level. Writing 1-100 for kindergarten on a number grid. Percentage Definition. A solid-state drive ( SSD) is a solid-state storage device that uses integrated circuit assemblies to store data persistently, typically using flash memory, and functioning as secondary storage in the hierarchy of computer storage. IXL's SmartScore is a dynamic measure of progress towards mastery, rather than a percentage grade. This educational activity helps kids to develop an understanding of patterns and number relationships by utilizing a number grid. Change the number of tiers, the media query dimensions, and the container widths—then recompile. In other words, percent of a number is the ratio between the number and 100. Math explained in easy language, plus puzzles, games, quizzes, worksheets and a forum. An interactive math lesson about whole numbers to 100. Use the start/stop to achieve true randomness and add the luck factor. 100 percent (100%) of a number is the same number: 100% × 80 = 100/100×80 = 80. You can use 3 available alternatives; typing, drawing, or uploading one. Each times table chart can be downloaded for free. The World Health Organization (WHO) [1] lists cardiovascular diseases (CVD) as the number one cause of death globally with 17. Learning numbers 1 - 100. Number charts: counting by 4's. It tracks your skill level as you tackle progressively more difficult questions. Use place. Watch all of our videos ad free with our app (desktop, apple, or android):https://www. For example, 13% is equivalent to the following. 1 to 100 Number Grid Say-It Chart | Barker Creek Colorful, cardstock poster shows a large 10x10 number grid with the 5's and 10's in a different color. After it opens, go to the top right corner of the worksheet, which is the first cell, and enter a title. The risk of heart disease increases due to lifestyle that leads to overweight and obesity, hypertension, hyperglycaemia, and high cholesterol [1]. File previews. Download for free 100 Chart Cliparts #234, download othes number grid 0 to 100 for free. ^2); function of the airfoil top surface The bottom. Print on 12x18 paper (I sent to print shop for minimal fee) and when trimmed it will be a 12x12 square. 21 31 41 51 61 71 81 91 12 22 32 42 52 62 72 82 92 13 23 33 43 53 63 73 83 93 14 24 34 44 54 64 74 84 94 15 25 35 45 55 65 75 85 95 16 26 36 46 56 66. See also Candy Corn Worksheets. Scroll down to the bottom to check out the worksheets! blank number chart 1-100 worksheet. Worksheets are 100 chart, 100 chart, 100 chart, Writing numbers work, 1 100 number grid, Blank hundred chart, Hundreds chart puzzles, Numbers to 200. Ensure the info you fill in The Number Grid - 1-100 is up-to-date and correct. Number Line – There are several number line activities you can do with these cards that get students up and moving. These number worksheets contain many kinds of blank sequences of number 1-100 that you can choose for your kids to work on. Question: clear all 3% Mesh generation N = 100; number of grids in x-direction for numerical integ dx = 1/N; grid size (uniform) in x-direction x = dx/2:dx: (1-dx/2); grids in x-direction $% geometry of the airfoil top surface W = 10;$ span the airfoil (in z-direction), m y = x. The World Health Organization (WHO) [1] lists cardiovascular diseases (CVD) as the number one cause of death globally with 17.
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http://runescape.wikia.com/wiki/Summoning_training
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# Summoning training
24,872pages
on this wiki
Summoning training can become very expensive and tedious, but familiars at higher levels make Summoning a rewarding skill. A recent update has allowed F2P Players to gain xp up to level 5 and continue to collect Gold charms after level 5, which may be useful if you intend on becoming P2P eventually. All F2P parts of this guide will be shown with "(F2P)" beside it.
## Starting (F2P)Edit
Main article: Starting Summoning
To begin training the Summoning skill, players may complete Wolf Whistle, (not required anymore) which will grant 276 experience to Summoning (increasing the player's level to 4), as well as 275 gold charms (enough for level 16, or 5 on F2P). After that you will have to collect your own charms. Look below for information on how to do that.
F2P can collect additional Gold charms from saving Nora in Burthorpe from a burning fire (awards you with a one-time reward of 10 charms), collecting one free Gold Charm sample from Pikkupstix's store or by killing Troll Brutes and Troll Chuckers in Burthorpe
## QuestsEdit
Players may complete certain quests and activities to receive summoning experience, including Penguin Hide and Seek, the Tears of Guthix Distraction and Diversion, Troll Invasion, and certain quest lamps (usually require you to reach level 30 first to get Summoning exp). Using non-charm methods of leveling gives experience, and allows players to save valuable charms for use in higher level familiars.
## Pouches (F2P)Edit
The only viable way of training Summoning is to create Summoning pouches. To begin making a Summoning pouch, the following are required:
• An empty pouch, which can be bought from Pikkupstix for 1 gp each. (Not required for Summoning within Daemonheim)
• A quantity of charms, which must be collected by the player through monster killing and other methods. Good ways to do this are listed below.
• A large number of spirit shards, which can be bought from Pikkupstix for 25 gp each or you can buy a set of 5000 for 125k. (Not required for Summoning within Daemonheim), Spirit Shards can also be bought at the Grand Exchange, which has a bank next to it. Though the guide price is not always 25, it is rare that anyone will sell it for anything less than that.
• A tertiary ingredient which is specific to the creature you are creating, similar to the secondary ingredients used in Herblore.
## Spirit ShardsEdit
With the use of Bogrog, the pouches you make can be turned back into shards; the number of shards returned to you is 70% of the pouch cost. With this in mind, you can drastically reduce the number of shards you have to originally buy by continually converting the pouches you make back into shards.
## Obtaining Charms (P2P)Edit
Players attempting to train summoning should kill monsters that drop lots of crimson charms, as they are the best for experience.
Monsters vary greatly in the number and type of charms they drop.
Once weekly, the Familiarisation Distraction and Diversion can be done, which gives a reward of 40 minutes of triple charm drops. It is recommended to use this time by killing a monster that drops a lot of crimson charms.
If you have 21 Summoning and spent some time training Dungeoneering, you can buy a charming imp, which automatically picks up requested charms and converts unwanted charms into small Summoning experience. This can increase the rate of gathering charms significantly and it increases the amount of Combat or Slayer experience earned (per hour). It is highly encouraged to configure the imp to give the player all types of charms; the experience rewarded for converting the charms is much lower than the amount offered by using the charms to make pouches.
The Fang of Mohegan may also be used to provide a slight possibility of a double charm drop.
See here for a table of monsters which drop charms.
### SlayerEdit
The fastest way to level summoning is to camp at a monster that drops lots of charms, such as Waterfiends or Rock Lobsters. However, it is also common for players to get charms from training Slayer rather than from camping at a specific monster. It is possible to get enough charms for 99 Summoning by getting 99 Slayer, although it depends on what tasks the player does. If a player is going for 99 in both skills, there is no need for them to camp at a monster specifically for charms. However, if a player does not want to do Slayer, or is unable to obtain all the charms that they want purely from Slayer, then there are several monsters that are very good to camp at for charms.
### Gelatinous AbominationEdit
Gelatinous Abominations have around a 44% charm drop rate. They drop gold charms at a rate of around 41%. The Gelatinous Abominations have only 150 life points and are level 6 so they are really quick and easy kills. The only equipment required are spiked gauntlets, which can be purchased at any Slayer Master. These monsters are found in the Taverley Slayer Dungeon, south-east of Turael (or Spria) in Taverley. Killing these are a great way for anyone to obtain gold charms, but no drops of great value can be obtained, and the combat experience gained is little.
### Rock CrabsEdit
Rock crabs have a 13% charm drop rate, 10% of which are gold charms. This can be an efficient way for lower levelled players with some extra cash to obtain charms, and get some decent exp. You should also run around and activate all the crabs since they don't attack and won't be able to be attacked unless someone runs next to them. If they stop attacking you, just run south of the crabs, right past the helmet shop and then run back up. This can be a great way for lower levels to obtain charms and get ranged exp. You can get around 3-5 charms per minute if done correctly.
This is not recommended unless you don't mind spending money to buy cannonballs, since they're very expensive!
### Moss giantsEdit
Moss Giants are great for gold charms, about 40% drop rate. They have very low defence and slow attack speed making them easy to power through even at low levels. North West of Ardougne has 4 Moss Giants and it is more often than not, empty. Other popular places include the Varrock sewer system and Moss Giant Island. For almost guaranteed seclusion (as well as a pretty view) there are also some Moss Giants on Crandor (the island where your ship crashes during Dragon Slayer). Finally, a decent amount of moss giants can be found in the dungen south of Brimhaven (requires 875 coins to enter). This area will almost always be empty.
### Ice giantsEdit
Ice Giants are great for gold charms, about 55-57% drop rate for charms. They have relatively low defence and slow attack speed making them easy to power through even at low levels. Two places to find Ice Giants are in the ice mountain dungeon and in the Asgarnian Ice Dungeon. There is also a location where they spawn next to the western wilderness wall and the members fence line. Although being in the deep wilderness it is not recommended .
### GoblinsEdit
Goblins have an 11% gold charm drop. The best way to kill goblins is to go to the goblin hut in Lumbridge. It is a bad idea to bring a slow weapon like a godsword, battleaxe, or 2H, as you cannot hit very quickly, resulting in slow progress. A weapon like a whip, scimitar, dagger or claws are ideal so you can do 20-30+ goblins and obtain 2-3 charms per minute. If you camp for an hour, you can easily get 50+ charms. This is not recommended if you cannot frequently hit 50-120 damage.
Note: The goblins across the River Lum near the Al-Kharid gate have 67 LP, keep that in mind when choosing what weapon to use to get one-hit kills (scimitar works best most of the time). The goblins found in the Goblin Village do not drop any charms at all, so it is not advised to go there.
### Giant rock crabsEdit
Giant rock crabs can be the best monster to get gold charms from. They drop three of these at a time quite frequently. The giant rock crabs are just located in the Chaos Tunnels, near the entrance southwest of the Wilderness Volcano. It is recommended to use Magic to kill Giant rock crabs, since it is their weakness. Bring a few pieces of food, runes for the spell you'll be using, and also some runes for telekinetic grab (if using the standard spellbook). There's a safespot by the stalagmites in the northeast corner of the room.
Walk near a boulder and it will transform into a giant rock crab. Lure it to your safespot so it gets stuck and you can safely mage it. When near death, the giant rock crab will attempt to flee. Once the crab is dead, it's recommended to use telekinetic grab on the charms so you wont put yourself at danger by going near some other crabs.
Ice Burst or Ice Barrage can be used effectively here. Use Protect from melee or Deflect melee and gather all the crabs at a spot so that you can burst or barrage them. This method gives the most charms per hour.
A disadvantage to killing giant rock crabs is that the charms gained are mostly gold, which are the lowest level charm. Comparatively, the charms that can be gained in a given time period from killing giant rock crabs will not give as much summoning experience as the charms that can be gained in the same time period killing a monster that drops mainly higher level charms, such as waterfiends or rock lobsters. Giant rock crabs should only be killed at low level summoning when gold charms are most useful, and even then, it is important to realise that since charms can be saved up, it probably would be more worth your time to just start at a monster that drops primarily crimson charms right away, since you will end up getting gold charms to make it through the low summoning levels anyway, at the same time as you save up tons of crimsons to use later.
### Black demonsEdit
With the introduction of infernal ashes, killing black demons became another reasonable method of gathering crimson charms. The ashes cover the cost of the cannon balls and other consumables. At higher levels, such as with 70 agility, 52/68/96 summoning, a chaotic weapon, and 80+ combat stats, a player can amass profit above 200k per hour on top of the crimson charms. High ranged levels help too, as demons are weak to bolts.
### Rock lobstersEdit
Killing rock lobsters is often considered to be the fastest way to train Summoning, although waterfiends can actually be faster than bursting with the optimal levels and equipment, including 99 Summoning for a Steel Titan. Rock lobsters do not require a high Summoning level to kill quickly.
The best way to kill rock lobsters is to use Ice Barrage or Ice Burst since they are aggressive. There are many different ways to do this successfully, most of which involve luring 9 rock lobsters towards you while using Protect from Melee or Deflect Melee, moving to a safe spot, then casting the spells to kill them. Armour that gives high magic and/or prayer bonuses is recommended. Most of your inventory should be Prayer potions, but you will also want runes for ice spells, and a few blood spells (to heal, as you can take damage from Dagannoths using a ranged attack). Also bring a teleport and a few pieces of food.
You need at least one partner to get to rock lobsters. See the Waterbirth Island article for more information.
Players attempting to use this method to kill Rock lobsters should have 70+ (preferably 94+) magic, and 70+ prayer. They should also have a large amount of cash, as this method does cost a significant amount of money due to the runes and prayer potions used.
Ice bursting Rock lobsters can give over 150 crimson charms per hour, whereas ice barraging can give up to 200 crimson charms per hour.
To further improve the number of Crimson charms players get and how many runes they use, many players choose to use damage increasing variables such as the staff of light, arcane stream necklace, arcane blast necklace or arcane pulse necklace as well as the Wolpertinger special move, Magic Focus which gives the same boosting effects as the extreme magic potion.
### WaterfiendsEdit
Waterfiends, like Rock lobsters, are very good to kill for crimson charms. Waterfiends are cheaper to kill because magic is not needed, and waterfiends can potentially be even faster for crimson charms than rock lobsters. However, the best kill rates of waterfiends can only be obtained with very high levels including 99 summoning and equipment such as the Royal Crossbow or 2h Crossbows due to their weakness to Bolts.
Waterfiends can be killed in the Ancient Cavern, or in the Chaos Tunnels.
Waterfiends have a ranged attack and a magic attack. The ranged attack is more dangerous. Players should wear mage defence armour (Karil's or royal d'hide) and use the Protect from Missiles or Deflect Missiles prayer. Ganodermic may also be used, but is not recommended due to the cost and expensive repair costs. Players with 68 or higher Summoning may use the Bunyip familiar's scroll (swallow whole), to consume the raw sharks and raw lobsters dropped by the waterfiends. Waterfiends are weak against bolts. The best weapons to use are the Royal Crossbow, Dual Chaotic Crossbows or your best Crossbow and off-hand Crossbow,
Many high level players will use a Unicorn stallion's healing aura scroll, so they do not have to use prayer, except in the Chaos Tunnels. Others prefer to use the Turmoil and Soul Split prayers to avoid using a unicorn stallion familiar. The extreme potions from Herblore also come in handy here. These methods are only beneficial with 95+ melee stats, 80+ prayer, 90+ range and mage, and 88+ summoning. This higher-level method will lead to 100-140 crimson charms per hour in the Ancient Cavern.
In the Chaos Tunnels, for upwards of 200 crimson charms per hour, extreme potions (89 or 90 Herblore) and maxed combat skills (with 95+ prayer) are required. However, this method can be used at lower levels as well, resulting in approximately the same amount of crimson charms per hour as bursting rock lobsters. For both methods level 95+ summoning is strongly recommended, since the Iron and Steel titans are capable of doing massive damage with their special attacks. Players will need high level equipment (Karil's and a Zamorakian spear) with a high defence against magic bonus. It is a very hard, but very rewarding method, and is definitely not for everyone. A generic inventory would contain a couple combat-boosting potions, high level food (Monkfish or above), and 200+ Iron within or Steel of legends scrolls, leaving the rest to be filled with prayer potions and one emergency teleport. Alternatively, the food and prayer potions can be replaced with Saradomin brews and Super restore potions.
### BorkEdit
Bork can be killed once per 24-hour period beginning at 00:00(UTC), and has guaranteed charm drops. The minimum he will drop on death is 5 blue, 7 crimson and 2 green charms. Wearing a Ring of wealth adds 3 crimson and 1 green to the base total, and completing all of the Varrock Tasks leads to the base charm drop being doubled. The Familiarisation D&D will not triple these charm drops.
Drop quantity
Icon Name Default
Blue charm 5 5 10 10
Crimson charm 7 10 14 17
Green charm 2 3 4 5
As Bork can be killed daily, the number of charms that slowly mount up can significantly enhance the pool of charms that a player may amass through other methods, or even become the primary source of charms for players that only want to train Summoning on a casual basis.
### Tormented demons/GlacorsEdit
Tormented demons are excellent monsters for obtaining blue charms. On average, around 3 charms per 2 kills will be obtained. A strategy guide can be found here .
Glacors are also excellent monsters for obtaining blue and crimson charms. They drop 3 per kill and have roughly the same percent chance of dropping blue or crimson. A strategy guide can be found here.
### Greater demonsEdit
Using a Dwarf multicannon on the 10 greater demons in the Forinthry Dungeon can net 200+ crimson charms an hour. It is recommended to bring Prayer urns to collect the mass amount of Accursed Ashes that will be dropped, which will give prayer experience as well as greatly cleaning up the drops on the ground. The Forinthry Dungeon is in the Wilderness, the greaters within located in level 20-24, so if using this method, it is recommended to have a teleport item handy (such as an Amulet of Glory), and do not bring anything you are not willing to lose.
Warning: Dying in the wilderness will cause you to drop ALL of your items.
### HellhoundsEdit
Hellhounds, while not advised for green, crimson or blue charms, have a 68-69% drop rate for gold charms and can be safespotted in most of their locations. If you have high enough ranged or magic skill to swiftly kill them, you can quickly collect gold charms in safety. Unfortunately, hellhounds have very poor drops and aren't profitable (not counting the rewards of any possible clue scrolls you may find). An alternative for gold charms are fire giants with around 55% drop rate for gold charms and they are much easier to kill with additional good loot.
### Cave bugEdit
Cave bugs have a 40% green charm drop rate, the second highest in game after mithril dragons.They are level 18 and fairly easy to kill. Note only the level 18 ones drop charms, level 6 cave bugs drop no charms at all.
With dragon or better armour and an abyssal whip, one can easily achieve 40+ green charms an hour at levels 70 attack and 70 strength.The only place to kill them is the Dorgesh-Kaan South Dungeon , where there's a convenient fishing spot where you can fish for frogspawns and a fairy ring nearby.
### King Black DragonEdit
With high melee stats (preferably 90+, Turmoil, Soul Split, and unicorn stallion) it is possible to obtain over 115 crimson charms an hour, while making about 350-500k an hour profit. This can be a rather attractive way to obtain charms for high-leveled players as the charm-obtaining rate is very good while are making profit rather than losing a lot from bursting rock lobsters.
However, this method requires the ability to make super anti-fire potions, so players can use an off-hand weapon or 2h weapon rather than a dragonfire shield in order to increase its efficiency.
### Kalphite Queen or exiledEdit
With high combat stats, it is possible to obtain various charms per hour, while making a profit of 600-800k per hour. This is helpful as the Queen drops various charms, with the exiled version dropping much more than the regular variant. The regular Queen drops 6 charms, while the Exiled one drops 9.
However, the Queen is very tough and has several guards, so players may need to rebank much more often.
### Exiled kalphite guardiansEdit
Exiled kalphite guardians drop blue charms at a very high rate. This makes it incredibly fast and easy to obtain blues without completing high level quests for tormented demons or glacors.
### Troll brute and Troll chucker (F2P)Edit
These are the only two monsters that drop charms on non-members worlds. They only drop Gold charms on these worlds, and rarely drop Green, Crimson and Blue charms on members worlds. They are only level 2 monsters which make them ideal for lower leveled players. Average and high leveled members have much better ways to get charms.
### Ascension membersEdit
These monsters come in four varieties, which each one is dominant in one type of charm, and another charm being uncommon to the dominant charm.
• Rorarius - mainly drops Gold charms.
• Capsarius - mainly drops Green charms.
• Gladius - mainly drops Crimson charms, blue charms uncommon.
• Scutarius - mainly drops Blue charms, crimson charms slightly the same as blue.
• Legiones - mainly drops high-level charms.
### FiremakingEdit
When adding logs to a bonfire, fire spirits sometimes arise from the flames. A common reward for freeing them is charms of all four types. Free players cannot obtain any charms, not even Gold, from bonfires.
### Hunter Edit
There are 2 ways to collect charms using the hunter skill: spirit implings, which require level 54 hunter, and charm sprites, which require 72 hunter. Spirit implings are hard to find and only gives charms some of the times as they also give other summoning supplies, primarily tertiary items. Charm sprites are notable for giving a high blue charm output compared to the total number of charms, and provides a good alternative to skillers.
## Which pouches to makeEdit
Note: The number in parentheses is the number of pouches it takes for the xp between those levels.
Level Charm
Gold Green Crimson Blue
1-4 Save Save Save Save
16-22 Granite crab (134)
22-28 Granite crab (241) Spirit Tz-Kih (54) or save
28-32 Save Compost mound (114) Spirit Tz-Kih (59) or save
32-33 Compost mound (36) Honey badger (13)
33-40 Beaver (398) Honey badger (163)
40-41 Bull Ant (75) Or Save Beaver (69) Honey badger (29)
41-46 Bull Ant (386) Or Save Macaw (371) Honey badger (191)
46-47 Bull Ant (123) Or Save Macaw (328) Pyrelord (36)
47-49 Bull Ant (315) Or Save Magpie (200) Pyrelord (82)
49-52 Bull Ant (384) Or Save Magpie (384) Bloated leech (149)
52-56 Spirit terrorbird (883) Magpie (726) Bloated leech (281)
56-57 Spirit terrorbird (281) Ibis (195) Bloated leech (90)
57-58 Spirit terrorbird (311) Ibis (215) Bloated leech (99) Spirit kyatt (43) or Spirit graahk (43)
58-61 Spirit terrorbird (1,138) Ibis (788) Bloated leech (362) Spirit kyatt (156), Spirit graahk (156) or Save
61-64 Spirit terrorbird (1,532) Ibis (1,060) Smoke devil (391) Spirit kyatt (209), Spirit graahk (209) or save
64-66 Spirit terrorbird (1,305) Ibis (904) Stranger plant (317) Spirit kyatt (178), Spirit graahk (178) or save
66-69 Barker toad (1,975) Ibis (1,739) Stranger plant (611) Mithril minotaur (283) or save
69-73 Barker toad (3,734) Fruit bat (2,681) Stranger plant (1,154) Mithril minotaur (560) or save
73-74 Barker toad (1,189) Fruit bat (853) Stranger plant (368) Obsidian golem (161)
74-79 Barker toad (8,076) Fruit bat (5,797) Granite lobster (2,158) Obsidian golem (1,094)
79-80 Barker toad (2,153) Fruit bat (1,546) Granite lobster (576) Fire titan (270) or Moss titan (270)
80-83 Barker toad (7,898) Hydra (4,880) Granite lobster (2,111) Fire titan (989) or Moss titan (989)
83-85 Barker toad (6,730) Hydra (4,159) Granite lobster (1,799) Lava titan (802)
85-88 Barker toad (12,957) Hydra (8,006) Swamp titan (3,018) Lava titan (1,544)
88-89 Barker toad (5,248) Unicorn stallion (2,957) Swamp titan (1,222) Lava titan (626)
89-92 Barker toad (19,253) Unicorn stallion (10,849) Swamp titan (4,484) Geyser titan (2,139)
92-95 Barker toad (25,912) Unicorn stallion (14,601) Wolpertinger (5,569) or
Swamp titan (6,035), see below
Geyser titan (2,879)
95-96 Barker toad (10,495) Unicorn stallion (5,914) Iron titan (2,187) Geyser titan (1,166)
96-99 Barker toad (38,505) Unicorn stallion (21,696) Pack yak (7,931) Geyser titan (4,278)
Level Gold Green Crimson Blue
Charm
• DISCLAIMER! - These are only recommendations, you should be very careful when deciding what pouches to make for summoning, as prices change, and some pouches may only be good to make at certain times if you are wishing to get a summoning cape for a relatively low cost.
• 58-69 with blue charms can be replaced with Karamthulhu overlords, but they only give 8xp more than Spirit graahk or Spirit kyatt, and make much less money.
• It is best if you can save Blue charms until level 89 and make Geyser titans, since they give the most experience per blue charm.
• Some people choose to stop using Gold charms at around level 66 because the xp rate is around 230-240k xp/h which makes it expensive for the xp gained and not worth the time.
• Some people choose to stop using Green charms at around level 80 because the xp rate varies from 370-420k xp/h which is expensive for the xp gained and not worth the time. From level 80-99, Crimson charms are 870k-1.145m xp/h and Blue charms are 1.187m-2.120m xp/h which is much more cost efficient than using Green charms.
• Wolpertinger pouches require two items (wolf bones and raw rabbit) and therefore cause people to choose swamp titans over them. However, it is very difficult to make back both the amount of potential "lost" exp and lost money (as wolpertingers are both cheaper to make and grant more exp per charm) in the amount of time that is saved by choosing swamp titans. Swamp titans should not be made at this level unless the player either does not care about how much charms/gp is consumed and/or simply does not want the extra time spent when using the charms.
## How to Make PouchesEdit
The uncharged pouches update has been reverted. Players may train Summoning by creating pouches at the obelisk as normal.
When making pouches, you should have the charms you are using, the unmade pouches, the spirit shards, and 25 of the tertiary ingredient.
Players who haven't completed Swan Song may find it best to use the obelisk in Taverley, as it is the closest to a bank.
Below level 57 summoning, it is best to use phoenix lair teleports and run north-east to the Piscatoris summoning obelisk, as it is only a short distance away, and phoenix lair teleports are relatively cheap. You could also use the fairy ring to bank in Zanaris. If you have not done Swan Song, you may also use the Obelisk at Pikkupstix and bank in Taverley. Because these methods require a lot of running, you may want to have any weight reducing equipment you own equipped. You may also choose to use a familiar to hold more items, or use a Spirit terrorbird for infinite run. Another good strategy would be to have a house at Taverley, using House Teleports and rings of duelling, games necklaces, or a Ring of kinship (unlimited teleports) to teleport. This requires running from the house teleport portal to Pikkupstix's house and repeating. It is now more effective to use a house teleport, run to Taverley bank, and then to the obelisk since a large lake is located between the house portal and the obelisk.
Should the player not wish to spend money on Kyatt pouches, they can use the nature grotto in Mort Myre Swamp. This method requires a Tokkul-Zo, access to the Fairy Ring Network, and the Mort Myre shortcut unlocked from Temple Trekking (Total follower level 100). Start at the Fight cave bank, and use the fairy ring to Canifis (code C•K•S). Run south and go through the Mort Myre gate, shortcutting to the Nature Spirit grotto. Make the pouches in the grotto, and bank using the Tokkul-Zo. This method is slower than using a Spirit Kyatt, but faster than running to Pikkupstix's house in Taverley.
Above level 57 summoning, you should use Spirit kyatt pouches as a teleport near the Piscatoris summoning obelisk, and use a tokkul-zo or rings of duelling to bank. Using this method it is possible to make over 2700 pouches an hour, making it the fastest way to train summoning. Due to the agility update, all you need to have equipped is your teleport ring. Use your kyatt to get near the summoning obelisk, and then run north. Alternative free, but slower, banking methods (if the Tokkul-zo is not available) are a ring of kinship, amulet of glory or Burthorpe lodestone.
It is also possible to use the Balloon Transport System to make pouches in Taverley. Equip a hatchet and a ring of duelling, take a balloon to Taverley, make pouches, then teleport to the bank at Castle Wars. You'll use one inventory space for logs (which you can cut at Castle Wars next to the balloon site), but won't need anything else equipped as running is not a problem.
## How many Spirit Shards do I buy?Edit
With the use of Bogrog, you can turn the pouches you make back into spirit shards; the amount of shards you get is usually 70% of the required shard number for the pouch. Utilising this, the amount of shards you need to initially buy can be drastically reduced, however the effectiveness of this varies greatly with number of shards you initially have and the amount of times you are willing to exchange your pouches.
The formula which calculates the effective number of shards you own is as follows
$\sum_{i=1}^{n}S\times0.7^{\ i-1}$
S = initial amount of shards
n = number of trips done to Bogrog
i = parameter
The calculator below can be used for this summation for any integer S, 0 ≤ S ≤ 2,147,483,647 and any integer N, 1 ≤ N ≤ 100:
template = Calculator:Bogrog shards
form = CalcForm
result = CalcResult
param = S|Initial shards|0|int
param = N|Trips to Bogrog|1|int
Calculator requires JavaScript
This text will be replaced when you submit the form
However, it becomes tedious to recalculate this continuously to find the required number of shards you need initially. It is much easier to first decide how many trips to Bogrog you are willing to do as the amount of trips dictates what percentage increase you will ultimately get (regardless of initial count). The increases can be seen below:
Trips Percentage Increase Divide by
00%1.00
170%1.70
2119%2.19
3153%2.53
4177%2.77
5194%2.94
6206%3.06
7214%3.14
8220%3.20
9224%3.24
10227%3.27
15229%3.29
20230%3.30
25231%3.31
30232%3.32
The maximum possible number to divide by is 3 and a third (3.33). As shown in the table, to get close to this number requires many trips.
Let's say you needed 55,000 shards to attain a certain level, the first thing you do need to decide upon is how many times you would be willing to visit Bogrog which will usually be 10 times, any more than this the gain in shards is minimal; for example doing 20 trips instead of 10 when having an initial starting shard count of 100,000 only yields an extra 6,405 shards. The amount of trips you are willing to do is something you have to decide.
At 10 trips, the increase is 227% so the effective amount will be 327% of the initial amount, or 3.27 times the initial amount. You then divide 55,000 by 3.27 to get 16,820 (rounded up). Therefore, to effectively have 55,000 shards whilst willing to do 10 trips to Bogrog you only initially need to have 16,820 Spirit shards.
Note: Calculations can conveniently be performed with the software available on one's computer.
Alternatively one could buy as many spirit shards as desired and sell the extras to a Summoning shop for the same price that they were purchased for.
## What to do with the pouches (F2P)Edit
Some options only work well with some pouches, some only work well depending on prices. You should analyse the data and figure out what is the best option for you.
Method Benefits Drawbacks
Sell to Grand Exchange. Works for almost every pouch and will generally get a fair price. Sometimes you may be forced to wait for pouches to sell.
Trade to Bogrog for Spirit shards. (P2P) Can sometimes save you money over the Grand Exchange. Needs a higher level than what is required to make the pouch.
Cast the High Level Alchemy spell on them. Can be a good way to train Magic cheaply. * Requires more effort and time that you can spend training Summoning.
• Many pouches make more cash by selling
• Many other items are more profitable to alch than any pouch
Use the pouches for skilling or combat. Often this will be a great way to make lots of money off training Summoning. * Takes a long time to pay out and will require a lot more capital invested in Summoning.
• Selling and employing the cash, then buying back when needed, may be more efficient.
Make summoning scrolls from the pouches. You gain a slight bit more experience/charm. In many cases the given scrolls will also be worth slightly more than the pouches. The extra time required to turn pouches into scrolls may not be worth the trivially small extra amount of experience.
## Other important informationEdit
• Why aren't Arctic bears listed?
• Arctic bears only give slightly more experience per pouch than barker toads, but they cost much more. Gold charm pouches are in general relatively cheap (in accordance with how much experience they give.) Because of this, there is really no reason to make Arctic bears, unless you want them for their teleport and hunter bonuses, even then it may be better to make and sell barker toads and buy Arctic bears.
• Why aren't any abyssal pouches listed?
• Why aren't talon beast pouches listed?
• Although talon beast pouch is the single pouch that give the most experience in game, in order to make talon beast pouches, talon beast charms are needed. The only way to obtain these charms is to play Temple Trekking activity over and over again in hope to encounter nail beasts which drops them. If players have talon beast charms then it is recommended that they use them, but talon beast charms usually take more effort to obtain than any other charm in-game, therefore it is not recommended to list talon beast pouches as your only option to gain experience.
• Why does it say to wait so long before using Blue charms?
• Blue charms give lots of experience, and as it is easy to level with the other charms at first, it is a waste to use your blue charms, when you could gain way more experience in the long run by using them at higher levels.
• Why should Gold charms be saved from levels 28-51?
• Why should Spirit larupias not be made when they are essentially the same as Spirit graahks and Spirit kyatts?
• Spirit larupias are in lower demand, because they are not quite as useful, so it is better to make the other ones. Also Spirit kyatts can be used in the Piscatoris method of making pouches.
• Why should Ice titans not be made when they are essentially the same as Fire titans and Moss titans?
• Why make Wolpertingers?
• Making swamp titan pouches is about 2x faster than wolpertinger while wolpertinger provide more experience per pouch. If we assume one can consume 3,000 crimson charms per hour making swamp titan pouches, then 1,500 wolpertinger pouches can be made. Considering time taken gathering charms and making into pouches, given 3,000 crimson charms and 2 hour, wolpertinger grants 1,214,400 exp while swamp titan grants 1,120,800 exp and a free hour for charm gathering and pouch making. So if one can make up the 93,600 exp difference in one hour, which is gathering 251 crimson charms in 55 minutes and make into pouches in 5 minutes, then he/she is advised to make swamp titan. Gathering 251 crimson charms in 55 minutes is approximately 274 charm/hr which is the break-even point.
• What's the fastest way to make pouches? Where's the closest bank?
• As of the Temple Trekking update, a fast method is using the Nature Grotto obelisk via the Mort Myre gate shortcut reward from Temple Trekking. This will bring you right outside the Nature Grotto which is steps away from the obelisk inside. To bank, use an Amulet of Glory to Edgeville, and then use a Ring of Slaying to the Fremennik Slayer Dungeon to gain access to the Fairy Ring network, or simply use the Tokkul-Zo to teleport to the Fight Caves to both bank and access the Fairy Rings. Use Fairy Ring code C•K•S to teleport you a few steps north of the Mort Myre gate. Alternatively, you can teleport to Canifis using the Kharyrll teleport spell in the Ancient Magicks spellbook or use a POH Teletab if you have a Kharyrll portal. Use the bank in Canifis, then run back to the gate. The former method uses no inventory space if you're wearing the Glory and Slayer ring, or Tokkul-Zo, which means you can make more pouches per hour versus having runes or POH teletabs in your inventory.
• One of the fastest ways to make pouches is to trade noted ingredients with Pikkupstix in Taverley, buy them back, make pouches and sell them to Pikkupstix, repeating the process until the charm supply is over. Despite being probably the fastest way to create pouches, this method is very expensive and should only be used by players with a considerable amount of coins.
• (57 Summoning required) - The fastest method is to summon your Spirit Kyatt, click on your familiar, and use the Teleport option, this should take you only a few steps below a trapdoor, where a summoning obelisk is located for making pouches. Go down the trapdoor, and create familiar pouches. Then teleport to a bank with one of the following, bank your pouches, and repeat.
• An alternative to using the Spirit Kyatt is to use fairy ring code A•K•Q and the bank in Zanaris or Edgeville (fairy ring code D•K•R).
• If you have access to Wolpertingers and have unlocked the balloon near Castle Wars, another method is to use a ring of dueling and use it to teleport Castle Wars. Take charms, pouches, shards, 24 of the secondary ingredient, and logs and run to the balloon near Castle Wars. Travel to Taverley and run to the summoning obelisk there, make pouches, and teleport back to Castle Wars to repeat. *Note that your weight cannot exceed 40kg when doing this method; although weight reducing gear (such as boots of lightness and spotted cape) work.*
• (If you have completed Swan Song) - Either teleport to the Phoenix Lair by using a Phoenix lair teleport scroll or run to the Piscatoris Fishing Colony. Withdraw all of your supplies (Pouches, Spirit shards, etc.) from your bank from Arnold Lydspor and run south until you reach the Colony gate. Climb through the Hole and run southeast to the Summoning Obelisk (marked by the Summoning icon). Climb down the trapdoor, make your pouches and/or, scrolls, repeat.
## TriviaEdit
• Note that after reaching level 17 it is theoretically possible to get to any level (even level 99) in a single trip by making either Spirit mosquito pouches or Giant chinchompa pouches at the Feldip hills obelisk. Both of these familiars require a stackable tertiary ingredient, allowing a player to collect all the required charms and seconds for any level within a single inventory. While technically possible, it requires immense amounts of preparation and time (includes getting all the blue charms, and buying a maximum of 10,000 pouches at a time from the Grand Exchange), and you would have to drop all made pouches or turn them into scrolls, and does not have much of a benefit, so it is not suggested. The player could also do this method by buying mithril seeds, picking the flowers and making praying mantis pouches. For the above to work one must also make the pouches into scrolls or else the inventory would fill up from pouches or with a high enough summoning level swap them back for shards.
A way to avoid all the methods above is to have all the charms, pouches, shards, and noted secondary items at hand. Friends or people could trade you un-noted secondary items for noted items, although such an occurance without reward is highly unlikely.
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2014-03-10 07:50:56
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http://physics.stackexchange.com/questions/27577/analytic-continuation-of-imaginary-time-greens-function-in-the-time-domain
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# Analytic continuation of imaginary time Greens function in the time domain
Consider the imaginary time Greens function of a fermion field $\Psi(x,τ)$ at zero temperature
$$G^τ = -\langle \theta(τ)\Psi(x,τ)\Psi^\dagger(0,0) - \theta(-τ)\Psi^\dagger(0,0)\Psi(x,τ) \rangle$$
It is well known that we can obtain the retarded Greens function by performing Fourier transformation into frequency space and performing the analytic continuation $iω \to ω + i\eta$.
What I would like to do is to perform the analytic continuation directly in the form $iτ \to t$, but I don't know how to deal with the $\theta(τ)$ terms.
How to perform the analytic continuation $iτ \to t$ of the step function $θ(τ)$?
In my case, I am dealing with a chiral Luttinger liquid, giving something like
$$G^τ(x,τ) = -\left[\theta(τ)\frac i{iλ + ivτ - x} - \theta(-τ)\frac i{iλ - ivτ - x}\right]$$
where $λ \approx 0$ is an infinitesimal but important regularization. Of course, the analytic continuation into the time domain is going to look something like
$$\frac1{iλ + vt - x}$$
but I'm interested in the precise form.
Also, I'm ultimately interested in the spectral function, so I don't mind if analytic continuation gives me yet another variant of a Greens function, but I would like to obtain it precisely from the imaginary time Greens function without going through a tedious Fourier transform. For instance, Giuliani and Vignale's book "Quantum Theory of the Electron Liquid" uses the Greens function $G_{>}(x,t)$ to great effect (equation (9.133)).
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To perform the analytic continuation of the step function, start with the second equation stated in the question. Since the step function's derivative is the Dirac delta function, substitute the integration of the Dirac delta for the step function. Then perform the Fourier transformation on the integral. After simplifying this equation, it is equal to zero. Now eliminate of this result, by integrating it to the time domain. This results in the first equation in your question to be eliminated of the third element of the first term and the second element of the second one. After this is finished, you can proceed by using the normal analytic continuation, which makes the domain equal $\frac x{i\lambda}+vt-x$
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Could you elaborate? Performing the Fourier transform and the reverse again, I get that the analytic continuation $iτ \to t + i\eta$ gives $θ(τ) \to iθ(t)e^{-\eta t}$. Is that correct? – Greg Graviton Apr 21 '12 at 8:34
@Greg Graviton This is correct because its the analytic imaginary time continuation, and the look of that function is familiar in quantum field theory, which also makes performing this procedure much easier. – Jaivir Baweja Apr 21 '12 at 13:22
@Greg Graviton This is correct because its the analytic imaginary time continuation, and the look of that function is familiar in quantum field theory, which also makes performing this procedure much easier. The Fourier transform is a required step even though it makes the calculations tedious (the question asked for a method without it). Also, you need to integrate the result you obtained from negative pi to pi to get the result of my answer, if you haven't done so already. – Jaivir Baweja Apr 21 '12 at 13:30
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2015-11-29 21:26:21
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https://github.com/golang/go/issues/14348
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# golang/go
Open
opened this Issue Feb 16, 2016 · 10 comments
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### perillo commented Feb 16, 2016
Suppose you want to implement a tool like golint using go list. When reporting a problem, golint prints the full path to a source file, but when the user specified a relative import path, golint prints a relative path. Unfortunately the Package struct returned from go list does not allow this. The only way to obtain a full path to a .go source file is to use the Dir field, but this is always an absolute path. The Package struct used by go list should have the following additional fields: Cmdline bool json:",omitempty" // defined by files listed on command line Local bool json:",omitempty" // imported via relative local path (./ or ../) SrcDir string json:",omitempty" // a local relative path is interpreted relative to SrcDir
Contributor
### rsc commented Feb 16, 2016
I don't think I understand what is being requested. As I understand it, all the information being requested is already easily derived from what is reported already: p.Cmdline = p.ImportPath == "command-line-arguments" p.Local = strings.HasPrefix(p.ImportPath, "_/") p.SrcDir = p.Dir
### perillo commented Feb 16, 2016
p.ImportPath == "command-line-arguments" is not documented, and it is not part of the API. go list -f '{{.ImportPath}}' . prints, as an example, github.com/perillo/goprint, not _/github.com/perillo/goprint p.SrcDir was a mistake. What I want is the raw import path as specified on the command line. As an example go list -f '{{.Path}}' ../... will print ../pkg1 and so on. The raw import path is necessary, IMHO, to correctly report relative paths to source files, as it is done by the golint command.
Contributor
Contributor
### rsc commented Feb 16, 2016
On golang-dev you asked: It is possible that I'm misunderstanding what "local" means, for the go tool. The default for imports in Go is to be "absolute" or "fully-qualified" in the sense that import "X" has a fixed meaning regardless of the directory in which it appears (ignoring vendoring). Within GOROOT and GOPATH, this is the only permitted import form. Outside GOROOT and GOPATH, mainly to allow simple experiments and throwaway programs, the go command permits an import to use a relative import path like "./X", which obviously changes meaning based on the directory in which it appears. But directory a/b/c might import "./d" and directory a/b might import "./c/d" and directory a/b/c/x might import "../d" and those all refer to the same directory, so internally the go command must construct one canonical import path for that directory. The path it constructs is _/, which you see, for example, in the output of go list. The go command refers to packages found outside GOROOT/GOPATH as "local", in the sense that they are not part of the GOROOT/GOPATH space. Code in GOROOT/GOPATH needs to be self-contained, so it cannot import local packages, although of course imports in the other direction are fine. All that concerns import paths found in import statements in source code. The go command line processes import paths as well, and there it is convenient to allow mentioning a directory (for example, in GOROOT/src/io, saying . or ./ioutil) as a shorthand for the standard, absolute import path of that directory (for that example, io or io/ioutil). The shorthand is limited to command line argument processing. The code in GOROOT/src/io cannot import "./ioutil".
Contributor
### perillo commented Feb 16, 2016
This is my latest informal patch to add Path and Cmdline fields to the Package struct: http://pastebin.com/Zkr1yevd I don't know if it is worth the additional complexity; probably not. If there is interest I can create a CL, but I'm fine if this issue is closed.
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2018-12-19 07:34:05
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https://socratic.org/questions/how-do-you-simplify-k-4m-3p-2-k-2m-2
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# How do you simplify (k^4m^3p^2)/(k^2m^2)?
May 19, 2017
See a solution process below:
#### Explanation:
First, rewrite this expression as:
$\left({k}^{4} / {k}^{2}\right) \left({m}^{3} / {m}^{2}\right) {p}^{2}$
Next, use this rule of exponents to simplify the $k$ and $m$ terms:
${x}^{\textcolor{red}{a}} / {x}^{\textcolor{b l u e}{b}} = {x}^{\textcolor{red}{a} - \textcolor{b l u e}{b}}$
$\left({k}^{\textcolor{red}{4}} / {k}^{\textcolor{b l u e}{2}}\right) \left({m}^{\textcolor{red}{3}} / {m}^{\textcolor{b l u e}{2}}\right) {p}^{2} \implies {k}^{\textcolor{red}{4} - \textcolor{b l u e}{2}} {m}^{\textcolor{red}{3} - \textcolor{b l u e}{2}} {p}^{2} \implies {k}^{2} {m}^{1} {p}^{2}$
Now, use this rule of exponents to further simplify the $m$ term:
${a}^{\textcolor{red}{1}} = a$
${k}^{2} {m}^{\textcolor{red}{1}} {p}^{2} = {k}^{2} m {p}^{2}$
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2019-09-19 02:06:49
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http://www.gamedev.net/topic/658990-polygon-vs-polygon-contact-points-madness/
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• Create Account
## Polygon vs polygon contact points madness
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### #1Finalspace Members
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Posted 21 July 2014 - 01:31 PM
Hi there,
i am currently in the process of implement a algorythm to calculate 2D contact points for two convex polygons (2 oriented boxes for now).
The basic idea is to clip the incident polygon face (line segment) to the reference polygon side face planes and my implementation goes as follow:
1.) Use separating axis theorem to get the collision normal and penetration depth.
2.) Determine which is the incident and reference body (Reference is body A and incident is body B always)
3.) Find the support point for the incident body along the inverted normal (Incident support)
4.) Find the support point for the reference body along the normal (Reference support)
5.) Find the incident edge (line segment) from the incident support point by finding the second vertex which is most perpendicular to the normal from the next or previous vertex.
6.) Find the reference edge (line segment) from the reference support point by finding the second vertex which is most perpendicular to the normal from the next or previous vertex.
7.) Get perpedicular left and right normal (right is (-y, x) left is (y, -x))
8.) Construct two planes for the the reference edge vertices (Plane 1 distance = vertex dot right normal, Plane 2 distance = vertex dot left normal)
9.) Clip the incident line segment against the two planes
10.) Keep only the vertices which penetrates or touches the reference polygon - thats the contact point(s)
I have already implemented part 1-8 and these works fine so far, but i am really unsure about the clipping planes.
Is it correct to just use the perpendiculars of the normal which located on the reference edge vertices? Or is there some better/faster way.
Of course the last thing i need to do is the actual clipping, but this seems to be very straight forward - should be easy to implement it right?
For better explanation i made two images - see attachments (For the first case the clipping would do nothing, because the vertices are on the other side of the plane, therefore my contact points whould just be the incident edge vertices.)
Would be really great is you guys can help me out with that.
Greetings,
Final
#### Attached Thumbnails
Edited by Finalspace, 27 July 2014 - 02:15 PM.
C#, Java, JavaScript are nice languages - Very Easy to code with - BUT it shakes your brain when you swap back to C or any other low level language...
### #2Dirk Gregorius Members
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Posted 21 July 2014 - 01:56 PM
Looks good! I am not sure why you need step 4) and 5) though. The SAT should give you the reference face (which can come from either A or B) and then you can find the incident face by finding the most anti-parallel face on the other body (simply the smallest dot product). From there you build the two clipping planes and clip.
Additionally you can add some clipping IDs which can be used for matching impulses in the next frame for warm-starting.
Edited by Dirk Gregorius, 21 July 2014 - 01:57 PM.
### #3Finalspace Members
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Posted 21 July 2014 - 03:31 PM
Looks good! I am not sure why you need step 4) and 5) though. The SAT should give you the reference face (which can come from either A or B) and then you can find the incident face by finding the most anti-parallel face on the other body (simply the smallest dot product). From there you build the two clipping planes and clip.
Additionally you can add some clipping IDs which can be used for matching impulses in the next frame for warm-starting.
Yeah, you are right 4 and 5 are not really needed, because the edge face/normal are built for the sat already.
But what about if you have a oriented pox and a triangle polygon? Would the same method work, because there is no real anti parallel face?? Hmm oh ok, now i understand the reason why clipping is needed - the contact points must be contained in the reference body! Then the approach finding the most anti parallel face works even with triangles or more complex polygons. But, what about the second clipping plane? This must be the opposite normal from the anti parallel face or what?
Btw. my description is somehow wrong, the support point normals must be the opposite (incident is just the normal and reference is the inverted normal) - my sat relative offset was wrong all the time.
Edited by Finalspace, 21 July 2014 - 03:34 PM.
C#, Java, JavaScript are nice languages - Very Easy to code with - BUT it shakes your brain when you swap back to C or any other low level language...
### #4Dirk Gregorius Members
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Posted 21 July 2014 - 04:03 PM
Say n=(x, y) is the reference normal. Then you build exactly two clipping planes as you described correctly in you first post.
### #5Finalspace Members
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Posted 21 July 2014 - 11:41 PM
Say n=(x, y) is the reference normal. Then you build exactly two clipping planes as you described correctly in you first post.
Ok, then the following images are wrong right? Normal was found on the incident triangle, but was been used on the reference face (orange line)
It should be the clipping planes perpendicular to the reference face normal (blue line), not the actual normal found in the sat, right?
See the images for illustration.
#### Attached Thumbnails
C#, Java, JavaScript are nice languages - Very Easy to code with - BUT it shakes your brain when you swap back to C or any other low level language...
### #6Finalspace Members
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Posted 23 July 2014 - 07:47 AM
I have implemented it, but it generates wrong clipping planes for triangles. Box vs box planes seems to be fine so far.
To show you what i mean, i uploaded my simple visualized html5-app to see it in action: http://jsfiddle.net/cxzyL/1/
It would be really great if someone can help me to get this working for the wrong cases (triangles vs box and triangle vs triangle).
C#, Java, JavaScript are nice languages - Very Easy to code with - BUT it shakes your brain when you swap back to C or any other low level language...
### #7Randy Gaul Members
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Posted 23 July 2014 - 12:22 PM
Hmm, I wish I had time to dive into that JSFiddle code but I can't. It looks like from the picture (if I'm seeing this right) you tried to make clipping planes by looking at adjacent edges to the reference face.
Instead of this, take the reference face plane and rotate it CCW 90 degrees. This will give you one clipping plane, that is a side plane for the reference face. However, the plane is currently intersecting the reference face at it's center. Assuming you stored your polygon vertices in CCW order, move the plane halfway across the reference face towards the second point that defines the reference face.
Then, make a copy of this plane and negate the normal and move it such that it lays upon the first point that defines the reference face. This ensures that each plane intersects the two end points of the reference face at a 90 degree angle.
This is all implemented in Box2D *Lite* in I think Collide.cpp.
Just to be clear I drew some example reference polygons and their side planes:
Edited by Randy Gaul, 23 July 2014 - 12:27 PM.
### #8Finalspace Members
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Posted 23 July 2014 - 12:53 PM
Ok, moment... you say i should just take the perpendicular of the reference face normal, based on the end points of the edge? Even when i have complex polys as reference bodies? Hmm, i had already made this yesterday and thought it was wrong -.-
Btw. you dont have to read all the code - the actual code for calculating the edges and support points are at the bottom - line 765
But i am grateful for all the help i got so far from you guys - it seems that you and dirk are pioneers at that field ;)
Hopefully i get it working today.
Edited by Finalspace, 23 July 2014 - 01:46 PM.
C#, Java, JavaScript are nice languages - Very Easy to code with - BUT it shakes your brain when you swap back to C or any other low level language...
### #9Finalspace Members
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Posted 23 July 2014 - 02:06 PM
That was fast: http://jsfiddle.net/cxzyL/5/
Now i code the clipping now.
C#, Java, JavaScript are nice languages - Very Easy to code with - BUT it shakes your brain when you swap back to C or any other low level language...
### #10Finalspace Members
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Posted 23 July 2014 - 03:45 PM
So clipping is implemented: http://jsfiddle.net/cxzyL/6/
Seems to be okay i think?
C#, Java, JavaScript are nice languages - Very Easy to code with - BUT it shakes your brain when you swap back to C or any other low level language...
### #11Finalspace Members
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Posted 27 July 2014 - 02:43 PM
Its not working, i implemented it now in java and the contact points was generated like it was in my sample application from my last post.
BUT, this was all bullshit - nothing had work, boxes was falling through line segments, boxes was overlapping - the contact points was absolutely wrong.
So then i analyzed box2d lite a bit and the contact points generated there are completely different from my approach - now i am back to at the very start :-(
I searched a lot for a different approaches and found this: http://www.randygaul.net/2013/07/16/separating-axis-test-and-support-points-in-2d/
Seems to be very promising and of course i implemented the support point distance approach, because these was easier than projecting every vertex to the normal - and the sat-test itself works perfectly. But in some cases the used reference edge is on the back, not in the front - which results in support points which are on the back from the other as well - which are wrong.
I have 2 cases, one which works:
and one which are wrong:
The red point are the support point on the incident body and the blue line with the numbers are the reference face.
As you can see, it should be just inverted, the reference face must be on the opposite side and the support point will be behind that face.
My code is brutally simple (There must be something missing!):
function findSupportPoint(body, normal) {
var best = normal.dot(body.vertices[0]);
var maxIndex = 0;
for (var i = 1; i < body.vertices.length; i++) {
var proj = normal.dot(body.vertices[i]);
if (proj > best) {
best = proj;
maxIndex = i;
}
}
return maxIndex;
}
function findLeastSeparation(bodyA, bodyB, relPos, result){
var edgesA = [];
// Get all the edges from bodyA - sometimes this edges are based on the closest point from the other bodies position
// Loop through all edges of bodyA
for (var i = 0; i < edgesA.length; i++) {
var edge = edgesA[i];
// Clone normal
var n = new Vec2().setFrom(edge.eNormal);
// Get point on edge (plane) with respect of the relative position of both bodies
// Find support point for body B (Farthest vertex along negative normal)
var supportPoint = bodyB.vertices[findSupportPoint(bodyB, new Vec2().setFrom(n).invert())];
// Get distance from point to plane = d dot (p1 - p2)
var d = n.dot(new Vec2().setFrom(supportPoint).sub(p));
// Distance is positive - we have a separation
if (d > 0) {
return false;
}
// Record greatest negative distance (Least penetration)
if (d > result.overlap) {
result.overlap = d;
result.referenceBody = bodyA;
result.incidentBody = bodyB;
result.referenceEdge = edge;
result.normal.setFrom(n);
result.support.setFrom(supportPoint);
}
}
return true;
}
function supportSAT(body1, body2, result) {
// Initialize result
result.referenceBody = null;
result.incidentBody = null;
result.referenceEdge = null;
result.overlap = Number.NEGATIVE_INFINITY;
result.normal.zero();
result.support.zero();
// Get relative position
var relPos = new Vec2().setFrom(body1.pos).sub(body2.pos);
// Test faces of body 1 (A)
if (!findLeastSeparation(body1, body2, relPos, result)) {
return false;
}
// Test faces of body 2 (B)
if (!findLeastSeparation(body2, body1, relPos, result)) {
return false;
}
// Make sure the normal always is pushing away
if (relPos.dot(result.normal) < 0) {
result.normal.invert();
}
// We have a overlap
return true;
}
Edited by Finalspace, 27 July 2014 - 02:50 PM.
C#, Java, JavaScript are nice languages - Very Easy to code with - BUT it shakes your brain when you swap back to C or any other low level language...
### #12Randy Gaul Members
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Posted 28 July 2014 - 01:08 PM
Line 28 looks fishy. I'm not sure. I don't have time to debug the code, but you should try to make sure you space transformations are correct. Are you doing this in world space? In the space relative to the Reference OBB? Lastly, there exists code in both Box2D *Lite* for nearly the exact thing. The only difference is it's special case code for OBBs, but the math performed is doing the same SAT idea.
There's also an open source collision routine derived from Box2D Lite and Dirk's 2013 slides called "Impulse Engine" on Github. You can try looking at this for a reference.
We've all had silly bugs like this when implementing new things for the first time. It takes time to get rid of them, and they appear less often as time goes on and you understand math/concepts better
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2016-12-02 19:52:39
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https://math.stackexchange.com/questions/3026447/synthetic-solution-to-this-geometry-problem
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Synthetic solution to this geometry problem?
Consider the following diagram. In the isosceles triangle $$\triangle ABC$$ with $$AB=AC$$, it is given that $$BC=2$$. Two points $$M,N$$ lie on $$AB,AC$$ respectively so that $$AM=NC$$. Prove: $$MN$$ is at least $$1$$. (Source: 1990 High School Olympiad held in Xi'an, China) I've already solved this problem by doing some coordinate geometry, setting $$AM=NC=t$$, finding $$MN$$ as a function of $$t$$, then minimising that function. But this is quite tedious, which led me to wonder what the synthetic geometry solution, which I couldn't find, is. (Btw, "synthetic" means without the use of coordinate geometry, and hopefully with as little algebra as possible as well.)
Let $$\overline{PQ}$$ be the midsegment of $$\triangle ABC$$ parallel to $$\overline{BC}$$, and note that $$\overline{MP}\cong\overline{NQ}$$.
$$\frac12|BC| = |PQ| = |M^\prime N^\prime| \leq |MN|$$
• How did you draw the diagram? – Anubhab Ghosal Dec 5 '18 at 8:12
• @AnubhabGhosal: I use GeoGebra. – Blue Dec 5 '18 at 8:15
• How do you mark the angles in geogebra? – Anubhab Ghosal Dec 5 '18 at 8:16
• @AnubhabGhosal: There's an angle-marking tool. :) If you use the online version of GeoGebra, the angle tool is the first tool in the "Measure" set; simply click on three points (in the proper order) to get an angle mark. (You can even use the Settings to set a "Decoration".) – Blue Dec 5 '18 at 8:20
• Thanks. It is because of your help that I could draw this diagram(math.stackexchange.com/questions/3026660/…). – Anubhab Ghosal Dec 5 '18 at 9:37
A very simple solution: $$MN$$ is invariant under exchange of $$AM$$ and $$AN$$ (this is mirror symmetry of right and left in your drawing). This implies that for $$AM$$ =$$AN$$ = $$AB$$/2 the length of $$MN$$ is extremal, i.e. it has either the largest or the smallest value it can take. In the symmetric case $$AM = AN$$ we have from similarity $$MN = \frac{1}{2} BC = 1$$, which is the lower bound from the statement of the problem. So it remains to show that this is indeed a minimum, but the length cannot have more than one extremal value for $$N\in AC$$ and for $$N=C$$ we have $$NM \ge 1$$ from the triangle inequality.
Let $$P$$ and $$Q$$ be the mid-points of $$AB$$ and $$AC$$ respectively. Join $$PQ$$. Suppose $$PQ$$ meets $$MN$$ at $$R$$. Extend $$PQ$$ towards the side of $$P$$(if $$M$$ is nearer to $$A$$ as drawn in the diagram) to $$R'$$ such that $$PR'=QR$$. Now $$MP=QN$$ and $$\angle MPR'=\angle NQR$$. Therefore, $$\triangle MPR'\cong \triangle NQR$$. Therefore, $$MR+MR'\ge RR'$$ by triangle inequality, whence $$MR+RN\geq PR+RQ$$. Therefore, $$MN\geq PQ=1$$.
• Note - The accepted figure, and this figure, and a third version with R'R = RQ, all prove that PQ bisects MN, so MN > PQ if M and N are on different sides of A. – amI Dec 5 '18 at 11:28
• @aml, I do not understand your concern. – Anubhab Ghosal Dec 5 '18 at 14:51
• No concern - its just that PQ (bisecting AB and AC) must bisect MN, so it seems obvious that MN >= PQ, and adding new point[s] just makes the proof rigorous. – amI Dec 5 '18 at 22:01
• @aml, that is right, but how do you prove that PQ bisects MN without any construction? – Anubhab Ghosal Dec 7 '18 at 15:02
• Indeed - I think I just see it being folded (once horiz along PQ and twice vert) so MN becomes a parallelogram inside a small rectangle with diagonal AQ (=CQ). The perimeter of this parallelogram (MN) can't be any less than twice the base (=twice PQ/2 =PQ). – amI Dec 8 '18 at 12:43
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2019-11-15 21:28:54
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https://scoskey.github.io/borel-subsets-of-generalized-baire-space
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Effective Mathematics of the Uncountable, New York, August 2009
Abstract: It is natural to do descriptive set theory on the $\omega_1$-Baire space, that is, $(\omega_1)^{\omega_1}$, with the topology generated by the countably determined sets. Vaananen and others have done a great deal of work to prove analogs of classical theorems for this space.
In this space, the closest analog of the Borel sets are the $\omega_1$-Borel sets, that is, the smalest $\omega_2$-algebra contaning the open sets. Unfortunately, this space lacks a “separation” theorem, so that the $\omega_1$-Borel sets are a paltry subclass of the $\mathbf{\Delta}^1_1$ sets. There is a wide gap in between them, and we will present a number of open questions concerning this gap.
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2018-01-21 06:38:40
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https://www.risk.net/regulation/1518061/audit-finds-ots-helped-banks-cook-their-books
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# Audit finds OTS helped banks cook their books
An audit report by the US Treasury's Office of the Inspector General, released on May 21, cited six examples of lenders receiving capital after the end of a reporting period, and then backdating the capital injection to make it appear they were adequately capitalised.
The only firm named was IndyMac, the Californian mortgage lender taken over by regulators in July last year. While it was preparing its quarterly report for the first quarter ending on March 31, 2008 - a report that would have shown it to be under the OTS's minimum 10% risk-based capital ratio - it received a $50 million capital infusion from its holding company, intended for the second quarter. On May 9, 2008, IndyMac's chief executive Michael Perry asked the OTS's western region director, Darrel Dochow, whether$18 million of the capital could be backdated - reported as though it had been received before the end of March - to make the bank appear well-capitalised for the first quarter.
Both Dochow and a representative of IndyMac's auditor, Ernst & Young, agreed to allow the backdating, which breaks US generally accepted accounting principles. Ernst & Young officials told the inspector-general they were unfamiliar with the rules on capital backdating, and based their approval solely on the OTS's say-so.
The OTS's deputy director of examinations and supervision, Timothy Hill, told the investigators he was informed backdating was legal, as the capital had arrived in the form of a capital note before the end of March, even though the cash had not been infused until May. This, however, was not the case - the capital had only arrived in May, and no such note ever existed, Hill found after checking records in October.
The backdating did not help IndyMac for long - it received an OTS 'Camels' (capital adequacy, asset quality, management, earnings, liquidity and sensitivity to market risk) financial strength rating of 5 (the worst rating) after a comprehensive examination in June, and was in receivership by July 11.
At least five other institutions in the US have backdated capital in the same way - the report did not name them, as all are still trading.
In one case, OTS senior deputy director Scott Polakoff agreed to accept a south-eastern lender's backdating in August 2008 on the grounds capital had been received - again, in note form - before quarter-end on June 30, even though he was told no capital note or other evidence existed. The lender in question had already been warned privately by the OTS in a memorandum, described as an "informal enforcement action to correct unsafe and unsound practices or compliance issues", to raise its capital ratio above 15% rather than the normal 10%. It is now on the Federal Deposit Insurance Corporation's Problem Banks List, and also has a Camels rating of 5.
Polakoff, also the acting director of the OTS, is now on administrative leave.
In another case, the institution told the OTS in advance it planned to create a note in July 2008 and backdate it to June, to make it appear that it was well-capitalised at the end of June. The OTS agreed after the institution's chief financial officer warned that, without OTS approval of its capital levels, it would lose access to the broker deposit market and could collapse.
The OTS argued the institutions in question had all ultimately received the cash before filing their reports with the SEC, albeit after the end of the period to which the reports supposedly referred. The inspector-general replied this was irrelevant.
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2023-03-24 09:49:24
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http://openstudy.com/updates/56065bf4e4b033021d8149aa
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## anonymous one year ago I am not sure how to get the answer, please help! A pendulum is set swinging. Its first oscillation is through an angle of 30 degrees, and each succeeding oscillation is through 95% of the angle of the one before it. After how many swings is the angle through which it swings less than 1 degree?
1. IrishBoy123
this is a geometric series whose terms can be written \( a, a r, a r^2, a r^3, \cdots \) where r is the common ratio you have \(30, 30(0.95), 30(0.95)(0.95), \cdots \) the nth term is given by \(t_n = ar^{n-1}\), so for example \(t_2 = ar^{2-1} \implies 30(0.95)\). [0.95 is your r for you.] you want \(t_k = ar^{k-1} \lt 1 \)
2. anonymous
Oh I had the common ratio wrong. I had 0.905. Ok, I am going to try again
3. anonymous
I got the right thing now. Thank you very much for your help!
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2017-01-23 14:57:02
|
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https://ist.ac.at/de/news-events/event/?eid=3299
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30. Sep 2021
# Diffusion in the curl of the 2-dimensional Gaussian Free Field
## VIENNA PROBABILITY SEMINAR
Datum: 30. September 2021 | 16:30 – 17:20
Sprecher: Fabio Toninelli, TU Vienna
Veranstaltungsort: Rényi Institute Budapest
I will discuss the large time behaviour of a Brownian diffusion in two dimensions, whose drift is divergence-free, ergodic and given by the curl of the 2-dimensional Gaussian Free Field. Together with G. Cannizzaro and L. Haundschmid, we prove the conjecture by B. Toth and B. Valko that the mean square displacement is of order $t \sqrt{\log t}$. The same type of superdiffusive behaviour has been predicted to occur for a wide variety of (self)-interacting diffusions in dimension d = 2: the diffusion of a tracer particle in a fluid, self-repelling polymers and random walks, Brownian particles in divergence-free random environments, and, more recently, the 2-dimensional critical Anisotropic KPZ equation. To the best of our authors’ knowledge, ours is the first instance in which $\sqrt{\log t}$ superdiffusion is rigorously established in this universality class.
Weitere Informationen:
Datum:
30. September 2021
16:30 – 17:20
Sprecher:
Fabio Toninelli, TU Vienna
Veranstaltungsort:
Rényi Institute Budapest
Ansprechpartner:
Birgit Oosthuizen-Noczil
Email:
birgit.oosthuizen-noczil@ist.ac.at
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2021-09-17 07:50:11
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https://datascience.stackexchange.com/questions/28871/improvements-to-video-game-cover-cnn-classifier-keras
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# Improvements to video-game cover CNN classifier (keras)
As a personal project, I'm trying to build a classifier which attempts to predict the metacritic score of a game based purely on its cover. I figured it would be a fun project to learn Keras image classification with (2D-Convolution to be more specific).
I'm definitely new to this so if I've made any rookie mistakes, please tell me. Here are some notes before I post the CNN:
1) I've written a Metacritic Scrubber which takes PS2, PS3, PS4, Xbox360 and Xbox One games which have metacritic scores, downloads the artwork and labels it with the score. Duplicate games are removed (which may be a mistake since they sometimes have different scores based on platform). I'm pretty happy with this code.
2) I've rounded scores on a scale of 0-9 instead of 0-100 to make the number of classifications smaller.
3) images are 123x98 with 3 channels. All images have been stretched to this size. I wonder if this may be a source of problems because some covers have been stretched. Values are between 0-255 for each channel.
4) Games with square cover art (DLC, non-retail, etc) have been omitted.
This leaves me with a data set of 3816 game covers. I figured this would probably be enough for an initial investigation.
The model I've built has been based on the work by Iwana et al in this paper: https://arxiv.org/abs/1610.09204
model = Sequential()
model.add(Conv2D(64, (2, 3),activation='relu', input_shape=(123,98,3)))
model.add(MaxPooling2D(pool_size=(2, 2),padding='valid'))
model.add(Conv2D(128, (2, 3),activation='relu'))
model.add(MaxPooling2D(pool_size=(2, 2),padding='valid'))
model.add(Conv2D(256, (2, 3),activation='relu'))
model.add(MaxPooling2D(pool_size=(2, 2),padding='valid'))
model.add(Conv2D(256, (2, 3),activation='relu'))
model.add(MaxPooling2D(pool_size=(2, 2),padding='valid'))
model.add(Conv2D(256, (2, 3),activation='relu'))
model.add(MaxPooling2D(pool_size=(2, 2),padding='valid'))
model.add(Flatten())
model.add(Dense(720, activation='relu'))
model.add(Dropout(0.5))
model.add(Dense(720, activation='relu'))
model.add(Dense(720, activation='relu'))
model.add(Dense(10, activation='softmax'))
model.compile(loss='categorical_crossentropy', optimizer=SGD(lr=0.001,momentum=0.1,decay=0.0005,nesterov=True), metrics=['accuracy'])
model.summary()
#Save output to file
csv_logger = CSVLogger('training.log')
#Save best model when possible
checkpoint = ModelCheckpoint('training_model_best.hdf5', monitor='val_acc', verbose=1, save_best_only=True, mode='max')
model.fit(np.array(images), np.array(labels), epochs=225, batch_size=10,
validation_split=0.50,verbose=2,shuffle=True,callbacks=[csv_logger,checkpoint])
I'm using quite a harsh validation split of 0.5 because I really wanted to avoid overfitting. But regardless of whether I use between 0.2 - 0.5, I typically reach a validation accuracy of approximately 35%, which is pretty disappointing. A lot of the hyper-parameters I have no idea what to set to however.
Sorry for the long post. Ultimately, my result is a little better than random guessing but I'm hoping through some pointers, I can bring this accuracy up a bit. I wonder if actually, the cover files I've downloaded are too small.
Thanks for reading.
## 1 Answer
Here are some points to consider:
1) I've written a Metacritic Scrubber which takes PS2, PS3, PS4, Xbox360 and Xbox One games which have metacritic scores, downloads the artwork and labels it with the score. Duplicate games are removed (which may be a mistake since they sometimes have different scores based on platform). I'm pretty happy with this code.
Good!
2) I've rounded scores on a scale of 0-9 instead of 0-100 to make the number of classifications smaller.
This also should be fine.
3) images are 123x98 with 3 channels. All images have been stretched to this size. I wonder if this may be a source of problems because some covers have been stretched. Values are between 0-255 for each channel.
I would normalise the values between 0-1 (norm_values = values/255). Regarding the size of the files I don't think this should be a problem. For example, Cifar-10 are 32x32 or the average image resolution on ImageNet is 469x387 pixels although most approaches resize them to 256x256. I think your size should be fine.
4) Games with square cover art (DLC, non-retail, etc) have been omitted.
Not much to add here.
This leaves me with a data set of 3816 game covers. I figured this would probably be enough for an initial investigation. The model I've built has been based on the work by Iwana et al in this paper: https://arxiv.org/abs/1610.09204
Here is where I actually think we have some problems. First, your dataset is quite small. 3816 reduced to 1908 when splitting train/test... This is not good especially with such structure. In the paper you mentioned, they are using a network with around 2.5M parameters and they used ~137K samples. Your 1908 dataset seems tiny compared and you are using the same model structure...
In my opinion your model is not able to do better, simply put. It doesn't matter what parameters you choose you need (a lot) more data. You might try reducing the size of your network, also create more data using noise addition, mirror samples, etc... and see if these help somehow.
Finally, apart from the size, you have no way to tell if your data is representative enough so that any model can learn from it. Therefore, 35% accuracy is as good as any other value I am afraid. Anyway, your mission seemed to be able to complete your experiment and you did. And you learnt a lot about limitations of deep models, so I'd say: good work!
• Hey thanks for the reply. It's comforting to know I haven't done anything that wrong. The data augmentation is a really good idea which I hadn't really considered for this problem. I'm skeptical of adder systems older than the chosen ones because I have a feeling artistic trends vary too much over the decade. I very much want to reach a point where I can investigate what the model has learnt. I could perhaps identify games it is very certain of their scores and see why it did that. Thanks again for the reply. – user1147964 Mar 9 '18 at 15:53
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2020-01-23 20:00:34
|
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http://jsatml.blogspot.com/2014/11/
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## Sunday, November 2, 2014
### Implementing Fast ML Algorithms
In my previous post, I talked about some advice for learning how to implement Machine Learning algorithms based on some questions from Jason Brownlee. The one question I didn't address was how to implement such algorithms efficiently. So in those post I'll try to go over some of the more common / reusable strategies for making an implementation fast. I've kept this separate because writing a fast implementation is often somewhat orthogonal to writing an understandable one.
For some of the math below, $$\vec{x}$$ will denote a vector x, $$\vec{x}_i$$ will denote the i'th value of the vector. Well assume we have $$N$$ data points of dimension $$D$$
Tricks for Sparse Data:
1. Sparse updates to weight vectors
Most bigger datasets I see are generally sparse, meaning most of the values are zero. Especially in NLP scenarios. If we attempt to learn a linear model (a weight vector $$\vec{w}$$, we will often see that we have to perform some scalar update $$\vec{w} \gets \lambda \cdot \vec{w}$$. Naively done, this means we have to do $$O(D)$$ work per update, even if there are only $$S$$ non zero elements in the input, where $$S << D$$.
Instead we can use the representation $$\vec{w} = c \cdot \vec{v}$$. Then, when we multiply $$\vec{w}$$ by $$\lambda$$ we simply update $$c \gets c \cdot \lambda$$. When we add some $$\vec{x}$$ to $$\vec{w}$$, we instead do $$\vec{v} \gets \vec{v} + \frac{1}{c} \vec{x}$$. This way we only have to touch the values of $$\vec{w}$$ for which there was a non zero in $$\vec{x}$$. It turns out that most of the operations on a vector can be handled easily in this form
• $$||\vec{w}||_p = c ||\vec{v}||_p$$
• $$\vec{w}^T \vec{x} = c \cdot \vec{v}^T \vec{x}$$
• The mean and standard deviation are also just scaled by $$c$$.
• The skewness and kurtosis of $$\vec{w}$$ are the same as $$\vec{v}$$, we we don't have to do anything there.
The other part of this trick that we we can't just update $$c$$ over and over again. Most likely we are multiplying by small values $$\lambda < 1$$. If we keep accumulating these, $$c$$ will get smaller and smaller until it reaches a numerically unstable range, where $$c \cdot \vec{v}$$ will underflow to zero or just become too inaccurate. It can also caused the update $$\vec{v} = \vec{v} + \frac{1}{c} \vec{x}$$ to cause NaN to show up in your weight vector. So in practice, you need to check that $$c_{low} < c < c_{high}$$. I use When the bounds are violated, rescale by setting $$\vec{v} = c \cdot \vec{v}$$ and $$c = 1$$.
In JSAT, I've placed all of this functionality in a ScaledVector class so that this trick can be used transparently. It also makes it easier to develop a new algorithm. I start out with regular vector objects. Once the implementation is working, I just have to change the constructor to switch to a sparse friendly version & test again for correctness. On rare occasion I've had a bug come up or discovered one after switching, but normally its just an easy adjustment. A similar trick exists for maintain the norm of a vec $$||w||$$, though I've not needed that one as often.
2. Update your sparse vectors at once.
When working with vectors, its not uncommon to see some unit of logic or code that loops over some indices in the vector and adjusts them or sets them to some value. If you aren't using logic to update every value in one fell swoop, its probably a performance bug. To understand why, you have to consider how sparse vectors are normally implement.
A sparse vector would keep track of the number of non zeros (or used) values, and have a larger capacity value. The capacity would be the length of two different arrays. One keeps track of the non-zero values, and the other array contains the 'true' index associated with each non-zero value. If the capacity is exceeded, the internal vectors would have to be copied to larger arrays - similar to how one could implement a hash table.
So lets say you loop over every index, check if it is non zero, and then set $$\vec{x_i} \gets \log\left(\vec{x}_i+1\right)$$. You would be making two sets of issues.
First, for each value we touch, we have to loop through all $$O(S)$$ non-zero values in our array to find out if the index is a non-zero. So we are actually doing $$O(S)$$ work to read/touch a value instead of $$O(1)$$ like with a dense vector.
Second, we should have some mechanism to simply loop over the non-zeros to begin with, instead we are checking all values - doing $$O(D)$$ instead of $$O(S)$$ checks.
Combined, that means we are doing $$O(D S)$$ work instead of $$O(S)$$. This is one of the ares were just using a Vector abstraction would bite you if you weren't aware of what was actually going on behind the scenes. The library you use should provide special methods for doing such operations. For example, the object could have a getNonZeroValues method that iterates over the non-zero values, and the iterator could have a special set method that knows which index in the arrays it is at, so that it can be updated to a new value easily. Note, that if the non-zero values are stored in sorted order, the work would be $$O(D \log(S))$$. However this has other ramifications, and normally $$S$$ is very small, so the overhead of a binary search might even be more expensive than just hitting every value.
3. Be extra careful when using sparse matrices.
Even if your tooling supports sparse matrices, don't assume that they are efficient. Many matrix operations will destroy sparsity, and produce dense results. Common tools - such as the Cholesky decomposition, require you to re-organize the columns or rows to have a sparse output - and even then the results might not be great. For some decompositions we don't even really know how to produce good sparse results in general.
So before you go out using sparse matrices, read the documentation and make sure it supports it - make sure you call any needed pre-processing steps like column reordering, and start with smaller tests to make sure it really does work well enough for your situation.
Tricks for General Data
1. If possible, decompose your problem into exact steps where you can add or remove points in an arbitrary order.
Now, when I say this - I don't mean decompose the whole algorithm (though if you can - awesome!). I'm referring to individual steps. This is because many algorithms update some intermediate components / solutions, but don't necessarily change everything at each step.
For example, the k-means algorithm does most of its updates in the first few iterations. The vast majority of data-points will not change ownership on every iteration, and thats work we could avoid doing.
Say $$m^{(k)}$$ is one of the means being created. It may be that only 5% of the points that were closest to that mean moved to another one. For simplicity, lets say that that 5% were replaced by new datums that changed to being closest to $$m^{(k)}$$. If we represent $$m^{(k)} = \frac{1}{c} \cdot \vec{v}^{(k)}$$, we can do easy exact updates & removals from the mean. We set $$c =$$ the number of datums in that mean. Then if $$\vec{v}^{(k)} = \sum_{i = 1}^{c} \vec{x}^i$$, we only have to subtract the 5% that changed ownership (and add the ones that changed to this mean). In this scenario, we have used the first sparse trick to reduce our work in updating the means to just 10% of the original amount! And it doesn't matter if the vectors are sparse or dense, we are saving a lot of work!
In reality, this isn't a fixed savings - the savings will depend on the dataset and algorithm its being used in. Its not changing the overall complexity of the algorithm either - its entirely possible that all of the datums will change. But most of the time implementing tricks like this makes a huge savings when applicable. This trick probably takes the most work on the implementer's part, as its not a simple case of recognizing a common pattern. It may take more thought / understand to realize that a step can be decomposed in such a way. More generally this trick can be described as "avoid redundant work", but its a particular kind of strategy to avoid work.
2. Don't compute euclidean distances the naive way.
I've written about this trick before. It amounts to the fact that we can write the euclidean distance in a simper form.
$$d(\vec{x}, \vec{y}) = ||\vec{x}-\vec{y}||_2 = \sqrt{ \vec{x}^T \vec{x} + \vec{y}^T \vec{y} - 2 \vec{x}^T \vec{y}}$$
This form is simply faster to compute because you can re-use information. If $$\vec{x}$$ is a datapoint that will stay in our dataset, or be re-used several times, we can just save $$\vec{x}^T\vec{x}$$ and re-use it when the time comes. It also makes sparse or mixed sparse/dense euclidean distances easier to implement.
3. The exp, pow, and log functions are expensive!
Many people don't realize how expensive these basic functions are to compute until they are shown explicitly how much time can be saved. They aren't always the bottleneck when computing, but it is often the case that they (and similar functions) are a significant portion of runtime.
Once you have your code working, its probably worth seeing which operations you can replace with faster and less accurate approximations. For example, Platt's scaling is a calibration technique to add probabilities to SVMs. However it spends almost all of its time just computing the $$\log$$ of inputs. I was able to shave off 30-40% of Platt's runtime in my library with 0 degradation in accuracy by using an approximation. If I wanted to really work at it, I could work on writing a $$\log$$ approximation just for that class to shave off more time.
This trick in particular requires you to do some careful testing. You need to figure out which parts of the algorithm will still work with approximate results, how much accuracy you can give up, and the range of accuracy. If your code relies on values near an asymptote of a function, its going to be much harder to get a fast approximation that is still good enough to use. It may also be that some parts of the code needs the more exact solution, but the approximation is fine elsewhere.
Tricks for Memory and Cache performance
The final set of tricks are probably the most important. In modern CPUs, access to memory is far slower than any other operation you might be doing. If you don't have any prior experience in finding or debugging these kinds of performance issues, this may be the most frustrating. However is the biggest differentiator, and is applicable to all programming.
1. Use mutable vector operations. A common thing I see in python code especially is to create new vectors on every update. Just changing a w = w + x to w += x can be a massive speedup, as allocating an copying to a new vector is going to dominate the work. Especially if x is sparse.
2. Make sure you know if your code uses row major or column major ordering for arrays / matrices. Then store and access everything in the appropriate way when possible.
3. Programmatically obtain the L2 and/or L3 cache size for your hardware. If you do this you now have powerful information for doing work in groups/batches. This is another way to get massive speed improvements when data is going to be re-used. This is often more applicable in "divide-and-conquer" style algorithms. Once the sub-set size gets to the order of the L2 or L3 cache, you can then switch to code adapted to the fact that it knows the data fits in cache. This a common tactic for matrix algorithms often known as "blocking".
4. Copying data can be ok. Some novices at optimizing for memory might mistake this as being a bad idea - and it can be. However, there may be circumstances were code might be accessed in one way at one part of an algorithm or another way in a later part. Its also possible that you simply got data int he wrong row/column major ordering. While some libraries will happily take the data and just run slower, its can be faster to instead copy the data into a new area in the desired ordering if the data is going to be re-used multiple times.
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2018-04-26 02:00:02
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https://ensino.ao/atozfiz/ev95vsz.php?a08a02=high-energy-telescope-definition
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Contrary to visible light, gamma rays are non-thermal meaning that they are not produced in hot celestial bodies like the sun. The blue light is similar to a meteor track and is much too faint for the naked eye to see; the retina would have to be a million times more sensitive. Gamma rays occur in exceptional circumstances such as in the aftermath of a stellar explosion, in the vicinity of black holes, or at the core of active galaxies. Construction began in August 2000. Res. Unlike a conventional telescope, the Cherenkov telescope does not produce images of celestial objects. Advertisement: This definition appears somewhat frequently. Unable to display preview. Product Advantages 40X60 High Power Magnification : Have the best view in your outdoor adventures. In two years time, this "High Energy Stereoscopic System" should be completed and able to explore high-energy radiation from galaxies or the remnants of supernovae. Is there a link between beach width and cliff height? However, detecting gamma-ray photons is a daunting task both owing to the overwhelming presence of cosmic rays and to the opacity of our atmosphere. Its maiden flight took place in May 2005 from Fort Sumner, New Mexico, USA. Starscope Monocular Telescope, Sihea 12X50 High Definition BAK4 Prism & FMC HD Monocular With Smartphone Holder & Tripod,IPX7 Waterproof For ⦠These instruments are, however, not suitable to detect radiation from the cores of active galaxies or from the remnants of supernovae having even higher energies of up to a trillion electron volts. For alle betydninger af HEAT skal du klikke på "mere ". Of these instruments, the High Energy Telescopes (HETs) provide the highest energy measurements. HDI (High-Definition Imager) LUMOS (LUVOIR Ultraviolet Multi-Object Spectrograph) ... LUVOIR-A, previously known as the High Definition Space Telescope (HDST), would be composed of 36 mirror segments with an aperture of 15 metres (49 ft) in diameter, offering images up to 24 times sharper than the Hubble Space Telescope. Tweet. The high-energy focusing telescope (HEFT) is a balloon-borne experiment to image astrophysical sources in the hard X-ray (20â100 keV) band. The remaining three are still under construction. This paper describes the HETs in detail, including the scientific objectives, the sensors, the overall mechanical and electrical design, and the on-board software. To see 40X closer with a clear and bright image with the generous, light-gathering 60mm objective wide lens. Abbreviation to define. In order to obtain the image of the gamma’s origin, a computer on the HESS site combines up to four images and determines the position as well as the energy of the air shower. How does wave action vary along this length of coast? high-energy - of or relating to elementary particles having energies of hundreds of thousands of electron volts. Rev. HET stands for High Energy Telescopes. Reames, Space Sci. (2005). Lannacombe Bay by John Bointon / CC BY. © 2020 Springer Nature Switzerland AG. Reames, S.W. It's like photographing a beach and resolving individual grains of sand. Meteors and the space shuttle frequently create sonic booms. (2005). High Energy Telescope: HET: Hudson Essex Terraplane (club) HET: Humane Experimental Technique: HET: Hollow Earth Theory: HET: Hemorrhoid Energy Therapy (treatment for internal hemorrhoids) HET: Health Education Trust: HET: Hydrate Equilibrium Temperature: HET: Hypothèse Extraterrestre (French) HET: Harvest Energy Technology, Inc. HET: Humoristisch Elektronisch Tijdschrift: HET doi: R.A. Mewaldt et al., J. Geophys. Cline, U.D. Die Entscheidung wird für 2021 erwartet. Such a Cherenkov flash illuminates an area of roughly 250 meters (1/10th of a mile) in diameter. Many have heard such sonic booms or air pressure shock waves, but few have actually ‘seen’ much less photographed them. Tylka, M.A. Menu Search "AcronymAttic.com. The mirrors, which are arranged in the shape of a honeycomb, detect weak light flashes that develop when cosmic gamma quanta penetrate the earth’s atmosphere. Part of Springer Nature. a *, J. E. Grindlaya, B. Allena, S. D. Barthelmyb, G. K. Skinnerb, N. Gehrelsb, and the EXIST HET Working Gro. The Federal Republic of Germany is represented by the Max Planck Institute for Nuclear Physics in Heidelberg, the Humboldt University in Berlin, the Ruhr University in Bochum, the University in Hamburg, the University in Kiel, and the Heidelberg Observatory. If you are visiting our English version, and want to see definitions of High-Energy Antimatter Telescope in other languages, please click the language menu on the right bottom. Work Benches. Wissenschaftliche Ziele von ATHENA sind vor ⦠Rev. It will have about 20 times more sensitivity, cover a much broader energy range, have considerably better energy resolution and provide a significantly improved angular resolution. Atmospheric flashes of Cherenkov light result when gamma particles cascade to create air and particle showers. X-ray and Gamma-ray telescopes are usually installed on Earth-orbiting satellites or high-flying balloons since the Earth's atmosphere is opaque to this part of the electromagnetic spectrum. Download preview PDF. See all departments. Because the particle moves faster than the speed of light in air, there is a sonic boom or shock wave, which sends out a flash of blue light in the direction of the primary gamma quantum and lasts a few billionths of a second. Runcorn (Akademie-Verlag, Berlin, 1973), p. 713, A.J. 2. high-energy - providing a relatively large amount of energy upon undergoing a chemical reaction. doi: D.V. HET stands for High Energy Telescopes. High-energy cosmic gamma rays produce e ± pairs in the atmosphere over $$\frac{9}{7}X_0$$ (with the radiation length $$X_0 \approx 36.5~\rm{g/cm}^2$$ in air). Das Advanced Telescope for High Energy Astrophysics ist das Projekt eines Weltraum-Röntgenteleskops der Europäischen Weltraumorganisation ESA. Fortunately for life on earth, a gamma particle from the universe does not penetrate to the earth’s surface, but if it flies past an atomic nucleus within the earth’s atmosphere, the gamma particle can transform itself into an electron and its (positive) antiparticle, a positron. Hobby Tools. It will have about 20 times more sensitivity, cover a much broader energy range, have considerably better energy resolution and provide a significantly improved angular resolution. Moreover, the instruments are weatherproof and are not protected by any sort of roofing or building. Sie sind auf der linken Seite unten aufgeführt. (2007, this issue). What does HET stand for? (2007, this issue). What does THE stand for? Find. Stone et al., Space Sci. The IMPACT investigation for the STEREO Mission includes a complement of Solar Energetic Particle instruments on each of the two STEREO spacecraft. Andere Bedeutungen von HET Neben Hochenergie-Teleskop hat HET andere Bedeutungen. Definition; HEAT: Hostile Environment Awareness Training: HEAT: High-Explosive Anti-Tank (munition) HEAT: Heat Escape Lessening Posture: HEAT: Highway Enforcement of Aggressive Traffic (various locations) HEAT: Higher Education and Training (IBM) HEAT: Higher Education Assistive Technology (various locations) HEAT: High-Energy Antimatter Telescope: HEAT: Help End Auto Theft The high energy γ-ray telescope selected for definition studies on the Gamma Ray Observatory provides a substantial improvement in observational capability over earlier instruments. pp 391-435 | In 1936, he was awarded the Nobel Prize for Physics. HEFT - High-Energy Focusing Telescope. 36, 1534 (2005). MAGIC detected [5] very high energy cosmic rays from the quasar 3C 279, which is 5 billion light years from Earth. The four telescopes are positioned to form a square; the distance between each telescope is 120 meters (~400 ft.). Bryant, T.L. (2007b, this issue). In order to detect the weak blue light, the HESS telescopes operate during nights when the moon is not visible. Im Falle einer positiven Entscheidung zur Realisierung des Projekts soll das Teleskop 2031Vorlage:Zukunft/In 5 Jahren gestartet werden. Therefore, the telescopes have to be directed at the same location in the sky for many hours. Mason et al., Space Sci. 113.10.144.2. 67, 4983 (1962), CCSDS Packet Telemetry Recommendation for Space Data System Standards, 102.0-B-5, Blue Book, Issue 5 (CCSDS, Washington, DC, 2000). Not logged in Videos. McDonald, J. Geophys. Then you'll be happy to know your kid's high energy is perfectly normal. Cliff erosion on Holderness by Bryan Ledgard / CC BY. J. doi: W.R. Cook, STEREO PHASIC Userâs Manual, Caltech internal document, 2002, J. Luhmann et al., Adv. During its journey through the air, this pair comes across more atomic nuclei and a gamma quantum is generated which then once again hits atomic nuclei. More than seventy scientists from Germany, France, England, Ireland, the Czech Republic, Armenia, Namibia, and the Republic of South Africa are involved in the HESS project. Printer friendly. Supernovae like this one can be used as cosmic tape measures, allowing astronomers to calculate the distance to their galaxies. The HETs are designed to measure the abundances and energy spectra of electrons, protons, He, and heavier nuclei up to Fe in interplanetary space. Kahler, C.K. Res. You will see meanings of High-Energy Antimatter Telescope in many other languages such as Arabic, Danish, Dutch, Hindi, Japan, Korean, Greek, Italian, Vietnamese, etc. J. For protons and He that stop in the HET, the kinetic energy range corresponds to â¼13 to 40 MeV/n. Showers. Protons that do not stop in the telescope (referred to as penetrating protons) are measured up to â¼100 MeV/n, as are penetrating He. Trending Products . 491, 414 (1997), STEREO Mission Operations Center (MOC) to Payload Operations Center (POC) and to STEREO Science Center (SSC) Interface Control Document (ICD), Applied Physics Laboratory internal document 7381-9045A (2002), E.C. This paper describes the HETs in detail, including the scientific objectives, the sensors, the overall mechanical and electrical design, and the on-board software. ft.). 90, 413 (1999), D.V. The camera enables exposure times of a mere one hundred millionth of a second. Thus, a single cosmic gamma particle creates roughly a thousand secondary particles in a cascade-like process or sub-atomic shower. Stopping electrons are measured in the energy range â¼0.7â6 MeV. Skal du rulle ned og klik for at se hver af dem. I voted for him, by the way. The HESS acronym alludes to the Austrian physicist Viktor Franz Hess (1883-1964) who discovered cosmic rays during ten balloon flights between 1911 and 1913. The Energetic X-ray Imaging Survey Telescope (EXIST) is a proposed next generation multi-wavelength survey mission. In order to study in more details the Pyramid and its Big Void, the team has then built in 2018 two new telescopes, which have been installed inside the Grand Gallery of the pyramid, where they faced additional challenges: ⦠The telescopes deployed by the CEA team has shown a remarkable robustness in harsh conditions, and opened the way to real-time, high-definition muography. The focus of the telescope has a French-built electronic camera with 960 photo tube detectors, which are mounted on an area of about 1.4 meters (~4 ft) in diameter. Given that the history of astronomy goes back centuries, the observations in the gamma spectrum are really among the newest areas in celestial research. The high energy gamma-ray telescope selected for definition studies on the Gamma Ray Observatory provides a substantial improvement in observational capability over earlier instruments. Res. See other definitions of HET. The year 2002 marked the completion of the first of four telescopes officially inaugurated. [1] Looking for abbreviations of HEFT? The work on the definition and technological preparation of the ATHENA (Advanced Telescope for High ENergy Astrophysics) mission continues to progress. C.M.S. Andre betydninger af HEAT Ud over Højenergi antistof teleskop har HEAT andre betydninger. Wall Cabinets. What coastal processes are happening here? Cohen et al., J. Geophys. If all four telescopes detect a flash simultaneously, a stereoscopic observation can be made. Andersen, J.F. Space Res. ten times more sensitive than earlier Cherenkov telescopes. The NASA/ESAâs Hubble Space Telescope has tracked the fading light of a supernova in the spiral galaxy NGC 2525, located 70 million light years away. Price Drops. The IMPACT investigation for the STEREO Mission includes a complement of Solar Energetic Particle instruments on each of the two STEREO spacecraft. This doubles the previous record distance from which very high energy cosmic rays have been detected. This is a preview of subscription content, H.H. Rev. Die Entscheidung wird für 2021 erwartet. My Collections. Examples: NFL, NASA, PSP, HIPAA. Beschreibung in Englisch: High Energy Telescope. The concept for the project crystallized in 1996 at the Max Planck Institute for Nuclear Physics in Heidelberg. The Telescope Array (TA) experiment, located in midwest Utah, USA(39.3N, 112.9W, Alt 1382 m), consists of two types of detector ().Both methods observe the high energy phenomenon known as an âair showerâ, which is generated by an ultra high energy cosmic ray. Ng, Astrophys. Worklights. Each telescope has a diameter of twelve meters (~40 feet) and 380 individual round mirrors that make up a light-collecting surface area of 108 square meters (~1000 sq. THE abbreviation stands for Telescope for High Energy. Not affiliated Das Advanced Telescope for High Energy Astrophysics ist das Projekt eines Weltraum-Röntgenteleskops der Europäischen Weltraumorganisation ESA. Special detectors in satellites and high altitude research rockets register gamma rays with energies of up to around ten billion electron volts. High energy coast. High-energy astronomy requires specialized telescopes to make observations since most of these particles go through most metals and glasses.. HET stands for High Energy Telescope Suggest new definition This definition appears somewhat frequently and is found in the following Acronym Finder categories: LUVOIR-A would be large enough to find and study the dozens of ⦠Cosmic Ray Conf., Mérida, México (2007a). Workshop Safety. Of these instruments, the High Energy Telescopes (HETs) provide the highest energy measurements. High-Energy Focusing Telescope listed as HEFT Faucets. The CAT imaging telescope provides a very high image definition (546 pixels with 0.12° spacing in the central part of the field of view), resulting in improved background rejection and energy resolution. The high-energy light is swallowed by the earth’s atmosphere yet the light cannot be captured with conventional lenses or mirrors. What is the abbreviation for Telescope for High Energy? Though the galaxy is over 2 million light-years away, the Hubble telescope is powerful enough to resolve individual stars in a 61,000-light-year-long section of the galaxy's pancake-shaped disk. J. Luhmann et al., Space Sci. Sometime around this era (2002) he won the David Schramm award for best writing in high-energy astronomy, for his article about the Chandra X-ray Observatory titled "Superman's Telescope". The parameters for international cooperation were set in January 1998 and a year and a half later, the Goellschau farm located in the Khomas Highland in Namibia was chosen as the location. The HESS telescopes are ten times more sensitive than earlier Cherenkov telescopes. Large BAK-4 Prism inside and Fully Multi-coated Lens Coating : Strengthen the key function of ⦠R.A. Mewaldt et al., Space Sci. Bitte scrollen Sie nach unten und klicken Sie, um jeden von ihnen zu sehen. Thus, direct detection of gamma-ray photons must be done in space, while instruments on the ground can see very high ⦠Searching for gamma quanta is tedious because they hit the earth with much less frequency than optical photons. Ziegler, Hydrogen Stopping Powers and Ranges in All Elements (Pergamon, New York, 1977), R. Baker, CPU24 Microprocessor Userâs Manual, GSFC internal document, 2003, D.A. doi: R.A. Mewaldt et al., 30th Int. The High Energy Telescope on EXIST. Rev. Why is the shore platform neither smooth nor flat? 504, 1002 (1998), D.V. Lee, Astrophys. Der kan være mere end én definition af HEAT, så tjek det ud på vores ordbog for alle betydninger af HEAT en efter en. Deals. Athena â Advanced Telescope for High Energy Astrophysics Athena Science Objectives Diagnose hot cosmic plasmas on all astrophysical environments via spatially resolved high resolution X ⦠De er listet til venstre nedenfor. 646, 1319 (2006), © Springer Science+Business Media, BV 2008, https://doi.org/10.1007/978-0-387-09649-0_14. Every night, the astronomers can sight up to a dozen different objects. Based on a Max Planck Institute Press Release, Stellar winds, the source material for the universe, are clumpy, GAIA REVEALS HOW SUN-LIKE STARS TURN SOLID AFTER THEIR DEMISE, Astronomers discover the giant that shaped the early days of our Milky Way, Short-lived light sources discovered in the sky. To perform research, the astronomers use a trick enabling them to make observations from the ground. High energy coast. Images. Ng, Astrophys. Higher energy X-ray and Gamma-ray telescopes refrain from focusing completely and use coded aperture masks: the patterns of the shadow the mask creates can be reconstructed to form an image. Only a large number of points produced in this way show a galaxy or the remnants of a supernova. Desai, F.B. On 3rd September 2002, a new window on the universe was opened in Namibia, Africa. Für alle Bedeutungen von HET klicken Sie bitte auf "Mehr". Solar Energy Supplies. This happens about ten kilometers (6.3 miles) above the earth’s surface. Where Is the Habitable Zone for M-Dwarf Stars? High-energy photons are a powerful tool to understand the most violent phenomena in our Universe. Cite as. Cherenkov light develops within this air shower. This service is more advanced with JavaScript available, The STEREO Mission by M.J. Rycroft, S.K. What coastal processes are happening here? This is where HESS comes into play. 86, 357 (1998), R. Tousey, in Space Research XIII, ed. The first of four telescopes of the HESS experiment was inaugurated at its location on Goellschau farm, which is at an altitude of 1800 meters. J. Up until now, this interesting spectral range has not been investigated in detail. doi: R. Müller-Mellin et al., Space Sci. HET is defined as High Energy Telescopes somewhat frequently. Rev. doi: G.M. The result is plotted as a point on a map of the sky. Reames, C.K. The signal indicated that the universe is more transparent than previously thought based on data from optical and infrared telescopes. Sinks. For stopping He, the individual isotopes 3He and 4He can be distinguished. High-Energy Antimatter Telescope. Holidays. Instead, it only detects the air shower in the earth’s atmosphere. Im Falle einer positiven Entscheidung zur Realisierung des Projekts soll das Teleskop 2031Vorlage:Zukunft/In 5 Jahren gestartet werden. (2007, this issue). J. Hong. Rev. Plumbing Parts & Tools. Tidal Locking Could Render Habitable Planets Inhospitable. The astronomers reckon that they will collect up to one thousand observation hours (~40 days) per year. It is High-Energy Focusing Telescope. Other ⦠Peter Gruss, President of the Max Planck Society, emphasized that the HESS telescopes are "in friendly competition with similar facilities in Australia, the USA, and on the Canary Islands". HESS is also a good example of "efficient international scientific collaboration". Over 10 million scientific documents at your fingertips. Heat Ud over Højenergi antistof Teleskop har HEAT andre betydninger af HEAT Ud over Højenergi antistof har... The energy range corresponds to â¼13 to 40 MeV/n af HEAT skal du rulle ned klik. New Mexico, USA, gamma rays with energies of up to one thousand observation hours ( ~40 ). Of up to around ten billion electron volts sight up to around ten billion electron volts year marked... Hot celestial bodies like the sun abbreviation for Telescope for High energy telescopes HETs. Zu sehen W.R. Cook, STEREO PHASIC Userâs Manual, Caltech internal document, 2002, J. Geophys miles... The same location in the hard X-ray ( 20â100 keV ) band pp 391-435 Cite. Ledgard / CC by from which very High energy Astrophysics ist das eines. Special detectors in satellites and High altitude research rockets register gamma rays with energies of up to around billion! X-Ray Imaging Survey Telescope ( EXIST ) is a preview of subscription content, H.H (,... Content, H.H telescopes ( HETs ) provide the highest energy measurements to â¼13 to 40 MeV/n lenses! ( ~40 days ) per year this length of coast more sensitive than earlier Cherenkov telescopes can. The year 2002 marked the completion of the sky for many hours from optical and infrared high energy telescope definition conventional! ( EXIST ) is a preview of subscription content, H.H to one thousand observation hours ( days!, Caltech internal document, 2002, J. Geophys 2007a ) by Bryan Ledgard / CC by a relatively amount... This doubles the previous record distance from which very High energy cosmic rays from the ground von! Flash simultaneously, a stereoscopic observation can be used as cosmic tape measures, allowing to... Enables exposure times of a mile ) in diameter this way show a galaxy or the remnants of mere! Bedeutungen von HET klicken Sie bitte auf Mehr '' Sie nach unten klicken... Content, H.H sources in the sky for many hours data from optical and infrared telescopes a of!, NASA, PSP, HIPAA, HIPAA will collect up to one thousand observation hours ( ~40 ). 5 ] very High energy Astrophysics ist das Projekt eines Weltraum-Röntgenteleskops der Europäischen Weltraumorganisation ESA this... With JavaScript available, the astronomers can sight up to around ten billion electron.. Galaxy or the remnants of a supernova to 40 MeV/n observations since most of these,! A Cherenkov flash illuminates an area of roughly 250 meters ( 1/10th of a mere one hundred millionth a. To a dozen different objects the earth ’ s surface telescopes are to... Caltech internal document, 2002, J. Geophys a relatively large amount of energy upon undergoing a chemical reaction in! Shower in the earth ’ s atmosphere yet the light can not be captured with conventional lenses mirrors. Undergoing a chemical reaction Luhmann et al., 30th Int understand the most violent phenomena our..., Adv Planck Institute for Nuclear Physics in Heidelberg - of or relating to elementary particles having energies of to... J. Luhmann et al., J. Geophys alle Bedeutungen von HET klicken Sie bitte auf ''... This interesting spectral range has not been investigated in detail about ten kilometers ( 6.3 miles above... Energy cosmic rays from the ground astronomers reckon that they are not produced in hot celestial bodies the! By Bryan Ledgard / CC by up to a dozen different objects swallowed by the earth s! Is more Advanced with JavaScript available, the kinetic energy range â¼0.7â6 MeV energy! Data from optical and infrared telescopes: Zukunft/In 5 Jahren gestartet werden with conventional lenses or.! Du klikke på mere the concept for the STEREO Mission includes a complement of Solar Energetic particle on... Doubles the previous record distance from which very High energy telescopes ( HETs ) provide the highest energy measurements Nuclear. Not produced in hot celestial bodies like the sun to be directed at the same in... More Advanced with JavaScript available, the STEREO Mission includes a complement Solar. Positioned to form a square ; the distance to their galaxies a conventional Telescope, the High telescopes. The four telescopes detect a flash simultaneously, a stereoscopic observation can be used as cosmic tape measures allowing... Of these instruments, the STEREO Mission includes a complement of Solar Energetic particle instruments on each of the of... Zu sehen bitte scrollen Sie nach unten und klicken Sie, um jeden von ihnen zu sehen Hochenergie-Teleskop HET.
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2021-04-21 14:45:57
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https://www.whsmith.co.uk/products/galois-teichmueller-theory-and-arithmetic-geometry-advanced-studies-in-pure-mathematics-63/9784864970143
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# Galois-teichmUEller Theory And Arithmetic Geometry (Advanced Studies in Pure Mathematics 63)
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### Description
From the 1980's, Grothendieck's "Esquisse d'un Programme" triggered tremendous developments in number theory and arithmetic geometry, extending from the studies of anabelian geometry and related Galois representations to those of polylogarithms and multiple zeta values, motives, rational points on arithmetic varieties, and effectiveness questions in arithmetic geometry. The present volume collects twenty-four articles written by speakers (and their coauthors) of two international meetings focused on the above themes held in Kyoto in October 2010. It includes both survey articles and research papers which provide useful information about this area of investigation.Published by Mathematical Society of Japan and distributed by World Scientific Publishing Co. for all markets except North America
### Other formatsView More
Hiroaki Nakamura, Okayama University, Japan Florian Pop, University of Pennsylvania, Philadelphia, PA, USA Leila Schneps, University of Paris VI, France Akio Tamagawa, Kyoto University, Japan
### Contents
Remarks on the Milnor conjecture over schemes by A. Auel On the decomposition of motivic multiple zeta values by F. C. S. Brown Combinatorics of the double shuffle Lie algebra by S. Carr and L. Schneps On the double zeta values by P. Cartier Harmonic Galois theory for finite graphs by S. Corry Twisted covers and specializations by P. Debes and F. Legrand Geometric interpretation of double shuffle relation for multiple $L$-values by H. Furusho Noether's problem for transitive permutation groups of degree $6$ by K. Hashimoto and H. Tsunogai Comparison of some quotients of fundamental groups of algebraic curves over $p$-adic fields by Y. Ihara Dimensions of moduli spaces of finite flat models by N. Imai Results and conjectures in profinite Teichmuller theory by P. Lochak Galois actions on complex braid groups by I. Marin The (local) lifting problem for curves by A. Obus Some remarks on profinite completion of spaces by G. Quick An abelian surface with constrained $3$-power torsion by C. Rasmussen Fake liftings of Galois covers between smooth curves by M. Saidi Motivic aspects of anabelian geometry by A. Schmidt On cuspidal sections of algebraic fundamental groups by J. Stix A note on quadratic residue curves on rational ruled surfaces by H. Tokunaga $n$-nilpotent obstructions to $\pi 1$ sections of $\mathbb P^1 - \{0,1,\infty \}$ and Massey products by K. Wickelgren Lie algebras of Galois representations on fundamental groups by Z. Wojtkowiak $p$-adic multiple zeta values, $p$-adic multiple $L$-values, and motivic Galois groups by G. Yamashita Topics surrounding the combinatorial anabelian geometry of hyperbolic curves I: Inertia groups and profinite Dehn twists by Y. Hoshi and S. Mochizuki Some congruence properties of Eisenstein invariants associated to elliptic curves by H. Nakamura
### Product Details
• ISBN13: 9784864970143
• Format: Hardback
• Number Of Pages: 850
• ID: 9784864970143
• ISBN10: 4864970149
### Delivery Information
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• 1st Class Delivery: Yes
• Courier Delivery: Yes
• Store Delivery: Yes
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2018-11-17 11:42:09
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https://socratic.org/questions/which-of-these-equations-represent-functions-where-x-is-the-input-and-y-is-the-o
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# Which of these equations represent functions where x is the input and y is the output?
## Select each answer that applies A. X=2 B. Y=2 C. Y=2x D. X=2y E. X+y=2
Jul 3, 2018
everybody, except (a)
#### Explanation:
a(x) = ? Vertical line
$b \left(x\right) = 2$
$c \left(x\right) = 2 x$
$d \left(x\right) = \frac{x}{2}$
$e \left(x\right) = 2 - x$
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2018-12-10 02:47:09
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https://www.techyv.com/questions/could-not-set-adobe-elements-file-resolution-correctly/
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## Could Not Set Adobe Elements File Resolution Correctly
Asked By 1010 points N/A Posted on -
I have created a file in Adobe Photoshop Element 6. When trying to set the size of my document to 350ppi, I am having an error stating:
Image Size
A number between 1000 and 256.565 is required. Closest value inserted.
And then it automatically becomes 256.565 but the file became too big. And when trying to edit the file, my computer froze and shows an error
Move command could not be completed as file is too large.
How can I set the resolution correctly?
SHARE
Answered By 30 points N/A #90224
## Could Not Set Adobe Elements File Resolution Correctly
Hi Joseph,
The problem you encountered is more that the height and width you use are 10x A4, so you're file is really bigger. The sizes are set to centimeters, not millimeters.
You need to try in PSE 6.0, with 21cm and 29,7cm instead, and it defaulted to 300 ppi, no problem. So I think once you corrected the sizes, it should be fine.
If you use consistently sizes and resolutions in either inches, or cm, this could be easier, as the PS will do the computation for you. I guess you want to preserve 300 dpi for printing (using inches), but you are printing on a A4 paper (using cm).
-Zorian
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2020-01-17 18:20:16
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https://www.physicsforums.com/threads/derivative-of-a-convolution.632856/
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# Derivative of a Convolution
1. Sep 2, 2012
### Shaybay92
Hi,
I want to verify that the form of a particular solution satisfies the following ODE:
v' + (b/m)v = u/m
with
vpart= ∫e-(b/m)(t-r) (u(r)/m) dr
where the limits are from 0 to t
So I tried to differentiate v with respect to t, in order to substitute it back into the equation. But, how do you do that when the integral is with respect to r? Is there a need to change variables? How can you do this?
Cheers
2. Sep 2, 2012
### haruspex
v(t) = ∫tf(t, r).dr
v(t+δt)= ∫t+δtf(t+δt, r).dr
= ∫tf(t+δt, r).dr + ∫tt+δtf(t+δt, r).dr
So v' = ∫t(d/dt)f(t, r).dr + f(t, t)
3. Nov 8, 2012
### Shaybay92
Sorry I'm not familiar with your method. I don't understand why you substitute "t+δt" for t. What approach are you using here? Could you elaborate or direct me to some further reading?
Cheers :)
4. Nov 8, 2012
### haruspex
The equation you posted for v(t) is generic - i.e. it's true for all t. So it's true both for a given t and for a later time t+δt. So you can write a second equation substituting t+δt for t consistently. Taking the difference, diving by δt, then letting δt tend to zero gives you v'. That is the standard process of differentiation.
5. Nov 8, 2012
### Shaybay92
Oh I see what you mean. Thanks for the clarification. I'm just not use to this notation :)
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2018-05-25 17:03:46
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https://tel.archives-ouvertes.fr/tel-00002457
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# Fonctions de corrélation en théorie des champs à température finie : aspects formels et applications au plasma de quarks et de gluons
Abstract : The general framework of this work is thermal field theory, and more precisely the perturbative calculation of thermal Green's
functions. In a first part, I consider three problems closely related to the formalism itself. After two introductory chapters
devoted to set up the framework and the notations used afterwards, a chapter is dedicated to a clarification of certain aspects
of the justification of the Feynman rules of the real time formalism. Then, I consider in chapter 4 the problem of cutting rules in
the real time formalisms. In particular, after solving a controversy on this subject, I generalize these cutting rules to the
retarded-advanced'' version of this formalism. Finally, the last problem considered in this part is that of the pion decay into
two photons in a thermal bath. I show that the discrepancies found in the literature are due to peculiarities of the analytical
properties of the thermal Green's functions. The second part deals with the calculation of the photon or dilepton (virtual
photon) production rate by a quark gluon plasma. The framework of this study is the effective theory based on the
resummation of hard thermal loops. The first aspect of this study is related to the production of virtual photons, where we show
that important contributions arise at two loops, completing the result already known at one loop. In the case of real photon
production, we show that extremely strong collinear singularities make two loop contributions dominant compared to one loop
ones. In both cases, the importance of two loop contributions can be interpreted as weaknesses of the hard thermal loop
approximation.
Mots-clés :
Document type :
Theses
Domain :
Cited literature [90 references]
https://tel.archives-ouvertes.fr/tel-00002457
Contributor : Marc Gingold <>
Submitted on : Monday, February 24, 2003 - 11:01:05 AM
Last modification on : Friday, November 6, 2020 - 3:35:03 AM
Long-term archiving on: : Friday, April 2, 2010 - 8:05:44 PM
### Identifiers
• HAL Id : tel-00002457, version 1
### Citation
François Gelis. Fonctions de corrélation en théorie des champs à température finie : aspects
formels et applications au plasma de quarks et de gluons. Physique des Hautes Energies - Expérience [hep-ex]. Université de Savoie, 1998. Français. ⟨tel-00002457⟩
Record views
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2020-11-25 14:54:19
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http://gmatclub.com/forum/if-number-n-is-randomly-drawn-from-a-set-of-all-non-negative-117593.html?sort_by_oldest=true
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# If number N is randomly drawn from a set of all non-negative
Author Message
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If number N is randomly drawn from a set of all non-negative [#permalink] 24 Jul 2011, 00:40
00:00
Difficulty:
5% (low)
Question Stats:
0% (00:00) correct 0% (00:00) wrong based on 0 sessions
If number N is randomly drawn from a set of all non-negative single-digit integers, what is the probability that 5N^3/8 is an integer?
[Reveal] Spoiler:
1/2
Last edited by Bunuel on 10 Jul 2013, 14:22, edited 2 times in total.
Renamed the topic, edited the question, added the OA and moved to PS forum.
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Re: Probability - Integer [#permalink] 24 Jul 2011, 03:58
DeeptiM wrote:
If number N is randomly drawn from a set of all non-negative single-digit integers, what is the probability that 5N^3/8 is an Integer?
is it (5N^3)/8 or 5N^(3/8)???
For the first case we need to check for the values of N for which 5(N/2)^3 is integer or N/2 is integer
therefore all even number satisfies this condition i.e. 2,4,6,8.
Number of terms = from 0 to 9. i.e. 10 numbers
Hence probability should be = 4/10 = 2/5
For the second case i guess the probability will be 1/5. as the only case when 5N^(3/8) is an integer when N is either 0 or 1. hence 2/10 = 1/5
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Re: Probability - Integer [#permalink] 24 Jul 2011, 20:07
1
KUDOS
Expert's post
DeeptiM wrote:
If number N is randomly drawn from a set of all non-negative single-digit integers, what is the probability that 5N^3/8 is an Integer?
Set of all non-negative single digit integers - 0 to 9 i.e. 10 numbers
For 5N^3/8 to be an integer, 5N^3 should be completely divisible by 8. This will happen only if N^3 is divisible by 8 i.e. if N has 2 as a factor. So for 5N^3/8 to be an integer, N should be even.
Out of the 10 numbers 0 - 9, 5 are even and 5 are odd.
Probability that 5N^3/8 is an integer is 1/2.
_________________
Karishma
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Save $100 on Veritas Prep GMAT Courses And Admissions Consulting Enroll now. Pay later. Take advantage of Veritas Prep's flexible payment plan options. Veritas Prep Reviews Manager Joined: 31 May 2011 Posts: 96 Location: India Concentration: Finance, International Business GMAT Date: 12-07-2011 GPA: 3.22 WE: Information Technology (Computer Software) Followers: 1 Kudos [?]: 24 [0], given: 4 Re: Probability - Integer [#permalink] 25 Jul 2011, 00:51 VeritasPrepKarishma wrote: DeeptiM wrote: If number N is randomly drawn from a set of all non-negative single-digit integers, what is the probability that 5N^3/8 is an Integer? Set of all non-negative single digit integers - 0 to 9 i.e. 10 numbers For 5N^3/8 to be an integer, 5N^3 should be completely divisible by 8. This will happen only if N^3 is divisible by 8 i.e. if N has 2 as a factor. So for 5N^3/8 to be an integer, N should be even. Out of the 10 numbers 0 - 9, 5 are even and 5 are odd. Probability that 5N^3/8 is an integer is 1/2. Ahh yes. I missed 0 in the first case Senior Manager Joined: 08 Nov 2010 Posts: 426 WE 1: Business Development Followers: 6 Kudos [?]: 28 [0], given: 161 Re: Probability - Integer [#permalink] 25 Jul 2011, 22:54 If i can give my 2 cents tip. When you see the words non-negative you should immediately think about zero in your solution. Same way about non-positive of course. Manager Joined: 30 Sep 2009 Posts: 111 Followers: 0 Kudos [?]: 12 [0], given: 77 Re: Probability - Integer [#permalink] 25 Jul 2011, 23:05 As per my understanding zero is neither postive nor negitive integer. why should zero be included in this question as the question is asking about the non-negative single-digit. Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 3780 Location: Pune, India Followers: 813 Kudos [?]: 3240 [0], given: 138 Re: Probability - Integer [#permalink] 26 Jul 2011, 00:16 Expert's post abhi398 wrote: As per my understanding zero is neither postive nor negitive integer. why should zero be included in this question as the question is asking about the non-negative single-digit. Since 0 is neither positive nor negative, 'non negative' means 'positive and 0' _________________ Karishma Veritas Prep | GMAT Instructor My Blog Save$100 on Veritas Prep GMAT Courses And Admissions Consulting
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Re: Probability - Integer [#permalink] 26 Jul 2011, 00:20
thanks karishma....
Never thought of that .....
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Re: Probability - Integer [#permalink] 27 Jul 2011, 09:11
VeritasPrepKarishma wrote:
abhi398 wrote:
As per my understanding zero is neither postive nor negitive integer.
why should zero be included in this question as the question is asking about the non-negative single-digit.
Since 0 is neither positive nor negative, 'non negative' means 'positive and 0'
Absolutely! I've fell to this GMAT's one of the favorite traps so many times that I had this pinned on my wall (the real one. not FB's :D
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Re: Probability - Integer [#permalink] 30 Jul 2011, 17:56
VeritasPrepKarishma wrote:
DeeptiM wrote:
If number N is randomly drawn from a set of all non-negative single-digit integers, what is the probability that 5N^3/8 is an Integer?
Set of all non-negative single digit integers - 0 to 9 i.e. 10 numbers
For 5N^3/8 to be an integer, 5N^3 should be completely divisible by 8. This will happen only if N^3 is divisible by 8 i.e. if N has 2 as a factor. So for 5N^3/8 to be an integer, N should be even.
Out of the 10 numbers 0 - 9, 5 are even and 5 are odd.
Probability that 5N^3/8 is an integer is 1/2.
The probability is not 1/2. You have assumed that the result would be an integer with n=0.
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Re: Probability - Integer [#permalink] 30 Jul 2011, 19:20
0 is an integer.
When N=0, then 5(N^3)/8 = 0
So the answer will be 5/10 or 1/2
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Re: Probability - Integer [#permalink] 30 Jul 2011, 19:26
Sudhanshuacharya wrote:
DeeptiM wrote:
If number N is randomly drawn from a set of all non-negative single-digit integers, what is the probability that 5N^3/8 is an Integer?
is it (5N^3)/8 or 5N^(3/8)???
For the first case we need to check for the values of N for which 5(N/2)^3 is integer or N/2 is integer
therefore all even number satisfies this condition i.e. 2,4,6,8.
Number of terms = from 0 to 9. i.e. 10 numbers
Hence probability should be = 4/10 = 2/5
For the second case i guess the probability will be 1/5. as the only case when 5N^(3/8) is an integer when N is either 0 or 1. hence 2/10 = 1/5
Whenever the question says
5N^3/8 first evaluate the exponent and then the Multiply/Divide
So you only need to consider 5N^3/8 will always be (5N^3)/8 or 5(N^3/8) or (5/8)N^3...All are the same
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Re: Probability - Integer [#permalink] 30 Jul 2011, 19:31
Sudhanshuacharya wrote:
DeeptiM wrote:
If number N is randomly drawn from a set of all non-negative single-digit integers, what is the probability that 5N^3/8 is an Integer?
is it (5N^3)/8 or 5N^(3/8)???
For the first case we need to check for the values of N for which 5(N/2)^3 is integer or N/2 is integer
therefore all even number satisfies this condition i.e. 2,4,6,8.
Number of terms = from 0 to 9. i.e. 10 numbers
Hence probability should be = 4/10 = 2/5
For the second case i guess the probability will be 1/5. as the only case when 5N^(3/8) is an integer when N is either 0 or 1. hence 2/10 = 1/5
5N^3^3^3/8
Now first calculate the exponents : 5N^3^3^3 from Right to Left the exponents and not Left to Right
So 5N^3^3^3/8 = 5N^3^27/8 and not 5N^27^3/8....In other words 3^27 <> 27^3
Hope you got my point.
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Re: Probability - Integer [#permalink] 31 Jul 2011, 21:10
krishp84 wrote:
Sudhanshuacharya wrote:
DeeptiM wrote:
If number N is randomly drawn from a set of all non-negative single-digit integers, what is the probability that 5N^3/8 is an Integer?
is it (5N^3)/8 or 5N^(3/8)???
For the first case we need to check for the values of N for which 5(N/2)^3 is integer or N/2 is integer
therefore all even number satisfies this condition i.e. 2,4,6,8.
Number of terms = from 0 to 9. i.e. 10 numbers
Hence probability should be = 4/10 = 2/5
For the second case i guess the probability will be 1/5. as the only case when 5N^(3/8) is an integer when N is either 0 or 1. hence 2/10 = 1/5
5N^3^3^3/8
Now first calculate the exponents : 5N^3^3^3 from Right to Left the exponents and not Left to Right
So 5N^3^3^3/8 = 5N^3^27/8 and not 5N^27^3/8....In other words 3^27 <> 27^3
Hope you got my point.
Hi..I didnt get thi spart..can u help elaborate pls..
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Re: Probability - Integer [#permalink] 01 Aug 2011, 08:33
ok - you have a question asking - what is the value of 2^3^4 ?
Is it 2^81 or 8^4 ?
It will be 2^81
This was easy...
Now let us say the question asked - what is the value of 2^3^4^2 ?
Will it be 2^3^16 or 8^4^2 ?
It will be 2^3^16
Takeaway - Always evaluate exponents from Right to Left if they are of the form - x^y^z^....
DeeptiM - Hope you got the point. Kudos are welcome if the post is useful.
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Re: Probability - Integer [#permalink] 04 Aug 2011, 16:47
VeritasPrepKarishma wrote:
DeeptiM wrote:
If number N is randomly drawn from a set of all non-negative single-digit integers, what is the probability that 5N^3/8 is an Integer?
Set of all non-negative single digit integers - 0 to 9 i.e. 10 numbers
For 5N^3/8 to be an integer, 5N^3 should be completely divisible by 8. This will happen only if N^3 is divisible by 8 i.e. if N has 2 as a factor. So for 5N^3/8 to be an integer, N should be even.
Out of the 10 numbers 0 - 9, 5 are even and 5 are odd.
Probability that 5N^3/8 is an integer is 1/2.
how'd you determine in detail that as long as N has 2 as a factor it is divisible by 8? I'm assuming the 5 in front means nothing except that it is not divisible by 8, so that the next item in the product must be. I thought it must be that N^3 has 2 as a factor and is >= 8
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Re: Probability - Integer [#permalink] 04 Aug 2011, 20:27
1
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Expert's post
pinchharmonic wrote:
how'd you determine in detail that as long as N has 2 as a factor it is divisible by 8? I'm assuming the 5 in front means nothing except that it is not divisible by 8, so that the next item in the product must be. I thought it must be that N^3 has 2 as a factor and is >= 8
Yes, 5 doesn't have any role to play except to tell you that it is not divisible by 8 and has no 2s in it. If instead of 5, we had 4, things would have been different (but let's not get into it right now).
So N^3/8 should be an integer. This means N^3 should be divisible by 8. But we have values for N, not N^3 so how do we figure what it means for N?
Let's say, N has 2 as a factor. Then N = 2a
N^3 = (2a)^3 = 8a^3
This number is divisible by 8 since it has 8 as a factor. Since N gets cubed, if it has a 2 in it, that becomes 8 and N^3 is divisible by 8.
If we consider that N^3 has 2 as a factor and is greater than 8 (as you suggested above), is it essential that N^3 will be divisible by 8?
Consider N^3 = 10 (has 2 as a factor and is >8) This is not divisible by 8 and for that matter, N is a not an integer in this case.
Consider now, what would have happened if instead of 5, we had 4.
We would need 4N^3/8 to be an integer i.e. N^3/2 to be an integer. Yet again, we need N^3 to be even, so N should be even too i.e. it must have a 2 in it.
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Re: Probability - Integer [#permalink] 05 Aug 2011, 09:59
VeritasPrepKarishma wrote:
pinchharmonic wrote:
how'd you determine in detail that as long as N has 2 as a factor it is divisible by 8? I'm assuming the 5 in front means nothing except that it is not divisible by 8, so that the next item in the product must be. I thought it must be that N^3 has 2 as a factor and is >= 8
Yes, 5 doesn't have any role to play except to tell you that it is not divisible by 8 and has no 2s in it. If instead of 5, we had 4, things would have been different (but let's not get into it right now).
So N^3/8 should be an integer. This means N^3 should be divisible by 8. But we have values for N, not N^3 so how do we figure what it means for N?
Let's say, N has 2 as a factor. Then N = 2a
N^3 = (2a)^3 = 8a^3
This number is divisible by 8 since it has 8 as a factor. Since N gets cubed, if it has a 2 in it, that becomes 8 and N^3 is divisible by 8.
If we consider that N^3 has 2 as a factor and is greater than 8 (as you suggested above), is it essential that N^3 will be divisible by 8?
Consider N^3 = 10 (has 2 as a factor and is >8) This is not divisible by 8 and for that matter, N is a not an integer in this case.
Consider now, what would have happened if instead of 5, we had 4.
We would need 4N^3/8 to be an integer i.e. N^3/2 to be an integer. Yet again, we need N^3 to be even, so N should be even too i.e. it must have a 2 in it.
karishma, thanks.
So is there a way to make this factoring process generic? It seems like your thought process is to find the lowest factor of the denominator, which is 2, ie not 4 or 8. Then make sure you have that same factor in the numerator. Then the product of all factors in the numerator must be >= than the denominator. Meaning if in numerator there are other products that are not factors you just ignore them. like 5 below
(5x2^3)/8
and if you happeneted to choose 3
(5x3^3)/8 = (5x3x3x3)/8
since 5 and 3 are both not factors you still dont have the factor portion of the numerator >=8
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Re: Probability - Integer [#permalink] 10 Jul 2013, 14:10
VeritasPrepKarishma wrote:
DeeptiM wrote:
If number N is randomly drawn from a set of all non-negative single-digit integers, what is the probability that 5N^3/8 is an Integer?
Set of all non-negative single digit integers - 0 to 9 i.e. 10 numbers
For 5N^3/8 to be an integer, 5N^3 should be completely divisible by 8. This will happen only if N^3 is divisible by 8 i.e. if N has 2 as a factor. So for 5N^3/8 to be an integer, N should be even.
Out of the 10 numbers 0 - 9, 5 are even and 5 are odd.
Probability that 5N^3/8 is an integer is 1/2.
what if N=8 or 4 rather than 2 ?
i think 8 should be a factor of n^3 means either N=8, or NxN=8 or NxNxN=8
hence we have 3 fav conditions
..... now i got stuck here....
can not decide total cond.
m i going wrong way ? pls explain
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Re: Probability - Integer [#permalink] 10 Jul 2013, 14:30
Expert's post
jimmy0220 wrote:
VeritasPrepKarishma wrote:
DeeptiM wrote:
If number N is randomly drawn from a set of all non-negative single-digit integers, what is the probability that 5N^3/8 is an Integer?
Set of all non-negative single digit integers - 0 to 9 i.e. 10 numbers
For 5N^3/8 to be an integer, 5N^3 should be completely divisible by 8. This will happen only if N^3 is divisible by 8 i.e. if N has 2 as a factor. So for 5N^3/8 to be an integer, N should be even.
Out of the 10 numbers 0 - 9, 5 are even and 5 are odd.
Probability that 5N^3/8 is an integer is 1/2.
what if N=8 or 4 rather than 2 ?
i think 8 should be a factor of n^3 means either N=8, or NxN=8 or NxNxN=8
hence we have 3 fav conditions
..... now i got stuck here....
can not decide total cond.
m i going wrong way ? pls explain
If number N is randomly drawn from a set of all non-negative single-digit integers, what is the probability that 5N^3/8 is an integer?
Set of non-negative single-digit integers = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}, 10 elements.
Now, \frac{5N^3}{8}=\frac{5N^3}{2^3} to be an integer N should be any even number. There are 5 even numbers in the set (0, 2, 4, 6, and 8), so the probability is 5/10=1/2.
Hope it's clear.
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Re: Probability - Integer [#permalink] 10 Jul 2013, 14:30
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2013-12-20 21:07:04
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http://www.caspase1inhibitor.com/2018/01/04/is-often-approximated-either-by-usual-asymptotic-hgola-et-al/
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# Is often approximated either by usual asymptotic h|Gola et al.
May be approximated either by usual asymptotic h|Gola et al.calculated in CV. The statistical significance of a model is often assessed by a permutation tactic based around the PE.Evaluation of your classification resultOne necessary part from the original MDR will be the evaluation of factor combinations relating to the appropriate classification of circumstances and EW-7197 custom synthesis controls into high- and low-risk groups, respectively. For every model, a 2 ?2 contingency table (also referred to as confusion matrix), summarizing the true negatives (TN), true positives (TP), false negatives (FN) and false positives (FP), could be produced. As mentioned prior to, the energy of MDR is often improved by implementing the BA as opposed to raw accuracy, if coping with imbalanced data sets. Within the study of Bush et al. [77], 10 unique measures for classification were compared with all the standard CE utilized in the original MDR approach. They encompass precision-based and receiver operating characteristics (ROC)-based measures (Fmeasure, geometric imply of sensitivity and precision, geometric mean of sensitivity and specificity, Euclidean distance from an ideal classification in ROC space), diagnostic testing measures (Youden Index, Predictive Summary Index), statistical measures (Pearson’s v2 goodness-of-fit statistic, likelihood-ratio test) and data theoretic measures (Normalized MedChemExpress FG-4592 Mutual Data, Normalized Mutual Facts Transpose). Based on simulated balanced data sets of 40 diverse penetrance functions when it comes to variety of disease loci (2? loci), heritability (0.five? ) and minor allele frequency (MAF) (0.two and 0.4), they assessed the power with the distinctive measures. Their benefits show that Normalized Mutual Details (NMI) and likelihood-ratio test (LR) outperform the regular CE and also the other measures in most of the evaluated circumstances. Both of these measures take into account the sensitivity and specificity of an MDR model, therefore need to not be susceptible to class imbalance. Out of these two measures, NMI is easier to interpret, as its values dar.12324 range from 0 (genotype and disease status independent) to 1 (genotype totally determines disease status). P-values might be calculated from the empirical distributions with the measures obtained from permuted information. Namkung et al. [78] take up these outcomes and compare BA, NMI and LR having a weighted BA (wBA) and numerous measures for ordinal association. The wBA, inspired by OR-MDR [41], incorporates weights based on the ORs per multi-locus genotype: njlarger in scenarios with smaller sample sizes, larger numbers of SNPs or with tiny causal effects. Among these measures, wBA outperforms all others. Two other measures are proposed by Fisher et al. [79]. Their metrics don’t incorporate the contingency table but make use of the fraction of cases and controls in each and every cell of a model straight. Their Variance Metric (VM) to get a model is defined as Q P d li n 2 n1 i? j = ?nj 1 = n nj ?=n ?, measuring the difference in case fracj? tions among cell level and sample level weighted by the fraction of people in the respective cell. For the Fisher Metric n n (FM), a Fisher’s exact test is applied per cell on nj1 n1 ?nj1 ,j0 0 jyielding a P-value pj , which reflects how uncommon every cell is. For any model, these probabilities are combined as Q P journal.pone.0169185 d li i? ?log pj . The greater each metrics will be the extra likely it really is j? that a corresponding model represents an underlying biological phenomenon. Comparisons of these two measures with BA and NMI on simulated information sets also.Could be approximated either by usual asymptotic h|Gola et al.calculated in CV. The statistical significance of a model is usually assessed by a permutation approach primarily based around the PE.Evaluation of your classification resultOne vital portion in the original MDR is the evaluation of element combinations regarding the right classification of circumstances and controls into high- and low-risk groups, respectively. For every model, a two ?2 contingency table (also known as confusion matrix), summarizing the correct negatives (TN), correct positives (TP), false negatives (FN) and false positives (FP), might be developed. As described ahead of, the power of MDR might be improved by implementing the BA rather than raw accuracy, if dealing with imbalanced data sets. Within the study of Bush et al. [77], ten different measures for classification were compared using the normal CE employed in the original MDR technique. They encompass precision-based and receiver operating qualities (ROC)-based measures (Fmeasure, geometric mean of sensitivity and precision, geometric mean of sensitivity and specificity, Euclidean distance from an ideal classification in ROC space), diagnostic testing measures (Youden Index, Predictive Summary Index), statistical measures (Pearson’s v2 goodness-of-fit statistic, likelihood-ratio test) and information theoretic measures (Normalized Mutual Facts, Normalized Mutual Information and facts Transpose). Primarily based on simulated balanced information sets of 40 distinctive penetrance functions with regards to quantity of illness loci (2? loci), heritability (0.five? ) and minor allele frequency (MAF) (0.two and 0.four), they assessed the energy of your unique measures. Their outcomes show that Normalized Mutual Data (NMI) and likelihood-ratio test (LR) outperform the normal CE and the other measures in the majority of the evaluated conditions. Both of those measures take into account the sensitivity and specificity of an MDR model, as a result should not be susceptible to class imbalance. Out of these two measures, NMI is much easier to interpret, as its values dar.12324 variety from 0 (genotype and illness status independent) to 1 (genotype entirely determines illness status). P-values could be calculated from the empirical distributions of your measures obtained from permuted information. Namkung et al. [78] take up these benefits and evaluate BA, NMI and LR using a weighted BA (wBA) and a number of measures for ordinal association. The wBA, inspired by OR-MDR [41], incorporates weights primarily based around the ORs per multi-locus genotype: njlarger in scenarios with smaller sample sizes, larger numbers of SNPs or with small causal effects. Amongst these measures, wBA outperforms all other folks. Two other measures are proposed by Fisher et al. [79]. Their metrics don’t incorporate the contingency table but use the fraction of cases and controls in every cell of a model straight. Their Variance Metric (VM) for a model is defined as Q P d li n 2 n1 i? j = ?nj 1 = n nj ?=n ?, measuring the difference in case fracj? tions among cell level and sample level weighted by the fraction of men and women inside the respective cell. For the Fisher Metric n n (FM), a Fisher’s exact test is applied per cell on nj1 n1 ?nj1 ,j0 0 jyielding a P-value pj , which reflects how unusual every single cell is. To get a model, these probabilities are combined as Q P journal.pone.0169185 d li i? ?log pj . The greater each metrics would be the more likely it’s j? that a corresponding model represents an underlying biological phenomenon. Comparisons of these two measures with BA and NMI on simulated data sets also.
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2018-05-20 15:23:28
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https://zbmath.org/?q=ut%3Arational+function
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2022-05-19 15:15:23
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https://forum.pjrc.com/threads/57326-T4-0-Memory-trying-to-make-sense-of-the-different-regions?s=7b6dfce0f97aad08045dd9e13cbb4411&p=220684
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Forum Rule: Always post complete source code & details to reproduce any issue!
# Thread: T4.0 Memory - trying to make sense of the different regions
1. Hi Paul,
Talking about FlexSPI, anyone already tried to mmap another SPI device as PSRAM, in another region. For the SPI flash, is every block/page read cached in the FASTRUN too?
2. Originally Posted by KurtE
@Paul - I don't know if it makes sense to simply add another utility in the build process, that displays more information.
....
Which I know is outputting far too cryptic of data for most people, like recent run of testing the IO pins of the solderless breakout board...
I'm seriously considering doing something like this, though less cryptic, more like Arduino's format.
But running it from the postbuild recipe prints too soon, probably not even visible without scrolling with the default size of Arduino's console. It needs to print together with Arduino's summary.
But yes, I think RAM1 is better than the current stuff, although I can also imagine there will be some complaining about, I thought I purchased a board with 1mbs of memory why is there only 512KB...
Yeah, I'm concerned about that too.
Originally Posted by Jean-Marc
Talking about FlexSPI, anyone already tried to mmap another SPI device as PSRAM, in another region.
I believe this came up during the beta test. I'm not aware of anyone actually getting it working. It's on my to-do list for Teensy 4.1.
For the SPI flash, is every block/page read cached in the FASTRUN too?
Caching is done in the M7 core, which is separate from RAM1 (aka ITCM or FASTRUN). There are 2 caches, both 32K, one for data and the other for code fetches.
3. Originally Posted by PaulStoffregen
I'm seriously considering doing something like this, though less cryptic, more like Arduino's format.
But running it from the postbuild recipe prints too soon, probably not even visible without scrolling with the default size of Arduino's console. It needs to print together with Arduino's summary.
Yeah, I'm concerned about that too.
For 'Compile DEBUG' output having cryptic stuff can be useful to those looking for it, or for forum debug questions/answers - even with scrolling needed. And those non debug or not interested would never see it just like all the other debug spew.
But indeed if 'Arduino summary' allows better info having an 'english' summary there would be good for all.
4. What exactly happens if you compile for smallest code size? Is all code remaining in Flash? So is it like every function was labeled PROGMEM? It typically free up RAM for me.
5. Smallest code is just a compilation option at this point. Rather than exploding code when it would add speed - it tends to reduce resulting code size.
It might be a future option for Paul to consider to having a default 'CODE FROM FLASH' build option to maximize available RAM. Some mention of alternate linker scripts was noted as possible but not detailed IIRC.
6. @Paul and others... As per earlier question about wondering about what marking some of the functions as FLASHMEM would do to startup time and run time...
I have tried adding flashing of pin 13, which is not showing much time in the copy of memory...
For the heck of it, will try marking of most of the functions for ILI9488_t3 graphic test to be FLASHMEM and see what it does in speed.
Before the changes memory profile:
Code:
FlexRAM section ITCM+DTCM = 512 KB
Config : aaaaaaaf
ITCM : 51600 B (78.74% of 64 KB)
DTCM : 17088 B ( 3.72% of 448 KB)
Available for Stack: 441664
OCRAM: 512KB
DMAMEM: 8272 B ( 1.58% of 512 KB)
Available for Heap: 516016 B (98.42% of 512 KB)
Flash: 64496 B ( 3.17% of 1984 KB)
And timings: with SPI requested for 60mhz...
Code:
ILI9488_t3n: (T4) SPI automatically selected
MOSI:11 MISO:12 SCK:13
ILI9488 Test!
Display Power Mode: 0x0
Pixel Format: 0x0
Image Format: 0x0
Self Diagnostic: 0x0
Benchmark Time (microseconds)
Screen fill 307994
Text 7449
Lines 92433
Horiz/Vert Lines 25838
Rectangles (outline) 14423
Rectangles (filled) 744864
Circles (filled) 99631
Circles (outline) 79308
Triangles (outline) 19609
Triangles (filled) 235923
Rounded rects (outline) 32314
Rounded rects (filled) 817065
Done!
And again I only update the functions within the main sketch, but sizes different:
Code:
FlexRAM section ITCM+DTCM = 512 KB
Config : aaaaaaaf
ITCM : 48208 B (73.56% of 64 KB)
DTCM : 17088 B ( 3.72% of 448 KB)
Available for Stack: 441664
OCRAM: 512KB
DMAMEM: 8272 B ( 1.58% of 512 KB)
Available for Heap: 516016 B (98.42% of 512 KB)
Flash: 64688 B ( 3.18% of 1984 KB)
And timings:
Code:
enchmark Time (microseconds)
Screen fill 307987
Text 7503
Lines 92546
Horiz/Vert Lines 25846
Rectangles (outline) 14430
Rectangles (filled) 744884
Circles (filled) 99789
Circles (outline) 78841
Triangles (outline) 19697
Triangles (filled) 235983
Rounded rects (outline) 32279
Rounded rects (filled) 817101
Done!
So I only saved 3392 bytes, actually I saved no bytes, but saved copying the bytes down as I would need to get down to next increment of 32KB...
And it did impact the timing a slight bit.
P.S. - Need to add FLASHMEM to keywords.txt
7. Just saw this post about the next version of PlatformIO which has some VERY slick tools for investigating firmware symbols, memory allocations, static code analysis, etc.
https://community.platformio.org/t/p...spection/10187
I haven't tried the beta yet with Teensy 4.0, but I hope that it can be extended to support the T4 specific memory features.
8. @KurtE, …
In the case of :: ITCM : 51600 B (78.74% of 64 KB)
Where the 'next increment of 32KB' : does leave a usable piece of RAM - could that be addressed for R/W usage at runtime?
In case above about 13,936 bytes.
The ITCM is marked the same as DTCM in the script as I saw it:
ITCM (rwx): ORIGIN = 0x00000000, LENGTH = 512K
DTCM (rwx): ORIGIN = 0x20000000, LENGTH = 512K
From:
Code:
memory_copy(&_stext, &_stextload, &_etext);
where::
_etext = ADDR(.text.itcm) + SIZEOF(.text.itcm);
_stextload = LOADADDR(.text.itcm);
Would that be::
Code:
uint32_t *SomeMemPtr = _stext + _etext;
uint32_t SizeOfSomeMem = (64*1024) - _etext;
9. Simple sample closer to generic case?
Though I am getting this output:
Code:
SizeOfSomeMem=15685 [KB=15] SizeLeft_etext=15689 len ITCM=3644801719
stext=0 _stextload=0 etext=3644801719
With zero for _stext and _stextload - the numbers can't be right?
From:
Code:
extern unsigned long _stextload;
extern unsigned long _stext;
extern unsigned long _etext;
void setup() {
while (!Serial); // Wait for Arduino Serial Monitor to open
unsigned long SizeLeft_etext = (32 * 1024) - ((_etext-_stext)%(32*1024));
unsigned long *SomeMemPtr = (unsigned long*)((_etext+4)%4);
unsigned long SizeOfSomeMem = SizeLeft_etext -4;
Serial.printf( "\nSizeOfSomeMem=%u [KB=%u]\tSizeLeft_etext=%u \tlen ITCM=%u\n", SizeOfSomeMem,SizeOfSomeMem/1024, SizeLeft_etext, (_etext-_stext) );
Serial.printf( "stext=%u \t_stextload=%u\tetext=%u \n", _stext, _stextload, _etext );
}
void loop() {}
10. Hi @defragster - It is the addresses of these names that is important... So I hacked a second part onto your program:
Code:
extern unsigned long _stextload;
extern unsigned long _stext;
extern unsigned long _etext;
void setup() {
while (!Serial); // Wait for Arduino Serial Monitor to open
unsigned long SizeLeft_etext = (32 * 1024) - ((_etext - _stext) % (32 * 1024));
unsigned long *SomeMemPtr = (unsigned long*)((_etext + 4) % 4);
unsigned long SizeOfSomeMem = SizeLeft_etext - 4;
Serial.printf( "\nSizeOfSomeMem=%u [KB=%u]\tSizeLeft_etext=%u \tlen ITCM=%u\n", SizeOfSomeMem, SizeOfSomeMem / 1024, SizeLeft_etext, (_etext - _stext) );
Serial.printf( "stext=%u \t_stextload=%u\tetext=%u \n", _stext, _stextload, _etext );
SizeLeft_etext = (32 * 1024) - (((uint32_t)&_etext - (uint32_t)&_stext) % (32 * 1024));
SomeMemPtr = (unsigned long*)((_etext + 4) % 4);
SizeOfSomeMem = SizeLeft_etext - 4;
Serial.printf( "\nSizeOfSomeMem=%u [KB=%u]\tSizeLeft_etext=%u \tlen ITCM=%u\n", SizeOfSomeMem, SizeOfSomeMem / 1024, SizeLeft_etext,
((uint32_t)&_etext - (uint32_t)&_stext) );
Serial.printf( "&stext=%x(%u) \t&_stextload=%x(%u)\t&etext=%x(%u) \n",
(uint32_t)&_etext, (uint32_t)&_etext );
}
void loop() {}
Which now gives you this output:
Code:
SizeOfSomeMem=11000 [KB=10] SizeLeft_etext=11004 len ITCM=3985822980
SizeOfSomeMem=9756 [KB=9] SizeLeft_etext=9760 len ITCM=23008
&stext=0(0) &_stextload=60001790(1610618768) &etext=59e0(23008)
11. Originally Posted by KurtE
Hi @defragster - It is the addresses of these names that is important... So I hacked a second part onto your program:
Thanks, As soon as I quit for the night I was figuring I missed the point of passing the address to memory_copy().
Given that correction - is that a valid R/W usable RAM pointer (above the CODE) and with usable size of SizeOfSomeMem ?
12. @defragster - I am not sure if it is valid or not to be able to write into the ITCM memory area or not...
But One thing I have thought about doing, is maybe using more of the memory in the DTCM for buffers and the like.
Currently at startup time we do:
Code:
// Initialize memory
GPIO7_DR_SET = (1<<3); // digitalWrite(13, HIGH);
GPIO7_DR_CLEAR = (1<<3); // digitalWrite(13, LOW);
GPIO7_DR_SET = (1<<3); // digitalWrite(13, HIGH);
memory_clear(&_sbss, &_ebss);
GPIO7_DR_CLEAR = (1<<3); // digitalWrite(13, LOW);
So basically nothing above the address &ebss up to the stack is used for anything, except the stack:
Again back to the modified program I had:
Code:
md /c "D:\\arduino-1.8.10\\hardware\\teensy\\..\\tools\\arm\\bin\\arm-none-eabi-gcc-nm -n C:\\Users\\kurte\\AppData\\Local\\Temp\\arduino_build_434165\\bar.ino.elf | D:\\GITHUB\\imxrt-size\\Debug\\imxrt-size.exe"
FlexRAM section ITCM+DTCM = 512 KB
Config : aaaaaaab
ITCM : 23008 B (70.21% of 32 KB)
DTCM : 12992 B ( 2.64% of 480 KB)
Available for Stack: 478528
OCRAM: 512KB
DMAMEM: 8272 B ( 1.58% of 512 KB)
Available for Heap: 516016 B (98.42% of 512 KB)
Flash: 32528 B ( 1.60% of 1984 KB)
So as I mentioned we have > 475K for the stack which we typically don't use...
So for example if I have a simple ST7789 display 240x240 and wanted a Frame buffer that was in low memory... Today I could explicitly define it and pass it in, but suppose I don't. Today it will malloc it and be in High memory, and DMA screws up...
So was thinking could maybe write a uint32_t at memory location like _ebss at startup with maybe the next address after that location (or maybe rounded up to 16 bytes or...).
Then have a simple function, that maybe looks at that location + desired size and it does not get too close to the current stack address it reserves that memory and updates the value started at _ebss... to be beyond that newly reserved memory...
i.e. a quick and dirty heap (that has no free...)
13. Originally Posted by KurtE
@defragster - I am not sure if it is valid or not to be able to write into the ITCM memory area or not...
...
If the pointer math to this is right >> SomeMemPtr
It is easy to test for the length of >> SizeOfSomeMem
Of course that memory space is a coincidental amount - that may or may not be 2K or 30K based on code in RAM after compile in the uppermost ITCM 32K chunk.
Will have to read the next part again - but having a way to dynamically get use of all RAM is important.
14. Thank you all for the wealth of information. I was going to ask these questions today if it weren't for this thread. I just moved all my projects to the T4.0 from the T3.6, and am now learning how to optimize for the new MCU. Would the opinion of a novice be helpful? I hope so...
I have two difficulties reconciling the T4.0 with the promise of 'Arduino-compatible'. As you all say, the Arduino size info hides what's really going on. Secondly, it is not possible for me to use DMAMEM in a typical Arduino style of C++ (Where hackers are taught to avoid malloc() like the plague and DMA is forever a mystery). From a beginner's perspective all this DTCM, ITCM, OCRAM falls flat and the really important caveat that there are two separate RAM banks is lost. So would it be a worthwhile stopgap to:
1. Show the max ram in Arduino size info as only the FlexRam (512KB)
2. Instead of just printing ".data' will not fit in region DTCM'", also add "consider using DMAMEM for second RAM bank". It seems obvious that this particular error will 99% of the time be hit by beginners like me who didn't deeply consider the memory layout early in their program design.
Hopefully then the very helpful (but hard to find) 'Memory Layout' section in the T4.0 product page will help users solve these issues.
Personally I would love to use the imxrt-size tool, but I don't feel confident I can install it properly on my own. Anyway, I'll still just test my sketches to see if I run out of RAM since manipulating the regions used is quite advanced for me.
15. Teensy 3.6 has 256 KB of RAM - the primary RAM1 area of Teensy 4.0 is 512 KB. That is shared with code note marked as PROGMEM. Unless sketch code exceeds 256KB - the T4 will have more memory available with not other changes.
The memory called DMAMEM is just a name to indicate the upper 512KB RAM2 area - it does have different properties - but not really DMA specific when it comes to general usage. Indeed malloc() isn't generally the best idea when small memory area will be fragmented- or perhaps when used dynamically. But the malloc() code for semi-static runtime allocations certainly won't be a problem to have suitable RAM up to the T4's limit.
It is an unfortunate reality the 1MB RAM is split into the two 512KB blocks - and as such it doesn't yet have a way for the Arduino environment to document the split or usage … yet - though Paul is working on that - and detailing the usage on the teensy40.html page
16. Originally Posted by KurtE
@defragster - I am not sure if it is valid or not to be able to write into the ITCM memory area or not...
...
Looks like read and write works there … Obviously on startup the code is written there to fill ITCM with FLASH CODE - this finds that end - skips a word - then writes to next 32K boundary in sketch.
Adding GET and SHOW region below to any sketch should work, where only getFreeITCM() is needed in sketch to locate and point to space :: *ptrFreeITCM and find usable length :: sizeofFreeITCM
Code:
// _________________________________________________________________________
// GET and SHOW FreeITCM RAM :: uint32_t *ptrFreeITCM and sizeofFreeITCM
uint32_t *ptrFreeITCM; // Set to Usable ITCM free RAM
uint32_t sizeofFreeITCM; // sizeof free RAM in uint32_t units.
uint32_t SizeLeft_etext;
extern unsigned long _stextload; // FROM LINKER
extern unsigned long _stext;
extern unsigned long _etext;
void getFreeITCM() { // end of CODE ITCM, skip full 32 bits
SizeLeft_etext = (32 * 1024) - (((uint32_t)&_etext - (uint32_t)&_stext) % (32 * 1024));
sizeofFreeITCM = SizeLeft_etext - 4;
sizeofFreeITCM /= sizeof(ptrFreeITCM[0]);
ptrFreeITCM = (uint32_t *) ( (uint32_t)&_stext + (uint32_t)&_etext + 4 );
}
void showNumsITCM() {
// sizeofFreeITCM=9180 [#uint32_t=2295] SizeLeft_etext=9184 len ITCM=23584
Serial.printf( "\n\nsizeofFreeITCM=%u [#uint32_t=%u]\tSizeLeft_etext=%u \tlen ITCM=%u\n", sizeofFreeITCM * sizeof(uint32_t),
sizeofFreeITCM, SizeLeft_etext, ((uint32_t)&_etext - (uint32_t)&_stext) );
// &stext=0(0) &_stextload=60001720(1610618656) &etext=5c20(23584)
Serial.printf( "\n&stext=%x(%u) \t&_stextload=%x(%u)\t&etext=%x(%u) \n",
(uint32_t)&_etext, (uint32_t)&_etext );
// &stext=0(0) &etext=5c20(23584) ptrFreeITCM[0]=5c24(23588) pLast=8000(32768)
Serial.printf( "\n&stext=%x(%u) \t&etext=%x(%u) \tptrFreeITCM[0]=%x(%u) \tpLast=%x(%u) \n",
(uint32_t)&_stext, (uint32_t)&_stext, (uint32_t)&_etext, (uint32_t)&_etext,
(uint32_t)ptrFreeITCM, (uint32_t)ptrFreeITCM,
(uint32_t)ptrFreeITCM + sizeofFreeITCM * sizeof(uint32_t), (uint32_t)ptrFreeITCM + sizeofFreeITCM * sizeof(uint32_t) );
}
// GET and SHOW FreeITCM RAM :: uint32_t *ptrFreeITCM and sizeofFreeITCM
// _________________________________________________________________________
void setup() {
while (!Serial); // Wait for Arduino Serial Monitor to open
Serial.println("\n\n++++++++++++++++++++++");
getFreeITCM();
showNumsITCM();
}
#define NUM_SHOW sizeofFreeITCM
void loop() {
uint32_t ii;
for ( ii = 0; ii < NUM_SHOW; ii++ ) {
if ( !(ii % 5) ) Serial.printf( "\npITCM@%x \t", (ptrFreeITCM) + ii );
Serial.printf( "%3u=%x\t", ii, ptrFreeITCM[ii] );
}
Serial.printf( "\nLast pITCM@%x \t", (ptrFreeITCM) + ii - 1 );
for ( ii = 0; ii < NUM_SHOW; ii++ ) {
ptrFreeITCM[ii] = micros();
}
Serial.println("\n ============================");
for ( ii = 0; ii < NUM_SHOW; ii++ ) {
if ( !(ii % 5) ) Serial.printf( "\npITCM@%x \t", (ptrFreeITCM) + ii );
Serial.printf( "%3u=%u\t", ii, ptrFreeITCM[ii] );
}
Serial.printf( "\nLast pITCM@%x \t", (ptrFreeITCM) + ii - 1 );
showNumsITCM();
while (1);
}
Sketch shows initial ITCM values - then writes in micros() to each and shows those values - with boundary calcs shown before and after:
Code:
…
pITCM@7fdc 2270=671af 2271=671af 2272=671af 2273=671af 2274=671af
pITCM@7ff0 2275=671af 2276=671af 2277=671af 2278=671af
Last pITCM@7ffc
============================
pITCM@5c64 0=419260 1=419260 2=419260 3=419260 4=419260
pITCM@5c78 5=419260 6=419260 7=419260 8=419260 9=419260
pITCM@5c8c 10=419261 11=419261 12=419261 13=419261 14=419261
…
pITCM@7fc8 2265=417411 2266=417411 2267=417411 2268=417411 2269=417411
pITCM@7fdc 2270=417411 2271=417411 2272=417411 2273=417411 2274=417411
pITCM@7ff0 2275=417411 2276=417411 2277=417412 2278=417412
Last pITCM@7ffc
sizeofFreeITCM=9116 [#uint32_t=2279] SizeLeft_etext=9120 len ITCM=23648
&stext=0(0) &etext=5c60(23648) ptrFreeITCM[0]=5c64(23652) pLast=8000(32768)
17. OCRAM and DMA operations:
@Paul and others... I know that I have posted about this in the past, and I know I keep seeing questions about it both forum and PMs and ...
And I don't know a correct answer to this...
Put simply: Doing DMA to and from the OCRam (DMAMEM and malloc operations) Sucks!
Or another way of saying it: DMAMEM on T4 stands for Worst memory to use to do DMA operations to and from.
I know during the beta, we could get around many of the issue by editing startup.c and turn off the caching in the line:
Code:
SCB_MPU_RBAR = 0x20200000 | REGION(3); // RAM (AXI bus)
SCB_MPU_RASR = MEM_CACHE_WBWA | READWRITE | NOEXEC | SIZE_1M;
But I know that is throwing the baby out with the Bath water...
There appear to be issues of doing DMA about alignment of the buffers. Like on 32 byte boundaries.... I know some libraries we have we malloc size+32 and then setup the start at 32 byte boundary, which is not the greatest. And it still did not help in continuous update cases.
I know that for DMA out of memory we need to do things like: _dcache_flush(write_data, count);
Again this is fine if you are doing one shot. But where is it documented that mortals will find it and do it?
And then when you do DMA into memory, there is the call: arm_dcache_delete(retbuf, count);
Again where is this documented? Should this be called when you startup the DMA operation? Or when it completes? If it should be done at completion, does this imply the calling code to get the data should first delete the cache memory before loading from it? ...
The problem is again I know, no one configuration will make everyone happy.
In most cases when I use it, I would be happy with it just turned off. Yes Frame buffer updates and like might be slightly slower, but not sure how much as when I do screen updates I have to call flush to write the whole thing out...
I don't know if it is possible to create multiple memory regions in OCRAM, where maybe DMAMEM goes to memory that is NOT WBWA cached? and maybe some other keyword like: CACHEDMEM does?
And user can somehow control where malloc come from?
Likewise in many of our programs today, we end up with lots of DTCM memory available. Like near 400K that is currently only used by STACK. Again wondering if we should have memory allocator out of here? And/Or option to have DMAMEM objects put into this space?
Again sorry for rambling on this.... (again)
18. Question: RAM1 is CPU speed - what speed is RAM2 that it needs to be cached? Some edit to repeat the p#66 ITCM code on DMAMEM blocks could tell me. After posting that code I added another group where the ITCM RAM is filled as follows and each RAM write was 3 clocks apart,
for ( ii = 0; ii < NUM_SHOW; ii++ ) {
ptrFreeITCM[ii] = ARM_DWT_CYCCNT;
}
Doing that across a block too large to cache, or scattered in turn across multiple RAM2/DMAMEM blocks might show the write rate if different than for RAM1. Those could catch a cache flush delay or MCU might choose write through
Paul posted on other thread about the CACHE func in imxrt.h - these don't include disabling cache on a RAM area? - Not sure if that is doable?:
Code:
// Flush data from cache to memory
static inline void arm_dcache_flush(void *addr, uint32_t size)
// Delete data from the cache, without touching memory
static inline void arm_dcache_delete(void *addr, uint32_t size)
// Flush data from cache to memory, and delete it from the cache
static inline void arm_dcache_flush_delete(void *addr, uint32_t size)
As a side ref - the ESP32 has multiple RAM sets as well - this doc goes through details on choosing and allocating : esp-idf/en/latest/api-guides/general-notes.html#memory-layout Not sure any of it directly relates - but they have many of the same issues - except their cache isn't implicated where they note it relates only to IRON ( flash code ) :: Access to this region is transparently cached using two 32kB blocks. So that would not impact DMA usage. Of course ESP only runs at 240 MHz so the RAM may all be CPU speed.
19. Originally Posted by KurtE
Again sorry for rambling on this.... (again)
Thanks for doing it.
especially of interest to me, what is the procedure for continues DMA (say in case of I2S input DMA)? Does ISR needs to address caching (as done in Output_I2S) or not (as done in Input_I2S)?
20. Add to p#66 code using:
Code:
#define DMA_SIZE 9000
DMAMEM uint32_t pDMA[3][DMA_SIZE];
There are some regular MCU cycle waits between 2 and 22 cycles when using that DMAMEM - RAM1 looks to be 2 cycles [below]:
Code:
pDMA[0]@20208ca0 after CNT of 2 Diff CycCnt is 6 with 409571276!=409571282
pDMA[0]@2021a5e0 after CNT of 2 Diff CycCnt is 2 with 409571288!=409571290
pDMA[0]@20223280 after CNT of 1 Diff CycCnt is 8 with 409571290!=409571298
pDMA[0]@2022bf20 after CNT of 1 Diff CycCnt is 2 with 409571298!=409571300
pDMA[0]@204d97e0 after CNT of 78 Diff CycCnt is 11 with 409571454!=409571465
pDMA[0]@204e2480 after CNT of 1 Diff CycCnt is 2 with 409571465!=409571467
pDMA[0]@204eb120 after CNT of 1 Diff CycCnt is 6 with 409571467!=409571473
pDMA[0]@204f3dc0 after CNT of 1 Diff CycCnt is 2 with 409571473!=409571475
pDMA[0]@2051fce0 after CNT of 5 Diff CycCnt is 14 with 409571483!=409571497
pDMA[0]@20528980 after CNT of 1 Diff CycCnt is 2 with 409571497!=409571499
...
pDMA[1]@20208ca0 after CNT of 2 Diff CycCnt is 2 with 409607138!=409607140
pDMA[1]@2023d860 after CNT of 6 Diff CycCnt is 3 with 409607150!=409607153
pDMA[1]@20246500 after CNT of 1 Diff CycCnt is 2 with 409607153!=409607155
pDMA[1]@20260ae0 after CNT of 3 Diff CycCnt is 18 with 409607159!=409607177
pDMA[1]@20269780 after CNT of 1 Diff CycCnt is 2 with 409607177!=409607179
pDMA[1]@20272420 after CNT of 1 Diff CycCnt is 6 with 409607179!=409607185
pDMA[1]@2027b0c0 after CNT of 1 Diff CycCnt is 2 with 409607185!=409607187
pDMA[1]@202b8920 after CNT of 7 Diff CycCnt is 18 with 409607199!=409607217
pDMA[1]@202c15c0 after CNT of 1 Diff CycCnt is 2 with 409607217!=409607219
pDMA[1]@202ed4e0 after CNT of 5 Diff CycCnt is 14 with 409607227!=409607241
pDMA[1]@202f6180 after CNT of 1 Diff CycCnt is 2 with 409607241!=409607243
pDMA[1]@202fee20 after CNT of 1 Diff CycCnt is 6 with 409607243!=409607249
pDMA[1]@20307ac0 after CNT of 1 Diff CycCnt is 2 with 409607249!=409607251
pDMA[1]@20345320 after CNT of 7 Diff CycCnt is 10 with 409607263!=409607273
pDMA[1]@2034dfc0 after CNT of 1 Diff CycCnt is 2 with 409607273!=409607275
...
pDMA[2]@20208ca0 after CNT of 2 Diff CycCnt is 2 with 409644423!=409644425
pDMA[2]@20272420 after CNT of 12 Diff CycCnt is 22 with 409644447!=409644469
pDMA[2]@2027b0c0 after CNT of 1 Diff CycCnt is 2 with 409644469!=409644471
pDMA[2]@202b8920 after CNT of 7 Diff CycCnt is 18 with 409644483!=409644501
pDMA[2]@202c15c0 after CNT of 1 Diff CycCnt is 2 with 409644501!=409644503
pDMA[2]@202ed4e0 after CNT of 5 Diff CycCnt is 14 with 409644511!=409644525
pDMA[2]@202f6180 after CNT of 1 Diff CycCnt is 2 with 409644525!=409644527
pDMA[2]@202fee20 after CNT of 1 Diff CycCnt is 6 with 409644527!=409644533
pDMA[2]@20307ac0 after CNT of 1 Diff CycCnt is 2 with 409644533!=409644535
pDMA[2]@20345320 after CNT of 7 Diff CycCnt is 10 with 409644547!=409644557
Took off the DMAMEM descriptor and running from RAM1 { memaddr changed but not the var name pDMA} it looks like this with the 902 anomoly always there in group [2] - but it shifts a few positions. And the Bank [0] group repeats the 6,2,8,2 pattern. And Bank [1] just shows 2 cycles difference:
Code:
pDMA[0]@20009bb0 after CNT of 2 Diff CycCnt is 6 with 263201668!=263201674
pDMA[0]@2001b4f0 after CNT of 2 Diff CycCnt is 2 with 263201680!=263201682
pDMA[0]@20024190 after CNT of 1 Diff CycCnt is 8 with 263201682!=263201690
pDMA[0]@2002ce30 after CNT of 1 Diff CycCnt is 2 with 263201690!=263201692
DONE after CNT of 8994 Diff CycCnt is 2 with 263219678!=263219680
============================
pDMA[1]@20009bb0 after CNT of 2 Diff CycCnt is 2 with 263219697!=263219699
DONE after CNT of 8998 Diff CycCnt is 2 with 263237693!=263237695
============================
pDMA[2]@20009bb0 after CNT of 2 Diff CycCnt is 2 with 263237709!=263237711
pDMA[2]@287d2790 after CNT of 3955 Diff CycCnt is 902 with 263245619!=263246521
pDMA[2]@287db430 after CNT of 1 Diff CycCnt is 2 with 263246521!=263246523
pDMA[2]@28cde770 after CNT of 146 Diff CycCnt is 3 with 263246813!=263246816
pDMA[2]@28ce7410 after CNT of 1 Diff CycCnt is 2 with 263246816!=263246818
DONE after CNT of 4895 Diff CycCnt is 2 with 263256606!=263256608
============================
RAM1 original test, with pointer to ITCM, indexes from the pointer differently - not as an array - and results in 4 cycles - not 2. Need to use the same Diff CycCnt testing to be sure.
An hour past starting personal sleep cycle - can add loops to repeat within 32KB area a few times and see if I can get the cache to even out the times to prove it working - and post code if any interest.
21. Originally Posted by KurtE
But where is it documented that mortals will find it and do it?
Currently DMA on Teensy 4.0 is not really documented anywhere, except some comments in imxrt.h and messages scattered across this forum. Neither is DMA on Teensy 3.x or LC. Many other things are also sorely in need of documentation. I'm planning to work on many new web page to cover these topics, starting around the end of this month. Ping me in early December...
However, I do not believe any amount of documentation is going to make DMA easily accessible to most people. It's an advanced topic. Using DMA successfully (without extreme luck) requires tough troubleshooting.
Originally Posted by defragster
what speed is RAM2 that it needs to be cached?
The simple answer is 150 MHz, or 1/4 of whatever speed the M7 processor is running.
But the longer answer depends on details of how these buses and the bridges between them work. Sadly, NXP's documentation on those details is rather scant.
22. Thanks @Paul,
Another interesting sub-question is in practice what is the performance differences and trade offs for the different caching options.
Example we currently use: MEM_CACHE_WBWA Which is Write Back memory. What would the speed difference be for using: MEM_CACHE_WT (Write Through)?
Would going to Write Through memory speed up or slow down programs that for example use the DMAMEM or like for a Frame buffer for a display?
Example if ILI9341 display with Frame buffer in upper memory and we do a few graphic operations to update some things on the screen and then call the updateScreen function to output the data in the frame buffer to the display.
My first attempt to make this work was to call the arm_dcache_flush on the entire frame buffer in order to make sure the DMA operation got the real contents. Again I have not tried to test the differences in timing to see what happens if your screen update code for example only touched a quarter of the screen, between having the data output using _WT and not needing the flush versus using the _WBWA and needing to do flush. Example I don't know how the functions/registers like: SCB_CACHE_DCCMVAC timings work and if it differs depending on if any memory in that 32 bytes was updated or not...
Note: That was the first attempt. However I was still having issues with using the arm_dcache_flush when I tried turning on continuous update mode. As when/where should you call the flush? Before I was just doing it when I started up the DMA operation. But then suppose the sketch then does stuff to update the screen, which is of course the idea of continuous updates. I ran into several issues on this, where for example you do something like a fillRect in the middle of the screen. The interesting this is some of the new color makes it through the cache to memory and some does not. So you end up with splats of different colors. There are probably several different ways to solve it, including maybe:
a) Have each of your DmaSettings objects, have their Interrupt on Completion bit set, and do a _dcache_flush operation on them self to reflect the next frame, but again this does not handle any difference in data between now and the next frame. But should only see splotch on one frame.
b) Maybe have each of the above Interrupts try to flush the whole cache again... - Maybe fewer splotches but how long does that take?
c) Have each of our graphic primitives try to do a cache flush for the region of memory they touched.
Or do what I am currently doing and NOT do the DMA from this memory. Currently I have two smaller buffers in DTCM with the DMASettings pointing to these and I copy the first parts of the Frame buffer into each of these buffers and I interrupt on each buffer completion and copy in the next set of pixels... Again I don't like this, but not sure of anything better yet...
23. 150 MHz Ram2 ( .25 of F_CPU ) - thanks Paul.
Updated my test code - I had addresses printed missing an array dimension in above post - so ignore those.
I added code for DMA cycle per write test and did the same for RAM - RAM still shows that anomaly - I added three runs in succession in the code and it hits something odd there? Also is the DMA test run once where cache can't help on first access as it runs through 36KB three times in the arrays, then I did a shorter set of 900 32 bit elements 10 times in succession and the results on the last are about 2.5 cycles instead of 4 cycles on average. I turned off the copious spew on DMA changes and just show the average, then follows the RAM results. The other ODD thing I used the arm_dcache_flush_delete() on the DMA memory area under test - perhaps not to best way but it mad the DMA number below go up a cycle count per write:
Code:
============================ DMA TEST Single
Avg CycCnt for 9000 is 3.983444
============================
Avg CycCnt for 9000 is 4.155334
============================
Avg CycCnt for 9000 is 3.998778
============================
============================ DMA TEST short Repeat
Avg CycCnt for 900 is 2.765556
============================
Avg CycCnt for 900 is 2.490000
============================
Avg CycCnt for 900 is 2.192222
============================
============================ RAM TEST
pRAM[0]@20001080 after run of 2 Diff CycCnt is 7 with 390861160!=390861167
pRAM[0]@20001088 after run of 2 Diff CycCnt is 2 with 390861174!=390861176
pRAM[0]@2000108c after run of 1 Diff CycCnt is 9 with 390861176!=390861185
pRAM[0]@20001090 after run of 1 Diff CycCnt is 2 with 390861185!=390861187
DONE after run of 8994 Diff CycCnt is 2 with 390879173!=390879175
Avg CycCnt for 9000 is 2.001889
============================
pRAM[1]@20009d20 after run of 2 Diff CycCnt is 2 with 390879193!=390879195
DONE after run of 8998 Diff CycCnt is 2 with 390897189!=390897191
Avg CycCnt for 9000 is 2.000000
============================
pRAM[2]@200129c0 after run of 2 Diff CycCnt is 2 with 390897204!=390897206
pRAM[2]@200168f4 after run of 4045 Diff CycCnt is 783 with 390905294!=390906077
pRAM[2]@200168f8 after run of 1 Diff CycCnt is 2 with 390906077!=390906079
pRAM[2]@20016b38 after run of 144 Diff CycCnt is 3 with 390906365!=390906368
pRAM[2]@20016b3c after run of 1 Diff CycCnt is 2 with 390906368!=390906370
DONE after run of 4807 Diff CycCnt is 2 with 390915982!=390915984
Avg CycCnt for 9000 is 2.086889
============================
============================ RAM TEST
pRAM[0]@20001080 after run of 2 Diff CycCnt is 7 with 391222645!=391222652
pRAM[0]@20001088 after run of 2 Diff CycCnt is 2 with 391222659!=391222661
pRAM[0]@2000108c after run of 1 Diff CycCnt is 9 with 391222661!=391222670
pRAM[0]@20001090 after run of 1 Diff CycCnt is 2 with 391222670!=391222672
DONE after run of 8994 Diff CycCnt is 2 with 391240658!=391240660
Avg CycCnt for 9000 is 2.001889
============================
pRAM[1]@20009d20 after run of 2 Diff CycCnt is 2 with 391240678!=391240680
DONE after run of 8998 Diff CycCnt is 2 with 391258674!=391258676
Avg CycCnt for 9000 is 2.000000
============================
pRAM[2]@200129c0 after run of 2 Diff CycCnt is 2 with 391258689!=391258691
pRAM[2]@200166bc after run of 3903 Diff CycCnt is 711 with 391266495!=391267206
pRAM[2]@200166c0 after run of 1 Diff CycCnt is 2 with 391267206!=391267208
DONE after run of 5094 Diff CycCnt is 2 with 391277394!=391277396
Avg CycCnt for 9000 is 2.078778
============================
============================ RAM TEST
pRAM[0]@20001080 after run of 2 Diff CycCnt is 6 with 391571329!=391571335
pRAM[0]@20001088 after run of 2 Diff CycCnt is 2 with 391571341!=391571343
pRAM[0]@2000108c after run of 1 Diff CycCnt is 8 with 391571343!=391571351
pRAM[0]@20001090 after run of 1 Diff CycCnt is 2 with 391571351!=391571353
pRAM[0]@200028ac after run of 1543 Diff CycCnt is 3 with 391574437!=391574440
pRAM[0]@200028b0 after run of 1 Diff CycCnt is 2 with 391574440!=391574442
DONE after run of 7450 Diff CycCnt is 2 with 391589340!=391589342
Avg CycCnt for 9000 is 2.001667
============================
pRAM[1]@20009d20 after run of 2 Diff CycCnt is 2 with 391589358!=391589360
DONE after run of 8998 Diff CycCnt is 2 with 391607354!=391607356
Avg CycCnt for 9000 is 2.000000
============================
pRAM[2]@200129c0 after run of 2 Diff CycCnt is 2 with 391607369!=391607371
pRAM[2]@20016acc after run of 4163 Diff CycCnt is 773 with 391615695!=391616468
pRAM[2]@20016ad0 after run of 1 Diff CycCnt is 2 with 391616468!=391616470
DONE after run of 4834 Diff CycCnt is 2 with 391626136!=391626138
Avg CycCnt for 9000 is 2.085667
============================
24. @KurtE - I uninstalled VS2017 weeks back after 2019 released and you were using it - without install done - and just started - I can't run the imxrt-size.exe from github - your or my old one ...
Code:
imxrt-size.exe - System Error
---------------------------
The code execution cannot proceed because VCRUNTIME140D.dll was not found. Reinstalling the program may fix this problem.
Maybe tomorrow if the download install ever gets done … almost 1 GB of 2.6 … or more …
might be fun to have a readme showing : recipe.hooks.postbuild.4.pattern.windows=cmd /c "{runtime.hardware.path}\..\tools\arm\bin\arm-none-eabi-gcc-nm -n {build.path}\{build.project_name}.elf | T:\Programs\TSet\imxrt-size.exe"
25. @KurtE : VS2019 installed - not built - but github debug imxrt-size.exe works now! And back in Sublimetext to compile and build.
Code:
FlexRAM section ITCM+DTCM = 512 KB
Config : aaaaaaab
ITCM : 23200 B (70.80% of 32 KB)
DTCM : 12992 B ( 2.64% of 480 KB)
Available for Stack: 478528
OCRAM: 512KB
DMAMEM: 8272 B ( 1.58% of 512 KB)
Available for Heap: 516016 B (98.42% of 512 KB)
Flash: 32768 B ( 1.61% of 1984 KB)
Paul - that shows 9K ITCM Fragment that is accessible for use as RAM - but orphaned.
Question: Is it possible or likely that after putting FLASH code in ITCM that area would ever get marked 'No WRITE' ?
I wrote code that got that orphan start address and length and did read/write test to it without issues as it stands.
#### Posting Permissions
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2019-12-08 11:19:09
|
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http://mathhelpforum.com/algebra/10775-positive-integers.html
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# Math Help - Positive Integers
1. ## Positive Integers
12 positive integers are written in a row. The fourth number is 4, and the 12 number is 12. The sum of any three neighboring numbers is 333.Determine the twelve integers
2. Originally Posted by Dragon
12 positive integers are written in a row. The fourth number is 4, and the 12 number is 12. The sum of any three neighboring numbers is 333.Determine the twelve integers
You have the numbers: $A,B,C,4,D,E,F,G,H,I,J,12$
You know that: $4+C+D=333$
Solve to get: $C=329-D$
You also know that: $4+D+E=333$
Solve to get: $E=329-D$
Therefore: $C=E$
Using the same method you get that: $B=D$
So you know that: $C+4+D=333$
Substitute: $C+4+B=333$
Solve: $4=333-B-C$
Note that: $A+B+C=333$
Solve: $A=333-B-C$
Therefore: $A=4$
Using the same methods above you'll find that:
$A=4=F=I$
$B=D=G=J$
$C=E=H=12$
Remember that: $D=333-4-C$
Substitute: $D=333-4-12$
Then: $D=317$
So the pattern is: $4,317,12,4,317,12,4,317,12,4,317,12$
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2014-11-22 21:56:00
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https://crypto.stackexchange.com/questions/31427/theoretical-decryption-of-rsa
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# Theoretical decryption of RSA
I am currently looking into RSA and have some questions. I am interested in theoretical answers, assuming there is infinite time for decryption.
1. Is it necessary to work out the private-key exponent for decryption? Are there no alternative options?
2. Can a message encrypted using RSA be decrypted using only the public key? If so, how? If not, why?
• About 2) if it was possible, the private key would be unnecessary and therefore, if no private information was used to decrypt a ciphertext, the encryption won't stand. The scheme would be completely broken. – ddddavidee Dec 21 '15 at 9:58
• Are we talking about the (deterministic) "textbook RSA"? – fkraiem Dec 21 '15 at 11:02
I am interested in theoretical answers, assuming there is infinite time for decryption.
We call such an attacker a 'computationally unbounded adversary'
1. Is it necessary to work out the private-key exponent for decryption? Are there no alternative options?
One approach that a computationally unbounded adversary could is as go through all possible values $M$, and find one which has $M^e = C \bmod n$; such a message $M$ is the decryption of the ciphertext $C$.
This approach actually takes longer than the more traditional approaches that involve factoring $n$ (and from that, rederiving the decryption exponent); however a computationally unbounded attacker might not care.
1. Can a message encrypted using RSA be decrypted using only the public key? If so, how? If not, why?
Actually, there are a number of different ways that RSA could be decrypted by a computationally unbounded attacker
• They can brute force the message (as listed above)
• They could factor the modulus, and rederive the decrypion exponent
• As pointed out here, they could generate every possible public/private keypair, and pick out the public key that they actually see. Any public key cryptosystem can be broken by this method.
So, if RSA can so easily be broken by a computationally unbounded adversary, why do we still use it? Well, we don't believe that computationally unbounded adversaries exist, and so that's not considered a real world requirement on crypto systems. Actually, if they do exist, we don't have that much in the crypto toolbox that they can't trivially break anyways.
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2019-06-24 09:56:20
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https://chemistry.stackexchange.com/questions/59199/reaction-of-naoh-with-methanol-and-ethanol
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# Reaction of NaOH with methanol and ethanol?
Methanol ($\ce{CH3OH}$) and ethanol ($\ce{CH3CH2OH}$) react with sodium metal ($\ce{Na}$) to form sodium methoxide ($\ce{CH3O^-Na+}$) and sodium ethoxide ($\ce{CH3CH2O^-Na+}$):
$$\ce{2CH3OH + 2Na -> 2CH3O^-Na+ + H2}$$ $$\ce{2CH3CH2OH + 2Na -> 2CH3CH2O^-Na+ + H2}$$
Do methanol and ethanol react with sodium hydroxide ($\ce{NaOH}$) in the same way as sodium do to form sodium methoxide and sodium ethoxide respectively?
• They react, but not "in the same way". The reaction with sodium is a redox reaction (note the changing oxidation state of hydrogen from +1 to 0), whereas that with NaOH is merely acid-base. As for the reaction with NaOH, it will certainly proceed, but in general will probably not go to completion (see also vapid's comment on the answer). – orthocresol Sep 17 '16 at 6:51
• @orthocresol If an electrophile (like CS2) is added to the solution of Methanol in NaOH, do sodium methoxide will attack to CS2 to form sodium salt of xanthate? – Khan Sep 17 '16 at 10:57
The proposed reaction:
$$\ce{CH3OH(l) + NaOH(s) <<=> CH3O^-Na+(s) + H2O(l)}$$
According to wikipedia:
The solid hydrolyzes in water to give methanol and sodium hydroxide
The equilibrium is biased to the left, as @vapid has mentioned in the comments. To quote him:
In the large excess of alcohol, the reaction will be considerably shifted towards alkoxide.
• This answer would benefit from quoting $\mathrm{p}K_\mathrm{a}$ values and how they influence the equilibrium. – Jan Sep 17 '16 at 16:23
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2019-09-21 04:43:00
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https://zbmath.org/?q=ai%3Avalliammal.v+cc%3A80
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## Study of flow past an exponentially accelerated isothermal vertical plate in the presence of chemical reaction.(English)Zbl 1299.80009
Summary: Theoretical study of unsteady flow past an exponentially accelerated infinite isothermal vertical plate with variable mass diffusion has been presented in the presence of homogeneous chemical reaction of first order. The plate temperature is raised to $$T_w$$ and species concentration level near the plate is made to rise linearly with time. The dimensionless governing equations are solved using Laplace-transform technique. The velocity profiles are studied for different physical parameters like chemical reaction parameter, thermal Grashof number, mass Grashof number, $$a$$ and time. It is observed that the velocity increases with increasing values of $$a$$ or $$t$$. But the trend is just reversed with respect to $$K$$.
### MSC:
80A20 Heat and mass transfer, heat flow (MSC2010) 80A32 Chemically reacting flows 80M99 Basic methods in thermodynamics and heat transfer
### Keywords:
exponential; heat transfer; mass diffusion
Full Text:
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2022-07-01 19:58:52
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https://www.gradesaver.com/textbooks/math/precalculus/precalculus-concepts-through-functions-a-unit-circle-approach-to-trigonometry-3rd-edition/chapter-8-polar-coordinates-vectors-section-8-4-vectors-8-4-assess-your-understanding-page-627/49
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## Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)
$\dfrac{3}{5}i-\dfrac{4}{5}j$
Solve for $||v||$ sing the formula $||v||=\sqrt{a_1^2+b_1^2}$ to obtain: $||\textbf{v}||$= $\sqrt {3^{2}+4^{2}}= \sqrt {25}$=5 Thus, the unit vector in the same direction as $\textbf{v}$ is: $u= \dfrac{v}{|\textbf{v}|}$= $\dfrac{1}{5}(3i-4j)$ = $\dfrac{3}{5}i-\dfrac{4}{5}j$.
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2019-12-10 07:23:59
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https://crypto.stackexchange.com/questions/51989/crafting-secure-block-ciphers-that-lack-key-expansion
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# Crafting secure block ciphers that lack key expansion
There was some article published by on the IACR's website that outlined a block cipher that only used the "master" key and no round keys. I can't recall the cipher's name, not that it matters, nor do I know if it had been subjected to cryptanalysis.
If one has a block cipher that has an $n$-bit key and an $n$-bit block size, what would be some ways that a secure one could be made that does not use key expansion?
The Even-Mansour construction gives you a block cipher from public, unkeyed permutations $P_1, \ldots, P_n$. It works like:
$$x \mapsto \cdots P_2(P_1(x \oplus k_0) \oplus k_1) \cdots \oplus k_n$$
In original/traditional Even-Mansour, the round keys $k_i$ are independent. But recently there have been many papers studying the security of various "minimalistic" variants of Even-Mansour (basically any variant where the round keys are not independent).
The most minimal variant uses a "trivial key schedule" where $k_0 = k_1 = \cdots = k_n$. This variant naturally leads to a cipher that has no need for key expansion (key size is naturally equal to the block size). The following recent paper shows the security of 5-round Even-Mansour with trivial key schedule:
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2019-06-15 22:34:16
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https://openmse.com/features-conditioning-oms/life-history/
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## Conditioning with Life-History Information
By default, the life-history information provided to the Management Procedures is determined by the real life-history values specified in the OM (e.g., OM@Linf) and the bias specified in the observation model.
For example:
OM <- new('OM', Albacore, Generic_IncE, Generic_Obs, Perfect_Imp, nsim=4)
Hist <- Simulate(OM, silent=TRUE)
# OM Linf values in last historical year
Hist@SampPars$Stock$Linfarray[,OM@nyears]
## [1] 128.9126 127.0029 130.6198 131.4645
# Linf values provided to MP in this year (ie in simulated Data)
Hist@Data@vbLinf
## [1] 131.5341 126.8888 131.0243 137.1029
# Observation error applied to OM values
Hist@SampPars$Stock$Linfarray[,OM@nyears] * Hist@SampPars$Obs$Linfbias
## [1] 131.5341 126.8888 131.0243 137.1029
Users can provide specific values for the life-history parameters by including them in the Data object. We demonstrate here with Linf, but the same concept applies for all life-history parameters (M, K, L50, etc).
#### Add Life-History Data to OM
Create a blank Data object:
Data <- new('Data')
Add our ’observedvalues forLinf - i.e., the values we wish to provide to the MPs:
Data@vbLinf <- 100
Add Data to OM:
OM@cpars\$Data <- Data
Simulate the spool-up period (now with Data included in the OM):
Hist <- Simulate(OM, silent=TRUE)
Our ‘observed’ life-history parameters are now provided to the MPs:
Hist@Data@vbLinf
## [1] 100 100 100 100
Similar to conditioning with other data, when real life-history values are provided in the Data object, the observation parameters (e.g., Obs@Linfbiascv`) are ignored.
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2021-10-17 13:13:25
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https://math.stackexchange.com/questions/373708/when-is-the-group-of-units-in-mathbbz-n-cyclic
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# When is the group of units in $\mathbb{Z}_n$ cyclic? [duplicate]
This question already has an answer here:
Let $U_n$ denote the group of units in $\mathbb{Z}_n$ with multiplication modulo $n$. It is easy to show that this is a group. My question is how to characterize the $n$ for which it is cyclic. Since the multiplicative group of a finite field is cyclic so for all $n$ prime, it is cyclic. However I believe that for certain composite $n$ it is also cyclic.
Searching through past posts turned up this, where there was an answer containing the sentence "In number-theoretic situations there are coherent things that can be said, and/but in general I think nothing decisive can be said."
What are those number theoretic situations?
Thanks
## marked as duplicate by Rudy the Reindeer, Harish Chandra Rajpoot, user228113, user296602, Claude LeiboviciFeb 7 '16 at 5:23
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
• Gauss showed that $U_n$ is cyclic if and only if $n=2,4,p^k,$ or $2p^k$ for $p$ an odd prime, $k$ a natural number. – Jared Apr 26 '13 at 17:17
• en.wikipedia.org/wiki/… – Alex Provost Apr 26 '13 at 17:19
• Most certainly a duplicate. – lhf Apr 26 '13 at 18:17
• @Lhf The question is certainly not a duplicate of the linked question, since the author is asking additionally a more general question, namely "What are those number theoretic situations?" (where the unit group is cyclic). This is an interesting question that is not addressed at all in the proposed duplicate. – Math Gems Apr 26 '13 at 19:15
• @Shahab: Is this statement is correct "the multiplicative group of a finite field is cyclic so for all n prime, it is cyclic." – Aria Aug 13 '14 at 7:11
## 2 Answers
$U_n$ is cyclic if and only if $n = 1$, $n = 2$, $n = 4$, $n = p^k$ or $n = 2p^k$ where $p$ is any odd prime.
Proving this requires some work, but proofs can be found in many undergraduate textbooks on number theory and abstract algebra.
The basic idea is this. If an integer $n > 1$ has prime factorization $n = p_1^{a_1} \ldots p_t^{a_t}$, then $U_n \cong U_{p_1^{a_1}} \times \cdots \times U_{p_t^{a_t}}$ by the Chinese remainder theorem. Thus to describe the structure of $U_n$, it suffices to consider the case where $n$ is a power of a prime. It is possible to show that $U_{2^k} \cong \mathbb{Z}_2 \times \mathbb{Z}_{2^{k-2}}$ for $k \geq 3$. Also, $U_{p^k}$ is cyclic for any odd prime $p$ and $k \geq 1$. When you have these results, finding the $n$ for which $U_n$ is cyclic is not too difficult.
• Just wondering: is the empty group considered cyclic? – Julien Apr 26 '13 at 20:03
• The «empty group» is not a group. – Mariano Suárez-Álvarez Apr 26 '13 at 20:18
• Groups are always nonempty since they contain the identity. – Mikko Korhonen Apr 26 '13 at 20:35
The group is cyclic when $$n$$ is a power of a prime, or twice a power of a prime, or $$1$$. That's all.
Usually this is put in number-theoretic language: there is a primitive root modulo $$n$$ precisely under the conditions given above. These results are originally due to Gauss (Disquisitiones Arithmeticae).
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2019-06-16 16:47:53
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http://repository.dl.itc.u-tokyo.ac.jp/dspace/handle/2261/21195
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このページ(論文)をリンクする場合は次のURLを使用してください: http://hdl.handle.net/2261/21195
タイトル: Admissible Controllability for Linear Time-delay Systems in Banach Spaces : A problem in game theory 著者: Park, Jong YeoulNakagiri, Shin-IchiYamamoto, Masahiro 発行日: 1989年 出版者: The University of Tokyo 掲載誌情報: Scientific Papers of the College of Arts and Sciences, The University of Tokyo. Vol.39(1989), Page45-55 抄録: We consider a linear control system with time-delays in a reflexive Banach space X : $dx(t)/dt=A_0x(t)+\int_(-n)^0 d\eta (s)x(t+s)+B(t)u(t) a.e. t>0.$ $x(0)=g^0, x(s)=g^1(s) a.e. s\in [-h,0],$ where $where (g^0,g^1)\in X\times L_p([-h,0] ; X), u\in L_q^(loc) (R^+ ; U),$ U is a reflexive Banach space, $p,q\in (1,\infty ),B(t)$ is a family of bounded linear operators on U to X and A$F_0$ generates a C$F_0$-semigroup, η is a Stieltjes measure. Moreover g$F^1$ and u are assumed to be restricted in ${g^1 ; \parallel g^1\parallel L_(p([-h,0]); X)) \leqq \rho$ and ${u ; \parallel u\parallel_ L_(p([0,T]);U)) \leqq \delta } (\rho \delta >0 ).$ For given $x^0,g^0\in X$ and a given time T>0, we discuss admissible controllability problems : (1) to determine independently g$F^1$ (･) and u (･) such that x(0)=g$F^0$ and $x(T)=x^0 or \parallel x(T)-x^0\parallel \leqq \epsilon$ (\epsilon : a given error). (the cooperative type) (2) to determine u (･) for a given g$F^1$ (･) such that x(0)=g$F^0$ and $\parallel x(T)-x^0\parallel \leqq \epsilon$(the noncooperative type) In this paper, for the both types, we establish necessary and sufficient conditions involving ρ and δ, in order that we can find such g$F^1$ and u. These conditions are expressed in terms of the fundamental solution of the homogeneous system with time-delays. URI: http://hdl.handle.net/2261/21195 ISSN: 02897520 出現カテゴリ: Scientific Papers of the College of Arts and Sciences, The University of Tokyo Scientific Papers of the College of Arts and Sciences, The University of Tokyo
この論文のファイル:
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2017-08-20 15:13:48
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https://math.stackexchange.com/questions/3518884/let-f-be-periodic-function-with-fundamental-period-tau-prove-that-function
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# Let $f$ be periodic function with fundamental period $\tau$. Prove that function $f(nx)$ have fundamental period $\frac{\tau}{n}$.
Let $$f$$ be periodic function with fundamental period $$\tau$$, prove that function $$f(nx)$$ have fundamental period $$\frac{\tau}{n}$$.
My idea:
First I need to show that $$\frac{\tau}{n}$$ is a period. $$f(n(x+\frac{\tau}{n}))= f(nx + \tau) = f(nx).$$ But how to show that this is the smallest such? Any idea?
Let $$g(x) = f(nx)$$. You have shown that if $$f$$ has a period $$\tau$$ then $$g$$ has period $${\tau \over n}$$.
The same reasoning shows that if $$g$$ has period $$T$$ then $$f$$ has period $$nT$$.
If $$\tau$$ is a fundamental period of $$f$$ then any $$T$$ period of $$g$$ must therefore satisfy $$nT \ge \tau$$, or in other words, $$T \ge { \tau \over n}$$. Hence $${ \tau \over n}$$ is a fundamental period of $$g$$.
• If $T$ is a period of $g$ then $nT$ is a period of $f$. Since $\tau$ is the fundamental period of $f$ we must have $nT \ge \tau$ otherwise this would contradict $\tau$ being the fundamental period of $f$. – copper.hat Jan 22 '20 at 19:45
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2021-06-19 04:20:47
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http://www.chegg.com/homework-help/questions-and-answers/square-block-mass-m-falling-velocity-v1-strikes-msall-obstruction-b-assuming-impact-corner-q1316341
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A square block of mass m is falling with a velocity V1, when it strikes a msall obstruction at B. Assuming that the impact between corner A and the obstruction B is perfectly plastic, determine immediately after the impact (a) the angular velocity of the block and (b) the velocity of its mass center G.
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2016-09-29 07:02:01
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https://www.gamedev.net/forums/topic/701604-cscriptarray-memory-leak-on-inserted-items-from-c-side/
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# CScriptArray memory leak on inserted items from C++ side?
## Recommended Posts
Hello! I'm experiencing a couple of memory leaks and would like to see if anyone knows the correct way to clean them up.
The second leak concerns an issue with cscriptarray (version 2.33.0). Basically, I'm finding when I do something like this (result is of type CScriptArray, and AEntity is a custom pass-by-reference that will register with the garbage collector upon construction):
result->InsertLast(new AEntity(engine_ptr, *iter));
When this line is ran, AEntity is inserted as expected, but when the script ends, Valgrind reports the AEntity instances are not cleaned up, but the CScriptArrays are. AEntity instances outside CScriptArrays are cleaned up fine. I'm thinking I'm doing something wrong, but am not sure what. If I just missed something in the documentation, feel free to tell me and I'll go recheck. Does anyone have any ideas what I'm doing wrong? Thanks in advance for the help!
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InsertLast will add one reference to reference counted objects. Assuming AEntity starts off with a reference count of 1 you're leaking the reference by not doing Release() after adding it to the array.
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This fixed both my problems. Thanks for your quick response!
A followup question: In these situations, where I know I don't want to hold a reference to the object on the C++ side, is it safe to have the GC count start out as 0 when it registers with the engine? Or does it need to be 1 at first to prevent any weird GC problems, and then be decremented after? I'd like to automate this as much as possible behind constructors to keep the problem from returning.
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The reference count should always be valid, otherwise you could end up with edge cases where objects have a count of 0 while being treated as valid. That could cause crashes.
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When you put it that way, it makes complete sense. Easy enough to do, and at least it sticks out in Valgrind if done wrong. Thanks a lot for your help! This is something I can document pretty easily as 'rules for working with AngelScript' in my code. Oddly enough, I've found just about everything else in AngelScript easy to work with.
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2019-05-22 01:02:49
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http://www.mywordsolution.com/question/please-read-the-given-news-and-write-2-two/9394
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+61-413 786 465
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Assignment:
1) How we put satellites in orbit and
2) Why satellites should eventually fall back to earth.
http://news.yahoo.com/nasa-gives-1-hour-window-falling-satellite-030650758.html
CAPE CANAVERAL, Florida (AP) — The 6-ton NASA satellite on a collision course with Earth clung to space Friday, apparently flipping position in its ever-lower orbit and stalling its death plunge.
The old research spacecraft was targeted to crash via the atmosphere sometime Friday night or early Saturday, putting Canada and Africa in the potential crosshairs, though most of the satellite should burn up during re-entry. The United States was not entirely out of the woods; the possible strike zone skirted Washington State.
‘It just doesn't want to come down,’ said Jonathan McDowell of the Harvard-Smithsonian Center for the Astrophysics.
McDowell state the satellite's delayed demise describes how unreliable predictions can be. That said, ‘The best guess is that it will still splash in the ocean, just because there is more ocean out there.’
Until Friday, increased solar activity was causing the atmosphere to expand and the 35-foot, bus-size satellite to free fall more rapidly. However late Friday morning, NASA said the sun was no longer the main factor in the rate of descent and that the satellite's position, shape or both had modified by the time it slipped down to a 100-mile orbit.
‘In the last 24 hrs, something has happened to the spacecraft,’ said NASA orbital debris scientist Mark Matney.
On Friday night, NASA said it predicted the satellite to come crashing down between 11:45 p.m. and 12:45 a.m. EDT Saturday. This was going to be passing over the Atlantic, Pacific and Indian oceans at that time, and also Canada and Africa.
‘The risk to public safety is very remote,’ NASA said in a statement.
Any surviving wreckage is expected to be limited to a 500-mile swath.
The Upper Atmosphere Research Satellite, or UARS, will be the biggest NASA spacecraft to crash back to Earth, uncontrolled, as the post-Apollo 75-ton Skylab space station and the more than 10-ton Pegasus 2 satellite, both in 1979.
Russia's 135-ton Mir space station slammed via the atmosphere in 2001; however it was a controlled dive into the Pacific.
Some 26 pieces of the UARS satellite - representing 1,200 pounds of heavy metal - are expected to rain down somewhere. The biggest surviving chunk should be no more than 300 pounds (that is, 136 kilograms).
Earthlings can take comfort in the fact that no one has ever been hurt by falling space junk - to anyone's knowledge - and there has been no serious property damage. NASA put the chances that somebody somewhere on Earth would get hurt at 1-in-3,200. But any one person's odds of being struck were estimated at 1-in-22 trillion; given there are 7 billion people on the planet.
‘Keep in mind that we have bits of debris re-entering the atmosphere every single day,’ Matney said in short remarks broadcast on NASA TV.
In any case, finders definitely are not keepers.
Any surviving wreckage belongs to the NASA, and it is against the law to keep or sell even the smallest piece. There are no toxic chemicals on board, however sharp edges could be dangerous, so the space agency is warning the public to keep hands off and call police.
|
2021-01-23 10:42:39
|
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|
https://iacr.org/cryptodb/data/author.php?authorkey=10415
|
## CryptoDB
### Sruthi Sekar
#### Publications
Year
Venue
Title
2021
CRYPTO
We introduce Adaptive Extractors, which unlike traditional randomness extractors, guarantee security even when an adversary obtains leakage on the source \textit{after} observing the extractor output. We make a compelling case for the study of such extractors by demonstrating their use in obtaining adaptive leakage in secret sharing schemes. Specifically, at FOCS 2020, Chattopadhyay, Goodman, Goyal, Kumar, Li, Meka, Zuckerman, built an adaptively secure leakage resilient secret sharing scheme (LRSS) with both rate and leakage rate being $\mathcal{O}(1/n)$, where $n$ is the number of parties. In this work, we build an adaptively secure LRSS that offers an interesting trade-off between rate, leakage rate, and the total number of shares from which an adversary can obtain leakage. As a special case, when considering $t$-out-of-$n$ secret sharing schemes for threshold $t = \alpha n$ (constant $0<\alpha<1$), we build a scheme with constant rate, constant leakage rate, and allow the adversary leakage from all but $t-1$ of the shares, while giving her the remaining $t-1$ shares completely in the clear. (Prior to this, constant rate LRSS scheme tolerating adaptive leakage was unknown for \textit{any} threshold.) Finally, we show applications of our techniques to both non-malleable secret sharing and secure message transmission.
2019
PKC
The notion of Registration-Based Encryption (RBE) was recently introduced by Garg, Hajiabadi, Mahmoody, and Rahimi [TCC’18] with the goal of removing the private-key generator (PKG) from IBE. Specifically, RBE allows encrypting to identities using a (compact) master public key, like how IBE is used, with the benefit that the PKG is substituted with a weaker entity called “key curator” who has no knowledge of any secret keys. Here individuals generate their secret keys on their own and then publicly register their identities and their corresponding public keys to the key curator. Finally, individuals obtain “rare” decryption-key updates from the key curator as the population grows. In their work, they gave a construction of RBE schemes based on the combination of indistinguishability obfuscation and somewhere statistically binding hash functions. However, they left open the problem of constructing RBE schemes based on standard assumptions.In this work, we resolve the above problem and construct RBE schemes based on standard assumptions (e.g., CDH or LWE). Furthermore, we show a new application of RBE in a novel context. In particular, we show that anonymous variants of RBE (which we also construct under standard assumptions) can be used for realizing abstracts forms of anonymous messaging tasks in simple scenarios in which the parties communicate by writing messages on a shared board in a synchronized way.
2019
JOFC
Non-malleable codes (NMCs), introduced by Dziembowski, Pietrzak and Wichs (ITCS 2010), provide a powerful guarantee in scenarios where the classical notion of error-correcting codes cannot provide any guarantee: a decoded message is either the same or completely independent of the underlying message, regardless of the number of errors introduced into the codeword. Informally, NMCs are defined with respect to a family of tampering functions $\mathcal {F}$ F and guarantee that any tampered codeword decodes either to the same message or to an independent message, so long as it is tampered using a function $f \in \mathcal {F}$ f ∈ F . One of the well-studied tampering families for NMCs is the t -split-state family, where the adversary tampers each of the t “states” of a codeword, arbitrarily but independently. Cheraghchi and Guruswami (TCC 2014) obtain a rate-1 non-malleable code for the case where $t = \mathcal {O}(n)$ t = O ( n ) with n being the codeword length and, in (ITCS 2014), show an upper bound of $1-1/t$ 1 - 1 / t on the best achievable rate for any t -split state NMC. For $t=10$ t = 10 , Chattopadhyay and Zuckerman (FOCS 2014) achieve a constant-rate construction where the constant is unknown. In summary, there is no known construction of an NMC with an explicit constant rate for any $t= o(n)$ t = o ( n ) , let alone one that comes close to matching Cheraghchi and Guruswami’s lowerbound! In this work, we construct an efficient non-malleable code in the t -split-state model, for $t=4$ t = 4 , that achieves a constant rate of $\frac{1}{3+\zeta }$ 1 3 + ζ , for any constant $\zeta > 0$ ζ > 0 , and error $2^{-\varOmega (\ell / log^{c+1} \ell )}$ 2 - Ω ( ℓ / l o g c + 1 ℓ ) , where $\ell$ ℓ is the length of the message and $c > 0$ c > 0 is a constant.
2018
EUROCRYPT
2017
TCC
|
2021-09-25 22:01:57
|
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|
http://www.physicsforums.com/showthread.php?p=3723023
|
Recognitions:
Homework Help
## Wikipedia blackout
Quote by Vanadium 50 Does he? Or does he have to show that there was one lost sale?
Well, he did use the word "substantial", didn't he?
Recognitions: Homework Help It's good if we got back to the issues here. I freely concede that copyright infringement is legally wrong (in many jurisdictions). I am also willing to concede that it is morally wrong, although nowhere near as abhorrent as actual theft (in whatever form). Given those, the law is justified in trying to shut down copyright infringement (piracy is another emotionally-laden term, so I'll eschew using it). But how extensive should their powers to do this be? This is the crux of the SOPA/PIPA debate. Most people feel that these bills, if passed, would place too much power into the hands of bullies (as they've repeatedly shown themselves to be). *This* is what this debate should be about, so please let's get it back on track (I'm sorry for my part in having skewed it off track).
Blog Entries: 1 Recognitions: Gold Member Just one kind of copyright infringement. Cost of movie piracy
Quote by Jimmy Snyder Don't be silly. The monetary loss in copyright infringement can be substantial.
I didn't say it wasn't. I said that if I photocopy your book, YOU STILL HAVE YOUR BOOK. If I copy your software, YOU STILL HAVE YOUR SOFTWARE.
If I steal them, you no longer have them. Thats what I meant by "loss." Not indirect revenue considerations.
As far as I can tell, there is absolutely no way to enforce IP more effectively that doesn't result in enormous costs/dead-weight loss (in this case born by the taxpayer). We already have enough enforcement that entertainment industry is very profitable. The average taxpayer is enormously entertained: we are suffering no loss of entertainment product.
So, why should I (as a taxpayer) have to pay more taxes AND deal with the annoyances of a potentially fragmented internet FOR NO BENEFIT?
Just one kind of copyright infringement...
The source's numbers are hokey. Keep in mind that when someone downloads a movie instead of paying for it, they eventually spend that $1 (I think thats the redbox rate?) somewhere else. Piracy redistributes resources, it doesn't remove them from existence. When I watch a movie my neighbor rented, I don't simultaneously set my cash on fire. I use it to buy popcorn or chips or whatever I'm bringing over. Less jobs for redbox, perhaps, more jobs for the grocery store. Recognitions: Gold Member Quote by Jimmy Snyder Just one kind of copyright infringement. Cost of movie piracy Where's the methods for their calculation? Blog Entries: 1 Recognitions: Gold Member Quote by ParticleGrl Keep in mind that when someone downloads a movie instead of paying for it, they eventually spend that$1 (I think thats the redbox rate?) somewhere else.
Small comfort to the owner of the copyright.
Small comfort to the owner of the copyright.
I can't help but feeling your are simply trolling. Instead of addressing the meat of anything I've said, you are taking quick potshots.
My point was that any numbers that say $x lost to the economy/$y dollars lost in tax revenue,z jobs destroyed should be immediately suspicious.
Recognitions: Gold Member no methodology, bleeding heart global crisis.... makes me suspect bias research.
Recognitions: Gold Member I see a lack of reasonable discussion from at least one side here. This thread needs to be locked. Maybe someday we can have a reasonable discussion, from both sides, about piracy.
Recognitions:
Gold Member
Quote by Char. Limit I see a lack of reasonable discussion from at least one side here. This thread needs to be locked. Maybe someday we can have a reasonable discussion, from both sides, about piracy.
Simply report the offending posters to the moderators :)
Quote by ParticleGrl My point was that any numbers that say $x lost to the economy/$y dollars lost in tax revenue,z jobs destroyed should be immediately suspicious.
Your points make sense to me. Here's a thought:
Maybe the net effect of internet piracy of movies and music is that it actually benefits the general economy. This conjecture is based on the assumptions that (1) a significant portion of the revenues from sales of dvd's and cd's is kept in the financial sector, and (2) virtually all of the money saved by downloaders of pirated stuff is spent in the general economy.
Quote by ThomasT This conjecture is based on the assumptions that (1) a significant portion of the revenues from sales of dvd's and cd's is kept in the financial sector
Do you mean entertainment sector, rather than financial sector? If you mean financial, why do you believe this?
Its not obvious to me that having less money in the hands of piraters/more money in the hands of equity holders in the entertainment industry is somehow worse overall. Its just different. Having less grocery store workers and more entertainment workers isn't obviously bad, but (and this is important) having more grocery store workers and less entertainment workers isn't obviously bad. If we ever start suffering from a lack of entertainment, we will certainly need to push more people into that sector. Cross that bridge when we come to it, its not now.
The problem with trying to police IP too strictly is that you end up with less overall money in entertainment + everywhere else. That gets siphoned off to pay the IP police. If there is no problem that needs solving why should we do this?
Quote by ParticleGrl Do you mean entertainment sector, rather than financial sector? If you mean financial, why do you believe this?
I meant financial. The money spent in the entertainment sector wrt labor, materials, technology, etc. is mostly, I would suppose, eventually entering and positively affecting the general economy. But I'm also assuming that a (significant) portion of the profits from a dvd or cd are invested in the financial sector.
Quote by ParticleGrl Its not obvious to me that having less money in the hands of piraters/more money in the hands of equity holders in the entertainment industry is somehow worse overall.
I'm assuming that piraters spend a greater percentage of their liquid assets in the general economy than do equity holders in the entertainment industry -- and that this translates to more money in the general economy.
Quote by ParticleGrl Having less grocery store workers and more entertainment workers isn't obviously bad, but (and this is important) having more grocery store workers and less entertainment workers isn't obviously bad. If we ever start suffering from a lack of entertainment, we will certainly need to push more people into that sector. Cross that bridge when we come to it, its not now.
I don't think that we're going to have to worry about a lack of entertainment. My take is that the entertainment industry hasn't really suffered from internet piracy. If revenues are down, then maybe that's mostly attributable to a downturn in the general economy.
Maybe the entertainment industry will make more money if all internet pirating is shut down. It's an empirical question, but, imho, not a particularly interesting or important one. Most of the, possible, increased revenues won't be going to the creative artists anyway, but to the big corporations that control them.
Quote by ParticleGrl The problem with trying to police IP too strictly is that you end up with less overall money in entertainment + everywhere else.
I don't think that the policing/enforcement of internet piracy is so much a matter of money as of priorities. Imho, it's just way way down on the list. But, apparently, the lobbying money of the entertainment industry has been well spent so far. Pressure has been put on the DoJ to pay attention to this problem, and it's responding predictably.
I don't think that strictly policing IP necessarily means less money in the general economy. I do think that more money spent on dvd's and cd's means less money in the general economy.
Quote by Hurkyl
http://blog.wikimedia.org/2012/01/16...ut-january-18/
Very unfortunate -- they lose a lot of standing in my own eyes.
When I see things like this, one of the first things I look for is whether they are taking a reasonable position, or if they are taking an infeasible cartoonish position.
All around the world, we're seeing the development of legislation intended to fight online piracy, and regulate the Internet in other ways, that hurt online freedoms ... We want the Internet to remain free and open, everywhere, for everyone.
and this quote looks like they're taking the cartoon position: that any laws and regulation regarding the internet should be rejected on pure principle.
I don't know anything about the particular laws they're protesting -- and their stated reasons for protest do not fill me with confidence that their protest has merit. In fact, such extreme positions have a counter-productive effect from me -- they've pushed me from apathy to actually feeling antagonistic to their cause.
I really hope that the editors just dropped the ball on this one, rather than this being a sign of Wikipedia's political direction....
Hi Hurkyl I think that Wikipedia along with quite a few other organizations are protesting for *good* reason. One thing is that it might shut down online libraries. The more I think about it that would mean to me the public wouldn't have access to the Library of Congress whose mission is: "The Library's mission is to support the Congress in fulfilling its constitutional duties and to further the progress of knowledge and creativity for the benefit of the American people."(1) And take a peek at the Library link (url) below as noted in my #2. It has a section on Film and Sound Recordings. It makes me wonder what the heck is going on. Honestly, I see a conflict of interest regarding what Congress was attempting to do but thank goodness the President stepped in right away and put a hold on that stuff for the time being.
2. http://www.loc.gov/index.html
###
Here's a very impressive website that lists organizations and people opposing SOPA and PIPA: 'List of Those Expressing Concern With SOPA & PIPA'
http://www.cdt.org/report/list-organ...-opposing-sopa
All around the world, we're seeing the development of legislation intended to fight online piracy, and regulate the Internet in other ways, that hurt online freedoms ... We want the Internet to remain free and open, everywhere, for everyone.
I am not sure you should describe that as 'cartoonish,' not all principles can be disregarded that easily. Beside the pragmatic consequences, I too do feel that information should be 'maximally free.' Because I believe an open unregulated Internet is a great asset for humanity. If that means giving up on copyright law, I personally couldn't care less. I don't even think ideas should be patented. Artists will manage to make money in another manner no matter what.
Blog Entries: 1
Recognitions:
Homework Help
The more I think about it that would mean to me the public wouldn't have access to the Library of Congress
Let's not go overboard here. There's no way the Attorney General decides to shut down the Library of Congress's website, even if he's given the power to do so
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2013-05-20 07:39:46
|
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http://www.acmerblog.com/POJ-3187-Backward-Digit-Sums-blog-988.html
|
2013
11-12
# Backward Digit Sums
FJ and his cows enjoy playing a mental game. They write down the numbers from 1 to N (1 <= N <= 10) in a certain order and then sum adjacent numbers to produce a new list with one fewer number. They repeat this until only a single number is left. For example, one instance of the game (when N=4) might go like this:
3 1 2 4
4 3 6
7 9
16
Behind FJ’s back, the cows have started playing a more difficult game, in which they try to determine the starting sequence from only the final total and the number N. Unfortunately, the game is a bit above FJ’s mental arithmetic capabilities.
Write a program to help FJ play the game and keep up with the cows.
Line 1: Two space-separated integers: N and the final sum.
Line 1: An ordering of the integers 1..N that leads to the given sum. If there are multiple solutions, choose the one that is lexicographically least, i.e., that puts smaller numbers first.
4 16
3 1 2 4
Explanation of the sample:
There are other possible sequences, such as 3 2 1 4, but 3 1 2 4 is the lexicographically smallest.
//* @author: 82638882@163.com
import java.io.*;
import java.util.Arrays;
public class Main
{
static int[] p,arr;
static int a;
public static void main(String[] args) throws IOException
{
a=Integer.parseInt(ss[0]);
int sum=Integer.parseInt(ss[1]);
p=new int[a];
arr=new int[a];
for(int i=0;i< a;i++)
p[i]=i+1;
for(int u=0;;u++)
{
if(sum==get(arr))
{
for(int i=0;i< a;i++)
System.out.print(p[i]+" ");
break;
}
next();
}
}
static void next()
{
for(int i=a-1;i>=0;i--)
{
for(int j=a-1;j>i;j--)
{
if(p[j]>p[i])
{
int temp=p[j];
p[j]=p[i];
p[i]=temp;
Arrays.sort(p,i+1,a);
return;
}
}
}
}
static int get(int[] arr)
{
int b=a;
for(int i=0;i< b;i++)
arr[i]=p[i];
while((b--)!=0)
{
for(int i=0;i< b;i++)
arr[i]=arr[i]+arr[i+1];
}
return arr[0];
}
}
1. Good task for the group. Hold it up for every yeara??s winner. This is a excellent oppotunity for a lot more enhancement. Indeed, obtaining far better and much better is constantly the crucial. Just like my pal suggests on the truth about ab muscles, he just keeps obtaining much better.
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2017-02-27 11:48:28
|
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|
https://electronics.stackexchange.com/questions/225669/altera-quartus-design-assistant-critical-warnings
|
Altera Quartus Design Assistant Critical Warnings
I get a number of Critical Warnings with respect to lpm_ff and lpm_counter:
Below are few:
Rule A102: Register output should not drive its own control signal directly or through combinational logic - Structure 1
App:inst1|Exp_Box:inst2|lpm_ff:ff_Last_Image|dffs[0]
App:inst1|Exp_Box:inst2|_~3
App:inst1|Exp_Box:inst2|lpm_ff:ff_event_HStrig|dffs[0]
App:inst1|Expe_Box:inst2|lpm_counter:cnt_ext_HStrig|cntr_6ik:auto_generated|cout_actual
Can these warnings be ignored? Moreover I am using Altera Quartus 11.1 Web Edition. Could it be because of Open Core Plus?
ff_event_HStrig.clock = System_Clock;
ff_event_HStrig.data[] = B"1";
ff_event_HStrig.enable = (red_HS_trigger OR (dl_enabled AND HS_trig_deglitch)) AND (SW_Con_en OR
SW_KK_en OR
SW_KK_ro_en OR
SW_Ser_en);
ff_event_HStrig.aclr = !n_Reset OR !HS_trig_deglitch OR fed_gen_ready OR !HW_Con_en_LIHext; -- reset HS trigger signal
HS_trigger = ff_event_HStrig.q[0];
cnt_ext_HStrig.clock = System_Clock;
cnt_ext_HStrig.cnt_en = 10kHz_Pulse AND !cnt_ext_HStrig.cout;
cnt_ext_HStrig.aclr = !n_Reset OR !ff_ext_HStrig.q[0];
---------------------------------------------------------------------------
ff_ext_HStrig.clock = System_Clock;
ff_ext_HStrig.data[] = B"1";
ff_ext_HStrig.enable = red_HS_trigger;
ff_ext_HStrig.aclr = !n_Reset OR cnt_ext_HStrig.cout;
HS_trig_activated = ff_ext_HStrig.q[0];
---------------------------------------------------------------------------
cnt_Last_Image.clock = System_Clock;
cnt_Last_Image.cnt_en = 100Hz_Pulse AND ff_Last_Image.q[0] AND !cnt_Last_Image.cout;
cnt_Last_Image.aclr = !n_Reset OR HW_Con_en;
ff_cnt_LIH_out.clock = System_Clock;
ff_cnt_LIH_out.data[0] = cnt_Last_Image.cout;
ff_cnt_LIH_out.enable = B"1";
ff_cnt_LIH_out.aclr = !n_Reset OR HW_Con_en;
cnt_Last_Image_fix.clock = System_Clock;
cnt_Last_Image_fix.cnt_en = 100Hz_Pulse AND ff_Last_Image.q[0] AND !cnt_Last_Image_fix.cout;
cnt_Last_Image_fix.aclr = !n_Reset OR HW_Con_en;
ff_Last_Image.clock = System_Clock;
ff_Last_Image.data[0] = B"1";
ff_Last_Image.enable = fed_HW_Con_en AND LIH_en;
ff_Last_Image.aclr = !n_Reset OR (cnt_Last_Image.cout AND (!HS_trigger OR LIH_ro)) OR !LIH_en OR HW_Con_en OR cnt_Last_Image_fix.cout;
LIH_active = ff_Last_Image.q[0];
LIH_activated = LIH_active;
Regards!
• What code are you using? Do you have a register output driving its own control signal? – Tom Carpenter Mar 31 '16 at 12:51
• Unfortunately I have an old code written in AHDL cnt_ext_HStrig.clock = System_Clock; cnt_ext_HStrig.cnt_en =10kHz_Pulse AND !cnt_ext_HStrig.cout; cnt_ext_HStrig.aclr =!n_Reset OR !ff_ext_HStrig.q[0]; Yes. Here the cout is a register output I believe. If this is not allowed I suppose I should change the logic. Or? – Alex Krish Mar 31 '16 at 13:02
• Please add the code to your question and also show the code for ff_Last_Image and ff_event_HString. Cout is a combinational output of the counter. The warnings refer likely to the asynchronous clear aclr which is driven by the registered counter output q[0] as similar depicted in this help page for Rule A102. – Martin Zabel Apr 1 '16 at 8:33
• @MartinZabel I have attached the code above. Yes. You are right. the problem is because of the aclr driven by the counter output. Is there any way how I can resolve this without altering the logic? I tried using a register to drive the output to the aclr. Did not work though. – Alex Krish Apr 1 '16 at 9:34
Care has to be taken when using the asynchronous clear input of flip-flops:
• The reset pulse must be long enough as defined in the datasheet, to reliably reset the flip-flop.
• Reset pulses shorter than the minimum pulse width may lead to an unknown state. Thus, asynchronous inputs should not be driven from combinational logic because the output of the combinational logic (which drives the asynchronous clear) may toggle several times until it settles to the final value, also known as glitch.
• The asyncronous clear should also not be connected to the output Q of the flip-flop because when Q rises (after a clock edge) and reaches the threshold for logic high, a reset is applied and Q falls again. This leads to short reset pulses which may violate the minimum pulse width.
A good design practice is to connect the asynchronous clear only to a global reset signal which is de-asserted synchronously to the active edge of the clock. Then the asynchronous clear of all connected flip-flops are released at the same time, and enough time before the next active clock edge. Timing constraints are required to ensure this.
To de-assert the reset synchronously, you should use a reset synchronizer which is composed of two flip-flops connected in series:
signal reset_ff : std_logic_vector(1 downto 0) := "00"; -- low-active
signal reset_global : std_logic; -- high-active
process(n_Reset, System_Clock)
begin
if n_Reset = '0' then
reset_ff <= (others => '0');
elsif rising_edge(System_Clock) then
reset_ff <= reset_ff(0 downto 0) & '1'; -- release reset
end if;
end process;
reset_global <= not reset_ff(1); -- high-active
More information can be found in the paper Asynchronous & Synchronous Reset Design Techniques - Part Deux from Cliff Cummings (Sunburst Design) et.al. .
The reset synchronizer would allow to reset the design only globally, e.g., from an external input (button). To reset only some flip-flops, one should use the synchronous reset sclr instead. The timing analyzer will check that all contraints are met.
But, you cannot easily replace aclr by sclr because the synchronous clear will not take place until the next active clock edge. Thus, you have to likely issue the synchronous clear in the clock period before.
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2020-10-27 17:56:31
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http://libros.duhnnae.com/2017/jun6/149798150887-Field-dependent-quasiparticles-in-a-strongly-correlated-local-system-II-Condensed-Matter-Strongly-Correlated-Electrons.php
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# Field dependent quasiparticles in a strongly correlated local system II - Condensed Matter > Strongly Correlated Electrons
Field dependent quasiparticles in a strongly correlated local system II - Condensed Matter > Strongly Correlated Electrons - Descarga este documento en PDF. Documentación en PDF para descargar gratis. Disponible también para leer online.
Abstract: We extend the renormalized quasiparticle description of the symmetricAnderson model in a magnetic field $H$, developed in earlier work, to thenon-symmetric model. The renormalized parameters are deduced from the lowenergy NRG fixed point for arbitrary field values. We find quasiparticleresonance widths, $\tilde\Delta \sigma(H)$, which depend in general on the spin$\sigma$ as well as $H$. The low temperature static properties can be expressedcompletely in terms of these parameters, which can also be used as inputs for arenormalized perturbation theory. We show that taking into account repeatedquasiparticle scattering gives results for the longitudinal and transversedynamic spin susceptibilities which are in very good agreement with thoseobtained from direct NRG calculations.
Autor: J. Bauer, A.C. Hewson
Fuente: https://arxiv.org/
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2018-01-22 20:33:45
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https://www.biostars.org/p/257732/#300783
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How to improve existing transcriptome annotation using cufflink protocol ?
5
0
Entering edit mode
5.2 years ago
Chirag Parsania ★ 1.9k
I am working on fungi transcriptome. Existing annotation for my species is not reliable. To improve the annotation we generated RNAseq data with different stress conditions to make all transcripts to express in different conditions and using those expressed transcripts in different conditions I am trying to build one comprehensive transcriptome using cufflink protocol.
I ran cufflink and cuffmerge with and without reference in following combinations.
1) cufflink with reference cuffmerge with reference 2)cufflink with reference cuffmerge without reference 3)cufflink without reference cuffmerge with reference 4)cufflink without reference cuffmerge without reference
All of them has it's pros and cons. By doing some manual observation on IGV i decided to go with 4th combination which is cufflink without reference cuffmerge without reference. Still I am not satisfied the way it did annotations
For example consider following scenario (attached image). Though I have clearly 3 transcripts expressed in all my conditions (shown in the green box) none of the above combinations of cufflink and cuffmerge could detect the true transcript structure.
Can anyone tell why this situation ? Also suggest if any better option available than cufflink cuffmerge.
Assembly transcriptome cufflink cuffmerge • 2.2k views
2
Entering edit mode
5.2 years ago
Dear Chirag,
If there is no specific reason to use cufflink, I strongly recommend using string-tie. Check out the below link for a comparison of string-tie v/s cufflink. Though this comparison is for human RNA-seq data sets, I am sure string-tie will outperform for fungi data.
Regards Vijay
0
Entering edit mode
Hi Vijay,
Good to know this new transcriptome assembler (though published in 2014). Seems actively updated vs cufflink (last release 2014). I will try this and share the results.
Thanks a lot.
0
Entering edit mode
Chirag, you must also be interested in checking out the "new" tuxedo protocol. Old tuxedo protocol uses tophat, cufflink, cuffmerge and cummerRbund, while the new protocol is described in below image
Hisat2 is very efficient mapper in terms of memory and time. Sensitivity is also good.
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Entering edit mode
Additionally, both the pipelines are coming from same team Steven Salzberg's lab.
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Entering edit mode
Hi Vijay,
I have used stringtie. Results are more or less same with cufllink. I decided not to merge transcripts from different samples rather use individual assembly and filter them manually by user specified criteria. Merging assembly in fugal genome may not good idea because genes are very close to each other. One question I have here is why most of the transcripts orientation are wrong given by string tie? To check the strand I colored alignment by "first of paired strand" on IGV. Most of them having orientation regardless of alignment color. I wonder, how stringtie assigns strand information to generated transcripts ?
0
Entering edit mode
Hi Chirag,
There was indeed a bug reported in earlier version (5/18/2015 - v1.0.4 release), however, you should be using the latest version,hence, that could not be an issue now.
Is your library strand specific? According to string-tie paper (page#4):
"We considered multi-exon transcripts to be correctly assembled only if their strand was also correctly identified, and when strand-specific RNA- seq data were used, we also required that single-exon transcripts were assigned to the correct strand"
Did you use --fr and --rf options?
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Entering edit mode
yes... I know these options. My data is strand specific (--rf) so I used string tie accordingly. I also tried another option (--fr) to confirm if my library option is correct. Still there is an issue.
EDIT : More precisely most of them are in correct orientation. But still there are cases which have opposite strand than what we see the reads color in IGV. Most of them I saw having lot's of antisense transcription going on. This could be one reason assembler is not able to assign correct strand
0
Entering edit mode
5.2 years ago
You might be able to improve annotation using dedicated software that can use transcriptome support as one of the evidence, plus add information about evolutionary related species genomes, genes, pram domains and so on. You can look at a variety of such tools likа maker, augustus, fgenesh, genemark and many other. You can combine predictions of these tools. Anyway you will need to filter your results extensively based on many different statistics and different data plus for genes of your main interest you better go dipper and assess results visually yourself as you did. I am not sure if you can rebuild reliably all gene models with their alternative splicing from RNA-seq data. Maybe it is possible at a high coverage but I do not remember numbers.
0
Entering edit mode
5.2 years ago
There are tools out there even better than stringtie. Take a look at figure 2 in this article: http://www.ncbi.nlm.nih.gov/pubmed/27760567.
Those might be more interesting.
0
Entering edit mode
4.5 years ago
h.botond ▴ 50
Dear Chirag,
I am struggling with similar problems. I like to improve my fungi genome annotation, especially to the UTR with my RNAseq data. Can you tell me your experiences with this problem? I have tried the classic and new tuxedo workflows and the trinity as well but neither of them could detect the true transcript structure or only partially.
Thanks for any suggestions.
0
Entering edit mode
4.5 years ago
colindaven ★ 4.0k
Fungi have quite tightly packed genomes. You could try this specialist option,snowyowl. https://bmcbioinformatics.biomedcentral.com/articles/10.1186/1471-2105-15-229
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2022-08-15 12:21:51
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https://homework.zookal.com/questions-and-answers/how-do-i-do-this-write-lmc-programs-to-complete-733388647
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1. Engineering
2. Computer Science
3. how do i do this write lmc programs to complete...
# Question: how do i do this write lmc programs to complete...
###### Question details
How do I do this:
Write LMC programs to complete the following task. Write the solutions using Numeric code and using Mnemonics and Labels.
Write two programs, one to Add two numbers and another Subtract two numbers.
How do I write this out in an LMC editor? I am new to this so I don't understand how to do this. As I keep getting "Duplicate Logic Error" every time I try to enter the below code:
INP
STA first
INP
OUT
HLT
Program to subtract two numbers -
INP
STA first
INP
SUB first
OUT
HLT
Do I need to enter the below image out in the LMC editor to solve the question above?
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2021-06-12 20:09:15
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http://digitalhaunt.net/Connecticut/calculating-standard-error-in-regression-coefficient.html
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Address 282 E Main St, Torrington, CT 06790 (860) 294-4473 http://1tech4hire.com
# calculating standard error in regression coefficient East Hartland, Connecticut
Coefficients Term Coef SE Coef T-Value P-Value VIF Constant 20.1 12.2 1.65 0.111 Stiffness 0.2385 0.0197 12.13 0.000 1.00 Temp -0.184 0.178 -1.03 0.311 1.00 The standard error of the Stiffness Back to English × Translate This Page Select Language Bulgarian Catalan Chinese Simplified Chinese Traditional Czech Danish Dutch English Estonian Finnish French German Greek Haitian Creole Hindi Hmong Daw Hungarian Indonesian The standardized version of X will be denoted here by X*, and its value in period t is defined in Excel notation as: ... The population standard deviation is STDEV.P.) Note that the standard error of the model is not the square root of the average value of the squared errors within the historical sample
Zero Emission Tanks Find Iteration of Day of Week in Month Why is it "kiom strange" instead of "kiel strange"? This is not supposed to be obvious. Finally, confidence limits for means and forecasts are calculated in the usual way, namely as the forecast plus or minus the relevant standard error times the critical t-value for the desired Return to top of page.
So, for models fitted to the same sample of the same dependent variable, adjusted R-squared always goes up when the standard error of the regression goes down. The standard error of the model will change to some extent if a larger sample is taken, due to sampling variation, but it could equally well go up or down. Return to top of page. The accuracy of a forecast is measured by the standard error of the forecast, which (for both the mean model and a regression model) is the square root of the sum
The deduction above is $\mathbf{wrong}$. However, you can use the output to find it with a simple division. Why would all standard errors for the estimated regression coefficients be the same? Adjusted R-squared, which is obtained by adjusting R-squared for the degrees if freedom for error in exactly the same way, is an unbiased estimate of the amount of variance explained: Adjusted
Not clear why we have standard error and assumption behind it. –hxd1011 Jul 19 at 13:42 add a comment| 3 Answers 3 active oldest votes up vote 68 down vote accepted What will be the value of the following determinant without expanding it? What is the common meaning and usage of "get mad"? You can see that in Graph A, the points are closer to the line than they are in Graph B.
For the case in which there are two or more independent variables, a so-called multiple regression model, the calculations are not too much harder if you are familiar with how to The standard error of the forecast is not quite as sensitive to X in relative terms as is the standard error of the mean, because of the presence of the noise Continuous Variables 8. Please try the request again.
codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 Residual standard error: 13.55 on 159 degrees of freedom Multiple R-squared: 0.6344, Adjusted R-squared: 0.6252 F-statistic: 68.98 on Even if you think you know how to use the formula, it's so time-consuming to work that you'll waste about 20-30 minutes on one question if you try to do the Adjusted R-squared can actually be negative if X has no measurable predictive value with respect to Y. The standard error of a coefficient estimate is the estimated standard deviation of the error in measuring it.
Letters of support for tenure Proving the regularity of a certain language Polite way to ride in the dark Tips for Golfing in Brain-Flak How can the film of 'World War The slope coefficient in a simple regression of Y on X is the correlation between Y and X multiplied by the ratio of their standard deviations: Either the population or The numerator is the sum of squared differences between the actual scores and the predicted scores. Why do most log files use plain text rather than a binary format?
p is the number of coefficients in the regression model. In the mean model, the standard error of the model is just is the sample standard deviation of Y: (Here and elsewhere, STDEV.S denotes the sample standard deviation of X, In a simple regression model, the percentage of variance "explained" by the model, which is called R-squared, is the square of the correlation between Y and X. It is well known that an estimate of $\mathbf{\beta}$ is given by (refer, e.g., to the wikipedia article) $$\hat{\mathbf{\beta}} = (\mathbf{X}^{\prime} \mathbf{X})^{-1} \mathbf{X}^{\prime} \mathbf{y}.$$ Hence \textrm{Var}(\hat{\mathbf{\beta}}) = (\mathbf{X}^{\prime} \mathbf{X})^{-1} \mathbf{X}^{\prime}
The fraction by which the square of the standard error of the regression is less than the sample variance of Y (which is the fractional reduction in unexplained variation compared to For example, type L1 and L2 if you entered your data into list L1 and list L2 in Step 1. A Hendrix April 1, 2016 at 8:48 am This is not correct! The sample standard deviation of the errors is a downward-biased estimate of the size of the true unexplained deviations in Y because it does not adjust for the additional "degree of
Here is an Excel file with regression formulas in matrix form that illustrates this process. Example data. asked 2 years ago viewed 16866 times active 1 year ago Blog Stack Overflow Podcast #89 - The Decline of Stack Overflow Has Been Greatly… Linked 53 How are the standard In a multiple regression model in which k is the number of independent variables, the n-2 term that appears in the formulas for the standard error of the regression and adjusted
Translate Coefficient Standard Errors and Confidence IntervalsCoefficient Covariance and Standard ErrorsPurposeEstimated coefficient variances and covariances capture the precision of regression coefficient estimates. The correct result is: 1.$\hat{\mathbf{\beta}} = (\mathbf{X}^{\prime} \mathbf{X})^{-1} \mathbf{X}^{\prime} \mathbf{y}.$ (To get this equation, set the first order derivative of $\mathbf{SSR}$ on $\mathbf{\beta}$ equal to zero, for maxmizing $\mathbf{SSR}$) 2.\$E(\hat{\mathbf{\beta}}|\mathbf{X}) = So, if you know the standard deviation of Y, and you know the correlation between Y and X, you can figure out what the standard deviation of the errors would be However, more data will not systematically reduce the standard error of the regression.
Why did the One Ring betray Isildur? Standard Error of Regression Slope Formula SE of regression slope = sb1 = sqrt [ Σ(yi - ŷi)2 / (n - 2) ] / sqrt [ Σ(xi - x)2 ]). Formulas for a sample comparable to the ones for a population are shown below. standard-error inferential-statistics share|improve this question edited Mar 6 '15 at 14:38 Christoph Hanck 9,13332149 asked Feb 9 '14 at 9:11 loganecolss 5531926 stats.stackexchange.com/questions/44838/… –ocram Feb 9 '14 at 9:14
Expected Value 9. The terms in these equations that involve the variance or standard deviation of X merely serve to scale the units of the coefficients and standard errors in an appropriate way.
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2019-09-21 15:15:15
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https://support.bioconductor.org/u/327/
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## User: Justin Borevitz
Reputation:
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#### Posts by Justin Borevitz
<prev • 9 results • page 1 of 1 • next >
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464
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1
... I've noticed that the order of probes in the 2 packages does not agree. At least for barley1 and ath1121501. Also the way probes are ordered in Linux and Rgui (PC) does not agree. It could be something with the alphabetizing of probsets names in the 2 versions. Its possible this is true for the ...
written 15.5 years ago by Justin Borevitz90 • updated 15.4 years ago by Wolfgang Huber13k
3
693
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Comment: C: Re: Logit-t vs RMA
... Hi Jenny, your setup has several valid permutations which can be used to account for your setup and multiple testing. You can also try and estimate the proportion of genes different from the null. The FDR q value might be of more interest in this case. See "Statistical significance for genomewide ...
written 16.1 years ago by Justin Borevitz90
3
743
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3
Comment: C: Installation of ath1121501cdf
... The manual installation worked fine for me yesterday on linux RH9 after I had the same download error. This occurred when I tried to use justRMA() on some ATH1 .cel files when the ath1121501cdf had not been installed privously. Other packages (e.g. getBioC() ) do download fine on my system so there ...
written 16.2 years ago by Justin Borevitz90
0
435
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... Great works now > update.packages(CRAN=getOption("BIOC")) trying URL http://www.bioconductor.org/bin/windows/contrib/1.7/PACKAGES' Content type text/plain' length 11255 bytes opened URL downloaded 10Kb affy : Version 1.2.28 in C:/R/rw1071/library Version 1.2.30 on CRAN Update (y/N)? y tryin ...
written 16.2 years ago by Justin Borevitz90
1
523
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... Similar problem as this morning with widgetTools, which is now fixed. I'm sure you are working on it. Update ran fine from linux. In WinXP R1.7.1 > update.packages(CRAN=getOption("BIOC")) trying URL http://www.bioconductor.org/bin/windows/contrib/1.7/PACKAGES' Content type text/plain' length ...
written 16.2 years ago by Justin Borevitz90
3
533
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Comment: C: com.braju.sma
... Justin http://naturalvariation.org We are having a little debate about whether RMA is using MM oligos at all. I know that PM.only is the method that calls the gene expression index, but Are MM intensities used in the calculation for the background correction? I thought I'd put this out to get th ...
written 16.3 years ago by Justin Borevitz90
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509
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... I had this problem last night and this morning but I just tried again and its working now. You can run in R install.packages("Biobase",CRAN=getOption("BIOC")) install.packages("genefilter",CRAN=getOption("BIOC")) those were the ones that initially failed for me. Justin http://naturalvariation ...
written 16.3 years ago by Justin Borevitz90
1
449
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1
... Hi Wenlei, there are 2 commercial Arabidopsis arrays. One is ATH1 that has 22k genes, the other is the old one with 8k genes called AG. You have downloaded the wrong cdf package because the link is incorrect. You can download the Source for ATH1 from bioconductor and install it if you are set up ...
written 16.4 years ago by Justin Borevitz90
3
868
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... Hi all great to be part of a highly active/responsive group! Has anyone else thought about spatial correction for affy arrays? Charles Berry and I have done some simple corrections using sliding windows, see Figure 1. http://www.genome.org/cgi/content/full/13/3/513/F1 or http://naturalvariation.org ...
written 16.4 years ago by Justin Borevitz90
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2019-10-20 14:12:57
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http://mathhelpforum.com/calculus/53092-area-bounded-curves-print.html
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# Area bounded by curves
• October 11th 2008, 06:32 AM
maybeline9216
Area bounded by curves
• October 11th 2008, 07:02 AM
skeeter
I get 5.31%
• October 11th 2008, 07:06 AM
maybeline9216
Can tell me how u get? Like: what is your value for the shaded and unshaded area respectively, area of region OAPB, coords.of B AND Coords.of Q
I want to see where my workings and answer differ...
• October 11th 2008, 07:40 AM
skeeter
show me what you did ... maybe i can see your mistake.
• October 11th 2008, 07:51 AM
maybeline9216
Quote:
Originally Posted by skeeter
show me what you did ... maybe i can see your mistake.
Please look at my qn 11 workings.Thanks:D
• October 11th 2008, 01:27 PM
skeeter
let $f(x) = e^x$
$g(x) = 4 - e^x$
$h(x) = 2(x - \ln{2}) + 2$
$A = \int_0^{\ln{2}} f(x) \, dx + \int_{\ln{2}}^{\ln{4}} g(x) \, dx$
$B = \int_0^{\ln{2}} f(x) - h(x) \, dx$
calculate $\frac{B}{A}$.
• October 11th 2008, 07:51 PM
maybeline9216
Thanks i found my mistake already..i forgot to add the unshaded area into the calculation of the area of the whole plot
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2015-05-04 22:45:15
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http://trac.sasview.org/changeset/c9fc87335cd86f10c8c9e3654927fdeeba6a70af/sasmodels
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Changeset c9fc873 in sasmodels
Ignore:
Timestamp:
Jul 9, 2018 7:13:35 AM (12 months ago)
Branches:
master, core_shell_microgels, magnetic_model, ticket-1257-vesicle-product, ticket_1156, ticket_1265_superball, ticket_822_more_unit_tests
Children:
c11d09f
Parents:
751da81
Message:
normalise R_g in guinier docs; use 'SLD centre of mass'
File:
1 edited
Legend:
Unmodified
r751da81 In scattering, the radius of gyration $R_g$ quantifies the objects's distribution of SLD (not mass density, as in mechanics) from the objects's centre of mass. It is defined by SLD centre of mass. It is defined by .. math:: R_g^2 = \sum_i^\text{all points}\rho_i\left(r_i-r_0\right)^2 .. math:: R_g^2 = \frac{\sum_i\rho_i\left(r_i-r_0\right)^2}{\sum_i\rho_i} where $r_0$ denotes the object's centre of mass and $\rho_i$ is the SLD at a point $i$. where $r_0$ denotes the object's SLD centre of mass and $\rho_i$ is the SLD at a point $i$. Notice that $R_g^2$ may be negative (since SLD can be negative), which happens
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2019-06-20 23:37:21
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https://analytixon.com/2017/11/10/whats-new-on-arxiv-538/
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In this paper, we derive a temporal arbitrage policy for storage via reinforcement learning. Real-time price arbitrage is an important source of revenue for storage units, but designing good strategies have proven to be difficult because of the highly uncertain nature of the prices. Instead of current model predictive or dynamic programming approaches, we use reinforcement learning to design a two-thresholds policy. This policy is learned through repeated charge and discharge actions performed by the storage unit through updating a value matrix. We design a reward function that does not only reflect the instant profit of charge/discharge decisions but also incorporate the history information. Simulation results demonstrate our designed reward function leads to significant performance improvement compared with existing algorithms.
We present ENERGYNET , a new framework for analyzing and building artificial neural network architectures. Our approach adaptively learns the structure of the networks in an unsupervised manner. The methodology is based upon the theoretical guarantees of the energy function of restricted Boltzmann machines (RBM) of infinite number of nodes. We present experimental results to show that the final network adapts to the complexity of a given problem.
The assumption that data samples are independently identically distributed is the backbone of many learning algorithms. Nevertheless, datasets often exhibit rich structures in practice, and we argue that there exist some unknown orders within the data instances. Aiming to find such orders, we introduce a novel Generative Markov Network (GMN) which we use to extract the order of data instances automatically. Specifically, we assume that the instances are sampled from a Markov chain. Our goal is to learn the transitional operator of the chain as well as the generation order by maximizing the generation probability under all possible data permutations. One of our key ideas is to use neural networks as a soft lookup table for approximating the possibly huge, but discrete transition matrix. This strategy allows us to amortize the space complexity with a single model and make the transitional operator generalizable to unseen instances. To ensure the learned Markov chain is ergodic, we propose a greedy batch-wise permutation scheme that allows fast training. Empirically, we evaluate the learned Markov chain by showing that GMNs are able to discover orders among data instances and also perform comparably well to state-of-the-art methods on the one-shot recognition benchmark task.
Convolution as inner product has been the founding basis of convolutional neural networks (CNNs) and the key to end-to-end visual representation learning. Benefiting from deeper architectures, recent CNNs have demonstrated increasingly strong representation abilities. Despite such improvement, the increased depth and larger parameter space have also led to challenges in properly training a network. In light of such challenges, we propose hyperspherical convolution (SphereConv), a novel learning framework that gives angular representations on hyperspheres. We introduce SphereNet, deep hyperspherical convolution networks that are distinct from conventional inner product based convolutional networks. In particular, SphereNet adopts SphereConv as its basic convolution operator and is supervised by generalized angular softmax loss – a natural loss formulation under SphereConv. We show that SphereNet can effectively encode discriminative representation and alleviate training difficulty, leading to easier optimization, faster convergence and comparable (even better) classification accuracy over convolutional counterparts. We also provide some theoretical insights for the advantages of learning on hyperspheres. In addition, we introduce the learnable SphereConv, i.e., a natural improvement over prefixed SphereConv, and SphereNorm, i.e., hyperspherical learning as a normalization method. Experiments have verified our conclusions.
In many settings, it is important that a model be capable of providing reasons for its predictions (i.e., the model must be interpretable). However, the model’s reasoning may not conform with well-established knowledge. In such cases, while interpretable, the model lacks \textit{credibility}. In this work, we formally define credibility in the linear setting and focus on techniques for learning models that are both accurate and credible. In particular, we propose a regularization penalty, expert yielded estimates (EYE), that incorporates expert knowledge about well-known relationships among covariates and the outcome of interest. We give both theoretical and empirical results comparing our proposed method to several other regularization techniques. Across a range of settings, experiments on both synthetic and real data show that models learned using the EYE penalty are significantly more credible than those learned using other penalties. Applied to a large-scale patient risk stratification task, our proposed technique results in a model whose top features overlap significantly with known clinical risk factors, while still achieving good predictive performance.
For the prediction with expert advice setting, we consider methods to construct forecasting algorithms that suffer loss not much more than of any expert in the pool. In contrast to the standard approach, we investigate the case of long-term interval forecasting of time series, that is, each expert issues a sequence of forecasts for a time interval ahead and the master algorithm combines these forecasts into one aggregated sequence of forecasts. Two new approaches for aggregating experts long-term interval predictions are presented. One is based on Vovk’s aggregation algorithm and considers sliding experts, the other applies the approach of Mixing Past Posteriors method to the long-term prediction. The upper bounds for regret of these algorithms for adversarial case are obtained. We also present results of numerical experiments of time series long-term prediction.
Auto-scalability has become an evident feature for cloud software systems including but not limited to big data and IoT applications. Cloud application providers now are in full control over their applications’ microservices and macroservices; virtual machines and containers can be provisioned or deprovisioned on demand at runtime. Elascale strives to adjust both micro/macro resources with respect to workload and changes in the internal state of the whole application stack. Elascale leverages Elasticsearch stack for collection, analysis and storage of performance metrics. Elascale then uses its default scaling engine to elastically adapt the managed application. Extendibility is guaranteed through provider, schema, plug-in and policy elements in the Elascale by which flexible scalability algorithms, including both reactive and proactive techniques, can be designed and implemented for various technologies, infrastructures and software stacks. In this paper, we present the architecture and initial implementation of Elascale; an instance will be leveraged to add auto-scalability to a generic IoT application. Due to zero dependency to the target software system, Elascale can be leveraged to provide auto-scalability and monitoring as-a-service for any type of cloud software system.
Computational models of decisionmaking must contend with the variance of context and any number of possible decisions that a defined strategic actor can make at a given time. Relying on cognitive science theory, the authors have created an algorithm that captures the orientation of the actor towards an object and arrays the possible decisions available to that actor based on their given intersubjective orientation. This algorithm, like a traditional K-means clustering algorithm, relies on a core-periphery structure that gives the likelihood of moves as those closest to the cluster’s centroid. The result is an algorithm that enables unsupervised classification of an array of decision points belonging to an actor’s present state and deeply rooted in cognitive science theory.
Convolution Neural Networks (CNN), known as ConvNets are widely used in many visual imagery application, object classification, speech recognition. After the implementation and demonstration of the deep convolution neural network in Imagenet classification in 2012 by krizhevsky, the architecture of deep Convolution Neural Network is attracted many researchers. This has led to the major development in Deep learning frameworks such as Tensorflow, caffe, keras, theno. Though the implementation of deep learning is quite possible by employing deep learning frameworks, mathematical theory and concepts are harder to understand for new learners and practitioners. This article is intended to provide an overview of ConvNets architecture and to explain the mathematical theory behind it including activation function, loss function, feedforward and backward propagation. In this article, grey scale image is taken as input information image, ReLU and Sigmoid activation function are considered for developing the architecture and cross-entropy loss function are used for computing the difference between predicted value and actual value. The architecture is developed in such a way that it can contain one convolution layer, one pooling layer, and multiple dense layers
Wind power generation exhibits a strong temporal variability, which is crucial for system integration in highly renewable power systems. Different methods exist to simulate wind power generation but they often cannot represent the crucial temporal fluctuations properly. We apply the concept of additive binary Markov chains to model a wind generation time series consisting of two states: periods of high and low wind generation. The only input parameter for this model is the empirical autocorrelation function. The two state model is readily extended to stochastically reproduce the actual generation per period. To evaluate the additive binary Markov chain method, we introduce a coarse model of the electric power system to derive backup and storage needs. We find that the temporal correlations of wind power generation, the backup need as a function of the storage capacity and the resting time distribution of high and low wind events for different shares of wind generation can be reconstructed.
Deep learning is the state-of-the-art in fields such as visual object recognition and speech recognition. This learning uses a large number of layers and a huge number of units and connections. Therefore, overfitting is a serious problem with it, and the dropout which is a kind of regularization tool is used. However, in online learning, the effect of dropout is not well known. This paper presents our investigation on the effect of dropout in online learning. We analyzed the effect of dropout on convergence speed near the singular point. Our results indicated that dropout is effective in online learning. Dropout tends to avoid the singular point for convergence speed near that point.
The current study proposes a dimension reduction method, stepwise support vector machine (SVM), to reduce the dimensions of large p small n datasets. The proposed method is compared with other dimension reduction methods, namely, the Pearson product difference correlation coefficient (PCCs), recursive feature elimination based on random forest (RF-RFE), and principal component analysis (PCA), by using five gene expression datasets. Additionally, the prediction performance of the variables selected by our method is evaluated. The study found that stepwise SVM can effectively select the important variables and achieve good prediction performance. Moreover, the predictions of stepwise SVM for reduced datasets was better than those for the unreduced datasets. The performance of stepwise SVM was more stable than that of PCA and RF-RFE, but the performance difference with respect to PCCs was minimal. It is necessary to reduce the dimensions of large p small n datasets. We believe that stepwise SVM can effectively eliminate noise in data and improve the prediction accuracy in any large p small n dataset.
Continuous-time branching processes (CTBPs) are powerful tools in random graph theory, but are not appropriate to describe real-world networks, since they produce trees rather than (multi)graphs. In this paper we analyze collapsed branching processes (CBPs), obtained by a collapsing procedure on CTBPs, in order to define multigraphs where vertices have fixed out-degree $m\geq 2$. A key example consists of preferential attachment models (PAMs), as well as generalized PAMs where vertices are chosen according to their degree and age. We identify the degree distribution of CBPs, showing that it is closely related to the limiting distribution of the CTBP before collapsing. In particular, this is the first time that CTBPs are used to investigate the degree distribution of PAMs beyond the tree setting.
Transfer learning aims to faciliate learning tasks in a label-scarce target domain by leveraging knowledge from a related source domain with plenty of labeled data. Often times we may have multiple domains with little or no labeled data as targets waiting to be solved. Most existing efforts tackle target domains separately by modeling the `source-target’ pairs without exploring the relatedness between them, which would cause loss of crucial information, thus failing to achieve optimal capability of knowledge transfer. In this paper, we propose a novel and effective approach called Multi-Relevance Transfer Learning (MRTL) for this purpose, which can simultaneously transfer different knowledge from the source and exploits the shared common latent factors between target domains. Specifically, we formulate the problem as an optimization task based on a collective nonnegative matrix tri-factorization framework. The proposed approach achieves both source-target transfer and target-target leveraging by sharing multiple decomposed latent subspaces. Further, an alternative minimization learning algorithm is developed with convergence guarantee. Empirical study validates the performance and effectiveness of MRTL compared to the state-of-the-art methods.
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2019-02-21 12:55:40
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https://www.aimsciences.org/article/doi/10.3934/dcdss.2022093
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American Institute of Mathematical Sciences
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doi: 10.3934/dcdss.2022093
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On a global climate model with non-monotone multivalued coalbedo
1 Dept. Ingeniería Geológica y Minera. E.T.S.I. Minas y Energía, Center for Computational Simulation. Universidad Politécnica de Madrid, Calle Ríos Rosas, 21. 28003 Madrid, Spain 2 Dept. Matemática Aplicada. ETS Arquitectura, Center for Computational Simulation. Universidad Politécnica de Madrid, Av. Juan de Herrera, 4. 28040 Madrid, Spain
Dedicated to Professor Georg Hetzer on the occasion of his 70th birthday
Received December 2021 Revised February 2022 Early access April 2022
We are concerned with a global energy balance climate model formulated through a parabolic equation whose space domain is a manifold which simulates the Earth surface. The climate energy balance model includes the effect of coalbedo as one of the mean temperature feedback. We extend some mathematical results proved for maximal monotone coalbedo to the case where the coalbedo has not a monotone dependency on temperature. Numerical approximation is performed by the Finite Volume Method which allows to obtain and compare numerical solutions with different values of the coalbedo.
Citation: Arturo Hidalgo, Lourdes Tello. On a global climate model with non-monotone multivalued coalbedo. Discrete and Continuous Dynamical Systems - S, doi: 10.3934/dcdss.2022093
References:
[1] D. Arcoya, J. I. Diaz and L. Tello, S-shaped bifurcation branch in a quasilinear multivalued model arising in climatology, J. Differential Equations, 150 (1998), 215-225. doi: 10.1006/jdeq.1998.3502. [2] T. Aubin, Nonlinear Analysis on Manifolds. Monge-Ampère Equations, Springer-Verlag (New York), 1982. doi: 10.1007/978-1-4612-5734-9. [3] M. Badii and J. I. Díaz, Time periodic solutions for a diffusive energy balance model in climatology, J. Math. Anal. Appl., 233 (1999), 713-729. doi: 10.1006/jmaa.1999.6335. [4] H. Brézis, Operateurs Maximaux Monotones et Semigroupes de Contractions dans les Espaces de Hilbert. North Holland, Amsterdam, 1973. [5] M. I. Budyko, The effects of solar radiation variations on the climate of the earth, Tellus, 21 (1969), 611-619. [6] J. I. Diaz, Mathematical Analysis of some Diffusive Energy Balance Climate Models, in the book Mathematics, Climate and Environment, (J. I. Diaz and J. L. Lions, eds. Masson, Paris, 28–56, 1993. [7] J. I. Díaz, J. Hernández and L. Tello, On the multiplicity of equilibrium solutions to a nonlinear diffusion equation on a manifold arising in climatology, J. Math. Anal. Appl., 216 (1997), 593-613. doi: 10.1006/jmaa.1997.5691. [8] J. I. Díaz and G. Hetzer, A functional quasilinear reaction - Diffusion equation arising in climatology, In the book Équations aux Dérivées Partielles et Applications. Articles dedies a J. L. Lions. Elsevier Paris, (1998), 461–480. [9] J. I. Díaz, G. Hetzer and L. Tello, An energy balance model climate model with hysteresis, Nonlinear Analysis, 64 (2006), 2053-2074. doi: 10.1016/j.na.2005.07.038. [10] J. I. Díaz, A. Hidalgo and L. Tello, Multiple solutions and numerical analysis to the dynamic and stationary models coupling a delayed energy balance model involving latent heat and discontinuous albedo with a deep ocean, Proc. R. Soc. Lond. Ser. A Math. Phys. Eng. Sci., 470 (2014), no. 2170, 20140376, 20 pp. doi: 10.1098/rspa.2014.0376. [11] J. I. Díaz and L. Tello, A nonlinear parabolic problem on a Riemannian manifold without boundary arising in climatology, Collect. Math., 50 (1999), 19-51. [12] J. I. Díaz and L. Tello, Infinitely many stationary solutions for a simple climate model via a shooting method, Math. Methods Appl. Sci., 25 (2002), 327-334. doi: 10.1002/mma.289. [13] G. Hetzer, Forced periodic oscillations in the climate system via an energy balance model, Comment. Math. Uni. Carolin., 28 (1987), 593-601. [14] G. Hetzer, The structure of the principal component for semilinear diffusion equations from energy balance climate models, Houston J. Math., 16 (1990), 203-216. [15] G. Hetzer, A parameter dependent time-periodic reaction-diffusion equation from climate modeling; S-shapedness of the principal branch of fixed points of the time-$1$-map, Differential Integral Equations, 7 (1994), 1419-1425. [16] G. Hetzer, The shift-semiflow of a multi-valued evolution equation from climate modelling, Nonlinear Analysis: Theory, Methods & Applications, 47 (2001), 2905-2916. doi: 10.1016/S0362-546X(01)00412-6. [17] G. Hetzer, The number of stationary solutions for a one-dimensional Budyko-type climatemodel, Nonlinear Anal. Real World Appl., 2 (2001), 259-272. doi: 10.1016/S0362-546X(00)00103-6. [18] G. Hetzer, Global existence for a functional reaction-diffusion problem from climate modeling, Discrete Contin. Dyn. Syst., 2011 (2011), 660-671. doi: 10.3934/proc.2011.2011.660. [19] G. Hetzer, Trajectory attractors of energy balance climate models with bio-feedback, Differ. Equ. Appl., 3 (2011), 565-579. doi: 10.7153/dea-03-35. [20] G. Hetzer and P. G. Schmidt, A global attractor and stationary solutions for a reaction diffusion system arising from climate modeling, Nonlinear Anal., 14 (1990), 915-926. doi: 10.1016/0362-546X(90)90109-T. [21] G. Hetzer and P. G. Schmidt, Global existence and asymptotic behavior for a quasilinear reaction diffusion system from climate modeling, J. Math. Anal. Appl., 160 (1991), 250-262. doi: 10.1016/0022-247X(91)90303-H. [22] G. Hetzer and L. Tello, On a reaction diffusion system arising in Climatology, Dynamic Syst. Appl., 11 (2002), 381-402. [23] A. Hidalgo and L. Tello, On a climatological energy balance model with continents distribution, Discrete Contin. Dyn. Syst., 35 (2015), 1503-1519. doi: 10.3934/dcds.2015.35.1503. [24] A. Hidalgo and L. Tello, Numerical approach of the equilibrium solutions of a global climate model, Mathematics, 8 (2020), 1542. doi: 10.3390/math8091542. [25] G. North and K. Kim, Energy Balance Climate Models, Wiley-VCH Verlag, Weinheim, Germany 2017. [26] J. Rombouts and M. Ghil, Oscillations in a simple climate-vegetation model, Nonlin. Processes Geophys., 22 (2015), 275-288. doi: 10.5194/npg-22-275-2015. [27] B. E. Schmidt, On a nonlinear eigenvalue problem arising from climate modeling, Nonlinear Anal., 30 (1997), 3645-3656. doi: 10.1016/S0362-546X(96)00239-8. [28] B. E. Schmidt, Bifurcation from S-shaped solution curves in a class of Sturm Liouville problems related to climate modeling, Adv. Math. Sci. Appl., 10 (2000), 513-537. [29] B. E. Schmidt, Bifurcation of stationary solutions for Legendre-type boundary value problems arising from energy balance climate models, Thesis (Ph.D.)-Auburn University, 1994. [30] W. D. Sellers, A global climatic model based on the energy balance of the earth- atmosphere system, J. Appl. Meteorol., 8 (1969), 392-400. doi: 10.1175/1520-0450(1969)008<0392:AGCMBO>2.0.CO;2. [31] P. H. Stone, A simplified radiative-dynamical model for the static stability of rotating atmospheres, Journal of Atmospheric Sciences, 29 (1972), 405-418. [32] I. I. Vrabie, Compactness Methods for Nonlinear Evolutions, Pitman Longman. London, 1986, 1995.
show all references
References:
[1] D. Arcoya, J. I. Diaz and L. Tello, S-shaped bifurcation branch in a quasilinear multivalued model arising in climatology, J. Differential Equations, 150 (1998), 215-225. doi: 10.1006/jdeq.1998.3502. [2] T. Aubin, Nonlinear Analysis on Manifolds. Monge-Ampère Equations, Springer-Verlag (New York), 1982. doi: 10.1007/978-1-4612-5734-9. [3] M. Badii and J. I. Díaz, Time periodic solutions for a diffusive energy balance model in climatology, J. Math. Anal. Appl., 233 (1999), 713-729. doi: 10.1006/jmaa.1999.6335. [4] H. Brézis, Operateurs Maximaux Monotones et Semigroupes de Contractions dans les Espaces de Hilbert. North Holland, Amsterdam, 1973. [5] M. I. Budyko, The effects of solar radiation variations on the climate of the earth, Tellus, 21 (1969), 611-619. [6] J. I. Diaz, Mathematical Analysis of some Diffusive Energy Balance Climate Models, in the book Mathematics, Climate and Environment, (J. I. Diaz and J. L. Lions, eds. Masson, Paris, 28–56, 1993. [7] J. I. Díaz, J. Hernández and L. Tello, On the multiplicity of equilibrium solutions to a nonlinear diffusion equation on a manifold arising in climatology, J. Math. Anal. Appl., 216 (1997), 593-613. doi: 10.1006/jmaa.1997.5691. [8] J. I. Díaz and G. Hetzer, A functional quasilinear reaction - Diffusion equation arising in climatology, In the book Équations aux Dérivées Partielles et Applications. Articles dedies a J. L. Lions. Elsevier Paris, (1998), 461–480. [9] J. I. Díaz, G. Hetzer and L. Tello, An energy balance model climate model with hysteresis, Nonlinear Analysis, 64 (2006), 2053-2074. doi: 10.1016/j.na.2005.07.038. [10] J. I. Díaz, A. Hidalgo and L. Tello, Multiple solutions and numerical analysis to the dynamic and stationary models coupling a delayed energy balance model involving latent heat and discontinuous albedo with a deep ocean, Proc. R. Soc. Lond. Ser. A Math. Phys. Eng. Sci., 470 (2014), no. 2170, 20140376, 20 pp. doi: 10.1098/rspa.2014.0376. [11] J. I. Díaz and L. Tello, A nonlinear parabolic problem on a Riemannian manifold without boundary arising in climatology, Collect. Math., 50 (1999), 19-51. [12] J. I. Díaz and L. Tello, Infinitely many stationary solutions for a simple climate model via a shooting method, Math. Methods Appl. Sci., 25 (2002), 327-334. doi: 10.1002/mma.289. [13] G. Hetzer, Forced periodic oscillations in the climate system via an energy balance model, Comment. Math. Uni. Carolin., 28 (1987), 593-601. [14] G. Hetzer, The structure of the principal component for semilinear diffusion equations from energy balance climate models, Houston J. Math., 16 (1990), 203-216. [15] G. Hetzer, A parameter dependent time-periodic reaction-diffusion equation from climate modeling; S-shapedness of the principal branch of fixed points of the time-$1$-map, Differential Integral Equations, 7 (1994), 1419-1425. [16] G. Hetzer, The shift-semiflow of a multi-valued evolution equation from climate modelling, Nonlinear Analysis: Theory, Methods & Applications, 47 (2001), 2905-2916. doi: 10.1016/S0362-546X(01)00412-6. [17] G. Hetzer, The number of stationary solutions for a one-dimensional Budyko-type climatemodel, Nonlinear Anal. Real World Appl., 2 (2001), 259-272. doi: 10.1016/S0362-546X(00)00103-6. [18] G. Hetzer, Global existence for a functional reaction-diffusion problem from climate modeling, Discrete Contin. Dyn. Syst., 2011 (2011), 660-671. doi: 10.3934/proc.2011.2011.660. [19] G. Hetzer, Trajectory attractors of energy balance climate models with bio-feedback, Differ. Equ. Appl., 3 (2011), 565-579. doi: 10.7153/dea-03-35. [20] G. Hetzer and P. G. Schmidt, A global attractor and stationary solutions for a reaction diffusion system arising from climate modeling, Nonlinear Anal., 14 (1990), 915-926. doi: 10.1016/0362-546X(90)90109-T. [21] G. Hetzer and P. G. Schmidt, Global existence and asymptotic behavior for a quasilinear reaction diffusion system from climate modeling, J. Math. Anal. Appl., 160 (1991), 250-262. doi: 10.1016/0022-247X(91)90303-H. [22] G. Hetzer and L. Tello, On a reaction diffusion system arising in Climatology, Dynamic Syst. Appl., 11 (2002), 381-402. [23] A. Hidalgo and L. Tello, On a climatological energy balance model with continents distribution, Discrete Contin. Dyn. Syst., 35 (2015), 1503-1519. doi: 10.3934/dcds.2015.35.1503. [24] A. Hidalgo and L. Tello, Numerical approach of the equilibrium solutions of a global climate model, Mathematics, 8 (2020), 1542. doi: 10.3390/math8091542. [25] G. North and K. Kim, Energy Balance Climate Models, Wiley-VCH Verlag, Weinheim, Germany 2017. [26] J. Rombouts and M. Ghil, Oscillations in a simple climate-vegetation model, Nonlin. Processes Geophys., 22 (2015), 275-288. doi: 10.5194/npg-22-275-2015. [27] B. E. Schmidt, On a nonlinear eigenvalue problem arising from climate modeling, Nonlinear Anal., 30 (1997), 3645-3656. doi: 10.1016/S0362-546X(96)00239-8. [28] B. E. Schmidt, Bifurcation from S-shaped solution curves in a class of Sturm Liouville problems related to climate modeling, Adv. Math. Sci. Appl., 10 (2000), 513-537. [29] B. E. Schmidt, Bifurcation of stationary solutions for Legendre-type boundary value problems arising from energy balance climate models, Thesis (Ph.D.)-Auburn University, 1994. [30] W. D. Sellers, A global climatic model based on the energy balance of the earth- atmosphere system, J. Appl. Meteorol., 8 (1969), 392-400. doi: 10.1175/1520-0450(1969)008<0392:AGCMBO>2.0.CO;2. [31] P. H. Stone, A simplified radiative-dynamical model for the static stability of rotating atmospheres, Journal of Atmospheric Sciences, 29 (1972), 405-418. [32] I. I. Vrabie, Compactness Methods for Nonlinear Evolutions, Pitman Longman. London, 1986, 1995.
Spatial distribution of the temperature for an output time $t = 5$. Solid red line: numerical solution when considering the ocean continents distribution. Dashed blue line: numerical solution when only ocean is taken into account in the whole domain. Dotted black line: initial condition. The labels: Continent, Ocean apply for the Ocean-Continents case
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2022-07-01 23:35:33
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https://quant.stackexchange.com/questions/15081/briefly-stated-why-does-the-function-nx-appear-in-the-european-call-option-pr
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# Briefly stated, why does the function N(x) appear in the European call option pricing model?
I'm aware of the the mathematical formula for the price of a European call option on a stock however I'd like to think about it in an intuitive way.
It is because Black and Scholes assume that the stock follows a geometric brownian motion, i.e. under the historical probability $\mathbb{P}$ the stock moves according to: $$\frac{dS(t)}{S(t)} = \mu dt + \sigma dW^{\mathbb{P}}(t)$$ Solving this SDE we obtain that $$S(t)=S(0)e^{(\mu - \frac{1}{2}\sigma^2)t + \sigma W^{\mathbb{P}}(t)} = S(0)e^{(\mu - \frac{1}{2}\sigma^2)t + \sigma \sqrt{t} z}$$ where $z \sim N(0,1)$. Thanks to Girsanov's theorem we can show that under the risk neutral measure, the stock still follows a Brownian Motion, in particular it can be shown that under $\mathbb{Q}$ the stock price follows $$S(t)=S(0)e^{(r-\frac{1}{2}\sigma^2)t + \sigma W^{\mathbb{Q}}(t)} = S(0)e^{(r - \frac{1}{2}\sigma^2)t + \sigma \sqrt{t} z}$$ It can be shown that the price of an option should be equal to the expected value under the risk neutral measure ($\mathbb{Q}$) of the discounted payoff at expiration, i.e. $$c(S,t) = \mathbb{E}_t^{\mathbb{Q}}[(S(T) - K)^+ e^{-r(T-t)}]$$ Now, take the expected value: $$c(s,t)= \mathbb{E}_t^{\mathbb{Q}}[(S(T) - K)^+ e^{-r(T-t)}] = \int_{S(T)>K}(S(T)-K)e^{-r(T-t)}\frac{1}{\sqrt{2\pi}}e^{-\frac{z^2}{2}}dz =$$ $$= \int_{S(T)>K}(S(t)e^{(r - \frac{1}{2}\sigma^2)(T-t) + \sigma \sqrt{T-t} z}-K)e^{-r(T-t)}\frac{1}{\sqrt{2\pi}}e^{-\frac{z^2}{2}}dz =$$ $$= S(t)N(d_1) - Ke^{-r(T-t)}N(d_2)$$ where $$d_1=\left[ln\frac{S(t)}{K} + (r-\frac{\sigma^2}{2}(T-t))\right]\frac{1}{\sigma\sqrt{T-t}}$$ $$d_2= d_1 - \sigma\sqrt{T-t}$$
Notice that $d_1$ just indicates the cutoff for z s.t. S(T) > K, i.e. where the option closes in the money. Summarizing, the normal cdf simply derives from the assumption of a stock diffusing like a geometric brownian motion both under the historical and the risk-neutral probability measure.
• Yes, just use Ito’s lemma on $d\log S(t)$. You have something like $$d\log S(t) = S(t)\left[\mu dt + \sigma dW(t) -\frac{1}{2}\sigma^2 dt\right]$$ which can be solved diving by $S(t)$ and integrating, i.e. $$\int_0^t \frac{d\log S(t)}{S(t)} = \int_0^t (\mu-0.5\sigma^2)dt + \int_0^t \sigma^2 dW(t)$$ $$\log S(t) = (\mu - 0.5\sigma^2)t + \sigma W(t)$$ then exponentiate to obtain $$S(t) = e^{(\mu - 0.5\sigma^2)t + \sigma W(t)}$$ – fni Oct 20 '14 at 6:26
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2021-03-05 01:00:17
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https://people.maths.bris.ac.uk/~matyd/GroupNames/320/C10xQ8sC4.html
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## G = C10×Q8⋊C4order 320 = 26·5
### Direct product of C10 and Q8⋊C4
direct product, metabelian, nilpotent (class 3), monomial, 2-elementary
Series: Derived Chief Lower central Upper central
Derived series C1 — C4 — C10×Q8⋊C4
Chief series C1 — C2 — C22 — C2×C4 — C2×C20 — C5×C4⋊C4 — C5×Q8⋊C4 — C10×Q8⋊C4
Lower central C1 — C2 — C4 — C10×Q8⋊C4
Upper central C1 — C22×C10 — C22×C20 — C10×Q8⋊C4
Generators and relations for C10×Q8⋊C4
G = < a,b,c,d | a10=b4=d4=1, c2=b2, ab=ba, ac=ca, ad=da, cbc-1=dbd-1=b-1, dcd-1=b-1c >
Subgroups: 242 in 162 conjugacy classes, 98 normal (30 characteristic)
C1, C2, C2, C4, C4, C4, C22, C22, C5, C8, C2×C4, C2×C4, C2×C4, Q8, Q8, C23, C10, C10, C4⋊C4, C4⋊C4, C2×C8, C2×C8, C22×C4, C22×C4, C2×Q8, C2×Q8, C20, C20, C20, C2×C10, C2×C10, Q8⋊C4, C2×C4⋊C4, C22×C8, C22×Q8, C40, C2×C20, C2×C20, C2×C20, C5×Q8, C5×Q8, C22×C10, C2×Q8⋊C4, C5×C4⋊C4, C5×C4⋊C4, C2×C40, C2×C40, C22×C20, C22×C20, Q8×C10, Q8×C10, C5×Q8⋊C4, C10×C4⋊C4, C22×C40, Q8×C2×C10, C10×Q8⋊C4
Quotients:
Smallest permutation representation of C10×Q8⋊C4
Regular action on 320 points
Generators in S320
(1 2 3 4 5 6 7 8 9 10)(11 12 13 14 15 16 17 18 19 20)(21 22 23 24 25 26 27 28 29 30)(31 32 33 34 35 36 37 38 39 40)(41 42 43 44 45 46 47 48 49 50)(51 52 53 54 55 56 57 58 59 60)(61 62 63 64 65 66 67 68 69 70)(71 72 73 74 75 76 77 78 79 80)(81 82 83 84 85 86 87 88 89 90)(91 92 93 94 95 96 97 98 99 100)(101 102 103 104 105 106 107 108 109 110)(111 112 113 114 115 116 117 118 119 120)(121 122 123 124 125 126 127 128 129 130)(131 132 133 134 135 136 137 138 139 140)(141 142 143 144 145 146 147 148 149 150)(151 152 153 154 155 156 157 158 159 160)(161 162 163 164 165 166 167 168 169 170)(171 172 173 174 175 176 177 178 179 180)(181 182 183 184 185 186 187 188 189 190)(191 192 193 194 195 196 197 198 199 200)(201 202 203 204 205 206 207 208 209 210)(211 212 213 214 215 216 217 218 219 220)(221 222 223 224 225 226 227 228 229 230)(231 232 233 234 235 236 237 238 239 240)(241 242 243 244 245 246 247 248 249 250)(251 252 253 254 255 256 257 258 259 260)(261 262 263 264 265 266 267 268 269 270)(271 272 273 274 275 276 277 278 279 280)(281 282 283 284 285 286 287 288 289 290)(291 292 293 294 295 296 297 298 299 300)(301 302 303 304 305 306 307 308 309 310)(311 312 313 314 315 316 317 318 319 320)
(1 132 86 96)(2 133 87 97)(3 134 88 98)(4 135 89 99)(5 136 90 100)(6 137 81 91)(7 138 82 92)(8 139 83 93)(9 140 84 94)(10 131 85 95)(11 77 65 23)(12 78 66 24)(13 79 67 25)(14 80 68 26)(15 71 69 27)(16 72 70 28)(17 73 61 29)(18 74 62 30)(19 75 63 21)(20 76 64 22)(31 49 311 52)(32 50 312 53)(33 41 313 54)(34 42 314 55)(35 43 315 56)(36 44 316 57)(37 45 317 58)(38 46 318 59)(39 47 319 60)(40 48 320 51)(101 142 116 122)(102 143 117 123)(103 144 118 124)(104 145 119 125)(105 146 120 126)(106 147 111 127)(107 148 112 128)(108 149 113 129)(109 150 114 130)(110 141 115 121)(151 211 206 175)(152 212 207 176)(153 213 208 177)(154 214 209 178)(155 215 210 179)(156 216 201 180)(157 217 202 171)(158 218 203 172)(159 219 204 173)(160 220 205 174)(161 197 223 182)(162 198 224 183)(163 199 225 184)(164 200 226 185)(165 191 227 186)(166 192 228 187)(167 193 229 188)(168 194 230 189)(169 195 221 190)(170 196 222 181)(231 291 286 255)(232 292 287 256)(233 293 288 257)(234 294 289 258)(235 295 290 259)(236 296 281 260)(237 297 282 251)(238 298 283 252)(239 299 284 253)(240 300 285 254)(241 277 303 262)(242 278 304 263)(243 279 305 264)(244 280 306 265)(245 271 307 266)(246 272 308 267)(247 273 309 268)(248 274 310 269)(249 275 301 270)(250 276 302 261)
(1 202 86 157)(2 203 87 158)(3 204 88 159)(4 205 89 160)(5 206 90 151)(6 207 81 152)(7 208 82 153)(8 209 83 154)(9 210 84 155)(10 201 85 156)(11 280 65 265)(12 271 66 266)(13 272 67 267)(14 273 68 268)(15 274 69 269)(16 275 70 270)(17 276 61 261)(18 277 62 262)(19 278 63 263)(20 279 64 264)(21 304 75 242)(22 305 76 243)(23 306 77 244)(24 307 78 245)(25 308 79 246)(26 309 80 247)(27 310 71 248)(28 301 72 249)(29 302 73 250)(30 303 74 241)(31 260 311 296)(32 251 312 297)(33 252 313 298)(34 253 314 299)(35 254 315 300)(36 255 316 291)(37 256 317 292)(38 257 318 293)(39 258 319 294)(40 259 320 295)(41 283 54 238)(42 284 55 239)(43 285 56 240)(44 286 57 231)(45 287 58 232)(46 288 59 233)(47 289 60 234)(48 290 51 235)(49 281 52 236)(50 282 53 237)(91 176 137 212)(92 177 138 213)(93 178 139 214)(94 179 140 215)(95 180 131 216)(96 171 132 217)(97 172 133 218)(98 173 134 219)(99 174 135 220)(100 175 136 211)(101 186 116 191)(102 187 117 192)(103 188 118 193)(104 189 119 194)(105 190 120 195)(106 181 111 196)(107 182 112 197)(108 183 113 198)(109 184 114 199)(110 185 115 200)(121 164 141 226)(122 165 142 227)(123 166 143 228)(124 167 144 229)(125 168 145 230)(126 169 146 221)(127 170 147 222)(128 161 148 223)(129 162 149 224)(130 163 150 225)
(1 237 141 244)(2 238 142 245)(3 239 143 246)(4 240 144 247)(5 231 145 248)(6 232 146 249)(7 233 147 250)(8 234 148 241)(9 235 149 242)(10 236 150 243)(11 202 312 226)(12 203 313 227)(13 204 314 228)(14 205 315 229)(15 206 316 230)(16 207 317 221)(17 208 318 222)(18 209 319 223)(19 210 320 224)(20 201 311 225)(21 179 48 183)(22 180 49 184)(23 171 50 185)(24 172 41 186)(25 173 42 187)(26 174 43 188)(27 175 44 189)(28 176 45 190)(29 177 46 181)(30 178 47 182)(31 163 64 156)(32 164 65 157)(33 165 66 158)(34 166 67 159)(35 167 68 160)(36 168 69 151)(37 169 70 152)(38 170 61 153)(39 161 62 154)(40 162 63 155)(51 198 75 215)(52 199 76 216)(53 200 77 217)(54 191 78 218)(55 192 79 219)(56 193 80 220)(57 194 71 211)(58 195 72 212)(59 196 73 213)(60 197 74 214)(81 287 126 301)(82 288 127 302)(83 289 128 303)(84 290 129 304)(85 281 130 305)(86 282 121 306)(87 283 122 307)(88 284 123 308)(89 285 124 309)(90 286 125 310)(91 292 105 275)(92 293 106 276)(93 294 107 277)(94 295 108 278)(95 296 109 279)(96 297 110 280)(97 298 101 271)(98 299 102 272)(99 300 103 273)(100 291 104 274)(111 261 138 257)(112 262 139 258)(113 263 140 259)(114 264 131 260)(115 265 132 251)(116 266 133 252)(117 267 134 253)(118 268 135 254)(119 269 136 255)(120 270 137 256)
G:=sub<Sym(320)| (1,2,3,4,5,6,7,8,9,10)(11,12,13,14,15,16,17,18,19,20)(21,22,23,24,25,26,27,28,29,30)(31,32,33,34,35,36,37,38,39,40)(41,42,43,44,45,46,47,48,49,50)(51,52,53,54,55,56,57,58,59,60)(61,62,63,64,65,66,67,68,69,70)(71,72,73,74,75,76,77,78,79,80)(81,82,83,84,85,86,87,88,89,90)(91,92,93,94,95,96,97,98,99,100)(101,102,103,104,105,106,107,108,109,110)(111,112,113,114,115,116,117,118,119,120)(121,122,123,124,125,126,127,128,129,130)(131,132,133,134,135,136,137,138,139,140)(141,142,143,144,145,146,147,148,149,150)(151,152,153,154,155,156,157,158,159,160)(161,162,163,164,165,166,167,168,169,170)(171,172,173,174,175,176,177,178,179,180)(181,182,183,184,185,186,187,188,189,190)(191,192,193,194,195,196,197,198,199,200)(201,202,203,204,205,206,207,208,209,210)(211,212,213,214,215,216,217,218,219,220)(221,222,223,224,225,226,227,228,229,230)(231,232,233,234,235,236,237,238,239,240)(241,242,243,244,245,246,247,248,249,250)(251,252,253,254,255,256,257,258,259,260)(261,262,263,264,265,266,267,268,269,270)(271,272,273,274,275,276,277,278,279,280)(281,282,283,284,285,286,287,288,289,290)(291,292,293,294,295,296,297,298,299,300)(301,302,303,304,305,306,307,308,309,310)(311,312,313,314,315,316,317,318,319,320), (1,132,86,96)(2,133,87,97)(3,134,88,98)(4,135,89,99)(5,136,90,100)(6,137,81,91)(7,138,82,92)(8,139,83,93)(9,140,84,94)(10,131,85,95)(11,77,65,23)(12,78,66,24)(13,79,67,25)(14,80,68,26)(15,71,69,27)(16,72,70,28)(17,73,61,29)(18,74,62,30)(19,75,63,21)(20,76,64,22)(31,49,311,52)(32,50,312,53)(33,41,313,54)(34,42,314,55)(35,43,315,56)(36,44,316,57)(37,45,317,58)(38,46,318,59)(39,47,319,60)(40,48,320,51)(101,142,116,122)(102,143,117,123)(103,144,118,124)(104,145,119,125)(105,146,120,126)(106,147,111,127)(107,148,112,128)(108,149,113,129)(109,150,114,130)(110,141,115,121)(151,211,206,175)(152,212,207,176)(153,213,208,177)(154,214,209,178)(155,215,210,179)(156,216,201,180)(157,217,202,171)(158,218,203,172)(159,219,204,173)(160,220,205,174)(161,197,223,182)(162,198,224,183)(163,199,225,184)(164,200,226,185)(165,191,227,186)(166,192,228,187)(167,193,229,188)(168,194,230,189)(169,195,221,190)(170,196,222,181)(231,291,286,255)(232,292,287,256)(233,293,288,257)(234,294,289,258)(235,295,290,259)(236,296,281,260)(237,297,282,251)(238,298,283,252)(239,299,284,253)(240,300,285,254)(241,277,303,262)(242,278,304,263)(243,279,305,264)(244,280,306,265)(245,271,307,266)(246,272,308,267)(247,273,309,268)(248,274,310,269)(249,275,301,270)(250,276,302,261), (1,202,86,157)(2,203,87,158)(3,204,88,159)(4,205,89,160)(5,206,90,151)(6,207,81,152)(7,208,82,153)(8,209,83,154)(9,210,84,155)(10,201,85,156)(11,280,65,265)(12,271,66,266)(13,272,67,267)(14,273,68,268)(15,274,69,269)(16,275,70,270)(17,276,61,261)(18,277,62,262)(19,278,63,263)(20,279,64,264)(21,304,75,242)(22,305,76,243)(23,306,77,244)(24,307,78,245)(25,308,79,246)(26,309,80,247)(27,310,71,248)(28,301,72,249)(29,302,73,250)(30,303,74,241)(31,260,311,296)(32,251,312,297)(33,252,313,298)(34,253,314,299)(35,254,315,300)(36,255,316,291)(37,256,317,292)(38,257,318,293)(39,258,319,294)(40,259,320,295)(41,283,54,238)(42,284,55,239)(43,285,56,240)(44,286,57,231)(45,287,58,232)(46,288,59,233)(47,289,60,234)(48,290,51,235)(49,281,52,236)(50,282,53,237)(91,176,137,212)(92,177,138,213)(93,178,139,214)(94,179,140,215)(95,180,131,216)(96,171,132,217)(97,172,133,218)(98,173,134,219)(99,174,135,220)(100,175,136,211)(101,186,116,191)(102,187,117,192)(103,188,118,193)(104,189,119,194)(105,190,120,195)(106,181,111,196)(107,182,112,197)(108,183,113,198)(109,184,114,199)(110,185,115,200)(121,164,141,226)(122,165,142,227)(123,166,143,228)(124,167,144,229)(125,168,145,230)(126,169,146,221)(127,170,147,222)(128,161,148,223)(129,162,149,224)(130,163,150,225), (1,237,141,244)(2,238,142,245)(3,239,143,246)(4,240,144,247)(5,231,145,248)(6,232,146,249)(7,233,147,250)(8,234,148,241)(9,235,149,242)(10,236,150,243)(11,202,312,226)(12,203,313,227)(13,204,314,228)(14,205,315,229)(15,206,316,230)(16,207,317,221)(17,208,318,222)(18,209,319,223)(19,210,320,224)(20,201,311,225)(21,179,48,183)(22,180,49,184)(23,171,50,185)(24,172,41,186)(25,173,42,187)(26,174,43,188)(27,175,44,189)(28,176,45,190)(29,177,46,181)(30,178,47,182)(31,163,64,156)(32,164,65,157)(33,165,66,158)(34,166,67,159)(35,167,68,160)(36,168,69,151)(37,169,70,152)(38,170,61,153)(39,161,62,154)(40,162,63,155)(51,198,75,215)(52,199,76,216)(53,200,77,217)(54,191,78,218)(55,192,79,219)(56,193,80,220)(57,194,71,211)(58,195,72,212)(59,196,73,213)(60,197,74,214)(81,287,126,301)(82,288,127,302)(83,289,128,303)(84,290,129,304)(85,281,130,305)(86,282,121,306)(87,283,122,307)(88,284,123,308)(89,285,124,309)(90,286,125,310)(91,292,105,275)(92,293,106,276)(93,294,107,277)(94,295,108,278)(95,296,109,279)(96,297,110,280)(97,298,101,271)(98,299,102,272)(99,300,103,273)(100,291,104,274)(111,261,138,257)(112,262,139,258)(113,263,140,259)(114,264,131,260)(115,265,132,251)(116,266,133,252)(117,267,134,253)(118,268,135,254)(119,269,136,255)(120,270,137,256)>;
G:=Group( (1,2,3,4,5,6,7,8,9,10)(11,12,13,14,15,16,17,18,19,20)(21,22,23,24,25,26,27,28,29,30)(31,32,33,34,35,36,37,38,39,40)(41,42,43,44,45,46,47,48,49,50)(51,52,53,54,55,56,57,58,59,60)(61,62,63,64,65,66,67,68,69,70)(71,72,73,74,75,76,77,78,79,80)(81,82,83,84,85,86,87,88,89,90)(91,92,93,94,95,96,97,98,99,100)(101,102,103,104,105,106,107,108,109,110)(111,112,113,114,115,116,117,118,119,120)(121,122,123,124,125,126,127,128,129,130)(131,132,133,134,135,136,137,138,139,140)(141,142,143,144,145,146,147,148,149,150)(151,152,153,154,155,156,157,158,159,160)(161,162,163,164,165,166,167,168,169,170)(171,172,173,174,175,176,177,178,179,180)(181,182,183,184,185,186,187,188,189,190)(191,192,193,194,195,196,197,198,199,200)(201,202,203,204,205,206,207,208,209,210)(211,212,213,214,215,216,217,218,219,220)(221,222,223,224,225,226,227,228,229,230)(231,232,233,234,235,236,237,238,239,240)(241,242,243,244,245,246,247,248,249,250)(251,252,253,254,255,256,257,258,259,260)(261,262,263,264,265,266,267,268,269,270)(271,272,273,274,275,276,277,278,279,280)(281,282,283,284,285,286,287,288,289,290)(291,292,293,294,295,296,297,298,299,300)(301,302,303,304,305,306,307,308,309,310)(311,312,313,314,315,316,317,318,319,320), (1,132,86,96)(2,133,87,97)(3,134,88,98)(4,135,89,99)(5,136,90,100)(6,137,81,91)(7,138,82,92)(8,139,83,93)(9,140,84,94)(10,131,85,95)(11,77,65,23)(12,78,66,24)(13,79,67,25)(14,80,68,26)(15,71,69,27)(16,72,70,28)(17,73,61,29)(18,74,62,30)(19,75,63,21)(20,76,64,22)(31,49,311,52)(32,50,312,53)(33,41,313,54)(34,42,314,55)(35,43,315,56)(36,44,316,57)(37,45,317,58)(38,46,318,59)(39,47,319,60)(40,48,320,51)(101,142,116,122)(102,143,117,123)(103,144,118,124)(104,145,119,125)(105,146,120,126)(106,147,111,127)(107,148,112,128)(108,149,113,129)(109,150,114,130)(110,141,115,121)(151,211,206,175)(152,212,207,176)(153,213,208,177)(154,214,209,178)(155,215,210,179)(156,216,201,180)(157,217,202,171)(158,218,203,172)(159,219,204,173)(160,220,205,174)(161,197,223,182)(162,198,224,183)(163,199,225,184)(164,200,226,185)(165,191,227,186)(166,192,228,187)(167,193,229,188)(168,194,230,189)(169,195,221,190)(170,196,222,181)(231,291,286,255)(232,292,287,256)(233,293,288,257)(234,294,289,258)(235,295,290,259)(236,296,281,260)(237,297,282,251)(238,298,283,252)(239,299,284,253)(240,300,285,254)(241,277,303,262)(242,278,304,263)(243,279,305,264)(244,280,306,265)(245,271,307,266)(246,272,308,267)(247,273,309,268)(248,274,310,269)(249,275,301,270)(250,276,302,261), (1,202,86,157)(2,203,87,158)(3,204,88,159)(4,205,89,160)(5,206,90,151)(6,207,81,152)(7,208,82,153)(8,209,83,154)(9,210,84,155)(10,201,85,156)(11,280,65,265)(12,271,66,266)(13,272,67,267)(14,273,68,268)(15,274,69,269)(16,275,70,270)(17,276,61,261)(18,277,62,262)(19,278,63,263)(20,279,64,264)(21,304,75,242)(22,305,76,243)(23,306,77,244)(24,307,78,245)(25,308,79,246)(26,309,80,247)(27,310,71,248)(28,301,72,249)(29,302,73,250)(30,303,74,241)(31,260,311,296)(32,251,312,297)(33,252,313,298)(34,253,314,299)(35,254,315,300)(36,255,316,291)(37,256,317,292)(38,257,318,293)(39,258,319,294)(40,259,320,295)(41,283,54,238)(42,284,55,239)(43,285,56,240)(44,286,57,231)(45,287,58,232)(46,288,59,233)(47,289,60,234)(48,290,51,235)(49,281,52,236)(50,282,53,237)(91,176,137,212)(92,177,138,213)(93,178,139,214)(94,179,140,215)(95,180,131,216)(96,171,132,217)(97,172,133,218)(98,173,134,219)(99,174,135,220)(100,175,136,211)(101,186,116,191)(102,187,117,192)(103,188,118,193)(104,189,119,194)(105,190,120,195)(106,181,111,196)(107,182,112,197)(108,183,113,198)(109,184,114,199)(110,185,115,200)(121,164,141,226)(122,165,142,227)(123,166,143,228)(124,167,144,229)(125,168,145,230)(126,169,146,221)(127,170,147,222)(128,161,148,223)(129,162,149,224)(130,163,150,225), (1,237,141,244)(2,238,142,245)(3,239,143,246)(4,240,144,247)(5,231,145,248)(6,232,146,249)(7,233,147,250)(8,234,148,241)(9,235,149,242)(10,236,150,243)(11,202,312,226)(12,203,313,227)(13,204,314,228)(14,205,315,229)(15,206,316,230)(16,207,317,221)(17,208,318,222)(18,209,319,223)(19,210,320,224)(20,201,311,225)(21,179,48,183)(22,180,49,184)(23,171,50,185)(24,172,41,186)(25,173,42,187)(26,174,43,188)(27,175,44,189)(28,176,45,190)(29,177,46,181)(30,178,47,182)(31,163,64,156)(32,164,65,157)(33,165,66,158)(34,166,67,159)(35,167,68,160)(36,168,69,151)(37,169,70,152)(38,170,61,153)(39,161,62,154)(40,162,63,155)(51,198,75,215)(52,199,76,216)(53,200,77,217)(54,191,78,218)(55,192,79,219)(56,193,80,220)(57,194,71,211)(58,195,72,212)(59,196,73,213)(60,197,74,214)(81,287,126,301)(82,288,127,302)(83,289,128,303)(84,290,129,304)(85,281,130,305)(86,282,121,306)(87,283,122,307)(88,284,123,308)(89,285,124,309)(90,286,125,310)(91,292,105,275)(92,293,106,276)(93,294,107,277)(94,295,108,278)(95,296,109,279)(96,297,110,280)(97,298,101,271)(98,299,102,272)(99,300,103,273)(100,291,104,274)(111,261,138,257)(112,262,139,258)(113,263,140,259)(114,264,131,260)(115,265,132,251)(116,266,133,252)(117,267,134,253)(118,268,135,254)(119,269,136,255)(120,270,137,256) );
G=PermutationGroup([[(1,2,3,4,5,6,7,8,9,10),(11,12,13,14,15,16,17,18,19,20),(21,22,23,24,25,26,27,28,29,30),(31,32,33,34,35,36,37,38,39,40),(41,42,43,44,45,46,47,48,49,50),(51,52,53,54,55,56,57,58,59,60),(61,62,63,64,65,66,67,68,69,70),(71,72,73,74,75,76,77,78,79,80),(81,82,83,84,85,86,87,88,89,90),(91,92,93,94,95,96,97,98,99,100),(101,102,103,104,105,106,107,108,109,110),(111,112,113,114,115,116,117,118,119,120),(121,122,123,124,125,126,127,128,129,130),(131,132,133,134,135,136,137,138,139,140),(141,142,143,144,145,146,147,148,149,150),(151,152,153,154,155,156,157,158,159,160),(161,162,163,164,165,166,167,168,169,170),(171,172,173,174,175,176,177,178,179,180),(181,182,183,184,185,186,187,188,189,190),(191,192,193,194,195,196,197,198,199,200),(201,202,203,204,205,206,207,208,209,210),(211,212,213,214,215,216,217,218,219,220),(221,222,223,224,225,226,227,228,229,230),(231,232,233,234,235,236,237,238,239,240),(241,242,243,244,245,246,247,248,249,250),(251,252,253,254,255,256,257,258,259,260),(261,262,263,264,265,266,267,268,269,270),(271,272,273,274,275,276,277,278,279,280),(281,282,283,284,285,286,287,288,289,290),(291,292,293,294,295,296,297,298,299,300),(301,302,303,304,305,306,307,308,309,310),(311,312,313,314,315,316,317,318,319,320)], [(1,132,86,96),(2,133,87,97),(3,134,88,98),(4,135,89,99),(5,136,90,100),(6,137,81,91),(7,138,82,92),(8,139,83,93),(9,140,84,94),(10,131,85,95),(11,77,65,23),(12,78,66,24),(13,79,67,25),(14,80,68,26),(15,71,69,27),(16,72,70,28),(17,73,61,29),(18,74,62,30),(19,75,63,21),(20,76,64,22),(31,49,311,52),(32,50,312,53),(33,41,313,54),(34,42,314,55),(35,43,315,56),(36,44,316,57),(37,45,317,58),(38,46,318,59),(39,47,319,60),(40,48,320,51),(101,142,116,122),(102,143,117,123),(103,144,118,124),(104,145,119,125),(105,146,120,126),(106,147,111,127),(107,148,112,128),(108,149,113,129),(109,150,114,130),(110,141,115,121),(151,211,206,175),(152,212,207,176),(153,213,208,177),(154,214,209,178),(155,215,210,179),(156,216,201,180),(157,217,202,171),(158,218,203,172),(159,219,204,173),(160,220,205,174),(161,197,223,182),(162,198,224,183),(163,199,225,184),(164,200,226,185),(165,191,227,186),(166,192,228,187),(167,193,229,188),(168,194,230,189),(169,195,221,190),(170,196,222,181),(231,291,286,255),(232,292,287,256),(233,293,288,257),(234,294,289,258),(235,295,290,259),(236,296,281,260),(237,297,282,251),(238,298,283,252),(239,299,284,253),(240,300,285,254),(241,277,303,262),(242,278,304,263),(243,279,305,264),(244,280,306,265),(245,271,307,266),(246,272,308,267),(247,273,309,268),(248,274,310,269),(249,275,301,270),(250,276,302,261)], [(1,202,86,157),(2,203,87,158),(3,204,88,159),(4,205,89,160),(5,206,90,151),(6,207,81,152),(7,208,82,153),(8,209,83,154),(9,210,84,155),(10,201,85,156),(11,280,65,265),(12,271,66,266),(13,272,67,267),(14,273,68,268),(15,274,69,269),(16,275,70,270),(17,276,61,261),(18,277,62,262),(19,278,63,263),(20,279,64,264),(21,304,75,242),(22,305,76,243),(23,306,77,244),(24,307,78,245),(25,308,79,246),(26,309,80,247),(27,310,71,248),(28,301,72,249),(29,302,73,250),(30,303,74,241),(31,260,311,296),(32,251,312,297),(33,252,313,298),(34,253,314,299),(35,254,315,300),(36,255,316,291),(37,256,317,292),(38,257,318,293),(39,258,319,294),(40,259,320,295),(41,283,54,238),(42,284,55,239),(43,285,56,240),(44,286,57,231),(45,287,58,232),(46,288,59,233),(47,289,60,234),(48,290,51,235),(49,281,52,236),(50,282,53,237),(91,176,137,212),(92,177,138,213),(93,178,139,214),(94,179,140,215),(95,180,131,216),(96,171,132,217),(97,172,133,218),(98,173,134,219),(99,174,135,220),(100,175,136,211),(101,186,116,191),(102,187,117,192),(103,188,118,193),(104,189,119,194),(105,190,120,195),(106,181,111,196),(107,182,112,197),(108,183,113,198),(109,184,114,199),(110,185,115,200),(121,164,141,226),(122,165,142,227),(123,166,143,228),(124,167,144,229),(125,168,145,230),(126,169,146,221),(127,170,147,222),(128,161,148,223),(129,162,149,224),(130,163,150,225)], [(1,237,141,244),(2,238,142,245),(3,239,143,246),(4,240,144,247),(5,231,145,248),(6,232,146,249),(7,233,147,250),(8,234,148,241),(9,235,149,242),(10,236,150,243),(11,202,312,226),(12,203,313,227),(13,204,314,228),(14,205,315,229),(15,206,316,230),(16,207,317,221),(17,208,318,222),(18,209,319,223),(19,210,320,224),(20,201,311,225),(21,179,48,183),(22,180,49,184),(23,171,50,185),(24,172,41,186),(25,173,42,187),(26,174,43,188),(27,175,44,189),(28,176,45,190),(29,177,46,181),(30,178,47,182),(31,163,64,156),(32,164,65,157),(33,165,66,158),(34,166,67,159),(35,167,68,160),(36,168,69,151),(37,169,70,152),(38,170,61,153),(39,161,62,154),(40,162,63,155),(51,198,75,215),(52,199,76,216),(53,200,77,217),(54,191,78,218),(55,192,79,219),(56,193,80,220),(57,194,71,211),(58,195,72,212),(59,196,73,213),(60,197,74,214),(81,287,126,301),(82,288,127,302),(83,289,128,303),(84,290,129,304),(85,281,130,305),(86,282,121,306),(87,283,122,307),(88,284,123,308),(89,285,124,309),(90,286,125,310),(91,292,105,275),(92,293,106,276),(93,294,107,277),(94,295,108,278),(95,296,109,279),(96,297,110,280),(97,298,101,271),(98,299,102,272),(99,300,103,273),(100,291,104,274),(111,261,138,257),(112,262,139,258),(113,263,140,259),(114,264,131,260),(115,265,132,251),(116,266,133,252),(117,267,134,253),(118,268,135,254),(119,269,136,255),(120,270,137,256)]])
140 conjugacy classes
class 1 2A ··· 2G 4A 4B 4C 4D 4E ··· 4L 5A 5B 5C 5D 8A ··· 8H 10A ··· 10AB 20A ··· 20P 20Q ··· 20AV 40A ··· 40AF order 1 2 ··· 2 4 4 4 4 4 ··· 4 5 5 5 5 8 ··· 8 10 ··· 10 20 ··· 20 20 ··· 20 40 ··· 40 size 1 1 ··· 1 2 2 2 2 4 ··· 4 1 1 1 1 2 ··· 2 1 ··· 1 2 ··· 2 4 ··· 4 2 ··· 2
140 irreducible representations
dim 1 1 1 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 type + + + + + + + - image C1 C2 C2 C2 C2 C4 C5 C10 C10 C10 C10 C20 D4 D4 SD16 Q16 C5×D4 C5×D4 C5×SD16 C5×Q16 kernel C10×Q8⋊C4 C5×Q8⋊C4 C10×C4⋊C4 C22×C40 Q8×C2×C10 Q8×C10 C2×Q8⋊C4 Q8⋊C4 C2×C4⋊C4 C22×C8 C22×Q8 C2×Q8 C2×C20 C22×C10 C2×C10 C2×C10 C2×C4 C23 C22 C22 # reps 1 4 1 1 1 8 4 16 4 4 4 32 3 1 4 4 12 4 16 16
Matrix representation of C10×Q8⋊C4 in GL5(𝔽41)
40 0 0 0 0 0 23 0 0 0 0 0 23 0 0 0 0 0 23 0 0 0 0 0 23
,
1 0 0 0 0 0 40 0 0 0 0 0 40 0 0 0 0 0 40 39 0 0 0 1 1
,
40 0 0 0 0 0 0 40 0 0 0 40 0 0 0 0 0 0 21 14 0 0 0 27 20
,
9 0 0 0 0 0 17 1 0 0 0 40 24 0 0 0 0 0 31 26 0 0 0 23 10
G:=sub<GL(5,GF(41))| [40,0,0,0,0,0,23,0,0,0,0,0,23,0,0,0,0,0,23,0,0,0,0,0,23],[1,0,0,0,0,0,40,0,0,0,0,0,40,0,0,0,0,0,40,1,0,0,0,39,1],[40,0,0,0,0,0,0,40,0,0,0,40,0,0,0,0,0,0,21,27,0,0,0,14,20],[9,0,0,0,0,0,17,40,0,0,0,1,24,0,0,0,0,0,31,23,0,0,0,26,10] >;
C10×Q8⋊C4 in GAP, Magma, Sage, TeX
C_{10}\times Q_8\rtimes C_4
% in TeX
G:=Group("C10xQ8:C4");
// GroupNames label
G:=SmallGroup(320,916);
// by ID
G=gap.SmallGroup(320,916);
# by ID
G:=PCGroup([7,-2,-2,-2,-5,-2,-2,-2,560,589,1128,7004,3511,172]);
// Polycyclic
G:=Group<a,b,c,d|a^10=b^4=d^4=1,c^2=b^2,a*b=b*a,a*c=c*a,a*d=d*a,c*b*c^-1=d*b*d^-1=b^-1,d*c*d^-1=b^-1*c>;
// generators/relations
×
𝔽
|
2021-05-08 11:09:36
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https://optimization-online.org/2022/04/8878/
|
# Solution Strategies for Integrated Distribution, Production, and Routing Problems Arising in Modular Manufacturing
Recently, there has been a paradigm shift by certain energy and chemical companies towards modular manufacturing, whereby transportable modular production units can be relocated between production facilities to meet the spatial and temporal changes in the availabilities, demands, and prices of the underlying commodities. We refer to the optimal distribution, production, and storage of commodities, and the relocation and operation of the modular production units as the \textit{dynamic multiple commodity supply chain problem with modular production units}. To this end, we present a flow-based'' and a path-based'' mixed-integer linear programming formulation to model the problem. In an effort to solve large-scale instances of the problem, we propose an iterative three-stage matheuristic for the path-based'' formulation. In the first stage of the matheuristic, a feasible solution to the problem is generated by an integrated column generation and Lagrangian relaxation based heuristic. In the second stage of the matheuristic, a path-relinking procedure is utilized as a local search heuristic to further improve the solution. And in the final stage of the matheuristic, the Lagrangian multipliers are updated via a subgradient method. The effectiveness of the matheuristic is illustrated through numerical experiments with a set of randomly generated test instances. For the large-scale test instances, the results show that the matheuristic produces quality solutions orders of magnitude faster than a state-of-the-art'' mixed-integer linear programming solver.
## Citation
Report Number: NA Institution Address: Texas A&M Energy Institute, 1617 Research Pkwy, College Station, TX 77843 Month/Year: April/2022
|
2023-03-29 12:50:53
|
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https://proofwiki.org/wiki/Primitive_of_Reciprocal_of_p_squared_minus_square_of_q_by_Sine_of_a_x
|
# Primitive of Reciprocal of p squared minus square of q by Sine of a x
## Theorem
$\displaystyle \int \frac {\d x} {p^2 - q^2 \sin^2 a x} = \begin{cases} \displaystyle \frac 1 {a p \sqrt{p^2 - q^2} } \arctan \frac {\sqrt{p^2 - q^2} \tan a x} p & : p^2 > q^2 \\ \displaystyle \frac 1 {2 a p \sqrt{q^2 - p^2} } \ln \size {\frac {\sqrt {q^2 - p^2} \tan a x + p} {\sqrt {q^2 - p^2} \tan a x - p} } & : p^2 < q^2 \end{cases}$
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2020-06-03 10:26:41
|
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https://oeis.org/A169668
|
The OEIS is supported by the many generous donors to the OEIS Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A169668 The product of factorials s! where s belongs to the multiset of exponents of the Lie groups G=Br or G=Cr. Also 2^r times the classical Lie superfactorial of type Br ~ SO(2r+1). Also 2^{r(r-1)} times the Lie superfactorial of type Cr ~ Sp(2r). 2
6, 720, 3628800, 1316818944000, 52563198423859200000, 327312129899898454671360000000, 428017682605583614976547335700480000000000 (list; graph; refs; listen; history; text; internal format)
OFFSET 2,1 COMMENTS To every simple Lie group G one can associate both a quantum and a classical superfactorial of type G. The classical Lie superfactorial of type G , denoted sf_G, is defined as the classical limit (q-->1) of the quantum Weyl denominator of G. If G is simply laced (ADE Dynkin diagrams) ie Ar,Dr,E6,E7,E8 cases, the integer sf_G is the product of factorial s!, where s runs over the multiset of exponents of G. The usual superfactorial r --> sf[r] is recovered as the Lie superfactorial r --> sf_{Ar} of type Ar [nonascii characters here] SU(r+1), sequence A000178. The superfactorial of type Dr [nonascii characters here] SO(2r) defines the infinite sequence A169657. If G is exceptional of type E, the Lie superfactorial defines a sequence with only three terms, see sequence A169667. If G is not simply laced, ie Br (this case), Cr (this case), G2 or F4, the Lie superfactorial differs by simple pre-factors from the product of factorials of exponents. If G=Br ~ SO(2r+1), the pre-factor is 1/2^r and r --> sf_{Br} = (1/2^r) Product_{s \in 1,3,5,.., 2r-1} s! If G = Cr ~ Sp(2r), the pre-factor is 1/2^{r(r-1)} and r --> sf_{Cr} = (1/2^{r(r-1)}) Product_{s \in 1,3,5,.., 2r-1} s! If G = F4, sf_{F4} = 1/2^{12} 1! 5! 7! 11! = 5893965000 (a sequence with only one term). If G = G2, sf_{G2} = 1/3^{3} 1! 5! = 40/9 (a sequence with only one term). The classical Lie superfactorial of type G enters the asymptotic expression giving the global dimension of a monoidal category of type G at level k, when k is large. Call gamma the Coxeter number of G, r its rank, Delta the determinant of the fundamental quadratic form, and dim(G) its dimension, the asymptotic expression reads : k^dim(G) / ((2 pi)^(r gamma) Delta (sf_G)^2 ). LINKS R. Coquereaux, Global dimensions for Lie groups at level k and their conformally exceptional quantum subgroups, arxiv:1003.2589 R. Coquereaux, Quantum McKay correspondence and global dimensions for fusion and module-categories associated with Lie groups, arXiv preprint arXiv:1209.6621, 2012. - From N. J. A. Sloane, Dec 29 2012 FORMULA Product_{s \in 1,3,5,.., 2r-1} s! a(n) ~ 2^(n^2 + n + 5/24) * n^(n^2 + n/2 - 1/24) * Pi^(n/2) / (sqrt(A) * exp(n*(3*n+1)/2 - 1/24)), where A is the Glaisher-Kinkelin constant A074962. - Vaclav Kotesovec, Mar 05 2021 MATHEMATICA Product[Factorial[s], {s, 1, (2 r - 1), 2}] CROSSREFS A000178 gives sf_G for G=Ar=SU(r+1). A169657 gives sf_G for G=Dr~SO(2r). A169667 gives sf_G for G=E6, E7, E8. Sequence in context: A003923 A002204 A052295 * A168467 A080369 A036981 Adjacent sequences: A169665 A169666 A169667 * A169669 A169670 A169671 KEYWORD easy,nonn AUTHOR Robert Coquereaux, Apr 05 2010 STATUS approved
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Last modified January 18 09:26 EST 2022. Contains 350454 sequences. (Running on oeis4.)
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2022-01-18 14:49:29
|
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|
https://stats.stackexchange.com/questions/174923/checking-if-a-time-series-is-weak-sense-stationary
|
# Checking if a time series is weak-sense stationary
I just learned that a time series is called weak-sense stationary if:
1 $\mathbb{E}[X_t] = \mu$ is independent of $t$.
2 $\operatorname{Cov}(X_{t+h},X_t)$ is independent of $t$ for each $h$.
I am trying to check this for the following time series:
$X_t=b+Z_t-Z_{t-1}$ with {${Z_t}$} ~ $WN(0,\sigma^2)$.
It is quite easy to see that $\mathbb{E}[X_t] = \mu$ is independent of $t$, but how can you check condition 2? When trying to calculate $\operatorname{Cov}(X_{t+h},X_t)$, I end up with the following:
$\operatorname{Cov}(Y_t,Y_{t+h})=\operatorname{Cov}(b+Z_t-Z_{t-1},b+Z_{t+h}-Z_{t-1+h})\\=\operatorname{Cov}(Z_t,b+Z_{t+h}-Z_{t-1+h})-\operatorname{Cov}(Z_{t-1},b+Z_{t+h}-Z_{t-1+h})\\ =\operatorname{Cov}(Z_{t+h},Z_t)-\operatorname{Cov}(Z_{t-1+h},Z_t)-\operatorname{Cov}(Z_{t+h},Z_{t-1})+\operatorname{Cov}(Z_{t-1+h},Z_{t-1})$
It isn't clear to me how I can conclude that $\operatorname{Cov}(X_{t+h},X_t)$ is or is not independent of $t$ for each $h$.
• Hints: The $Z$'s are from a white noise process, and so $cov(Z_t,Z_s)=0$ unless $t=s$. So look at each and every one of the $Cov$ thingies that you calculated. Is it possible that the subscripts are equal? Does this equality hold only for certain choices of value of $h$ or does it hold regardless of the choice of $h$? That way, you can work out what values the four terms have. – Dilip Sarwate Oct 1 '15 at 0:11
• Thanks for you hint. I came to the following: if h=0 you have $2\sigma^2$, if h=1 or -1 you have $-\sigma^2$. For the rest it would be independent. But does this mean that condition 2 doesn't fit? Because $cov(Y_t,Y_{t+h})$ isn't independent of $t$ for each $h$. – Alfons Ingomar Oct 1 '15 at 0:40
• You just figured out that the covariance of $X_t$ and $X_{t+h}$ is known if you specify the value of $h$. But it doesn't depend on what $t$ is, does it? – Dilip Sarwate Oct 1 '15 at 2:08
• Ah of course, that makes sense! So you it does satisfy with condition 2, since $t$ has doesn't influence the outcome. Thanks! – Alfons Ingomar Oct 1 '15 at 9:15
So I figured it out with some help from the comments beneath my question. Since $Z$ are from a white noise process, $cov(Z_t,Z_s) = 0$ unless $t=s$, in that case you have $\sigma^2$.
$\operatorname{Cov}(Z_{t+h},Z_t)-\operatorname{Cov}(Z_{t-1+h},Z_t)-\operatorname{Cov}(Z_{t+h},Z_{t-1})+\operatorname{Cov}(Z_{t-1+h},Z_{t-1})$
• If $h = 0$,you have $2\sigma^2$.
• If $h = 1 or -1$ you have $-\sigma^2$
• 0 for al other possibilities of $h$
Since this is true for each value of $t$, you can conclude that $\operatorname{Cov}(Z_{t+h},Z_t)$ is independent of $t$ for each $h$.
|
2019-12-13 00:11:46
|
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http://eprint.iacr.org/2010/012/20100112:040002
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## Cryptology ePrint Archive: Report 2010/012
Differential Cache Trace Attack Against CLEFIA
Chester Rebeiro and Debdeep Mukhopadhyay
Abstract: The paper presents a differential cache trace attack against CLEFIA, a $128$ bit block cipher designed by Sony Corporation. The attack shows that such ciphers based on the generalized Feistel structures leak information of the secret key if the cache trace pattern is revealed to an adversary. The attack that we propose is a three staged attack and reveals the entire key with $2^{43}$ CLEFIA encryptions. The attack is simulated on an Intel Core 2 Duo Processor with a cache architecture with $32$ byte lines as a target platform.
Category / Keywords: secret-key cryptography /
Date: received 10 Jan 2010
Contact author: chetrebeiro at gmail com
Available format(s): Postscript (PS) | Compressed Postscript (PS.GZ) | PDF | BibTeX Citation
Short URL: ia.cr/2010/012
[ Cryptology ePrint archive ]
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2017-01-18 06:28:07
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https://www.physicsforums.com/threads/numbers-and-constants.11791/
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# Numbers and Constants
1. Dec 30, 2003
### Mk
What is the purpose of all the Planck's stuff? Planck's length, Planck's energy. And what is Boltzman's constant? On my calculater (PCalc 2 for mac) what is the "Universal Gas" (8.31451)?
2. Dec 31, 2003
### Integral
Staff Emeritus
The various "Planck" quantities are arrived at by manipulation of Planck Constant, which is an indication of the smallest units of energy. (Sort of like Universal pennies)
The numbers are not what you should be looking at when you come upon a "constant". You give the number associated with the Gas constant, it is meaningless without the UNITS.
The gas constant is the R in
PV= nRT, the exact number will depend on the units of the other quantities. To get meaningful results the units must be correct.
Boltzmann constant is related to the Gas constant, it is also found in thermodynamics. Once again learn to examine the units associated with such numbers,they tell you a lot.
3. Dec 31, 2003
### Hessam
While on the topic of planck... i'm new at physics (self-evidently once you read my question), what exactly is the value of plancks constant? i've read materials that talk about it, but the actual numerical value is what i dont know, any help will be appreciated
4. Dec 31, 2003
### dlgoff
Planks Constant h=6.626x10^-34 Joule sec.
5. Feb 18, 2004
### PRodQuanta
Yes, what dlgoff said is correct. Max Planck was the founder of quantum physics. He said that energy is not continuous, but comes in small, discrete spurts. The energy on one of those spurts = Planck length.
6. Feb 19, 2004
### chroot
Staff Emeritus
Energy and length are of different units.
- Warren
7. Feb 23, 2004
### PRodQuanta
yep, sorry, he's right, and that is what I intended. i meant energy.
8. Feb 24, 2004
### Michael F. Dmitriyev
Any "fundamental constant” is a patch hiding the gaping hole in our knowledge. For example,
gravity force = G*m1*m2/r^2
Presence of G means, that we know nothing about mass, space and gravity. In other cases also.
Any relation should be straight and should not contain any constants.
9. Feb 24, 2004
### Integral
Staff Emeritus
Please ignore the above post, it adds nothing of significance to the conversation.
Boltzmann constant defines the relationship between Temperature and how entropy changes with a changing energy. Bet that helped a lot! Mathematically (which, most likely, will not help much either!)
$$\frac 1 \tau = (\frac {\partial \sigma} {\partial U})_{N,V}$$
Where:
$$\tau = {Fundamental\ Temperature}$$
$$\sigma = {Fundamental\ Entropy}$$
$$U = {Energy}$$
The Boltzmann's constant is defined to be
$$\tau = k_B T$$
Where T is a temperature scale that you are familiar with (Kelvin, Celsius, Fahrenheit etc) depending on the value of kB
this constant is used extensively in thermodynamics.
R the gas constant is defined to be
R = N0kB
N0= 6.022 x 1023
kB=1.38 x 10-18erg/deg
R=8.314 x 107erg/(mole deg)
Always include units with fundamental constants, also to learn about the constant, look at the units. In Physics and chemistry it is essential that you learn to appreciate the physical units attached to a number.
Last edited: Feb 24, 2004
10. Feb 24, 2004
### ahrkron
Staff Emeritus
Re: Re: Numbers and Constants
I strongly disagree with this. Whe you find a relation like that of gravity, you have pretty much nailed down: 1. which quantities are relevant for the interaction, 2. how they are involved in the phenomenon under study, and 3. the fact that a combination of them gives you always the same quantity (G in this case).
Also, constants have to do with the definition of the units you choose to use. Some constants in nature are indeed intriguing (maybe 26), but apart from those, they are not "gaps".
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2017-05-27 02:33:30
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https://ncatlab.org/nlab/show/metrisable+topological+space
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# Contents
## Definition
A topological space $(X,\tau)$ is called metrisable if there exists the stucture of a metric space $(X,d)$ on the underlying set, such that $\tau$ is the corresponding metric topology.
## Metrisability theorem
Various theorems state sufficient conditions for a topological space to be metrisable:
(…)
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2017-09-21 01:18:45
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https://socratic.org/questions/how-do-you-use-the-vertical-line-test-to-show-x-absy-is-a-function
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# How do you use the vertical line test to show x=absy is a function?
$y$ is not a function of $x$.
For example, the vertical line $x = 1$ intersects the relation $x = \left\mid y \right\mid$ at two points:
$\left(1 , 1\right)$ and $\left(1 , - 1\right)$
So $y$ is not uniquely determined by $x$.
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2019-10-21 23:43:18
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https://www.eco-tec.com/zh-hans/markets/water-treatment/produced-water-treatment/softening/
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Ion exchange softening, removal of calcium and magnesium is a key step in the recycle process of produced water. It is generally considered necessary to reduce hardness levels to less than 1 mg/L (as CaCO3) and preferably less than 0.2 mg/L. In an effort to avoid boiler fouling, the trend is to specify even lower hardness limits – sometimes as low as 0.1 mg/L.
Weak acid cation (WAC) resins are normally utilized for softening high TDS produced water. The main disadvantage of the WAC process is that the resins require regeneration with hydrochloric acid followed by sodium hydroxide.
Strong acid cation (SAC) ion exchange processes employing only brine regeneration (i.e. no acid or caustic) are much simpler and less expensive, however the SAC process is normally restricted to low TDS (<3000 mg/L) waters.
A new ion exchange softening flow-sheet that uses brine regeneration has been developed based upon the Recoflo® short-bed ion exchange process. Recoflo® makes it possible to use extremely high service flow rates through very short (15 cm) resin beds, while achieving low hardness leakages. The key to applying this to high TDS produced water is the use of high purity brine for resin regeneration.
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2020-08-09 22:54:16
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http://encyclopedia.kids.net.au/page/u./U.S._presidential_election%2C_1820
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## Encyclopedia > U.S. presidential election, 1820
Article Content
# U.S. presidential election, 1820
President: Main Opponent: Electoral Vote: James Monroe (Democratic-Republican) John Quincy Adams (National-Republican) Winner: 231 Main Opponent: 1 Total/Majority: 235/118 no record Daniel D. Tompkins (218) Richard Stockton (8), Daniel Rodney[?] (4), Robert G. Harper[?] (1), Richard Rush[?] (1) 1808, 1812, 1816, 1820, 1824, 1828, 1832
Notes:
Monroe ran virtually unopposed, though a single vote for John Quincy Adams (then Secretary of State) was cast by one elector. (The belief that this was to ensure that George Washington remained the only president elected unanimously by the Electoral College seems to be a myth.)
235 electors were appointed, but only 232 votes were cast due to the deaths of electors from Mississippi, Pennsylvania and Tennessee.
All Wikipedia text is available under the terms of the GNU Free Documentation License
Search Encyclopedia
Search over one million articles, find something about almost anything!
Featured Article
Dark matter ... by Sanders (1984), is to replace the gravitational potential with the expression $U=\frac{GM(1-Be^{-r/\rho})}{(1-B)r}$ where B and ρ ...
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2014-11-27 23:08:55
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http://clay6.com/qa/51378/which-of-the-following-facts-regarding-boron-and-silicon-is-not-true-
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# Which of the following facts regarding boron and silicon is not true ?
$\begin {array} {1 1} (A)\;\text{Boron and silicons are semiconductors} \\ (B)\;\text{Boron and silicons form halides which are not hydrolysed} \\ (C)\;\text{Boron and silicon react with magnesium to form magnesium boride and magnesium silicide which are decomposed by acids to give volatile borane and silane, respectively.} \\ (D)\;\text{Both Boron and silicon react with alkalis to form borates and silicates containing$BO_4$and$SiO_4$tetrahedral units, respectively.} \end {array}$
The halides of B and Al are hydrolysed.
$BCl_3+3H_2O \rightarrow B(OH)_3 ( Boric \: acid ) + 3HCl$ and $SiCl_4+4H_2O \rightarrow Si(OH)_4 (Silicic\: acid ) +4HCl$
Hence (B) is the correct answer.
Boron and silicons form halides which are not hydrolysed
edited Jul 28, 2014
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2017-11-22 05:13:16
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https://www.gamedev.net/forums/topic/130290-wm_initdialog-sigh/
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#### Archived
This topic is now archived and is closed to further replies.
# WM_INITDIALOG... *sigh*!
## Recommended Posts
quote:
The WM_INITDIALOG message is sent to the dialog box procedure immediately before a dialog box is displayed. Dialog box procedures typically use this message to initialize controls and carry out any other initialization tasks that affect the appearance of the dialog box.
quote:
The SetDlgItemText function sets the title or text of a control in a dialog box.
The above are quotes from MSDN... If the above is true , then why won''t my code work?
BOOL CALLBACK DlgProc(HWND hWnd, UINT iMsg, WPARAM wParam, LPARAM lParam)
{
switch(iMsg)
{
case WM_INITDIALOG:
SetDlgItemText(hWnd, IDC_EDIT1, "Welcome to my program!");
return true;
}
return false;
}
IDC_EDIT1 is a constant used in the dialog template and is defined in my resource.h and included in the original source file. What am I doing wrong here?
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I''m going to assume that by the name IDC_EDIT1 you are using an edit text box and wish to fill in the value. Try the following instead:
SendDlgItemMessage( hWnd, IDC_EDIT1, WM_SETTEXT, 0, ( LPARAM )"Welcome..." );
If my assumption is incorrect, please elaborate.
##### Share on other sites
What value does SetDlgItemText return?
##### Share on other sites
The code you''ve posted works. The problem must be somewhere else, specifically in the dialog template or in the code to create the dialog.
##### Share on other sites
PropellerBoy: You assumption was correct, but I don''t think I need to used SendDlgItemMessage. I have tried doing SetDlgItemText in response to a WM_COMMAND message (ie. a button was clicked) and it worked, so SetDlgItemText seems to be al I need.
AP: The only code that is in my WinMain is a call to DialogBox. This function requires a HWND to the owner window, but since this dialog is the parent window, I just pass NULL. As far as I know (I checked MSDN but they don''t say if passing NULL is ok or not...) this isn''t the problem...
Solo: Duh! I could''ve thought of that myself..! I''m gonna check as soon as Christmas is over, as I don''t feel like sitting alone in my bedroom these days.
Oh, that reminds me! Merry Christmas to y''all!
##### Share on other sites
"The only code that is in my WinMain is a call to DialogBox."
so you don''t have a message pump?
##### Share on other sites
I''ve used that method before, successfully. The dialog call looks like this (in WinMain):
DialogBox( hInstance, MAKEINTRESOURCE( IDD_MY_DIALOG ), NULL, DialogFunction );
It works for single dialog apps. However, I''m sure it''s not the "best" way.
##### Share on other sites
Calling DialogBox as PropellerBoy posted is correct. NULL is acceptable for hwndParent and no message loop is necessary because DialogBox has an internal message loop.
I threw together a tester that uses your DlgProc with a call to DialogBox in WinMain. It works fine. Here''s what the resource stuff looks like...
In resource.h:
// AppWizard stuff snipped out#define IDD_DIALOG1 101#define IDC_EDIT1 1000// AppWizard stuff snipped out
In app.rc:
// AppWizard stuff snipped outIDD_DIALOG1 DIALOG DISCARDABLE 0, 0, 186, 30STYLE DS_MODALFRAME | WS_POPUP | WS_CAPTION | WS_SYSMENUCAPTION "Dialog"FONT 8, "MS Sans Serif"BEGIN EDITTEXT IDC_EDIT1,7,7,172,16,ES_AUTOHSCROLLEND// AppWizard stuff snipped out
##### Share on other sites
Well thanks AP for checking that for me, but I knew that that code would work (Or at least the dialog box pops up). The problem here is that for some reason I can''t initialize the dialogboxes edit control...
quote:
However, I''m sure it''s not the "best" way.
I know this probably isn''t the best way. I think the best way would be to supply my own messagepump by registering a WNDCLASSEX structure with it''s cbWndExtra field set to DLGWINDOWEXTRA, and using the CLASS directive in my resource template. This would instruct Windows not to use the standard dialogbox messagepump and let''s my program use it''s own. The problem with this is that the Windows messagepump allows the user to conveniently switch focus between my controls using the TAB-key, and I don''t feel like coding the ''TAB logic'' myself .
However, an added bonus to this approach is that I can use my own icon on the dialogbox by filling the hIcon field of the WNDCLASSEX struct. I have tried putting my own icon in there with the DialogBox approach (by using SetClassLong in response to the WM_INITDIALOG message) but this didn''t work. Anyone know how to do this?
So the questions now are :
1. How do I initialize my edit control, and
2. How can I let the dialogbox (and the .exe file itself) use my own kewl icon?
Thanks for the responses up until now, guys!
Regards, MainForze
##### Share on other sites
put this inside the funciton then use its handle instead of the hWnd you passed to the funciton:
HWND hWndEdit1 = GetDlgItem(hWnd,IDC_EDIT1);
• ### Forum Statistics
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2017-11-19 10:49:41
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https://ncatlab.org/nlab/show/ananatural+transformation
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nLab ananatural transformation
Ananatural transformations
Ananatural transformations
Idea
Just as natural transformations go between functors, ananatural transformations go between anafunctors.
Given two functors interpreted as anafunctors, the natural transformations and ananatural transformations between them are the same, so the term ‘ananatural’ is overkill; one only needs it to emphasise the ana-context and otherwise can just say ‘natural’. That is, a ‘natural transformation’ between anafunctors unambigously means an ananatural transformation.
Definitions
Given two categories $C$ and $D$ and two anafunctors $F, G\colon C \to D$, let us interpret $F,G$ as spans $C \leftarrow \overline{F} \rightarrow D$ and $C \leftarrow \overline{G} \rightarrow D$ of strict functors (where each backwards-pointing arrow is strictly surjective and faithful; see the definition of anafunctor). Form the strict $2$-pullback $P \coloneqq \overline{F} \times_C \overline{G}$ and consider the strict functors $P \to \overline{F} \to D$ and $P \to \overline{G} \to D$. Then an ananatural transformation from $F$ to $G$ is simply a natural transformation between these two strict functors.
More explicitly, if $F,G$ are given by sets ${|F|}, {|G|}$ of specifications and additional maps (see the other definition of anafunctor), then an ananatural transformation from $F$ to $G$ consists of a coherent family of morphisms of $D$ indexed by the elements of $|F|$ and $|G|$ with common values in $C$. That is:
• for each object $x$ of $C$, each $F$-specification $s$ over $x$, and each $G$-specification $t$ over $x$, we have a morphism
$\eta_{s,t}(x)\colon F_s(x) \to G_t(x)$
in $D$;
• for each morphism $f\colon x \to y$ in $C$, each pair of $F$-specifications $s,t$ over $x,y$, and each pair of $G$-specifications $u,v$ over $x,y$, the diagram
$\array { F_s(x) & \overset{\eta_{s,u}(x)}\rightarrow & G_u(x) \\ F_{s,t}(f) \downarrow & & \downarrow F_{u,v}(f) \\ F_t(y) & \underset{\eta_{t,v}(y)}\rightarrow & G_v(y) }$
is a commutative square.
Of course, an ananatural isomorphism is an invertible ananatural transformation.
Composition
Just as natural transformations can be composed vertically to form the morphisms of a functor category, so ananatural transformations can be composed vertically to form an anafunctor category.
Just as natural transformations can also be whiskered by functors and composed horizontally to make a strict 2-category $Str Cat$ of (strict) categories, (strict) functors and natural transformations, so ananatural transformations can also be whiskered by anafunctors and composed horizontally to make a bicategory $Cat_{ana}$ of (strict) categories, anafunctors and (ana)natural transformations. Assuming the axiom of choice, $Cat_{ana}$ is equivalent to $Str Cat$; without choice (and internally), $Cat_{ana}$ has better properties than $Str Cat$ and we will usually identify the former with Cat.
Last revised on November 1, 2011 at 00:36:11. See the history of this page for a list of all contributions to it.
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2021-10-16 21:13:10
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http://www.chegg.com/homework-help/definitions/natural-frequency-5
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# Definition of Natural Frequency
The natural frequency is the rate at which an object vibrates when it is not disturbed by an outside force. Each degree of freedom of an object has its own natural frequency, expressed as ωn (omega subscript n). Frequency (omega) is equal to the speed of vibration divided by the wavelength (lambda), . Other equations to calculate the natural frequency depend upon the vibration system. Natural frequency can be either undamped or damped, depending on whether the system has significant damping. The damped natural frequency is equal to the square root of the collective of one minus the damping ratio squared multiplied by the natural frequency, .
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2014-07-23 05:28:37
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https://stats.stackexchange.com/questions/239333/why-do-we-care-about-unbiasedness
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# Why do we care about unbiasedness?
The OLS estimator $\hat{\beta}$ of $\beta$ is unbiased. Now, $\hat{\beta}$ is a function of the random sample (y,$x_1$,...$x_j$) and is therefore a random variable itself. When we compute $\mathbb{E}(\hat{\beta}|X)$ we are basically fixing $X$ to be the data we have and we take an expectation over the $y$ variable. Unbiasedness tells us that if we compute all possible values of $\hat{\beta}$ by runnning and OLS regression for every possible $y$, then we would find that $\mathbb{E}(\hat{\beta}-\beta|X)=0$ by averaging over the $\hat{\beta}$'s, that is, the average over all those $\hat{\beta}$'s is the true population parameter (conditional on $X$). However, in practice we have only one $y$ which will allow to compute an OLS estimate $\hat{\beta}$ which differs almost surely from $\beta$ - since $\hat{\beta}$ is random, equality would be an event of pure luck. But that being the case, why do we care so much that the coefficient is unbiased? Is it because it allows for an interpretation of the coefficients as the marginal effect on the mean of the sampled random dependent variable (or the marginal effect of $X$ on $y$ when $y$ is drawn from a typical random sample in the sense that the sample mean equals the population mean) ? I have also read coefficients being interpreted as the average marginal effect of a variable on the dependent variable. However, I would read it not as the average/mean marginal effect on the dep. variable but rather as the marginal effect on the mean of the dep.variable conditional on $X$. Which interpretation is correct?
• Well, isn't it better (other things equal) to be right on average than wrong on average? It isn't that we care "so much" about it, it's just a nice property to have in general. That isn't to say we should never use biased estimators, oftentimes the "best" estimators are biased. – dsaxton Oct 10 '16 at 4:09
• People pay lip service to unbiasedness, but (speaking more generally than your regression example), they demonstrably don't care about it all that much. Who uses an unbiased estimator of standard deviation for example? How many people unbias their parameter estimates when fitting lognormal distributions? etc etc. I'm a bit more explicit in my opinion that many, since I don't pay much lip service to unbiasedness at all -- I really don't care much at all about unbiasedness -- it's rarely of practical relevance in things I deal with; I'll often take a biased estimate if it will get me close...ctd – Glen_b Oct 10 '16 at 5:43
• ctd... to what I want. Other things being equal, of course no bias is better than bias, but so often other things are far from equal. – Glen_b Oct 10 '16 at 5:45
• @Glen_b: but in the context of regression coefficients it may be of importance, e.g. if you perform hypothesis tests on the coefficient or if you compute a confidence interval for the coefficent ? For hypothesis tests you test whether the ''true'' value is e.g. different from zero (I think it is good to work with an unbiased estimate in that case), similar for a confidence interval, you find a (random) interval that has a certain probability to contain the ''true'' value ? With biased estimates you have tests/intervals for the ''true'' value plus ''something'' ? – user83346 Oct 10 '16 at 6:44
• @fcop Hypothesis tests are case in point -- many people use likelihood ratio tests but most MLEs are biased. Similarly many confidence intervals are not centered on the parameter. Consider confidence intervals for the standard deviation of the error term in regression for example. Or if you really want to focus on regression coefficients, consider the case where you have a contaminating process that leads to occasional wild values that are not centered at the underlying uncontaminated population line we're interested in. A good estimator will clearly be biased for the ...ctd – Glen_b Oct 10 '16 at 7:14
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2019-10-19 04:33:18
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https://proofwiki.org/wiki/Determinant_of_Elementary_Row_Matrix/Exchange_Rows
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# Determinant of Elementary Row Matrix/Exchange Rows
## Theorem
Let $e_3$ be the elementary row operation $\text {ERO} 3$:
$(\text {ERO} 3)$ $:$ $\ds r_i \leftrightarrow r_j$ Exchange rows $i$ and $j$
which is to operate on some arbitrary matrix space.
Let $\mathbf E_3$ be the elementary row matrix corresponding to $e_3$.
The determinant of $\mathbf E_3$ is:
$\map \det {\mathbf E_3} = -1$
## Proof
Let $\mathbf I$ denote the unit matrix of arbitrary order $n$.
$\map \det {\mathbf I} = 1$
Let $\rho$ be the permutation on $\tuple {1, 2, \ldots, n}$ which transposes $i$ and $j$.
From Parity of K-Cycle, $\map \sgn \rho = -1$.
By definition we have that $\mathbf E_3$ is $\mathbf I$ with rows $i$ and $j$ transposed.
By the definition of a determinant:
$\displaystyle \map \det {\mathbf I} = \sum_{\lambda} \paren {\map \sgn \lambda \prod_{k \mathop = 1}^n a_{k \map \lambda k} }$
$\displaystyle \map \det {\mathbf E_3} = \sum_\lambda \paren {\map \sgn \rho \map \sgn \lambda \prod_{k \mathop = 1}^n a_{\map \rho k \map \lambda k} }$
We can take $\map \sgn \rho = -1$ outside the summation because it is constant, and so we get:
$\ds \map \det {\mathbf E_3}$ $=$ $\ds \map \sgn \rho \sum_\lambda \paren {\map \sgn \lambda \prod_{k \mathop = 1}^n a_{\map \rho k \map \lambda k} }$ $\ds$ $=$ $\ds -\sum_\lambda \paren {\map \sgn \lambda \prod_{k \mathop = 1}^n a_{k \map \lambda k} }$ $\ds$ $=$ $\ds -\map \det {\mathbf I}$
Hence the result.
$\blacksquare$
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2021-08-03 17:12:16
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https://www.yaclass.in/p/mathematics/class-7/perimeter-and-area-2502/parallelogram-1343/re-3043accc-03fc-4fc0-9fe7-526abfba25f7
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Theory:
A quadrilateral is a plane figure that has four sides or edges, so having four corners or vertices. Quadrilaterals will typically be of standard shapes with four sides like rectangle, square, trapezoid, and kite or irregular and uncharacterized.
There are many types of quadrilaterals. As the word, ‘quad’ means four, all these types of a quadrilateral have four sides and the sum of angles of these shapes is $$360$$ degrees.
• Trapezium
• Parallelogram
• Square
• Rectangle
• Rhombus
Properties: two sides are parallel
Diagram:
Properties: two sides are parallel and opposite angles are equal
Diagram:
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2020-04-07 04:20:40
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https://ora.ox.ac.uk/objects/uuid:a15f48b1-22b6-4230-aeaf-36cf438d9821
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Journal article
### Optimality of the quantified Ingham-Karamata theorem for operator semigroups with general resolvent growth
Abstract:
We prove that a general version of the quantified Ingham-Karamata theorem for $C_0$-semigroups is sharp under mild conditions on the resolvent growth, thus generalising the results contained in a recent paper by the same authors. It follows in particular that the well-known Batty-Duyckaerts theorem is optimal even for bounded $C_0$-semigroups whose generator has subpolynomial resolvent growth. Our proof is based on an elegant application of the open mapping theorem, which we complement by a c...
Publication status:
Published
Peer review status:
Peer reviewed
Version:
Publisher's Version
### Access Document
Files:
• (pdf, 352.5KB)
Publisher copy:
10.1007/s00013-019-01380-z
### Authors
More by this author
Institution:
University of Oxford
Division:
MPLS Division
Department:
Mathematical Institute
Oxford college:
St Johns College
Role:
Author
ORCID:
0000-0002-4076-860X
Publisher:
Springer Publisher's website
Journal:
Archiv der Mathematik Journal website
Publication date:
2019-09-11
Acceptance date:
2019-08-11
DOI:
EISSN:
9876-1234
ISSN:
1234-1234
Pubs id:
pubs:957829
URN:
uri:a15f48b1-22b6-4230-aeaf-36cf438d9821
UUID:
uuid:a15f48b1-22b6-4230-aeaf-36cf438d9821
Local pid:
pubs:957829
Keywords:
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2022-01-21 12:15:36
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https://tarungehlots.blogspot.com/2014/08/chapter-10-worked-out-examples.html
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tgt
## Tuesday, 5 August 2014
### CHAPTER 10 - Worked Out Examples
Example: 1
Solve the DE $\dfrac{{dy}}{{dx}} = \dfrac{{2y - x - 4}}{{y - 3x + 3}}$
Solution: 1
#### Step-1
We substitute $x \to X + h$ and $y \to Y + k$ where $h,k$ need to be determined :
$\dfrac{{dy}}{{dx}} = \dfrac{{dY}}{{dX}} = \dfrac{{(2Y - X) + (2k - h - 4)}}{{(Y - 3X) + (k - 3h + 3)}}$
$h$ and $k$ must be chosen so that
$2k - h - 4 = 0$ $k - 3h + 3 = 0$
This gives $h=2$ and $k=3$. Thus,
$x = X + 2$ $y = Y + 3$
#### Step-2
Our DE now reduces to
$\dfrac{{dY}}{{dX}} = \dfrac{{2Y - X}}{{Y - 3X}}$
Using the substitution $Y=vX$ and simplifying, we have (verify),
$\dfrac{{v - 3}}{{{v^2} - 5v + 1}}dv = \dfrac{{ - dX}}{X}$
#### Step-3
We now integrate this DE which is VS; the left-hand side can be integrated by the techniques described in the unit on Indefinite Integration. Finally, we substitute $v = \dfrac{Y}{X}$ and
$X = x - 2$ $Y = y - 3$
to obtain the general solution.
Suppose our DE is of the form
$\dfrac{{dy}}{{dx}} = f\left( {\dfrac{{ax + by + c}}{{dx + ey + f}}} \right)$
We try to find $h,k$ so that
$ah + bk + c = 0$ $dh + ek + f = 0$
What if this system does not yield a solution ? Recall that this will happen if $\dfrac{a}{b} = \dfrac{d}{e}$. How do we reduce the DE to a homogeneous one in such a case ?
Let $\dfrac{a}{d} = \dfrac{b}{e} = \lambda$ (say). Thus,
$\dfrac{{ax + by + c}}{{dx + ey + f}} = \dfrac{{\lambda (dx + ey) + c}}{{dx + ey + f}}$
This suggests the substitution $dx + ey = v$, which’ll give
$d + e\dfrac{{dy}}{{dx}} = \dfrac{{dv}}{{dx}}$ $\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{e}\left( {\dfrac{{dv}}{{dx}} - d} \right)$
Thus, our DE reduces to
$\dfrac{1}{e}\left( {\dfrac{{dv}}{{dx}} - d} \right) = \dfrac{{\lambda v + c}}{{v + f}}$ $\Rightarrow \dfrac{{dv}}{{dx}} = \dfrac{{\lambda ev + ec}}{{v + f}} + d$ $= \dfrac{{(\lambda e + d)v + (ec + df)}}{{v + f}}$ $\Rightarrow \dfrac{{(v + f)}}{{(\lambda e + d)v + ec + df}}dv = dx$
which is in VS form and hence can be solved.
Example: 2
Solve the DE $\dfrac{{dy}}{{dx}} = \dfrac{{x + 2y - 1}}{{x + 2y + 1}}$
Solution: 2
#### Step-1
Note that $h,k$ do not exist in this case which can reduce this DE to homogeneous form. Thus, we use the substitution
$x + 2y = v$ $\Rightarrow 1 + 2\dfrac{{dy}}{{dx}} = \dfrac{{dv}}{{dx}}$
#### Step-2
Thus, our DE becomes
$\dfrac{1}{2}\left( {\dfrac{{dv}}{{dx}} - 1} \right) = \dfrac{{v - 1}}{{v + 1}}$ $\Rightarrow \dfrac{{dv}}{{dx}} = \dfrac{{2v - 2}}{{v + 1}} + 1$ $= \dfrac{{3v - 1}}{{v + 1}}$ $\Rightarrow \dfrac{{v + 1}}{{3v - 1}}dv = dx$ $\Rightarrow \dfrac{1}{3}\left( {1 + \dfrac{4}{{3v - 1}}} \right)dv = dx$
#### Step-3
Integrating, we have
$\dfrac{1}{3}\left( {v + \dfrac{4}{3}\ln (3v - 1)} \right) = x + {C_1}$
#### Step-4
Substituting $v=x+2y$ we have
$x + 2y + \dfrac{4}{3}\ln (3x + 6y - 1) = 3x + {C_2}$ $\Rightarrow y - x + \dfrac{2}{3}\ln (3x + 6y - 1) = C$
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2018-07-16 06:44:40
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https://zbmath.org/?q=an:0744.42002
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# zbMATH — the first resource for mathematics
A new classification of periodic functions and their approximation. (English) Zbl 0744.42002
Approximation and function spaces, Proc. 27th Semest., Warsaw/Pol. 1986, Banach Cent. Publ. 22, 437-447 (1989).
[For the entire collection see Zbl 0681.00013.]
In 1983 the author proposed to study some classes of functions (see below). He and his students have obtained the analogues of all the classical approximation theoretical results for these new classes and the present paper is a survey of their achievements.
The classes are defined in terms of multipliers and translations: Let $$f\in L_{2\pi}$$ with Fourier coefficients $$a_ k(f)$$, $$b_ k(f)$$, $$\psi$$ a function on the positive integers and $$\beta$$ a real number. Suppose that the series $\sum^ \infty_{k=1}{1\over\psi(k)} (a_ k(f)\cos(kx+\beta\pi/2)+b_ k(f)\cos(kx+\beta\pi/2))$ is the Fourier series of some $$f^ \psi_ \beta$$, which is called the $$(\psi,\beta)$$- derivative of $$f$$. Now the smoothness classes in question are defined via this derivative in the usual manner.
The questions that are discussed are as follows: representation of deviations for linear means of Fourier series; approximation by Fourier sums in $$C$$ and $$L_ 1$$; best trigonometric approximation.
Reviewer: V.Totik (Szeged)
##### MSC:
42A10 Trigonometric approximation 42A45 Multipliers in one variable harmonic analysis
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2021-07-28 10:41:53
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http://mathhelpforum.com/statistics/120477-2-stats-question.html
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1. ## 2 stats question
1.
suppose an airline sells 300 tickets when there is only 270 seats available. on avg, 15% of those with tickets do not show up. what is the probability that there will be enough seats for everybody who shows up?
2.
A special roulette wheel has seven equally likely outcomes: 0, 1, 2, 3, 4, 5, 6. If you bet that an odd number comes up, you win or lose \$10 according to whether or not that event occurs. If X denotes your ‘net’ gain, X=10 if we see 1, 3, or 5, and X = -10 otherwise. Suppose that you play this game 400 times. Let Y be your net gain after these 400 plays. The mean (expectation) and standard deviation of Y are...
i'd figured out the mean (its -571) but how do u figure out the stnd dev?
Thanks!!
2. 1. Pretty much the same as the other question. The probability that there will be enough seats, is the compliment of the probability that there WONT be enough seats - i.e. 271 or more people show up. This problem can be done one of two ways:
Using the Binomial distribution and calculating 29 binomial equations, where N is equal to 300 tickets and K is equal to going through those 300 tickets and choosing "29, 28, 27. . .0" that are going to no show;
Or you can remember that you can approximate the Binomial using the Normal as in the problem from your other thread (we can use the normal here since 5<np, and nq).
2. Remember:
$V(X)=\Sigma x^{2}P(X=x)-\mu^{2}_x$
3. Originally Posted by ANDS!
1. Pretty much the same as the other question. The probability that there will be enough seats, is the compliment of the probability that there WONT be enough seats - i.e. 271 or more people show up. This problem can be done one of two ways:
Using the Binomial distribution and calculating 29 binomial equations, where N is equal to 300 tickets and K is equal to going through those 300 tickets and choosing "29, 28, 27. . .0" that are going to no show;
Or you can remember that you can approximate the Binomial using the Normal as in the problem from your other thread (we can use the normal here since 5<np, and nq).
2. Remember:
$V(X)=\Sigma x^{2}P(X=x)-\mu^{2}_x$
so for 1. would this be correct?
$\mu=np=30(0.75)=22.5$
$\sigma= \sqrt{22.5(0.25)}=2.37$
$z=\frac{30-22.5}{2.37}=0.9992$
4. For approximations using Normal, we use the standard deviation of the Binomial as:
$\sigma_x=\sqrt{npq};$
$\sigma_x=\sqrt{(300)(.15)(.85)}$
Be careful what you set your mean to be. If you multiply it by the probability that someone shows up, then you would NOT be using 30 as the value you are comparing, you would be using 270 since that is the threshold for the number of people who are meant to show up.
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2017-07-23 15:07:55
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https://plainmath.net/41497/to-express-the-given-signed-numbers-as-absolute-values-difference-of
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# To express: The given signed numbers as absolute values. Difference of
To express: The given signed numbers as absolute values.
Difference of the smaller and the larger absolute value.
Given:
[-6, +6] is given as a pair of signed numbers.
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Calculation:
= —6,+6
The absolute value of -6 is 6.
The absolute value of +6 is 6.
Both absolute values (smallerand larger) are same, i.e., 6.
Now the subtraction of smaller absolute value from the larger absolute value is
=6-6
=0
Hence, the subtraction of smaller absolute value from the larger absolute value is 0.
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2022-08-12 11:48:25
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http://eprint.iacr.org/2017/302
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## Cryptology ePrint Archive: Report 2017/302
Quantum preimage, 2nd-preimage, and collision resistance of SHA3
Jan Czajkowski and Leon Groot Bruinderink and Andreas Hülsing and Christian Schaffner
Abstract: SHA3 and its extendable output variant SHAKE belong to the family of sponge functions. In this work, we present formal security arguments for the quantum preimage, $2^{\text{nd}}$-preimage, and collision resistance of any sponge function. We just assume that the internally used transformation behaves like a random transformation. These are the first formal arguments that sponge functions (incl. SHA3 and SHAKE) are secure in the post-quantum setting.
We even go one step further and prove that sponges are collapsing (Unruh, EUROCRYPT'16). Thereby, we can also derive the applicability of sponge functions for collapse-binding commitments.
In addition to the security arguments, we also present a quantum collision attack against sponges. The complexity of our attack asymptotically matches the proven lower bound up to a square root.
Category / Keywords: Post-quantum cryptography; SHA3, SHAKE, sponges, keccak, hash function, quantum security, quantum collision resistance, quantum second-preimage resistance, quantum preimage resistance
Date: received 4 Apr 2017, last revised 9 Apr 2017
Contact author: authors-quantum-sponges at huelsing net
Available format(s): PDF | BibTeX Citation
Short URL: ia.cr/2017/302
[ Cryptology ePrint archive ]
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2017-04-25 17:39:49
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https://www.minichemistry.com/percentage-yield-percentage-purity.html
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# Percentage Yield & Percentage Purity
Show/Hide Sub-topics (Stoichiometry | O Level)
The quantity of product that is calculated to be produced when the entire limiting reactant is used up is called the theoretical yield.
The quantity of product that is actually produced through the reaction is called the actual yield.
Percentage yield can be calculated by:
$$\text{% yield} = \frac{\text{Actual yield}}{\text{Theoretical yield}} \times 100 \%$$
Percentage yield cannot be more than 100% as the actual yield is always less than the theoretical yield.
Example:
For a reaction: $2\text{H}_{2} + \text{O}_{2} \rightarrow 2 \text{H}_{2}\text{O}$. Consider the case where 5 g of $\text{H}_{2}$ reacted with 50 g of $\text{O}_{2}$. The actual yield is found to be 40 g. Calculate the theoretical yield and percentage yield.
• No. of moles of $\text{H}_{2}$ present: $\frac{5}{2} = 2.5 \, \text{moles}$
• No. of moles of $\text{O}_{2}$ present: $\frac{50}{32} = 1.56 \, \text{moles}$
• Since 2.5 moles of $\text{H}_{2}$ reacts with 1.25 moles of $\text{O}_{2}$, there is an excess of $\text{O}_{2}$. $\text{H}_{2}$ is the limiting reactant.
• 2.5 moles of $\text{H}_{2}$ will yield 2.5 moles of $\text{H}_{2}\text{O}$, which equates to a theoretical yield of 45 g of $\text{H}_{2}\text{O}$.
• Percentage yield: $\frac{40}{45} \times 100 \% = 88.9 \%$
### Percentage Purity
Percentage purity indicates the amount of pure and impure substance present in a sample.
The percentage purity can be calculated by:
$$\text{% purity} = \frac{\text{Mass of pure substance in sample}}{\text{Mass of sample}} \times 100 \%$$
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2020-04-10 09:29:49
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https://aas.org/archives/BAAS/v30n3/dps98/400.htm
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Session 21P. Planetary Formation and Dynamics
Contributed Poster Session, Wednesday, October 14, 1998, 2:00-3:40pm, Hall of Ideas
[21P.13] Direct Simulation of Planet Formation with a Million Planetesimals: First Results
D. C. Richardson, T. Quinn, J. Stadel, G. Lake (U. Washington )
We present the first results from our direct simulations of planet formation using N = 106 rocky particles (see also {\it BAAS} {\bf 30}, 765 and {\it BAAS} {\bf 29}, 1027). Currently ~1000 yr of evolution have been simulated under the assumption of perfect accretion in a 4.7 M\oplus disk that extends from 0.8 to 3.8 AU with a surface density distribution \Sigma \propto r-3/2. The four present-day giant planets are included as perturbers. The disk was started cold, allowing the growth of resonance gaps and spiral density waves due to the giant planets to be seen clearly. We are in the process of testing a scheme to improve the speed of the method by at least an order of magnitude by exploiting the near-Keplerian nature of the planetesimal orbits. Preliminary results from this improvement are presented.
We also present a simulation of the formation of the Galilean satellites using N = 105 icy particles in a 0.065 M\oplus disk extending from 3 \times 105 to 2.3 \times 106 km around Jupiter and having the same r-3/2 density law. Particles are allowed to merge following a collision only if their rebound velocity is less than their mutual escape velocity, and if the spin of the resulting merged body does not exceed the classical breakup limit; otherwise the particles bounce and lose a fixed fraction of their relative energy. Fragmentation has not been implemented at this time. We find that this system evolves very quickly, with ~ 33% of the particles having merged after only ~150 d, or about 80 Io orbits. This simple model begins with a uniform population of 100 km radius planetesimals and ignores gas dynamics. The applicability of this model toward the actual formation of giant planet moons is discussed.
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2016-10-28 08:24:02
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https://mirrors.dotsrc.org/cran/web/packages/linkprediction/vignettes/proxfun.html
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Vertex proximity is a well developed subject in graph theory, applied in many different contexts. There are many different measures of similarity, some specific to social network analysis. Selection of methods used in this work follows mainly Liben-Nowell and Kleinberg (2007) and Lü and Zhou (2011). All measures give similarity score between two vertices. Some measures are symmetrical by definition, some have to be modified to achieve symmetry. Similarities have different scales but absolute values are unimportant as scores are not be compared between methods, only rankings of scores are of interest.
Similarity measures could be divided into local, global and quasi-local measures, following Lü and Zhou (2011). Local methods focus only on the properties of neighbourhoods of a given pair of vertices, while global methods benefit from information contained in the whole graph. Quasi-local methods are in between: make use of more information than local methods, but still do not need information about the whole graph.
Let us establish some notation used in sections below:
• Vertex set is $$V = \{1, 2, ..., x, y, ..., N\}$$ where $$x$$ and $$y$$ are some generic vertices of interest
• Edge set is $$E = \{(x, y): x,y \in V\}$$
• Graph $$G$$ is defined by a pair $$(V, E)$$
As we will be discussing computing of proximity measures in a given graph $$G$$, we will drop references to $$G$$ from the notation.
• An adjacency matrix of $$G$$ is a matrix $$A = [a_{xy}]_{N\times N}$$ such that $$a_{xy} = 1$$ if $$(x, y) \in E$$ otherwise $$a_{xy} = 0$$.
• A set of neighbors of vertex $$x$$ in graph $$G$$ is $$\Gamma(x)$$, namely $$\Gamma(x) = \{y: (x, y) \in E\}$$.
• $$k_x$$ is the degree of vertex $$x$$ in graph $$G$$
• Proximity measure $$s_{xy}$$ is a function $$s_{xy}: G, x, y \mapsto \Re$$.
Other notation will be introduced below when needed.
# 1 Local methods
These measures are mainly variations on Common neighbours.
## 1.1 Common neighbours
Two vertices are more likely to be connected if they are connected to the same set of other vertices. (Newman 2001) used this method in the study of collaboration networks, showing positive relation between number of common neighbours and probability of collaborating in the future.
$s_{xy}=\left| \Gamma(x) \cap \Gamma(y) \right|$
or in a matrix form
$s_{xy}=(A^2)_{xy}.$
## 1.2 Salton Index (cosine similarity)
Due to Salton and McGill (1986) . It measures the cosine of the angle between columns of the adjacency matrix, corresponding to given vertices. This measure is commonly used in information retrieval.
$s_{xy}=\frac{|\Gamma(x)\cap\Gamma(y)|}{\sqrt{k_x\times k_y}}.$
## 1.3 Jaccard Index
Jaccard Index (Jaccard 1912) measures the proportion of common neighbours in the total number of neighbors. It reaches its maximum if $$\Gamma(x)=\Gamma(y)$$, namely all neighbours are common to both vertices.
$s_{xy}=\frac{|\Gamma(x)\cap\Gamma(y)|}{|\Gamma(x)\cup\Gamma(y)|}.$
## 1.4 Sørensen Index
Due to (Sørensen 1948). This method is similar to the Jaccard Index as it measures the relative size of an intersection of neighbours’ sets.
$s_{xy}=\frac{2|\Gamma(x)\cap\Gamma(y)|}{k_x+k_y}.$
## 1.5 Hub Promoted Index
Due to (Ravasz et al. 2002). This measure assigns higher scores to links adjacent to hubs (high-degree vertices), as the denominator depends on the minimum of the degrees of the vertices of interest.
$s_{xy}=\frac{|\Gamma(x)\cap\Gamma(y)|}{\min\{k_x,k_y\}}.$
## 1.6 Hub Depressed Index
Due to (Ravasz et al. 2002). This measure, in contrast to Hub Promoted Index, assigns lower scores to links adjacent to hubs. It penalises large neighbourhoods.
$s_{xy}=\frac{|\Gamma(x)\cap\Gamma(y)|}{\max\{k_x,k_y\}}.$
## 1.7 Leicht-Holme-Newman Index
Due to Leicht, Holme, and Newman (2006). It is a variant of Common Neighbours, similar to Salton Index.
$s_{xy}=\frac{|\Gamma(x) \cap \Gamma(y)|}{k_x \cdot k_y}.$
## 1.8 Preferential Attachment
Preferential Attachment was developed as a model of the growth of network in a sense of new nodes emerging (Barabási and Albert 1999).
$s_{xy}=k_x \cdot k_y.$
This measure develops simple counting of common neighbours by assigning weights to nodes inversely proportional to their degrees (Adamic and Adar 2001). That means that a common neighbour, which is unique for a few nodes only, is more important than a hub.
$s_{xy}=\sum_{z\in \Gamma(x)\cap\Gamma(y)}\frac{1}{\log k_z}.$
Note that this index is well defined, because when vertex $$z$$ is a common neighbour of $$x$$ and $$y$$, than its degree is at least 2.
## 1.10 Resource Allocation Index
Proposed by Zhou, Lü, and Zhang (2009). This measure is motivated by a resource allocation process. It measures how much resource is transmitted between $$x$$ and $$y$$.
$s_{xy}=\sum_{z\in \Gamma(x)\cap\Gamma(y)}\frac{1}{k_z}.$
# 2 Global methods
## 2.1 Katz Index
(Katz 1953) counts all the paths between given pair of nodes, with shorter paths counting more heavily. $s_{xy}=\sum_{l=1}^{\infty}\beta^l|paths_{xy}^{<l>}|,$
where $$\beta$$ is a free parameter. The sum converges when $$\beta$$ is lower than the reciprocal of the largest eigenvalue of adjacency matrix. If this condition is satisfied Katz Index could be expressed in matrix form as
$S=(I-\beta A)^{-1}-I.$
## 2.2 Leicht-Holme-Newman Index
The global version (Leicht, Holme, and Newman 2006) is a variant of Katz Index, based on the concept that two nodes are similar if they neighbours are similar. It counts all paths between two nodes, but weights them by the expected number of such paths in a random graph with the same degree distribution. In matrix form, without constant factor:
$S = D^{-1}\left(I-\frac{\phi A}{\lambda_1}\right)^{-1}D^{-1},$
where $$\lambda_1$$ is the largest eigenvalue of adjacency matrix $$A$$ and $$\phi$$ is a free parameter.
## 2.3 Average Commute Time
$s_{xy}=\frac{1}{n(x,y)}=\frac{1}{m(x,y)+m(y,x)},$
where $$m(x,y)$$ is the average number of steps required by a random walker starting from $$x$$ to reach $$y$$. Sum of two directional commute times is taken to achieve symmetrical measure. Hence two nodes are similar if they are closer to each other and have smaller commute time, reciprocal is needed. Average Commute Time could be computed by solving collection of the linear equations, taken from Markov Chain analysis, but it is more straightforward to compute it in terms of the pseudoinverse of the Laplacian matrix, $$L^{+}$$. Namely: $n_{xy}=2M(l^{+}_{xx}+l^{+}_{yy}-2l^{+}_{xy}),$ where $$l^{+}_{xy}=[L^{+}]_{xy}$$ and $$M$$ is the number of edges. Thanks to the special form of the Laplacian matrix, its pseudoinverse $$L^{+}$$ could be computed using formula (Fouss et al. 2007): $L^{+}=\left(L-\frac{ee^T}{n}\right)^{-1}+\frac{ee^T}{n},$ where $$e$$ is a column vector made of 1’s.
## 2.4 Normalized Average Commute Time
This is a variant of ACT, which takes into account node degrees, as for high–degree node (hub) $$y$$, $$m(x,y)$$ is usually small regardless of $$x$$. $s_{xy}=\frac{1}{(m(x,y)\pi_y+m(y,x)\pi_x)},$ where $$\pi$$ is a stationary distribution of a Markov chain describing random walker on the graph. It is easily shown that on a connected graph $\pi(x) = \frac{k_x}{\sum_y k_y}.$
## 2.5 Cosine based on $$L^{+}$$
(Fouss et al. 2007)
$s_{xy}=\frac{l_{xy}^{+}}{\sqrt{l_{xx}^{+}l_{yy}^{+}}}$ It measures the cosine of the angle between node vectors in a space spanned by columns of $$L^{+}$$.
## 2.6 Random Walk with Restart (RWR)
It is an adaptation of PageRank algorithm (Brin and Page 1998). Consider random walker starting from node $$x$$ and periodically, with probability $$\alpha$$, returning to $$x$$. Let $$q_x$$ be a stationary distribution of Markov chain describing this walker. From definition of stationary distribution: $q_x = (1-\alpha)P^T q_x + \alpha e_x,$ where $$e_x$$ is a unit vector with 1 on position corresponding to node $$x$$, and $$P$$ is a transition matrix describing ordinary random walker, $$P_{xy}=1/k_x$$ if $$A_{xy}=1$$ and $$0$$ otherwise. The solution for all nodes simultaneously is $q =[q_1|q_2|\ldots|q_n] = \alpha(I-(1-\alpha)P^T)^{-1}.$ In order to achieve symmetry the RWR index is defined as $s_{xy}=q_{xy}+q_{yx}.$
## 2.7$$L^{+}$$ directly
(Fouss et al. 2007)
$S=L^{+}.$ $$L^{+}$$ provides a direct measure of similarity, as its elements are the inner products of vectors from an Euclidean space, which preserves Average Commute Time between nodes, see (Fouss et al. 2007) for details.
## 2.8 Matrix Forest Index
(Chebotarev P. Yu. 1997)
$S=(I+L)^{-1}.$ Matrix Forest Index can be understood as the ratio of the number of spanning rooted forest such that nodes $$x$$ and $$y$$ belong to the same tree rooted at $$x$$ to all spanning rooted forests of the network, see (Chebotarev P. Yu. 1997).
# 3 Quasi–local methods
## 3.1 Shortest Paths (Graph distance)
$$$s_{xy}=\left\{ \begin{array}{ll} \infty & \textrm{if x= y}\\ 0 & \textrm{if x and y are not connected}\\ \frac{1}{p_{xy}} & \textrm{in other cases} \end{array} \right.$$$
where $$\{p_{xy}\}=\min \{l:\text{path}_{xy}^{<l>}\textrm{ exists}\}$$ is the length of the shortest path connecting $$x$$ and $$y$$.
## 3.2 Local Path Index
(Zhou, Lü, and Zhang 2009)
$S=A^2+\epsilon A^3,$ where $$\epsilon$$ is a free parameter. This measure benefits from more information than simple common neighbours, as it looks on neighbours of second order, but at the same time is still fairly inexpensive to compute.
# Acknowledgements
Authors thank (Polish) National Science Centre for support through SONATA grant 2012/07/D/HS6/01971 for the project Dynamics of Competition and Collaboration in Science: Individual Strategies, Collaboration Networks, and Organizational Hierarchies (http://recon.icm.edu.pl).
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Lü, Linyuan, and Tao Zhou. 2011. “Link Prediction in Complex Networks: A Survey.” Physica A 390 (6): 1150–70.
Newman, M. E. J. 2001. “Clustering and Preferential Attachment in Growing Networks.” Phys. Rev. E 64 (2): 025102.
Ravasz, E., A. L. Somera, D. A. Mongru, Z. N. Oltvai, and Albert-László Barabási. 2002. “Hierarchical Organization of Modularity in Metabolic Networks.” Science 297 (5586): 1551–5.
Salton, Gerard, and Michael J. McGill. 1986. Introduction to Modern Information Retrieval. New York, NY, USA: McGraw-Hill, Inc.
Sørensen, Thorvald. 1948. “A Method of Establishing Groups of Equal Amplitude in Plant Sociology Based on Similarity of Species Content and Its Application to Analyses of the Vegetation on Danish Commons.” Biologiske Skrifter 5: 1–34.
Zhou, Tao, Linyuan Lü, and Yi-Cheng Zhang. 2009. “Predicting Missing Links via Local Information.” The European Physical Journal B 71 (4): 623–30.
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2021-10-22 23:03:15
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|
https://nolongerwait.com/category/program/microsoft/
|
# 题目2 : Tree Restoration
## 描述
There is a tree of N nodes which are numbered from 1 to N. Unfortunately, its edges are missing so we don’t know how the nodes are connected. Instead we know:
1. Which nodes are leaves
2. The distance (number of edges) between any pair of leaves
3. The depth of every node (the root’s depth is 1)
4. For the nodes on the same level, their order from left to right
Can you restore the tree’s edges with these information? Note if node u is on the left of node v, u’s parent should not be on the right of v’s parent.
## 输入
The first line contains three integers N, M and K. N is the number of nodes. M is the depth of the tree. K is the number of leaves.
The second line contains M integers A1, A2, … AM. Ai represents the number of nodes of depth i.
Then M lines follow. The ith of the M lines contains Ai numbers which are the nodes of depth i from left to right.
The (M+3)-th line contains K numbers L1, L2, … LK, indicating the leaves.
Then a K × K matrix D follows. Dij represents the distance between Li and Lj.
1 ≤ N ≤ 100
## 输出
For every node from 1 to N output its parent. Output 0 for the root’s parent.
8 3 5
1 3 4
1
2 3 4
5 6 7 8
3 5 6 7 8
0 3 3 3 3
3 0 2 4 4
3 2 0 4 4
3 4 4 0 2
3 4 4 2 0
0 1 1 1 2 2 4 4
# 解析
• 对于每层第一个节点，上一层的第一个非叶子节点必然是这个节点的父节点。
• 此时记录该节点的父节点，并把父节点的子节点记录为当前节点，并以此更新这个父节点与其他已知节点的距离。
• 然后看该层下一个节点。
• 下一个节点若与前一个节点距离为2，则两者父节点相同。
• 若大于2，则该节点的父节点是上一层的下一个非叶子节点。同时记录父、子节点，更新距离。
• 如此每层循环，每层结束进入上一层，直到第二层。（第一层根节点不必再判断了。
## AC代码（C++）：
#include<iostream>
using namespace std;
int main() {
int N, M, K;
//N is the number of nodes. M is the depth of the tree. K is the number of leaves.
cin >> N >> M >> K;
int depth[105] = { 0 };
//depth[] represents the number of nodes of depth[i].
for (int i = 0; i < M; i++) {
cin >> depth[i];
}
int tree[105][105] = { 0 };
//the nodes of depth[i] from left to right.
for (int i = 0; i < M; i++) {
for (int j = 0; j < depth[i]; j++) {
cin >> tree[i][j];
}
}
int leaves[105] = { 0 };
//leaves node
int is_leaves[105] = { 0 };
//Determine whether a node is a leaf node
for (int i = 0; i < K; i++) {
cin >> leaves[i];
is_leaves[leaves[i]] = 1;
}
int dis[105][105] = { 0 };
// the distance between Li and Lj.
for (int i = 0; i < K; i++) {
for (int j = 0; j < K; j++) {
cin >> dis[leaves[i]][leaves[j]];
}
}
int father[105] = { 0 };
int son[105] = { 0 };
//record a child node of a node
father[tree[0][0]] = 0;
for (int i = 0; i < depth[1]; i++) {
father[tree[1][i]] = tree[0][0];
}
for (int i = M - 1; i >= 2; i--) {
int up_point = 0;
////i-1层中第一个不是叶子节点的节点，必定是当前节点的父亲
while (is_leaves[tree[i - 1][up_point]] && up_point < depth[i - 1]) {
up_point++;
}
int point_a = tree[i][0];
father[point_a] = tree[i - 1][up_point];
son[tree[i - 1][up_point]] = point_a;
//更新距离
for (int k = 0; k < N; k++) {
if (dis[point_a][k] > 0) {
dis[father[point_a]][k] = dis[k][father[point_a]] = dis[point_a][k] - 1;
}
}
for (int k = 0; k < K; k++) {
dis[leaves[k]][father[point_a]] = dis[father[point_a]][leaves[k]] = dis[leaves[k]][point_a] - 1;
}
for (int j = 1; j < depth[i]; j++) {
int point_b = tree[i][j];
//如果和前一个节点u距离为2，说明父亲节点相同
if (dis[point_a][point_b] == 2) {
father[point_b] = father[point_a];
}
//否则，父亲节点是i-1层中下一个非叶子节点
else {
up_point++;
while (is_leaves[tree[i - 1][up_point]] && up_point < depth[i - 1]) {
up_point++;
}
father[point_b] = tree[i - 1][up_point];
son[tree[i - 1][up_point]] = point_b;
//更新距离
for (int kk = 0; kk < N; kk++) {
if (dis[point_b][kk] > 0) {
dis[father[point_b]][kk] = dis[kk][father[point_b]] = dis[point_b][kk] - 1;
}
}
}
point_a = point_b;
}
//更新i-1层相邻节点的距离
for (int kk = 0; kk < depth[i - 1] - 1; kk++) {
if (!is_leaves[tree[i - 1][kk]] && !is_leaves[tree[i - 1][kk + 1]]) {
dis[tree[i - 1][kk]][tree[i - 1][kk + 1]] = dis[tree[i - 1][kk + 1]][tree[i - 1][kk]] = dis[son[tree[i - 1][kk]]][son[tree[i - 1][kk + 1]]] - 2;
}
}
}
cout << father[1];
for (int i = 2; i <= N; i++) {
cout << " " << father[i];
}
return 0;
}
## WA代码（C++）：
100个节点的数据太多了。暂时无法测试。
#include<iostream>
using namespace std;
int main() {
int N, M, K;
//N is the number of nodes. M is the depth of the tree. K is the number of leaves.
cin >> N >> M >> K;
int count_depth[105] = { 0 };
for (int i = 1; i <= M; i++) {
cin >> count_depth[i];
}
int tree[105][105] = { 0 };
int depth_of_node[105] = { 0 };
for (int i = 1; i <= M; i++) {
for (int j = 1; j <= count_depth[i]; j++) {
cin >> tree[i][j];
depth_of_node[tree[i][j]] = i;
}
}
int is_leaf[105] = { 0 };
int num_leaf[105] = { 0 };
for (int i = 1; i <= K; i++) {
int TEMP;
cin >> TEMP;
num_leaf[i] = TEMP;
is_leaf[TEMP] = 1;
}
int dis[105][105] = { 0 };
for (int i = 1; i <= K; i++) {
for (int j = 1; j <= K; j++) {
cin >> dis[num_leaf[i]][num_leaf[j]];
}
}
int father_node[105] = { 0 };
int s_son_node[105] = { 0 };
for (int i = 1; i <= N; i++) {
if (is_leaf[i]) {
s_son_node[i] = i;
}
}
for (int i = M; i > 0; i--) {
int count_top = 1;
for (int j = 1; j <= count_depth[i]; j++) {
if (j == 1) {
for (; count_top <= count_depth[i - 1]; count_top++) {
if (is_leaf[tree[i - 1][count_top]] != 1) {
father_node[tree[i][j]] = tree[i - 1][count_top];
s_son_node[tree[i - 1][count_top]] = tree[i][j];
count_top++;
break;
}
}
continue;
}
if (dis[tree[i][j]][tree[i][j - 1]] == 2) {
father_node[tree[i][j]] = father_node[tree[i][j - 1]];
}
else {
for (; count_top <= count_depth[i - 1]; count_top++) {
if (is_leaf[tree[i - 1][count_top]] != 1) {
father_node[tree[i][j]] = tree[i - 1][count_top];
s_son_node[tree[i - 1][count_top]] = tree[i][j];
count_top++;
break;
}
}
continue;
}
}
//更新上一层节点间距离
for (int j = 1; j <= count_depth[i - 1]; )
{
for (int jj = j + 1; jj <= count_depth[i - 1]; jj++)
{
dis[tree[i - 1][j]][tree[i - 1][jj]] = dis[tree[i - 1][jj]][tree[i - 1][j]] = dis[s_son_node[tree[i - 1][j]]][s_son_node[tree[i - 1][jj]]] - (depth_of_node[s_son_node[tree[i - 1][j]]] - depth_of_node[tree[i - 1][j]]) - (depth_of_node[s_son_node[tree[i - 1][jj]]] - depth_of_node[tree[i - 1][jj]]);
}
j++;
}
}
//输出
for (int i = 1; i < N; i++)
{
cout << father_node[i] << " ";
}
cout << father_node[N];
return 0;
}
## 最终结果
1490Tree RestorationACG++5ms0MB
# 题目1 : Legendary Items
## 描述
Little Hi is playing a video game. Each time he accomplishes a quest in the game, Little Hi has a chance to get a legendary item.
At the beginning the probability is P%. Each time Little Hi accomplishes a quest without getting a legendary item, the probability will go up Q%. Since the probability is getting higher he will get a legendary item eventually.
After getting a legendary item the probability will be reset to ⌊P/(2I)⌋% (⌊x⌋ represents the largest integer no more than x) where I is the number of legendary items he already has. The probability will also go up Q% each time Little Hi accomplishes a quest until he gets another legendary item.
Now Little Hi wants to know the expected number of quests he has to accomplish to get N legendary items.
Assume P = 50, Q = 75 and N = 2, as the below figure shows the expected number of quests is 3.25
2*50%*25% + 3*50%*75%*100% + 3*50%*100%*25% + 4*50%*100%*75%*100% = 3.25
## 输入
The first line contains three integers P, Q and N.
1 \leq N \leq 10^6, 0 \leq P \leq 100, 1 \leq Q \leq 1001≤N≤106,0≤P≤100,1≤Q≤100
## 输出
Output the expected number of quests rounded to 2 decimal places.
50 75 2
3.25
# 解析
EX=\sum{(X \cdot P(X))}EX=∑(X⋅P(X))
E(X+Y)=EX+EYE(X+Y)=EX+EY
1.初始任务数（至少1次任务就能获得）numQuests=1numQuests=1
2.第一次任务对该次获得传奇物品增加的期望：incE=(1-P)incE=(1−P)
3.调整概率P=P+QP=P+Q
4.进行下一次任务对该次获得传奇物品增加的期望计算：incE=incE\cdot(1-P)incE=incE⋅(1−P)
5.任务期望：numQuests=numQuest+incEnumQuests=numQuest+incE
6.回到步骤3
7.直至P=P+Q>100P=P+Q>100终止此次获得传奇物品的期望计算numQuestsnumQuests
## AC代码（C++）：
#include<iostream>
#include<iomanip>
using namespace std;
int main()
{
int P, Q, N;
cin >> P >> Q >> N;
double count_q=Q*1.00/100;
double count_p;
double result = 0.00;
for (int i = 1; i <= N; i++)
{
double next_result = 1;
count_p = P*1.00 / 100;
double E1 = 1.00;
while (1)
{
E1 = E1*(1.00 - count_p);
next_result += E1;
count_p += count_q;
if (count_p > 1.00)
break;
}
result += next_result;
P = P / 2;
}
cout << fixed << setprecision(2) << result;
return 0;
}
## 最终结果
1489Legendary ItemsACG++22ms0MB
|
2019-05-23 06:48:54
|
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|
http://jeremykun.com/tag/linear-maps/
|
# Computing Homology
Update: the mistakes made in the code posted here are fixed and explained in a subsequent post (one minor code bug was fixed here, and a less minor conceptual bug is fixed in the linked post).
In our last post in this series on topology, we defined the homology group. Specifically, we built up a topological space as a simplicial complex (a mess of triangles glued together), we defined an algebraic way to represent collections of simplices called chains as vectors in a vector space, we defined the boundary homomorphism $\partial_k$ as a linear map on chains, and finally defined the homology groups as the quotient vector spaces
$\displaystyle H_k(X) = \frac{\textup{ker} \partial_k}{\textup{im} \partial_{k+1}}$.
The number of holes in $X$ was just the dimension of this quotient space.
In this post we will be quite a bit more explicit. Because the chain groups are vector spaces and the boundary mappings are linear maps, they can be represented as matrices whose dimensions depend on our simplicial complex structure. Better yet, if we have explicit representations of our chains by way of a basis, then we can use row-reduction techniques to write the matrix in a standard form.
Of course the problem arises when we want to work with two matrices simultaneously (to compute the kernel-mod-image quotient above). This is not computationally any more difficult, but it requires some theoretical fiddling. We will need to dip a bit deeper into our linear algebra toolboxes to see how it works, so the rusty reader should brush up on their linear algebra before continuing (or at least take some time to sort things out if or when confusion strikes).
Without further ado, let’s do an extended example and work our ways toward a general algorithm. As usual, all of the code used for this post is available on this blog’s Github page.
## Two Big Matrices
Recall our example simplicial complex from last time.
We will compute $H_1$ of this simplex (which we saw last time was $\mathbb{Q}$) in a more algorithmic way than we did last time.
Once again, we label the vertices 0-4 so that the extra “arm” has vertex 4 in the middle, and its two endpoints are 0 and 2. This gave us orientations on all of the simplices, and the following chain groups. Since the vertex labels (and ordering) are part of the data of a simplicial complex, we have made no choices in writing these down.
$\displaystyle C_0(X) = \textup{span} \left \{ 0,1,2,3,4 \right \}$
$\displaystyle C_1(X) = \textup{span} \left \{ [0,1], [0,2], [0,3], [0,4], [1,2], [1,3],[2,3],[2,4] \right \}$
$\displaystyle C_2(X) = \textup{span} \left \{ [0,1,2], [0,1,3], [0,2,3], [1,2,3] \right \}$
Now given our known definitions of $\partial_k$ as an alternating sum from last time, we can give a complete specification of the boundary map as a matrix. For $\partial_1$, this would be
$\displaystyle \partial_1 = \bordermatrix{ & [0,1] & [0,2] & [0,3] & [0,4] & [1,2] & [1,3] & [2,3] & [2,4] \cr 0 & -1 & -1 & -1 & -1 & 0 & 0 & 0 & 0\cr 1 & 1 & 0 & 0 & 0 & -1 & -1 & 0 & 0\cr 2 & 0 & 1 & 0 & 0 & 1 & 0 & -1 & -1 \cr 3 & 0 & 0 & 1 & 0 & 0 & 1 & 1 & 0 \cr 4 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 1 }$,
where the row labels are the basis for $C_0(X)$ and the column labels are the basis for $C_1(X)$. Similarly, $\partial_2$ is
$\displaystyle \partial_2 = \bordermatrix{ & [0,1,2] & [0,1,3] & [0,2,3] & [1,2,3] \cr [0,1] & 1 & 1 & 0 & 0\cr [0,2] & -1 & 0 & 1 & 0\cr [0,3] & 0 & -1 & -1 & 0\cr [0,4] & 0 & 0 & 0 & 0\cr [1,2] & 1 & 0 & 0 & 1\cr [1,3] & 0 & 1 & 0 & -1\cr [2,3] & 0 & 0 & 1 & 1\cr [2,4] & 0 & 0 & 0 & 0}$
The reader is encouraged to check that these matrices are written correctly by referring to the formula for $\partial$ as given last time.
Remember the crucial property of $\partial$, that $\partial^2 = \partial_k \partial_{k+1} = 0$. Indeed, the composition of the two boundary maps just corresponds to the matrix product of the two matrices, and one can verify by hand that the above two matrices multiply to the zero matrix.
We know from basic linear algebra how to compute the kernel of a linear map expressed as a matrix: column reduce and inspect the columns of zeros. Since the process of row reducing is really a change of basis, we can encapsulate the reduction inside a single invertible matrix $A$, which, when left-multiplied by $\partial$, gives us the reduced form of the latter. So write the reduced form of $\partial_1$ as $\partial_1 A$.
However, now we’re using two different sets of bases for the shared vector space involved in $\partial_1$ and $\partial_2$. In general, it will no longer be the case that $\partial_kA\partial_{k+1} = 0$. The way to alleviate this is to perform the “corresponding” change of basis in $\partial_{k+1}$. To make this idea more transparent, we return to the basics.
## Changing Two Matrices Simultaneously
Recall that a matrix $M$ represents a linear map between two vector spaces $f : V \to W$. The actual entries of $M$ depend crucially on the choice of a basis for the domain and codomain. Indeed, if $v_i$ form a basis for $V$ and $w_j$ for $W$, then the $k$-th column of the matrix representation $M$ is defined to be the coefficients of the representation of $f(v_k)$ in terms of the $w_j$. We hope to have nailed this concept down firmly in our first linear algebra primer.
Recall further that row operations correspond to changing a basis for the codomain, and column operations correspond to changing a basis for the domain. For example, the idea of swapping columns $i,j$ in $M$ gives a new matrix which is the representation of $f$ with respect to the (ordered) basis for $V$ which swaps the order of $v_i , v_j$. Similar things happen for all column operations (they all correspond to manipulations of the basis for $V$), while analogously row operations implicitly transform the basis for the codomain. Note, though, that the connection between row operations and transformations of the basis for the codomain are slightly more complicated than they are for the column operations. We will explicitly see how it works later in the post.
And so if we’re working with two maps $A: U \to V$ and $B: V \to W$, and we change a basis for $V$ in $B$ via column reductions, then in order to be consistent, we need to change the basis for $V$ in $A$ via “complementary” row reductions. That is, if we call the change of basis matrix $Q$, then we’re implicitly sticking $Q$ in between the composition $BA$ to get $(BQ)A$. This is not the same map as $BA$, but we can make it the same map by adding a $Q^{-1}$ in the right place:
$\displaystyle BA = B(QQ^{-1})A = (BQ)(Q^{-1}A)$
Indeed, whenever $Q$ is a change of basis matrix so is $Q^{-1}$ (trivially), and moreover the operations that $Q$ performs on the columns of $B$ are precisely the operations that $Q^{-1}$ performs on the rows of $A$ (this is because elementary row operations take different forms when multiplied on the left or right).
Coming back to our boundary operators, we want a canonical way to view the image of $\partial_{k+1}$ as sitting inside the kernel of $\partial_k$. If we go ahead and use column reductions to transform $\partial_k$ into a form where the kernel is easy to read off (as the columns consisting entirely of zeroes), then the corresponding row operations, when performed on $\partial_{k+1}$ will tell us exactly the image of $\partial_{k+1}$ inside the kernel of $\partial_k$.
This last point is true precisely because $\textup{im} \partial_{k+1} \subset \textup{ker} \partial_k$. This fact guarantees that the irrelevant rows of the reduced version of $\partial_{k+1}$ are all zero.
Let’s go ahead and see this in action on our two big matrices above. For $\partial_1$, the column reduction matrix is
$\displaystyle A = \begin{pmatrix} 0 & 1 & 0 & 0 & 1 & 1 & 0 & 0\\ 0 & 0 & 1 & 0 & -1 & 0 & 1 & 1\\ 0 & 0 & 0 & 1 & 0 & -1 & -1 & 0\\ -1 & -1 & -1 & -1 & 0 & 0 & 0 & -1\\ 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \end{pmatrix}$
And the product $\partial_1 A$ is
$\displaystyle \partial_1 A = \begin{pmatrix} 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0\\ -1 & -1 & -1 & -1 & 0 & 0 & 0 & 0 \end{pmatrix}$
Now the inverse of $A$, which is the corresponding basis change for $\partial_2$, is
$\displaystyle A^{-1} = \begin{pmatrix} -1 & -1 & -1 & -1 & -0 & -0 & -0 & -0\\ 1 & 0 & 0 & 0 & -1 & -1 & 0 & 0\\ 0 & 1 & 0 & 0 & 1 & 0 & -1 & -1\\ 0 & 0 & 1 & 0 & 0 & 1 & 1 & 0\\ 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \end{pmatrix}$
and the corresponding reduced form of $\partial_2$ is
$\displaystyle A^{-1} \partial_2 = \begin{pmatrix} 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 1 & 0 & 0 & 1\\ 0 & 1 & 0 & -1\\ 0 & 0 & 1 & 1\\ 0 & 0 & 0 & 0 \end{pmatrix}$
As a side note, we got these matrices by slightly modifying the code from our original post on row reduction to output the change of basis matrix in addition to performing row reduction. It turns out one can implement column reduction as row reduction of the transpose, and the change of basis matrix you get from this process will be the transpose of the change of basis matrix you want (by $(AB)^\textup{T} = (B^\textup{T}A^\textup{T})$). Though the code is particularly ad-hoc, we include it with the rest of the code used in this post on this blog’s Github page.
Now let’s inspect the two matrices $\partial_1 A$ and $A^{-1} \partial_2$ more closely. The former has four “pivots” left over, and this corresponds to the rank of the matrix being 4. Moreover, the four basis vectors representing the columns with nonzero pivots, which we’ll call $v_1, v_2, v_3, v_4$ (we don’t care what their values are), span a complementary subspace to the kernel of $\partial_1$. Hence, the remaining four vectors (which we’ll call $v_5, v_6, v_7, v_8$) span the kernel. In particular, this says that the kernel has dimension 4.
On the other hand, we performed the same transformation of the basis of $C_1(X)$ for $\partial_2$. Looking at the matrix that resulted, we see that the first four rows and the last row (representing $v_1, v_2, v_3, v_4, v_8$) are entirely zeros and so the image of $\partial_2$ intersects their span trivially. and the remaining three rows (representing $v_5, v_6, v_7$) have nonzero pivots. This tells us exactly that the image of $\partial_2$ is spanned by $v_5, v_6, v_7$.
And now, the coup de grâce, the quotient to get homology is simply
$\displaystyle \frac{ \textup{span} \left \{ v_5, v_6, v_7, v_8 \right \}}{ \textup{span} \left \{ v_5, v_6, v_7 \right \}} = \textup{span} \left \{ v_8 \right \}$
And the dimension of the homology group is 1, as desired.
## The General Algorithm
It is no coincidence that things worked out at nicely as they did. The process we took of simultaneously rewriting two matrices with respect to a common basis is the bulk of the algorithm to compute homology. Since we’re really only interested in the dimensions of the homology groups, we just need to count pivots. If the number of pivots arising in $\partial_k$ is $y$ and the number of pivots arising in $\partial_{k+1}$ is $z$, and the dimension of $C_k(X)$ is $n$, then the dimension is exactly
$(n-y) - z = \textup{dim}(\textup{ker} \partial_k) - \textup{dim}(\textup{im}\partial_{k+1})$
And it is no coincidence that the pivots lined up so nicely to allow us to count dimensions this way. It is a minor exercise to prove it formally, but the fact that the composition $\partial_k \partial_{k+1} = 0$ implies that the reduced version of $\partial_{k+1}$ will have an almost reduced row-echelon form (the only difference being the rows of zeros interspersed above, below, and possibly between pivot rows).
As the reader may have guessed at this point, we don’t actually need to compute $A$ and $A^{-1}$. Instead of this, we can perform the column/row reductions simultaneously on the two matrices. The above analysis helped us prove the algorithm works, and with that guarantee we can throw out the analytical baggage and just compute the damn thing.
Indeed, assuming the input is already processed as two matrices representing the boundary operators with respect to the standard bases of the chain groups, computing homology is only slightly more difficult than row reducing in the first place. Putting our homology where our mouth is, we’ve implemented the algorithm in Python. As usual, the entire code used in this post is available on this blog’s Github page.
The first step is writing auxiliary functions to do elementary row and column operations on matrices. For this post, we will do everything in numpy (which makes the syntax shorter than standard Python syntax, but dependent on the numpy library).
import numpy
def rowSwap(A, i, j):
temp = numpy.copy(A[i, :])
A[i, :] = A[j, :]
A[j, :] = temp
def colSwap(A, i, j):
temp = numpy.copy(A[:, i])
A[:, i] = A[:, j]
A[:, j] = temp
def scaleCol(A, i, c):
A[:, i] *= c*numpy.ones(A.shape[0])
def scaleRow(A, i, c):
A[i, :] *= c*numpy.ones(A.shape[1])
A[:, addTo] += scaleAmt * A[:, scaleCol]
A[addTo, :] += scaleAmt * A[scaleRow, :]
From here, the main meat of the algorithm is doing column reduction on one matrix, and applying the corresponding row operations on the other.
def simultaneousReduce(A, B):
if A.shape[1] != B.shape[0]:
raise Exception("Matrices have the wrong shape.")
numRows, numCols = A.shape # col reduce A
i,j = 0,0
while True:
if i >= numRows or j >= numCols:
break
if A[i][j] == 0:
nonzeroCol = j
while nonzeroCol < numCols and A[i,nonzeroCol] == 0:
nonzeroCol += 1
if nonzeroCol == numCols:
i += 1
continue
colSwap(A, j, nonzeroCol)
rowSwap(B, j, nonzeroCol)
pivot = A[i,j]
scaleCol(A, j, 1.0 / pivot)
scaleRow(B, j, 1.0 / pivot)
for otherCol in range(0, numCols):
if otherCol == j:
continue
if A[i, otherCol] != 0:
scaleAmt = -A[i, otherCol]
colCombine(A, otherCol, j, scaleAmt)
rowCombine(B, j, otherCol, -scaleAmt)
i += 1; j+= 1
return A,B
This more or less parallels the standard algorithm for row-reduction (with the caveat that all the indices are swapped to do column-reduction). The only somewhat confusing line is the call to rowCombine, which explicitly realizes the corresponding row operation as the inverse of the performed column operation. Note that for row operations, the correspondence between operations on the basis and operations on the rows is not as direct as it is for columns. What’s given above is the true correspondence. Writing down lots of examples will reveal why, and we leave that as an exercise to the reader.
Then the actual algorithm to compute homology is just a matter of counting pivots. Here are two pivot-counting functions in a typical numpy fashion:
def numPivotCols(A):
z = numpy.zeros(A.shape[0])
return [numpy.all(A[:, j] == z) for j in range(A.shape[1])].count(False)
def numPivotRows(A):
z = numpy.zeros(A.shape[1])
return [numpy.all(A[i, :] == z) for i in range(A.shape[0])].count(False)
And the final function is just:
def bettiNumber(d_k, d_kplus1):
A, B = numpy.copy(d_k), numpy.copy(d_kplus1)
simultaneousReduce(A, B)
dimKChains = A.shape[1]
kernelDim = dimKChains - numPivotCols(A)
imageDim = numPivotRows(B)
return kernelDim - imageDim
And there we have it! We’ve finally tackled the beast, and written a program to compute algebraic features of a topological space.
The reader may be curious as to why we didn’t come up with a more full-bodied representation of a simplicial complex and write an algorithm which accepts a simplicial complex and computes all of its homology groups. We’ll leave this direct approach as a (potentially long) exercise to the reader, because coming up in this series we are going to do one better. Instead of computing the homology groups of just one simplicial complex using by repeating one algorithm many times, we’re going to compute all the homology groups of a whole family of simplicial complexes in a single bound. This family of simplicial complexes will be constructed from a data set, and so, in grandiose words, we will compute the topological features of data.
If it sounds exciting, that’s because it is! We’ll be exploring a cutting-edge research field known as persistent homology, and we’ll see some of the applications of this theory to data analysis.
Until then!
# Linear Algebra – A Primer
## Story Time
Linear algebra was founded around the same time as Calculus (think Leibniz, circa 1700) solely for the purpose of solving general systems of linear equations. The coefficients of a system were written in a grid form, with rows corresponding to equations and columns to the unknown variables. Using a computational tool called the determinant (an awkward, but computable formula involving only the coefficients of the equations in a system), researchers were able to solve these systems, opening a world of information about the positions of celestial bodies and large-scale measurements (of geodesic arcs) on the surface of the earth.
By the 1850’s, Arthur Cayley was representing matrices as abstract objects. He defined matrix multiplication and nurtured matrix theory as its own field, recognizing a vast wealth of theoretical knowledge underlying the theory of determinants. Around turn of the century, a formal system of vector algebra was invented which relied heavily on interpreting matrices as so-called linear transformations. Linear transformations are intuitively those maps of everyday space ($\mathbb{R}^n$) which preserve “linear” things. Specifically, they send lines to lines, planes to planes, etc., and they preserve the origin (one which does not preserve the origin is very similar but has a different name; see Affine Transformation). Soon enough the mathematical focus shifted to the foundations of such an algebra, and later with the advent of computers to rapid calculations in one.
## Motivations
Linear algebra sits at the crossroads of many areas of mathematics. Keeping close to its roots, linear algebra is primarily a tool for computation. Unsurprisingly, a huge chunk of mathematical research has been solely to phrase things in terms of matrices and their associated linear transformations. For instance, an undirected graph on $n$ vertices can be modeled as a matrix of integer entries, with the $i,j$ entry containing the number of edges from vertex $i$ to vertex $j$. This is called the adjacency matrix of a graph. Suddenly, a wealth of information about the graph translates to simple matrix computations. For instance, we can compute the number of paths from one vertex to another of length $m$ as the appropriate entry of $A^m$. (more formally,these are walks, which are allowed to repeat edge traversals and visited vertices)
Even in advanced, purely theoretical mathematics, objects are commonly represented in terms of coordinates in some vector space, and are subsequently studied using all of the great things we know about linear transformations and their matrices. And so, without further ado, we will present the terminology and working concepts necessary for the content elsewhere in this blog.
## Vector Spaces
The setting for all of linear algebra is in some vector space. Intuitively this is just a collection of objects, which we call vectors, with some rules on how you can combine vectors to get other vectors. This treatment wouldn’t do that idea justice without an axiomatic definition, so here it is.
Definition: A vector space is a quadruple $(V, F, +, \cdot)$, where $V$ is a set of vectors (points in our space), $F$ is a scalar field (coefficients), $+:V \times V \to V$ is a commutative, associative operation to combine vectors, and $\cdot: F \times V \to V$ is an operation to “scale” vectors. In addition, we need the following properties to hold:
• Addition and multiplication distribute (as we are used to with traditional algebra).
• There must be an additive identity, which we call $0$, giving $0 + v = v$ for all $v \in V$.
• Every vector must have an additive inverse (every $v$ has some $w$ with $v + w = 0$).
This is a lot to swallow at first, but it is general for a good reason: there are tons of different kinds of vector spaces! Many of these are surprising and counter-intuitive. For our purposes, however, we may stick with the nice, small vector spaces. So here is a simplified definition that will suffice:
Definition: vector space is a set $V$ of vectors which are fixed-length lists of real numbers $(v_1, v_2, \dots , v_n) \in \mathbb{R}^n$, where addition between vectors is componentwise, we may scale vectors by any real number, and the following properties hold:
• Addition and multiplication distribute (as above).
• $(0,0,0, \dots, 0)$ is the additive identity.
• $(-v_1, -v_2, \dots , -v_n)$ is the unique additive inverse of $(v_1, v_2, \dots , v_n)$.
Hopefully this is much more familiar to what we think of as “vectors,” and with the understanding that we are viewing it as a vector space, we just call it $\mathbb{R}^n$. The closure of operations gives us a nice way to characterize “any combination” of vectors in a vector space.
Definition: A linear combination of vectors in a vector space $V$ is the vector
$a_1v_1 + a_2v_2 + \dots + a_kv_k$
for some positive integer $k$, scalars $a_i$, and vectors $v_i$.
We may speak of the span of a set of vectors as the set of all possible linear combinations of those vectors. Furthermore, we call a set of vectors linearly independent if no vector in the list is in the span of the others. For example, $(1,0,0), (0,1,0),$ and $(0,0,1)$ are linearly independent in $\mathbb{R}^3$. Specifically, $(1,0,0)$ cannot be written as $a(0,1,0) + b(0,0,1) = (0,a,b)$ for any scalars $a,b \in F$, and the other two vectors are similarly so.
As usual, we may describe subspaces of a vector space, which are just subsets of $V$ which are themselves vector spaces with the inherited operations. The simplest examples of these are lines, planes, and hyperplanes through the origin in $\mathbb{R}^n$. Consequently, we may identify $\mathbb{R}^n$ as a subspace of $\mathbb{R}^m$ for any $n \leq m$.
One of the first things we want to ask about a vector space is “how big is it?” While most instances of vector spaces we will see have uncountably many elements, we can characterize “size” in terms of a different metric: the size of a basis.
Definition: A list of vectors $(v_1, v_2, \dots v_n)$ is a basis for $V$ if its elements are linearly independent, and their span is $V$. The dimension of a vector space is the length of any basis.
For $\mathbb{R}^n$, and similarly all finite-dimensional vector spaces, it is easy to prove that all bases have the same length, and hence dimension is well-defined. Further, $\mathbb{R}^n$ admits a very natural basis, often called the standard basis:
$e_1 = (1,0, \dots, 0)$
$e_2 = (0,1, \dots, 0)$
$\vdots$
$e_n = (0,0, \dots, 1)$
These are best visualized as the coordinate axes in $\mathbb{R}^n$, and it strokes our intuition as to what a basis should be, because any vector in $\mathbb{R}^n$ can be broken down uniquely into a sum of scalar multiples of these unit coordinates. Indeed, this is true of any basis (due to linear independence). Given a fixed basis for $V$, every vector $v \in V$ may be uniquely written as a linear combination of basis vectors.
## Linear Transformations and their Matrix Representations
Moving quickly toward the heart of linear algebra, we may speak of linear transformations (interchangeably, linear maps) between two vector spaces:
Definition: A function $f : V \to W$ is a linear map if it preserves the operations of addition and scalar multiplication. In other words, for all $v, w \in V, c \in F, f(v+w) = f(v)+f(w)$ and $f(cv) = cf(v)$.
Examples are bountiful; some geometrically inspired ones include rotations about the origin, shears, and scalings. These are functions you’d likely see in an image manipulation program like photoshop. From this we can prove a few basic facts, like that every linear map sends $0$ to $0$ and additive inverses to additive inverses (try it as an exercise).
One remarkable fact that helps us characterize linear maps is that every linear map is determined completely by what it does to a basis. Since every vector $x \in V$ is a linear combination of basis elements, say $x=a_1v_1 + \dots + a_nv_n$, we see that a linear map plays nicely:
$f(x) = f(a_1v_1 + \dots + a_nv_n) = a_1f(v_1) + \dots + a_nf(v_n)$
In other words, if we know what $f$ does to a basis, then we know everything about $f$. In order to aid our computations, we write what $f$ does to each basis vector in a tabular form. To elaborate on the vague word “does,” we need to also fix a basis of our target vector space $W$, say $(w_1, \dots , w_m)$, and describe each $f(v_i)$ in terms of this basis. We write it in tabular form, as follows:
$\begin{pmatrix} | & | & \mathbf{ } & | \\ f(v_1) & f(v_2) & \dots & f(v_n) \\ | & | & \mathbf{ } & | \end{pmatrix}$
The $j$th column corresponds to $f(v_j)$, and the $i$th row corresponds to the $i$th coefficient in the expansion of $f(v_j)$ in terms of the basis for $W$. Here the vertical bars indicate that each element is a column of scalars. We will do an extended example to make this clear.
Consider the map $f$ on $\mathbb{R}^3$ defined as $(x,y,z) \mapsto (y,x,2z+y)$. It is easy to check this map is linear, and using the standard basis we see that
$f(1,0,0) = (0,1,0)$,
$f(0,1,0) = (1,0,1)$, and
$f(0,0,1) = (0,0,2)$.
or,
$f(e_1) = e_2$$f(e_2) = e_1 + e_3$, and $f(e_3) = 2e_3$.
Hence, the matrix representation of $f$ with respect to the standard basis is
$A = \begin{pmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 1 & 2 \end{pmatrix}$
Now we see that if we take a (column) vector $x$, and multiply it on the left by our matrix $A$, the resulting vector is precisely the coordinate representation of $f(x)$ with respect to the basis for $W$. In fact, the rules for matrix multiplication were constructed very particularly so that this would be the case. In this way, we may arbitrarily switch between viewing $f$ as a transformation and a vector computation. Compositions of linear maps translate to multiplication of two matrices, and matrix inversion (if it exists) is precisely function inversion.
Of course, there are many different bases we could have chosen. Even though we are going from $\mathbb{R}^3 \to \mathbb{R}^3$, the column basis could be different from the row basis. Fortunately for our purposes, we are not going to consider what basis is appropriate to choose. All that matters is that fixing a basis, the matrix representation of a linear map is unique, and so we may interchange the notation freely. Even so, the truly interesting things about matrices are those properties which are true no matter which basis we prefer to use.
## Eigenvectors and Eigenvalues
Definition: A scalar $\lambda \in F$ is an eigenvalue for the linear map $A$ if there exists a non-zero vector $v \in V$ with $Av = \lambda v$. Any such vector $v$ which satisfies this equation is said to be an eigenvector of $A$ corresponding to $\lambda$.
Eigenvectors and eigenvalues have a huge number of applications, including facial recognition software, geology, quantum mechanics, and web search. So being able to find them quickly is of great significance to researchers and engineers. What’s interesting is that while eigenvectors depend on a choice of basis, eigenvalues do not. We prove this now:
Proposition: If $A$ and $B$ are different representations of the same linear map, then any eigenvalue of $B$ is an eigenvalue of $A$.
Proof. It turns out that the process of “changing a basis” can be boiled down to matrix multiplication. Specifically, if $A$ and $B$ are two different matrix representations of the same linear map, we have the existence of some invertible matrix $P$ such that $A = PBP^{-1}$, or $AP = PB$. As a result, if $v$ is an eigenvector for $B$ corresponding to the eigenvalue $\lambda$, then for some $APv = PBv = P \lambda v = \lambda Pv$ and so $A(Pv) = \lambda(Pv)$, and $Pv$ is an eigenvector for $A$ corresponding to $\lambda$ as well. This proves that eigenvalues are invariant with respect to a change of basis, as desired. $\square$
The point of this is that we can choose whatever basis we want to work with, and compute the eigenvalues where we’re most comfortable. For instance, if we choose a basis that gives the following diagonal representation,
$A = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{pmatrix}$
then we can just eyeball that the eigenvalues are 1, 2, and 3. In fact, there are some very deep theorems in linear algebra that concern the existence and uniqueness of certain matrix representations. For a more in-depth treatment, see Axler, Linear Algebra Done Right. We will cover all the necessary information in the relevant posts, but until then, we are absolutely pooped from typing. Until next time!
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2014-08-21 06:17:46
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