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Find $p$ for which all solutions of system/equation are real There is system of $5$ equations
$$
a+b+c+d+e = p; \\
a^2+b^2+c^2+d^2+e^2 = p; \\
a^3+b^3+c^3+d^3+e^3 = p; \\
a^4+b^4+c^4+d^4+e^4 = p; \\
a^5+b^5+c^5+d^5+e^5 = p, \\
\tag{1}
$$
where $p\in\mathbb{R}$.
One can prove (like here) that $a,b,c,d,e$ are roots of equation
$$
x^5-\binom{p}{1}x^4+\binom{p}{2}x^3-\binom{p}{3}x^2 + \binom{p}{4}x - \binom{p}{5} = 0.
\tag{2}
$$
I want to create related task for my son: to find polynomial $(2)$ for given system $(1)$. But it would be great to create task, where all $a,b,c,d,e$ are real and distinct. To have possibility to check manually all the sums (without complex numbers).
Question: Which values of $p$ provide $5$ real (pairwise) distinct $a,b,c,d,e$? And if there exists such one at all (for $5$ variables)?
|
One can also use inequalities to give some restrictions on the solution.
We have $$ \sum a^2(a-1)^2 = \sum a^2 - 2 \sum a^3 + \sum a^4 = p - 2p + p = 0 $$ any solution $(a,b,c,d,e)$. Since $x^2(x-1)^2 \ge 0$ for all real $x$, each of $a, b, c, d, e$ must be either $0$ or $1$. Obviously, they cannot be all distinct.
|
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|
Finding a recurrence relation, first few terms of power series solution to differential equation I'm attempting to find a recurrence relation and the first few terms of a power series solution for the differential equation:
$$(1-x^2)y'' - 2xy' + \lambda y = 0$$
Where $\lambda$ is some integer.
So I've assumed a solution $y = \sum_{n=0}^\infty a_nx^n$, and so forth for $y'$ and $y''$.
However, whenever I plug in and try to derive the recurrence relation, I'm not able to reduce it nicely, and as such I can't figure the first few terms of the series.
I'm sort of new to differential equations in general so please bear with my naiveté.
|
For the differential equation $(1-x^{2}) y'' - 2 x y' + \lambda y = 0$ it is seen that if $y$ is of the form $\sum a_{n} x^{n}$ the following holds.
\begin{align}
y(x) &= \sum_{n=0}^{\infty} a_{n} \, x^{n} \\
y'(x) &= \sum_{n=0}^{\infty} n \, a_{n} \, x^{n-1} \\
y''(x) &= \sum_{n=0}^{\infty} n(n-1) \, a_{n} \, x^{n-2}
\end{align}
\begin{align}
0 &= (1 - x^{2} ) \, \sum_{n=0}^{\infty} n(n-1) \, a_{n} x^{n-2} - 2 x \, \sum_{n=0}^{\infty} n \, a_{n} \, x^{n-1} + \lambda \sum_{n=0}^{\infty} a_{n} \, x^{n} \\
&= \sum_{n=2}^{\infty} n(n-1) \, a_{n} \, x^{n-2} - \sum_{n=0}^{\infty} (n(n+1) - \lambda) \, a_{n} \, x^{n} \\
&= \sum_{n=0}^{\infty} (n+2)(n+1) \, a_{n+2} \, x^{n} - \sum_{n=0}^{\infty} (n(n+1) - \lambda) \, a_{n} \, x^{n} \\
&= \sum_{n=0}^{\infty} \left[ (n+2)(n+1) \, a_{n+2} - (n(n+1) - \lambda) \, a_{n} \right] \, x^{n}
\end{align}
From this equation the coefficient equation can be obtained. It is
\begin{align}
a_{n+2} = \frac{n(n+1) - \lambda}{(n+1)(n+2)} \, a_{n} \hspace{10mm} n \geq 0.
\end{align}
Now that the recurrence relation has been obtained. Try a few values of $n$ to obtain the first few terms. The first two terms are defined as $a_{0}, a_{1}$ and the remaining are to follow.
\begin{align}
a_{2} &= \frac{- \lambda }{2!} \, a_{0} \\
a_{3} &= \frac{2-\lambda}{2 \cdot 3} \, a_{1} = \frac{(-1) (\lambda - 2)}{3!} \, a_{1} \\
a_{4} &= \frac{6 - \lambda}{3 \cdot 4} \, a_{2} = \frac{(-1)^{2} \lambda (\lambda - 6)}{4!} \, a_{0}
\end{align}
and so on. The solution for $y(x)$ is of the form
\begin{align}
y(x) &= a_{0} \left[1 - \frac{\lambda}{2!} \, x^{2} + \frac{(-1)^{2} \lambda (\lambda -6)}{4!} \, x^{4} + \frac{(-1)^{3} \lambda (\lambda -6)(\lambda -20)}{6!} \, x^{6} + \cdots \right] \\
& \hspace{5mm} + a_{1} \left[x + \frac{(-1)(\lambda - 2)}{ 3!} \, x^{3} + \frac{(-1)^{2} (\lambda - 2)(\lambda - 12)}{5!} \, x^{5} + \cdots \right]
\end{align}
|
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|
Problem in permutation groups involving conjugates I have to find a permutation $a$ satisfying $ a xa^{-1}=y$ where $ x=(12) (34)$ and $y=(56) (13)$
My attempt in solving the problem was-
$$ a(12)(34)a^{-1}= a(12)(a^{-1}a)(34)a^{-1}= (a(12)a^{-1})(a(34)a^{-1})=(56)(13)$$
$$\implies (a(1) \ a(2)) (a(3) \ a(4))= (56)(13)$$
Comparing both sides, $$ a(1)=5,\\ a(2)=6, \\ a(3)=1, \\ a(4)=3$$
Based on this information, I can write $a=\left(\begin{matrix}1 & 2 & 3 & 4 & 5 & 6\\ 5 & 6 & 1 & 3& 4& 2\end{matrix}\right)$, which satisfies above four equations.
Is my solution correct or not? If yes, thanks for confirming it and if not, then can someone please tell where I did go wrong?
|
Your solution is correct! One way to confirm this would be to write out
$$
a^{-1}=
\begin{pmatrix}
5 & 6 & 1 & 3& 4& 2\\1 & 2 & 3 & 4 & 5 & 6
\end{pmatrix}
$$
and
$$
x=(12)(34)
$$
and
$$
a=
\begin{pmatrix}
1 & 2 & 3 & 4 & 5 & 6\\5 & 6 & 1 & 3& 4& 2
\end{pmatrix}
$$
and simply follow the "route" of each element:
$$
5\overset{a^{-1}}{\longmapsto}1\overset{x}{\longmapsto}2\overset{a}{\longmapsto}6\\
6\overset{a^{-1}}{\longmapsto}2\overset{x}{\longmapsto}1\overset{a}{\longmapsto}5\\
1\overset{a^{-1}}{\longmapsto}3\overset{x}{\longmapsto}4\overset{a}{\longmapsto}3\\
3\overset{a^{-1}}{\longmapsto}4\overset{x}{\longmapsto}3\overset{a}{\longmapsto}1\\
4\overset{a^{-1}}{\longmapsto}5\overset{x}{\longmapsto}5\overset{a}{\longmapsto}4\\
2\overset{a^{-1}}{\longmapsto}6\overset{x}{\longmapsto}6\overset{a}{\longmapsto}2
$$
so $axa^{-1}$ acts exactly like $y$.
|
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|
Proving $(n+1)!>2^{n+3}$ for all $n\geq 5$ by induction I am stuck writing the body a PMI I have been working on for quite some time.
Theorem: $∀n∈N ≥ X$, $(n+1)!>2^{n+3}$
I will first verify that the hypothesis is true for at least one value of $n∈N$.
Consider $n=3$: (not valid)
$$(3+1)!>2^{3+3} \implies 4!>2^{6} \implies 24>64$$
Consider $n=4$: (not valid)
$$(4+1)!>2^{4+3} \implies 5!>2^{7} \implies 120>128$$
Consider $n=5$: (valid)
$$(5+1)!>2^{5+3} \implies 6!>2^{8} \implies 720>256$$
Consider $n=6$: (valid)
$$(6+1)!>2^{6+3} \implies 7!>2^{9} \implies 5040>512$$
So clearly $X$ is $5$.
For the inductive assumption, we will assume the hypothesis holes from $n=5$ up to some arbitrary values $k$: $(k+1)!>2^{k+3}$
This is where I am lost. Originally I had written:
Now I will prove true for $k+1$ showing that: $(k+2)!>2^{k+4}$. Consider the $k+1$ term:
$$(k+2)(k+1)k! = (k+2)(k+1)2^k = (k+2)(k+1)2^k = (2)(2)(2^k) = (4)(2^k) = 2^{k+4}$$
by the inductive assumption since $k>5$ so $k+1>2$ and $k+2>2$ but I know this isn't correct.
I know the proof should look something like this, but I have no idea why:
$$(k+2)(k+1)!>(k+2)(2^{k+3})>2(2^{k+3})=2^{k+4}$$
|
Your first two steps are correct. You have established that:$$(n+1)!\gt2^{n+3}$$is first true when $n=5$. You then correctly assumed that it holds for some $n=k\ge5$, giving you:$$(k+1)!\gt2^{k+3}\tag{1}$$And, finally you also proceeded in the right direction by then saying that you now need to prove that this implies it is also true for $n=k+1$. If we look at $n=k+1$ we get:$$(k+2)!=(k+2)(k+1)!$$Now use the result from (1) to deduce that:$$(k+2)!\gt(k+2)2^{k+3}$$You can then show that:$$(k+2)2^{k+3}=k.2^{k+3}+2.2^{k+3}=k.2^{k+3}+2^{k+4}\gt2^{k+4}$$
|
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|
Find minimum of $P=\frac{\sqrt{3(2x^2+2x+1)}}{3}+\frac{1}{\sqrt{2x^2+(3-\sqrt{3})x +3}}+\frac{1}{\sqrt{2x^2+(3+\sqrt{3})x +3}}$ For $x\in\mathbb{R}$ find minimum of $P$.
$P=\dfrac{\sqrt{3(2x^2+2x+1)}}{3}+\dfrac{1}{\sqrt{2x^2+(3-\sqrt{3})x +3}}+\dfrac{1}{\sqrt{2x^2+(3+\sqrt{3})x +3}}$
Source : Viet Nam national test for high school student.
I can't slove this problem.
|
note when $x=0$, the three items are equal which hints the min is $\sqrt{3}$.
first take $\dfrac{1}{\sqrt{a}}+\dfrac{1}{\sqrt{b}} \ge \sqrt{\dfrac{8}{a+b}}$ which is easy to prove by AM-GM.
$P \ge \dfrac{\sqrt{3(2x^2+2x+1)}}{3}+\sqrt{\dfrac{8}{4x^2+6x+6}} \ge \dfrac{\sqrt{3(2x^2+2x+1)}}{3}+\sqrt{\dfrac{8}{6x^2+6x+6}} $, when $x=0$ get "="
let $t=\sqrt{x^2+x+1},t\ge \dfrac{\sqrt{3}}{2} \implies P\ge \sqrt{\dfrac{2}{3}}\left( \sqrt{t^2-\dfrac{1}{2}}+\dfrac{\sqrt{2}}{t} \right)$
now we prove $ \sqrt{t^2-\dfrac{1}{2}}+\dfrac{\sqrt{2}}{t} \ge \dfrac{3}{\sqrt{2}} \iff \sqrt{t^2-\dfrac{1}{2}}\ge \dfrac{3}{\sqrt{2}}-\dfrac{\sqrt{2}}{t} >0 \iff t^2-\dfrac{1}{2} \ge \dfrac{9}{2}-\dfrac{6}{t}+\dfrac{2}{t^2} \iff t^2(t^2-1) \ge 4t^2-6t+2 \iff (t-1)(t^2(t+1)-2(2t-1)) \ge 0 \iff (t-1)(t(t^2+t-2)-2(t-1))\iff (t-1)^2(t^2+2t-2) \ge 0 \iff (t-1)^2(t+1+\sqrt{3})(t+1-\sqrt{3})\ge 0$
it is trivial $(t+1+\sqrt{3})>0,(t+1-\sqrt{3}) \ge \dfrac{\sqrt{3}}{2}+1-\sqrt{3} =1-\dfrac{\sqrt{3}}{2}>0 $
so when $t=1, P_{min}= \sqrt{3}$
$t=1 \implies x=0 or x=-1$,since first two "=" are all take $x=0$ which mean $P$ get min when $x=0$.
BTW, it may be hard to prove a stronger one:
$P \ge \sqrt{\dfrac{x^2}{3}+3}$
|
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|
Residue at essential singularity I need a little help with the following problem. I've tried many ways, but i didnt succeed. I think there needs to be a trick or something, some transformation. The task is to find the residue of the function at its singularity e.g. z=-3
\begin{equation}
f(z)=\cos\left(\frac{z^2+4z-1}{z+3}\right)
\end{equation}
I tried to write it as
\begin{align}
\cos\left(\frac{z^2+4z-1}{z+3}\right)=1-\frac{1}{2!}\left((z+1)-\frac{4}{z+3}\right)^2+\frac{1}{4!}\left((z+1)-\frac{4}{z+3}\right)^4-\frac{1}{6!}\left((z+1)-\frac{4}{z+3}\right)^6+\ldots
\end{align}
and collect the coefficients at $\frac{1}{z+3}$ using binomial expansion of the brackets, but it seems to be a dead end, because there is to much of them and well hidden. If somebody could give me a hint, that would be great. Thanks.
|
$$
\begin{align}
&\cos\left(\frac{z^2+4z-1}{z+3}\right)\\
&=\cos\left((z+3)-2-\frac4{z+3}\right)\\
&=\cos\left((z+3)-\frac4{z+3}\right)\cos(2)+\sin\left((z+3)-\frac4{z+3}\right)\sin(2)\tag{1}
\end{align}
$$
Since $\cos\left((z+3)-\frac4{z+3}\right)$ is an even function of $z+3$, its residue at $z=-3$ is $0$.
The $(z+3)^{-1}$ term of
$$
\frac{(-1)^n}{(2n+1)!}\left((z+3)-\frac4{z+3}\right)^{2n+1}\tag{2}
$$
is
$$
\begin{align}
&\frac{(-1)^n}{(2n+1)!}\binom{2n+1}{n}(z+3)^n\left(-\frac4{z+3}\right)^{n+1}\\[6pt]
&=\frac{-1}{(2n+1)!}\binom{2n+1}{n}\frac{4^{n+1}}{z+3}\tag{3}
\end{align}
$$
Summing and multiplying by $\sin(2)$, we get the residue to be
$$
\begin{align}
-\sin(2)\sum_{n=0}^\infty\frac{4^{n+1}}{n!(n+1)!}
&=-2\sin(2)I_1(4)\\
&=-17.7485131\tag{4}
\end{align}
$$
where $I_n(z)$ is the modified Bessel function of the first kind.
|
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|
limit of $\frac{xy-2y}{x^2+y^2-4x+4}$ as $(x,y)$ tends to $(2,0)$ Am I able to substitute $x$ by ($k$+$2$) with $k$ tending to $0$, then using polar coordinates to deduce its limit?!
\begin{equation*}
\lim_{(x,y)\to(2,
0)}{xy-2y\over x^2+y^2-4x+4}.
\end{equation*}
|
Set $z=x-2$ then $z\to 0$ as $x\to 2$ and
$\lim\limits_{(x,y)\to (2,0)}\frac{(x-2)y}{(x-2)^2+y^2}=\lim\limits_{(z,y)\to (0,0)}\frac{zy}{z^2+y^2}=\lim\limits_{(my,y)\to(0,0)}\frac{my^2}{(1+m^2)y^2}=\frac{m}{(1+m^2)}$
Thus there is no limit.
|
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|
Proving $ 1+\frac{1}{4}+\frac{1}{9}+\cdots+\frac{1}{n^2}\leq 2-\frac{1}{n}$ for all $n\geq 2$ by induction Question:
Let $P(n)$ be the statement that $1+\dfrac{1}{4}+\dfrac{1}{9}+\cdots +\dfrac{1}{n^2} <2- \dfrac{1}{n}$. Prove by mathematical induction. Use $P(2)$ for base case.
Attempt at solution:
So I plugged in $P(2)$ for the base case, providing me with $\dfrac{1}{4} < \dfrac{3}{2}$ , which is true.
I assume $P(n)$ is true, so I need to prove $P(k) \implies P(k+1)$.
So $\dfrac{1}{(k+1)^2} < 2 - \dfrac{1}{k+1}$.
I don't know where to go from here, do I assume that by the Inductive hypothesis that it's true?
|
Hint: You assume that the statement is true for $n=k$. In other words,
$$
1+\frac{1}{4}+\frac{1}{9}+\cdots+\frac{1}{k^2}<2-\frac{1}{k}.
$$
Now, add $\frac{1}{(k+1)^2}$ to both sides to get
$$
1+\frac{1}{4}+\frac{1}{9}+\cdots+\frac{1}{k^2}+\frac{1}{(k+1)^2}<2-\frac{1}{k}+\frac{1}{(k+1)^2}.
$$
What you would really like is that
$$
2-\frac{1}{k}+\frac{1}{(k+1)^2}<2-\frac{1}{k+1}
$$
because then, by transitivity, your result would hold. So, can you prove that?
|
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|
Norm of the sum of two vectors This problem has two parts.
Part a): $x$ and $y$ are vectors. If $||x|| = 7, ||y|| = 11$, what is the smallest value possible for $||x+y||$? (Note: the || || denotes the norm of a vector).
This is what I have tried so far: I put vector $x$ equal to $\begin{pmatrix} a \\ b \end{pmatrix}$ and vector $y$ equal to $\begin{pmatrix} c \\ d \end{pmatrix}$. $||x|| = 7$ would then be, after simplification, $a^2+b^2 = 49$. Similarly, for $||y|| = 11$, after simplification, $c^2+d^2 = 121$.
Then, $||x+y|| = \sqrt{(a+c)^2 + (b+d)^2}$. Expanding gives us $\sqrt{(a^2+b^2) + (c^2+d^2) + 2(ac+bd)} = \sqrt{49+121+2(ac+bd)} = \sqrt{170+2(ac+bd)}$. That is where I was stuck--any hints for the next few steps?
Part b): $x$ and $y$ are vectors (these are not the same vectors as in part a). If $||x|| = 4, ||y|| = 5, ||x+y|| = 7,$ what is $||2x-3y||$?
Using the same approach as in part a), where vector $x$ is equal to $\begin{pmatrix} a \\ b \end{pmatrix}$ and vector $y$ is equal to $\begin{pmatrix} c \\ d \end{pmatrix}$, $a^2+b^2 = 16$ and $c^2+d^2 = 25$. Similarly, for $||x+y||$, after simplification, it equals $41+2(ac+bd) = 49$ -> $ac+bd = 4$. I'm not sure what to do next after this part too. Any hints?
|
for the second part it is best to use vectors
$|\vec{x} + \vec{y}|^2 = (\vec{x} + \vec{y})\cdot ( \vec{x} + \vec{y})=
|\vec{x}|^2 + |\vec{y}|^2 +2\vec{x} \cdot \vec{y}$
solve the equation
$|7|^2 =
|5|^2 + |4|^2 +2(\vec{x} \cdot \vec{y})$
for $\vec{x} \cdot \vec{y}$
which you can use to evaluate
$|2\vec{x} -3 \vec{y}|^2 =
4|\vec{x}|^2 + 9|\vec{y}|^2 -6 (\vec{x} \cdot \vec{y})$
|
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|
$ax^2+bx+c=0$ has roots $x_1,x_2$. what are the roots of $cx^2+bx+a=0$. Given solution: Dividing the first equation by $x^2$ we get
$c(\frac{1}{x^2})+b(\frac{1}{x})+a=0$
so $(\frac{1}{x_1}),(\frac{1}{x_2})$ are the roots of $cx^2+bx+a=0$.{How?It is not obvious to me.}
The answers so far are proving retrospectively that the roots are indeed those given above.I would like to know how the relation of the roots is derived.
|
Another way to see it is to look at sum and product of roots. $x_1$ and $x_2$ are roots of $ax^2+bx+c=0$ so $x_1+x_2=-\frac{b}{a}$ and $x_1\cdot x_2=\frac{c}{a}$.
$$\frac{a}{c}=\frac{1}{x_1}\cdot \frac{1}{x_2}$$
$$\frac{1}{x_1}+\frac{1}{x_2}=\frac{x_1+x_2}{x_1x_2}=-\frac{b}{c}$$
and this shows that $\frac{1}{x_1}$ and $\frac{1}{x_2}$ are the roots of
$$X^2+\frac{b}{c}X+\frac{a}{c}=0$$
Equivalently
$$cX^2+bX+a=0$$
|
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|
Prove that if $k \in \mathbb{N}$, then $k^4+2k^3+k^2$ is divisble by $4$ I am trying to solve by induction and have established the base case (that the statement holds for $k=1$).
For the inductive step, I tried showing that the statement holds for $k+1$ by expanding $(k+1)^4+2\cdot(k+1)^3+(k+1)^2$, but this equals $16k^4+34k^3+3k^2+16k+4$, and $4$ cannot be factored out.
|
$\underline{\text{Proof by induction:}}$
First, show that this is true for $k=1$:
$1^4+2\cdot1^3+1^2=4$
Second, assume that this is true for $k$:
$k^4+2k^3+k^2=4n$
Third, prove that this is true for $k+1$:
$(k+1)^4+2(k+1)^3+(k+1)^2=$
$4(k+1)^3+\color{red}{k^4+2k^3+k^2}=$
$4(k+1)^3+\color{red}{4n}=$
$4[(k+1)^3+n]$
Please note that the assumption is used only in the part marked red.
$\underline{\text{Proof by modular-arithmetic:}}$
Consider the following cases:
*
*$k\equiv0\pmod4 \implies k^4+2k^3+k^2\equiv0+0+0\equiv0\pmod4$
*$k\equiv1\pmod4 \implies k^4+2k^3+k^2\equiv1+2+1\equiv4\equiv0\pmod4$
*$k\equiv2\pmod4 \implies k^4+2k^3+k^2\equiv16+16+4\equiv36\equiv0\pmod4$
*$k\equiv3\pmod4 \implies k^4+2k^3+k^2\equiv81+54+9\equiv144\equiv0\pmod4$
Please note that this method is handy only when dealing with a relatively small divisor.
|
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|
Maximum value of $ x^2 + y^2 $ given $4 x^4 + 9 y^4 = 64$ It is given that $4 x^4 + 9 y^4 = 64$.
Then what will be the maximum value of $x^2 + y^2$?
I have done it using the sides of a right-angled triangle be $2x , 3y $ and hypotenuse as 8 .
|
Since you tagged Lagrange Multipliers, I will use that method.
Allow that $F(x,y) = 4x^4+9y^4-64$ and $G(x,y)=x^2+y^2-C$, where $C$ is treated as a constant and is the maximum you wish to find.
We allow that $\nabla F(x,y) = \lambda \nabla G(x,y)$ and find two new equations:
$16x^3 = \lambda 2x$ and
$36y^3 = \lambda 2y$
$\implies x^4 = \frac{\lambda ^2}{64}$ and $y^4 = \frac{\lambda ^2}{324}$.
Plugging those into $F(x,y)$ we find that:
$$\lambda = 26.6256...$$
We now go back to our equations after applying the gradient and solve for $x_{max}$ and $y_{max}$. We get:
$x_{max}=1.8245...$ and $y_{max}=1.21622...$
Plugging this into $x^2+y^2$ we obtain the maximum, $4.80727$.
|
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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|
Difficulty turning a quadratic equation to "vertex"-form I'm having difficulty reducing a quadratic equation to its "vertex-form" by following my textbook and nearly every tutorial I can find online.
The starting equation is:
$$f(x) = -2x^2 + 16x - 24$$
Next, I divided each term by $-2$ (to leave the first term as $x^2$):
$$ = x^2 - 8x + 12$$
Next, to complete the square, I divide $x$'s coefficient by $2$, then square it ($(b/2)^2$):
$$\left(\frac{8}{2}\right)^2 = 16$$
Which gives me $16$. I add $16$ to complete the square, but also subtract it so it doesn't affect the value:
$$ = x^2 - 8x + 16 + 12 - 16$$
Finally, I complete the square:
$$ = (x - 4)^2 + 12 - 16$$
Then simplify:
$$= (x - 4)^2 - 4$$
According to by book though,the answer is:
$$-2(x - 4)^2 + 8$$
Unfortunately, the text skips over steps and makes it very unclear how they got the answer that they did. Can anyone point out where I went wrong above? I compiled the above steps from several videosand tutorials, but I must have gone wrong somewhere.
|
You made your error when you divided by $-2$. When you divided $f(x) = -2x^2 + 16x - 24$ by $-2$, you obtained $\color{red}{f(x) = x^2 - 8x + 12}$. The expression in red is not equal to the original function since the $y$-intercept of the original function is $-24$, while the $y$-intercept of the function in red is $12$.
You want to transform the equation $f(x) = -2x^2 + 16x - 24$ into the form $f(x) = a(x - h)^2 + k$. We first extract a factor of $-2$ from the quadratic and linear terms so that the expression in parentheses has the form $x^2 + 2kx$. We do this since we want to complete the square on $x^2 + kx$ to form the perfect square $x^2 + 2kx + k^2 = (x + k)^2$.
\begin{align*}
f(x) & = -2x^2 + 16x - 24\\
& = -2(x^2 - 8x) - 24
\end{align*}
The term in parentheses now has the form $x^2 + 2kx$, where $2k = -8$, so $k = -4$, and $k^2 = 16$. If we add $16$ inside the parentheses, we will obtain the perfect square $x^2 - 8x + 16 = (x - 4)^2$. However, adding $16$ inside the parentheses adds $-2 \cdot 16 = -32$ to the original expression, we must add $32$ to the expression in order to balance the equation.
\begin{align*}
f(x) & = -2(x^2 - 8x) - 24\\
& = -2(x^2 - 8x + 16) - 24 + 32\\
& = -2(x - 4)^2 + 8
\end{align*}
which tells us that the graph of $f(x)$ has vertex $(4, 8)$ and opens downwards. Note that the $y$-intercept is
$$f(0) = -2(0 - 4)^2 + 8 = -2(16) + 8 = -32 + 8 = -24$$
as we would expect.
Check: We verify that the vertex form is equal to the original function.
\begin{align*}
f(x) & = -2(x - 4)^2 + 8\\
& = -2(x^2 - 8x + 16) + 8\\
& = -2x^2 + 16x - 32 + 8\\
& = -2x^2 + 16x - 24
\end{align*}
Completing the square refers to the process of adding $k^2$ to an expression of the form $x^2 + 2kx$ to form the perfect square $x^2 + 2kx + k^2$. One purpose of completing the square is to transform a quadratic function into vertex form. What confused you is that it can also be used to solve for the $x$-intercepts. When we solve for the $x$-intercepts, we set $f(x) = 0$ and solve for $x$. It is here that we divide by $-2$ (or, more generally, the leading coefficient). Continuing from above yields
\begin{align*}
f(x) & = 0\\
-2(x - 4)^2 + 8 & = 0\\
(x - 4)^2 - 4 & = 0\\
(x - 4)^2 & = 4\\
x - 4 & = \pm 2\\
x & = 4 \pm 2
\end{align*}
\begin{align*}
x & = 4 + 2 & x & = 4 - 2\\
x & = 6 & x & = 2
\end{align*}
You can verify that these solutions are correct by substituting them into the equation $f(x) = -2x^2 + 16x - 24$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
linear mapping question If $f$ is a linear mapping from $R^3$ $\rightarrow$ $R^3$, and $fof$=$2$$f$.
Suppose that $v$ $\in$ $Imf$
Can i say that $fof$($v$)=$2$$f$($v$) $\Rightarrow$ $fof$($v$)=$f$($2$$v$) $\Rightarrow$ $f^{-1}$($f$($f$($v$))=$f^{-1}$($f$($2$$v$)) $\Rightarrow$ $f$($v$)=$2$$v$ ? i know the other way by considering an element $u$ $\in$ $R^3$ s.t $f$($u$)=$v$ and its done. but am i able to use the method i wrote before?
($N.B$: since $v$ $\in$ $Imf$ $\Rightarrow$ $v$ $\in$ $R^3$ so that $f$($v$) exists.)
|
Take any basis in $\Bbb R^3$ and write the matrix $A$ of the map $f$ in the basis. YOu initial condition essentially says that $$A^2=2A.$$This implies that the eigenvalues of $A$ lie in the set $\{0,2\}$.
Without losing generality we can suppose that $A$ is already in its Jordan normal form. There are several possibilities for the JNF of this matrix:
$$\begin{pmatrix}
2&1&0\\0&2&1\\0&0&2
\end{pmatrix}, \begin{pmatrix}
2&1&0\\0&2&0\\0&0&2
\end{pmatrix}\\
\begin{pmatrix}
2&1&0\\0&2&0\\0&0&0
\end{pmatrix},\begin{pmatrix}
2&0&0\\0&2&0\\0&0&2
\end{pmatrix}\\
\begin{pmatrix}
0&1&0\\0&0&0\\0&0&2
\end{pmatrix},\begin{pmatrix}
0&1&0\\0&0&0\\0&0&0
\end{pmatrix}\\
\begin{pmatrix}
2&0&0\\0&0&0\\0&0&0
\end{pmatrix},\begin{pmatrix}
0&0&0\\0&0&0\\0&0&0
\end{pmatrix},\begin{pmatrix}
2&0&0\\0&2&0\\0&0&0
\end{pmatrix}
$$
It is easy to see that any JNF with a Jordan cell of order $2$ or $3$ does not fit the description $A^2=2A$. One can argue that this is the consequence of the fact that all roots of the polynomial $x^2-2x$ are simple.
Therefore, the only options we have are
$$
\begin{pmatrix}
2&0&0\\0&2&0\\0&0&2
\end{pmatrix},
\begin{pmatrix}
2&0&0\\0&0&0\\0&0&0
\end{pmatrix},\begin{pmatrix}
0&0&0\\0&0&0\\0&0&0
\end{pmatrix},\begin{pmatrix}
2&0&0\\0&2&0\\0&0&0
\end{pmatrix}
$$
Finally, in the general case the matrix $A$ is described as $P^{-1}JP$ where $P$ is an arbitrary invertible matrix and $J$ is one of the four options above.
In terms of linear operators it means that there exists a basis $v_1,v_2,v_3$ such that $f(v_i)= l_iv_i$ with $l_i\in\{0,2\}$. Essentilly, we have only four such operators. The operator $\forall w\,f(w)=2w$ is only one of them.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Prove (or disprove): $a\times b=c\times d$, the solutions for $x$ in the equation $\frac {a^x+b^x}{c^x+d^x} = \frac{a+b}{c+d}$ is only $\pm 1$. Prove (or disprove): If $a,b,c,d$ are positive real numbers with $a\times b=c\times d$, then the only solutions for $x$ in the equation $$\frac {a^x+b^x}{c^x+d^x} = \frac{a+b}{c+d}$$ are $x = \pm 1$.
Other than the obvious $a=b=c=d$ solution.
|
Clearly, if $a = c, b = d$ or $a = d, b = c$, then all $x$ satisfy the equation. Hence the statement is false. However, one can show that this is the only exception, namely, if $a,b,c,d > 0$ such that $ab = cd$ and $\{a,b\}\neq\{c,d\}$, then $x = \pm 1$ is the only solution to $$\frac{a^x+b^x}{a+b} = \frac{c^x+d^x}{c+d}.$$
Suppose $ab = cd = s^2$. Without loss of generality, we may assume $s = 1$ since otherwise we can divide $a,b,c,d$ by $s$ and the solution set will not change. We may also assume $a \ge b$ and $c \ge d$. Now $b = 1/a, d = 1/c$ and $a,c \ge 1$.
Consider $f\colon [0,\infty)\times \mathbb{R}\to\mathbb{R}$ defined by $$f(z, x) = \frac{e^{zx} + e^{-zx}}{e^{z} + e^{-z}}.$$
Claim $f(z,x)$ is decreasing in $z$ if $|x| < 1$ and is increasing in $z$ if $|x| > 1$. Therefore $f$ is injective in $z$ if $x\neq \pm 1$.
Suppose the claim holds. We know that $f(\log a,x)=f(\log c,x)$ implies either $x = \pm 1$ or $a = c$.
Proof of Claim Observe that $$\partial_z f(z,x) = (-1 + x)\left(e^{(1+x)z}-e^{-(1+x)z}\right) + (1+x)\left(e^{(x-1)z}-e^{(1-x)z}\right).$$
If $|x| < 1$, then all terms above are negative and so $f$ is decreasing in $z$. If $|x| > 1$, then all terms above are positive, and so $f$ is increasing in $z$.
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/1237382",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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|
Find $\lim\limits_{x \to 0} \frac{(1+3x)^{1/3}-\sin(x)-1}{1-\cos(x)}$ I would like to find using Taylor series :
$$\lim\limits_{x \to 0} \frac{(1+3x)^{1/3}-\sin(x)-1}{1-\cos(x)}$$
So I compute the taylor series of the terms at the order $1$ :
$(1+3x)^{1/3}=1+x+o(x)$ and $-\sin x -1=-1-x+o(x)$ and $1-\cos(x)$ does not have a taylor series at the order $1$ so we have $0$ at the numerator and denominator when we search the limit for $x=0$, according to wolfram we should find $-2$, how is it possible ?
Thank you.
|
I will show another way to compute this limit. Let us transform the original expression as follows:
\begin{eqnarray*}
\frac{(1+3x)^{\frac{1}{3}}-1-(\sin x)}{1-(\cos x)} &=&\frac{\left( (1+3x)^{%
\frac{1}{3}}-1-x\right) -\left( \sin x-x\right) }{1-(\cos x)} \\
&=&\frac{9\times \left( \frac{(1+3x)^{\frac{1}{3}}-1-\frac{1}{3}(3x)}{%
(3x)^{2}}\right) -\left( \frac{\sin x-x}{x^{2}}\right) }{\left( \frac{1-\cos
x}{x^{2}}\right) }.
\end{eqnarray*}
Using standard limits
\begin{equation*}
\lim_{u\rightarrow 0}\frac{(1+u)^{\frac{1}{3}}-1-\frac{1}{3}u}{u^{2}}=-\frac{%
1}{9},\ \ \ and\ \ \ \ \lim_{x\rightarrow 0}\frac{\sin x-x}{x^{2}}=0,\ \ \ \
and\ \ \ \ \lim_{x\rightarrow 0}\frac{1-\cos x}{x^{2}}=\frac{1}{2},
\end{equation*}
the required limit follows
\begin{eqnarray*}
\lim_{x\rightarrow 0}\frac{(1+3x)^{\frac{1}{3}}-1-(\sin x)}{1-(\cos x)} &=&%
\frac{9\times \lim\limits_{x\rightarrow 0}\left( \frac{(1+3x)^{\frac{1}{3}%
}-1-\frac{1}{3}(3x)}{(3x)^{2}}\right) -\lim\limits_{x\rightarrow 0}\left(
\frac{\sin x-x}{x^{2}}\right) }{\lim\limits_{x\rightarrow 0}\left( \frac{%
1-\cos x}{x^{2}}\right) } \\
&=&\frac{9\times \left( -\frac{1}{9}\right) -\left( 0\right) }{\left( \frac{1%
}{2}\right) }=-2.
\end{eqnarray*}
The standard limits can be computed using Taylor series, or l'Hospital's rule (or none of them!).
|
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|
solving given linear equation So before you guys judge me, I honestly am so clueless with this so please bare with my dumb questions.
I have been at this equation that I've been going for an hour now,
$\frac{(3x-1)}{2} -2 = 5-\frac{4x}{3+1}$ and I'm trying to solve for $x$.
So I go with
$6\cdot(3x-1)/2-2=(5-4x)/3 +1$ to cancel out the denominators, and I get
$9x+3-6=10-8x+2$
then I move all the like terms,
$9x+8x=12+9$
then my final answer is
$x=21/17$
that didn't seem correct with me so then I tried,
cancelling out the denominators by multiplying each side on their own number, 3 and 2.
then I got $3x-1-2=5-4x+1$
and my answer is
$x=9/7$
How do I go about solving this?.
|
Your Equation looks like:
$$\frac {3x+1} {2} - 2 = 5-\frac{4x}{3+1}$$
Adding in the denominator on the right.
$$\frac {3x+1} {2} - 2 = 5-\frac{4x}{4}$$
Which simplifies to
$$\frac {3x+1} {2} - 2 = 5-x$$
We want to eliminate the denominator so we multiply by $2$.
$$ 2\frac {3x+1} {2} - (2)2 = (2)5-(2)x$$
Simplifying
$$3x+1 - 4 = 10-2x$$
Adding on the right.
$$ 3x- 3 = 10-2x$$
Add $2x$ to both sides and add $3$ to both sides.
$$ 5x = 13$$
$$x=\frac{13}{5}$$
|
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|
Prove by mathematical induction: $\frac{1}{n}+\frac{1}{n+1}+\dots+\frac{1}{n^2}>1$ Could anybody help me by checking this solution and maybe giving me a cleaner one.
Prove by mathematical induction:
$$\frac{1}{n}+\frac{1}{n+1}+\dots+\frac{1}{n^2}>1; n\geq2$$.
So after I check special cases for $n=2,3$, I have to prove that given inequality holds for $n+1$ case by using the given $n$ case. Ok, so this is what I've got by now:
$$\frac{1}{n+1}+\frac{1}{n+2}+\dots+\frac{1}{n^2}+\frac{1}{n^2+1}+\dots+\frac{1}{(n+1)^2}\overset{?}{>}1$$
$$\frac{1}{n+1}+\frac{1}{n+2}+\dots+\frac{1}{n^2}+\frac{1}{n^2+1}+\dots+\frac{1}{n^2+2n+1}\overset{?}{>}1$$
$$\frac{1}{n}+\frac{1}{n+1}+\frac{1}{n+2}+\dots+\frac{1}{n^2}+\frac{1}{n^2+1}+\dots+\frac{1}{n^2+2n+1}\overset{?}{>}1+\frac{1}{n}$$
From the $n$ case we know that:
$$\frac{1}{n}+\frac{1}{n+1}+\frac{1}{n+2}+\dots+\frac{1}{n^2}+\frac{1}{n^2+1}+\dots+\frac{1}{n^2+2n+1}>1+\frac{1}{n^2+1}+\dots+\frac{1}{n^2+2n+1}$$
So we basically have to prove that:
$$1+\frac{1}{n^2+1}+\dots+\frac{1}{n^2+2n+1}\overset{?}{>}1+\frac{1}{n}$$
$$\frac{1}{n^2+1}+\dots+\frac{1}{n^2+2n+1}\overset{?}{>}\frac{1}{n}$$
Since
$$n^2+1<n^2+2<\dots<n^2+2n+1<2n^2+n$$
for $n\geq2$, then also:
$$\frac{1}{n^2+1}>\frac{1}{2n^2+n}$$
$$\frac{1}{n^2+2}>\frac{1}{2n^2+n}$$
.
.
$$\frac{1}{n^2+2n+1}>\frac{1}{2n^2+n}$$
so then we have:
$$\frac{1}{n^2+1}+\dots+\frac{1}{n^2+2n+1}>\frac{1}{2n^2+n}+\dots+\frac{1}{2n^2+n}=(2n+1)\frac{1}{2n^2+n}=\frac{1}{n}$$
|
Here's a "cleaner solution" that's an improvement of the result, though it doesn't use Induction.
Strengthened claim:
$$ \frac{1}{n} + \frac{1}{n+1} + \ldots + \frac{1}{3n-2} \geq 1.$$
Proof: Since $ \frac{ 1}{ 2n-1 - k} + \frac{1}{2n-1 +k} = \frac{2(2n-1)}{(2n-1 - k)(2n-1+k)} \geq \frac{2(2n-1)}{(2n-1)^2} = \frac{2}{2n-1}$, the claim follows by pairing up the extreme ends.
Corollary: For $n \geq 2$, since $ 3n - 2 \leq n^2$ , hence the original statement is true.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Simultaneous Quadratic Equations: $x^2 + y ^ 2 - 2 x + 6y - 35 = 0$ and $2x + 3y = 5$ I've been given the task to simultaneously solve:
$$x^2 + y ^ 2 - 2 x + 6y - 35 = 0$$
$$2x + 3y = 5$$
I've tried applying the substitution method by reordering the second equation to both $x$ and $y$, but both times I have got a surd.
According to the answers I should be getting $(-2, 3)$ and $(\frac{100}{13}, -\frac{45}{13})$, so evidently I am doing something wrong.
This is in the chapter Equations Reducible to Quadratics and I can not see how the material in the intro applies to the question.
|
From the second equation, we have that $x=\frac{5-3y}{2}$. Plugging this into the first equation,
$$\left(\frac{5-3y}{2}\right)^2+y^2-(5-3y)+6y-35=0\\
\frac{9y^2-30y+25}{4}+y^2+9y-40=0$$
Multiplying by $4$,
$$9y^2-30y+25+4y^2+36y-160=0\\
13y^2+6y-135=0$$
Applying the quadratic formula,
$$y=\frac{-6\pm\sqrt{6^2-4(13)(-135)}}{26}\\
=\frac{-6\pm84}{26}=\frac{-3\pm42}{13}=3,-\frac{45}{13}$$
Plugging these values back into our equation $x=\frac{5-3y}{2}$, we get solutions of $(2,-3)$, and $(\frac{100}{13},-\frac{45}{13}).$
|
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"url": "https://math.stackexchange.com/questions/1242861",
"timestamp": "2023-03-29T00:00:00",
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|
Method to integrate $\cos^4(x)$ Here my attempts for integrating $\cos^4(x)$ in few methods.
1st method.
$(\cos^2x)^2=(\frac{1}{2})^2(1+\cos2x)^2$
$=\frac{1}{4}(1+2\cos2x+\cos^22x)=\frac{1}{4}(1+2\cos2x)+\frac{1}{4}(\cos^22x)$
$=\frac{1}{4}(1+2\cos2x)+\frac{1}{4}(\frac{1}{2}(1+\cos4x))$
$=\frac{3}{8}+\frac{1}{2}\cos2x+\frac{1}{8}\cos4x$
So, $\int\cos^4xdx=\int(\cos^2x)^2dx$
$=\int\frac{3}{8}dx+\frac{1}{2}\int\cos2xdx+\frac{1}{8}\int\cos4xdx$
$=\frac{1}{32}(12x+8\sin2x+\sin4x)+c$
2nd method.
By using reduction formula, $\cos^4xdx=\frac{1}{4}\sin x\cos^3x+\frac{3}{4}\int\cos^2xdx$
$=\frac{1}{4}\sin x\cos^3x+\frac{3}{4}\int(\frac{1}{2}\cos2x+\frac{1}{2})dx$
$=\frac{1}{4}\sin x\cos^3x+\frac{3}{8}\int\cos2x+\frac{3}{8}\int1dx$
$=\frac{1}{4}\sin x\cos^3x+\frac{3}{16}\sin 2x+\frac{3}{8}x$
$=\frac{1}{4}\sin x \cos^3x+\frac{3}{8}\sin x \cos x+\frac{3}{8}x$
3rd method, By using De Moivre's formula.
4th method
$\int \cos^4xdx=\int (\frac{e^{ix}+e^{-ix}}{2})^4dx$
$=\frac{1}{16}\int (e^{4ix}+4e^{2ix}+6+4e^{-2ix}+e^{-4ix})dx$
$=\frac{1}{16}\int (e^{4ix}+e^{-4ix}+4(e^{2ix}+e^{-2ix})+6)dx$
$=\frac{1}{16}(\frac{1}{4i}(e^{4ix}-e^{-4ix})+\frac{2}{i}(e^{2ix}-e^{-2ix})+6x)+c$
$\frac{1}{16}(\frac{1}{2}(\frac{e^{4ix}-e^{-4ix}}{2i})+4(\frac{e^{2ix}-e^{-2ix}}{2i})+6x)+c$
$\frac{1}{16}(\frac{1}{2}\sin 4x+4\sin 2x+6x)+c$
$\frac{1}{8}(\frac{1}{4}\sin 4x+2\sin 2x+3x)+c$
I want to ask is there more method to integrate it? Thanks.
|
I think you gave the most straightforward methods, including the sure-to-success replacement by $e^{ix}+e^{-ix}$. Basically any other method is going to be a more obfuscated form of this one.
You could also give a shot to the rational parametrization of the unit circle, by using $\cos x = \frac{1-t^2}{1+t^2}$ and $dx = \frac{2\,dt}{1+t^2}$,
so that $\int \cos^4 x dx = \int \frac {2(1-t^2)^4dt}{(1+t^2)^5}$, expanding this ugly rational fraction at the poles, and then reusing the rational parametrization to fall back on the trigonometric result.
There might also be a way to do this using Fourier expansion, but I'm not sure I want to see this.
|
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|
Evaluating a double integral with Fubini's theorem We want to evaluate the double integral $$ \int \int_{R} f(x,y) ~ \text{d}x \text{d}y $$ over the rectangle $R$ given by $0 \leq x \leq 2, 0 \leq y \leq 1$, with the function $f$ defined by $$ f(x,y) = \dfrac{xy(x^2 - y^2)}{(x^2 + y^2)^3} \quad \text{ for } (x, y) \neq (0,0) $$ and $f(0,0) = 0$. We integate first in the $y$-direction. For any $x \in [0,2]$, we define $A(x) = \int^{1}_{0} f(x,y) ~ \text{d}y$. For now, we assume that $x \neq 0$. We make the substitutions $u = x^2 + y^2, \text{d}u = 2y \text{d}y$, use $x^2 - y^2 = x^2 - (u - x^2) = 2x^2 - u$, and compute \begin{align*}
A(x) &= \int^{x^2 + 1}_{x^2} \dfrac{x(2x^2 - u)}{2u^3} ~ \text{d}u = \int^{x^2 + 1}_{x^2} \left( \dfrac{x^3}{u^3} - \dfrac{x}{2u^2} \right) ~ \text{d}u = \left. - \dfrac{x^3}{2u^2} + \dfrac{x}{2u} \right|^{u = x^2 + 1}_{u = x^2} \\
&= - \dfrac{x^3}{2(x^2 + 1)^2} + \dfrac{x}{2(x^2 + 1)} + \dfrac{1}{2x} - \dfrac{1}{2x} = \dfrac{x}{2(x^2 + 1)^2}
\end{align*} We note that this formula remains valid for $x = 0$, as $f$ vanishes on the whole $y$-axis. Then we integrate in the $x$-direction $$ \int^{2}_{0} A(x) ~ \text{d}x = \int^{2}_{0} \dfrac{x}{2(x^2 + 1)^2} ~ \text{d}x = \left. \dfrac{-1}{4(x^2 + 1)} \right|^{x = 2}_{x = 0} = - \dfrac{1}{20} + \dfrac{1}{4} = \dfrac{1}{5} $$ In the next step we switch the order of integration, i.e. we integrate first in the $x$-direction. For any $y \in [0,1]$, we define $B(y) = \int^{2}_{0} f(x,y) ~ \text{d}x $. If $y \neq 0$, we again make the substitution $u = x^2 + y^2$, except now $\text{d}u = 2x \text{d}x$. We obtain $$ B(y) = \left. - \dfrac{y}{2u} + \dfrac{y^3}{2u^2} \right|^{u = 4 + y^2}_{u = y^2} = - \dfrac{y}{2(4 + y^2)} + \dfrac{y^3}{2(4 + y^2)^2} = \dfrac{-2y}{(4+y^2)^2} $$ This formula too remains valid for $y = 0$, as $f$ vanishes on the $x$-axis. Then we integrate in the $y$-direction $$ \int^{1}_{0} B(y) ~ \text{d}y = \int^{1}_{0} \dfrac{-2y}{(4+y^2)^2} ~ \text{d}y = \left. \dfrac{1}{4 + y^2} \right|^{y=1}_{y=0} = \dfrac{1}{5} - \dfrac{1}{4} = - \dfrac{1}{20} $$ which is not the same as our first answer.
What happened here? The fact that all the functions we integrated
are continuous functions of one variable offers no clue that anything is wrong. The
point is that Fubini's Theorem $\textbf{does not apply}$, because the function $f$ is not integrable over $R$, indeed, it is not even bounded on $R$.
At the end we should conclude that $f$ is not continuous in $(0,0)$. I don't see how to do that. Can anyone give me a hint?
|
using polar coordinates f(x,y)=f(r,a)=(sin4a)/r^2 if a is not zero ans 0 otherwise.If we take the ray a=pi/8 we can see that f is not bounded near the origen and so it is not continuous: f(0,0)=0
|
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Is this a good proof of the binomial identity? Prove that the binomial identity ${n\choose k} = {n-1\choose k-1} + {n-1\choose k}$ is true using the following expression: $(1+x)^n = (1+x)(1+x)^{n−1}$ and the binomial theorem.
What I have:
We know from the binomial theorem that:
$$(x+1)^n= {n\choose 0} x^0 + {n\choose 1} x^1+\cdots+{n\choose k} x^k+\cdots+ {n\choose n-1} x^{n-1}+ {n\choose n} x^n.$$
By using the property that $(1+x)^{n} = (1+x)(1+x)^{n−1}$, we can take out a factor of $x$ to get:
$$x(x+1)^{n-1}={n-1\choose 0} x^0+\cdots+{n-1\choose k-1} x^{k-1}+\cdots+{n+1\choose n-1} x^{n-1}$$
If we then divide by a factor of $x$ we get:
$$1(x+1)^{n-1}={n-1\choose 0} x^0+\cdots +{n-1\choose k-1} x^k+\cdots+{n-1\choose n-1} x^{n-1}$$
Substituting $(x+1)^{n-1}={n-1\choose 0} x^0 +\cdots+{n-1\choose k-1} x^k+\cdots+{n-1\choose n-1} x^{n-1}$ into the equation $x(x+1)^{n-1}= {n-1\choose 0} x^0+\cdots+{n-1\choose k-1} x^{k-1}+\cdots+ {n+1\choose n-1}x^{n-1}$, the equation can be reduced to:
$${n\choose k}x^k = {n-1\choose k-1} x^{k-1}x+{n-1 \choose k}x^k 1$$
Dividing by $x$ we get: ${n\choose k} = {n-1\choose k-1} + {n-1\choose k}$
Is this a good proof? What can I do to improve it? Is there a better way to solve this problem?
|
Your proof can be slightly modified as follows:
Note that
$x(x+1)^{n-1}=x^n+{n-1 \choose 1}x^{n-1}+{n-1 \choose 2}x^{n-2}+...+{n-1 \choose n-2}x^2+x$
$(x+1)^{n-1}=x^{n-1}+{n-1 \choose 1}x^{n-2}+{n-1 \choose 2}x^{n-3}+...+{n-1 \choose n-2}x+1$
Add the 2 equations and look at the addition of like terms.
$(x+1)^n=x^n + ({n-1 \choose 1}+{n-1 \choose 0})x^{n-1} + ({n-1 \choose 2}+{n-1 \choose 1})x^{n-2}+...+({n-1 \choose n-1}+{n-1 \choose n-2})x+1$
Expand $(x+1)^n$ and see how the identity can be proved.
|
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|
How to Find the Function of a Given Power Series? (Please see edit below; I originally asked how to find a power series expansion of a given function, but I now wanted to know how to do the reverse case.)
Can someone please explain how to find the power series expansion of the function in the title? Also, how would you do it in the reverse case? That is, given the power series expansion, how can you deduce the function $\frac{x}{1-x-x^3}$?
I've also rewritten the function as $$\frac{x}{1-x-x^3} = x \cdot \frac{1}{1-x(1-x^2)}$$ which is of the form of the Maclaurin series $\frac{1}{1-x} = 1+x+x^2+x^3+\ldots$
So the series expansion would then be $$x \cdot (1+(x-x^3)+(x-x^3)^2+(x-x^3)^3+\ldots)$$
However, expanding this to find the simplified power series expansion becomes complicated. According to WolframAlpha (link: http://www.wolframalpha.com/input/?i=power+series+of+x%2F%281-x-x%5E3%29) it eventually works out to $$x+x^2+x^3+2x^4+3x^5+4x^6+6x^7+9x^8+13x^9+19x^{10} +\ldots$$ which is a function defined recursively by $f_n = f_{n-1} + f_{n-3}$ for all $n\gt 3$ with the initial condition that $f_1=f_2=f_3=1$
Appreciate any and all help!
Thanks for reading,
A
Edit: I actually want to find the reverse of the original question. Given the recursively defined function (see above) $$x+x^2+x^3+2x^4+3x^5+4x^6+6x^7+9x^8+13x^9+19x^{10} +\ldots ,$$ how can I show that the function of this power series expansion is $\frac{x}{1-x-x^3}$?
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$$\frac{1}{1-x}=1+x+x^2+x^3+x^4+...$$
$$x(\frac{1}{1-(x+x^3)})=x(1+(x+x^3)+(x+x^3)^2+(x+x^3)^3+.....)$$
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Could you explain the expansion of $(1+\frac{dx}{x})^{-2}$? Could you explain the expansion of $(1+\frac{dx}{x})^{-2}$?
Source: calculus made easy by S. Thompson.
I have looked up the formula for binomial theorem with negative exponents but it is confusing. The expansion stated in the text is:
$$\left[1-\frac{2\,dx}{x}+\frac{2(2+1)}{1\cdot2}\left(\frac{dx}{x}\right)^2 - \text{etc.}\right] $$
Please explain at a high school level.
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Consider the expression $$A=(1+a)^{-2}=\frac 1{(1+a)^2}=\frac 1 {1+2a+a^2}$$ and, hoping you already know it, perform the long division. Limiting to first terms, you will arrive to $$A=1-2 a+3 a^2-4 a^3+5 a^4+\cdots$$ Now, replace in this last expression $a$ by $\frac{dx}x$; this will lead to $$A=1 - 2\left(\frac{dx}{x}\right) +3\left(\frac{dx}{x}\right)^2 - 4\left(\frac{dx}{x}\right)^3 + 5\left(\frac{dx}{x}\right)^4+\cdots$$
Another way, would be to consider $$B=(1+a)^{-1}=\frac 1{1+a}$$ and perform the long division again. This will give $$B=1-a+a^2-a^3+a^4-a^5+\cdots$$ But $$\frac{dB}{da}=-\frac 1{(1+a)^2}=-A$$ which makes $$-A=\frac{d}{da}\Big(1-a+a^2-a^3+a^4-a^5+\cdots \Big)=-1+2 a-3 a^2+4 a^3-5 a^4+\cdots$$ Multiplying both sides by $-1$ leads to the previous result.
|
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Precalculus unit circle with imaginary axis. (a) Suppose $p$ and $q$ are points on the unit circle such that the line through $p$ and $q$ intersects the real axis. Show that if $z$ is the point where this line intersects the real axis, then $z = \dfrac{p+q}{pq+1}$.
(b) Let $P_1 P_2 \dotsb P_{18}$ be a regular 18-gon. Show that $P_1 P_{10}$, $P_2 P_{13}$, and $P_3 P_{15}$ are concurrent.
I have gotten nowhere on this problem, but I have a hint:
One of those three segments is more interesting than the other two. Which one, and why? And how can you use that fact to make part (a) relevant?
Any help is appreciated!
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There probably is an easy, intuitive way to solve part (a), but apparently neither of us has found it. Therefore we try the straightforward but tedious ways: analytic geometry for the intersection of the line and the $x$-axis, and real- and imaginary-parts of the complex numbers for the expression.
Let $p=a+bi$ and $q=c+di$.
Intersection of the line through $p$ and $q$ and the $x$-axis: the slope of the line through $p$ and $q$ is $m=\frac{d-b}{c-a}$. The point-slope form of the line through $p$ with that slope is then
$$y-b=\frac{d-b}{c-a}(x-a)$$
Setting $y=0$ and solving for $x$ we get
$$x=\frac{ad-bc}{d-b}$$
Thus our formula for $z_1$, the complex number at the intersection with the $x$-axis, is
$$z_1=\frac{ad-bc}{d-b}$$
Simplifying the expression: Leaving out some of the tedious simplification steps, we get
$$\begin{align}
z_2 &= \frac{p+q}{pq+1} \\[1em]
& = \frac{a+bi+c+di}{(a+bi)(c+di)+1} \\[1em]
& = \frac{(a+c)+(b+d)i}{(ac-bd+1)+(ad+bc)i} \\[1em]
& = \frac{[(a+c)+(b+d)i]\cdot[(ac-bd+1)-(ad+bc)i]}
{[(ac-bd+1)+(ad+bc)i]\cdot[(ac-bd+1)-(ad+bc)i]} \\[1em]
& = \frac{[c(a^2+b^2)+a(c^2+d^2)+a+c]+[-d(a^2+b^2)-b(c^2+d^2)+b+d]i}
{(a^2+b^2)(c^2+d^2)+2(ac-bd)+1}
\end{align}$$
Since $p$ and $q$ are on the unit circle, $a^2+b^2=1$ and $c^2+d^2=1$. Substituting those we get
$$\begin{align}
z_2 &= \frac{[c+a+a+c]+[-d-b+b+d]i}{1+2(ac-bd)+1} \\[1em]
&= \frac{a+c}{ac-bd+1}
\end{align}$$
Now we check if those expressions for $z_1$ and $z_2$ are equal.
$$\frac{ad-bc}{d-b} \stackrel{?}{=} \frac{a+c}{ac-bd+1}$$
Cross-multiplying,
$$ad-bc+cd(a^2+b^2)-ab(c^2+d^2) \stackrel{?}{=} ad-bc+cd-ab$$
Again using $a^2+b^2=1$ and $c^2+d^2=1$, we get equality.
Thus the statement in part (a) is proved.
As for part (b), points $P_1$ and $P_{10}$ are opposite vertices of the regular $18$-gon. Therefore segment $\overline{P_1P_{10}}$ is a diameter of the circumcircle of the $18$-gon.
You could choose the particular regular $18$-gon to be inscribed the unit circle, with $P_1$ at unity. $P_{10}$ would then be at $-1$ and the segment between them is on the $x$-axis. If we let $r$ be the first complex $18$th root of $1$, namely $r=e^{i\pi/9}$, then $P_2$ is at $r$, $P_3$ at $r^2$, $P_4$ at $r^3$, and so on.
The diagonal between $P_2$ and $P_{13}$ then meets the requirements of part (a) and you can use the formula to find where that diagonal intersects the $x$-axis, namely at $\frac{r+r^{12}}{r^{13}+1}$. You can do the same for the diagonal between $P_3$ and $P_{15}$ and get $\frac{r^2+r^{14}}{r^{16}+1}$. Show that those two $x$-coordinates are the same, and part (b) is done.
|
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Proving that $\frac{1}{n}+\frac{1}{n+1}+\cdots+\frac{1}{2n}>\frac{13}{24}$ by induction. Where am I going wrong? I have to prove that
$$\frac{1}{n}+\frac{1}{n+1}+\dots+\frac{1}{2n}>\frac{13}{24}$$ for every positive integer $n$.
After I check the special cases $n=1,2$, I have to prove that the given inequality holds for the $n+1$ case by using the $n$ case. So, I have to prove:
$$\frac{1}{n+1}+\frac{1}{n+2}+\dots+\frac{1}{2n}+\frac{1}{2n+1}+\frac{1}{2n+2}\overset{?}{>}\frac{13}{24}$$
what is equivavlent to
$$\frac{1}{n}+\frac{1}{n+1}+\frac{1}{n+2}+\dots+\frac{1}{2n}+\frac{1}{2n+1}+\frac{1}{2n+2}\overset{?}{>}\frac{13}{24}+\frac{1}{n}$$
By using the $n$ case, I know that
$$\frac{1}{n}+\frac{1}{n+1}+\frac{1}{n+2}+\dots+\frac{1}{2n}+\frac{1}{2n+1}+\frac{1}{2n+2}>\frac{13}{24}+\frac{1}{2n+1}+\frac{1}{2n+2}$$
so I basically have to prove that
$$\frac{13}{24}+\frac{1}{2n+1}+\frac{1}{2n+2}\overset{?}{>}\frac{13}{24}+\frac{1}{n}$$
what is equivavlent to
$$\frac{1}{2n+1}+\frac{1}{2n+2}\overset{?}{>}\frac{1}{n}$$
$$n(2n+2)+n(2n+1)\overset{?}{>}(2n+1)(2n+2)$$
$$4n^2+3n\overset{?}{>}4n^2+6n+2$$
$$-3n\overset{?}{>}2$$
$$n\overset{?}{<}-\frac{2}{3}$$
which can not be because $n$ is a positive integer. What am I doing wrong? Tnx!
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Let
$$ A_n = \frac{1}{n}+\frac{1}{n+1}+\ldots+\frac{1}{2n}=H_{2n}-H_{n-1}.\tag{1}$$
Then:
$$ A_n-A_{n+1} = \frac{1}{n}-\frac{1}{2n+1}-\frac{1}{2n+2}>0 \tag{2} $$
hence the sequence $\{A_n\}$ is decreasing and to prove our claim it is enough to show that:
$$ \lim_{n\to +\infty}A_n >\frac{13}{24}=0.541666\ldots.\tag{3}$$
However, by a Riemann sum argument:
$$ \lim_{n\to +\infty}A_n = \lim_{n\to +\infty}\frac{1}{n}\left(\frac{1}{1}+\frac{1}{1+\frac{1}{n}}+\frac{1}{1+\frac{2}{n}}+\ldots+\frac{1}{2}\right)\\=\int_{1}^{2}\frac{dx}{x}=\log 2=0.69314718\ldots\tag{4}$$
and we are done. We may also notice that from the upper bound:
$$ A_n-A_{n-1} = \frac{2+3n}{2n(n+1)(2n+1)}\leq\frac{3}{(2n)(2n+1)}\tag{5}$$
it follows that:
$$ A_n \geq \frac{3}{2}-\sum_{n\geq 1}\frac{3}{(2n)(2n+1)}=\log 8-\frac{3}{2}=0.579441\ldots\tag{6}$$
without using any integral.
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Evaluate the integral $\int \frac{x}{(x^2 + 4)^5} \mathrm{d}x$
Evaluate the integral $$\int \frac{x}{(x^2 + 4)^5} \mathrm{d}x.$$
If I transfer $(x^2 + 4)^5$ to the numerator, how do I integrate?
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with $u = x^2 + 4$:
$$ \int (x^2 + 4)^{-5 } {x dx}
= \frac 12 \int u^{-5} {du}
= - \frac 1{8} u^{-4}
= - \frac 1{8} (x^2 + 4)^{-4}
$$
|
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Prove that $f : [-1, 1] \rightarrow \mathbb{R}$, $x \mapsto x^2 + 3x + 2$ is strictly increasing.
Prove that $f : [-1, 1] \rightarrow \mathbb{R}$, $x \mapsto x^2 + 3x + 2$ is strictly increasing.
I do not have use derivatives, so I decided to apply the definition of being a strictly increasing function, which should be:
If we pick 2 numbers $a$ and $b$ from the domain of a function $f$, where $a < b$, then $f(a) < f(b)$.
Now (if that is a correct definition), I have tried to apply it to my case:
Let $a$, $b \in [-1, 1]$ and $a < b$. We want to show that $f(a) < f(b)$ or that $f(b) - f(a) > 0$.
We know that $x^2 + 3x + 2 = (x + 2)(x + 1)$, thus we have that $f(a) = (a + 2)(a + 1)$ and $f(b) = (b + 2)(b + 1)$ , therefore we need to show that:
$$(b + 2)(b + 1) - (a + 2)(a + 1) > 0$$
We can see that $(b + 2)$ and $(a + 2)$ will always be positive, and that $(b + 2) > (a + 2)$ , since $b > a$ (by assumption).
Since $b > a$, we know that $b > -1$ (otherwise $a \geq b$), thus $(b + 1) > 0$ (so we know that $(b + 2)(b + 1) > 0$ . We also have at most $(a + 1) = 0$, and thus we have that $(a + 2)(a + 1) \ge 0$.
So, here is the proof that $f(b) - f(a) > 0$ .
Am I correct? If yes, what can I improve it? If not, where are the erros and possible solutions?
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The proof given looks fine to me, modulo my comment (see above).
Here's another way to see it, sans calculus:
Choose $x_1, x_2 \in [-1, 1]$ with $x_1 > x_2$. Then
$f(x_1) - f(x_2) = x_1^2 + 3x_1 + 2 - x_2^2 - 3x_2 - 2 = x_1^2 - x_2^2 + 3(x_1 - x_2)$
$= (x_1 + x_2)(x_1 - x_2) + 3(x_1 - x_2) = (x_1 + x_2 + 3)(x_1 - x_2); \tag{1}$
note that $x_1 - x_2 > 0$ since $x_1 > x_2$ and that $x_1 + x_2 + 3 > 0$ since $x_1 + x_2 > -2$ since $x_1 > x_2 \ge -1$. Thus
$f(x_1) -f(x_2) > 0, \tag{2}$
or
$f(x_1) > f(x_2), \tag{3}$
and we're done!
Note that all we really needed was $x_1 > x_2 \ge -1$.
Of course, with calculus we see that $f'(x) = 2x + 3 >0$ for $x \ge -1$, etc. etc. etc. And since the minimum occurs at $x = -3/2$, Mark Bennet's comment is corroborated.
|
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Using implicit differentiation with a fraction How do I solve this? What steps? I have been beating my head into the wall all evening.
$$ x^2 + y^2 = \frac{x}{y} + 4 $$
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We can multiply both sides of the equation
$$x^2 + y^2 = \frac{x}{y} + 4 \tag{1}$$
by $y$ to obtain
$$x^2y + y^3 = x + 4y \tag{2}$$
Differentiating equation 2 implicitly with respect to $x$ yields
\begin{align*}
2xy + x^2y' + 3y^2y' & = 1 + 4y'\\
(x^2 + 3y^2 - 4)y' & = 1 - 2xy\\
y' & = \frac{1 - 2xy}{x^2 + 3y^2 - 4} \tag{3}
\end{align*}
Let's see why this answer is equivalent to that given by @CivilSigma. If we solve equation 1 for $x/y$, we obtain
$$\frac{x}{y} = x^2 + y^2 - 4$$
Hence,
\begin{align*}
y' & = \frac{1 - 2xy}{x^2 + 3y^2 - 4}\\
& = \frac{1 - 2xy}{x^2 + y^2 - 4 + 2y^2}\\
& = \frac{1 - 2xy}{\dfrac{x}{y} + 2y^2}\\
& = \frac{y - 2xy^2}{x + 2xy^3}
\end{align*}
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Finding the sum of the trigonometric serie: There are two series:
$$1) 1+\dfrac{\cos{x}}{p}+\dfrac{\cos{2x}}{p^2}+...+\dfrac{\cos{nx}}{p^n}=\sum_{k=0}^{n}{\dfrac{\cos{kx}}{p^k}}$$
$$2) \dfrac{\sin{x}}{p}+\dfrac{\sin{2x}}{p^2}+...+\dfrac{\sin{nx}}{p^n}=\sum_{k=0}^{n}{\dfrac{\sin{kx}}{p^k}}$$
Where $p$ its a real constant with absolute value greater than 1.
How can i find the sum of each series?
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Without complex numbers:
$$C_n=\sum_{k=0}^n\frac{\cos(kx)}{p^k}=\sum_{k=1}^{n+1}\frac{\cos(k-1)x}{p^{k-1}}=\sum_{k=1}^n\frac{\cos(k-1)x}{p^{k-1}}+\frac{\cos nx}{p^n}\\
=p\sum_{k=1}^{n}\frac{\cos kx\cos x+\sin kx\sin x}{p^k}+\frac{\cos nx}{p^n}\\
=p(C_n-1)\cos x+pS_n\sin x+\frac{\cos nx}{p^n}.$$
$$S_n=\sum_{k=0}^n\frac{\sin(kx)}{p^k}=\sum_{k=1}^{n+1}\frac{\sin(k-1)x}{p^{k-1}}=\sum_{k=1}^n\frac{\sin(k-1)x}{p^{k-1}}+\frac{\sin nx}{p^n}\\
=p\sum_{k=1}^{n}\frac{\sin kx\cos x-\cos kx\sin x}{p^k}+\frac{\sin nx}{p^n}\\
=pS_n\cos x-p(C_n-1)\sin x+\frac{\sin nx}{p^n}.$$
Then,
$$(1-p\cos x)C_n-p\sin xS_n=-p\cos x+\frac{\cos nx}{p^n}\\
p\sin xC_n+(1-p\cos x)S_n=p\sin x+\frac{\sin nx}{p^n}.$$
Now, solve this $2\times2$ linear system of equations in $C_n$ and $S_n$. The discriminant is
$$\Delta=1-2p\cos x+p^2,$$
and
$$C_n\Delta=\left(-p\cos x+\frac{\cos nx}{p^n}\right)(1-p\cos x)+\left(p\sin x+\frac{\sin nx}{p^n}\right)p\sin x\\
=-p\cos x+p^2-\frac{\cos nx-p\cos(n-1)x}{p^n}$$
$$S_n\Delta=\left(p\sin x+\frac{\sin nx}{p^n}\right)(1-p\cos x)-\left(-p\cos x+\frac{\cos nx}{p^n}\right)p\sin x\\
=p\sin x+\frac{\sin nx-p\sin(n-1)x}{p^n}.$$
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Prove that $n^2(n^2+1)(n^2-1)$ is a multiple of $5$ for any integer $n$. Prove that $n^2(n^2+1)(n^2-1)$ is a multiple of $5$ for any integer $n$.
I was thinking of using induction, but wasn't really sure how to do it.
|
If $n$ is divisible by five, then $n^2(n^2+1)(n^2-1)$ is divisible by $5$, since $n^2$ is a factor.
Now suppose that $n$ is not divisble by $5$. Then $n=5m+r$ for some $r=1,2,3,4$. Note that $(n^2 + 1)(n^2-1) = n^4-1$.
$$n^4 = (5m+r)^4 = (5m)^4 + 4 (5m)^3r + 6 (5m)^2 r^2 + 4 (5m) r^3 + r^4$$ by the binomial theorem. We see then that $n^4 = 5\cdot M + r^4$ for some $M$. Now we just need to show that $r^4-1$ is divisible by $5$ to establish that $n^4-1$ is divisible by $5$.
This does follow by Fermat's little theorem. If that sounds foreign to you, it's easy enough to check by hand. $r$ can take four different values, $1,2,3,4$ (remember we already assumed that $r\neq 0$ since we are assuming $n$ is not divisble by $5$).
$$1^4 -1 = 0$$
$$2^4 -1= 15$$
$$3^4 - 1 = 80$$
$$4^4 - 1 = 255$$
Thus we see that, if $n$ is not divisible by $5$, then $$n^4 -1 = 5 M + r^4 - 1$$ and both $5M$ and $r^4 - 1$ are divisible by five. Therefore, $n^4-1$ is divisible by five, and $$n^2(n^2+1)(n^2-1) = n^2(n^4-1)$$ is divisible by $5$.
|
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Compare $\sum_{k=1}^n \left\lfloor \frac{k}{\varphi}\right\rfloor$ ... Given two integer sequences
\begin{equation*}
\displaystyle A_n=\sum_{k=1}^n \left\lfloor \frac{k}{\varphi}\right\rfloor ,
\end{equation*}
\begin{equation*}
B_n=\left\lfloor\dfrac{n^2}{2\varphi}\right\rfloor-\left\lfloor \dfrac{n}{2\varphi^2}\right\rfloor
\end{equation*}
here: $\quad\varphi=\dfrac{1+\sqrt{5}}{2}\quad$ (golden ratio)
Prove that: $|A_n-B_n|\leq 1.$
I realized that the difference between $A_n$ and $B_n$ is very small but the failure in finding an exact formula for $A_n$
Could you help me?
|
The difference $A_n-B_n$ can be proven to have no fixed upper or lower bound.
Let $S_n = \sum_{k=1}^n \lfloor k\varphi \rfloor$
To calculate $S_n$ let $n'=\lfloor n\varphi\rfloor$ and $T_n = \sum_{k=1}^n \lfloor k\varphi^2 \rfloor$
$\lfloor k\varphi\rfloor$ and $\lfloor k\varphi^2 \rfloor$ are complementary Beatty sequences so $S_n+T_{n-n'}=\sum_{k=1}^{n'} k=\frac {n'}2(n'+1)$
Since $\lfloor k\varphi^2\rfloor=\lfloor k\varphi\rfloor+k$ so $T_n-S_n=\sum_{k=1}^n(\lfloor k\varphi^2\rfloor-\lfloor k\varphi\rfloor)=\frac n2(n+1)$
giving a recursive formula to find $S_n$.
Let $n=n_m=\lfloor\dfrac{L_m}{5}\rfloor$ where $L_m$ is the Lucas sequence $L_n=\varphi^n+\bar\varphi^n$ and $\bar\varphi=-1/\varphi$ .
Then for $m\ge0$ we have, by inspection, $\lfloor n_m/\varphi\rfloor=n_{m-1}-1$ when $m\equiv 2\mod 4$ and $=n_{m-1}$ otherwise, which allows us to use the recursive formula to show by induction on m that:
if $m\equiv0\mod4$ then $5n=L_m-2$ and $50S_n = L_{2m+1}+L_{m+1}-5L_m-\frac{5m}{2}+8$
if $m\equiv1\mod4$ then $5n=L_m-1$ and $50S_n = L_{2m+1}+3L_{m+1}-5L_m+\frac{5(m-1)}{2}-8$
if $m\equiv2\mod4$ then $5n=L_m-3$ and $50S_n = L_{2m+1}-L_{m+1}-5L_m-\frac{5(m-2)}{2}+8$
if $m\equiv3\mod4$ then $5n=L_m-4$ and $50S_n = L_{2m+1}-3L_{m+1}-5L_m+\frac{5(m-3)}{2}+12$
Let $U_n$ be an approximation for $S_n$ in some sense, specifically with $U_n=\sum_{k=1}^n (k\varphi -\frac12)=\dfrac{\varphi n(n+1)-n}2$ and we can see that the difference $S_n-U_n$ is dominated by a term linear in $m$, with sign depending on whether $m$ is even. For example, in the case $m\equiv 0\mod 4$, then $5n=\varphi^m+\bar\varphi^m-2$ and it follows that
$50S_n=\varphi^{2m+1}+\varphi^{m+1}-5\varphi^m-\frac{5m}2+8-5\bar\varphi^m+\bar\varphi^{m+1}+\bar\varphi^{2m+1}$
$50U_n=\varphi^{2m+1}+\varphi^{m+1}-5\varphi^m+10-6\varphi-\bar\varphi^{m-1}-5\bar\varphi^m-\bar\varphi^{2m-1}$
$S_n-U_n=-\frac m{20}+$...(terms of lower degree)
If we define $C_n=\dfrac{n^2}{2\varphi}-\dfrac{n}{2\varphi^2}$ then $C_n$ differs from $B_n$ by no more than 1 and $C_n=\dfrac{\varphi(n^2+n)-(n^2+2n)}2$.
Now $A_n-C_n=S_n-U_n$ which has no upper or lower bound. $\Box$
|
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|
Prove the Apollonius' theorem. Let in a $\Delta ABC$, D is the midpoint of $BC$.Prove that:
$AB^2+AC^2=2CD^2+2AD^2$
MY ATTEMPT :
Given that $BD=DC$ and we construct $E \ such \ that\ AE=EC\implies AC=\frac{EC}{2} \ and \ DE||AB \implies DE=\frac{AB}{2}$
For the $\Delta DEC$ we have $DC^2=DE^2+EC^2 \implies 4DC^2=AC^2+AB^2 $
we have $AB^2+AC^2=2CD^2+2CD^2 \tag{1}$
In $\Delta ADE$ we have, $AD^2=AE^2+DE^2 \implies AE^2+DC^2-EC^2 \implies
AD^2=DC^2$
Hence $2CD^2=2AD^2$
Thus, we have
$AB^2+AC^2=2CD^2+2AD^2$
I am not sure of this proof. Though this proof is well explained in wikipedia.I tried to check if this can be solved using elementary geometry.
|
Use the cosine rule twice:
$$\cos B=\dfrac {c^2+\frac {a^2}{4}-m^2}{2\times c\times \frac {a}{2}}\implies ac\cos B=c^2+\dfrac {a^2}{4}-m^2$$
$$\cos C=\dfrac {b^2+\frac {a^2}{4}-m^2}{2\times b\times \frac {a}{2}}\implies ab\cos C=b^2+\dfrac {a^2}{4}-m^2$$
Adding the above two and using $a=b\cos C+c\cos B$, we get $$a^2=b^2+c^2+\dfrac {a^2}{2}-2m^2$$ or $$AB^2 +AC^2 =2CD^2 +2AD^2$$(on rearranging the terms)
|
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|
What is the value of $\frac{a^3}{a^6+a^5+a^4+a^3+a^2+a+1}$ If $\frac{a}{a^{2}+1} = \frac{1}{3}$ Then find the value of $$\frac{a^3}{a^6+a^5+a^4+a^3+a^2+a+1}$$
Any hints to help me?
|
Using $a^2=3a-1$ we can compute $a^3=8a-3,\; a^4=21a-8, ... , a^6=144a-55$ by iteration. Then your term equals
$$
\frac{8a-3}{29(8a-3)}=\frac{1}{29}.
$$
|
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Proof that $(n^7-n^3)(n^5+n^3)+n^{21}-n^{13}$ is a multiple of $3$. I proved that $$(n^7-n^3)(n^5+n^3)+n^{21}-n^{13}$$ is a multiple of $3$ through the use of Little Fermat's theorem but i want to know if there exist other proofs(maybe for induction). How can I demonstrate it?
This my proof:
$$n^3(n^4-1)(n^5+n^3)+n^{13}(n^8-1)$$
Now i know that $$n^4-1\equiv 0\pmod 3 (n\neq 3k)$$ and $$n^8-1\equiv 0\pmod 3 (n\neq 3k).$$
Therefore I proved that $(n^7-n^3)(n^5+n^3)+n^{21}-n^{13}$ is a multiple of $3$.
|
I will prove the induction method here for $n \geq 0$
Base case : $n=0,1$
Clearly $0$ works.
now $(1^7-1^3)(1^5 + 1^3) + 1^{21} - 1^13 = 0$ which is also a multiple of $3$
Now assume it works for a value $k \geq 0$ that is $$(k^7-k^3)(k^5 + k^3) + k^{21} - k^{13}$$ is a mulitple of $3$
That means that $3m = (k^7-k^3)(k^5 + k^3) + k^{21} - k^{13} = k^{21} -k^{13} + k^{12} + k^{10} -k^{8} -k^{6} $ for some integer $m$
Now we just need to show that this holds for $(k+1)$. That is we need to show that $(k+1)^{21} -(k+1)^{13} + (k+1)^{12} + (k+1)^{10} -(k+1)^{8} -(k+1)^{6}$ is a multiple of $3$
Can you continue from here ??
|
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|
Find the least nonnegative residue of $3^{1442}$ mod 700 So I have that $700=7\cdot2^2\cdot5^2$ and I got that $3^2\equiv1\pmod2$ so then $3^{1442}\equiv1\pmod2$ also $3^2\equiv1\pmod{2^2}$ so $3^{1442}\equiv1\pmod{2^2}$ which covers one of the divisors of $700$. Im not sure if I'm supposed to use $2$ or $2^2$ and I was able to find that $3^2\equiv-1\pmod5$ so $3^{1442}\equiv-1\pmod5$, For mod $7$ I wasn't able to come up with an answer in a way like the other two, and I'm not really sure how to do this to find the least non negative residue
|
Go $\pmod4$, $\pmod7$ and $\pmod{25}$. We have
\begin{align}
3^2 \equiv 1\pmod4\\
3^6 \equiv 1\pmod7\\
3^{20} \equiv 1\pmod{25}
\end{align}
This gives us that
\begin{align}
3^{60} \equiv 1\pmod4\\
3^{60} \equiv 1\pmod7\\
3^{60} \equiv 1\pmod{25}
\end{align}
This means
$$3^{60} \equiv 1\pmod{700}$$
Note that $3^{1442} = 3^{24\cdot60+2} = \left(3^{60}\right)^{24} \cdot 3^2$. Hence, we obtain
$$3^{1442} \equiv 3^2\pmod{700} \equiv9\pmod{700}$$
|
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What is the connection between the discriminant of a quadratic and the distance formula? The $x$-coordinate of the center of a parabola $ax^2 + bx + c$ is $$-\frac{b}{2a}$$
If we look at the quadratic formula
$$\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$
we can see that it specifies two points at a certain offset from the center
$$-\frac{b}{2a} \pm \frac{\sqrt{b^2 - 4ac}}{2a}$$
This means that $\frac{\sqrt{b^2 - 4ac}}{2a}$ is the (horizontal) distance from the vertex to the roots. If I squint, the two squared-ish quantities being subtracted under a square root sign reminds me of the Euclidean distance formula
$$\sqrt{(x_0 - x_1)^2 + (y_0 - y_1)^2}$$
Is there a connection? If not, is there any intuitive or geometric reason why $\frac{\sqrt{b^2 - 4ac}}{2a}$ should be the horizontal distance from the vertex to the roots?
|
For the equation $ax^2 + bx + c = 0$ the second coefficients represents the average of the two roots:
$$ \frac{1}{2}(r_1 + r_2) = -\frac{b}{2a} $$
Instead of using the quadratic formula directly we can just observe that:
$$ (x - r_1)(x - r_2) = x^2 - (r_1 + r_2) + r_1 r_2$$
This also recovers the product mean of the two roots is $r_1 r_2 = \frac{c}{a}$.
In order to get the distance between the two roots we need to estimate the distance $|r_1 - r_2|$:
$$ |r_1 - r_2| = \sqrt{(r_1 + r_2)^2 - 4 r_1 r_2} = \sqrt{\frac{b^2}{a^2} - \frac{4c}{a}} = \frac{\sqrt{b^2 - 4ac}}{a}$$
|
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Ways to place 3 red, 4 blue and 5 green wagons such that no 2 blue wagons were standing next to each other As the title says I need to find the number of ways to to place 3 red, 4 blue and 5 green wagons such that no 2 blue wagons were standing next to each other. The wagons of the same color are considered completely identical.
I think I solved it, but would like to have somebody to check it and tell if it's correct.
So my approach is to first count all possible combinations of these 12 wagons and then subscribe the combinations in which there are at least 2 blue wagons next to each other.
The total number of combinations is:
$C_{12}^{3} * C_{9}^{4} * C_{5}^{5} = 27720$
The number of combinations in which at least 2 blue wagons are standing next to each other (that's the part I have the most doubts about):
There are 11 ways to place 2 blue wagons next to each other if we have 12 places for wagons (1 and 2, 2 and 3, etc.). Once we've placed them this way, it doesn't matter where we place another 2 blue wagons. So we have $C_{10}^{2} $ ways to do it. And the number of combinations with blue wagons next to each other is 11 * 45 = 495
And the final answer would be 27720 - 495 = 27225.
|
Here is a different approach.
We first line up the three red and five green wagons (leaving enough space between each pair of wagons so that another wagon can be parked between them). This can be done by choosing which three of the eight positions will be reserved for the red wagons, which can be done in $\binom{8}{3}$ ways. We now choose four of the nine available spaces (seven between successive pairs of the eight wagons that have already been parked and two at the end) in which to place the blue wagons, which can be done in $\binom{9}{4}$ ways. Therefore, the number of ways that we can arrange three red, four blue, and five green wagons so that no two blue wagons are consecutive is
$$\binom{8}{3} \cdot \binom{9}{4} = 7056$$
To make your approach work, we have to use the Inclusion-Exclusion Principle, with the cases being at least two consecutive blue wagons, at least three consecutive blue wagons, and all four blue wagons being consecutive. The case of two consecutive blue wagons is tricky since counting the cases in which there are at least two consecutive blue wagons count cases in which there are two disjoint pairs of blue wagons twice.
As you realized, the number of arrangements of three red, four blue, and five red wagons is
$$\binom{12}{3}\binom{9}{4}\binom{5}{5} = \frac{12!}{3!4!5!}$$
We must subtract from these the number of arrangements in which at least two blue wagons are consecutive.
The number of ways in which at least two blue wagons are consecutive can be found by thinking of the twelve wagons as a pair of blue wagons, two additional blue wagons, three red wagons, and five green wagons. This gives us a total of eleven objects. We can place the pair of blue wagons in $\binom{11}{1}$ ways, the other two blue wagons in $\binom{10}{2}$ ways since ten places remain after the pair is placed, the three red wagons in the eight remaining spaces in $\binom{8}{3}$ ways, and the five green wagons in $\binom{5}{5}$ ways. This yields
$$\binom{11}{1}\binom{10}{2}\binom{8}{3}\binom{5}{5} = \frac{11!}{1!2!3!5!}$$
arrangements in which two blue wagons are consecutive. However, we have counted each case in which there are two disjoint pairs of blue wagons twice. Since a pair can begin in any of the first eleven places, there are $\binom{11}{2}$ places to start a pair. However, $\binom{10}{1}$ of these are consecutive, so the pair would not be disjoint. Thus, there are $\binom{11}{2} - \binom{10}{1} = \binom{10}{2}$ (by Pascal's Identity) ways to place two disjoint pairs of blue wagons. Once they have been placed, there are eight spaces remaining. We can place the three red wagons in these eight spaces in $\binom{8}{3}$ ways and the five green wagons in the five remaining spaces in $\binom{5}{5}$ ways. Thus, there are
$$\binom{10}{2}\binom{8}{3}\binom{5}{5} = \frac{10!}{2!3!5!}$$
arrangements in which there are two disjoint pairs of blue wagons. Thus, the number of arrangements in which at least two blue wagons are consecutive is
$$\frac{11!}{1!2!3!5!} - \frac{10!}{2!3!5!}$$
The number of ways in which at least three blue wagons are consecutive can be found by thinking of the twelve wagons as ten objects, namely the three consecutive blue wagons, the other blue wagon, the three red wagons, and the five green wagons. We can place the trio of consecutive blue wagons in $\binom{10}{1}$ ways, the other blue wagon in one of the nine remaining spaces in $\binom{9}{1}$ ways, the three red wagons in the eight remaining spaces in $\binom{8}{3}$ ways, and the five green wagons in the last five spaces in $\binom{5}{5}$ ways. Hence, there are
$$\binom{10}{1}\binom{9}{1}\binom{8}{3}\binom{5}{5} = \frac{10!}{1!1!3!5!}$$
ways of arranging the wagons so at least three blue wagons are consecutive.
The number of ways we can arrange the wagons in which all four blue wagons are consecutive can be found by thinking of the four blue wagons as one object. Then we have nine objects to place, the block of four consecutive blue wagons, the three red wagons, and the five green wagons. We can place the four consecutive blue wagons in $\binom{9}{1}$ ways, the three red wagons in $\binom{8}{3}$ ways, and the five green wagons in $\binom{5}{5}$ ways. Thus, there are
$$\binom{9}{1}\binom{8}{3}\binom{5}{5} = \frac{9!}{1!3!5!}$$
arrangements in which all four blue wagons are consecutive.
By the Inclusion-Exclusion Principle, the number of arrangements of three red wagons, four blue wagons, and five green wagons in which no two blue wagons are consecutive is
$$\frac{12!}{3!4!5!} - \left[\frac{11!}{1!2!3!5!} - \frac{10!}{2!3!5!}\right] + \frac{10!}{1!1!3!5!} - \frac{9!}{1!3!5!} = 7056$$
which agrees with the result obtained above.
|
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Solving an equation with LambertW function I have function $f(x)$ which has a maximum at $x^*$ ($\frac{d^2f(x)}{dx^2}<0$ can be shown):
$$f(x)=\frac{1}{2} (1-x) \log \left(\frac{2 d g x}{1-x}+1\right)$$
In the research article, the solution for $x^*$ is given as
$$x^*=\frac{(2 d g-1)-Q(2)}{(2 d g-1) \left(Q(2)+1\right)};~\text{where}~Q(t)=W\left(\frac{d g t-1}{\exp (1)}\right)$$
where $W$ is the LambertW function.
When I get the first derivative, and make it to zero, I have
$$\frac{d g}{2 d g x-x+1}-\frac{1}{2} \log \left(1-\frac{2 d g x}{x-1}\right)=0\rightarrow exp\left(\frac{2 d g}{2 d g x-x+1}\right)= \left(\frac{2 d g x-x+1}{1-x}\right)$$
But I am not sure how I can reduce this to get $x^*$.
|
Start with
$$\exp\left(\frac{2 d g}{2 d g x-x+1}\right)= \left(\frac{2 d g x-x+1}{1-x}\right)$$
Next, divide both sides by $e$ to obtain
$$\exp\left(\frac{(2 d g-1)(1-x)}{2 d g x-x+1}\right)= e^{-1}\left(\frac{2 d g x-x+1}{1-x}\right)$$
Then, multiply both sides by $2dg-1$ to reveal
$$ze^z=(2dg-1)/e$$
where $z=\frac{(2 d g-1)(1-x)}{2 d g x-x+1}$. Thus,
$$z=W\left(\frac{2dg-1}{e}\right)$$
whereupon solving for $x$ yields
$$x=\frac{(2dg-1)-W\left(\frac{2dg-1}{e}\right)}{(2dg-1)\left(1+W\left(\frac{2dg-1}{e}\right)\right)}$$
|
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How to prove a Fibonacci inequality using Strong Induction? Using strong induction I am trying to prove that
$$F_n \geq \left(\frac{1+\sqrt{5}}{2}\right)^{n-2} \text{ for all } n \geq 2$$
for the Fibonacci Sequence defined by: $F_0 = 0$, $F_1 = 1$, and $F_n = F_{n-1} + F_{n-2}$ for $n \geq 2$.
I know how to do strong induction for normal sequences, but not for inequalities (I have never liked inequalities).
If someone wouldn't mind pointing me in the right direction regarding what to do, I would be very grateful.
|
\begin{align}
F_n & = F_{n-1}+F_{n-2} \\[8pt]
& \ge \left(\frac{1+\sqrt{5}}{2}\right)^{n-3} + \left( \frac{1+\sqrt{5}}{2} \right)^{n-4} & & \text{(by the induction hypothesis)} \\[8pt]
& = \left(\frac{1+\sqrt{5}}{2}\right)^{n-4} \left( \frac{1+\sqrt{5}}{2} + 1 \right) & & \text{(Here we just pulled out a common factor.)} \\[8pt]
& = \left(\frac{1+\sqrt{5}}{2}\right)^{n-4} \left( \frac{1+\sqrt{5}}{2} \right)^2 & & \text{Where did this come from? See below.} \\[8pt]
& = \left(\frac{1+\sqrt{5}}{2}\right)^{n-2}
\end{align}
We used the equality $\dfrac{1+\sqrt{5}}{2} + 1 = \left( \dfrac{1+\sqrt{5}}{2} \right)^2$.
To see this, just simplify both sides completely. In both cases, it comes to $\dfrac{3+\sqrt 5} 2$.
|
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Solvability of $a_{1}^{2} = a_{2}^{n} + a_{3}^{n} + a_{4}^{n}$ when $n \geq 5$ is prime? Aware of a Darmon-Merel theorem that asserts that if $n \geq 5$ is prime then the equation $a_{1}^{2} = a_{2}^{n} + a_{3}^{n}$ has no solution in relatively prime integers $a_{1}, a_{2}, a_{3},$ I wonder if this is also true, under the same conditions, for the equation
$$a_{1}^{2} = a_{2}^{n} + a_{3}^{n} + a_{4}^{n}?$$
If there is already any known result in existence, bringing it in is more than welcome.
|
For $0<a\leq b\leq c\leq 100$ with $\gcd(a,b,c)=1$, we have the sums including the one by user R. Israel,
$$3^5+ \color{brown}{49}^5+ 69^5 = 42971^2$$
$$20^5+ 68^5+ \color{brown}{81}^5 = 70313^2$$
$$19^5+ 80^5+ 97^5 = 108934^2$$
$$ \color{brown}{16}^5+ 83^5+ 97^5 = 111926^2$$
While your question was on prime $n\geq 5$, there is also for $n=6$,
$$42^6+ \color{brown}{81}^6+ \color{brown}{100}^6 = 1134865^2$$
(It's quite interesting that a lot of the addends are squares.)
For $n=7,8$, there are none in that range (anyone can extend it?), though there is the near-miss,
$$(a^2-b^2)^8+(a^2+b^2)^8+(2ab)^8 = 2(a^8+14a^4b^4+b^8)^2$$
|
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"url": "https://math.stackexchange.com/questions/1269338",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Finding this weird limit involving periodic functions with periods 5 and 10. If $f(x)$ and $g(x)$ are two periodic functions with periods 5 and 10 respectively, such that:
$$\lim_{x\to0}\frac{f(x)}x=\lim_{x\to0}\frac{g(x)}x=k;\quad k>0$$
then for $n\in\mathbb N$, the value of :
$$\lim_{n\to\infty}\frac{f(5(4+\sqrt{15})^n)}{g(10(4+\sqrt{14})^n)}$$.
Note: Everything should be done by hand, any claculating/plotting, etc. device/software is not allowed.
What I've thought is make up such a function:
$$f(x)\propto\sin\frac{2\pi x}{5}\qquad g(x)\propto\sin\frac{2\pi x}{10}$$
But the condition for the limit at zero cna not be satisfied in anyways from here.
Also:
$$5(4+\sqrt{15})^n=5\times4^n(1+\sqrt{15}/4)^n\sim 5\times4^n+5\times4^n\times n\sqrt{15}/4+...$$
wheer $5\times4^n\in\mathbb Z$
So, by periodicity:
$$f(5\alpha+\beta)=f(\beta)\qquad \alpha\in\mathbb Z,\beta\in\mathbb R_{|\beta|<5}$$
And then:
$$L=\lim_{n\to\infty}\frac{f(5\times4^nn\sqrt{15}/4+\cdots)}{g(10\times4^nn\sqrt{14}/4+\cdots)}$$
But this proceeds nowhere, as I was wshing to apply L'Hospital or retain only the $(\beta, |\beta|<5)$ part, but :D.
|
The trick is that the conjugates [Galois conjugates, not complex conjugates] of $4+\sqrt{15}$ and $4+\sqrt{14}$ have modulus less than $1$, and if $a,b\in \mathbb{Z}$ then
$$(a+\sqrt{b})^n + (a-\sqrt{b})^n \in \mathbb{Z}$$
for all $n\in \mathbb{N}$.
So we have
\begin{align}
f\bigl(5(4+\sqrt{15})^n\bigr) &= f\Bigl(5\bigl((4+\sqrt{15})^n + (4-\sqrt{15})^n\bigr) - 5(4-\sqrt{15})^n\Bigr) = f\bigl(-5(4-\sqrt{15})^n\bigr),\\
g\bigl(10(4+\sqrt{14})^n\bigr) &= g\Bigl(10\bigl((4+\sqrt{14})^n + (4-\sqrt{14})^n\bigr) - 10(4-\sqrt{14})^n\Bigr) = g\bigl(-10(4-\sqrt{14})^n\bigr).
\end{align}
Now the arguments of $f$ resp. $g$ converge to $0$ for $n\to \infty$, and
\begin{align}
\frac{f\bigl(5(4+\sqrt{15})^n\bigr)}{g\bigl(10(4+\sqrt{14})^n\bigr)}
&= \frac{f\bigl(-5(4-\sqrt{15})^n\bigr)}{g\bigl(-10(4-\sqrt{14})^n\bigr)}\\
&= \underbrace{\frac{f\bigl(-5(4-\sqrt{15})^n\bigr)}{-5(4-\sqrt{15})^n}}_{\to k}\cdot \underbrace{\frac{-10(4-\sqrt{14})^n}{g\bigl(-10(4-\sqrt{14})^n\bigr)}}_{\to \frac{1}{k}}\cdot \underbrace{\frac{5(4-\sqrt{15})^n}{10(4-\sqrt{14})^n}}_{\to 0}\\
&\to 0
\end{align}
since $0 < 4-\sqrt{15} < 4-\sqrt{14}$. If there was a typo and the power in the argument should have been $(4+\sqrt{15})^n$ for both, $f$ and $g$, then the limit would be $\frac{1}{2}$.
|
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"url": "https://math.stackexchange.com/questions/1269729",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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|
To find jordan canonical form Which of the following matrices have Jordan canonical form of equal to the $3\times 3$ matrix
$$
\begin{pmatrix}
0 & 1 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{pmatrix}$$
a)$
\begin{pmatrix}
0 & 0 & 1 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{pmatrix}$
b)$
\begin{pmatrix}
0 & 0 & 1 \\
0 & 0 & 1 \\
0 & 0 & 0
\end{pmatrix}$
c)$
\begin{pmatrix}
0 & 1 & 1 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{pmatrix}$
d)$
\begin{pmatrix}
0 & 1 & 1 \\
0 & 0 & 1 \\
0 & 0 & 0
\end{pmatrix}$
Here characteristic equation of the matrix is $x^3$.Hence the 3 eigenvalues of the matrix are zero.
Do we want to find the eigenvalues of all the matrices in the options?Is there any other way?
|
Finding the eigenvalues of any of these matrices is not particularly difficult; they are all upper-triangular.
Hint: Suppose that $A$ is a $3 \times 3$ matrix with zero as its only (complex) eigenvalue. Note that $A$ has the desired J-C form if and only if $A \neq 0$ but $A^2 = 0$ (why?).
|
{
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"url": "https://math.stackexchange.com/questions/1270082",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
The limit of this function when x goes to minus infinity? I'm looking for the limit of $$\mathop {\lim }\limits_{x \to - \infty } \left( {\sqrt {1 + x + {x^2}} - \sqrt {1 - x + {x^2}} } \right)$$
I know it should be -1, but for some reason I always get to 1.
I'm not sure where the difference between $$- \infty$$ and $$\infty$$ is:
$$\sqrt {1 + x + {x^2}} - \sqrt {1 - x + {x^2}} {{\sqrt {1 + x + {x^2}} + \sqrt {1 - x + {x^2}} } \over {\sqrt {1 + x + {x^2}} + \sqrt {1 - x + {x^2}} }} = {{1 + x + {x^2} - \left( {1 - x + {x^2}} \right)} \over {\sqrt {1 + x + {x^2}} + \sqrt {1 - x + {x^2}} }} = {{2x} \over {\sqrt {1 + x + {x^2}} + \sqrt {1 - x + {x^2}} }}$$
$$ = {{2x} \over {\sqrt {1 + x + {x^2}} + \sqrt {1 - x + {x^2}} }}{{{1 \over x}} \over {{1 \over x}}} = {2 \over {\sqrt {{1 \over {{x^2}}} + {1 \over x} + 1} + \sqrt {{1 \over {{x^2}}} - {1 \over x} + 1} }} = {2 \over {\sqrt {0 + 0 + 1} + \sqrt {0 - 0 + 1} }} = 1$$
|
The problem is when you move $1/x$ inside the square roots! $1/x$ is negative but pulling it into the square roots like that makes it positive.
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
How does $-\frac{1}{x-2} + \frac{1}{x-3}$ become $\frac{1}{2-x} - \frac{1}{3-x}$ I'm following a solution that is using a partial fraction decomposition, and I get stuck at the point where $-\frac{1}{x-2} + \frac{1}{x-3}$ becomes $\frac{1}{2-x} - \frac{1}{3-x}$
The equations are obviously equal, but some algebraic manipulation is done between the first step and the second step, and I can't figure out what this manipulation could be.
The full breakdown comes from this solution
$$
\small\begin{align}
\frac1{x^2-5x+6}
&=\frac1{(x-2)(x-3)}
=\frac1{-3-(-2)}\left(\frac1{x-2}-\frac1{x-3}\right)
=\bbox[4px,border:4px solid #F00000]{-\frac1{x-2}+\frac1{x-3}}\\
&=\bbox[4px,border:4px solid #F00000]{\frac1{2-x}-\frac1{3-x}}
=\sum_{n=0}^\infty\frac1{2^{n+1}}x^n-\sum_{n=0}^\infty\frac1{3^{n+1}}x^n
=\bbox[4px,border:1px solid #000000]{\sum_{n=0}^\infty\left(\frac1{2^{n+1}}-\frac1{3^{n+1}}\right)x^n}
\end{align}
$$
Original image
|
I am a grade 8 student, so I may not be able to explain really well.
First, I need to prove that $-\frac {1} {x-2}=\frac {1} {2-x}$
To prove, let's assume that "$x$" can be any number, for instance, I take $x$=8.
So by substituting,
\begin{align}
-\frac {1} {x-2} & = -\frac {1} {8-2}\\
& = -\frac {1} {6}
\end{align}
And same for this,
\begin{align}
\frac {1} {2-8} & =\frac {1} {-6}\\
& = -\frac {1} {6}
\end{align}
Therefore, we have proven that $-\frac {1} {x-2}=\frac {1} {2-x}$
I also need to prove that $\frac {1} {x-3}=-\frac {1} {3-x}$
So by substituting,
\begin{align}
\frac {1} {8-3} & =\frac {1} {5}\\
\end{align}
and the same for this,
\begin{align}
-\frac {1} {3-8} & =-\frac {1} {-5}\\
& = \frac {-1} {-5}\\
& = \frac {1} {5}\\
\end{align}
Therefore, we have proven that $\frac {1} {x-3}=-\frac {1} {3-x}$
By why it worked? The truth is, it is just having -1÷(-1)=1 (negative$\times$negative=positive)(And anything times 1 is the same number)
So, from $-\frac {1} {x-2}$ to $\frac {1} {2-x}$, they inserted both -1 for numerator and denominator as the following below.
\begin{align}
-\frac {1} {x-2} & = \frac {-1} {x-2}\\
& = \frac {-1(-1)} {-1(x-2)}\\
& = \frac {1} {-x+2}\\
& = \frac {1} {2-x}\\
\end{align}
same goes to $\frac {1} {x-3}=-\frac {1} {3-x}$
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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|
How to compute $\sum_{n=0}^{\infty}{\frac{3^n(n + \frac{1}{2})}{n!}}$? I have to compute the series $\displaystyle\sum_{n=0}^{\infty}{\frac{3^n(n + \frac{1}{2})}{n!}}$.
$$\displaystyle\sum_{n=0}^{\infty}{\frac{3^n(n + \frac{1}{2})}{n!}} = \sum_{n=0}^{\infty}{\frac{3^n\frac{1}{2}}{n!}} + \sum_{n=0}^{\infty}{\frac{3^nn}{n!}} = \frac{e^3}{2} + \sum_{n=0}^{\infty}{\frac{3^nn}{n!}},$$ but how to compute the $\displaystyle\sum_{n=0}^{\infty}{\frac{3^nn}{n!}}$?
|
$$\begin{align}\sum\limits_{n=0}^\infty \frac{n\cdot 3^n}{n!}&=0+\sum\limits_{n=1}^\infty \frac{3^n}{(n-1)!}\\&=3\sum\limits_{n=0}^\infty \frac{3^n}{n!}\\&=3\cdot e^3\end{align}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1275769",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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|
$5(a^2+b^2)$ covers all numbers $=a_2^2+a_2^2=b_1^2+b_2^2$? I start by noting that 4a*2b=2a*4b
I write 4a*2b as $((2a+b)+(2a-b))*((2a+b)-(2a-b)) = (2a+b)^2-(2a-b)^2$
I follow a similar principle for 2a*4b which I write as $(a+2b)^2-(a-2b)^2$
to arrive at
$(2a+b)^2-(2a-b)^2= (a+2b)^2-(a-2b)^2$ or
$(2a+b)^2+(a-2b)^2= (a+2b)^2+(2a-b)^2=5(a^2+b^2)$
The expression $5(a^2+b^2)$ thus gives numbers which can be represented as a sum of two squares in two different ways, and I observe that if a=b, a=2b (or b=2a) we have simpler relations such as (a,b)=(1,1) which gives us $5*2=3^ 2+1 $
I observe that we cover $50=7^2+1^2=5^5+5^5$, $65=8^2+1^1=7^2+4^2$ etc.
I now wonder does this simple formula cover all such numbers?
This doesn’t necessarily follow from my simple calculation.
|
$$ 13 (a^2 + b^2) = (3a+2b)^2 + (2a-3b)^2 = (3a-2b)^2 + (2a+3b)^2$$
$$ 17 (a^2 + b^2) = (4a+b)^2 + (a-4b)^2 = (4a-b)^2 + (a+4b)^2$$
$$ 29 (a^2 + b^2) = (5a+2b)^2 + (2a-5b)^2 = (5a-2b)^2 + (2a+5b)^2$$
$$ 37 (a^2 + b^2) = (6a+b)^2 + (a-6b)^2 = (6a-b)^2 + (a+6b)^2$$
$$ 41 (a^2 + b^2) = (5a+4b)^2 + (4a-5b)^2 = (5a-4b)^2 + (4a+5b)^2$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Prove series convergence The problem states:
a) If $\sum_1^\infty {a_n}$ converges and ${b_n}=n^\frac{1}{n}{a_n}$, then $ \sum_1^\infty{b_n}$ converges, and
b) If $\sum_1^\infty {a_n}$ converges and ${b_n}=\frac{{a_n}}{(1+|{a_n}|)}$, then $\sum_1^\infty {b_n}$ converges.
For a) We have that ${c_n}=n^\frac{1}{n}$ is a stictly decreasing sequence (for $n>e$) and that it's bounded from above by $e^\frac{1}{e}$, so $\sum_1^\infty {a_n}{c_n}=\sum_1^\infty {b_n}$ converges( Abel's test).
For b) We have that ${c_n}=\frac{1}{(1+|{a_n}|)}$ is a monotone decreasing sequence, and it's bounded from above by 1, so $\sum_1^\infty {a_n}{c_n}=\sum_1^\infty {b_n}$ converges.
That's it, I would like to know if my proof is right, any help will be appreciated.
|
In (a) it's correct with one detail: since $c_n$ is decreasing, you should check whether it is bounded from below.
Problem (b) is a bit funny since it isn't true if we don't assume anything else, e.g. that $\displaystyle \sum_{n=1}^{\infty} a_n$ converges absolutely. Here's a counterexample: $a_n$ is the sequence of all terms in the following sum:
$$\sum_{k=1}^{\infty} \sum_{n=1}^{k^2} \left[ -\frac{1}{k} + \frac{1}{2k} + \frac{1}{2k} \right] = -1 + \frac{1}{2} + \frac{1}{2} \underbrace{- \frac{1}{2} + \frac{1}{4} + \frac{1}{4}}_{\text{4 times}} \underbrace{ - \frac{1}{3} + \frac{1}{6} + \frac{1}{6} }_{\text{9 times}} - \ldots $$
It clearly converges to $0$. The corresponding sum of $\displaystyle \sum_{n=1}^{\infty} b_n$ is
$$\sum_{k=1}^{\infty} \sum_{n=1}^{k^2} \left[ \frac{ -\frac{1}{k} }{1+\frac{1}{k}} + \frac{ \frac{1}{2k} }{1+\frac{1}{2k}} + \frac{\frac{1}{2k}}{1+\frac{1}{2k}} \right] = \sum_{k=1}^{\infty} \sum_{n=1}^{k^2} \frac{1}{k} \left[ \frac{1}{1+\frac{1}{2k}} - \frac{1}{1+\frac{1}{k}}\right] \\ = \sum_{k=1}^{\infty} \sum_{n=1}^{k^2} \frac{1}{2k^2} \cdot \frac{1}{\left( 1+\frac{1}{k} \right) \left( 1+\frac{1}{2k} \right)} = \sum_{k=1}^{\infty} \frac{1}{2} \cdot \frac{1}{\left( 1+\frac{1}{k} \right) \left( 1+\frac{1}{2k} \right)}$$
which clearly diverges.
|
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An interesting point of a triangle. (Help needed to prove a statement.) Consider a triangle whose sides are segments of $\color{red}{\text{line}}$, $\color{blue}{\text{line}}$, $\color{green}{\text{line}}$ falling in the circum-circle $c$. Let $\color{red}{\text{P}}$,$\color{green}{\text{P}}$, $\color{blue}{\text{P}}$ be the poles (with respect to $c$) of the corresponding sides of the triangle.
Now, take a point $P$ different from the poles. Connect the poles with $P$. The connecting lines will intersect the corresponding edges or the elongations of theses edges mentioned above (perhaps in the $\infty$). (Corresponding means: $\color{red}{\text{ red broken line}}$ with $\color{red}{\text{ red edge line }}$, etc.
Then connect the vertices of the triangle with the the opposite intersection points mentioned above as shown in the figure below (white lines). The white lines will meet in one point. (Perhaps in the infinity; then the white lines are parallel.)
I call this point the $P$-pole point of the triangle with respect to its circum-circle and point $P$. I cannot prove that the pole point always exists. (It exists even if the white lines are parallel.) Any help, please? Any known results?
The same statement can be told easier in the language of hyperbolic geometry:
Take an ideal triangle and a point $P$ not on the sides. Drop perpendiculars from $P$ to the sides of the triangle. Consider the intersection points. Then connect these intersection points with the opposite vertices with suitable parallels. These parallels will meet in one point, the "pole point of the ideal triangle-with respect to $P$. (See the figure below.)
To be honest I don't have a clue as to how to prove the statements given above. I found the "pole point" in the clear blue.
|
Any two ideal triangles are congruent, or in other words, any non-degenerate triangle inscribed into the unit circle can be mapped to any other by a projective transformation which fixes the unit circle. So without loss of generality you can restrict your considerations to one special case.
I like coordinates, so I'd start with $\triangle A_1A_2A_3$, the matrix $M$ of its circumcircle, and point $P$ chosen as follows (in homogeneous coordinates):
$$P=\begin{pmatrix}x\\y\\z\end{pmatrix} \qquad
M=\begin{pmatrix}1&&\\&1&\\&&-1\end{pmatrix}$$
\begin{align*}
A_1&=\begin{pmatrix}1\\0\\1\end{pmatrix} &
A_2&=\begin{pmatrix}0\\1\\1\end{pmatrix} &
A_3&=\begin{pmatrix}0\\-1\\1\end{pmatrix} \\
a_1=A_2\vee A_3&=\begin{pmatrix}1\\0\\0\end{pmatrix} &
a_2=A_3\vee A_1&=\begin{pmatrix}1\\-1\\-1\end{pmatrix} &
a_3=A_1\vee A_2&=\begin{pmatrix}1\\1\\-1\end{pmatrix} \\
B_1=Ma_1&=\begin{pmatrix}1\\0\\0\end{pmatrix} &
B_2=Ma_2&=\begin{pmatrix}1\\-1\\1\end{pmatrix} &
B_3=Ma_3&=\begin{pmatrix}1\\1\\1\end{pmatrix} \\
b_1=P\vee B_1&=\begin{pmatrix}0\\z\\-y\end{pmatrix} &
b_2=P\vee B_2&=\begin{pmatrix}y+z\\z-x\\-x-y\end{pmatrix} &
b_3=P\vee B_3&=\begin{pmatrix}y-z\\z-x\\x-y\end{pmatrix} \\
C_1=a_1\wedge b_1&=\begin{pmatrix}0\\y\\z\end{pmatrix} &
C_2=a_2\wedge b_2&=\begin{pmatrix}z+y\\x-z\\2z-x+y\end{pmatrix} &
C_3=a_3\wedge b_3&=\begin{pmatrix}z-y\\z-x\\2z-x-y\end{pmatrix} \\
c_1=A_1\vee C_1&=\begin{pmatrix}y\\z\\-y\end{pmatrix} &
c_2=A_2\vee C_2&=\begin{pmatrix}3z-2x+y\\y+z\\-y-z\end{pmatrix} &
c_3=A_3\vee C_3&=\begin{pmatrix}3z-2x-y\\y-z\\y-z\end{pmatrix}
\end{align*}
$A_i$ are the corners of the triangle. $a_i$ its edges, your solid colored lines. $B_i$ their poles, your $\color{red}P,\color{green}P,\color{blue}P$. $b_i$ the connection of these to $P$, i.e. the perpendiculars, which you drew using dashed lines. $C_i$ are the points where these intersect the edges. Finally, $c_i$ are the lines connecting these, which you drew in white.
These three lines $c_i$ are concurrent if their determinant is zero.
$$\begin{vmatrix}
y&3z-2x+y&3z-2x-y\\
z&y+z&y-z\\
-y&-y-z&y-z
\end{vmatrix}=0$$
The point of concurrency is
$$Q=c_1\wedge c_2=
\begin{pmatrix}y^2- z^2 \\ 2xy - 2yz \\ y^2 + 2xz - 3z^2\end{pmatrix}$$
There are some degenerate situations, where that point becomes undefined, i.e. the null vector. Computing all solutions to this set of three quadratic homogeneous equations, you find the degenerate situations to be
$$P\in\left\{
\begin{pmatrix}1\\0\\0\end{pmatrix},
\begin{pmatrix}1\\-1\\1\end{pmatrix},
\begin{pmatrix}1\\1\\1\end{pmatrix}
\right\}=\{B_1,B_2,B_3\}$$
Those are the cases you explicitely excluded, so in all other situations, everything is fine.
|
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|
Minimum value of trigonometric equation Find the minimum value of the expression $$y=\frac{16-8\sin^{2} 2x +8\cos^{4} x}{\sin^{2} 2x} .$$ When I convert the expression completely into $2x$, cross multiply and make the discriminant of the quadratic equation greater than $0$, I get the minimum value $-\infty$. I know it is wrong, but why?
|
$$8\cos^4x = 8\left(\dfrac{1+\cos (2x)}{2}\right)^2 = 8\left(\dfrac{1+2\cos (2x)+ \cos^2(2x)}{4}\right)=2+4\cos (2x)+2\cos^2(2x) = 2+4t+2t^2, t = \cos (2x) \Rightarrow y = \dfrac{16-8(1-t^2)+2+4t+2t^2}{1-t^2} = \dfrac{10+10t^2+4t}{1-t^2}=f(t), -1 \leq t \leq 1$$. Can you take it from here?
|
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|
Computing a limit similar to the exponential function I want to show the following limit:
$$
\lim_{n \to \infty}
n
\left[
\left( 1 - \frac{1}{n} \right)^{2n}
- \left( 1 - \frac{2}{n} \right)^{n}
\right]
= \frac{1}{e^{2}}.
$$
I got the answer using WolframAlpha, and it seems to be correct numerically, but I am having trouble proving the result. My first instinct was to write the limit as
$$
\lim_{n \to \infty}
\frac
{
\left( 1 - \frac{1}{n} \right)^{2n}
- \left( 1 - \frac{2}{n} \right)^{n}
}
{1/n}.
$$
Then, I tried applying l'Hopital's rule, and I got
$$
\lim_{n \to \infty}
\frac
{
\left( 1 - \frac{1}{n} \right)^{2n}
\left( 2 \log\left( 1 - \frac{1}{n} \right) + \frac{2}{n-1} \right)
-
\left( 1 - \frac{2}{n} \right)^{n}
\left( \log\left( 1 - \frac{2}{n} \right) + \frac{2}{n-2} \right)
}
{-1/n^{2}}.
$$
This does not seem to have gotten me anywhere. My second attempt was to use the binomial theorem:
$$
\begin{align*}
n
\left[
\left( 1 - \frac{1}{n} \right)^{2n}
- \left( 1 - \frac{2}{n} \right)^{n}
\right]
&
=
n
\left[
\sum_{k=0}^{2n} \binom{2n}{k} \frac{(-1)^{k}}{n^{k}}
- \sum_{k=0}^{n} \binom{n}{k} \frac{(-1)^{k} 2^{k}}{n^{k}}
\right]
\\ &
=
\sum_{k=2}^{n} \left[ \binom{2n}{k} - \binom{n}{k} 2^{k} \right] \frac{(-1)^{k}}{n^{k-1}}
+ \sum_{k=n+1}^{2n} \binom{2n}{k} \frac{(-1)^{k}}{n^{k-1}}.
\end{align*}
$$
At this point I got stuck again.
|
We can proceed in the following manner
\begin{align}
L &= \lim_{n \to \infty}n\left\{\left(1 - \frac{1}{n}\right)^{2n} - \left(1 - \frac{2}{n}\right)^{n}\right\}\notag\\
&= \lim_{n \to \infty}n\left(1 - \frac{2}{n}\right)^{n}\left\{\dfrac{\left(1 - \dfrac{1}{n}\right)^{2n}}{\left(1 - \dfrac{2}{n}\right)^{n}} - 1\right\}\notag\\
&= \frac{1}{e^{2}}\lim_{n \to \infty}n\left\{\dfrac{\left(1 - \dfrac{2}{n} + \dfrac{1}{n^{2}}\right)^{n}}{\left(1 - \dfrac{2}{n}\right)^{n}} - 1\right\}\notag\\
&= \frac{1}{e^{2}}\lim_{n \to \infty}n\left\{\left(1 + \frac{1}{n^{2} - 2n}\right)^{n} - 1\right\}\notag\\
&= \frac{1}{e^{2}}\lim_{n \to \infty}f(n)\tag{1}
\end{align}
Now note that
\begin{align}
n\left(1 + \frac{n}{n^{2} - 2n} - 1\right) < f(n) &= n\left[\left\{\left(1 + \frac{1}{n^{2} - 2n}\right)^{n^{2} - 2n}\right\}^{n/(n^{2} - 2n)} - 1\right]\notag\\
&< n\left\{\exp\left(\frac{n}{n^{2} - 2n}\right) - 1\right\}\notag\\
&= n\cdot\frac{n}{n^{2} - 2n}\cdot\frac{n^{2} - 2n}{n}\left\{\exp\left(\frac{n}{n^{2} - 2n}\right) - 1\right\}\notag\\
\end{align}
and we get $$\frac{n^{2}}{n^{2} - 2n} < f(n) < \frac{n^{2}}{n^{2} - 2n}\cdot\frac{n^{2} - 2n}{n}\left\{\exp\left(\frac{n}{n^{2} - 2n}\right) - 1\right\}$$ Taking limits as $n \to \infty$ and using squeeze theorem we see that $f(n) \to 1$ as $n \to \infty$. From equation $(1)$ we can see that $L = 1/e^{2}$.
In order to derive the inequalities we have used $(1 + x)^{n} > 1 + nx$ for $x > 0$ and $n$ a positive integer and $(1 + (1/n))^{n} < e$ for all positive integers $n$. Both these inequalities are pretty standard and can be easily proved. Also note that if we put $t = n/(n^{2} - 2n)$ then $t \to 0$ as $n \to \infty$ and hence $(e^{t} - 1)/t \to 1$ as $n \to \infty$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Diagonal-free Sudoku grid I have a Sudoku grid with the property that diagonally adjacent elements are distinct (it is also a torus under the same property).
My question is up to isomorphism, is the grid unique?
Here's the grid:
$$\begin{array}{|c|c|c|c|c|c|c|c|c|}
\hline
6& 5& 7& 3& 4& 2& 1& 9& 8\\ \hline
9& 8& 1& 5& 6& 7& 4& 3& 2\\ \hline
3& 2& 4& 8& 9& 1& 6& 5& 7\\ \hline
5& 7& 6& 2& 3& 4& 9& 8& 1\\ \hline
8& 1& 9& 7& 5& 6& 3& 2& 4\\ \hline
2& 4& 3& 1& 8& 9& 5& 7& 6\\ \hline
7& 6& 5& 4& 2& 3& 8& 1& 9\\ \hline
1& 9& 8& 6& 7& 5& 2& 4& 3\\ \hline
4& 3& 2& 9& 1& 8& 7& 6& 5\\ \hline
\end{array}$$
To explain the diagonal-free property, if we have:
\begin{array}{c|c}
a&b\\ \hline
c&d\\
\end{array}
then $a\ne d$ and $b\ne c$. This does NOT imply the the whole diagonal is distinct (as in X-factor).
If we wrap the grid into a cylinder, and then bend the tube into a torus, the diagonal property still holds, so, for example, the $9$ on the base row is considered to be diagonally adjacent to the $4$ and $7$ on the top row.
There are two links on OEIS:
*
*De-diagonalized grid
*Diagonalized grid
Isomorphism in this instance implies we can change the orientation, change the permutations of the numbers, perform row/column swaps as long as the conditions still hold. This can create grids with no internal diagonals but that are not torii, for example (this is my original grid by the way!):
$$\begin{array}{|c|c|c|c|c|c|c|c|c|}
\hline
1&2&3&7&8&9&4&5&6\\ \hline
7&8&9&4&5&6&1&2&3\\ \hline
4&5&6&1&2&3&7&8&9\\ \hline
6&1&2&3&7&8&9&4&5\\ \hline
9&4&5&6&1&2&3&7&8\\ \hline
3&7&8&9&4&5&6&1&2\\ \hline
5&6&1&2&3&7&8&9&4\\ \hline
8&9&4&5&6&1&2&3&7\\ \hline
2&3&7&8&9&4&5&6&1\\ \hline
\end{array}$$
We can see that the $2$ on the base line is diagonally related to the $2$ on the top line, and so this grid is not a torus. The torus example is derived from this one using only the operations defined above.
So I am asking if there is a diagonal-free grid with a quintessentially different structure to either of the two grids given here.
ADDENDUM
For reference here is a Sudoku grid with diagonals, from Conceptis Puzzles:
The diagonally adjacent givens are highlighted - there may be more!
|
In the general case, a grid will be in an isomorphism equivalence class of $9!\cdot 3!^4 \cdot 2$ corresponding to permutations of the symbols, permutations of the rows and columns, and the symmetry of the square which is not already covered by permutation of the rows and columns. I'm only counting $3!^2$ permutations of the rows rather than $9!$ because most of those $9!$ will break one or more of the $3\times 3$ squares with a Sudoku constraint.
I've set a computer program to generate solutions which match the constraints (and the additional constraint that the first row is $123456789$); so far it has found more than $20000$. The first two are
$$\begin{array}{|c|c|c|c|c|c|c|c|c|}
\hline
1& 2& 3& 4& 5& 6& 7& 8& 9\\ \hline
4& 7& 8& 1& 3& 9& 2& 5& 6\\ \hline
\textbf{5}& \textbf{9}& 6& 7& 2& 8& 4& 1& 3\\ \hline
6& 3& 4& 8& 1& 5& 9& 7& 2\\ \hline
\textbf{9}& \textbf{5}& 2& 3& 4& 7& 1& 6& 8\\ \hline
7& 8& 1& 9& 6& 2& 5& 3& 4\\ \hline
2& 4& 7& 5& 8& 3& 6& 9& 1\\ \hline
3& 1& 9& 6& 7& 4& 8& 2& 5\\ \hline
8& 6& 5& 2& 9& 1& 3& 4& 7\\ \hline
\end{array}$$
$$\begin{array}{|c|c|c|c|c|c|c|c|c|}
\hline
1& 2& 3& 4& 5& 6& 7& 8& 9\\ \hline
4& 7& 8& 1& 3& 9& 2& 5& 6\\ \hline
\textbf{9}& \textbf{5}& 6& 7& 2& 8& 4& 1& 3\\ \hline
6& 3& 4& 8& 1& 5& 9& 7& 2\\ \hline
\textbf{5}& \textbf{9}& 2& 3& 4& 7& 1& 6& 8\\ \hline
7& 8& 1& 9& 6& 2& 5& 3& 4\\ \hline
2& 4& 7& 5& 8& 3& 6& 9& 1\\ \hline
3& 1& 9& 6& 7& 4& 8& 2& 5\\ \hline
8& 6& 5& 2& 9& 1& 3& 4& 7\\ \hline
\end{array}
$$
which, while clearly closely related, I don't believe to be actually isomorphic. And they're certainly not isomorphic to the grid in your question, which has no rectangle whose four corners match diagonally in this way.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Cannot understand an Integral $$\displaystyle \int _{ \pi /6 }^{ \pi /3 }{ \frac { dx }{ \sec x+\csc x } } $$
I had to solve the integral and get it in this form.
My attempt:
$$\int _{ \pi /6 }^{ \pi /3 }{ \frac { dx }{ \sec x+\csc x } } $$
$$=\int _{\frac{\pi}{6}}^{ \frac{\pi}{3}} \dfrac{\sin x \cos x }{ \sin x+\cos x }dx $$
Substituting $t=\tan(\frac{x}{2})$,
$$\int_{\tan(\frac{\pi}{12})}^{\tan(\frac{\pi}{6})} \dfrac{2t}{1+t^2}\times\dfrac{1-t^2}{1+t^2}\times\dfrac{2}{1+t^2}dt$$
$$2\int_{2-\sqrt{3}}^{\frac{1}{\sqrt{3}}} \dfrac{2t(1-t^2)}{(1+t^2)^3}dt$$
Substituting $u=1+t^2$, $2t dt=du$, $1-t^2 = 2-u$
$$2\int_{8-4\sqrt{3}}^\frac{4}{3} \dfrac{(2-u)}{u^3}du$$
$$\displaystyle 4\int_{8-4\sqrt{3}}^\frac{4}{3} \dfrac{du}{u^3} \displaystyle -2\int_{8-4\sqrt{3}}^\frac{4}{3} \dfrac{du}{u^2}$$ Could somebody please tell me where I have gone wrong? Also could someone please tell me how to change the limits of the definite integral throughout?
|
In general if you have an integral
$$\int_a^b f(x)dx$$
Upon making the substitution $t=g(x)$ the limits of the integral become $g(a)$ and $g(b)$. In your case you made the substitution $t=\tan \frac x2$ so the lower limit becomes $\tan \frac{\pi}{12}$ and the upper limit becomes $\tan \frac{\pi}{6}$. Thus the integral becomes
$$\int_{\tan\frac{\pi}{12}}^{\tan\frac{\pi}{6}} \dfrac{2t}{1+t^2}\times\dfrac{1-t^2}{1+t^2}\times\dfrac{2}{1+t^2}dt$$
And continue as you have otherwise.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Proving $x\ln(\frac{x}{a})+y\ln(\frac{y}{b})\geq (x+y)\ln(\frac{x+y}{a+b})$
Let a,b,x,y be positive reals. Prove $x\ln(\frac{x}{a})+y\ln(\frac{y}{b})\geq (x+y)\ln(\frac{x+y}{a+b})$
I don't have any olympic background, so I may be missing some standard trick.
The inequality looks closely related to the concavity of $\ln$.
EDIT: the following is wrong as stated by Macavity.
Indeed using points $\frac{x}{a}$,$\frac{y}{b}$, and $\frac{x}{x+y}$,$\frac{y}{x+y}$ as weights, one has $$\frac{x}{x+y}\ln(\frac{x}{a})+\frac{y}{x+y}\ln(\frac{y}{b})\geq \ln(\frac{x^2}{x+y}\frac{1}{a}+\frac{y^2}{x+y}\frac{1}{b})$$
But I can't prove that $\displaystyle\frac{x^2}{x+y}\frac{1}{a}+\frac{y^2}{x+y}\frac{1}{b}\geq \frac{x+y}{a+b}$
|
use Cauchy-Schwarz inequality we have
$$\left(\dfrac{x^2}{a}+\dfrac{y^2}{b}\right)(a+b)\ge (x+y)^2$$
|
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|
Demonstrate that the two formulae for a scalar product are equivalent. In the figure below, three vectors are joined together to form a triangle.
The name of each vector is a single letter in boldface,
each vector is specified by three lengths in an $xyz$ coordinate system,
and vectors $\boldsymbol{a}$ and $\boldsymbol{b}$ are separated by angle $\theta$.
Let $a$, $b$, and $c$ (with neither boldface nor subscripts) represent the
magnitudes (lengths) of vectors $\boldsymbol{a}$, $\boldsymbol{b}$, and $\boldsymbol{c}$.
Note that the Pythagorean theorem is used.
\begin{eqnarray*}
a &=& \sqrt{a_x^2 + a_y^2 + a_z^2} \\
b &=& \sqrt{b_x^2 + b_y^2 + b_z^2} \\
c &=& \sqrt{c_x^2 + c_y^2 + c_z^2}
\end{eqnarray*}
The expression $\boldsymbol{a} \cdot \boldsymbol{b}$
signifies the $scalar\ product$ (a.k.a. the $dot\ product$)
of vectors $\boldsymbol{a}$ and $\boldsymbol{b}$.
There are two common formulas for computing $\boldsymbol{a} \cdot \boldsymbol{b}$.
The first is the sum of the products of the $xyz$ components.
The second is the product of the magnitudes and the cosine of angle $\theta$.
\begin{eqnarray*}
\boldsymbol{a} \cdot \boldsymbol{b} &=& a_x b_x + a_y b_y + a_z b_z \\
\boldsymbol{a} \cdot \boldsymbol{b} &=& a b \cos \theta
\end{eqnarray*}
It's not obvious that these two formulas for the scalar product
are equivalent. Therefore, demonstrate that they are ---
demonstrate that
\begin{eqnarray*}
a_x b_x + a_y b_y + a_z b_z &=& a b \cos \theta
\end{eqnarray*}
|
Vector $\boldsymbol{a}$ is the sum of vectors $\boldsymbol{b}$ and $\boldsymbol{c}$.
I.e., $\boldsymbol{a} = \boldsymbol{b} + \boldsymbol{c}$. Solving for $\boldsymbol{c}$ we have
\begin{eqnarray*}
\boldsymbol{c} &=& \boldsymbol{a} - \boldsymbol{b} \\
&=& (a_x, a_y, a_z) - (b_x, b_y, b_z) \\
&=& \big( (a_x-b_x),(a_y-b_y),(a_z-b_z) \big) \\
&=& \big( c_x , c_y , c_z \big)
\end{eqnarray*}
We calculate magnitude $c$
using the $xyz$ components of vectors $\boldsymbol{a}$ and $\boldsymbol{b}$.
\begin{eqnarray*}
c &=& \sqrt{c_x^2 + c_y^2 + c_z^2} \\
&=& \sqrt{(a_x-b_x)^2 + (a_y-b_y)^2 + (a_z-b_z)^2}
\end{eqnarray*}
We start with the squared magnitude $c^2$, expand it, rearrange it,
then reduce it with the squared magnitudes $a^2$ and $b^2$.
\begin{eqnarray*}
c^2 &=& (a_x-b_x)^2 + (a_y-b_y)^2 + (a_z-b_z)^2 \\
&=& (a_x^2-2a_xb_x+b_x^2) + (a_y^2-2a_yb_y+b_y^2) + (a_z^2-2a_zb_z+b_z^2) \\
&=& (a_x^2+a_y^2+a_z^2) + (b_x^2+b_y^2+b_z^2) - 2(a_xb_x+a_yb_y+a_zb_z) \\
&=& a^2 + b^2 - 2(a_xb_x+a_yb_y+a_zb_z)
\end{eqnarray*}
The Law of Cosines asserts that $c^2 = a^2 + b^2 - 2ab\cos\theta$.
We replace $c^2$ with the result from above, subtract $a^2 + b^2$ from both sides,
divide both sides by $-2$, and we're done.
\begin{eqnarray*}
\mathrm{Law\ of\ Cosines:\phantom{X}} c^2 &=& a^2 + b^2 - 2ab\cos\theta \\
a^2 + b^2 - 2(a_xb_x+a_yb_y+a_zb_z) &=& a^2 + b^2 - 2ab\cos\theta \\
-2(a_xb_x+a_yb_y+a_zb_z) &=& -2ab\cos\theta \\
a_xb_x + a_yb_y + a_zb_z &=& ab\cos\theta
\end{eqnarray*}
QED
|
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|
Find the generating function of this sequence I need to find the generating function of the sequence $c_n = (a_0, a_1, a_2, \ldots)$, where:
$$a_n =
\begin{cases}
2^{n/2} & \text{if $n$ is even,} \\
1 & \text{if $n$ is odd.}
\end{cases}$$
I have written out the first few terms of the sequence:
$$(1, 1, 2, 1, 4, 1, 8, 1, 16, 1, \ldots)$$ and have noticed that it seems to be a combination of the sequences $a_n = (1, 0, 2, 0, 4, 0, 8, 0, 16, 0, ...)$ and $b_n = (0, 1, 0, 1, 0, 1, ...).$
The first sequence, $a_n$, has the generating function
$$x\sum_{n = 0}^\infty(2x)^n = \frac{x}{1 - 2x}$$
and $b_n$ has the generating function
$$x\sum_{n = 0}^\infty x^{2n} = \frac{x}{1 - x^2}.$$
Therefore, I intuitively thought that $c_n = a_n + b_n$ and that the generating function of $c_n$ was equal to the sum of the generating functions of $a_n$ and $b_n$, which is equal to:
$$\frac{x}{1 - 2x} + \frac{x}{1 - x^2} = \frac{-x^3 - 2x^2 + 2x}{(1-2x)(1-x^2)}. \space\space (*)$$
However, when I tried to convert $(*)$ back to a form which involves an infinite sum, it did not give me the sequence that I expected ($c_n$).
I would appreciate help with the solution of this problem.
|
Let us set $A(X)$ the generating function whose coefficients are $(a_n)$. I claim that :
$$A(X)-\frac{1}{1-X}=\sum_{n=1}^{\infty}(2^n-1)X^{2n}=\sum_{n=1}^{\infty}\sum_{k=0}^{n-1}2^kX^{2n} $$
$$\sum_{n=1}^{\infty}\sum_{k=0}^{n-1}2^kX^{2n}=X^2\sum_{n=1}^{\infty}\sum_{k=0}^{n-1}2^kX^{2(n-1)}=X^2\sum_{n=0}^{\infty}\sum_{k=0}^{n}2^kX^{2n} $$
Now the last generating function appears as a Cauchy product in the variable $Y:=X^2$ so :
$$X^2\sum_{n=0}^{\infty}\sum_{k=0}^{n}2^kX^{2n}=X^2\sum_{n=0}^{\infty}2^nX^{2n}\sum_{n=0}^{\infty}X^{2n}=X^2\frac{1}{1-X^2}\frac{1}{1-2X^2}$$
Now :
$$A(X)=\frac{1}{1-X}+\frac{X^2}{(1-X^2)(1-2X^2)}=\frac{(1+X)(1-2X^2)+X^2}{(1-X^2)(1-2X^2)} $$
$$A(X)=\frac{1+X-2X^2-2X^3+X^2}{(1-X^2)(1-2X^2)}=\frac{1+X-X^2-2X^3}{(1-X^2)(1-2X^2)} $$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Proving that $\frac{1}{a^3(b+c)}+\frac{1}{b^3(c+a)}+\frac{1}{c^3(a+b)}\ge\frac32$ using derivatives
Let $a,b,c\in\mathbb{R}^+$ and $abc=1$. Prove that
$$\frac{1}{a^3(b+c)}+\frac{1}{b^3(c+a)}+\frac{1}{c^3(a+b)}\ge\frac32$$
This isn't hard problem. I have already solved it in following way:
Let $x=\frac1a,y=\frac1b,z=\frac1c$, then $xyz=1$. Now, it is enought to prove that
$$L\equiv\frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y}\ge\frac32$$
Now using Cauchy-Schwarz inequality on numbers $a_1=\sqrt{y+z},a_2=\sqrt{z+x},a_3=\sqrt{x+y},b_1=\frac{x}{a_1},b_2=\frac{y}{a_2},b_3=\frac{z}{a_3}$ I got
$$(x+y+z)^2\le((x+y)+(y+z)+(z+x))\cdot L$$
From this
$$L\ge\frac{x+y+z}2\ge\frac32\sqrt[3]{xyz}=\frac32$$
Then I tried to prove it using derivatives. Let $x=a,y=b$ and
$$f(x,y)=\frac1{x^3\left({y+\frac1{xy}}\right)}+\frac1{y^3\left({x+\frac1{xy}}\right)}+\frac1{\left({\frac1{xy}}\right)^3(x+y)}$$
So, I need to find minimum value of this function. It will be true when
$$\frac{df}{dx}=0\land\frac{df}{dy}=0$$
After simplifying $\frac{df}{dx}=0$ I got
$$\frac{-y(3xy^2+2)}{x^3\left({xy^2+1}\right)^2}+\frac{1-x^2y}{y^2\left({x^2y+1}\right)^2}+\frac{x^2y^3(2x+3y)}{\left({x+y}\right)^2}=0$$
Is there any easy way to write $x$ in term of $y$ from this equation?
|
You can use this way to do. Your inequality $L\ge \frac32$ is equivalent to
$$ 2[x^2(x+y)(x+z)+y^2(x+y)(y+z)+z^2(x+z)(y+z)]\ge 3(x+y)(x+z)(y+z). $$
Let
$$ F(x,y,z)=2[x^2(x+y)(x+z)+y^2(x+y)(y+z)+z^2(x+z)(y+z)]-3(x+y)(x+z)(y+z)-\lambda(xyz-1). $$
Then set
$$ \frac{\partial F}{\partial x}=0, \frac{\partial F}{\partial y}=0,\frac{\partial F}{\partial z}=0, \frac{\partial F}{\partial \lambda}=0. $$
Easy calculation shows that, $\frac{\partial F}{\partial x}=\frac{\partial F}{\partial y}$ gives $(x-y)[3(x+y)+\lambda z]=0.$
So $x=y$. Similarly $x=y=z$. But $xyz=1$ and hence $x=y=z=1$. So $f(x,y,z)$ reaches its minimum $0$ when $x=y=z=1$ or $f(x,y,z)\ge0$. Thus $L\ge\frac32$.
|
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|
Verifying $\frac{\cos{2\theta}}{1 + \sin{2\theta}} = \frac{\cot{\theta} - 1}{\cot{\theta} + 1}$ My math teacher gave us an equality involving trigonometric functions and told us to "verify" them. I tried making the two sides equal something simple such as "1 = 1" but kept getting stuck. I would highly appreciate if someone could show me (step by step) how to verify or solve this problem.
$$\frac{\cos{2\theta}}{1 + \sin{2\theta}} = \frac{\cot{\theta} - 1}{\cot{\theta} + 1}$$
|
Use
$\cos(2x)=\cos^2(x) - \sin^2(x)$ and
$1+\sin(2x) = 1+2\sin(x)\cos(x) = \sin^2(x)+\cos^2(x)+2\sin(x)\cos(x)=(\sin(x)+\cos(x))^2$
So we have
$$\frac{\cos(2x)}{1+\sin(2x)}=\frac{\cos^2(x) - \sin^2(x)}{(\sin(x)+\cos(x))^2}=\frac{\cos(x) - \sin(x)}{\cos(x) + \sin(x)}=\frac{\cot(x)-1}{\cot(x)+1}$$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
About right identity which is not left identity in a ring Let $S$ be the subset of $M_2(\mathbb{R})$ consisting of all matrices of the form
$\begin{pmatrix}
a & a \\
b & b
\end{pmatrix}$
The matrix $\begin{pmatrix}
x & x \\
y & y
\end{pmatrix}$ is right identity in $S$ if and only if $x+y=1$. Fine, I can see that.
But I cannot see why "If $x+y=1$ , then $\begin{pmatrix}
x & x \\
y & y
\end{pmatrix}$ is not a left identity in $S$".
I have tried that, if $\begin{pmatrix}
x & x \\
y & y
\end{pmatrix}$ is a left inverse then : $\begin{pmatrix}
x & x \\
y & y
\end{pmatrix}\begin{pmatrix}
a & a \\
b & b
\end{pmatrix}=\begin{pmatrix}
x(a+b) & x(a+b) \\
y(a+b) & y(a+b)
\end{pmatrix}=\begin{pmatrix}
a & a \\
b & b
\end{pmatrix}$ in which case we have $x(a+b)=a$ and $y(a+b)=b$. What can i do with $x+y=1$?
|
Your result:
$\begin{pmatrix}
x & x \\
y & y
\end{pmatrix}\begin{pmatrix}
a & a \\
b & b
\end{pmatrix}=\begin{pmatrix}
x(a+b) & x(a+b) \\
y(a+b) & y(a+b)
\end{pmatrix}$
shows that in $S$ does not exists a left identity.
|
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|
what is the decomposition of $(x^2+y^2+z^2)(x+y+z)(x+y-z)(-x+y+z)(x-y+z)-8x^2y^2z^2$? I want to have a decomposition of this :
$$(x^2+y^2+z^2)(x+y+z)(x+y-z)(-x+y+z)(x-y+z)-8x^2y^2z^2$$
I have tried all possible calculation which came to my mind,I will describe one of it which is better but no result:
I put $(x+y-z)=a$ ,$ (-x+y+z)=b$ and $(x-y+z)=c$ and we have $(x+y+z)=a+b+c$ so we have $$(x^2+y^2+z^2)(a+b+c)abc-8x^2y^2z^2$$
but I couldn't make it simpler, it will be great if you help me about it,thanks.
|
Maple does the job by
factor(expand((x^2+y^2+z^2)*(x+y+z)*(x+y-z)*(-x+y+z)*(x-y+z)-8*x^2*y^2*z^2));
$$ - \left( {x}^{2}-{y}^{2}-{z}^{2} \right) \left( {x}^{2}+{y}^{2}-{z}^{
2} \right) \left( {x}^{2}-{y}^{2}+{z}^{2} \right)
.$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find the length of chord $BC$.
On a semicircle with diameter $AD$. Chord $BC$ is parallel to the diameter.Further each of the chords $AB$ and $CD$ has length of $2$ cm while $AD$ has the length $8$ cm.Find the length of $BC$.
$a.)7.5\quad cm\\
\color{green}{b.)7\quad cm}\\
c.)7.75\quad cm\\
d.)\text{cannot be determined}$
I constructed $BO$ and with the help of cosine rule i found $\angle AOB$.then i found $\angle BOC$
and then again applying cosine rule in $\triangle BOC$ i found $BC$ ,
but i m looking for a more simple short way.
I have studied maths up to $12th$ grade.
|
Join the points B & D to obtain right $\Delta ABD$. Thus, we get $$BD=\sqrt{(AD)^2-(AB)^2}=\sqrt{8^2-2^2}=2\sqrt{15}$$ Now, draw a perpendicular say BM from the point B to the hypotenuse AD in right $\Delta ABD$. Then the normal distance between the parallel chords BC & AD is equal to BM i.e. length of perpendicular drawn from the vertex B to the hypotenuse AD of right $\Delta ABD$
$$BM=\frac{(AB)\times(BD)}{\sqrt{(AB)^2+(BD)^2}}=\frac{(2)\times(2\sqrt{15})}{\sqrt{(2)^2+(2\sqrt{15})^2}}=\frac{4\sqrt{15}}{8}=\frac{\sqrt{15}}{2}$$
Let the length of BC be $x$. Now, consider a right $\Delta AMB$ with hypotenuse $AB=2 cm$ & legs $BM=\frac{\sqrt{15}}{2}$ & $$AM=\frac{AD-BC}{2}=\frac{8-x}{2}$$ Applying pythagorus theorem in right $\Delta AMB$ as follows $$(AB)^2=(BM)^2+(AM)^2 \implies (2)^2=\left(\frac{\sqrt{15}}{2}\right)^2+\left(\frac{8-x}{2}\right)^2 \implies 16=15+(8-x)^2$$ $$\implies 8-x=\pm 1 \implies x=7 \space \text{or} \space x=9$$ But $BC<AD$ hence, we have $BC=7$
|
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|
Is this true that $(\cos^2A+\cos^2B+\cos^2C+2\cos A\cos B\cos C=1 \implies A+B+C=\pi)$? Assume that $A,B,C$ are positive real numbers and $A,B,C \in (0,\frac{\pi}{2}]$ and we have
$$\cos^2A+\cos^2B+\cos^2C+2\cos A\cos B\cos C = 1 $$
prove or disprove that $$A+B+C=\pi$$
|
With an obvious shorthand notation, we can solve the equation for $\cos C$:
$$c^2+2abc+a^2+b^2-1=0,$$
$$c=-ab\pm\sqrt{a^2b^2-a^2-b^2+1}=-ab\pm\sqrt{(1-a^2)(1-b^2)}.$$
Then
$$\cos C=-\cos A\cos B\pm \sin A\sin B=-\cos(A\pm B)$$
and
$$\pm C=\pi\pm A\pm B.$$
|
{
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|
Solving the Diophantine equation $x^n-y^n=1001$ For all $n \in \mathbb{N}$, solve the Diophantine equation $x^n-y^n=1001$, where $x,y \in \mathbb{N}$.
The cases $n=1,2$ are trivial ones. But for $n>2$ I can't find any solutions. How could I prove that there are no integer solutions for $n>2$?
|
Since $x,y \in \mathbb{N}$ and $x^4$ - $y^4$= $(y + h)^4$ - $y^4 = 4y^3h + 6y^2h^2 + 4yh^3 + h^4$
we have x - y = 1 for all n≥4 because $7^4$ = 2041 > 1001 then 7 and greater than 7 candidates (11, 13, 77,…) must be discarded. Therefore ${n\choose1}y^{n-1} +{n\choose2}y^{n-2} + … + {n\choose1}y + 1$ = 1001; hence
y[${n\choose1}y^{n-2} +{n\choose2}y^{n-3} + … + {n\choose1}$] = 1000 = $2^35^3$ and y = 1, 2, 4, 8, 5, 10, 25,….We discard these candidates to factors by the following calculation which shows that 1000 is strictly between the given powers:
$2^9 – 1 < 1000 < 2^{10} – 1$
$3^6 – 2^6 < 1000 < 3^7 – 2^7$
$5^4 – 4^4 < 1000 < 5^5 – 4^5$
$6^4 – 5^4 < 1000 < 6^5 – 5^5$
$9^3 – 8^3 < 1000 < 9^4 – 8^4$
$11^3 – 10^3 < 1000 < 11^4 – 10^4$
$26^2 – 25^2 < 1000 < 26^3 – 25^3$.
This finish the proof for all n >3.
The case n = 3 can be solve by a quadratic equation and careful calculation about impossibility of discriminant as a (necessary) square.
|
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|
how to parameterize the ellipse $x^2 + xy + 3y^2 = 1$ with $\sin \theta$ and $\cos \theta$ I am trying draw the ellipse $x^2 + xy + 3y^2 = 1$ so I can draw it. Starting from the matrix:
$$ \left[ \begin{array}{cc} 1 & \frac{1}{2} \\ \frac{1}{2} & 3 \end{array}\right]$$
I computed the eigenvalues $2 \pm \frac{1}{2}\sqrt{5}$ and the eigenvectors (not normalized):
$$\left[ \begin{array}{c} x\\ y \end{array}\right]
= \left[ \begin{array}{c} 1\\ 2\pm \sqrt{5} \end{array}\right] $$
So then I tried writing down some combination of the data I generated:
$$ \left[ \begin{array}{c} x(\theta)\\ y(\theta) \end{array}\right]
= \cos \theta \left[ \begin{array}{c} 1\\ 2+ \sqrt{5} \end{array}\right] + \sin \theta \left[ \begin{array}{c} 1\\ 2- \sqrt{5} \end{array}\right]
$$
However, I have a hard time checking the ellipse equation holds true for all $\theta$:
$$x(\theta)^2 + x(\theta)y(\theta) + 3y(\theta)^2 = 1$$
What are the correct functions $x(\theta), y(\theta)$ ?
Following the comments, rescaling the eigenvectors and multiplying the eigenvalues:
$$ \left[ \begin{array}{c} x(\theta)\\ y(\theta) \end{array}\right]
= \frac{2 + \frac{1}{2}\sqrt{5}}{\sqrt{10 + 4 \sqrt{5}}} \left[ \begin{array}{c} 1\\ 2+ \sqrt{5} \end{array}\right]\cos \theta +
\frac{2 - \frac{1}{2}\sqrt{5}}{\sqrt{10 - 4 \sqrt{5}}}\left[ \begin{array}{c} 1\\ 2- \sqrt{5} \end{array}\right]\sin \theta
$$
Is it clear that the ellipse equation is satisfied? I am not sure how to check this.
|
let us define $\theta$ by $$\cos\theta = \frac1{\sqrt{10+4\sqrt 5}}, \sin \theta=\frac{2+\sqrt 5}{\sqrt{10+4\sqrt 5}}.$$ then you can verify that
$$\pmatrix{1&1/2\\1/2&3}\pmatrix{\cos \theta&-\sin \theta\\\sin \theta&\cos \theta} = \pmatrix{\cos \theta&-\sin \theta\\\sin \theta&\cos \theta}\pmatrix{2+\sqrt5/2&0\\0&2-\sqrt5/2}$$ that is
$$\pmatrix{1&1/2\\1/2&3}=
\pmatrix{\cos \theta&-\sin \theta\\\sin \theta&\cos \theta}
\pmatrix{2+\sqrt5/2&0\\0&2-\sqrt5/2}
\pmatrix{\cos \theta&\sin \theta\\-\sin \theta&\cos \theta}$$ that is $$A = U^\top DU, U^\top U = I \text{ where } U = \pmatrix{\cos \theta&\sin \theta\\-\sin \theta&\cos \theta} . $$
you can now define the new coordinate transformation by $\xi, \eta$ by the relation
$$\pmatrix{\xi\\\eta} = U\pmatrix{x\\y}, \pmatrix{x\\y} = U^\top \pmatrix{\xi\\\eta}.$$
with this we get
$$\begin{align}x^2 + xy + 3y^2 &=\pmatrix{x&y} A\pmatrix{x\\y}\\ &=
\pmatrix{x&y} U^\top D U \pmatrix{x\\y}\\
&= \pmatrix{\xi & \eta}D\pmatrix{\xi\\\eta} \\
&= (2 + \sqrt5/2)\xi^2 + (2 - \sqrt5/2)\eta^2\end{align}$$
|
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|
Finding the real root of the polynomial $2x^3-3x^2+2 $ I want to get exactly roots of this equation...
$2x^3-3x^2+2 = 0$
I try to solve it but can not find the solution.
wolframealpha just give me aproximation..
I know the real root is $-1< root <-1/2$.
|
Here are the three exact solutions, which you can find by cubic root methods such as Cardano's:
$\left\{\frac{1}{2} \left(1-\frac{1}{\sqrt[3]{3-2 \sqrt{2}}}-\sqrt[3]{3-2
\sqrt{2}}\right),\frac{1}{2}+\frac{1}{4} \sqrt[3]{3-2 \sqrt{2}} \left(1-i
\sqrt{3}\right)+\frac{1+i \sqrt{3}}{4 \sqrt[3]{3-2 \sqrt{2}}},\frac{1}{2}+\frac{1-i
\sqrt{3}}{4 \sqrt[3]{3-2 \sqrt{2}}}+\frac{1}{4} \sqrt[3]{3-2 \sqrt{2}} \left(1+i
\sqrt{3}\right)\right\}$
The first root is real, and has a decimal approximation of $-0.677651...$.
|
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|
The lines $x+2y+3=0$ , $x+2y-7=0$ and $2x-y+4=0$ are sides of a square. Equation of the remaining side is? I found out the area between parallel lines as $ \frac{10}{\sqrt{5}} $ and then I used
$ \frac{|\lambda - 4|}{\sqrt{5}} = \frac{10}{\sqrt{5}} $ to get the values as $-6$ and $14$ . I am getting the final equations as $2x-y-6=0$ and $2x-y+14=0$ but this answer is wrong. According to my book the correct equations are $2x-y+6=0$ and $2x-y-14=0$. Please tell me where I am wrong!
|
It is clear that the lines: $x+2y+3=0$ & $x+2y-7$ are parallel hence the forth line (side of square) must be parallel to third line: $2x-y+4=0$ Hence, let the forth line: $2x-y+c=0$ having slope $2$.
Let each side of square $a$ then it is equal to the distance between the first two (first & second) parallel lines calculated as follows $$a=\frac{\left|3-(-7)\right|}{\sqrt{1^2+2^2}}=\frac{10}{\sqrt{5}}=2\sqrt{5}$$
The side $a$ is also equal to the distance between the second two (third & fourth) parallel lines calculated as follows $$a=\frac{\left|c-4\right|}{\sqrt{(2)^2+(-1)^2}}=\frac{\left|c-4\right|}{\sqrt{5}}$$ Now, equating both the values of side $a$ of the square, we have $$\frac{\left|c-4\right|}{\sqrt{5}}=2\sqrt{5}\implies \left|c-4\right|=10 $$ $$ c-4=10 \implies c=14$$ & $$ c-4=-10 \implies c=-6$$ Thus, corresponding above two values of $c$, we get two lines: $2x-y+14=0$ & $2x-y-6=0$ representing fourth unknown side of the square lying on either side of third line: $2x-y+4=0$. Obviously, the answers are same as you have obtained. There is some printing mistake in the answers provided in your book.
According to your book, the answers are: $2x-y+6=0$ & $2x-y-14=0$ Then, note that the sides of the square are not equal as you have already mentioned that the figure is a square & first two parallel sides: $x+2y+3=0$ & $x+2y-7=0$ confirms that the square has its each side $2\sqrt{5}$. Hence the answers according to your book are wrong because those do not satisfy the given conditions. The correct answers are: $2x-y-6=0$ & $2x-y+14=0$
|
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|
What mistake have I made when trying to evaluate the limit $\lim \limits _ {n \to \infty}n - \sqrt{n+a} \sqrt{n+b}$? Suppose $a$ and $b$ are positive constants.
$$\lim \limits _ {n \to \infty}n - \sqrt{n+a} \sqrt{n+b} = ?$$
What I did first:
I rearranged $\sqrt{n+a} \sqrt{n+b} = n \sqrt{1+ \frac{a}{n}} \sqrt{1+ \frac{b}{n}}$ and so: $$\lim \limits _ {n \to \infty} n - n \sqrt{1+ \frac{a}{n}} \sqrt{1+ \frac{b}{n}} = 0$$
Because both $\frac{a}{n}$ and $\frac{b}{n}$ tend to $0$.
What would give a correct answer:
Plotting the function $$f(x) = x - \sqrt{x+a} \sqrt{x+b}$$
Clearly indicates that it has an asymptote in $- \frac{a+b}{2}$. This result can be obtained multiplying the numerator and the denominator by $n + \sqrt{n+a} \sqrt{n+b}$:
$$\lim \limits _ {n \to \infty}n - \sqrt{n+a} \sqrt{n+b} = $$
$$-\lim \limits _{n \to \infty} \frac {n(a+b)}{n + \sqrt{n+a} \sqrt{n+b}} - \lim \limits _{n \to \infty} \frac {ab}{n + \sqrt{n+a} \sqrt{n+b}}$$
The second limit is clearly $0$ and the first one gives the correct answer (dividing the numerator and denominator by $n$).
Why the first way I tried is wrong? I might have done something silly but I cannot find it.
|
Your error is in this equality:
\begin{equation*}
\lim\limits_{n\rightarrow \infty }n-n\sqrt{1+\frac{a}{n}}\sqrt{1+\frac{b}{n}}%
=0\ \ \ \ \ \ (Here\ it\ is).
\end{equation*}
Explanation. You known that
\begin{eqnarray*}
\lim_{n\rightarrow \infty }n &=&+\infty ,\ \ \ and \\
\lim_{n\rightarrow \infty }n\sqrt{1+\frac{a}{n}}\sqrt{1+\frac{b}{n}}
&=&+\infty ,\ \ \ too.
\end{eqnarray*}
However, it seems that you forget that $\infty -\infty $ is an indetermined
form, that is, when one have
\begin{equation*}
\lim_{n\rightarrow \infty }a_{n}=\infty ,\ \ \ \ \ and\ \ \ \ \ \
\lim_{n\rightarrow \infty }b_{n}=\infty
\end{equation*}
then one cannot give any conclusion about
\begin{equation*}
\lim_{n\rightarrow \infty }(a_{n}-b_{n})
\end{equation*}
because one can have many situations. In this case one should make some
further study. For this example you use the classic technique which consists
of multiplying and dividing by what is called the conjugate of the original
expression, as follows (step by step)
\begin{eqnarray*}
\left( n-n\sqrt{1+\frac{a}{n}}\sqrt{1+\frac{b}{n}}\right) &=&\left( n-n%
\sqrt{1+\frac{a}{n}}\sqrt{1+\frac{b}{n}}\right) \left( \frac{n+n\sqrt{1+%
\frac{a}{n}}\sqrt{1+\frac{b}{n}}}{n+n\sqrt{1+\frac{a}{n}}\sqrt{1+\frac{b}{n}}%
}\right) \\
&=&\frac{n^{2}-\left( n\sqrt{1+\frac{a}{n}}\sqrt{1+\frac{b}{n}}\right) ^{2}}{%
n+n\sqrt{1+\frac{a}{n}}\sqrt{1+\frac{b}{n}}},\ \ \ \ \ \
((x-y)(x+y)=x^{2}-y^{2}) \\
&=&\frac{n^{2}-n^{2}(1+\frac{a}{n})(1+\frac{b}{n})}{n+n\sqrt{1+\frac{a}{n}}%
\sqrt{1+\frac{b}{n}}} \\
&=&\frac{n^{2}(1-(1+\frac{a}{n})(1+\frac{b}{n}))}{n(1+\sqrt{1+\frac{a}{n}}%
\sqrt{1+\frac{b}{n}})} \\
&=&\frac{n(1-\left( 1+\frac{a}{n}+\frac{b}{n}+\frac{ab}{n^{2}}\right) )}{(1+%
\sqrt{1+\frac{a}{n}}\sqrt{1+\frac{b}{n}})} \\
&=&\frac{-(a+b+\frac{ab}{n})}{(1+\sqrt{1+\frac{a}{n}}\sqrt{1+\frac{b}{n}})}
\end{eqnarray*}
therefore, passing to the limit one gets
\begin{eqnarray*}
\lim_{n\rightarrow \infty }\left( n-n\sqrt{1+\frac{a}{n}}\sqrt{1+\frac{b}{n}}%
\right) &=&\lim_{n\rightarrow \infty }\frac{-(a+b+\frac{ab}{n})}{(1+\sqrt{1+%
\frac{a}{n}}\sqrt{1+\frac{b}{n}})} \\
&=&\frac{-(a+b+0)}{1+\sqrt{1+0}\sqrt{1+0}}=-\frac{(a+b)}{2}
\end{eqnarray*}
|
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|
Axis angle and length of ellipse For an ellipse defined by
$$x = a \cos(t + \alpha)$$
$$y = b \cos(t + \beta)$$
What are the angles and lengths of each axis?
I've tried to work backwards from the expression for a rotated ellipse but I can't seem to equate the trigonometric expressions.
|
You can write the coordinates at time $t$
\begin{eqnarray}
\left( \begin{array}{c}
x(t) \\
y(t)
\end{array}
\right) =
\left( \begin{array}{cc}
a_{11} & a_{12} \\
a_{21} & a_{22}
\end{array}
\right)\cdot
\left( \begin{array}{c}
\cos t \\
\sin t
\end{array}
\right)
\end{eqnarray}
Consider a singular value decomposition
$$\left( \begin{array}{cc}
a_{11} & a_{12} \\
a_{21} & a_{22}
\end{array} \right ) = \left( \begin{array}{cc}
\cos u & -\sin u \\
\sin u & \cos u
\end{array} \right ) \cdot
\left( \begin{array}{cc}
d_1 & 0 \\
0 & d_2
\end{array} \right )
\cdot
\left( \begin{array}{cc}
\cos v & -\sin v \\
\sin v & \cos v
\end{array} \right )
$$
So we get
$$\left( \begin{array}{c}
x(t) \\
y(t)
\end{array}
\right) = \left( \begin{array}{cc}
\cos u & -\sin u \\
\sin u & \cos u
\end{array} \right ) \cdot \left( \begin{array}{c}
d_1 \cos (t+v) \\
d_2 \sin (t+v)
\end{array}
\right) $$
So this is a rotation by angle $u$ of a common ellipse with semi-axes $|d_1|$ and $|d_2|$.
|
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|
How to solve certain types of integrals I'm asking for a walk through of integrals in the form:
$$\int \frac{a(x)}{b(x)}\,dx$$
where both $a(x)$ and $b(x)$ are polynomials in their lowest terms. For instance $$\int \frac{x^3+2x}{x^2+1}\,dx$$
Is there a trick to doing these? or will I have to integrate by a clever substitution?
|
The first step is generally to divide out the integrand so as to get a polynomial plus a rational function whose numerator has lower degree than its denominator. Here you get
$$\frac{x^3+2x}{1+x^2}=x+\frac{x}{x^2+1}\;.$$
Integrating the $x$ term (and, more generally, the polynomial quotient) is easy, so we’ve reduced the problem to integrating something of the form $\frac{p(x)}{q(x)}$, where $p(x)$ and $q(x)$ are polynomials, and the degree of $p(x)$ is less than the degree of $q(x)$. The general solution for such problems is partial fractions; here, however, we’re more fortunate, because the numerator $x$ is a constant multiple of the derivative of the denominator. If you substitute $u=x^2+1$, you find that $du=2x\,dx$, so that $x\,dx=\frac12du$, and
$$\int\frac{x}{x^2+1}\,dx=\frac12\int\frac{du}u\;,$$
which is a standard, basic integral. I would not call this a clever substitution: recognizing that the numerator of a fraction is a constant multiple of the derivative of the denominator is a standard technique.
Suppose that the original numerator had been $x^3+x+2$. Again we do the division to get a polynomial plus a ‘proper’ rational function:
$$\frac{x^3+x+2}{x^2+1}=x+\frac2{x^2+1}\;.$$
This time you should recognize that $\frac2{x^2+1}$ is just twice the derivative of $\tan^{-1}x$, again a standard integration.
Finally, suppose that the original fraction had been
$$\frac{x^3+x+1}{x^2+2x}=x+\frac{1-x}{x^2+2x}\;.$$
This time you might as well simply reduce the remaining fraction to partial fractions:
$$\frac{1-x}{x(x+2)}=\frac{A}x+\frac{B}{x+2}\;,$$
so $A(x+2)+Bx=1-x$, $(A+B)x+2A=1-x$, $A+B=-1$, and $2A=1$, so $A=\frac12$, and $B=-\frac32$. Thus,
$$\frac{1-x}{x^2+2x}=\frac12\left(\frac1x-\frac3{x+2}\right)\;,$$
leaving you with two easy integrations.
|
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|
Proof of sum in an inequality I was having hard time solving this one, any help will be greatly appreciated.
prove that:
$$
{39\over e^2}\le\sum_{n=1}^\infty {4n^2-1\over e^n}-{3\over e}\le{54\over e^2}
$$
|
We know $\sum_{n=0}^\infty x^n=\frac{1}{1-x}$. Taking derivatives $\sum_{n=1}^\infty nx^{n-1}=\frac1{(1-x)^2}$. Multiply both sides by $x$ and take another derivative to get $\sum_{n=1}^\infty n^2x^{n-1}=\frac{1-x^2}{(1-x)^4}$. Thus $\sum_{n=1}^\infty 4n^2x^n=\frac{4x(1-x^2)}{(1-x)^4}$. Thus $\sum_{n=1}^\infty(4n^2x^n-x^n)=\frac{4x(1-x^2)}{(1-x)^4}-\sum_{n=1}^\infty x^n=\frac{4x(1-x^2)}{(1-x)^4}- \frac{x}{1-x}$. Thus $\sum_{n=1}^\infty(4n^2-1)x^n=\frac{4x(1-x^2)}{(1-x)^4}- \frac{x}{1-x}$. Plug in $\frac1e$ to get $\sum_{n=1}^\infty\frac{4n^2-1}{e^n}=\frac{4e(e+1)}{(e-1)^3}-\frac1{e-1}$.
|
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|
how to determine the conditions that 2 vectors parallel? Given 2 vectors,$u=(3,5)$,$v=(s,s^2)$,in what situations do u and v parallel?$(s≠0)$
In order to be parallel,$u$ must be proportional to $v$,vice verse.Let $k$ be a scalar $neq 0$,then $ku=(3k,5k)=v=(s,s^2)$,which gives $3k=s$;$5k=s^2 \rightarrow 5k=9k^2 (k \neq 0) \rightarrow 5=9k→k=\frac{5}{9}$,plug $k=\frac{5}{9}$ in the above equation $3k=s,s=\frac{5}{3}$.
Actually $k$ can also be multiplied to $s$,which is $kv=(ks,ks^2)=(3,5)$.Following the similar procedure,we get :$ks=3$;$ks^2=5 \rightarrow 3s=5,s=\frac{5}{3}$.But if we plug $s=\frac{5}{3}$ in $ks=3$,we get $\frac{5}{3k}=3$,which gives $k=\frac{9}{5}$.
The two methods should give the same answer. I just don't know what's going wrong here.
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There's nothing wrong. Your first value of $k = \frac{5}{9}$ means that $\frac{5}{9} u = v$. Your second value of $k = \frac{9}{5}$ is for k multiplying the other vector, and means that $\frac{9}{5} v = u$. But that's exactly the same you got before.
$$\frac{5}{9} u = v \implies u = \frac{9}{5} v$$
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Prove that $\sin (\theta) + \cos(\theta) \ge 1$
Let $\theta$ be an arbitrary acute angle. Prove that $\sin (\theta) + \cos(\theta) \ge 1$.
$$\big(\sin (\theta) + \cos (\theta)\big)^2 = 1 + 2 \sin(\theta)\cos(\theta)\ge 0$$
so, \begin{align*}\big(\sin(\theta)) + \cos(\theta)\big)^2 &> 1\\
\big(\sin(\theta)+ \cos(\theta)\big)^2 &\ge 1\end{align*}
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This is a brute force approach.
There is an $R$ and $\alpha$ such that $\sin \theta + \cos \theta = R\sin(\theta + \alpha)$.
First, $R\sin(\theta + \alpha) = R\cos\alpha\sin\theta + R\sin\alpha\cos\theta$.
So $R\sin\alpha = R\cos\alpha = 1$. Since $\tan\alpha = 1$, $\alpha = 45^\circ$. Adding the squares of $R\sin\alpha$ and $R\cos\alpha$ gets $R^2 = 2 \therefore R = \sqrt 2$.
$$\sin\theta + \cos\theta = \sqrt{2}\sin(\theta + 45^\circ)$$
It should be clear now if you look at a graph of the $\sin$ function.
Using complex numbers:
$$e^{i\theta} - ie^{i\theta} = (\cos\theta + \sin\theta) + i(\sin\theta - \cos\theta) = (1-i)e^{i\theta} = \sqrt{2}e^{i(\theta - 45^\circ)}$$
Now trace the $90^\circ$ arc from $\theta = 0$ to $\theta = 90^\circ$, and it's clear that the real part is bigger than $1$.
|
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|
General solution of the system of equations It seems like an easy question to solve but could not figure it out:
If it is known that the following system of equations have a solution $x = x_1(t)$, where $x_1(t)$ is a second order polynomial and $y= y_1(t)$ where $y_1(t)$ is a first order polynomial.
$$x' = (t-t^2)x + (t^3-t^2+t+1)y$$
$$y' = (1-t)x + (t^2-t+1)y$$
What is the general solution?
I tried to plug in $x=ax^2+bx+c$ and $y=mx+n$. However could not end up with a solution. I will be glad if you can help.
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$$x' = (t-t^2)x + (t^3-t^2+t+1)y$$
$$y' = (1-t)x + (t^2-t+1)y$$
Letting $x = at^2+bt+c$ and $y = mt+n$, we get that
$$2at+b = (t-t^2)(at^2+bt+c) + (t^3-t^2+t+1)(mt+n) =$$
$$(m-a)t^4+(a-b-m+n)t^3+(b-c+m-n)t^2+(c+m+n)t+n$$
and
$$m = (1-t)(at^2+bt+c) + (t^2-t+1)(mt+n)$$
From the first equation, we get that $m-a=0$, $a-b-m+n=0$, $b-c+m-n=0$, $c+m+n=2a$, and $n=b$. From this, we get that $a = m = c$, and $b = n = 0$.
Plugging these back into the second equation, we get
$$a = (1-t)(at^2+a) + (t^2-t+1)at$$
which simplifies to
$$1 = t^2-t^3+1-t + t^3-t^2+t = 1$$
which is a degenerate equation.
So the general solution is $$x = k(t^2+1), y = kt$$ for all $k \ne 0$.
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How to evaluate $\lim\limits_{n\to\infty}\frac{1}{n}((x+\frac{a}{n})^2+(x+\frac{2a}{n})^2+...+(x+\frac{(n-1)a}{n})^2)$ I don't know how to transform the expression $\frac{1}{n}((x+\frac{a}{n})^2+(x+\frac{2a}{n})^2+...+(x+\frac{(n-1)a}{n})^2)$
The solution, after transformation is
$\frac{n-1}{n}x^2+2\frac{1+...+(n-1)}{n^2}ax+\frac{1^2+...+(n-1)^2}{n^3}a^2$
Thanks for replies.
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This is just the limit of the Riemann sum of $\displaystyle\int_{0}^1(x+a)^2da$, which is $\displaystyle x^2+x+\frac{1}{3}$.
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Both ways seems right to me The question: $$\int {2e^x \over e^{2x}-1}dx $$
Solution 1 (we solv it this way in the exam today)
$\int {2e^x \over (e^x-1)(e^x+1)}dx = \int {(\frac{A}{e^x-1} + \frac{B}{e^x+1})dx} \qquad $from this, $A=1$ and $B=1$. Then,
$\int {\frac{1}{e^x-1} }dx + \int {\frac{1}{e^x+1}}dx = \ln(e^x-1)+\ln(e^x+1)+c = \ln(e^{2x}-1)+c$
Solution 2 (teacher's way)
Let $\ e^x=u \ $ and $\ e^xdx=du \ $. So the integral changes to $\ \int{2\over u^2-1}du \ $. Then,
$\int{({A \over u-1}+{B \over u+1})}du \qquad$ from this, $A=1$ and $B=-1$. Then,
$\int{({1 \over u-1}-{1 \over u+1})}du = \int{1 \over u-1}du - \int{1 \over u+1}du = \ln(u-1)-ln(u+1)+c = \ln(\frac{u-1}{u+1})+c$
And if we replace $u$ with $e^x$: $\ln(\frac{e^x-1}{e^x+1})+c$
So, which one is the right answer?
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You can re-write you integrand as $$\frac{2}{e^{x}-e^{-x}}=\text{cosech} x.$$ Thus
\begin{eqnarray}
\int\text{cosech}xdx &=& \ln \text{tanh} \frac{x}{2}+c \\
&=& \ln \left(\frac{e^{\frac{x}{2}}-e^{\frac{x}{2}}}{e^{\frac{x}{2}}+e^{\frac{x}{2}}}\right)+c
\end{eqnarray}
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find the minimum value of $\sqrt{x^2+4} + \sqrt{y^2+9}$ Question:
Given $x + y = 12$, find the minimum value of $\sqrt{x^2+4} + \sqrt{y^2+9}$?
Key:
I use $y = 12 - x$ and substitute into the equation, and derivative it.
which I got this
$$\frac{x}{\sqrt{x^2+4}} + \frac{x-12}{\sqrt{x^2-24x+153}} = f'(x).$$
However, after that. I don't know how to do next in order to find the minimum value. Please help!
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Just another way:
$$\sqrt{x^2+4}+\sqrt{y^2+9} \ge \sqrt{(x+y)^2+(2+3)^2} = 13$$
where we have used the triangle inequality.
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Is this correct $(\cot x)(\sin x)=2(\cot x)^2$ $0≤x≤2\pi$ $x= \pi/2 , 3\pi/2$ $1.4371774 , 5.139467567$ Steps I took:
$$\cot x \sin x=2\cot^2 x$$
$$\cot x \sin x-2\cot^2 x=0$$
$$\cot x (\sin x-2\cot x)=0$$
$$\cot x \left(\frac{\sin^2 x}{\sin x} -2\frac{\cos x}{\sin x} \right)=0$$
$$\cot x \left(\frac{\sin^2 x-2\cos x}{\sin x} \right)=0$$
$$\cot x \left(\frac{1-\cos^2 x-2\cos x}{\sin x} \right)=0$$
$$\cot x \csc x (\cos x+2.414213562)(\cos -0.4142135624)=0$$
$$\cot x =0 \rightarrow x=\frac{\pi}{2},\frac{3\pi}{2}$$
$\csc x$ cannot be $0$
$\cos x$ cannot be $-2.414213562$
$$\cos x = 0.4142135624$$
$$x=1.4371774 , 5.139467567$$
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i will use $$x = \cos t, y = \sin t , x^2 + y^2 = 1.$$ you have $$\cot t\sin t = 2 \cot^2 t \to \frac x y y=2\frac {x^2}{y^2} \to 0=x(2x-y^2)=x(-x^2 + 2x + 1)$$ the solutions are $$x = 0, x = \frac{-2 \pm 2\sqrt 2}{-2} =1+\sqrt 2, \sqrt 2 - 1$$ since $x < 1,$ we have $$ x= 0, x = \sqrt 2 - 1$$
and the corresponding $t$ values are $$\pi/2, \cos^{-1}(\sqrt 2 - 1), 2\pi - \cos^{-1}(\sqrt 2 - 1), 3\pi/2.$$
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Roots of unity question. Question
Let $\omega=\cos\dfrac{4\pi}{7}+i\sin\dfrac{4\pi}{7}$. Show that
$\omega-1=2\sin\dfrac{2\pi}{7}\left(\cos\dfrac{11\pi}{14}+i\sin\dfrac{11\pi}{14}\right)$.
My attempt
Observe that $\omega$ is a seventh root of unity. Label the roots $1, \nu, \nu^2,\ldots,\nu^6$. Then $\omega=\nu^2$.
We have $1+ \nu+ \nu^2+\ldots+\nu^6=0$ and so $1+ \nu+ \omega+\ldots+\nu^6=0$.
Then $\omega-1=-2-(\nu+\nu^6)-(\nu^3+\nu^4)-\nu^5$.
But $\nu+\nu^6=2\cos\dfrac{2\pi}{7}$ and $\nu^3+\nu^4=2\cos\dfrac{6\pi}{7}$
I do not know how to continue.
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Using $$\cos(2\theta)-1=-2\sin^2\theta,\ \ \sin(2\theta)=2\sin\theta\cos\theta$$
$$-\sin\theta=\cos\left(\frac{\pi}{2}+\theta\right),\ \ \cos\theta=\sin\left(\frac{\pi}{2}+\theta\right)$$gives you$$\begin{align}\omega-1&=\cos\frac{4\pi}{7}+i\sin\frac{4\pi}{7}-1\\&=\cos\frac{4\pi}{7}-1+i\sin\frac{4\pi}{7}\\&=-2\sin^2\frac{2\pi}{7}+i\cdot 2\sin\frac{2\pi}{7}\cos\frac{2\pi}{7}\\&=2\sin\frac{2\pi}{7}\left(-\sin\frac{2\pi}{7}+i\cos\frac{2\pi}{7}\right)\\&=2\sin\frac{2\pi}{7}\left(\cos\frac{11\pi}{14}+i\sin\frac{11\pi}{14}\right)\end{align}$$
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For which values of a parameter an equation has one Real root The following equation is given $$\log_{x-1}(x^2+2ax) - \log_{x-1}(8x-6a-3)=0$$
And I am trying to find for which values of $a$ it has only one root, which is real.
It is obvious that $$x-1>0 \Rightarrow x>1, x\neq2 $$
Also $$x^2+2ax>0$$ $$8x-6a-3>0 \Rightarrow x>\frac{6a+3}8 $$
Now I combine the two logarithms $$\log_{x-1}\left(\frac{x(x+2a)}{8x-6a-3}\right)=0$$
and then $$\frac{x(x+2a)}{8x-6a-3}=1$$ $$\frac{x^2+2(a-4)x+6a+3}{8x-6a-3}=0$$ $$x^2+2(a-4)x+6a+3=0$$
The discriminant is $$D=(a-13)(a-1)$$
and the roots are $$x_1= -a+4-\sqrt{(a-13)(a-1)}$$ $$x_2= -a+4+\sqrt{(a-13)(a-1)}$$ so that $x_1<x_2$.
*
*Case 1:
For $D=0$ we have only one root for $x$. $D=0$ for $a=13$ or $a=1$.
For $a=13$, $x_1=x_2=-9\not\gt1$ so $a=13$ is not a solution.
For $a=1$, $x_1=x_2=3>1$. Also $x=3>\frac{6a+3}{8}=\frac98$ and $x^2+2ax=15>0$, so for $a=1$ we have only one root, therefore it is a solution.
*
*Case 2:
Now let $D>0$, so that $x_1 \ne x_2$. $D>0$ for $a\in(-\infty,1)\cup(13,+\infty)$
I consider five sub-cases:
A) When $x_1 = 2$ and $x_2^2+2ax_2>0, x_2>\frac{6a+3}8 $
B) When $x_2=2$ and $x_1>1, x_1^2+2ax_1>0, x_1>\frac{6a+3}8 $.
C) When $1\in[x_1,x_2)$ and $x_2^2+2ax_2>0, x_2>\frac{6a+3}8 $.
D) When $\frac{6a+3}8\in[x_1,x_2)$ and $x_2>1, x_2^2+2ax_2>0 $.
E) When $x_1^2+2ax_1\le0$ and $x_2>1, x_2^2+2ax_2>0, x_2>\frac{6a+3}8 $
The solutions I find are
A) $a=\frac9{10}, x_2=\frac{21}5$
B) none
C), D), E) It all comes up to solving inequalities involving polynomials for $a$ of degree 4, which makes me doubt I am doing it right.
Am I missing any cases or am I considering too many cases? How would you approach this problem?
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For $D=0$, $a=1$ is the only solution.
For $D=(a-1)(a-13)\gt 0$, note that $x_2\not=2$ and that $x_1=2$ for $a=\frac{9}{10}$ (and $a=\frac{9}{10}$ is sufficient). In the following, let us separate it into cases.
*
*For $a\lt -\frac 12$, we have $\frac{6a+3}{8}\lt 0\lt 1\lt -2a.$ So, the condition we need is $x_2\gt -2a$ and $x_1\le -2a$, i.e. $$\small a\lt -\frac 12\ \ \text{and}\ \ -a+4+\sqrt{(x-1)(x-13)}\gt -2a\ \ \text{and}\ \ -a+4-\sqrt{(a-1)(a-13)}\le -2a.$$Solving these gives you $a\lt -\frac 12$.
*For $-\frac 12\le a\lt 0$, we have $0\lt -2a\le 1$ and $0\le \frac{6a+3}{8}\lt 1$. So, the condition we need is $x_2\gt 1$ and $x_1\le 1$, i.e. $-\frac 12\le a\lt 0$.
*For $0\le a\lt\frac 56$, we have $-2a\le 0\lt\frac{6a+3}{8}\lt 1$. So, the condition we need is $x_2\gt 1$ and $x_1\le 1$, i.e. $0\le a\le \frac 12$.
*For $\frac 56\le a\lt \frac{9}{10}\ \text{or}\ \frac{9}{10}\lt a\lt 1\ \text{or}\ a\gt 13$, we have $-2a\lt 0\lt 1\le\frac{6a+3}{8}$. So, the condition we need is $x_2\gt\frac{6a+3}{8}$ and $x_1\le\frac{6a+3}{8}$. However, there is no such $a$.
Hence, the answer is
$$a\le\frac 12\ \ \ \text{or}\ \ \ a=\frac{9}{10}\ \ \ \text{or}\ \ \ a=1.$$
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What is the most unusual proof you know that $\sqrt{2}$ is irrational? What is the most unusual proof you know that $\sqrt{2}$ is irrational?
Here is my favorite:
Theorem: $\sqrt{2}$ is irrational.
Proof:
$3^2-2\cdot 2^2 = 1$.
(That's it)
That is a corollary of
this result:
Theorem:
If $n$ is a positive integer
and there are positive integers
$x$ and $y$ such that
$x^2-ny^2 = 1$,
then
$\sqrt{n}$ is irrational.
The proof is in two parts,
each of which
has a one line proof.
Part 1:
Lemma: If
$x^2-ny^2 = 1$,
then there are arbitrarily large integers
$u$ and $v$ such that
$u^2-nv^2 = 1$.
Proof of part 1:
Apply the identity
$(x^2+ny^2)^2-n(2xy)^2
=(x^2-ny^2)^2
$
as many times as needed.
Part 2:
Lemma: If
$x^2-ny^2 = 1$
and
$\sqrt{n} = \frac{a}{b}$
then
$x < b$.
Proof of part 2:
$1
= x^2-ny^2
= x^2-\frac{a^2}{b^2}y^2
= \frac{x^2b^2-y^2a^2}{b^2}
$
or
$b^2
= x^2b^2-y^2a^2
= (xb-ya)(xb+ya)
\ge xb+ya
> xb
$
so
$x < b$.
These two parts
are contradictory,
so
$\sqrt{n}$
must be irrational.
Two things to note about
this proof.
First,
this does not need
Lagrange's theorem
that for every
non-square positive integer $n$
there are
positive integers $x$ and $y$
such that
$x^2-ny^2 = 1$.
Second,
the key property of
positive integers needed
is that
if $n > 0$
then
$n \ge 1$.
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Using the rational root theorem on $x^2 - 2 = 0$ is a very simple and elegant way of proving the irrationality of $ \sqrt{2} $. Peersonally, I like it beacuse it can be explained easily to high school students
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|
Find probability of exactly one $6$ in first ten rolls of die, given two $6$s in twenty rolls I am trying to calculate the probability that, when rolling a fair die twenty times, I roll exactly one $6$ in the first ten rolls, given that I roll two $6$s in the twenty rolls.
My thoughts
Let $A = \{\text {Exactly one 6 in first ten rolls of a die} \}$ and $B = \{\text {Exactly two 6s in twenty rolls of a die} \}.$
Then I want to find
$$P[A\mid B] = \frac{P[A \cap B]}{P[B]}.$$
By the binomial distribution formula, we get that
$$P[B] = {20 \choose 2} \cdot \left(\frac{1}{6}\right)^2 \cdot \left(\frac{5}{6}\right)^{18}.$$
Furthermore I think that $P[A \cap B]$ is equal to the probability of rolling exactly one $6$ in ten rolls and then rolling exactly one $6$ in another set of ten rolls. That is,
$$P[A \cap B] = \left[{10 \choose 1} \cdot \left(\frac{1}{6}\right)^1 \cdot \left(\frac{5}{6}\right)^9\right]^2.$$
Am I correct in thinking this?
If so, then it follows that the required probability is
$$P[A \mid B] = \frac{\left[{10 \choose 1} \cdot \left(\frac{1}{6}\right)^1 \cdot \left(\frac{5}{6}\right)^9\right]^2}{{20 \choose 2} \cdot \left(\frac{1}{6}\right)^2 \cdot \left(\frac{5}{6}\right)^{18}},$$
which, I know, can be simplified further!
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((10C1)(1/6)(5/6)^5)(10C1)(1/6)(5/6)^5
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Prove a combinatorial identity: $ \sum_{n_1+\dots+n_m=n} \prod_{i=1}^m \frac{1}{n_i}\binom{2n_i}{n_i-1}=\frac{m}{n}\binom{2n}{n-m}$ Prove the combinatorial identity
$$
\sum_{n_1+\ldots+n_m=n} \;\;
\prod_{i=1}^m \frac{1}{n_i}\binom{2n_i}{n_i-1}=\frac{m}{n}\binom{2n}{n-m}, \enspace n_i>0,i=1,\ldots,m
$$
I "discovered" this equality during experiments with Maple, but I have no idea how to prove it. It may have a connection with Catalan numbers but that hasn't helped me.
UPDATE
Now we have brilliant proof of this equality. But may be it have purely combinatorial proof? Or proof with Catalan number's properties?
|
Using the generating function of the Catalan numbers the left is
$$[z^n] \left(-1 + \frac{1-\sqrt{1-4z}}{2z}\right)^m.$$
This is
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n+1}}
\left(-1 + \frac{1-\sqrt{1-4z}}{2z}\right)^m \; dz.$$
Using Lagrange inversion put $1-4z=w^2$ so that $1/4-z=1/4 \times w^2$
or $z=1/4\times(1-w^2)$ and $dz = -1/2 \times w\; dw.$ This gives for
the integral
$$\frac{1}{2\pi i}
\int_{|w-1|=\epsilon}
\frac{4^{n+1}}{(1-w^2)^{n+1}}
\left(-1 +\frac{1-w}{1/2\times(1-w^2)}\right)^m
\times\left(-\frac{1}{2} w\right) \; dw
\\ = \frac{1}{2\pi i}
\int_{|w-1|=\epsilon}
\frac{4^{n+1}}{(1-w^2)^{n+1}}
\left(\frac{w^2-1+2-2w}{1-w^2}\right)^m
\times\left(-\frac{1}{2} w\right) \; dw
\\ = \frac{1}{2\pi i}
\int_{|w-1|=\epsilon}
\frac{4^{n+1}}{(1-w^2)^{n+1}}
\frac{(w-1)^{2m}}{(1-w^2)^{m}}
\times\left(-\frac{1}{2} w\right) \; dw
\\ = \frac{1}{2\pi i}
\int_{|w-1|=\epsilon}
\frac{4^{n+1}}{(1-w^2)^{n+m+1}}
(1-w)^{2m}
\times\left(-\frac{1}{2} w\right) \; dw
\\ = - \frac{2^{2n+1}}{2\pi i}
\int_{|w-1|=\epsilon}
\frac{1}{(1-w)^{n+m+1}(1+w)^{n+m+1}}
(1-w)^{2m} \times w\; dw
\\ = - \frac{2^{2n+1}}{2\pi i}
\int_{|w-1|=\epsilon}
\frac{1}{(1-w)^{n-m+1}(1+w)^{n+m+1}}
\times w\; dw
\\ = \frac{(-1)^{n-m} 2^{2n+1}}{2\pi i}
\int_{|w-1|=\epsilon}
\frac{1}{(w-1)^{n-m+1}(1+w)^{n+m+1}}
\times w\; dw.$$
The integral has two pieces, the first is
$$\frac{(-1)^{n-m} 2^{2n+1}}{2\pi i}
\int_{|w-1|=\epsilon}
\frac{1}{(w-1)^{n-m}(2+w-1)^{n+m+1}}
\; dw
\\ = \frac{(-1)^{n-m} 2^{n-m}}{2\pi i}
\int_{|w-1|=\epsilon}
\frac{1}{(w-1)^{n-m}(1+(w-1)/2)^{n+m+1}}
\; dw.$$
This is
$$(-1)^{n-m} 2^{n-m}
(-1)^{n-m-1} {n-m-1+n+m\choose n+m}\frac{1}{2^{n-m-1}}
= -2{2n-1\choose n+m}.$$
The second piece is
$$\frac{(-1)^{n-m} 2^{2n+1}}{2\pi i}
\int_{|w-1|=\epsilon}
\frac{1}{(w-1)^{n-m+1}(2+w-1)^{n+m+1}}
\; dw
\\ = \frac{(-1)^{n-m} 2^{n-m}}{2\pi i}
\int_{|w-1|=\epsilon}
\frac{1}{(w-1)^{n-m+1}(1+(w-1)/2)^{n+m+1}}
\; dw.$$
This is
$$(-1)^{n-m} 2^{n-m}
(-1)^{n-m} {n-m+n+m\choose n+m}\frac{1}{2^{n-m}}
= {2n\choose n+m}.$$
Collecting the two pieces we obtain
$${2n\choose n+m}
-2{2n-1\choose n+m}
= {2n\choose n+m} - 2\frac{n-m}{2n}{2n\choose n+m}
\\= \frac{2n-2n+2m}{2n}{2n\choose n-m}
= \frac{m}{n}{2n\choose n-m}.$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Show that $a^{25} \pmod{88}$ is congruent with $ a^{5} \pmod {88}.$
Show that $a^{25} \pmod{88}$ is congruent with $ a^{5} \pmod {88}.$
I have proved it in the case that $\gcd(88,a)=1$, but in the other case , I don't know it. Any ideas?
|
Note that $88=11 \cdot 8$, so $\Bbb Z _{88} \simeq \Bbb Z _8 \times \Bbb Z _{11}$. Therefore, you must show that $a^{25} = a^5$ in both $\Bbb Z _8$ and $\Bbb Z _{11}$. We shall use Euler's theorem: in $\Bbb Z _n$, we have $a^{\varphi (n)} = 1$, where $\varphi$ is Euler's function. Note that $\varphi(8)=4$ and $\varphi (11)=10$.
In $\Bbb Z _{11}$, every $a \ne 0$ is coprime to $11$, so $a^{10} = a^{\varphi (11)} = 1$, so $a^{25} = a^{20+5} = (a^{10})^2 a^5 = a^5$.
In $\Bbb Z _8$, if $\gcd (a, 8) = 1$, then $a^4 = a^{\varphi (8)} = 1$, so $a^{25} = a^{24+1} = (a^4)^6 a = a$; on the other hand, $a^5 = a^{4+1} = a^4 a=a$, so $a^{25}=a^5 \mod 8$.
If $\gcd(a, 8) >1$ then $2|a$, say $a=2b$. Then $a^{25} = 2^{25} b^{25} =0$, because $8 | 2^{25}$. Similarly, $a^5 = 2^5 b^5 = 0$, so again $a^{25} = a^5$.
The last case to examine is $a=0$, but this is trivial.
|
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|
solving difficult complex number proving if $z= x+iy$ where $y \neq 0$ and $1+z^2 \neq 0$, show that the number $w= z/(1+z^2)$ is real only if $|z|=1$
solution :
$$1+z^2 = 1+ x^2 - y^2 +2xyi$$
$$(1+ x^2 - y^2 +2xyi)(1+ x^2 - y^2 -2xyi)=(1+ x^2 - y^2)^2 - (2xyi)^2$$
real component
$$(1+ x^2 - y^2)x - yi(2xyi) = x + x^3 + xy^2$$
imaginary component
$$-2yx^2 i +yi + x^2 yi - y^3 i=0i$$
$$-2yx^2 +y + x^2 y - y^3 =0$$
...
can't solve this question
|
A another approach is to utilise (as @mann and myself concluded) the fact that
$$
w = \bar{w}
$$
where bar represents the complex conjugate.
$$
\frac{z}{1+z^2} = \frac{\bar{z}}{1+\bar{z}^2}
$$
re-arrange to yeild
$$
z + z\bar{z}^2 = \bar{z} + \bar{z}z^2 \implies z-\bar{z} = \bar{z}z^2-z\bar{z}^2=z\bar{z}\left(z-\bar{z}\right)
$$
thus we get
$$
1=z\bar{z} = |z|^2\implies |z| = 1
$$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/1316782",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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|
How to integrate $\int_{-1}^1 \tan^{-1}\left(\frac{1}{\sqrt{1-x^2}}\right )\,dx$? Evaluate
$$\int_{-1}^1 \tan^{-1}\bigg (\dfrac{1}{\sqrt{1-x^2}}\bigg ) dx $$
Could somebody please help integrate this without using Differentiation under the Integral Sign?
|
$$
\begin{aligned}
I &=\int_{-1}^{1} \tan ^{-1}\left(\frac{1}{\sqrt{1-x^{2}}}\right) d x \\
&=2 \int_{0}^{1} \tan ^{-1}\left(\frac{1}{\sqrt{1-x^{2}}}\right) d x \\
& \stackrel{IBP}{=} 2\left[\tan ^{-1}\left(\frac{1}{\sqrt{1-x^{2}}}\right)\right]_{0}^{1}-2 \int_{0}^{1} x \cdot\frac{1}{1+\left(\frac{1}{\sqrt{1-x^{2}}}\right)^{2}}\left(\frac{x}{\left(-x^{2}\right)^{\frac{3}{2}}}d x \right)\\
&=\pi-2 \underbrace{\int_{0}^{1} \frac{x^{2}}{\left(2-x^{2}\right) \sqrt{1-x^{2}}} d x}_{J}
\end{aligned}
$$
Let $x=\sin \theta$, where $0 \leqslant \theta \leqslant \dfrac{\pi}{2},$ then
$$
\begin{aligned}
J &=\int_{0}^{\frac{\pi}{2}} \frac{\sin ^{2} \theta}{\left(2-\sin ^{2} \theta\right) \cos \theta} \cos \theta d \theta \\
&=\int_{0}^{\frac{\pi}{2}} \frac{\sin ^{2} \theta}{2-\sin ^{2} \theta} d \theta \\
&=\int_{0}^{\frac{\pi}{2}}\left(\frac{2}{2-\sin ^{2} \theta}-1\right) d \theta \\
&=2 \int_{0}^{\frac{\pi}{2}} \frac{\csc ^{2} \theta}{2 \csc ^{2} \theta-1} d \theta-\frac{\pi}{2} \\
&=-2 \int_{0}^{\frac{\pi}{2}} \frac{d(\cot \theta)}{1+2 \cot ^{2} \theta}-\frac{\pi}{2}\\
&=-2 \cdot \frac{1}{\sqrt{2}}\left[\tan ^{-1}(\sqrt{2} \cot \theta)\right]_{0}^{\frac{\pi}{2}}-\frac{\pi}{2} \\
&=\frac{\pi \sqrt{2}}{2}-\frac{\pi}{2}
\end{aligned}
$$
Now we can conclude that $I=\pi(2-\sqrt{2})$.
|
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|
Group isomorphism and matrices Let $\mathbb{F}$ be a field. Consider the following three groups-
$$G=\left\{\begin {pmatrix}
1&a&b\\
0&1&c\\
0&0&1\\
\end{pmatrix} : a,b,c\in \mathbb{F}\right\} $$
$$H=\left\{\begin {pmatrix}
1&0&d\\
0&1&0\\
0&0&1\\
\end{pmatrix} : d\in \mathbb{F}\right\} $$
$$T=\left\{\begin {pmatrix}
x\\
y
\end{pmatrix} : x,y\in \mathbb{F}\right\} $$
Prove that $G/H \cong T$.
First I showed that $H=Z(G)$.
Then I concluded that $G/H=G/Z(G)\cong Inn(G)$.
I thought I might construct an isomorphism between $T$ and $Inn(G)$ and then use transitivity of isomorphism.
I calculated what a function $\phi$ in $Inn(G)$ would look like, and got:
$\phi \left( \begin {pmatrix}
1&a&b\\
0&1&c\\
0&0&1\\
\end{pmatrix} \right)= \begin {pmatrix}
1&\tilde a&\tilde b\\
0&1&\tilde c\\
0&0&1\\
\end{pmatrix} \times \begin {pmatrix}
1&a&b\\
0&1&c\\
0&0&1\\
\end{pmatrix} \times \begin {pmatrix}
1&\tilde a&\tilde b\\
0&1&\tilde c\\
0&0&1\\
\end{pmatrix} ^{-1} = \begin {pmatrix}
1& a& b-a\tilde c+ c\tilde a\\
0&1&c\\
0&0&1\\
\end{pmatrix}$
So the result is depended only on $\tilde a$ and $\tilde c$ and not on $\tilde b$.
"Well, that's great!", I thought to myself. "I can now construct $\psi : Inn(G) \rightarrow T$ to be $\psi(\phi_g)=
\begin {pmatrix}
\tilde a\\
\tilde c\\
\end{pmatrix} $ when $g= \begin {pmatrix}
1&\tilde a&\tilde b\\
0&1&\tilde c\\
0&0&1\\
\end{pmatrix} $"
But unfortunately, I wasn't able to prove that $\psi$ is an isomorphism (it is not injective).
Then I came here and spent an hour learning lots of $\LaTeX$in order to write this.
Can anyone get me out of this conundrum? Many thanks!
|
Do the multiplication
$$
\begin{pmatrix}
1&a&b\\
0&1&c\\
0&0&1
\end{pmatrix}
\begin{pmatrix}
1&a'&b'\\
0&1&c'\\
0&0&1
\end{pmatrix}
=\begin{pmatrix}
1&a'+a&b'+ac'+b\\
0&1&c'+c\\
0&0&1
\end{pmatrix}
$$
which means that the map
$$
f\colon G\to T,\qquad
\begin{pmatrix}
1&a&b\\
0&1&c\\
0&0&1
\end{pmatrix}\mapsto\begin{pmatrix}a\\c\end{pmatrix}
$$
is a homomorphism. What's the kernel of $f$?
|
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"timestamp": "2023-03-29T00:00:00",
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|
Show if $\sum_{n=1}^{\infty} \frac{(-1)^n}{1+\sqrt{n}}$ is absolute convergent or divergent Show if $\sum_{n=1}^{\infty} \frac{(-1)^n}{1+\sqrt{n}}$ is absolute convergent or divergent
First i subbed numbers in
$$\lim_{n \to \infty} \frac{(-1)^n}{1+\sqrt{n}} = \frac{-1}{1+\sqrt{1}} + \frac{1}{1+\sqrt{2}} - \frac{-1}{1+\sqrt{3}}$$
So it's divergent
1 LIMIT
$$\lim_{n \to \infty} \frac{1}{1+\sqrt{n}}=0 \quad \text{hence divergent} $$
2 $a_{n}$ and $a_{n+1}$
$$ \frac{1}{1+\sqrt{n}}>\frac{1}{1+\sqrt{n+1}} $$
$$ \frac{1}{1+\sqrt{n}}-\frac{1}{1+\sqrt{n+1}}> 0 $$
hence increasing
or
$$ \frac{1}{1+\sqrt{n}}-\frac{1}{1+\sqrt{n+1}} = \frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n+1}\sqrt{n+1}} $$
$$ \therefore \sqrt{n+1}-\sqrt{n}> 0$$ hence decreasing
$ \sum_{n=1}^{\infty} \left\lvert \frac{-1^n}{1+\sqrt{n}} \right\rvert = \sum_{n=1}^{\infty} \frac{1}{1+\sqrt{n}} $
$$ \sum_{n=1}^{\infty} \frac{1}{1+\sqrt{n}} = \frac{1}{1+\sqrt{1}} + \frac{1}{1+\sqrt{2}} + \frac{1}{1+\sqrt{3}} $$
if this is convergent then the whole thing is absolutely convergent but i don't know how to prove this
UPDATE
I got something out for the second part
|
$$\sum_{n=1}^{\infty}\left|\frac{(-1)^n}{1+\sqrt{n}}\right|=\sum_{n=1}^{\infty} \frac1{1+\sqrt{n}}>\frac12\sum_{n=1}^{\infty} \frac1{\sqrt{n}}$$ is divergent.
|
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|
How to determine the Laurent expansion of $\tan{z}$ around $z=0$ that is convergent in $z=\pi$ I want to determine the Laurent expansion of $\tan{z}$ around $z=0$ that is convergent in $z=\pi$ (only the first couple of terms). Now I know that if $\sum_{n=-\infty}^{\infty}c_nz^n$ then
$$c_n=\frac{1}{2\pi i}\int_K \frac{\tan(z)}{z^{n+1}}$$
Where I need to choose $K$ in the annulus $\frac{\pi}{2}<|K(t)|<\frac{3\pi}{2}$. This means that
$$c_n=\text{Res}_{z=0}\frac{\tan(z)}{z^{n+1}}+\text{Res}_{z=-\frac{\pi}{2}}\frac{\tan(z)}{z^{n+1}}+\text{Res}_{z=\frac{\pi}{2}}\frac{\tan(z)}{z^{n+1}}$$
However I'm not sure how I can determine these residues, without already knowing the Laurent expansion of $\tan(z)$. Only the residue in $z=0$ is easy, since the 'standard' Taylor series of $\tan(z)$ is valid there. However, in $z=\pm\frac{\pi}{2}$ I have no clue of how I could approach this.
Thanks
Edit:
Using Mathematica I found out that
$$\text{Res}_{z=-\frac{\pi}{2}}\frac{\tan(z)}{z^{n+1}} = (-1)^{1-n}\left(\frac{2}{\pi}\right)^n$$
$$\text{Res}_{z=\frac{\pi}{2}}\frac{\tan(z)}{z^{n+1}} = -\left(\frac{2}{\pi}\right)^n$$
But I have no Idea how it finds this. Any help would be much appreciated.
|
Oke so I decided to just brute-force this for the residu in $z=\frac{\pi}{2}$ ($z=-\frac{\pi}{2}$ is similar)
$$\begin{align}
(z-\frac{\pi}{2})\tan{z-\frac{\pi}{2}}&=(z-\frac{\pi}{2})\frac{\sin{z-\frac{\pi}{2}}}{\cos{z-\frac{\pi}{2}}}\\
&=(z-\frac{\pi}{2})\frac{1-\frac{1}{2!}(z-\frac{\pi}{2})^2+\frac{1}{4!}(z-\frac{\pi}{2})^4\cdots}{-(z-\frac{\pi}{2})+\frac{1}{3!}(z-\frac{\pi}{2})^3-\frac{1}{5!}(z-\frac{\pi}{2})^5\cdots}\\
&=-\frac{1-\frac{1}{2!}(z-\frac{\pi}{2})^2+\frac{1}{4!}(z-\frac{\pi}{2})^4\cdots}{1-\lbrack\frac{1}{3!}(z-\frac{\pi}{2})^2-\frac{1}{5!}(z-\frac{\pi}{2})^4\cdots\rbrack}\\
&=-\lbrack 1-\frac{1}{2!}(z-\frac{\pi}{2})^2+\frac{1}{4!}(z-\frac{\pi}{2})^4\cdots\rbrack\lbrack 1+ \left(\frac{1}{3!}(z-\frac{\pi}{2})^2-\frac{1}{5!}(z-\frac{\pi}{2})^4\cdots\right)+\left(\frac{1}{3!}(z-\frac{\pi}{2})^2-\frac{1}{5!}(z-\frac{\pi}{2})^4\cdots\right)^2+\cdots\rbrack\\
&=-1+\left(\frac{1}{2!}-\frac{1}{3!}\right)(z-\frac{\pi}{2})^2\cdots
\end{align}$$
So we find that
$$\tan{z}=-\frac{1}{z-\frac{\pi}{2}}+\cdots$$
Which is all we really need to know, since it's easy to see that
$$\frac{1}{z^{n+1}}=\left(\frac{2}{\pi}\right)^{n+1}+\cdots$$
And so we can finally conclude that indeed
$$\text{Res}_{z=\frac{\pi}{2}}\frac{\tan{z}}{z^{n+1}}=-\frac{2^{n+1}}{\pi^{n+1}}$$
And from here is easily observed that
$$\text{Res}_{z=-\frac{\pi}{2}}\frac{\tan{z}}{z^{n+1}}=(-1)^n\frac{2^{n+1}}{\pi^{n+1}}$$
And
So we find for the first few coefficients:
$$c_{-3}=-\frac{\pi^2}{2}$$
$$c_{-1}=-2$$
$$c_{1}=1-\frac{8}{\pi^2}$$
$$c_{3}=\frac{1}{3}-\frac{32}{\pi^4}$$
$$c_{5}=\frac{2}{15}-\frac{128}{\pi^6}$$
Where the positive fractions come from the $\text{Res}_{z=0}\frac{\tan{z}}{z^{n+1}}$ term (which are easier to determine, since we can just use the standard taylor series for $\tan$ those terms).
|
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|
Mistake with Integration with Beta, Gamma, Digamma Fuctions Problem: Evaluate:
$$I=\int_0^{\pi/2} \ln(\sin(x))\tan(x)dx$$
I tried to attempt it by using the Beta, Gamma and Digamma Functions. My approach was as follows:
$$$$
Consider $$I(a,b)=\int_0^{\pi/2} \sin^a(x)\sin^b(x)\cos^{-b}(x)dx$$
$$$$
$$=\dfrac{1}{2}\beta\bigg (\dfrac{a+b+1}{2},\dfrac{1-b}{2}\bigg )= \dfrac{1}{2} \dfrac{\Gamma(\frac{a+b+1}{2})\Gamma(\frac{1-b}{2})}{\Gamma(\frac{a+2}{2})}$$ Now,
$$\dfrac{1}{2}\dfrac{\partial}{\partial a} \beta\bigg (\dfrac{a+b+1}{2},\dfrac{1-b}{2}\bigg )\bigg |_{a=0,b=1} = \int_0^{\pi/2}\ln(\sin(x))\sin^a(x)\tan^b(x)dx \bigg |_{a=0,b=1} = I$$
Now,
$$\dfrac{\partial}{\partial a} \beta\bigg (\dfrac{a+b+1}{2},\dfrac{1-b}{2}\bigg )$$
$$$$
$$=\dfrac{1}{2} \dfrac{\Gamma(\frac{1-b}{2})}{(\Gamma(\frac{a+2}{2}))^2}\bigg (\Gamma '\bigg (\frac{a+b+1}{2}\bigg )\Gamma \bigg ( \frac{a+2}{2}\bigg ) - \Gamma '\bigg ( \frac{a+2}{2}\bigg )\Gamma \bigg (\frac{a+b+1}{2}\bigg )\bigg ) $$
$$$$
$$= \dfrac{1}{2}\dfrac{\Gamma(\frac{1-b}{2})}{\Gamma(\frac{a+2}{2})}\bigg (\psi\bigg ( \frac{a+b+1}{2}\bigg )\Gamma \bigg (\frac{a+b+1}{2}\bigg ) - \psi\bigg ( \frac{a+2}{2}\bigg )\Gamma \bigg (\frac{a+b+1}{2}\bigg )\bigg )$$
$$$$
$$=\dfrac{1}{2} \dfrac{\Gamma(\frac{a+b+1}{2})\Gamma(\frac{1-b}{2})}{\Gamma(\frac{a+2}{2})}\bigg (\psi\bigg ( \frac{a+b+1}{2}\bigg )-\psi\bigg ( \frac{a+2}{2}\bigg )\bigg )$$
$$$$
$$\Rightarrow\dfrac{\partial}{\partial a} \beta\bigg (\dfrac{a+b+1}{2},\dfrac{1-b}{2}\bigg )=\dfrac{1}{2} \dfrac{\Gamma(\frac{a+b+1}{2})\Gamma(\frac{1-b}{2})}{\Gamma(\frac{a+2}{2})}\bigg (\psi\bigg ( \frac{a+b+1}{2}\bigg )-\psi\bigg ( \frac{a+2}{2}\bigg )\bigg )$$
$$$$
$$\Longrightarrow \dfrac{1}{2} \dfrac{\partial}{\partial a} \beta \bigg ( \dfrac{a+b+1}{2} ,\dfrac{1-b}{2} \bigg ) \bigg |_{a=0,b=1} = I $$
$$$$
$$=\dfrac{1}{2}\times \dfrac{1}{2} \dfrac{\Gamma(\frac{a+b+1}{2})\Gamma(\frac{1-b}{2})}{\Gamma(\frac{a+2}{2})}\bigg (\psi\bigg ( \frac{a+b+1}{2}\bigg )-\psi\bigg ( \frac{a+2}{2}\bigg )\bigg ) \bigg |_{a=0,b=1}$$
$$$$
$$ =\dfrac{1}{4}\dfrac{\Gamma(0)\Gamma(1)}{\Gamma(1)}\bigg (\psi\bigg ( 1 \bigg )-\psi\bigg (1\bigg )\bigg )$$
$$$$
Could somebody please be so kind as to tell me where I have gone wrong? I would be truly grateful for your assistance. Thanks very, very much in advance!
|
OK, I realize this does not really answer your question (but look at the update below), in the sense it does not point at your error. Anyways, I think that you should be flexible with your methods doing integrals, so here is how it can be done with the change of variables I suggested:
You end up with the integral
$$
I=\int_0^1 \log u\frac{u}{1-u^2}\,du.
$$
Now writing
$$
\frac{1}{1-u^2}=1+u^2+u^4+\cdots,
$$
and using that (integrating by parts)
$$
\begin{aligned}
\int_0^1 u^{2k+1}\log u\,du&=\bigl[\frac{u^{2k+2}}{2k+2}\log u\bigr]_0^1-\frac{1}{2k+2}\int_0^1 u^{2k+2}\frac{1}{u}\,du\\
&=-\frac{1}{4}\frac{1}{(k+1)^2}.
\end{aligned}
$$
Thus,
$$
I=-\frac{1}{4}\sum_{k=0}^{+\infty}\frac{1}{(k+1)^2}=-\frac{\pi^2}{24}.
$$
The last equality by the Basel problem.
Updated version
The problem in your calculation, by inserting $a=0$ and $b=1$, you get a $\Gamma(0)$ (which is undefined, or as its best $\pm\infty$ depending on if you let $b\to 1$ from left or right) times something that is zero. This is indeterminate, and you must look at limits. One way is as follows:
If you first let $a=0$, and write $b=1+\epsilon$, then you get
$$
\frac{1}{4}\Gamma(-\epsilon/2)\Gamma(1+\epsilon/2)\bigl(\Psi(1+\epsilon/2)-\Psi(1)\bigr)
$$
We multiply and divide by $\epsilon/2$ (lethal weapon number 2), to write this as
$$
\frac{1}{4}\frac{\epsilon}{2}\Gamma(-\epsilon/2)\Gamma(1+\epsilon/2)\frac{\Psi(1+\epsilon/2)-\Psi(1)}{\epsilon/2}
$$
By letting $\epsilon\to 0$, you will find that the limit is
$$
\frac{1}{4}\times (-1)\times 1 \times \frac{\pi^2}{6}=-\frac{\pi^2}{24}.
$$
Here, we have used the facts that $\lim_{x\to 0} x\Gamma(x)=1$ and $\Psi^{(1)}(1)=\pi^2/6$.
|
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|
Use Newton’s method to find the solution of $2x^3+5x-6=0$ by using the following steps Use Newton’s method to find the solution of $$2x^3+5x-6=0$$
by using the following steps
a) Find, $f(0)$ where $$f(x)=2x^3+5x-6$$
b) Find, $f(1)$ where $$f(x)=2x^3+5x-6$$
c) Choose a guess point $$x_0$$
d) Perform the iterations, until the values stabilize to the 8-digits after the decimal sign.
my work
for a)
newton formula
$$x_{n+1} = x_n -\frac{f(x_n )}{f'(x_n )} $$ , for $n = 0,1,2,3,...$
$$f' (x)=6x^2+5$$
$$x_1=x_0-\frac{-6}{5} $$
$$ F(0)= x_0- \frac{-6}{5} $$
for b)
$$x_2=x_1-\frac{-6}{5} $$
for c)
Let $$x_0=1$$
$$x_1=1-\frac{-6}{5}=\frac{11}{5}$$
for d)
$$x_1=x_0-\frac{-6}{5}$$
$$x_2=x_1-\frac{-6}{5}$$
$$x_3=x_2-\frac{-6}{5}$$
$$x_4=x_3-\frac{-6}{5}$$
$$x_5=x_4-\frac{-6}{5}$$
is that correct ???
|
As you wrote, Newton method updates the initial guess $x_0$ according to $$x_{n+1} = x_n -\frac{f(x_n )}{f'(x_n )}$$ So, if $f(x)=2 x^3+5 x-6$, $f'(x)=6 x^2+5$, this gives, after simplification $$x_{n+1} =\frac{4 x_n^3+6}{6 x_n^2+5}$$ As AjmalW answered, choose $x_0=1$ and start repeating the iterations. You will find $x_1=\frac{10}{11}$, $x_2=\frac{11986}{13255}$ and so on until the convergence criteria is met.
|
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|
High School Trigonometry ( Law of cosine and sine) I am preparing for faculty entrance exam and this was the question for which I couldn't find the way to solve (answer is 0). I guess they ask me to solve this by using the rule of sine and cosine:
Let $\alpha$, $\beta$ and $\gamma$ be the angles of arbitrary triangle with sides a, b and c respectively. Then $${b - 2a\cos\gamma \over a\sin\gamma} + {c-2b\cos\alpha \over b\sin\alpha} + {a - 2c\cos\beta \over c\sin\beta}$$ is equal to (answer is zero but I need steps).
|
The Law of Cosines is equivalent to (and is often proven via) the statements
$$a = b \cos\gamma + c\cos\beta \qquad b = c \cos\alpha + a \cos\gamma \qquad c = a \cos\beta + b \cos\alpha$$
so, your sum becomes
$$\frac{c \cos\alpha - a \cos\gamma}{a\sin\gamma} + \frac{a \cos\beta - b \cos\alpha}{b\sin\alpha} + \frac{b \cos\gamma - c \cos\beta}{c\sin\beta}$$
Further, the Law of Sines allows us to write
$$a = d \sin\alpha \qquad b = d\sin\beta \qquad c = d \sin\gamma$$
where $d$ is the triangle's circumdiameter. This gives
$$\frac{\sin\gamma \cos\alpha - \sin\alpha \cos\gamma}{\sin\alpha\sin\gamma} + \frac{\sin\alpha \cos\beta - \sin\beta \cos\alpha}{\sin\beta\sin\alpha} + \frac{\sin\beta \cos\gamma - \sin\gamma \cos\beta}{\sin\gamma\sin\beta}$$
$$= \left( \cot\alpha - \cot\gamma \right) + \left( \cot\beta - \cot\alpha \right) + \left( \cot\gamma - \cot\beta \right ) = 0 $$
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"timestamp": "2023-03-29T00:00:00",
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|
Probability that on three rolls of dice, there will be at least one 6 showing up? What is the probability that on three rolls of dice, there will be at least one 6 showing up?
Attempt:
Since there can be one six or two sixes or three sixes on three rolls, I considered separate cases and added them up.
So $(1/6)(5/6)(5/6) + (1/6)(1/6)(5/6) + (1/6)(1/6)(1/6) = 31/216$, but answer is incorrect as per book. Can anyone suggest where I am wrong ?
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You have forgotten to include th fact that any six can occur in any of the rolls, i.e. you have multiply the probabilities for exactly one and two sixs by three, giving
$$P=3\cdot\frac{1}{6}\cdot\frac{5}{6}\cdot\frac{5}{6}+3\cdot\frac{1}{6}\cdot\frac{1}{6}\cdot\frac{5}{6}+\frac{5}{6}\cdot\frac{5}{6}\cdot\frac{5}{6}=\frac{91}{216}.$$
It may be easier to compute the complementary probability of no six showing up: $$P^c=\left(\frac{5}{6}\right)^3=\frac{125}{216}$$
and using $$P=1-P^c=\frac{216-125}{216}=\frac{91}{216}$$
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/1327201",
"timestamp": "2023-03-29T00:00:00",
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|
Given $x^2 + 4x + 6$ as factor of $x^4 + ax^2 + b$, then $a + b$ is I got this task two days ago, quite difficult for me, since I have not done applications of Vieta's formulas and Bezout's Theorem for a while. If can someone solve this and add exactly how I am supposed to use these two theorem's on this task, I would be thankful.
Given $x^2 + 4x + 6$ as factor of $x^4 + ax^2 + b$, then $a + b$ is equal to?
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The first solution I came up with involves actually finding the roots of the quadratic, then employing Factor/Remainder theorem, so while it is a tad more tedious, it's also very direct. I will follow with an alternative solution that is more elegant.
By the quadratic formula, roots of $x^2 + 4x + 6$ are the complex conjugate pair $-2 \pm i\sqrt 2$.
By Factor Theorem, those will also be roots of the quartic (biquadratic), giving the linear simultaneous equations:
$(-2 + i\sqrt 2)^4 + a(-2 + i\sqrt 2)^2 + b = 0$
$(-2 - i\sqrt 2)^4 + a(-2 - i\sqrt 2)^2 + b = 0$
Now $(-2 \pm i\sqrt 2)^2 = 2 \mp 4i\sqrt 2$
Squaring again,
$(-2 \pm i\sqrt 2)^4 = -28 \mp 16i \sqrt 2$
which allows you to express the simultaneous equations as:
$(-28 - 16i\sqrt 2) + a(2 - 4i\sqrt 2) + b = 0$
$(-28 + 16i\sqrt 2) + a(2 + 4i\sqrt 2) + b = 0$
by subtraction, we almost immediately have $a(8i\sqrt 2) = -32i\sqrt 2 \implies a = -4$
By adding and back substitution, we get: $-56 + 4(-4) + 2b = 0 \implies b = 36$
And of course, to answer the original question, $a+b= 32$
Alternative solution:
Let the roots of the original quadratic be $x_1, x_2$.
Vieta's formulas give:
$x_1 + x_2 = -4$
$x_1x_2 = 6$
from which we can deduce that:
$x_1^2 + x_2^2 = (x_1 + x_2)^2 - 2x_1x_2 = 4$
and
$x_1^2x_2^2 = 36$
Now, by Factor Theorem, $x_1$ and $x_2$ are also roots of the biquadratic.
If we let $y = x^2$, the biquadratic can be written as:
$y^2 + ay + b = 0$
which will have roots $y_1$ and $y_2$ where $y_1 = x_1^2$ and $y_2 = x_2^2$
Applying Vieta's formula to the quadratic in $y$ allows us to deduce that:
$y_1 + y_2 = -a \implies x_1^2 + x_2^2 = -a$
and
$y_1y_2 = b \implies x_1^2x_2^2 = b$
and by reference to the above results, that allows us to immediately conclude that $a = -4, b = 36$ and $a+b = 32$.
This solution involves Vieta's formulas, and is less tedious than my original method.
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|
Differential equation $y'' \cdot y^3 = 1$ I use these substitutions $y'=p(y)$ and $y'' = p' \cdot p$ to solve the equation, thus I have the consequence of the solution's steps:
$$ p'py^3 = 1 \implies p'p = \frac{1}{y^3} \implies \frac {dp}{dy} p = \frac {1}{y^3} \implies \int p dp = \int \frac{1}{y^3} dy \implies \\ p = \sqrt{C_1 - \frac {1}{y^2}}$$
Then I try to reverse my substitution and get:
$$ y' = \sqrt{C_1 - \frac {1}{y^2}}$$
What kind of differential equations is it? How do I have to solve it?
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Multiply both sides by $2y'/y^3$. Then
$$ 2y'y'' = \frac{y'}{y^3}, $$
and we can recognise both sides as derivatives. Integrating once,
$$ y'^2 = A^2-\frac{1}{2y^2} = \frac{A^2 y^2-1}{y^2}, $$
and then you can rearrange to
$$ \frac{y^2y'^2}{A^2 y^2-1} = 1, $$
and then take a square root and integrate:
$$ x_0 \pm x = \int \frac{y \, dy}{\sqrt{A^2 y^2-1}} = \frac{1}{A^2} \sqrt{A^2y^2-1}, $$
which you can rearrange to find $A^4 (x_0 \pm x )^2 + 1 = A^2 y^2$, (which is a couple of hyperbolae).
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"timestamp": "2023-03-29T00:00:00",
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Find the cubic equation of $x=\sqrt[3]{2-\sqrt{3}}+\sqrt[3]{2+\sqrt{3}}$ Find the cubic equation which has a root $$x=\sqrt[3]{2-\sqrt{3}}+\sqrt[3]{2+\sqrt{3}}$$
My attempt is
$$x^3=2-\sqrt{3}+3\left(\sqrt[3]{(2-\sqrt{3})^2}\right)\left(\sqrt[3]{(2+\sqrt{3})}\right)+3\left(\sqrt[3]{(2-\sqrt{3})}\right)\left(\sqrt[3]{(2+\sqrt{3})^2}\right)+2-\sqrt{3}$$
$$x^3=4+\left(\sqrt[3]{(2-\sqrt{3})^2}\right)\left(\sqrt[3]{(2+\sqrt{3})}\right)+3\left(\sqrt[3]{(2-\sqrt{3})}\right)\left(\sqrt[3]{(2+\sqrt{3})^2}\right)$$
then what I will do??
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Identifying with Cardano's formula
$$x=\sqrt[3]{-\frac q2-\sqrt{\frac{q^2}4+\frac{p^3}{27}}}+\sqrt[3]{-\frac q2+\sqrt{\frac{q^2}4+\frac{p^3}{27}}},$$
we find $q=-4$, then $p=-3$.
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/1331417",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Maximum value of trigonometric expression If
$r=3+\tan c \tan a, \quad
q=5+\tan b \tan c, \quad
p=7+\tan a \tan b$
Provided $a,b,c$ are positive and
$a+b+c=\dfrac{\pi}2$
Find the maximum value of
$\sqrt p + \sqrt q + \sqrt r$ .
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Let $\sqrt p=P$ etc.
$$R^2-3+Q^2-5+P^2-7=1\iff P^2+Q^2+R^2=16$$
Let $P=4\cos A,Q=4\sin A\cos B,R=4\sin A\sin B$
$P+Q+R=4[\cos A+\sin A(\cos B+\sin B)]\le4[\cos A+\sqrt2\sin A]$ if $\sin A\ge0$
Now $\cos A+\sqrt2\sin A=\sqrt3\cos\left(A-\arccos\dfrac1{\sqrt3}\right)\le\sqrt3$
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"timestamp": "2023-03-29T00:00:00",
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|
Finding flux over a surface If $F = 6z\mathbf i + (2x+y)\mathbf j -x\mathbf k$, evaluate $\int_S \mathbf F \cdot \mathbf n ds$ over the surface bounded by the cylinder $x^2 + z^2 = 9. x=0, y=0, z=0 $ and $ y= 6$
Okay, so I know this is a quarter cylinder in the first octant. Now assuming the surface they're talking about is $z = \sqrt{ 9 - x^2 } $, here's what I've got so far.
$ n = \nabla f $ where $f(x,y,z)= x^2 + z^2 - 9 = 0$ which yields $<2x, 0, 2z>$
now for $\hat n$
$$ \hat n = \frac {<2x, 0, 2z>} {\sqrt {4x^2 + 4z^2 }}$$
and do get $ds$, I used $g(x,y) = \sqrt{ (9 - x^2) } $ where $g_x = \frac {-x}{\sqrt{9-x^2}}$ and $g_y= 0 $
therefore $$ds = \sqrt{ \frac {9}{9-x^2} } = \frac {3}{\sqrt{ {9-x^2} }} dA$$
Finally $$\int_S\mathbf F\cdot \hat n ds = <6z, 2x+4, -x> \cdot \frac {<2x, 0, 2z>} {\sqrt {4x^2 + 4z^2 }}\frac {3}{\sqrt{ {9-x^2} } } $$
$$ = \int\int_R \frac {10x \sqrt{ {9-x^2} } } {\sqrt {4x^2 + 4z^2 }}\frac {3}{\sqrt{ {9-x^2} } }dA $$ OR
$$ = \int\int_R \frac {15x } {\sqrt {x^2 + z^2 } }dxdy = \int_0^6\int_0^3 5x dx dy$$
My question is, is this setup correct or is any of this wrong?
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For a direct computation, note that your surface is the union of the five surfaces parameterized by
\begin{align*}
\mathbf r(x,y)&=\left\langle x,y,\sqrt{9-x^2}\right\rangle & (x,y)&\in[0,3]\times[0,6] \\
\mathbf s(x,y) &= \langle x,y,0\rangle & (x,y)&\in[0,3]\times[0,6] \\
\mathbf t(y,z) &=\langle0,y,z\rangle & (y,z)&\in[0,6]\times[0,3] \\
\mathbf u(r,\theta) &= \langle r\cos\theta,0,r\sin\theta\rangle & (r,\theta)&\in[0,3]\times[0,\pi/2] \\
\mathbf v(r,\theta) &= \langle r\cos\theta,6,r\sin\theta\rangle & (r,\theta)&\in[0,3]\times[0,\pi/2]
\end{align*}
Our computation is then
\begin{align*}
\iint_S(\mathbf F\cdot\mathbf n)\,dS
&= \int_0^6\int_0^3\mathbf F\bigl(\mathbf r(x,y)\bigr)\cdot(\mathbf r_x\times\mathbf r_y)\,dx\,dy \\
&\qquad+\int_0^6\int_0^3\mathbf F\bigl(\mathbf s(x,y)\bigr)\cdot(\mathbf s_y\times\mathbf s_x)\,dx\,dy \\
&\qquad\qquad+\int_0^3\int_0^6\mathbf F\bigl(\mathbf t(y,z)\bigr)\cdot(\mathbf t_z\times\mathbf t_y)\,dy\,dz \\
&\qquad\qquad\qquad+\int_0^{\pi/2}\int_0^3\mathbf F\bigl(\mathbf u(r,\theta)\bigr)\cdot(\mathbf u_r\times\mathbf u_\theta)\,dr\,d\theta \\
&\qquad\qquad\qquad\qquad+\int_0^{\pi/2}\int_0^3\mathbf F\bigl(\mathbf v(r,\theta)\bigr)\cdot(\mathbf v_\theta\times\mathbf u_r)\,dr\,d\theta \\
&= \int_0^6\int_0^3\left\langle6\sqrt{9-x^2},2\,x+y,-x\right\rangle\cdot\left\langle \frac{x}{\sqrt{9-x^2}},0,1\right\rangle\,dx\,dy \\
&\qquad+\int_0^6\int_0^3\langle0,2\,x+y,-x\rangle\cdot\langle0,0,-1\rangle\,dx\,dy \\
&\qquad\qquad+\int_0^3\int_0^6\langle6\,z,y,0\rangle\cdot\langle-1,0,0\rangle\,dy\,dz \\
&\qquad\qquad\qquad+\int_0^{\pi/2}\int_0^{3}\langle6r\sin\theta,2r\cos\theta,-r\cos\theta\rangle\cdot\langle0,-r,0\rangle\,dr\,d\theta \\
&\qquad\qquad\qquad\qquad+\int_0^{\pi/2}\int_0^{3}\langle6r\sin\theta,2r\cos\theta+6,-r\cos\theta\rangle\cdot\langle0,r,0\rangle\,dr\,d\theta \\
&= \underbrace{5\int_0^6\int_0^3x\,dx\,dy+\int_0^6\int_0^3x\,dx\,dy-6\int_0^3\int_0^6z\,dy\,dz}_{=0} \\
&\qquad+\int_0^{\pi/2}\int_0^3-2r^2\cos\theta\,dr\,d\theta+\int_0^{\pi/2}\int_0^3(2 r^2\cos\theta+6r)\,dr\,d\theta \\
&= 6\int_0^{\pi/2}\int_0^3r\,dr\,d\theta \\
&= 6\left(\frac{9}{2}\right)\int_0^{\pi/2}d\theta \\
&= 6\left(\frac{9}{2}\right)\left(\frac{\pi}{2}\right) \\
&=\frac{27}{2}\pi
\end{align*}
The problem is considerably easier if we use the divergence theorem. We are asked to compute
$$
\iint_{S}(\mathbf F\cdot\mathbf n)\,dS
$$
where $\mathbf F(x,y,z)=\langle 6\,z,2\,x+y,-x\rangle$ and $S$ is the quarter cylinder you describe. The divergence theorem states that
$$
\iint_{S}(\mathbf F\cdot\mathbf n)\,dS=\iiint_T(\nabla\cdot\mathbf F)\,dV
$$
where $T$ is the solid enclosed by $S$ and $\displaystyle\nabla=\left\langle\frac{\partial}{\partial x},\frac{\partial}{\partial y},\frac{\partial}{\partial z}\right\rangle$. Our computation is then
\begin{align*}
\iint_{S}(\mathbf F\cdot\mathbf n)\,dS
&=\iiint_T(\nabla\cdot\mathbf F)\,dV \\
&=\iiint_T\left\langle\frac{\partial}{\partial x},\frac{\partial}{\partial y},\frac{\partial}{\partial z}\right\rangle\cdot\langle 6\,z,2\,x+y,-x\rangle\,dV \\
&= \iiint_T(0+1+0)\,dV \\
&= \iiint_TdV \\
&= \text{volume of }T \\
&= \frac{1}{4}\pi\cdot 3^2\cdot 6 \\
&= \frac{27}{2}\pi
\end{align*}
This verifies that our direct computation was, indeed, correct.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Complex Integration with trignometric function
Verify that
$\int_0^{\frac{\pi}{2}}\frac{d\theta}{a+\sin^2\theta}=\frac{\pi}{2[(a(a+1)]^\frac{1}{2}}$
I know that $\sin\theta=\frac{e^{i\theta}-e^{-i\theta}}{2}$ then I did
$$\int_0^{\frac{\pi}{2}}\frac{d\theta}{a+\sin^2\theta}=\int_0^{\frac{\pi}{2}}\frac{d\theta}{a+\frac{1}{4}(e^{i\theta}-e^{-i\theta})^2}=\int_0^{\frac{\pi}{2}}\frac{4d\theta}{4a+(e^{i\theta}-e^{-i\theta})^2}$$
I do not know if this is right but I'm really stuck
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NOTE 1:
First note that then integral diverges if $-1\le a\le 0$. So, we will assume that either $a>0$ or $a<-1$.
NOTE 2:
We will simplify things quite a bit by first invoking the identity
$$\sin^2x=\frac{1-\cos 2x}{2} \tag 1$$
By simplifying first, we will reduce the denominator of a contour integral from a $4$th degree polynomial to a $2$nd degree one!
SIMPLIFYING THE INTEGRAL
Using the identity $(1)$, the integral of interest $I\equiv \int_0^{\pi/2}\frac{1}{a+\sin^2x}dx$ simplifies to
$$\begin{align}
I&\equiv \int_0^{\pi/2}\frac{1}{a+\sin^2x}dx\\\\
&=\int_0^{\pi/2}\frac{2}{(2a+1)-\cos 2x}dx\\\\
&=\int_0^{\pi}\frac{1}{(2a+1)-\cos x}dx\\\\
&=\frac12\int_0^{2 \pi}\frac{1}{(2a+1)-\cos x}dx
\end{align}$$
COMPLEX-PLANE ANALYSIS
Now, we will move to the complex plane. Let $C$ be the unit circle $|z|=1$ in the complex plane. Then, let $z=e^{ix}$ so that $dz=ie^{ix}dx$, and $\cos x = \frac12 (z+z^{-1})$. This gives
$$\begin{align}
I&=\frac12 \oint_C \frac{1}{(2a+1)-\frac12(z+z^{-1})}\frac{dz}{iz}\\\\
&=i \oint_C \frac{1}{z^2-2(2a+1)z+1}dz
\end{align}$$
The roots of $z^2-2(2a+1)+1=0$ are $z=(2a+1)\pm 2\sqrt{a(a+1)}$. If $a>0$, then the root $(2a+1)-2\sqrt{a(a+1)}<1$ and the root $(2a+1)+2\sqrt{a(a+1)}>1$. If $a<-1$, then $(2a+1)+2\sqrt{a(a+1)}<1$ and $(2a+1)-2\sqrt{a(a+1)}>1$.
CASE 1:
For $a>0$, the residue of $i \frac{1}{z^2-2(2a+1)z+1}$ is given by
$$\text{Res}\left(i \frac{1}{z^2-2(2a+1)z+1},z=(2a+1)-2\sqrt{a(a+1)}\right)=i\frac{1}{-4\sqrt{a(a+1)}}$$
and the value of the integral $I$ is
$$\bbox[5px,border:2px solid #C0A000]{I=\int_0^{\pi/2}\frac{1}{a+\sin^2x}dx=\frac{\pi}{2\sqrt{a(a+1)}}}$$
as expected!
CASE 2:
For $a<-1$, the residue of $i \frac{1}{z^2-2(2a+1)z+1}$ is given by
$$\text{Res}\left(i \frac{1}{z^2-2(2a+1)z+1},z=(2a+1)+2\sqrt{a(a+1)}\right)=i\frac{1}{4\sqrt{a(a+1)}}$$
and the value of the integral $I$ is
$$\bbox[5px,border:2px solid #C0A000]{I=\int_0^{\pi/2}\frac{1}{a+\sin^2x}dx=-\frac{\pi}{2\sqrt{a(a+1)}}}$$
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|
Compute definite integral Question: Compute
$$\int_0^1 \frac{\sqrt{x-x^2}}{x+2}dx.$$
Attempt: I've tried various substitutions with no success. Looked for a possible contour integration by converting this into a rational function of $\sin\theta$ and $\cos \theta$.
|
Let us make the problem a bit more general considering $$I=\int \frac{\sqrt{x-x^2}}{x+a}dx$$ and let us apply the same approach Mario G proposed. We then arrive to $$\begin{align*}
I=\int\frac{\sqrt{x-x^2}}{x+2}dx
&=\frac{1}{2}\int\frac{\cos^2( t)}{\sin t+(2a+1)}\,dt\\
\end{align*}$$ Now, let use the tangent half-angle substitution $y=\tan(\frac t2)$
$$I=\int\frac{ \left(y^2-1\right)^2}{\left(y^2+1\right)^2 \left((2 a+1) y^2+2
y+(2 a+1)\right)}\,dy$$ Now, partial fraction decomposition which gives for the integrand $$\frac{2 a+1}{y^2+1}-\frac{2
y}{\left(y^2+1\right)^2}-\frac{4 a\left(a+1\right)}{(2 a+1)y^2+2y+(2 a+1) }$$
Integrating the different species, we then arrive to $$I=(2 a+1) \tan ^{-1}(y)+\frac{1}{y^2+1}-2 \sqrt{a(a+1)} \tan ^{-1}\left(\frac{(2 a+1) y+1}{2 \sqrt{a(a+1)}}\right)$$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Arithmetc Progression containing odd terms Here is the question:
If for an AP of odd number of terms,the sum of all the terms is $\frac{15}{8}$ times the sum of the terms in odd places then find the number of terms in the AP.
my try:First of all i thought that in an odd AP there will be (2n+1) terms becuase (2n) terms will contain even number of terms where as there will be n+1 odd terms.
Then I equated the following equation
$\frac{2n+1}{2}[2a + 2nd]=\frac{15}{8}*\frac{n+1}{2}[2a+nd]$
But then I wondered there are three unknowns and one equation ,then how can i solve for $n$.So my question is that how can we solve for $n$ or $2n+1$ ?
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First consider a constant progression, $t_n=1$. Then the ratio of the sums is $\dfrac{2n}{n+1}=\dfrac{15}8$ so that $\color{green}{n=15}$.
$$\frac{1+1+1+1+1+1+1+1+1+1+1+1+1+1+1}{1+1+1+1+1+1+1+1}=\frac{15}8.$$
This is compatible with a linear progression $t_n=n$, as
$$\frac{1+2+3+4+5+6+7+8+9+10+11+12+13+14+15}{1+3+5+7+9+11+13+15}=\frac{120}{64}=\frac{15}8.$$
By linearity, the property extends to any $t_n=a+bn$.
|
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"timestamp": "2023-03-29T00:00:00",
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.