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Closed form for a sum involving binomial coefficient $\sum_{j=0}^n \binom{n}{j} \frac1{j+1} = \frac{2^{n+1}-1}{n+1}$
Possible Duplicate:
How can I compute $\sum\limits_{k = 1}^n \frac{1} {k + 1}\binom{n}{k} $?
How to derive the following equality?
$$\sum_{j=0}^n \binom{n}{j} \frac1{j+1} = \frac{2^{n+1}-1}{n+1}$$
|
Using the identity
$$
\binom{n}{j}\frac{n+1}{j+1}=\binom{n+1}{j+1}\tag{1}
$$
and summing
$$
\begin{align}
\sum_{j=0}^n\binom{n}{j}\frac{n+1}{j+1}
&=\sum_{j=0}^n\binom{n+1}{j+1}\\
&=2^{n+1}-1\tag{2}
\end{align}
$$
Dividing by $n+1$ yields
$$
\sum_{j=0}^n\binom{n}{j}\frac{1}{j+1}=\frac{2^{n+1}-1}{n+1}\tag{3}
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/87244",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
}
|
How to compute the characteristic polynomial of $A$ The matrix associated with $f$ is:
$$ \left(\begin{array}{rrr}
3 & -1 & -1 \\
-1 & 3 & -1 \\
-1 & -1 & 3
\end{array}\right) .
$$
First, I am going to find the characteristic equation of $\det(A- \lambda I)$. Please correct me if I'm wrong.
$$= (3-\lambda)(3-\lambda)(3-\lambda)-1-1-(3-\lambda)-(3-\lambda)-(3-\lambda)
=-\lambda^3+9\lambda^2-24\lambda+16 .$$
How to factor this? I know the $\lambda$'s should be $1$, $4$ and $4$. But how am I supposed to find these values?
|
This comes up pretty often, I think it is worth throwing in a bit more than you asked.
Suppose I have an $n$ by $n$ square matrix where all entries are equal to $1$:
$$ R_n \; \; = \; \;
\left( \begin{array}{ccccccc}
1 & 1 & 1 & 1 & 1 & 1 & 1 \\
1 & 1 & 1 & 1 & 1 & 1 & 1 \\
1 & 1 & 1 & 1 & 1 & 1 & 1 \\
1 & 1 & 1 & 1 & 1 & 1 & 1 \\
1 & 1 & 1 & 1 & 1 & 1 & 1 \\
1 & 1 & 1 & 1 & 1 & 1 & 1 \\
1 & 1 & 1 & 1 & 1 & 1 & 1
\end{array}
\right).
$$
What are the eigenvalues and eigenvectors? ( Here $n=7,$ I will keep typing $n$ because the exact value really does not matter).
Well, we get an automatic eigenvalue $n,$ with eigenvector (I use columns always)
$$
\left( \begin{array}{c}
1 \\
1 \\
1 \\
1 \\
1 \\
1 \\
1
\end{array}
\right).
$$
We then get $n-1$ genuine eigenvalues $0,$ with linearly independent eigenvectors
$$
\left( \begin{array}{c}
1 \\
-1 \\
0 \\
0 \\
0 \\
0 \\
0
\end{array}
\right) \;
\left( \begin{array}{c}
1 \\
0 \\
-1 \\
0 \\
0 \\
0 \\
0
\end{array}
\right) \;
\left( \begin{array}{c}
1 \\
0 \\
0 \\
-1 \\
0 \\
0 \\
0
\end{array}
\right) \;
\left( \begin{array}{c}
1 \\
0 \\
0 \\
0 \\
-1 \\
0 \\
0
\end{array}
\right) \;
\left( \begin{array}{c}
1 \\
0 \\
0 \\
0 \\
0 \\
-1 \\
0
\end{array}
\right) \;
\left( \begin{array}{c}
1 \\
0 \\
0 \\
0 \\
0 \\
0 \\
-1
\end{array}
\right) \;
$$
The complete list of eigenvalues is $\left\{n,0,\ldots,0 \right\}.$
What would I get for the eigenvalues of $k R_n,$ for a real number $k?$ With exactly the same eigenvectors, I would now have eigenvalues $\left\{kn,0,\ldots,0 \right\}.$ In your example the value of $k$ is $-1.$ All that has changed is the eigenvalue of the eigenvector with all 1's, which is now sent to $kn$ times itself.
What are the eigenvalues of $ w I + k R_n,$ with real numbers $w,k?$ With exactly the same (basis of) eigenvectors, we just add, for each eigenvector, $w$ times each. So the eigenvalues are $\left\{w + kn,w,\ldots,w \right\}.$ Again, these are genuine eigenvalues (the Jacobi normal form is diagonal, which we also know from symmetry). The determinant of the matrix is then $$ ( w + k n ) \; w^{n-1} = w^n + k \; n \, w^{n-1}.$$
When I have a basis of eigenvectors, and I know the list of eigenvalues, the characteristic polynomial is just the product of $n$ linear terms, each one given by $\left(\mbox{eigenvalue} - \lambda \right)$ in the order you are using. Here, your answer is just
$$ ( w + k n - \lambda) \; ( w - \lambda)^{n-1}.$$
Now, for your example, indeed $n=3.$
How to get the 3's on the diagonal? In order to get the $-1'$s off the diagonal, we must take $k=-1,$ as I mentioned. So we must add 4 to bump $-1$ up to 3, so that $w=4.$ Meanwhile $w + k n = 1$ and $n-1=2.$ Your characteristic polynomial is
$$ \det (A - \lambda I) = (1-\lambda) \; (4 - \lambda)^2.$$
Arturo and Didier (and all enlightened people) would want me to point out how spectacularly incorrect this method is for a matrix such as
$$
\left( \begin{array}{rr}
3 & 1 \\
0 & 3
\end{array}
\right) ,
$$
where the characteristic polynomial is easy enough to find but there is no basis of eigenvectors.
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/88324",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 1
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|
Simplifying rational fractions I can't get this one either for whatever reason, spent about 20 minutes on it and I have made no progress at all.
$$\frac{x^2}{(x^2-4)} - \frac{x+1}{x+2}.$$
I know that I can simplify this into one fraction so I make it $$\frac{x^2}{x+2}-\frac{(x+1)(x^2-4)}{(x^2-4)(x+2)}$$
I then can simplify it further making the $(x^2+4)$ into $(x-2)(x+2)$ and the $(x+2)$ into $(x-1)(x+1)$ but this does not help me get the answer. I know I have to manipulate it in some counter intuitive way but I can not make it work.
|
$$\frac{x^2}{x^2-4} - \frac{x+1}{x+2} = \frac{x^2}{(x-2)(x+2)}-\frac{x+1}{x+2}=\frac{x^2}{(x-2)(x+2)}-\frac{(x-2)(x+1)}{(x-2)(x+2)}.$$
Working with the numerators:
$$
x^2-(x-2)(x+1) = x^2 - (x^2 -x -2).
$$
Here's the easiest mistake to make (I've seen this happen zillions of times including in calculus courses):
Right: $x^2 - (x^2 -x -2) = x^2 - x^2 + x + 2$
Wrong: $x^2 - (x^2 -x -2) = x^2 - x^2 - x - 2$
The numerator ends up being $x+2$, so we get another simplification:
$$
\frac{x+2}{(x-2)(x+2)} = \frac{1}{x-2}.
$$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/92372",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
How can I evaluate $\int_{-\infty}^{\infty}\frac{e^{-x^2}(2x^2-1)}{1+x^2}dx$? How can I solve this integral: $$\int_{-\infty}^{\infty}\frac{e^{-x^2}(2x^2-1)}{1+x^2}dx.$$ Can I solve this problem using the Laplace transform? How can I do this?
|
To find $I=\int_{-\infty}^{\infty}\frac{e^{-x^2}(2x^2-1)}{1+x^2}dx$,
let us start by defining for $n \in \mathbb{N}$
$$
I_n = \int_{-\infty}^{\infty} x^{2n} e^{-x^2}\,dx
$$
and note that
$I_0=\sqrt{\pi}$, also called the Gaussian integral,
can be solved either by a beautiful polar coordinate trick
or by this technique, and that
$I_1=\frac{\sqrt{\pi}}{2}$ was shown
here. We can find the rest using
integration by parts. With
$$
\begin{array}{lcl}
v=-\frac{1}{2}e^{-x^2} && dv=xe^{-x^2}dx \\
u=x^{2n-1} && du=(2n-1)x^{2n-2}dx,
\end{array}
$$
for $n>1$ we have
$$
\begin{array}{lcl}
I_n &=& \int udv = uv - \int vdu \\
&=& -\frac{1}{2} \left[ x^{2n-1} e^{-x^2} \right]_{-\infty}^{\infty}
+ \frac{2n-1}{2} \int_{-\infty}^{\infty} x^{2n-2} e^{-x^2} \\
&=& \frac{2n-1}{2} I_{n-1}
\end{array}
$$
where the first right-hand term vanishes because
the exponential term dominates all polynomial terms.
This shows that, inductively,
$$
I_n = \frac{(2n)!}{2^{2n}n!} I_0
$$
But using $\frac{1}{1-t}=\sum_{n=0}^{\infty}t^n$,
we can expand the integrand of $I$:
$$
\frac{2x^2-1}{1+x^2} =
(2x^2-1) \sum_{n=0}^{\infty} (-1)^n x^{2n} =
2-3\sum_{n=0}^{\infty} (-1)^n x^{2n}
$$
and hence we get (noting that the integrals all
converge because of the dominating exponential term)
$$
\begin{array}{lcl}
I &=& 2I_0-3(I_0-I_1+\dots) = 2I_0-3 \sum_{n=0}^{\infty} (-1)^n I_n \\
&=& \left( 2-3 \sum_{n=0}^{\infty} (-1)^n \frac{(2n)!}{2^{2n}n!} \right) I_0
\end{array}
$$
where the series
$$
\sum_{n=0}^{\infty} (-1)^n \frac{(2n)!}{2^{2n}n!} =
1 - \frac{1}{2} + \frac{1 \cdot 3}{2 \cdot 2}
- \frac{1 \cdot 3 \cdot 5}{2 \cdot 2 \cdot 2}
+ \cdots = e \sqrt\pi\; \text{erfc}(1)
$$
can also be expressed using
$\Gamma(\frac{1}{2}-n)=(-1)^n\frac{(2n)!}{4^{n}n!}\sqrt\pi$
and a series expansion for the complementary error function
as
$$
I = 2\sqrt\pi - 3 \; \sum_{n=0}^{\infty} \; \Gamma(\tfrac{1}{2}-n)
= 2\sqrt\pi - 3e\pi\;\text{erfc}(1) \;.
$$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/93741",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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|
Determining $|z-1|$ when $z=\cos\theta +i\sin\theta$ and $\theta$ is acute As the question indicates we are supposed to find the modulus of z-1.
When trying to solve the problem I drew a diagram which you can see below:
The book I am working in solved a similar problem when showing examples, however the question was for $|z+1|$. There they proved that the modulus is given by $2\cos {\theta \over 2}$. In the case of $z-1$ though the shape is not a rhombus (which it is when you construct the diagram for $|z+1|$). As a result the diagonals are not perpendicular. The book however states that $|z-1|=2\sin {\theta \over 2}$
Proof that $|z+1|=2\cos {\theta \over 2}$
Using the diagram above one can see that $|z+1|=2\cos {\theta \over 2}$ as the length from the origin to the intersection of the diagonals is given by $\cos {\theta \over 2}$. Now, to get the length of the whole diagonal just multiply that by 2.
Does anyone know whether the book is right and I am missing something or if the answer is in fact different (in which case the correct answer would be appreciated).
|
If $z=x+iy$, then the modulus of $z$ is given by $\sqrt{x^2+y^2}$.
So with $z=\cos\theta +i\sin\theta$, we have $z-1=(\cos\theta-1)+i\sin\theta$. So the modulus is $$\sqrt{(\cos\theta-1)^2+\sin^2\theta}=\sqrt{\cos^2-2\cos\theta+1+\sin^2\theta}=\sqrt{2-2\cos\theta}$$
since $\cos^2\theta+\sin^2\theta=1$.
Using the identity $\cos\theta=\cos^2\frac{\theta}{2}-\sin^2\frac{\theta}{2}$ and $\cos^2\frac{\theta}{2}+\sin^2\frac{\theta}{2}=1$, we then get;
$$2-2\cos\theta=4\sin^2\frac{\theta}{2}$$ and so finding the square root gives the answer.
NOTE: Write $\theta=\frac{\theta}{2}+\frac{\theta}{2}$ and use the identity $\cos (A+B)=\cos A\cos B-\sin A\sin B$ to obtain the expression for $\cos\theta$ above
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/94356",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 2
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|
solve complex equation $x^8 = \frac{1+i}{\sqrt{3} - i} = \frac{\sqrt[8]{\frac{2}{\sqrt{2}}}(\cos \frac{\pi}{4} + i \sin{\frac{\pi}{4}})}{2 \cos \frac{\pi}{6} + i \sin \frac{3\pi}{2}}$
What's the way to solve this kind of equation? I think there must be 8 solutions.
I tried to solve the following two equations
$a^6 = 1+i$
$b^6 = \sqrt{3} - i$
I tried the following for the first equation
$-1 = i^2 = (a^6 - 1)^2 = a^{12} - 2a^6 +1$
Then I would need to solve this:
$a^{12} - 2a^6 +2 = 0$
Is this the right approach?
|
HINT:
$$x^8=\frac{1+i}{\sqrt{3}-i}=\frac{1+i}{\sqrt{3}-i}.\frac{\sqrt{3}+i}{\sqrt{3}+i}=\frac{\sqrt{3}-1}{4}+i\frac{1-\sqrt{3}}{4}$$
Then use polar coordinates, $x^8=r(cos\theta +i\sin\theta)$ to obtain all the 8 roots.
Think of the identity $(r(cos\theta +i\sin\theta))^n=r^n(\cos n\theta+i\sin n\theta)$
(This is De-Moivre's Theorem)
|
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"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
"answer_count": 4,
"answer_id": 3
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|
How to solve the recurrence $T(n) = \frac{n}{2}T(\frac{n}{2}) + \log n$ I am trying to solve the recurrence below but I find myself stuck.
$T(n) = \frac{n}{2}T(\frac{n}{2}) + \log n$
I have tried coming up with a guess by drawing a recurrence tree. What I have found is
number of nodes at a level: $\frac{n}{2^{i}}$
running time at each node: $\log \frac{n}{2^{i}}$
total running time at each level: $\frac{n}{2^{i}} \log \frac{n}{2^{i}}$
I try to sum this over through n > $\log n$ which is the height of the tree
$$\begin{align}
\sum\limits_{i=0}^n \frac{n}{2^{i}} \log \frac{n}{2^{i}}&=
n \sum\limits_{i=0}^n \frac{1}{2^{i}} \log \frac{n}{2^{i}}\\
&=n \sum\limits_{i=0}^n \frac{1}{2^{i}} (\log n - \log 2^{i})\\
&=n \sum\limits_{i=0}^n \frac{1}{2^{i}} \log n - \sum\limits_{i=0}^n \frac{1}{2^{i}} \log 2^{i}\\
&=n \sum\limits_{i=0}^n \frac{1}{2^{i}} \log n - \sum\limits_{i=0}^n \frac{ \log 2^{i}}{2^{i}}\\
&=n \sum\limits_{i=0}^n \frac{1}{2^{i}} \log n - d \sum\limits_{i=0}^n \frac{i}{2^{i}}\\
&=2 n \log n - d \sum\limits_{i=0}^n \frac{i}{2^{i}}
\end{align}$$
I am still trying to figure out the sum of the second summation above but I somehow feel that it will be bigger than $2n \log n$ which makes my whole approach wrong.
Is there another way to tackle this problem?
|
The summation
$$ \sum\limits_{i=0}^n \frac{i}{2^{i}} $$
is equal to
$$ \sum\limits_{i=0}^n i(\frac{1}{2})^i $$
which is of the form
$$ \sum\limits_{i=0}^n ix^i, x = \frac{1}{2} $$
Consider the summation $$ \sum\limits_{i=0}^n x^i = \frac{x^{n+1} - 1}{x - 1} $$
By differentiating with respect to $x$ we have
$$ \frac{d}{dx}\left( \sum\limits_{i=0}^n x^i \right) = \frac{d}{dx} \left( \frac{x^{n+1} - 1}{x - 1} \right)$$
$$ \Rightarrow \sum\limits_{i=0}^n ix^{i-1} = \frac{(x - 1)(n + 1)x^n - (x^{n+1} -1)(1)}{(x - 1)^2} $$
$$ \Rightarrow \sum\limits_{i=0}^n ix^{i-1} = \frac{(x^{n+1} - x^n)(n + 1) - x^{n+1} + 1}{(x - 1)^2} $$
$$ \Rightarrow \sum\limits_{i=0}^n ix^{i-1} = \frac{(nx^{n+1} - nx^n) + (x^{n+1} - x^n) - x^{n+1} + 1}{(x - 1)^2} $$
$$ \Rightarrow \sum\limits_{i=0}^n ix^{i-1} = \frac{nx^{n+1} - nx^n + x^{n+1} - x^n - x^{n+1} + 1}{(x - 1)^2} $$
$$ \Rightarrow \sum\limits_{i=0}^n ix^{i-1} = \frac{nx^{n+1} - nx^n - x^n + 1}{(x - 1)^2} $$
By multiplying $x$ to both sides,
$$ x\sum\limits_{i=0}^n ix^{i-1} = x\left( \frac{nx^{n+1} - nx^n - x^n + 1}{(x - 1)^2} \right)$$
$$ \Rightarrow \sum\limits_{i=0}^n ix^{i} = \frac{nx^{n+2} - nx^{n+1} - x^{n+1} + x}{(x - 1)^2} $$
By substitution $ \left( x = \frac{1}{2} \right) $,
$$ \sum\limits_{i=0}^n i\left(\frac{1}{2}\right)^{i} = \frac{n\left(\frac{1}{2}\right)^{n+2} - n\left(\frac{1}{2}\right)^{n+1} - \left(\frac{1}{2}\right)^{n+1} + \left(\frac{1}{2}\right)}{(\left(\frac{1}{2}\right) - 1)^2} $$
$$ \Rightarrow \sum\limits_{i=0}^n i\left(\frac{1}{2}\right)^{i} = \frac{n\left(\frac{1}{2^{n+2}}\right) - n\left(\frac{1}{2^{n+1}}\right) - \left(\frac{1}{2^{n+1}}\right) + \left(\frac{1}{2}\right)}{\left(-\frac{1}{2}\right)^2} $$
$$ \Rightarrow \sum\limits_{i=0}^n i\left(\frac{1}{2}\right)^{i} = \frac{\frac{n}{2^{n+2}} - \frac{n + 1}{2^{n+1}} + \frac{1}{2}}{\left(\frac{1}{4}\right)} $$
$$ \Rightarrow \sum\limits_{i=0}^n i\left(\frac{1}{2}\right)^{i} = 4 \left( \frac{n}{2^{n+2}} - \frac{n + 1}{2^{n+1}} + \frac{1}{2} \right) $$
$$ \Rightarrow \sum\limits_{i=0}^n i\left(\frac{1}{2}\right)^{i} = 4 \left( \frac{n}{2^{n+2}} - \frac{2(n + 1)}{2^{n+2}} + \frac{2^{n+1}}{2^{n+2}} \right) $$
$$ \Rightarrow \sum\limits_{i=0}^n i\left(\frac{1}{2}\right)^{i} = \frac{n - 2(n + 1) + 2^{(n+1)}}{2^{n}} $$
$$ \Rightarrow \sum\limits_{i=0}^n \frac{i}{2^{i}} = \frac{n - 2(n + 1) + 2^{(n+1)}}{2^{n}} $$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Showing $24|(n+1)\implies24|\sigma(n)$ Question:
Show that if $n$ is a positive integer such that $24$ divides into $n + 1$, then $24$ divides the sum of all divisors of $n$ (denoted in number theory by $\sigma(n)$ or $\sigma_1(n)$).
For example if $n = 95$, then $n + 1 = 96 = 4 \times 24$ and the sum of the divisors of $n$ is $$1 + 5 + 19 + 95 = 120 = 5 \times 24.$$ (Note that the number $n$ is included among its divisors.)
|
We use a pairing argument, working first modulo $3$ and then modulo $8$.
We are told that $n\equiv -1\pmod{24}$. It follows that $n\equiv -1\pmod{3}$. Split the set of divisors of $n$ into unordered pairs $\{a,b\}$ such that $ab=n$. (Since $n\equiv -1\pmod 3$, the number $n$ is not a perfect square, so every divisor of $n$ is taken care of.)
For any pair $\{a,b\}$ with $ab=n$, one of $a$ and $b$ is congruent to $1$ modulo $3$, and the other is congruent to $-1$. So all pair sums are congruent to $0$. Therefore the sum of all pair sums, that is, the sum of the divisors, is congruent to $0$ modulo $3$.
The same idea works modulo $8$. We are told that $n\equiv -1\pmod{8}$. If $\{a,b\}$ is an unordered pair with $ab=n$, then either (i) One of $a$ and $b$ is congruent to $1$ modulo $8$, and the other is congruent to $-1$, or (ii) One of $a$ and $b$ is congruent to $3$ modulo $8$, and the other is congruent to $-3$. In either case the sum is congruent to $0$ modulo $8$.
Thus the sum of the divisors of $n$ is divisible by $3$ and by $8$, and therefore by $24$.
|
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|
$(\cos \alpha, \sin \alpha)$ - possible value pairs We introduced the complex numbers as elements of $ \mathrm{Mat}(2\times 2, \mathbb{R})$ with
$$
\mathbb{C} \ni x =
\left(\begin{array}{cc}
a & -b \\
b & a \\
\end{array}\right) =
\frac{1}{\sqrt{a^2+b^2}}
\left(\begin{array}{cc}
\frac{a}{\sqrt{a^2+b^2}} & \frac{-b}{\sqrt{a^2+b^2}} \\
\frac{b}{\sqrt{a^2+b^2}} & \frac{a}{\sqrt{a^2+b^2}} \\
\end{array}\right)
$$
Then we concluded that $0 \leq \frac{a}{\sqrt{a^2+b^2}} \leq 1$ and $0 \leq \frac{b}{\sqrt{a^2+b^2}} \leq 1$ and therefore we could find an $\alpha$ so that $\cos \alpha = \frac{a}{\sqrt{a^2+b^2}}$ and $\sin \alpha = \frac{b}{\sqrt{a^2+b^2}}$. Then we can write the matrix with $\cos$ and $\sin$ and can write it as the Euler form as well. So far so good.
My question ist about the following:
Why is it that we cand find an $\alpha$ so that $\cos \alpha = \frac{a}{\sqrt{a^2+b^2}}$ and $\sin \alpha = \frac{b}{\sqrt{a^2+b^2}}$ for every possible value combination of $a$ and $b$?
Can't be there a combination of $a$ and $b$ where we can't find one and the same angle $\alpha$ so that the identites are true?
|
There are two key factors here: one is the one noted, that each of these quantities is between $0$ and $1$.
But the other, which is very important, is that the sum of their squares is equal to $1$:
$$\left(\frac{a}{\sqrt{a^2+b^2}}\right)^2 + \left(\frac{b}{\sqrt{a^2+b^2}}\right)^2 = 1.$$
Because the sum of their squares is equal to $1$, the point
$$\left(\frac{a}{\sqrt{a^2+b^2}}, \frac{b}{\sqrt{a^2+b^2}}\right)$$
satisfies the equation
$$x^2+y^2=1.$$
But this is the equation of the circle of radius $1$ with center at the origin. Every point on that circle is of the form $(\cos\alpha,\sin\alpha)$, where $\alpha$ is the angle between the positive $x$-axis and the line that goes from the origin to the point. So that means that $\alpha$ exists.
|
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|
Lim $x\to 0$ of $\frac{\sin(\pi x)}{\tan(\sqrt{3}x)}$ $$
\lim_{x \to 0} \frac{\sin(\pi x)}{\tan(\sqrt{3} x)}
$$
I need a step by step explanation. Thank you.
|
Results going to be used:
*
*$\displaystyle \lim_{x \to 0} \frac{\sin{x}}{x} =1 = \lim_{x \to 0} \frac{\tan{x}}{x}$
What you have is
\begin{align*}
\lim_{x \to 0}\: \frac{\sin(\pi{x})}{\tan\sqrt{3}{x}} &= \lim_{x \to 0} \: \frac{\sin\pi{x}}{\pi{x}} \times \frac{\sqrt{3}x}{\tan{\sqrt{3}x}} \times \frac{\pi{x}}{\sqrt{3}x} \\ &= \lim_{x \to 0} \: \frac{\pi{x}}{\sqrt{3}x} = \frac{\pi}{\sqrt{3}}
\end{align*}
|
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|
Find $\lim\limits_{(x,y) \to (0,0)} \frac{e^{-\frac{1}{x^2+y^2}}}{x^4+y^4} $ I'm trying to find $$\lim\limits_{(x,y) \to (0,0)} \frac{e^{-\frac{1}{x^2+y^2}}}{x^4+y^4} .$$
After I tried couple of algebraic manipulation, I decided to use the polaric method.
I choose $x=r\cos \theta $ , $y=r\sin \theta$, and $r= \sqrt{x^2+y^2}$, so I get
$$\lim\limits_{r \to 0} \frac{e^{-\frac{1}{r^2}}}{r^4\cos^4 \theta+r^4 \sin^4 \theta } $$
What do I do from here?
Thanks a lot!
|
Since $2(x^4+y^4)\geqslant(x^2+y^2)^2$, the ratio $r(x,y)=\dfrac{\mathrm e^{-1/(x^2+y^2)}}{x^4+y^4}$ is such that
$$
0\lt r(x,y)\leqslant u\left(\frac1{x^2+y^2}\right),
\quad\text{where}\ u:z\mapsto2z^2\mathrm e^{-z}.
$$
Now, $\frac1{x^2+y^2}\to+\infty$ when $(x,y)\to(0,0)$ and $u(z)\to0$ when $z\to+\infty$ (as the saying goes, at infinity the exponential prevails on the powers) hence $r(x,y)\to0$ when $(x,y)\to(0,0)$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Complex numbers equation: $z^4 = -16$ Probably dumb question, but I would like to ask it anyway.
I was to solve this equation:
$z^4 = -16$
At first glance, the way to solve it would be (as any other equation):
$z^2 = \sqrt{-1} * 4 \lor z^2 = -\sqrt{-1} * 4$
$z^2 = 4i \lor z^2 = -4i$
$z = 2\sqrt{i} \lor z = -2\sqrt{i} \lor z = 2i\sqrt{i} \lor z = -2i\sqrt{i}$
However, in suggested answers, there are $z_1, z_2, z_3$ and $z_4$ found using trigonometric formulas. Is there something particulary wrong in my solution and the only proper one uses trigonometric form of complex number?
|
We have
$$z^4 + 16 = (z^2 + 4)^2 - 8z^2 = (z^2 - 2\sqrt{2} z + 4)(z^2 + 2\sqrt{2} z + 4).$$
Anotherway $$z^2 - 2\sqrt{2} z + 4 = (z-\sqrt{2})^2 + 2 = (z-\sqrt{2})^2 -(\sqrt{2} i)^2 = (z-\sqrt{2} + \sqrt{2} i)(z-\sqrt{2} + \sqrt{2} i)).$$
The same for $z^2 + 2\sqrt{2} z + 4$
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Possibility to simplify $\sum\limits_{k = - \infty }^\infty {\frac{{{{\left( { - 1} \right)}^k}}}{{a + k}} = \frac{\pi }{{\sin \pi a}}} $ Is there any way to show that
$$\sum\limits_{k = - \infty }^\infty {\frac{{{{\left( { - 1} \right)}^k}}}{{a + k}} = \frac{1}{a} + \sum\limits_{k = 1}^\infty {{{\left( { - 1} \right)}^k}\left( {\frac{1}{{a - k}} + \frac{1}{{a + k}}} \right)}=\frac{\pi }{{\sin \pi a}}} $$
Where $0 < a = \dfrac{n+1}{m} < 1$
The infinite series is equal to
$$\int\limits_{ - \infty }^\infty {\frac{{{e^{at}}}}{{{e^t} + 1}}dt} $$
To get to the result, I split the integral at $x=0$ and use the convergent series in $(0,\infty)$ and $(-\infty,0)$ respectively:
$$\frac{1}{{1 + {e^t}}} = \sum\limits_{k = 0}^\infty {{{\left( { - 1} \right)}^k}{e^{ - \left( {k + 1} \right)t}}} $$
$$\frac{1}{{1 + {e^t}}} = \sum\limits_{k = 0}^\infty {{{\left( { - 1} \right)}^k}{e^{kt}}} $$
Since $0 < a < 1$
$$\eqalign{
& \mathop {\lim }\limits_{t \to 0} \frac{{{e^{\left( {k + a} \right)t}}}}{{k + a}} - \mathop {\lim }\limits_{t \to - \infty } \frac{{{e^{\left( {k + a} \right)t}}}}{{k + a}} = \frac{1}{{k + a}} \cr
& \mathop {\lim }\limits_{t \to \infty } \frac{{{e^{\left( {a - k - 1} \right)t}}}}{{k + a}} - \mathop {\lim }\limits_{t \to 0} \frac{{{e^{\left( {a - k - 1} \right)t}}}}{{k + a}} = - \frac{1}{{a - \left( {k + 1} \right)}} \cr} $$
A change in the indices will give the desired series.
Although I don't mind direct solutions from tables and other sources, I prefer an elaborated answer.
Here's the solution in terms of $\psi(x)$. By separating even and odd indices we can get
$$\eqalign{
& \sum\limits_{k = 0}^\infty {\frac{{{{\left( { - 1} \right)}^k}}}{{a + k}}} = \sum\limits_{k = 0}^\infty {\frac{1}{{a + 2k}}} - \sum\limits_{k = 0}^\infty {\frac{1}{{a + 2k + 1}}} \cr
& \sum\limits_{k = 0}^\infty {\frac{{{{\left( { - 1} \right)}^k}}}{{a - k}}} = \sum\limits_{k = 0}^\infty {\frac{1}{{a - 2k}}} - \sum\limits_{k = 0}^\infty {\frac{1}{{a - 2k - 1}}} \cr} $$
which gives
$$\sum\limits_{k = 0}^\infty {\frac{{{{\left( { - 1} \right)}^k}}}{{a + k}}} = \frac{1}{2}\psi \left( {\frac{{a + 1}}{2}} \right) - \frac{1}{2}\psi \left( {\frac{a}{2}} \right)$$
$$\sum\limits_{k = 0}^\infty {\frac{{{{\left( { - 1} \right)}^k}}}{{a - k}}} = \frac{1}{2}\psi \left( {1 - \frac{a}{2}} \right) - \frac{1}{2}\psi \left( {1 - \frac{{a + 1}}{2}} \right) + \frac{1}{a}$$
Then
$$\eqalign{
& \sum\limits_{k = - \infty }^\infty {\frac{{{{\left( { - 1} \right)}^k}}}{{a + k}}} = \sum\limits_{k = 0}^\infty {\frac{{{{\left( { - 1} \right)}^k}}}{{a + k}}} + \sum\limits_{k = 0}^\infty {\frac{{{{\left( { - 1} \right)}^k}}}{{a - k}}} - \frac{1}{a} = \cr
& = \left\{ {\frac{1}{2}\psi \left( {1 - \frac{a}{2}} \right) - \frac{1}{2}\psi \left( {\frac{a}{2}} \right)} \right\} - \left\{ {\frac{1}{2}\psi \left( {1 - \frac{{a + 1}}{2}} \right) - \frac{1}{2}\psi \left( {\frac{{a + 1}}{2}} \right)} \right\} \cr} $$
But using the reflection formula one has
$$\eqalign{
& \frac{1}{2}\psi \left( {1 - \frac{a}{2}} \right) - \frac{1}{2}\psi \left( {\frac{a}{2}} \right) = \frac{\pi }{2}\cot \frac{{\pi a}}{2} \cr
& \frac{1}{2}\psi \left( {1 - \frac{{a + 1}}{2}} \right) - \frac{1}{2}\psi \left( {\frac{{a + 1}}{2}} \right) = \frac{\pi }{2}\cot \frac{{\pi \left( {a + 1} \right)}}{2} = - \frac{\pi }{2}\tan \frac{{\pi a}}{2} \cr} $$
So the series become
$$\eqalign{
& \sum\limits_{k = - \infty }^\infty {\frac{{{{\left( { - 1} \right)}^k}}}{{a + k}}} = \frac{\pi }{2}\left\{ {\cot \frac{{\pi a}}{2} + \tan \frac{{\pi a}}{2}} \right\} \cr
& \sum\limits_{k = - \infty }^\infty {\frac{{{{\left( { - 1} \right)}^k}}}{{a + k}}} = \pi \csc \pi a \cr} $$
The last being an application of a trigonometric identity.
|
A related identity is proven in this answer using residue theory. Here is a real approach to that identity.
Convergence of the Principal Value
We will look at the principal value
$$
\begin{align}
f(x)
&=\lim_{n\to\infty}\sum_{k=-n}^n\frac1{k+x}\tag{1a}\\
&=\frac1x-\sum_{k=1}^\infty\frac{2x}{k^2-x^2}\tag{1b}
\end{align}
$$
$\text{(1b)}$ converges for all non-integer $x$.
Properties of $\boldsymbol{f(x)}$
$\bullet$ $f(x)$ has period $1$:
$$
\begin{align}
f(x)-f(x+1)
&=\lim_{n\to\infty}\left(\sum_{k=-n}^n\frac1{k+x}-\sum_{k=-n}^n\frac1{k+1+x}\right)\tag{2a}\\
&=\lim_{n\to\infty}\left(\sum_{k=-n}^n\frac1{k+x}-\sum_{k=-n+1}^{n+1}\frac1{k+x}\right)\tag{2b}\\
&=\lim_{n\to\infty}\left(\frac1{-n+x}-\frac1{n+1+x}\right)\tag{2c}\\[9pt]
&=0\tag{2d}
\end{align}
$$
Explanation:
$\text{(2a):}$ definition
$\text{(2b):}$ substitute $k\mapsto k-1$ in the right sum
$\text{(2c):}$ the sums telescope
$\text{(2d):}$ evaluate the limit
$\bullet$ $f(1/2)=0$:
$$
\begin{align}
f(1/2)
&=\lim_{n\to\infty}\sum_{k=-n}^n\frac1{k+1/2}\tag{3a}\\
&=\lim_{n\to\infty}\frac1{n+1/2}\tag{3b}\\[9pt]
&=0\tag{3c}
\end{align}
$$
Explanation:
$\text{(3a):}$ definition
$\text{(3b):}$ for $1\le j\le n$, the term with $k=j-1$ cancels the term with $k=-j$
$\phantom{\text{(3b):}}$ which leaves the term with $k=n$
$\text{(3c):}$ evaluate the limit
$\bullet$ $f(x)^2+\pi^2=-f'(x)$:
$$
\begin{align}
f(x)^2
&=\lim_{n\to\infty}\sum_{k=-n}^n\frac1{k+x}\sum_{j=-n}^n\frac1{j+x}\tag{4a}\\[3pt]
&=\sum_{k=-\infty}^\infty\frac1{(k+x)^2}+\lim_{n\to\infty}\sum_{\substack{|j|,|k|\le n\\j\ne k}}\left(\frac1{k+x}-\frac1{j+x}\right)\frac1{j-k}\tag{4b}\\
&=\sum_{k=-\infty}^\infty\frac1{(k+x)^2}+\lim_{n\to\infty}\sum_{\substack{|j|,|k|\le n\\j\ne k}}\frac2{k+x}\frac1{j-k}\tag{4c}\\
&=\sum_{k=-\infty}^\infty\frac1{(k+x)^2}+\lim_{n\to\infty}\sum_{k=1}^n\sum_{\substack{|j|\le n\\j\ne k}}\left(\frac2{k+x}+\frac2{k-x}\right)\frac1{j-k}\tag{4d}\\
&=\sum_{k=-\infty}^\infty\frac1{(k+x)^2}-\lim_{n\to\infty}\sum_{k=1}^n\left(\frac2{k+x}+\frac2{k-x}\right)(H_{n+k}-H_{n-k})\tag{4e}\\
&=-f'(x)-4\int_0^1\frac{\log\left(\frac{1+t}{1-t}\right)}{t}\,\mathrm{d}t\tag{4f}\\[12pt]
&=-f'(x)-\pi^2\tag{4g}
\end{align}
$$
Explanation:
$\text{(4a):}$ product of limits equals the limit of the products
$\text{(4b):}$ the left sum contains the terms with $j=k$
$\phantom{\text{(11):}}$ apply partial fractions to the right sum
$\text{(4c):}$ take advantage of the symmetry in $j$ and $k$
$\text{(4d):}$ for $k=0$, the sum in $j$ is $0$
$\phantom{\text{(4d):}}$ for $k\lt0$, if we substitute $(j,k)\mapsto(-j,-k)$,
$\phantom{\text{(4d):}}$ we get the sum with $k\gt0$ and $k+x\mapsto k-x$
$\text{(4e):}$ the sum in $j$ telescopes to $H_{n-k}-H_{n+k}$
$\text{(4f):}$ the left sum is $f'(x)$
$\phantom{\text{(4f):}}$ the right sum is a Riemann Sum with $\frac kn\mapsto t$ and $\frac1n\mapsto\mathrm{d}t$
$\phantom{\text{(4f):}}$ $H_{n+k}-H_{n-k}\mapsto\log\left(\frac{1+t}{1-t}\right)$ and $\frac2{k-x}+\frac2{k+x}\mapsto\frac{4\,\mathrm{d}t}t$
$\text{(4g):}$ $4\int_0^1\sum\limits_{k=0}^\infty\frac{2\,x^{2k}}{2k+1}\,\mathrm{d}x=4\sum\limits_{k=0}^\infty\frac2{(2k+1)^2}=\pi^2$
Conclude that $\boldsymbol{f(x)=\pi\cot(\pi x)}$
We can separate $(4)$ and integrate:
$$
\begin{align}
\int\frac{\pi\,\mathrm{d}f}{f^2+\pi^2}&=-\int\pi\,\mathrm{d}x\tag{5a}\\
\tan^{-1}(f/\pi)&=C-\pi x\tag{5b}\\[9pt]
f&=\pi\tan(C-\pi x)\tag{5c}
\end{align}
$$
$(3)$ allows us to compute $C=\pi/2$, giving
$$
f(x)=\pi\cot(\pi x)\tag6
$$
for $x\in(0,1)$. $(2)$ removes this restriction on $x$, validating $(6)$ for all non-integer $x$. That is,
$$
\sum_{k=-\infty}^\infty\frac1{k+x}=\pi\cot(\pi x)\tag7
$$
when taken in the principal value sense.
Answer to the Question
$$
\begin{align}
\sum_{k=-\infty}^\infty\frac{(-1)^k}{k+x}
&=\sum_{k=-\infty}^\infty\frac2{2k+x}-\sum_{k=-\infty}^\infty\frac1{k+x}\tag{8a}\\
&=\sum_{k=-\infty}^\infty\frac1{k+x/2}-\sum_{k=-\infty}^\infty\frac1{k+x}\tag{8b}\\[6pt]
&=\pi\cot(\pi x/2)-\pi\cot(\pi x)\tag{8c}\\[15pt]
&=\pi\csc(\pi x)\tag{8d}
\end{align}
$$
Explanation:
$\text{(8a):}$ the alternating sum is twice the sum of the even terms
$\phantom{\text{(8a):}}$ minus the sum of all the terms
$\text{(8b):}$ adjust the terms of the left sum to apply $(7)$
$\text{(8c):}$ apply $(7)$
$\text{(8d):}$ $\frac{1+\cos(\pi x)}{\sin(\pi x)}-\frac{\cos(\pi x)}{\sin(\pi x)}=\frac1{\sin(\pi x)}$
|
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|
Product Chain Power Rule： Either it's the book or I am wrong. The problem: $x^3\sqrt{2x+4}$
$f(x):= x^3$, $g(x):= \sqrt{2x+4}$
$(f\times g)' = f^{\prime}g+fg^{\prime}$ thus it should be
$3x^2\sqrt{2x+4} + (x^3)[\frac{1}{2}(2x+4)^{\frac{-1}{2}}(2)]$
which is $3x^2\sqrt{2x+4}+\frac{x^3}{\sqrt{2x+4}}$
The book gives: $3x^2\sqrt{2x+4}+\frac{x^3}{2\sqrt{2x+4}}$
I'm correct? I always get worry when my answers don't match the book.
$\frac{d}{dx}[f\times g(h(x))] = f^{\prime} \times g(h(x))+ f\times g^{\prime}(h(x))h^{\prime}$ right?
|
As my colleagues have astutely pointed out, the product rule states $(fg)^{\prime} = f^{\prime} g + f g^{\prime}$. Define $f(x)= x^3$ and $g(x)= \sqrt{2x+4}$. As $f^{\prime}(x) = 3 x^{2}$ and $g^{\prime}(x) = \frac{1}{\sqrt{2x+4}}$, the product rule gives
\begin{align}
(f(x)g(x))^{\prime} = f^{\prime}(x) g(x) + f(x) g^{\prime}(x) = 3x^{2} \sqrt{2x + 4} + \frac{x^3}{\sqrt{2x+4}}.
\end{align}
The answer in your book has an incorrect factor of $2$ in the denominator of the second term. This factor should cancel with the factor of $2$ coming from $2x$ (in $g(x)$) by the chain rule.
|
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|
Showing $ \sum_{n=0}^{\infty} \frac{1}{(3n+1)(3n+2)}=\frac{\pi}{3\sqrt{3}}$ I would like to show that:
$$ \sum_{n=0}^{\infty} \frac{1}{(3n+1)(3n+2)}=\frac{\pi}{3\sqrt{3}} $$
We have:
$$ \sum_{n=0}^{\infty} \frac{1}{(3n+1)(3n+2)}=\sum_{n=0}^{\infty} \frac{1}{3n+1}-\frac{1}{3n+2} $$
I wanted to use the fact that $$\arctan(\sqrt{3})=\frac{\pi}{3} $$ but $\arctan(x)$ can only be written as a power series when $ -1\leq x \leq1$...
|
Regularized the series:
$$ \begin{eqnarray}
\sum_{n=0}^m \frac{1}{(3n+1)(3n+2)} &=& \sum_{n=0}^m \left( \frac{1}{3n+1} - \frac{1}{3n+2} \right) = \sum_{n=0}^m \int_0^1 \left( x^{3n} - x^{3n+1} \right) \mathrm{d} x \\
&=& \int_0^1 \left( \frac{(1-x^{3m+3}) (1-x)}{1-x^3} \right) \mathrm{d} x =
\int_0^1 \frac{1-x^{3m+3}}{1+x + x^2} \mathrm{d} x
\end{eqnarray}
$$
Now we can take the limit by dominating convergence theorem:
$$
\sum_{n=0}^\infty \frac{1}{(3n+1)(3n+2)} = \int_0^1 \frac{\mathrm{d} x}{1+x+x^2} = \left.\frac{2 \sqrt{3}}{3} \arctan\left(\frac{2x+1}{\sqrt{3}}\right)\right|_0^1 = \frac{\pi}{3 \sqrt{3}}
$$
|
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|
If $f(x)f(y)=f(\sqrt{x^2+y^2})$ how to find $f(x)$ As we know, for the $$f(x)f(y)=f(x+y)$$ $f(x)=\mathrm e^{\alpha x}$ is a solution.
What about
$f(x)f(y)=f(\sqrt{x^2+y^2})$?
Does anybody know about the solution of the function equation?
I tried to find $f(x)$.
See my attempts below to find $f(x)$.
$$f(x)=a_0+a_1x+\frac{a_2x^2}{2!}+\frac{a_3x^3}{3!}+\cdots$$
$$f(y)=a_0+a_1y+\frac{a_2y^2}{2!}+\frac{a_3y^3}{3!}+\cdots$$
$$f(x)f(y)=a_0f(y)+a_1f(y)x+\frac{a_2f(y)x^2}{2!}+\frac{a_3f(y)x^3}{3!}+\cdots$$
$$f(\sqrt{x^2+y^2})=a_0+a_1\sqrt{x^2+y^2}+\frac{a_2(x^2+y^2)}{2!}+\frac{a_3(x^2+y^2)^{3/2}}{3!}+\cdots=$$
$$f(\sqrt{x^2+y^2})=a_0+a_1y\sqrt{1+(x/y)^2}+\frac{a_2(x^2+y^2)}{2!}+\frac{a_3y^2(1+(x/y)^2)^{3/2}}{3!}+\cdots=f(x)f(y)=a_0f(y)+a_1f(y)x+\frac{a_2f(y)x^2}{2!}+\frac{a_3f(y)x^3}{3!}+\cdots$$
if we use binom expansion for $(1+(x/y)^2)^{m}$
$$(1+(x/y)^2)^{m}=1+\frac{mx^2}{y^2}+\frac{m(m-1)x^4}{2!y^4}+\frac{m(m-1)(m-2)x^6}{3!y^6}+\cdots$$
Let's put the expansion to the equation $f(\sqrt{x^2+y^2})$
$$
\begin{align}
& f(\sqrt{x^2+y^2}) =a_0 + a_1 y \left( 1 + \frac{(1/2)x^2}{y^2} + \frac{(1/2)((1/2)-1)x^4}{2!y^4} \right. \\ \\
& \left. {} + \frac{(1/2)((1/2)-1)((1/2)-2)x^6}{3!y^6} + \cdots\right) + \frac{ a_2 (x^2+y^2)}{2!} \\ \\
& + \frac{a_3y^2 \left(1+\frac{(3/2)x^2}{y^2}+\frac{(3/2)((3/2)-1)x^4}{2!y^4}+\frac{(3/2)((3/2)-1)((3/2)-2)x^6}{3!y^6}+\cdots\right)}{3!} +\cdots \\ \\
& = a_0f(y)+a_1f(y)x+\frac{a_2f(y)x^2}{2!}+\frac{a_3f(y)x^3}{3!}+\cdots
\end{align}
$$
If we equal for all $x^n$ terms in both sides
we can see $a_{2n-1}=0$, but to find $a_{2n}$ seems hard for me.
Any idea to find $a_{2n}$
Thanks in advice.
|
The answer to this question is a well known result called Maxwell's theorem, after James Clerk Maxwell. This earlier question deals with it:
very elementary proof of Maxwell's theorem
|
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|
How do I find complex roots of a quartic polynomial using quadratic formula? For something like:
$$
z^4 + 8z^2 + 3
$$
how can I find all the complex roots using the quadratic equation, or is there a better method?
I tried a u substitution using $u = z^2$, but then when I applied the quadratic formula, I was getting real roots and there was no where to sub u back in.
|
Presumably you got $u=z^2=-4\pm\sqrt{13}$. Now you need to solve for $z$.
Both of our values of $z^2$ are negative. The solutions of $z^2=-4+\sqrt{13}$ are
$z=\pm i \sqrt{4-\sqrt{13}}$, and the solutions of $z^2=-4-\sqrt{13}$ are $z=\pm i\sqrt{4+\sqrt{13}}$.
Remark: Things get somewhat more complicated when you want to find the square roots of a general complex number. For example, suppose that we want to find the square roots of $3+4i$. One usual way is to first rewrite $z$ as
$$\sqrt{3^2+4^2} \left( \frac{3}{\sqrt{3^2+4^2}}+ \frac{4}{\sqrt{3^2+4^2}}i\right).$$
Note that $\sqrt{3^2+4^2}=5$.
Let $\theta$ be an angle whose cosine is $\frac{3}{5}$ and whose sine is $\frac{4}{5}$. Then the square roots of $z$ are
$$\pm \sqrt{5}\left(\cos\frac{\theta}{2}+\sin\frac{\theta}{2}i\right).$$
|
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|
Where have I gone wrong? Contour integration $\int_{-a}^a {u\over 1+u+u^2} du$ as $a\to \infty$ I would like to integrate $\int_{-a}^a {u\over 1+u+u^2} du$ as $a\to \infty$.
So I thought I might use the residue theorem. In the complex plane, the singularities occur at $z=e^{\pm i2\pi\over 3}$. So if we close the contour in the upper half plane we enclose $z=e^{i2\pi\over 3}$. This is a covert simple pole, so res$\left({z\over 1+z+z^2};e^{i2\pi\over 3}\right)={z\over 2z+1}|_{z=e^{i2\pi\over 3}}={1\over 2}+{i\over 2\sqrt3}$. So by the residue theorem, the integral is $2\pi i \left({1\over 2}+{i\over 2\sqrt3}\right)=\pi(i-{1\over\sqrt 3})$.
But this cannot be the answer since it contains $i$. Where have I gone wrong? Thanks.
|
You can use the usual integration:
$$\eqalign{
& \int\limits_{ - a}^a {\frac{u}{{{u^2} + u + 1}}du} = \frac{1}{2}\int\limits_{ - a}^a {\frac{{2u}}{{{u^2} + u + 1}}du} \cr
& = \frac{1}{2}\left\{ {\int\limits_{ - a}^a {\frac{{2u + 1}}{{{u^2} + u + 1}}du} - \int\limits_{ - a}^a {\frac{{du}}{{{u^2} + u + 1}}} } \right\} \cr
& = \frac{1}{2}\left\{ {\int\limits_{ - a}^a {\frac{{2u + 1}}{{{u^2} + u + 1}}du} - \int\limits_{ - a}^a {\frac{{du}}{{{{\left( {u + \frac{1}{2}} \right)}^2} + \frac{3}{4}}}} } \right\} \cr
& = \frac{1}{2}\left\{ {\log \left| {{u^2} + u + 1} \right| - \frac{2}{{\sqrt 3 }}\arctan \left( {\frac{2}{{\sqrt 3 }}\left( {x + \frac{1}{2}} \right)} \right)} \right\}_{ - a}^a \cr
& = \frac{1}{2}\log \left| {\frac{{{a^2} + a + 1}}{{{a^2} - a + 1}}} \right| - \frac{1}{{\sqrt 3 }}\arctan \left( {\frac{2}{{\sqrt 3 }}\left( {a + \frac{1}{2}} \right)} \right) + \frac{1}{{\sqrt 3 }}\arctan \left( {\frac{2}{{\sqrt 3 }}\left( { - a + \frac{1}{2}} \right)} \right) \cr} $$
This gives as a result $-\dfrac{\pi}{\sqrt 3}$
|
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|
Proving $x^n - y^n = (x-y)(x^{n-1} + x^{n-2} y + ... + x y^{n-2} + y^{n-1})$ In Spivak's Calculus 3rd Edition, there is an exercise to prove the following:
$$x^n - y^n = (x-y)(x^{n-1} + x^{n-2} y + ... + x y^{n-2} + y^{n-1})$$
I can't seem to get the answer. Either I've gone wrong somewhere, I'm overlooking something, or both. Here's my (non) proof:
$$\begin{align*}
x^n - y^n &= (x - y)(x^{n-1} + x^{n-2}y +\cdots+ xy^{n-2} + y^{n-1}) \\
&= x \cdot x^{n-1} + x \cdot x^{n-2} \cdot y + \cdots + x \cdot x \cdot y^{n-2} + x \cdot y^{n-1}\\
&\qquad + (-y) \cdot x^{n-1} + (-y) \cdot x^{n-2} \cdot y + \cdots + (-y) \cdot x \cdot y^{n-2} + (-y) \cdot y^{n-1}\\
&= x^n + x^{n-1} y + \cdots + x^2 y^{n-2} + x y^{n-1} - x^{n-1}y - y^2 x^{n-2} - \cdots- x y^{n-1} - y^n \\
&= x^n + x^2 y^{n-2} - x^{n-2} y^2 - y^n \\
&\neq x^n - y^n
\end{align*}$$
Is there something I can do with $x^n + x^2 y^{n-2} - x^{n-2} y^2 - y^n$ that I'm not seeing, or did I make a mistake early on?
EDIT:
I should have pointed out that this exercise is meant to be done using nine of the twelve basic properties of numbers that Spivak outlines in his book:
*
*Associate law for addition
*Existence of an additive identity
*Existence of additive inverses
*Commutative law for additions
*Associative law for multiplication
*Existence of a multiplicative identity
*Existence of multiplicative inverses
*Commutative law for multiplication
*Distibutive law
|
Since $x^{n-1} + x^{n-2} y + \dots + x y^{n-2} + y^{n-1}$ is a geometric series with $n$ terms and a common factor of $y/x$, it equals
$$
\frac{x^{n-1}\left(1-\left(\frac{y}{x}\right)^n\right)}{1-\frac{y}{x}}=\frac{x^{n}\left(1-\left(\frac{y}{x}\right)^n\right)}{x-y}=\frac{x^n-y^n}{x-y} \, ;
$$
multiplying through by $(x-y)$ gives the desired result.
|
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|
Prove $\left \lfloor \frac{1}2 \left( 2+\sqrt3 \right) ^{2002} \right\rfloor \equiv -1 \pmod7 $ Prove
$$ \large \left\lfloor \frac{1}2 \left( 2+\sqrt3 \right) ^{2002} \right\rfloor \equiv -1 \pmod7 $$
So far my intuion only tells me that this has something to do with $(2+\sqrt3)(2-\sqrt3)=1$, but I don't even know where to begin.
I'm looking for elegant solution.
|
$$(2+\sqrt 3)^2=7+2\sqrt3$$
Now, $$(7+4\sqrt3)^{2m+1}+(7-4\sqrt3)^{2m+1}$$
$$=2\left(7^{2m+1}+\binom {2m+1}27^{2m-1}\cdot4^2\cdot3+\binom{2m+1}47^{2m-3}\cdot4^4\cdot3^2+\cdots+\binom{2m+1}{2m}7\cdot4^{2m}\cdot3^m\right)$$ which is divisible by $2\cdot 7=14.$
Consequently, $$\frac12(7+4\sqrt3)^{2m+1}+\frac12(7-4\sqrt3)^{2m+1}\equiv0\pmod 7$$
Like Mathlover has observed, $0<\frac{7-4\sqrt3}2=\frac{7^2-(4\sqrt3)^2}{2(7+4\sqrt3)}<1\implies 0<\frac12(7-4\sqrt3)^{2m+1}<1$
Hence, $$7k-1<\frac12(7+4\sqrt3)^{2m+1}<7k$$ some integer $k.$
|
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|
Steps for solving this inequality: $ x \geq \frac{6}{x - 1} $ Solve:
$$
x \geq \frac{6}{x - 1}
$$
I figure that I can't really multiply by both sides by $(x - 1)$ since I'm not sure if it will be positive or negative. So I multiplied by $(x - 1)^2$ to get:
$$
x^3 - 2x^2 -5x \geq -6
$$
but wasn't sure how to proceed to an answer.
|
$$
x \ge \frac{6}{x-1}
$$
Move everything to one side:
$$
x-\frac{6}{x-1} \ge 0
$$
The common denominator is $x-1$, so we have:
$$
\frac{x(x-1)}{x-1} - \frac{6}{x-1} \ge 0
$$
Simplify:
$$
\frac{x^2-x-6}{x-1}\ge 0
$$
Factor:
$$
\frac{(x-3)(x+2)}{x-1} \ge 0
$$
This changes signs at $-2$, $1$, and $3$. It is positive if $x>3$, negative if $x$ is between $3$ and $1$, positive if $x$ is between $1$ and $-2$, and negative if $x<-2$. It is zero when the numerator is $0$, i.e. when $x=-2$ or $x=3$.
So the solution is
$$
x\ge3\text{ or } -2\le x<1.
$$
|
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|
How to evaluate $\displaystyle\int {1\over (1+kx^2)^{3/2}}dx$ What change of variable should I use to integrate $$\displaystyle\int {1\over (1+kx^2)^{3/2}}dx$$
I know the answer is $$\displaystyle x\over \sqrt{kx^2+1}.$$ Maybe a trig or hyperbolic function?
|
For clarity purposes
$$
\begin{align*}
\dfrac{1}{2\sqrt{k}}\int \dfrac{1}{t^{\frac{3}{2}} \sqrt{t-1}}dt &= \dfrac{1}{2\sqrt{k}}\int \dfrac{\sqrt{t}}{t^{2} \sqrt{t-1}}dt\\
&= \dfrac{1}{2\sqrt{k}}\int \dfrac{1}{t^{2} \sqrt{1-\frac{1}{t}}}dt \tag{A}
\end{align*}
$$
Now in this integral above substitute
$$ \sqrt{1-\frac{1}{t}} = u$$ which would imply
$$ \frac{1}{2t^2 \sqrt{(1-\frac{1}{t})}} dt = du$$
Thus, $(A)$ would simplify to
$$ \frac{1}{2\sqrt{k}} \int 2 du = \frac{1}{2\sqrt{k}} 2u = \frac{u}{\sqrt{k}} = \left(\sqrt{\frac{t-1}{tk}}\right) = \frac{x}{\sqrt{1+kx^2}}$$
|
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|
Find a three digit number ($\overline{xyz}$)? (Excuse me for my english: I'm spanish speaker)
"Find a three digit number $\overline{xyz}$ such that $x^2 +y^2 + z^2$ is equal to the number (xyz). "
I have this equation:
$$x^2 -100x +(y^2-10y+z^2-z)=0$$
And the discriminant (in $x$):
$$\Delta_x = -4y^2 +40y-4z^2+4z+10000$$
Solutions for $x_i$ are:
$$x_i = \frac{100 \pm \sqrt{\Delta_x}}{2}$$
in which ${\Delta_x} \geq 0$. The obvious conditions are:
i) $x \neq 0$ and $x=1,..., 9$;
ii) $y, z = 0, ..., 9$.
I need an orientation because pick $(y,z) \in \{0,...9\} \times \{0,...9\}$ I believe it's arduous task.
Another "attemp" from myself is to find $(x, y, z) \in \{1,...9\} \times \{0,...9\} \times \{0,...9\} $ such that
$$(x-50)^2 + (y-5)^2 +\left(z- \frac{1}{2}\right)^2 = \frac{10101}{4}$$
but I'm not skilfull in multivariable calculus :(
Any advice is welcome.
|
Since $x^2+y^2+z^2\leq 3\times 9^2 = 243$, we know that $x=0$, $x=1$, or $x=2$. But if $x=2$, then $x^2+y^2+z^2 \leq 2\times 9^2 + 4 = 166$, so we must definitely have $x=0$ or $x=1$.
But $x=1$ doesn't work: that would give us a maximum possible total of $1+9^2+9^2 = 163$; that would meant that $y\leq 6$; but then the best we can do is $1+6^2+9^2 = 118$, which means $y\leq 1$, but then we can't even get to $100$, since the maximum possible total would be $1+1+81$.
So we must have $x=0$.
Now we are down to $y^2+z^2 = 10y+z$, or $y^2 - 10y + (z^2-z) = 0$.
This means $y = 5\pm \frac{\sqrt{100 - 4(z^2-z)}}{2}$. So $100-4(z^2-z)$ must be an even square. So $z^2-z$ must be $0$, $9$, $16$, $21$, $24$, or $25$.
Since $z^2-z = z(z-1)$, with $0\leq z\leq 9$ integral, this means the total cannot be $9$, $21$, or $25$. If $z\gt 1$, then we cannot have a perfect square, which excludes $16$; and $24$ cannot be written as a product of consecutive integers, which only leaves $z^2-z=0$. Thus, $z=0$ or $z=1$. Both lead to $y=0$ or $y=10$, but the latter is impossible.
So the only possibilities are $000$ and $001$.
|
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|
Show that $2^{3^k}+1$ is divisible by $3^k$ for all positive integers $k$ I attempted this by induction: Here is what I did
For $k=1, 2^{3^{1}}+1 = 2^{3}+1 = 9 = 3^{2}$ is divisible by $3^{1}$ , so the result is true for $k=1$
Now I assume the result to be true for $k=m$,
$2^{3^{m}}+1$ is divisible by $3^m$. To show the result to be true for $k=(m+1)$,
$2^{3^{m+1}}+1 = 2^{3^m} \times 2^3+1$ and I was stuck here.
|
Note that $2^{3^{m+1}}+1 = \left(2^{3^m} \right)^3 + 1 = \left(2^{3^m} + 1 \right)\left(\left(2^{3^m} \right)^2 - 2^{3^m} + 1 \right)$. By induction hypothesis, $3^m$ divides $\left(2^{3^m} + 1 \right)$. All you need to show is that $3$ divides $\left(\left(2^{3^m} \right)^2 - 2^{3^m} + 1 \right)$.
Hint: $2^{3^m} \equiv -1 (\bmod 3)$.
In-fact, proceeding along the same lines as you have done you can show that $3^{k+1}$ divides $2^{3^k}+1$
|
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|
Sum of the reciprocal of sine squared I encountered an interesting identity when doing physics homework, that is,
$$ \sum_{n=1}^{N-1} \frac{1}{\sin^2 \dfrac{\pi n}{N} } = \frac{N^2-1}{3}. $$
How is this identity derived? Are there any more related identities?
|
Here it is another approach, just for fun. We may notice that $\cos\frac{\pi}{n+1},\cos\frac{2\pi}{n+1},\ldots,\cos\frac{n\pi}{n+1}$ are the roots of the polynomial
$$ U_n(x) = \sin((n+1)\arccos x)/\sqrt{1-x^2} $$
hence
$$ S_n=\sum_{k=1}^{n-1}\frac{1}{\sin^2\frac{\pi k}{n}}=\frac{1}{2\pi i}\oint \frac{1}{1-z^2}\cdot\frac{U_{n-1}'(z)}{U_{n-1}(z)}\,dx $$
where the integral is performed along a simple contour enclosing $[-1+\varepsilon,1-\varepsilon]$. By enforcing the substitution $z=\cos\theta$ we are left with
$$ S_n = -\frac{1}{2\pi i}\oint \frac{\cot\theta-n\cot(n\theta)}{\sin^2\theta}\,d\theta $$
where the integral is performed along a simple contour enclosing $[\varepsilon,\pi-\varepsilon]$. Due to the periodicity of the integrand function and the fact that the sum of the residues of such meromorphic function has to be zero we have
$$ S_n = \operatorname*{Res}_{z=0}\frac{\cot z-n\cot(n z)}{\sin^2 z} = [z^{2}]\frac{z\cot z-nz\cot(n z)}{\left(\frac{\sin z}{z}\right)^2}=(1-n^2)\cdot[z^2]z\cot z$$
and we may finish be considering that in a neighbourhood of the origin, up to a term $O(z^3)$,
$$ z\cot z = \frac{z\cos z}{\sin z} \equiv \frac{z\left(1-\frac{z^2}{2}\right)}{z\left(1-\frac{z^2}{6}\right)}\equiv 1-\frac{z^2}{3} $$
such that $S_n = \frac{n^2-1}{3}$.
|
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|
sequence $U_k=a_0U_{k-1}+a_1U_{k-2}+\cdots+a_{k-1}U_0$ Is there a general formula for $U_k$ defined by
$$U_k=a_0U_{k-1}+a_1U_{k-2}+\cdots+a_{k-1}U_0$$
where the $a_i$ are in arithmetic progression and $U_0=1$? Do there always exist $c,d$ such that $U_k\to cd^k$ as $k\to\infty$?
If $a_n=n+1$,
$U_k=F_{2k}$ where $F_n$ is the $n^{th}$ Fibonacci number.
If $a_n=bn-1 \ \land \ b>1$,
$U_k=b^2(b+1)^{k-2} \ \forall \ k\geq2$
|
The recurrence equation is written as
$$
U_k = \sum_{m=0}^{k-1} a_{k-1-m} U_m \tag{1}
$$
Let's form the generating function $f(x) = \sum_{k=0}^\infty x^k U_k$. Then, summing eq. 1, multiplied on both sides with $x^k$, from $k=1$ to $\infty$:
$$
f(x) - 1 = x \cdot f(x) \cdot \sum_{k=0}^\infty a_k x^k = x f(x) g(x)
$$
Since $a_k = a_0 + (a_1-a_0) k$, $g(x) = \sum_{k=0}^\infty a_k x^k = a_0 \frac{1-2x}{(1-x)^2} + a_1 \frac{x}{(1-x)^2}$. Hence
$$
f(x) = \frac{1}{1-x g(x)} = \frac{(1-x)^2}{(1-x)^2 - a_0 x + x^2 (2a_0 - a_1)}
$$
The case of $a_n = n+1$, that corresponds to $a_0=1$ and $a_1 = 2$, we recover the generating function of $F_{2k}$:
$$
f(x) = \frac{(1-x)^2}{1-3 x + x^2} = 1 + x + 3 x^2 + 8 x^3 + \mathcal{o}(x^3)
$$
The case of $a_n = b n -1$ maps to $a_0 =-1$ and $a_1 = b-1$ and generating function:
$$
f(x) = \frac{(1-x)^2}{1-x- b x^2}
$$
The above generating function does not correspond to $U_k = b^2 (b+1)^{k-2}$ for $k \geqslant 2$.
In[44]:= u[0] = 1;
In[45]:= u[k_Integer?Positive] :=
u[k] = Expand[Sum[a[k - 1 - m] u[m], {m, 0, k - 1}]]
In[49]:= (Sum[u[k] x^k, {k, 0, 12}] /. {a[m_] :> b m - 1}) -
Series[-((-1 + x)^2/(-1 + x + b x^2)), {x, 0, 12}]
Out[49]= SeriesData[x, 0, {}, 13, 13, 1]
The asymptotic growth of $U_k$ is determined by poles of $f(x)$.
|
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|
Finding $\int \frac{1}{x}\sqrt{1+\frac{1}{x^4}}\,dx$ Finding $$\int \frac{1}{x}\sqrt{1+\frac{1}{x^4}}\,dx$$
I suppose I need integration by parts and trigo substitution
Let $u=\frac{1}{x} \Rightarrow du = -\frac{1}{x^2} dx$
Let $dv = \sqrt{1+(\frac{1}{x^2})^2}$, $\frac{1}{x^2} = \tan{\theta}$. Is my substitution OK?
So $x=\frac{1}{\sqrt{\tan{\theta}}} \Rightarrow dx = -\frac{\sec^2{\theta}}{2\sqrt{\tan{\theta}}}\,d\theta$. But this will be very complicated to integrate later?
Am I supposed to be trying something else?
UPDATE: An attempt
$$\int \frac{1}{x} \sqrt{1+\frac{1}{x^4}} dx$$
Let $u = \frac{1}{x} \Rightarrow du = -\frac{1}{x^2} dx$
Let $dv = \sqrt{1+(\frac{1}{x^2})^2} dx$
Let $\alpha = \frac{1}{x^2} \Rightarrow d\alpha = -\frac{1}{2x^3}$
$dv=\sqrt{1+\alpha^2} d\alpha$
Let $\alpha = \tan{\theta} \Rightarrow d\alpha = \sec^2{\theta} d\theta$
$dv = \sqrt{1+\tan^2{\theta}} \sec^2{\theta} d\theta = \sec^3{\theta}$. Looks wrong here ?
|
$$ \begin{aligned}
\int\frac{1}{x}\sqrt{1+\frac{1}{x^4}}\,\mathrm{d}x&=\int \frac{\sqrt{x^4 +1}}{x^3}\,\mathrm{d}x\\
&=\int\frac{x^4 + 1}{x^3\sqrt{x^4+1}}\,\mathrm{d}x\\
&=\int\frac{x}{\sqrt{\left(x^2\right)^2+1}}\,\mathrm{d}x + \int \frac{1}{x^5}\frac{1}{\sqrt{1 + 1/x^4}}\,\mathrm{d}x\\
&=\frac{1}{2}\ln\Big(x^2 + \sqrt{x^4 + 1}\Big) - \frac{1}{2}\sqrt{1+1/x^4} + C\\
&=\frac{1}{2}\ln\Big(x^2 + \sqrt{x^4 + 1}\Big) - \frac{\sqrt{x^4 + 1}}{2x^2}+C
\end{aligned} $$
|
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|
Find The Number of Squares Consider the set of points $$S=\{(x,y):x,y\text{ are non-negative integers}\le n\}$$ Find the number of squares that can be formed with vertices belonging to $S$ and sides parallel to the axes.
|
I will need some drawing help. For an $n$ that is small but not too small, put dots at the $(n+1)\times (n+1)$ gridpoints with coordinates $(x,y)$, where $0\le x,y\le n$. Something like $n=5$ is good enough.
Now draw the diagonals that go in the Northwest to Southeast direction. The diagonal closest to the origin has $2$ gridpoints on it, the next diagonal has $3$ gridpoints, the next has $4$, and so on. This continues until we hit the main diagonal, which has $n+1$ gridpoints. As we go further up, the number of gridpoints decreases, to $n$, then $n-1$, and so on until our last diagonal, which has $2$ gridpoints.
Take any of these diagonals, and pick two gridpoints on it. Then there is a unique square which has these two points as its Northwest and Southeast corners. As we consider all of our diagonals, and all the ways to choose $2$ points, we produce in this way all possible squares, in exactly one way. It follows that the total number of squares is
$$\binom{2}{2}+\binom{3}{2}+\cdots +\binom{n}{2}+\binom{n+1}{2}+\binom{n}{2}+\cdots + \binom{3}{2}+\binom{2}{2}.$$
Nice, but not really simple. We now proceed to simplify. There is the middle term $\dbinom{n+1}{2}$ plus twice the quantity
$$\binom{2}{2}+\binom{3}{2}+\cdots +\binom{n}{2}.\tag{$\ast$}$$
We claim that the quantity $(\ast)$ is equal to $\dbinom{n+1}{3}$.
For $\dbinom{n+1}{3}$ is the number of ways of picking $3$ numbers from the $n+1$ numbers $0, 1, 2,\dots,n$. When we pick $3$ numbers, the smallest of the numbers picked could be $n-2$. Then the other two numbers can be picked in $\binom{2}{2}$ ways (well, $1$ way). Or else the smallest number picked could be $n-3$. Then the other two can be picked in $\binom{3}{2}$ ways. Or else the smallest is $n-4$, in which case the other two can be picked in $\binom{4}{2}$ ways. Continue on down. Finally, the smallest of the $3$ numbers picked could be $0$, in which case the other two can be picked in $\binom{n}{2}$ ways. We have proved that
$$\binom{2}{2}+\binom{3}{2}+\cdots +\binom{n}{2}=\binom{n+1}{3}.\tag{$\ast\ast$}$$
Now put things together. The total number of squares is therefore
$$\binom{n+1}{2}+2\binom{n+1}{3}.$$
Remark: Now for the interesting part! In the posted solutions, it was shown that the number of squares is equal to
$$1^2+2^2+3^2+\cdots +n^2.$$
We conclude that
$$1^2+2^2+3^2+\cdots +n^2=\binom{n+1}{2}+2\binom{n+1}{3}.$$
The expression on the right looks pretty nice as is. But if we really want to, we can expand it as
$$\frac{(n+1)(n)}{2}+2\frac{(n+1)(n)(n-1)}{3!}.$$
A little simplification (bring to a common denominator, factor out the common factors $n$ and $n+1$) gives us to the more familiar expression
$$\frac{n(n+1)(2n+1)}{6}.$$
So by counting the number of squares in a grid in two different ways, we can obtain a closed form formula for the sum of the first $n$ perfect squares.
|
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|
How do you integrate $\int \frac{1}{a + \cos x} dx$? How do you integrate $\int \frac{1}{a + \cos x} dx$? Is it solvable by elementary methods? I was trying to do it while incorrectly solving a homework problem but I couldn't find the answer.
Thanks!
|
Let $ y = \frac{x}{2}$.
$$\frac{1}{a + \cos 2y} = \frac{1}{a -1 + 2\cos ^2 y} = \frac{\sec^2 y}{(a-1)\sec^2 y + 2} = \frac{\sec^2 y}{a + 1 + (a-1)\tan^2 y} $$
Thus
$$\int \frac{1}{a + \cos x} \text{d}x = \int \frac{2}{a + \cos 2y} \text{d}y $$
$$ = \int \frac{ 2\sec^2 y}{ a + 1 + (a-1)\tan^2 y} \text{d} y$$
Now make the subsitution $t = \tan y$.
I remember having used the same trick before: Summing the series $ \frac{1}{2n+1} + \frac{1}{2} \cdot \frac{1}{2n+3} + \cdots \ \text{ad inf}$
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Improper Integral Question: $ \int_0 ^ \infty\frac{x\log x}{(1+x^2)^2} dx $ I have to test the convergence of the integral :
$$ \int_0 ^ \infty\frac{x\log x}{(1+x^2)^2} dx $$
Please suggest. Also, have to show that the value of the integral is zero ?
|
You want to find
$$ \int_0 ^ \infty\frac{x\log x}{(1+x^2)^2} dx $$
First, we find a primitive of the integrand, this is, we need
$$ \int\frac{x\log x}{(1+x^2)^2} dx =F(x)$$
Use integration by parts with $u = \log x$ and $dv = \dfrac{x}{(1+x^2)^2}$.
This makes $du = dx/x$ and $v=-\dfrac{1}{2(1+x^2)}$, from where
$$ \int\frac{x\log x}{(1+x^2)^2} dx = -\dfrac{\log x}{2(1+x^2)}+\int \dfrac{dx}{2x(1+x^2)}$$
To find the last integral, use partial fractions:
$$\frac{1}{{2x\left( {1 + {x^2}} \right)}} = \frac{1}{2}\left( {\frac{1}{x} - \frac{x}{{1 + {x^2}}}} \right)$$
This means
$$\int {\frac{{dx}}{{2x\left( {1 + {x^2}} \right)}}} = \frac{1}{2}\log x - \frac{1}{4}\log \left( {1 + {x^2}} \right)$$
So the primitive we're looking for is
$$ \int\frac{x\log x}{(1+x^2)^2} dx = -\dfrac{\log x}{2(1+x^2)}+\frac{1}{2}\log x - \frac{1}{4}\log \left( {1 + {x^2}} \right)$$
Writing this more tidily,
$$\int {\frac{{x\log x}}{{{{(1 + {x^2})}^2}}}} dx = - \frac{1}{2}\left[ {\frac{{\log x}}{{1 + {x^2}}} + \log \frac{{\sqrt {1 + {x^2}} }}{x}} \right]$$
Now we need to evaluate those expressions at $x \to 0$, and $x \to \infty$.
$$\int\limits_0^\infty {\frac{{x\log x}}{{{{(1 + {x^2})}^2}}}dx} = - \mathop {\lim }\limits_{x \to \infty } \frac{1}{2}\left[ {\frac{{\log x}}{{1 + {x^2}}} + \log \frac{{\sqrt {1 + {x^2}} }}{x}} \right] + \mathop {\lim }\limits_{x \to 0} \frac{1}{2}\left[ {\frac{{\log x}}{{1 + {x^2}}} + \log \frac{{\sqrt {1 + {x^2}} }}{x}} \right]$$
You can convince yourself pretty easily that
$$\mathop {\lim }\limits_{x \to \infty } \frac{{\log x}}{{1 + {x^2}}} = \mathop {\lim }\limits_{x \to \infty } \log \frac{{\sqrt {1 + {x^2}} }}{x} = 0$$
so all it is needed is to find
$$\mathop {\lim }\limits_{x \to 0} \frac{1}{2}\left[ {\frac{{\log x}}{{1 + {x^2}}} + \log \frac{{\sqrt {1 + {x^2}} }}{x}} \right]$$
which is an indeterminate form $\infty -\infty$.
This might be really tedious to solve, so I will use something that might be frowned upon:
$$\mathop {\lim }\limits_{x \to 0} \frac{1}{2}\left[ {\frac{{\log x}}{{1 + {x^2}}} + \log \frac{{\sqrt {1 + {x^2}} }}{x}} \right]=$$
$$\mathop {\lim }\limits_{x \to 0} \frac{1}{2}\left[ {\frac{{\log x}}{{1 + {x^2}}} + \frac{1}{2}\log \left( {\frac{1}{{{x^2}}} + 1} \right)} \right] = $$
Here's the catch: for $x \to 0$
$$\eqalign{
& 1 + {x^2} \sim 1 \cr
& \frac{1}{{{x^2}}} + 1 \sim \frac{1}{{{x^2}}} \cr} $$
so the limit can be simplified to
$$\mathop {\lim }\limits_{x \to 0} \frac{1}{2}\left[ {\log x + \frac{1}{2}\log \left( {\frac{1}{{{x^2}}}} \right)} \right]=$$
$$\mathop {\lim }\limits_{x \to 0} \frac{1}{2}\left( {\log x - \log x} \right) = 0$$
So
$$ \int_0 ^ \infty\frac{x\log x}{(1+x^2)^2} dx =0$$
This is why maybe using subtitutions is more recommended.
|
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|
Chain rule for $ y = (x^2 + x^3)^4$ I am trying to find the derivative of $ y = (x^2 + x^3)^4$
and it seems pretty simple I get
$4(x^2+x^3)^3 (2x+3x^2)$ This seems to be the proper answer to me but the book gets
$4x^7 (x+1)^3 (3x+2)$ and I have no idea how that happened, what process the author went through or why. My answer seems to be a better and more accurate answer since that is what the chain rule will give you.
Why is my answer wrong?
|
You did the differentiation correctly. The book's solution is the same as your answer, but in simplified form.
To obtain the book's answer from yours: factor $x^2$ from the $(x^2+x^3)$ term in your expression and apply the rule $(ab)^n=a^nb^n$, factor $x$ from the $(2x+3x^2)$ term, and finally combine the $x^6$ and $x$ terms.
$$\eqalign{
4(\color{maroon}{x^2+x^3})^3 (\color{darkgreen}{2x+3x^2})
&= 4 \bigl(\color{maroon}{ x^2(1+x)}\bigr)^3\cdot\color{darkgreen}{ x(2+3x)}\cr
&=4\cdot (x^2)^3(1+x)^3\cdot x(2+3x)\cr
&=4x^6(1+x)^3\cdot x(2+3x)\cr
&=4x^7(x+1)^3(3x+2).
}
$$
|
{
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
A trigonometric identity: $(\sin x)^{-2}+(\cos x)^{-2}=(\tan x+\cot x)^2$ I've been trying to prove it for a while, but can't seem to get anywhere.
$$\frac{1}{\sin^2\theta} + \frac{1}{\cos^2\theta} = (\tan \theta + \cot \theta)^2$$
Could someone please provide a valid proof?
I am not allowed to work on both sides of the equation.
Work so far:
RS:
$$
\begin{align}
& \frac{\sin^2\theta}{\cos^2\theta} + \frac{\cos^2\theta}{\sin^2\theta} + 2 \\[10pt]
& = \frac{\sin^4\theta}{(\cos^2\theta)(\sin^2\theta)} + \frac{\cos^4\theta}{(\sin^2\theta) (\cos^2\theta)} + \frac{(\sin^4\theta)(\cos^2\theta)}{(\sin^2\theta)(\cos^2\theta)} + \frac{(\sin^2\theta)(\cos^4\theta)}{(\sin^2\theta)(\cos^2\theta)} \\[10pt]
& = \frac{\sin^4\theta + \cos^4\theta + (\sin^4\theta)(\cos^2\theta) + (\sin^2\theta)(\cos^4\theta)}{(\cos^2\theta)(\sin^2\theta)}
\end{align}
$$
I am completely lost after this.
|
Hint :
$$\sin^2 \theta=\frac{\tan^2 \theta}{1+\tan^2 \theta}$$
$$\cos^2 \theta=\frac{1}{1+\tan^2 \theta}$$
$$\tan \theta = \frac{1}{\cot \theta}$$
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Proof that $\int_{-1}^{1}\sqrt{1-x^2}dx=\frac{\pi}{2}$? In my textbook appears that $\displaystyle\int_{-1}^{1}\sqrt{1-x^2}dx=\frac{\pi}{2}$
But where does this equation come from?
|
Not the "cleverest" method, like the above, - but works!
$$\int_{-1}^1 \sqrt{1-x^2} \ dx$$
To compute that integral, one may substitute $x=\sin{t}$, and get:
$$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sqrt{1-\sin^2{t}} \ d(\sin{t})=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos{t}\cos{t} \ dt=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos^2{t} \ dt$$
$$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos^2{t} \ dt=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{1+\cos{2t}}{2} \ dt=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{dt}{2}+\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{\cos{2t}}{2}=$$
$$= \frac{\pi}{2}+\frac{1}{2}\sin{t}\cos{t}{\huge{|}}_{-\frac{\pi}{2}}^{\frac{\pi}{2}}=\frac{\pi}{2}+0=\frac{\pi}{2}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/143287",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Prove $\gcd(a+b,a^2+b^2)$ is $1$ or $2$ if $\gcd(a,b) = 1$
Assuming that $\gcd(a,b) = 1$, prove that $\gcd(a+b,a^2+b^2) = 1$ or $2$.
I tried this problem and ended up with
$$d\mid 2a^2,\quad d\mid 2b^2$$
where $d = \gcd(a+b,a^2+b^2)$, but then I am stuck; by these two conclusions how can I conclude $d=1$ or $2$?
And also is there any other way of proving this result?
|
Since $\text{gcd}(a,b) = 1$, we have that there exists $x,y \in \mathbb{Z}$ such that $$ax+by = 1$$ Hence, we have that $$(a+b)x + b(y-x) = 1$$ and $$a(x-y) + (a+b)y = 1$$
Squaring and adding the two equations, we get that $$(a+b)^2 x^2 + b^2(y-x)^2 + 2b(a+b)x(y-x) + (a+b)^2 y^2 + a^2(x-y)^2 + 2a(a+b)y(x-y) = 2$$
Rearranging the above equation, we get that
$$(a+b) \left( (a+b) (x^2+y^2) + 2 (x-y) (ay-bx) \right) + (a^2 + b^2)(x-y)^2 = 2$$
Hence, we have that $$(a+b) X + (a^2 + b^2) Y =2$$
where $X = \left( (a+b) (x^2+y^2) + 2 (x-y) (ay-bx) \right)$ and $Y = (x-y)^2$. This implies that $\text{gcd}(a,b) \vert 2$.
Hence, $$\text{gcd}(a,b) = 1 \text{ or } 2$$
|
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"url": "https://math.stackexchange.com/questions/153125",
"timestamp": "2023-03-29T00:00:00",
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|
Using conjugates to find a limit with a cubic root: $\lim_{h\to 0}\frac{\sqrt[3]{h+1}-1}{h}$ I currently have $$\lim_{h\to 0}\frac{\sqrt[3]{h+1}-1}{h}$$
Now, I know that when you have square root instead of a cubic root it's easy. You just multiply by the conjugate and simplify afterwards. If it were a sqrt I know that the limit is 1/2. I know that the result here is 1/3 but I can't seem to get there. I always end up with 1/4 due to getting something like:
$$\frac{h+1-1}{h(\sqrt[3]{h+1}+1)(\sqrt[3]{h+1}+1)}$$
After simplifying and evaluating the limit you end up with
$$\frac{1}{(\sqrt[3]{0+1}+1)(\sqrt[3]{0+1}+1)}$$
which turns out to be $1/4$.
What am I doing wrong?
|
You can't conjugate cubic roots the same way you do square roots. You don't get the answer you think you get if you multiply $(\sqrt[3]{h+1}-1)(\sqrt[3]{h+1}+1)(\sqrt[3]{h+1}+1)$. Multiply it out and see that you don't get $h+1-1$:
$$\begin{align*}
\left(\sqrt[3]{h+1}-1\right)\left(\sqrt[3]{h+1}+1\right)\left(\sqrt[3]{h+1}+1\right) &= \left( \sqrt[3]{h+1}^2 - 1\right)\left(\sqrt[3]{h+1}+1\right)\\
&= \sqrt[3]{h+1}^3 + \sqrt[3]{h+1}^2 - \sqrt[3]{h+1} - 1\\
&= h+1 +(h+1)^{2/3} - (h+1)^{1/3} - 1.
\end{align*}$$
So you don't get the concellation you think you get.
If you don't know about derivatives yet, you can do a similar trick to the one used for square roots. When dealing with square roots, you are making use of the identity
$$(a+b)(a-b) = a^2-b^2.$$
Here, you want to get rid of a cubic root, so you should make use of the identity
$$(a-b)(a^2+ab+b^2) = a^3-b^3.$$
So what we want to do is multiply the numerator and denominator by the factor
$$\sqrt[3]{h+1}^2 + \sqrt[3]{h+1} + 1$$
(taking $a=\sqrt[3]{h+1}$ and $b=1$) to get the right cancellation. If we do, we get:
$$\begin{align*}
\lim_{h\to 0}\frac{\sqrt[3]{h+1} - 1}{h} &= \lim_{h\to 0}\frac{((h+1)^{1/3}-1)((h+1)^{2/3} + (h+1)^{1/3}+1)}{h((h+1)^{2/3} + (h+1)^{1/3} + 1)}\\
&= \lim_{h\to 0}\frac{h+1-1}{h((h+1)^{2/3} + (h+1)^{1/3}+1)}\\
&= \lim_{h\to 0}\frac{h}{h((h+1)^{2/3} + (h+1)^{1/3} + 1)}\\
&= \lim_{h\to 0}\frac{1}{(h+1)^{2/3} + (h+1)^{1/3} + 1}\\
&= \frac{1}{\sqrt[3]{1^2} + \sqrt[3]{1} + 1}\\
&= \frac{1}{1+1+1}\\
&= \frac{1}{3}.
\end{align*}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
How to find the closed form formula for this recurrence relation $ x_{0} = 5 $
$ x_{n} = 2x_{n-1} + 9(5^{n-1})$
I have computed: $x_{0} = 5, x_{1} = 19, x_{2} = 83, x_{3} = 391, x_{4} = 1907$, but cannot see any pattern for the general $n^{th}$ term.
|
If you're not familiar with the method that Phira explained, divide both sides by $2^n$:
$$
\dfrac{x_{n}}{2^n} = \dfrac{x_{n-1}}{2^{n-1}} + \dfrac{9}{2}\left(\dfrac{5}{2}\right)^{n-1}
$$
Call $\dfrac{x_{n}}{2^n} = s_n$:
$$
s_n = s_{n-1} + \dfrac{9}{2}\left(\dfrac{5}{2}\right)^{n-1}
$$
If we keep expanding $s_{n-1}$ in the RHS recursively, we get:
$$
s_n = s_0 + \dfrac{9}{2}\left(\dfrac{5}{2}\right)^{0} + \dfrac{9}{2}\left(\dfrac{5}{2}\right)^{1} \cdots + \dfrac{9}{2}\left(\dfrac{5}{2}\right)^{n-2} + \dfrac{9}{2}\left(\dfrac{5}{2}\right)^{n-1}
$$
Where $s_0 = \dfrac{x_0}{2^0} = x_0$.
This means that:
$$
s_n = x_0 + \sum_{k=1}^{n}\dfrac{9}{2}\left(\dfrac{5}{2}\right)^{k-1}
$$
This is a geometric series:
$$
s_n = 3 \left(\dfrac{5}{2}\right)^n + 2
$$
And therefore:
$$
x_n = 3 \cdot 5^n + 2 \cdot 2^n
$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Plotting a quadratic equation in the $\,xy\,$- plane My question is:
Represent the following set of points in the $\,xy\,$- plane:
$$\left\{ (x,y)\,\, |\,\, x^2 + y^2 - 2x - 2y + 1 = 0 \right\}$$
What i got: $\,\,(x-2)^2 + (y-2)^2 = 1\,\,$
I am not getting what to do next.
Any help to solve this question would be greatly appreciated.
Thank you,
|
The first problem is that you carried out the algebra incorrectly. When you complete the square with $x^2-2x$ you should get $(x-1)^2-1$, which you can verify by multiplying it out. Similarly, $y^2-2y=(y-1)^2-1$. Thus, $$\begin{align*}x^2+y^2-2x-2y+1&=(x-1)^2-1+(y-1)^2-1+1\\
&=(x-1)^2+(y-1)^2-1\;,
\end{align*}$$
and the points where $x^2+y^2-2x-2y+1=0$ are the points where $(x-1)^2+(y-1)^2-1$, i.e., where $(x-1)^2+(y-1)^2=1$.
What’s the distance between the points $(x,y)$ and $(1,1)$? It’s $\sqrt{(x-1)^2+(y-1)^2}$, right? And if $(x-1)^2+(y-1)^2=1$, then $\sqrt{(x-1)^2+(y-1)^2}=\sqrt1=1$, so your set contains the points whose distance from $(1,1)$ is $1$. What does that set of points look like?
|
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|
Evaluating $\int \frac{x^3}{(x^2 + 1)^\frac{1}{3}}dx$ $$\int \frac{x^3}{(x^2 + 1)^\frac{1}{3}}dx$$
I am suppose to make a $u$ substitution and to make this a rational integral and then evaluate it from there but I have no idea how to do that. There aren't any good examples of this in the book and I can not find any us that make this doable.
|
Let $x^2 + 1 = t^3$. Hence, we get that $2x dx = 3t^2dt$ and $xdx = \dfrac{3t^2}{2} dt$. Hence, we have that
\begin{align}
I & = \int \dfrac{x^3}{(x^2+1)^{1/3}} dx = \int \dfrac{x^2 (xdx)}{(x^2+1)^{1/3}}\\
& = \int \dfrac{t^3-1}{t} \dfrac{3t^2}{2} dt = \dfrac32 \int(t^4-t) dt\\
& = \dfrac32 \left(\dfrac{t^5}{5} - \dfrac{t^2}{2} \right) + C\\
& = \dfrac32 \left( \dfrac{(x^2+1)^{5/3}}{5} - \dfrac{(x^2+1)^{2/3}}{2}\right) + C
\end{align}
|
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|
Area of a surface of revolution of $y = \sqrt{4x+1}$ $y = \sqrt{4x+1}$ for $1 \leq x \leq 5$
I really have no idea what to do with this problem, I attempted something earlier which I will not type up because it took me two pages.
$$y = \sqrt{4x+1}$$
$$\int 2 \pi \sqrt{4x+1} \sqrt{1 + \frac{4}{1+4x}}dx$$
$$2 \pi \int \sqrt{4x+1} \sqrt{1 + \frac{4}{1+4x}}dx$$
Nothing really seems obvious at this point, I attempted a u substitution of $u = 1+4x$ but it does not help simplify this problem really.
$$ \pi /2 \int \sqrt{u} \sqrt{1 + \frac{4}{u}}du$$
I thought about making a wonky trig substitution but it didn't seem to help and was overly complicated.
|
From the last step::
$$ \frac{\pi}{2} \int \sqrt u \frac{\sqrt{4 +u}}{\sqrt u} du = \frac \pi 2 \int \sqrt{4 + u} du $$
substituting $4 + u = p \implies du = dp \;\;$, we get
$$ = \frac \pi 2 \int \sqrt p dp = \frac \pi 2 \frac{p^{3/2}}{3/2} = \frac \pi 3 (4+u)^{3/2} $$
|
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|
Positive definite quadratic form $\sum_{i=1}^n x_i^2 + \sum_{1\leq i < j \leq n} x_{i}x_j$ Is $\sum_{i=1}^n x_i^2 + \sum_{1\leq i < j \leq n} x_{i}x_j$ positive definite?
Approach:
The matrix of this quadratic form can be derived to be the following
$$M := \begin{pmatrix}
1 & \frac{1}{2} & \frac{1}{2} & \cdots & \frac{1}{2} \\
\frac{1}{2} & 1 & \frac{1}{2} & \cdots & \frac{1}{2} \\
\cdots & \cdots & \cdots & \cdots & \cdots \\
\frac{1}{2} & \frac{1}{2} & \frac{1}{2} & \cdots & \frac{1}{2} \\
\frac{1}{2} & \frac{1}{2} & \frac{1}{2} & \cdots & 1 \\
\end{pmatrix}$$
It suffices to show that $\operatorname{det}M > 0$, then the claim follows.
Any hints how to show the positivity of this determinant?
|
We can more generally show that the $n\times n$ determinant of $$\left|\begin{array}{cccc}a & b & \cdots & b \\ c & a & \ddots & \vdots\\ \vdots & \ddots & \ddots & b \\ c & \cdots & c & a\end{array}\right|$$
is $\left\{\begin{array}{ll}\frac{1}{c-b}(c(a-b)^n-b(a-c)^n) & \text{ if }b\neq c \\ (a-b+nb)(a-b)^{n-1}& \text{ if } b=c\end{array}\right.$
So in your case, the determinant is $\frac{n+1}{2^n}>0$.
Proof :
Let $$H(x)=\left|\begin{array}{cccc}a-x & b-x & \cdots & b-x \\ c-x & a-x & \ddots & \vdots\\ \vdots & \ddots & \ddots & b-x \\ c-x & \cdots & c-x & a-x\end{array}\right|$$
Do $L_2\leftarrow L_2-L_1,\ldots,L_n\leftarrow L_n-L_1$ :
$$H(x)=\left|\begin{array}{ccc}a-x & b-x & \cdots & \cdots & \cdots & b-x \\ \\ c-a & a-b & 0&\cdots&\cdots& 0\\\vdots & c-b & \ddots&\ddots&&\vdots\\\vdots&\vdots & \ddots&\ddots&\ddots&\vdots\\\vdots&\vdots&&\ddots&\ddots&0\\c-a&c-b&\cdots&\cdots&c-b&a-b\end{array}\right|$$
and $C_2\leftarrow C_2-C_1,\ldots,C_n\leftarrow C_n-C_1$ :
$$H(x)=\left|\begin{array}{cccccc}a-x&b-a&\cdots&\cdots&\cdots&b-a\\c-a&2a-b-c&a-c&\cdots&\cdots&a-c\\\vdots&a-b&\ddots&\ddots&&\vdots\\
\vdots&\vdots&\ddots&\ddots&\ddots&\vdots\\\vdots&\vdots&&\ddots&\ddots&a-c\\c-a&a-b&\cdots&\cdots&a-b&2a-b-c\end{array}\right|$$
So it exists $\alpha$ and $\beta$ such as $H(x)=\alpha x+\beta$.
If $b\neq c$ we can evaluate $H(b)$ and $H(c)$ with the first writing of $H(x)$ : $$\left\{\begin{array}{c}\alpha b+\beta=(a-b)^n \\ \alpha c+\beta=(a-c)^n\end{array}\right.$$
So you get $\alpha$ and $\beta$, and your determinant is $H(0)=\frac{1}{c-b}(c(a-b)^n-b(a-c)^n)$.
If $b=c$ : your determinant is $\lim_{c\rightarrow b}\frac{1}{c-b}(c(a-b)^n-b(a-c)^n)=-f'(b)$ where $f(x)=(c+b-x)(a-x)^n$.
We can easily improve this proof to the case where the elements on the diagonal are not equal
|
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|
How did Ramanujan get this result? We know Ramanujan got this result
$$\sqrt{1+2\sqrt{1+3\sqrt{1+\cdots }}}=3$$
and he used the formula
$$x+n+a=\sqrt{ax+{{(n+a)}^{2}}+x\sqrt{a(x+n)+{{(n+a)}^{2}}+(x+n)\sqrt{\cdots }}}$$
where $x=2,n=1,a=0$ ,we get the first result, but I don't know how to prove it, can you help me?
|
Just use the following formula repeatedly
$ n^2=1+(n-1) (n+1) $
then we have
$ 2=\sqrt{1+1\times 3} $
$ 2=\sqrt{1 +1 \sqrt{1+2\times 4}}$
$ 2=\sqrt{1 +1 \sqrt{1+2 \sqrt{1+3\times 5}}} $
$ 2=\sqrt{1 +1 \sqrt{1+ 2 \sqrt{1+ 3 \sqrt{1+4\times 6}}}} $
$ 2=\sqrt{1+1 \sqrt{1+2 \sqrt{1+ 3 \sqrt{1+4 \sqrt{1+5\times 7}}}}}$
$ 2=\sqrt{1+1 \sqrt{1+2 \sqrt{1+3 \sqrt{1+4 \sqrt{1+5 \sqrt{1+6\times 8}}}}}} $
$ \cdots $
By using
$ n+(n-1) n(n+1) =n^3 $
I found the below new results
$ \sqrt[3]{2+1\times2 \sqrt[3]{3+2\times 3 \sqrt[3]{4+3\times 4 \sqrt[3]{5+4\times 5 \sqrt[3]{6+5\times 6 \sqrt[3]{7+...\ }}}}}}=2 $
$ \sqrt[3]{3+2\times 3 \sqrt[3]{4+3\times 4 \sqrt[3]{5+4\times 5 \sqrt[3]{6+5\times 6 \sqrt[3]{7+6\times 7 \sqrt[3]{8+...\ }}}}}}=3 $
$ \sqrt[3]{4+3\times 4 \sqrt[3]{5+4\times 5 \sqrt[3]{6+5\times 6 \sqrt[3]{7+6\times 7 \sqrt[3]{8+7\times 8 \sqrt[3]{9+...\ }}}}}}=4 $
$ \cdots $
More result can be found below:
http://eslpower.org/Notebook.htm
|
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"timestamp": "2023-03-29T00:00:00",
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|
prove that $\frac{b+c}{a^2}+\frac{c+a}{b^2}+\frac{a+b}{c^2}\ge\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$ Assume: $a,b,c >0$ prove that :
$$\frac{b+c}{a^2}+\frac{c+a}{b^2}+\frac{a+b}{c^2}\ge\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$$
|
$$\sum_{cyc}\left(\frac{b+c}{a^2}-\frac{1}{a}\right)>\sum_{cyc}\left(\frac{b+c}{a^2}-\frac{2}{a}\right)=$$
$$=\sum_{cyc}\frac{c-a-(a-b)}{a^2}=\sum_{cyc}(a-b)\left(\frac{1}{b^2}-\frac{1}{a^2}\right)\sum_{cyc}\frac{(a-b)^2(a+b)}{a^2b^2}\geq0$$
Done!
|
{
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"url": "https://math.stackexchange.com/questions/161318",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Computing the derivative from the definition Using the limit definition of the derivative which I know is:
$$f'(x)=\lim_{h\to0}\left(\frac{f(x+h)-f(x)}{h}\right)$$
I am trying to solve this problem
$$f(x)= \frac{x}{x+2} $$
How do I go about properly solving this, I seemed to get
$$\frac{x}{x+2}\ $$
as my answer again? What are the steps I should follow?
I am trying to find the derivative of $f(x)= \frac{x}{x+2}$ using the definition of the derivative.
|
I think that you mean that you want to use the definition of the derivative to find the derivative of the function $$f(x)=\frac{x}{x+2}\;.$$ The definition is that $$f\,'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}h\;.$$ Note the limit: it’s essential.
For your function this is
$$\begin{align*}
\lim_{h\to 0}\frac{\frac{x+h}{(x+h)+2}-\frac{x}{x+2}}h&=\lim_{x\to 0}\left(\frac1h\left(\frac{x+h}{(x+h)+2}-\frac{x}{x+2}\right)\right)\\
&=\lim_{h\to 0}\left(\frac1h\cdot\frac{(x+2)(x+h)-x(x+2+h)}{(x+2)(x+2+h)}\right)\\
&=\lim_{h\to 0}\frac{x^2+2x+hx+2h-x^2-2x-hx}{h(x+2)(x+2+h)}\\
&=\lim_{h\to 0}\frac{2h}{h(x+2)(x+2+h)}\\
&=\lim_{h\to 0}\frac2{(x+2)(x+2+h)}\;,
\end{align*}$$
where I’ve simply put the original numerator $f(x+h)-f(x)$ over a common denominator and simplified. At this point we’ve managed to get rid of the factor of $h$ in the denominator, so we can take the limit: as $h\to 0$, the numerator just sits there at $2$, and the denominator approaches $(x+2)(x+2)$, or $(x+2)^2$. Thus,
$$f\,'(x)=\lim_{x\to 0}\frac2{(x+2)(x+2+h)}=\frac2{(x+2)^2}\;.$$
Added: By the way, if your algebra is good enough, you can notice that $$\frac{x}{x+2}=\frac{(x+2)-2}{x+2}=1-\frac2{x+2}\;.$$ Then
$$\begin{align*}
f(x+h)-f(x)&=\left(1-\frac2{(x+h)+2}\right)-\left(1-\frac2{x+2}\right)\\
&=\frac2{x+2}-\frac2{x+2+h}\\
&=\frac{2\big(x+2+h-(x+2)\big)}{(x+2)(x+2+h)}\\
&=\frac{2h}{(x+2)(x+2+h)}\;,
\end{align*}$$
and you reach the point of being able to take the limit a bit more easily.
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Slope of curve in $\mathbb{R}^3$ While doing revision, I came across this problem:
The surface given by $z=x^2-y^2$ is cut by the plane given by $y=3x$, producing a curve in the plane. Find the slope of this curve at the point $(1,3,-8)$.
I tried substituting $y=3x$ into $z=x^2-y^2$, yielding $z=-8x^2$. Then, $\frac{dz}{dx}=-16x=-16$.
However the answer is $-8\sqrt{\frac{2}{5}}$.
Thank you very much for any help.
|
This is a very badly posed question, and does not have an answer. (Read the comments.)
The following is a solution to a rephrased question which can be answered, however.
Question:
The surface given by $z=x^2−y^2$ is cut by the plane given by $y=3x$, producing a curve in the plane.
Treating the intersection as a curve in the said plane with vertical axis along $(0,0,1)$ and horizontal axis along $(1,3,0)/\sqrt{10}$, find the slope of this curve at the point (1,3,−8).
Solution:
Any point on the plane has Cartesian coordinates in the form $$\frac{a}{\sqrt{10}}\begin{pmatrix} 1\\3 \\0 \end{pmatrix} + b\begin{pmatrix} 0\\0\\1 \end{pmatrix}.$$ Substituting this into $z=x^2−y^2$, we get $b = -4a^2/5$.
So the "slope" at a point on this intersection, with $a$ and $b$ given, is $$\frac{db}{da} = -\frac{8a}{5}.$$
Setting $$\begin{pmatrix}1 \\3\\-8\end{pmatrix} = \frac{a}{\sqrt{10}}\begin{pmatrix} 1\\3 \\0 \end{pmatrix} + b\begin{pmatrix} 0\\0\\1 \end{pmatrix},$$ we get $a = \sqrt{10}$ and so the "slope" at this point is $-8\sqrt{\frac{2}{5}}$.
Solution using grad:
Let $f:=x^2-y^2-z$ and $g:=y-3x$. At the point $(1,3,-8)$, $\nabla f=(2,-6,-1)$ and $\nabla g=(-3,1,0)$. Their cross product, $(1,3,-16)$, is along the tangent direction of the intersecting curve produced by the surface and the plane, at the point $(1,3,-8)$.
Denote the angle between $(1,3,-16)$ and $(1,3,0)$ (i.e. the "horizontal") by $\theta$. Then, using the dot product, $\cos\theta = \sqrt{\frac{5}{133}}$. The "slope" is $$\tan \theta = - \sqrt{\frac{1}{\cos^2 \theta}-1} = -8\sqrt{\frac{2}{5}},$$ where the negative square root is taken because the "vertical" is along $(0,0,1)$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/161996",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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|
Integration Example How can i find the integration of this example
$$\int \frac{\sin x}{\sin x - \cos x } dx$$
I tried first add cos and then substracting cos but then what about $$\int \frac{\cos x}{\sin x - \cos x } dx\ ?$$
|
Now let
$$ I = \int \frac{\cos x}{\cos x + \sin x} \, \mathrm{d}x \quad \text{and} \quad J = \int \frac{\sin x}{\cos x + \sin x} \, \mathrm{d}x$$
Notice that
\begin{align*}
I + J & = \int \frac{\cos x + \sin x}{\cos x + \sin x} \, \mathrm{d}x = x + \mathcal{C} \\
I - J & = \int \frac{(\sin x + \cos x)'}{\cos x + \sin x} \, \mathrm{d}x = \ln\bigl( \cos x + \sin x \bigr) + \mathcal{C}
\end{align*}
Solving the system yields
$$I = \frac{1}{2}x + \frac{1}{2}\ln\bigl( \cos x + \sin x \bigr) + \mathcal{C}$$
and
$$J = \frac{1}{2}x - \frac{1}{2}\ln\bigl( \cos x + \sin x \bigr) + \mathcal{C}$$
as wanted.
|
{
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"url": "https://math.stackexchange.com/questions/162783",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
Integral of determinant Good evening. I need help with this task
$$
\int\limits_{-\pi}^\pi\int\limits_{-\pi}^\pi\int\limits_{-\pi}^\pi{\det}^2\begin{Vmatrix}\sin \alpha x&\sin \alpha y&\sin \alpha z\\\sin \beta x&\sin \beta y&\sin \beta z\\\sin \gamma x&\sin \gamma y&\sin \gamma z\end{Vmatrix} \text{d}x\,\text{d}y\,\text{d}z
$$
where $\alpha,\beta,\gamma$ are integers.
Computations are horrible, I gave up.
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Exact solution and Mathematica code to produce it (< 10 seconds computing time):
M = {{Sin[a x], Sin[a y], Sin[a z]},
{Sin[b x], Sin[b y], Sin[b z]},
{Sin[c x], Sin[c y], Sin[c z]}};
Expand[Det[M]^2] /. Plus -> List;
Total[Integrate[%, {x, -Pi, Pi}, {y, -Pi, Pi}, {z, -Pi, Pi}]]
It looks terrible but for a computer no problem:
$-\frac{8 \left(\pi -\frac{\sin (2 c \pi )}{2 c}\right) (b \cos (b \pi ) \sin (a \pi )-a \cos (a \pi ) \sin (b \pi ))^2}{\left(a^2-b^2\right)^2}-\frac{8 (2 c \pi -\sin (2 c \pi )) (b \cos (b \pi ) \sin (a \pi )-a \cos (a \pi ) \sin (b \pi ))^2}{\left(a^2-b^2\right)^2 c}+\frac{16 (a \cos (a \pi ) \sin (b \pi )-b \cos (b \pi ) \sin (a \pi )) (2 c \cos (c \pi ) \sin (a \pi )-2 a \cos (a \pi ) \sin (c \pi )) (b \cos (b \pi ) \sin (c \pi )-c \cos (c \pi ) \sin (b \pi ))}{\left(a^2-b^2\right) \left(a^2-c^2\right) \left(b^2-c^2\right)}+\frac{16 (2 b \cos (b \pi ) \sin (a \pi )-2 a \cos (a \pi ) \sin (b \pi )) (a \cos (a \pi ) \sin (c \pi )-c \cos (c \pi ) \sin (a \pi )) (c \cos (c \pi ) \sin (b \pi )-b \cos (b \pi ) \sin (c \pi ))}{\left(a^2-b^2\right) \left(a^2-c^2\right) \left(c^2-b^2\right)}+\frac{16 (a \cos (a \pi ) \sin (b \pi )-b \cos (b \pi ) \sin (a \pi )) (a \cos (a \pi ) \sin (c \pi )-c \cos (c \pi ) \sin (a \pi )) (2 c \cos (c \pi ) \sin (b \pi )-2 b \cos (b \pi ) \sin (c \pi ))}{\left(a^2-b^2\right) \left(a^2-c^2\right) \left(b^2-c^2\right)}-\frac{(\sin (2 a \pi )-2 a \pi ) (2 b \pi -\sin (2 b \pi )) \left(\pi -\frac{\sin (2 c \pi )}{2 c}\right)}{2 a b}-\frac{(\sin (2 a \pi )-2 a \pi ) \left(\pi -\frac{\sin (2 b \pi )}{2 b}\right) (2 c \pi -\sin (2 c \pi ))}{2 a c}-\frac{\left(\pi -\frac{\sin (2 a \pi )}{2 a}\right) (\sin (2 b \pi )-2 b \pi ) (2 c \pi -\sin (2 c \pi ))}{2 b c}-\frac{8 \left(\pi -\frac{\sin (2 b \pi )}{2 b}\right) (c \cos (c \pi ) \sin (a \pi )-a \cos (a \pi ) \sin (c \pi ))^2}{\left(a^2-c^2\right)^2}-\frac{8 (2 b \pi -\sin (2 b \pi )) (c \cos (c \pi ) \sin (a \pi )-a \cos (a \pi ) \sin (c \pi ))^2}{b \left(a^2-c^2\right)^2}-\frac{8 \left(\pi -\frac{\sin (2 a \pi )}{2 a}\right) (c \cos (c \pi ) \sin (b \pi )-b \cos (b \pi ) \sin (c \pi ))^2}{\left(b^2-c^2\right)^2}-\frac{8 (2 a \pi -\sin (2 a \pi )) (c \cos (c \pi ) \sin (b \pi )-b \cos (b \pi ) \sin (c \pi ))^2}{a \left(b^2-c^2\right)^2}$
Perhaps someone can simplify it from here?
Edit:
Confirmed
FullSimplify[%,
a \[Element] Integers &&
b \[Element] Integers &&
c \[Element] Integers]
outputs $6\pi^3$.
|
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|
If $n\ne 4$ is composite, then $n$ divides $(n-1)!$. I have a proof and need some feedback. It seems really obvious that the statement is true but it is always the obvious ones that are a little trickier to prove. So I would appreciate any feedback. Thank you!
Here is what I am asked to prove:
If $n$ is composite then $(n-1)! \equiv 0 \pmod n$.
Proof:
$n$ is composite $\implies n=ab$ where $a,b \in \mathbb{Z}$ and $0<a,b<n$.
Case 1:
If $a=b$ then $n=a^{2}$.
Now $n \mid (n-1)! \implies a \mid (n-1)!$, so
$$\begin{aligned}
(n-1)! &\equiv 1\times 2\times \dotsb \times a \times\dotsb\times (n-a)\times\dotsb\times (n-1) \\ &\equiv 1\times 2\times \dotsb\times a \times\dotsb\times -a\times\dotsb\times -1 \\ &\equiv 0 \pmod n
\end{aligned}$$
Case 2:
$0<a<b<n$.
Then, since $a \mid n$, $b \mid n$ and $n \mid (n-1)!$ we have that $a \mid (n-1)!$ and $b \mid (n-1)!$.
So this implies
$(n-1)! \equiv 1\times 2\times \dotsb\times a \times\dotsb\times b\times\dotsb\times (n-1) \equiv 0 \pmod n$,
Q.E.D.
|
We will assume that $n>4$, since $4\hspace{-3pt}\not|\,3!$.
Let $p$ be the smallest factor of $n$. Since $n$ is composite, $p\le\sqrt{n}$.
If $p=\sqrt{n}$, then since $n>4$, we must have $p>2$ so that $2p<p^2=n$. Thus, $p\le n-1$ and $2p\le n-1$, and therefore, $2n=p\cdot2p\,|\,(n-1)!$
If $p<\sqrt{n}$, then $n/p>\sqrt{n}$. Thus, $p\le n-1$ and $n/p\le n-1$, and therefore, $n=p\cdot n/p\,|\,(n-1)!$
In either case, $n|(n-1)!$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Given that $\tan^{-1}(x)+\tan^{-1}(y)+\tan^{-1}(xy)=11/12π$, prove that when $x=1, dy/dx=-1-\sqrt{3}/2$ Given that $x$ and $y$ satisfy the equation:
$$\arctan(x)+\arctan(y)+\arctan(xy)=11/12π$$
Prove that, when $x=1, dy/dx=-1-\sqrt{3}/2$.
I tried to differentiate both sides:
$$1/(1+x^2)+y/(1+y^2)+(y+x\,dy/dx)/(1+(xy)^2)=0$$
and I know that when $x=1, y=\sqrt{3}$ by putting $x=1$ into the given equation.
so I got $1/2+√3/4+(√3+dy/dx)/4=0$
$$\implies dy/dx=-2-2√3$$
Thanks for pointing out the mistake. but the answer is still wrong..
|
$$
\begin{align}
& \arctan x + \arctan y + \arctan(xy) = \arctan\left( \frac{x+y}{1-xy} \right) + \arctan(xy) \\
& = \arctan\left( \frac{\frac{x+y}{1-xy} + xy}{1-\left(\frac{x+y}{1-xy}\right)xy} \right) = \arctan\left( \frac{ x + y + xy - x^2 y^2}{1-xy -x^2 y - xy^2} \right) = \frac{11}{12} \pi.
\end{align}
$$
So
$$
\frac{ x + y + xy - x^2 y^2}{1-xy -x^2 y - xy^2} = \tan\left(\frac{11}{12}\pi\right) = -\tan\frac{\pi}{12} = \tan\left(\frac \pi 4 - \frac \pi 3\right)
$$
$$
x + y + xy - x^2 y^2 = \left(1-xy -x^2 y - xy^2\right) \tan\left(\frac \pi 4 - \frac \pi 3\right)
$$
To differentiate both sides you need the chain rule and the product rule, but not the quotient rule. And the tangent of that difference of two fractions is easy to find.
|
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|
Show that $\left(1+\dfrac{1}{n}\right)^n$ is monotonically increasing
Show that $U_n:=\left(1+\dfrac{1}{n}\right)^n$, $n\in\Bbb N$, defines a monotonically increasing sequence.
I must show that $U_{n+1}-U_n\geq0$, i.e. $$\left(1+\dfrac{1}{n+1}\right)^{n+1}-\left(1+\dfrac{1}{n}\right)^n\geq0.$$
I am trying to go ahead of this step.
|
Here's a proof I learned by solving problem 90 from Section 11.1 of James Stewart's Calculus: Early Transcendentals (8th edition). We'll need the following result:
Lemma: If $0\leq a<b$ and $n$ is a positive integer, then
$$\frac{b^{n+1}-a^{n+1}}{b-a}<(n+1)b^n$$
Proof: Define the function $f:[0,\infty)\rightarrow\mathbb{R}$ by $f(x)=x^{n+1}$, where $n$ is a positive integer. Then $f$ is differentiable (and hence continuous) over $[0,\infty)$. Moreover, by the power rule, $f'(x)=(n+1)x^n$ for all $x\geq 0$, which is strictly increasing because $n$ is a positive integer (this will be important later, so we'll tuck it in our back pocket).
Since $f$ is differentiable over $[0,\infty)$, and $0\leq a<b$, it is necessarily continuous over $[a,b]$ and differentiable over $(a,b)$, so the mean value theorem asserts that there exists $c\in(a,b)$ such that
$$f'(c)=\frac{f(b)-f(a)}{b-a}
=\frac{b^{n+1}-a^{n+1}}{b-a}$$
Since $f'$ is strictly increasing and $c<b$, it follows that $f'(c)<f'(b)=(n+1)b^n$, so
$$\frac{b^{n+1}-a^{n+1}}{b-a}=f'(c)<f'(b)=(n+1)b^n$$
It immediately follows that $\frac{b^{n+1}-a^{n+1}}{b-a}<(n+1)b^n$.
Simple algebraic manipulation shows that $\frac{b^{n+1}-a^{n+1}}{b-a}<(n+1)b^n$ is equivalent to $b^n[(n+1)a-nb]<a^{n+1}$. With this established, we choose $a=1+\frac{1}{n+1}$ and $b=1+\frac{1}{n}$ (this is justified since $\frac{1}{n+1}<\frac{1}{n}$ for all $n>0$, and hence $1+\frac{1}{n+1}<1+\frac{1}{n}$), which gives
$$\left(1+\frac{1}{n}\right)^n\left[(n+1)\left(1+\frac{1}{n+1}\right)-n\left(1+\frac{1}{n}\right)\right]<\left(1+\frac{1}{n+1}\right)^{n+1}$$
Therefore,
$$\left(1+\frac{1}{n}\right)^n(n+1+1-n-1)<\left(1+\frac{1}{n+1}\right)^{n+1}$$
and thus,
$$\left(1+\frac{1}{n}\right)^n(1)=\left(1+\frac{1}{n}\right)^n<\left(1+\frac{1}{n+1}\right)^{n+1}$$
This shows that $U_n=\left(1+\frac{1}{n}\right)^n$ is strictly increasing.
|
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"timestamp": "2023-03-29T00:00:00",
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|
How many $3\times 3$ binary matrices $X$ are there with determinant $0$ and $X^2=X^T$? How many $3 \times 3$ binary matrices $X$ are there with determinant as $0$ that also satisfy $X^2 = X^T$?
|
There are $2^9=512$ binary $3\times 3$ matrices. Of these, $7\times 6\times 4 = 168$ are invertible, so there are $344$ singular ones.
Here's a somewhat naive way of going about it; I suspect there must be a clever/elegant way of doing it, but I couldn't think of one and then started going down this path. All computations are over $\mathbb{F}_2$.
If the first row of $X$ is all zeros, then so is the first row of $X^2$, so the first column of $X$ must be all zeros. The matrix is of the form
$$\left(\begin{array}{ccc}
0 & 0 & 0\\
0 & a & b\\
0 & c & d
\end{array}\right).$$
Then
$$X^2 = \left(\begin{array}{ccc}
0 & 0 & 0\\
0 & a^2+bc & b(a+d)\\
0 & c(a+d) & bc+d^2
\end{array}\right).$$
Hence $a^2+bc = a+bc = a$, so $bc=0$. Since $b(a+d)=c$ and $c(a+d)=b$, they are both equal to $0$.
Similar computations hold if the second or third row are equal to $0$. So the matrices that have at least one row equal to $0$ and satisfy the property are diagonal, with arbitrary values in the diagonal, at least one equal to $0$ (we cannot have the identity, because it is nonsingular). That shows that there are exactly $7$ matrices in which a row is equal to $0$.
If no row is equal to $0$, but the matrix is singular, then either the matrix has a repeated row, or the third row is the sum (modulo $2$) of the first two rows. Suppose first that the matrix has the first and second row repeated; then the first two rows of $X^2$ are the same. Since the $(1,1)$ and $(2,1)$ entries of $X^2=X^T$ are equal, that means that the $(1,1)$ and $(1,2)$ entries of $X$ are equal. Since the $(1,2)$ and $(2,2)$ entries of $X^2$ are equal, that means that the $(2,1)$ and $(2,2)$ entries of $X$ are equal. And since the $(1,3)$ and $(2,3)$ entries of $X^2$ are equal, then the $(3,1)$ and $(3,2)$ entries of $X$ are equal. So we have:
$$X = \left(\begin{array}{ccc}
a & a & b\\
a & a & b\\
c & c & d
\end{array}\right).$$
Then
$$X^2 = \left(\begin{array}{ccc}
a+a+bc & a+a+bc & ab+ab+bd\\
a+a+bc & a+a+bc & ab+ab+bd\\
ac+ac+dc & ac+ac+dc & bc+bc+d
\end{array}\right).$$
Therefore, $bc=a$, $bd=c$, $dc=b$. So $c = bd = dcd = cd$. Therefore, either $c=0$ or $d=1$.
If $c=0$, then $a=b=0$, and we have a row of zeros, which we are assuming we do not have. So $c=d=1$. Then $a=b=1$, so we just have the matrix of all $1$s.
Likewise, if the first and third rows are equal, then the first and third rows of $X^2$ are equal, so the $(1,1)$ and $(3,1)$ entries of $X^2$ are equal, hence the $(1,1)$ and $(1,3)$ entries of $X$ are equal; the $(2,1)$ and $(2,3)$ entries are equal; and the $(3,1)$ and $(3,3)$ entries are equal, so $X$ would be
$$X = \left(\begin{array}{ccc}
a & b & a\\
c & d & c\\
a & b & a
\end{array}\right).$$
Similar computations follow, and likewise if the second and third row are equal. So there is only one such singular matrix.
Finally, if no row is zero and there are no repeated rows, then the third row must be the sum of the first two; hence, the same is true in $X^2$. That means that the third column of $X$ must be the sum of the first two, we we have:
$$X = \left(\begin{array}{ccc}
a & b & a+b\\
r& s & r+s\\
a+r & b+s & a+b+r+s
\end{array}\right).$$
Then the $(1,1)$ entry of $X^2$ is $a+rb+(a+r)(a+b) = a+rb+a+ab+ar+rb = a(b+r)$, so $b+r=1$.
The $(1,2)$ entry of $X^2$ is $ab+bs+(a+b)(b+s) = ab+bs + ab+as + b + bs = as+b$, and it must be equal to $r$. So $as+b=r$, hence $as=r+b$, so $as=1$. Therefore, $a=s=1$. Plugging in we have
$$X = \left(\begin{array}{ccc}
1 & b & r\\
r & 1 & b\\
b & r & 1
\end{array}\right)$$
with $b+r=1$. A quick check shows both choices work (namely, $b=0$, $r=1$; or $b=1$, $r=0$). This gives two possibilities.
So in total, if I didn't mess up, we have: seven possibilities with at least one row equal to $0$; one possibility with repeated rows; and two more possibilities with two linearly independent rows and the third row a linear combination of the other two, for a total of $10$ such matrices.
|
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Prove $\frac{\sec{A}+\csc{A}}{\tan{A} + \cot{A}} = \sin{A} + \cos{A}$ and $\cot{A} + \frac{\sin{A}}{1 + \cos{A}} = \csc{A}$ Can anyone help me solve the following trig equations.
$$\frac{\sec{A}+\csc{A}}{\tan{A} + \cot{A}} = \sin{A} + \cos{A}$$
My work thus far
$$\frac{\frac{1}{\cos{A}}+\frac{1}{\sin{A}}}{\frac{\sin{A}}{\cos{A}}+\frac{\cos{A}}{\sin{A}}}$$
$$\frac{\frac{\sin{A} + \cos{A}}{\sin{A} * \cos{A}}}{\frac{\sin{A}}{\cos{A}}+\frac{\cos{A}}{\sin{A}}}$$
But how would I continue?
My second question is
$$\cot{A} + \frac{\sin{A}}{1 + \cos{A}} = \csc{A}$$
My work is
$$\frac{\cos{A}}{\sin{A}} + \frac{\sin{A}}{1 + \cos{A}} = \csc{A}$$
I think I know how to solve this one by using a common denominator but I am not sure.
|
Solution 1:
$$\dfrac{\dfrac{\sin{A} + \cos{A}}{\sin{A} \cos{A}}}{\dfrac{\sin^2{A} + \cos^2{A}}{\sin{A} \cos{A}}}$$
$$ = \frac{\sin{A} + \cos{A}}{\sin^2{A} + \cos^2{A}}$$
$$ = \sin{A} + \cos{A}$$
Solution 2:
$$\frac{\cos{A}(1 + \cos{A}) + \sin^2{A}}{\sin{A} (1 + \cos{A})}$$
$$= \frac{\color{red}{\cos{A} + 1}}{\sin{A} (\color{red}{\cos{A} + 1})}$$
$$= \frac{1}{\sin{A}} = \csc{A}$$
PS: I don't know how to put those cross-marks(cancellations) on fractions, if someone knows, please comment it, I'll edit it.
|
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|
Find an equation to transform $\{1, 2, 3, 4, 5,6, 7, 8\}$ to $\{8,7,6,5,4,3,2,1\}$ Is there a way to transform $\{1, 2, 3, 4, 5,6, 7, 8\}$ to $\{8,7,6,5,4,3,2,1\}$ by applying a single function?
That is, transform $1$ to $8$, $2$ to $7$, $3$ to $6$ etc.
|
$f(x) = 9 - x$
$f(1) = 9 - 1= 8$
$f(3) = 9 - 3 = 6$
etc.
|
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Show that the curve $y^2 = x^3 + 2x^2$ has a double point, and find all rational points Show that the curve $y^2 = x^3 + 2x^2$ has a double point. Find all rational points on this curve.
By implicit differentiation of $x$, $-3x^2 - 4x$ vanishes iff $x = -4/3$ and $0$.
By implicit differentiation of $y$, $2y$ vanishes iff $y = 0$.
Taking the second derivative, I got $-6x-4$ and then using the point on the curve $(0,0)$ I got $-4$. Is this my double point?
Thanks for any help!
|
Since you are a visual person, you should try drawing it. It seems fairly obvious from the real picture that there is a double point at zero, and then you just use whatever methods your book usually employs to prove that it actually is a double point.
On the other hand, finding all rational solutions means finding all rational points such that
$$y = \pm\sqrt{x^3+2x^2} = \pm~ x\sqrt{x+2}.$$
I will omit the sign for the rest of this post. Note that $x+2$ must have a rational root. In other words, $x+2=\frac{p^2}{q^2}$ for integers $p$ and $q$, so $x=\frac{p^2-2q^2}{q^2}$. This yields
$$y = \frac{p^2-2q^2}{q^2} \cdot \frac{p}{q}$$
and multiplying by $q^3$, we obtain that
$$q^3y = p^3 - 2pq^2$$
over the integers. Hence, any rational solution of your equation is parametrized by $p,q\in\mathbb{Z}$ and then,
$$\begin{align*}
y &= \frac{p^3-2pq^2}{q^3} \\
x &= \frac{p^2-2q^2}{q^2}.
\end{align*}$$
|
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The minimum value of $(\frac{1}{x}-1)(\frac{1}{y}-1)(\frac{1}{z}-1)$ if $x+y+z=1$ $x, y, z$ are three distinct positive reals such that $x+y+z=1$, then the minimum possible value of $(\frac{1}{x}-1) (\frac{1}{y}-1) (\frac{1}{z}-1)$ is ?
The options are: $1,4,8$ or $16$
Approach: $$\begin{align*}
\left(\frac{1}{x} -1\right)\left(\frac{1}{y}-1\right)\left(\frac{1}{z}-1\right)&=\frac{(1-x)(1-y)(1-z)}{xyz}\\
&=\frac{1-(x+y+z)+(xy+yz+zx)-xyz}{xyz}\\
&=\frac{1-1+(xy+yz+zx)-xyz}{xyz}\\
&=\frac{xy+yz+zx}{xyz} - 1
\end{align*}$$
Now by applying $AM≥HM$, I got the least value of $(xy+yz+zx)/xyz$ as $9$, so I got final answer as $8$. Is it correct?
|
If you look at the statement of the AM-HM inequality correctly, you'd see that
$$
\frac 1x + \frac 1y + \frac 1z \ge 9
$$
and equality only happens when $x=y=z$. Therefore we can assume equality happens to find the minimum value, but $x+y+z = 1$ implies $x=y=z=1/3$. Therefore the minimum is $8$ and is attained uniquely at the point $(1/3, 1/3, 1/3)$. Does that clarify the doubts you had?
Hope that helps,
|
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Factorise the determinant $\det\Bigl(\begin{smallmatrix} a^3+a^2 & a & 1 \\ b^3+b^2 & b & 1 \\ c^3+c^2 & c &1\end{smallmatrix}\Bigr)$ Factorise the determinant $\det\begin{pmatrix} a^3+a^2 & a & 1 \\ b^3+b^2 & b & 1 \\ c^3+c^2 & c &1\end{pmatrix}$.
My textbook only provides two simple examples.
Really have no idea how to do this type of questions..
|
$$\begin{align*}
\det\begin{pmatrix} a^3+a^2 & a & 1 \\ b^3+b^2 & b & 1 \\ c^3+c^2 & c &1\end{pmatrix} & = \det\begin{pmatrix} a^2 & a & 1 \\ b^2 & b & 1 \\ c^2 & c &1\end{pmatrix}+\det\begin{pmatrix} a^3 & a & 1 \\ b^3 & b & 1 \\ c^3 & c &1\end{pmatrix}\\
& = -(a-b)(b-c)(c-a) -(a-b)(b-c)(c-a)(a+b+c)\\
&= (a-b)(b-c)(a-c)(a+b+c+1)
\end{align*}$$
see Vandermonde matrix
Hey guys, editing my post to address concerns. You're right, I didn't put in all the details because I thought the method was clear once you've seen Vandermonde Determinants. Here it is explicitly:
$$\begin{align*}
\det\begin{pmatrix} a^3 & a & 1 \\ b^3 & b & 1 \\ c^3 & c &1\end{pmatrix} & = \det\begin{pmatrix} a^3+(b+c)a^2 & a & 1 \\ b^3 +(a+c)b^2& b & 1 \\ c^3+(a+b)c^2 & c &1\end{pmatrix}\\
& = \sum_{cyc}\det\begin{pmatrix} a^3 & a & 1 \\ ab^2 & b & 1 \\ ac^2 & c &1\end{pmatrix} & \\
&= \sum_{cyc}a\det\begin{pmatrix} a^2 & a & 1 \\ b^2 & b & 1 \\ c^2 & c &1\end{pmatrix}
\end{align*}$$
(At this point we have shown the result.)
Okay, so I guessed intuitively that $$\det\begin{pmatrix} (b+c)a^2 & a & 1 \\ (a+c)b^2& b & 1 \\ (a+b)c^2 & c &1\end{pmatrix}=0$$But its easy to prove, by finding the eigenvector:
$$\begin{pmatrix}1 \\ -(ab+bc+ca) \\ abc\end{pmatrix}$$
|
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Evaluating $\int ^\frac{\pi}{2}_{0} \sin\left(2x+\frac{\pi}{4}\right)\ dx$
Find the exact value of the following definite integral:
$$\int ^\frac{\pi}{2}_{0} \sin\left(2x+\frac{\pi}{4}\right)\:dx=\left[-\frac{1}{2}(2x+\frac{\pi}{4})\right]^\frac{\pi}{2}_{0}$$
$$=-\frac{1}{2}\left(2\frac{\pi}{2}+\frac{\pi}{4}\right)+\frac{1}{2}\left(2\cdot 0+\frac{\pi}{4}\right)$$
$$=-\frac{1}{2}\left(\pi+\frac{\pi}{4}\right)+\frac{1}{2}\left(\frac{\pi}{4}\right)=-\frac{\pi}{2}$$
but the right answer is:
$$\int_{0}^{\frac{\pi}{2}}{\sin{\left(2x+\frac{\pi}{4}\right)\:dx}}=\frac{\sqrt{2}}{2}$$
Help me out! thanks!
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$$\int_0^{\pi/2}\sin(2x+\tfrac{\pi}{4})\,dx=[−\tfrac{1}{2}\cos(2x+\tfrac{π}{4})]_0^{\pi/2}=2\sqrt{2}$$
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|
hints on solving DE How to solve this DE?
$$ {dx \over x} = {dy \over y} = {dz \over z - a \sqrt{x^2+y^2+z^2}}$$
From the first part, I get $y = c_1x$. How to find the other solution?
The answer according to answer sheet is $ z + \sqrt{x^2 + y^2 + z^2} = c_2$. Thank you for help.
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$$ {dx \over x} = {dy \over y} = {dz \over z - a \sqrt{x^2+y^2+z^2}}$$ You get $y=c_1x$, so put it into the third fraction:
$$ {dx \over x} = {dz \over z - a \sqrt{x^2+c_1^2x^2+z^2}}$$ $$ {dx \over x} = {dz \over z - a \sqrt{(1+c_1^2)x^2+z^2}}={dz \over z - a \sqrt{Cx^2+z^2}}$$ which is homogeneous equation: $$(z - a \sqrt{Cx^2+z^2})dx=xdz, x\neq 0$$ by taking $u=\frac{z}{x}$, you get: $${-adx \over x} = {du \over \sqrt{C+u^2}}$$ then integrating from both sides gives: $$\ln|u+\sqrt{C+u^2}|=-a\ln|x|+c_2$$ or $$\ln|z+\sqrt{x^2+y^2+z^2}|=(1-a)\ln|x|+c_2$$ Are you sure, you don't have any information about that $a$?
|
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proof of a finite sum involving a binomial coefficient and a variable. I found that the following equation holds for integers $l$, $k$, and any $x \neq 0,1$,
$$\tag{1}
\sum\limits_{l = 0}^k {\left( { - 1} \right)^l } \left( {\begin{array}{*{20}c}
k \\
l \\
\end{array}} \right)\frac{{x^{l} }}{{\left( {\frac{{1 + l}}{x} + k - l} \right)\left( {\frac{l}{x} + 1 + k - l} \right)}} = \frac{x\left( {1 - x} \right)^{k}}{{k + 1}}
$$
both in numerically by Matlab and analytically by Mathematica.
So I think there is a reference proving the equation. I have searched equaitons in Wolfram, Wiki, and some tables of series, But I couldn't find any related one.
Actually there were some equations looks like this
$$\tag{2}
\sum\limits_{l = 0}^k {\left( { - 1} \right)^l } \left( {\begin{array}{*{20}c}
k \\
l \\
\end{array}} \right)f(k,l,x) = g(k,x),
$$
but no help.
Also I tried in this way: break the equation into two terms like this
$$\tag{3}
\sum\limits_{l = 0}^k {\left( { - 1} \right)^l } \left( {\begin{array}{*{20}c}
k \\
l \\
\end{array}} \right)\frac{{x^{l} }}{{\left( {\frac{{1 + l}}{x} + k - l} \right)}} + \sum\limits_{l = 0}^k {\left( { - 1} \right)^l } \left( {\begin{array}{*{20}c}
k \\
l \\
\end{array}} \right)\frac{{x^{l} }}{{\left( {\frac{l}{x} + 1 + k - l} \right)}},
$$
and ran it in Mathematica. But they result in Gauss hypergeometric functions, $_2 F_1 (-k,*,*,x)$, with some coefficients, respectively.
Also $\times2$, I tried to prove it by myself showing the equation holds for $k=0$ and any $x \neq0,1$, then it holds as well when $k+1$ by using the eq (1). but I couldn't...beacuse the $k+1$ case becomes a totally different equation compared to eq (1)...... lol
How can I prove it or find a proof?
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We have, $$\frac{1}{\left(\frac lx + k - l+1 \right) \left(\frac{l+1}{x} + k-l \right)} = \frac{x}{1-x} \left(\frac{1}{\frac lx + k - l+1} - \frac{1}{\frac{l+1}{x} + k-l} \right)$$
Therefore, $$\sum_{l=0}^{k} (-1)^l \binom{k}{l} \frac{x^l}{\left(\frac lx + k - l+1 \right) \left(\frac{l+1}{x} + k-l \right)} = \frac{x}{1-x} \left(\sum_{l=0}^{k} (-1)^l \binom{k}{l} \frac{x^l}{\frac lx + k - l+1} - \sum_{l=0}^{k} (-1)^l \binom{k}{l} \frac{x^l}{\frac{l+1}{x} + k-l} \right)$$
Call the term inside the bracket as $S$. So, we only need to show that $S = \displaystyle \frac{(1-x)^{k+1}}{k+1}$.
Now, $\displaystyle \sum_{l=0}^{k} (-1)^l \binom{k}{l} \frac{x^l}{\frac lx + k - l+1} = \frac 1{k+1} + \sum_{l=0}^{k-1} (-1)^{l+1} \binom{k}{l+1} \frac{x^{l+1}}{\frac {l+1}x + k - l} $, and
$\displaystyle \sum_{l=0}^{k} (-1)^l \binom{k}{l} \frac{x^l}{\frac {l+1}x + k - l} = \frac {(-1)^k x^{k+1}}{k+1} + \sum_{l=0}^{k-1} (-1)^{l} \binom{k}{l} \frac{x^{l}}{\frac {l+1}x + k - l} $
Thus, on subtracting, we get that
$$S = \frac{1}{k+1} + \frac {(-x)^{k+1}}{k+1} + \sum_{l=0}^{k-1} \left((-1)^{l+1} \binom{k}{l+1} \frac{x^{l+1}}{\frac {l+1}x + k - l} + (-1)^{l+1} \binom{k}{l} \frac{x^{l}}{\frac {l+1}x + k - l}\right)$$
Now, since $\binom{k}{l+1} = \frac{k-l}{l+1} \binom{k}{l}$, therefore,
$$(-1)^{l+1} \binom{k}{l+1} \frac{x^{l+1}}{\frac {l+1}x + k - l} + (-1)^{l+1} \binom{k}{l} \frac{x^{l}}{\frac {l+1}x + k - l} = (-1)^{l+1} \binom{k}{l} \frac{x^{l+1}}{l+1 +(k-l)x} \left(\frac{k-l}{l+1} x + 1\right) = (-1)^{l+1} \binom{k}{l} \frac{x^{l+1}}{l+1} = (-1)^{l+1} \binom{k+1}{l+1} \frac{x^{l+1}}{k+1}$$
Thus, $\displaystyle S = \sum_{l=0}^{k+1} \binom{k+1}{l} \frac{x^l}{k+1} = \frac{(1-x)^{k+1}}{k+1}$, which was what we needed.
|
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How many unique distances are there in a 5 x 5 grid? I cannot figure this out:
I have a square in the plane with side length $5$. $A$ and $B$ are points in the square. The coordinates of $A$ and $B$ are always integers.
I want to know how many unique Euclidean distances are possible between $A$ and $B$.
I thought $15$?
Editor note The original ambiguous phrasing of this question and the consequent edits to clarify resulted in conflicting solutions. Please take this into consideration as you vote on the answers.
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If you count vertical or horizontal distances the values are 0,1,2,3,4,5;
If the horizontal (or vertical) has length 1 and the vertical has length 1, 2, 3, 4, 5 then the distances are $\sqrt{1^2+1^2}=\sqrt{2}, \sqrt{1^2+2^2}=\sqrt{5}, \sqrt{1^2+3^2}=\sqrt{10}, \sqrt{1^2+4^2}=\sqrt{17}, \sqrt{1^2+5^2}=\sqrt{26}$,
If the horizontal (or vertical) has length 2 and the vertical has length 2, 3, 4, 5 then the distances are $\sqrt{2^2+2^2}=\sqrt{8}, \sqrt{2^2+3^2}=\sqrt{13}, \sqrt{2^2+4^2}=\sqrt{20}, \sqrt{2^2+5^2}=\sqrt{29}$,
If the horizontal (or vertical) has length 3 and the vertical has length 3, 4, 5 then the distances are $\sqrt{3^2+3^2}=\sqrt{18}, \sqrt{3^2+4^2}=\sqrt{25}=5, \sqrt{3^2+5^2}=\sqrt{34}$.
If the horizontal (or vertical) has length 4 and the vertical has length 4, 5 then the distances are $\sqrt{4^2+4^2}=\sqrt{32}, \sqrt{4^2+5^2}=\sqrt{41}$.
If the horizontal (or vertical) has length 5 and the vertical has length 5 then the distances are $\sqrt{5^2+5^2}=\sqrt{50}$.
So there are 20 distinct values for the distances between two points.
|
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For $A \in \mathbb{R}^{3 \times 3}$, find $P, Q \in \mathbb{R}^{3 \times 3}$ such that $A = P-Q$, where $P^2 = P$, $Q^2 = Q$ and $PQ = 0 = QP$ This is an exercise from a previous linear algebra exam:
The diagonalisable matrix $$A = \begin{pmatrix} 3 & -6 & 2\\ 4 & -7 & 2\\ 8 & -12 & 3 \end{pmatrix} \in \mathbb{R}^{3 \times 3}$$ has the eigenvalues 1 and -1. Find $P, Q \in \mathbb{R}^{3 \times 3}$ such that $A = P-Q$, where $P^2 = P$, $Q^2 = Q$ and $PQ = 0 = QP$.
We want $A = P-Q$, so we will have $A^2 = P+Q$ and by plugging in $Q = P - A$, we find
$$A^2 = 2P - A \iff P = \frac{1}{2} (A^2 + A).$$
This gives
$$P = \begin{pmatrix} 2 & -3 & 1\\ 2 & -3 & 1\\ 4 & -6 & 2 \end{pmatrix} \text{ and } Q = \begin{pmatrix} -1 & 3 & -1\\ -2 & 4 & -1\\ -4 & 6 & -1 \end{pmatrix}$$
and these two matrices satisfy all conditions.
Question: The problem states that $A$ is diagonalisable and also gives its eigenvalues. Is there another, maybe even faster way to find $P$ and $Q$ by diagonalising $A$?
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There's an invertible matrix $B$ such that $A=BDB^{-1}$, where $D$ is diagonal and has diagonal entries $\pm1$. $D=E-F$ where $E,F$ are diagonal, $E$ has 1 where $D$ does and 0 elsewhere, $F$ has 1 where $D$ has $-1$ and 0 elsewhere. Let $P=BEB^{-1}$, $Q=BFB^{-1}$.
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Probability without replacement question I have a bag full of 10 marbles: 2 black, 1 blue, 1 yellow, 3 green, 1 brown, and 2 purple. I draw 5 marbles one at a time without replacement. What is the probability of a black marble to be in the five drawn?
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At least one black: It is easier to find first the probability of no black.
There are $\binom{10}{5}$ ways to choose $5$ marbles, all equally likely. Note that there are $\binom{8}{5}$ ways to choose $5$ marbles from the $8$ non-black. So the probability that all the balls are non-black is
$$\frac{\binom{8}{5}}{\binom{10}{5}},$$
and therefore the probability of at least one black is
$$1-\frac{\binom{8}{5}}{\binom{10}{5}}.$$
B: Exactly one black: There are $\binom{2}{1}$ ways of choosing one black from the two available. For each such way, there are $\binom{8}{4}$ ways to choose the non-blacks to go with it. So the total number of ways to pick exactly one black, and the rest non-black, is $\binom{2}{1}\binom{8}{4}$. Thus our probability is
$$\frac{\binom{2}{1}\binom{8}{4}}{\binom{10}{5}}.$$
Remark: We used general techniques. For A, there is a simpler way. The probability that the first marble chosen is non-black is $\frac{8}{10}$. Given the first was non-black, the probability that the second is non-black is $\frac{7}{9}$, since there are $7$ non-blacks left in a total of $9$. So the probability the first two are non-black is $\frac{8}{10}\cdot\frac{7}{9}$. Continue in this way. The probability all five are non-black is
$$\frac{8}{10}\cdot\frac{7}{9}\cdot\frac{6}{8}\cdot\frac{5}{7}\cdot\frac{4}{6}.$$
As in the earlier discussion, subtract the above from $1$.
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How to evaluate $\lim\limits_{n\to+\infty} \prod\limits_{k=1}^n (1+k/n^2)$? I've got a limit which puzzle me several days. The question is
$$ \lim_{n\to+\infty} \prod_{k=1}^n\left(1+\frac{k}{n^2}\right).$$
Can you help me? Thank you in advance
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Intuitively, we have
$$\log\left( 1 + \frac{k}{n^2} \right) = \frac{k}{n^2} + O\left(\frac{1}{n^2}\right) \quad \Longrightarrow \quad \log \prod_{k=1}^{n} \left( 1 + \frac{k}{n^2} \right) = \frac{1}{2} + O\left(\frac{1}{n}\right)$$
and therefore the log-limit is $\frac{1}{2}$.
Here is a more elementary approach: Let $P_n$ denote the sequence inside the limit. Then just note that
$$ P_n^2 = \left[ \prod_{k=1}^{n} \left( 1 + \frac{k}{n^2} \right) \right]^2 = \prod_{k=1}^{n} \left( 1 + \frac{k}{n^2} \right)\left( 1 + \frac{n-k}{n^2} \right) = \prod_{k=1}^{n} \left( 1 + \frac{1}{n}+\frac{k(n-k)}{n^4} \right). $$
Now fix $m$ and let $n \geq m$. Since $k (n-k) \leq \frac{1}{4}n^2$, we have
$$ \frac{k(n-k)}{n^4} \leq \frac{1}{4n^2} \leq \frac{1}{4mn}.$$
Thus we have
$$ \left( 1 + \frac{1}{n} \right)^n \leq P_n^2 \leq \left( 1 + \frac{1+(1/4m)}{n} \right)^n. $$
Thus taking $n \to \infty$,
$$e \leq \liminf_{n\to\infty} P_n^2 \leq \limsup_{n\to\infty} P_n^2 \leq e^{1+1/(4m)}.$$
Since $m$ is now arbitrary, we have $P_n^2 \to e$, or equivalently, $P_n \to \sqrt{e}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/183061",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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|
How to find $x$ in some trigonometric equations How to solve these trigonometric equations?
$$\tan2x-\sin4x = 0$$
and
$$\tan2x = \sin x$$
I can't do this, please help me! I did this:
\begin{align}
\tan2x &= 2\sin x\\
\\
\frac{\sin2x}{\cos2x} &= \tan x
\end{align}
|
$\tan2x=\sin4x$
$=>\frac{\sin2x}{\cos2x}=2\sin2x \cos2x$
$=>\sin2x(2cos^22x-1)=0$
$=>\sin2x\cos4x=0$
$\sin2x=0=>2x=n\pi=>x=\frac{r\pi}{2}$ for some integer r.
$\cos4x=0=>4x=\frac{(2r+1)\pi}{2}=>x=\frac{(2r+1)\pi}{8}$ for some integer r.
Again, $\sin4x=\sin x$,
$4x=n\pi+(-1)^nx$ for some integer n.
If $n$ is even $=2m(say)=>3x=2m\pi=>x=\frac{2m\pi}{3}$
If n is odd$=2m+1(say)=>5x=(2m+1)\pi=>x=\frac{(2m+1)\pi}{5}$
So we need to resolve r, m in integers such that,
$\frac{r\pi}{2} or \frac{(2r+1)\pi}{8}$ equals to $\frac{2m\pi}{3} or \frac{(2m+1)\pi}{5} $
(1)If $\frac{r\pi}{2} = \frac{2m\pi}{3}=>3r=4m=>r|4=>r=4s$ for some integer s.
$=> x= \frac{r\pi}{2} =\frac{4s\pi}{2} =2s\pi $ .
(2)If $\frac{r\pi}{2} = \frac{(2m+1)\pi}{5}=>5r=(2m+1)2=>5r=4m+2=>5(r-2)=4(m-2)$
$=>4|(r-2)=>r=4s+2$ for some integer s.
$=> x= \frac{r\pi}{2} = \frac{(4s+2)\pi}{2}=(2s+1)\pi$
(3)If $\frac{(2r+1)\pi}{8}=\frac{2m\pi}{3}=>3(2r+1)=16m$ which is impossible as the LHS is odd, the RHS is even for integral r,m.
So, there will be no solution in this case.
(4)If $\frac{(2r+1)\pi}{8}=\frac{(2m+1)\pi}{5}=>5(2r+1)=8(2m+1)$ which is again unsolvable in integers.
So, combining all the 4 cases, $x = t\pi$ where t is any integer(as t is even in case(1), odd in (2)).
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/183619",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Calculating the residue of power towers I want to calculate the residue of a power tower. How do I do that?
For example, I want to know the answer to this:
$$2 \uparrow\uparrow 10 \pmod{10^9}$$
|
When dealing with power towers with bases not relatively prime to the modulus, it's useful to employ the Chinese Remainder Theorem. And then repeatedly apply the Euler's Theorem.
$2 \uparrow \uparrow 10 \pmod {2^9} = 0$, so we only need to calculate $2 \uparrow \uparrow 10 \pmod{5^9}$.
By Euler's Theorem, we need to first study $2 \uparrow \uparrow 9 \pmod{\phi (5^9)} = 2 \uparrow \uparrow 9 \pmod{4 \cdot 5^8}$. So, as $2 \uparrow \uparrow 9 \equiv 0 \pmod{4}$, so by Chinese Remainder Theorem, we only need to settle the case when $2 \uparrow \uparrow 9 \pmod {5^8}$
Similarly proceeding at every step, we go a few more levels deeper, to get that we need to settle the congruence $2 \uparrow \uparrow 4 \pmod{4 \cdot 5^3}$. As $2 \uparrow \uparrow 4 = 2^{16} = 256^2 \equiv 6^2 \equiv 36 \pmod{125}$, so we get that $2 \uparrow \uparrow 4 \equiv 36 \pmod{4 \cdot 5^3}$, as $4| 36$.
Now, we unwrap the calculations. $2 \uparrow \uparrow 5 = 36^2 \equiv 1296 \pmod{5^4}$, hence $2 \uparrow \uparrow 5 \equiv 1296 \pmod{4 \cdot 5^4}$ as $4|1296$.
At this stage the calculations become too tedious to perform, but I hope you get the idea.
|
{
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Finite summation
Possible Duplicate:
why is $\sum\limits_{k=1}^{n} k^m$ a polynomial with degree $m+1$ in $n$
What is the proof without induction for :
$(1)$ $\sum_{i=1}^n\ i^2= \frac{n(n+1)(2n+1)}{6}$
$(2)$ $\sum_{i=1}^n\ i^3=\frac{n^2(n+1)^2}{4}$
|
Let $S(n,t)=\sum_{1 ≤r ≤n} r^t$
We know,
$(r+1)^2-r^2=2\cdot r+1$
Putting r=1,2,3,...,n-1,n we get,
$(1+1)^2-1^2=2\cdot1+1$
$(2+1)^2-2^2=2\cdot2+1$
...
$(n-1+1)^2-(n-1)^2=2\cdot (n-1)+1$
$(n+1)^2-n^2=2\cdot n+1$
Adding both sides, $(n+1)^2-1=2S(n,1)+n$
=>$S(n,1)=\frac{n(n+1)}{2}$
We know, $(r+1)^3-r^3=3\cdot r^2+3\cdot r+1$
Putting r=1,2,3,...,n-1,n and adding them we get,
$(n+1)^3-1=3S(n,2)+3S(n,1)+n$
But $S(n,1)=\frac{n(n+1)}{2}$
$=>3S(n,2)=n^3+3n^2+3n+1-1-n-3\frac{n(n+1)}{2}$
$=>6S(n,2)=2n^3+6n^2+6n-2n-3n(n+1)$
$=2n^3+3n^2+n=n(2n^2+3n+1)=n(n+1)(2n+1)$
$=>S(n,2)=\frac{n(n+1)(2n+1)}{6}$
Now we know, $(r+1)^4-r^4=4\cdot r^3+6\cdot r^2+4\cdot r+1$
Putting r=1,2,3,...,n-1,n and adding them we get,
$(n+1)^4-1=4S(n,3)+6S(n,2)+4S(n,1)+n$
Putting the values $S(n,1)$ and $S(n,2)$
$n^4+4n^3+6n^2+4n+1-1=4S(n,3)+(2n^3+3n^2+n)+2(n^2+n)+n$
$=>4S(n,3)=n^4+2n^3+n^2$
$=>S(n,3)=(\frac{n(n+1)}{2})^2$
We can go as far as we like.
|
{
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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|
Proving that $\frac{\csc\theta}{\cot\theta}-\frac{\cot\theta}{\csc\theta}=\tan\theta\sin\theta$
Prove $\dfrac{\csc\theta}{\cot\theta}-\dfrac{\cot\theta}{\csc\theta}=\tan\theta\sin\theta$
So, LS=
$$\dfrac{\csc\theta}{\cot\theta}-\dfrac{\cot\theta}{\csc\theta}$$
$$\left(\dfrac{1}{\sin\theta}\cdot \dfrac{\tan\theta}{1}\right)-\left(\dfrac{1}{\tan\theta}\cdot \dfrac{\sin\theta}{1}\right)$$
$$\dfrac{\tan\theta}{\sin\theta}-\dfrac{\sin\theta}{\tan\theta}$$
Now, considering the fact that I must have a common denominator to subtract, would this be correct:
$$\dfrac{\tan^2\theta}{\tan\theta\sin\theta}-\dfrac{\sin^2\theta}{\tan\theta\sin\theta}\Rightarrow \dfrac{\tan^2\theta-\sin^2\theta}{\tan\theta\sin\theta}$$
I feel like I'm close to the answer because the denominator is the RS of the OP. Please help. Do not give me the answer.
|
$$\frac{\csc\theta}{\cot\theta}-\frac{\cot\theta}{\csc\theta}=\frac{\csc^2\theta-\cot^2\theta}{\cot\theta.\csc\theta}=\frac{1}{\cot\theta\csc\theta}=\frac{1}{\cot\theta}.\frac{1}{\csc\theta}=\tan\theta.\sin\theta$$
I have used the identity $\csc^2\theta-\cot^2\theta=1$ here.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/185205",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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|
How to solve $100x +19 =0 \pmod{23}$ How to solve $100x +19 =0 \pmod{23}$, which is $100x=-19 \pmod{23}$ ?
In general I want to know how to solve $ax=b \pmod{c}$.
|
$100x +19 =0 \pmod{23}$
$100x ≡ -19 \pmod{23}$
$92x+8x=4-23\pmod{23}$
$=>8x≡4 \pmod{23}$ dividing both sides by $23$ and taking resdiues,
So, $2x≡1 \pmod{23}$ dividing both sides by $4$ which is possible as $(4,23)=1,$
Now $\frac{23}{2}=11+\frac{1}{2}$ So, $23-2\cdot 11=1$
(Please refer to this and this, for the theorem used.)
So, $2x≡23-2\cdot 11 \pmod{23}=>2x≡-22\pmod{23}$
$x≡-11\pmod{23}$ dividing both sides by $2$ as $(2,23)=1$
$x≡-11\pmod{23}≡12\pmod{23}$
To solve the general problem, $ax≡b\pmod{c}$,
$d=(a,b)$ must divide $c$ to admit any solution(according to this or this).
If we divide either side by $d$, so that $Ax≡B\pmod{C}$ where $\frac{a}{A}=\frac{b}{B}=\frac{c}{C}=d$ and (A,B)=1.
Using convergent property of continued fraction, we can get integers $r,s$ such that $rA-sB=±1$
For example, $\frac{17}{13}=1+\frac{4}{13}=1+\frac{1}{\frac{13}{4}}=1+\frac{1}{3+\frac{1}{4}}$
So, the convergent here is $1+\frac{1}{3}=\frac{4}{3}$
So. $17\cdot 3 - 13\cdot 4$ must be $±1$ , and is $-1$.
Put $rA-sB$ with proper sign in place of $(1)$, $Ax≡B(1)\pmod{C}$
One example can be found here.
Please refer to this for the details.
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Proving $n^4 + 4 n^2 + 11$ is $16k\,$ for odd $n$
if $n$ is an odd integer, prove that $n^4 + 4 n^2 + 11$ is of the form
$16 k$.
And I went something like:
$$\begin{align*}
n^4 +4 n^2 +11
&= n^4 + 4 n^2 + 16 -5 \\
&= ( n^4 +4 n^2 -5) + 16 \\
&= ( n^2 +5 ) ( n^2-1) +16
\end{align*}$$
So, now we have to prove that the product of $( n^2 +5 )$ and $( n^2-1)$ is a multiple of 16.
But, how can we do this?
If anybody has any idea of how I can improve my solution, please share it here.
Edit updated to include the necessary hypothesis that $n$ is odd.
|
If $m$ is odd, then $m^2 \equiv 1$ (mod 8), since $(2a+1)^{2} = 4(a^{2}+a) +1$ and $a^2 +a$ is always even when $a$ is an integer. If $h$ is an integer congruent to $3$ (mod 8), then $h^{2}-9 = (h-3)(h+3)$ is divisible by $16$. Now when $n$ is odd, we have $n^{2}+2 \equiv 3$ (mod $8$), so $(n^{2}+2)^{2} \equiv 9$ (mod $16$), so $n^{4} + 4n^{2} + 11 \equiv 9 +7 \equiv 0$ (mod 16).
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
I know that, $S_{2n}+4S_{n}=n(2n+1)^2$. Is there a way to find $S_{2n}$ or $S_{n}$ by some mathematical process with just this one expression? $S_{2n}+4S_{n}=n(2n+1)^2$, where $S_{2n}$ is the Sum of the squares of the first $2n$ natural numbers, $S_{n}$ is the Sum of the squares of the first $n$ natural numbers.
when, $n=2$
$S_{2n}=S_{4}=1^2+2^2+3^2+4^2=30$
$S_{n}=S_{2}=1^2+2^2=5$
$S_{4}+4S_{2}=2(2*2+1)^2=50$
|
With the additional information you provided about $S_n$ (the sum of squares of the first $n$ integers), there's a neat solution (among other solutions) that uses the perturbation method described in Concrete Mathematics.
Let $C_n$ denote the sum of cubes of the first $n$ natural numbers. Then
\begin{equation}
\begin{split}
C_{n+1} =& C_n + (n+1)^3 = \sum_{k=1}^{n+1}k^3 = \sum_{k=0}^{n}(k+1)^3 = \sum_{k=0}^{n} k^3 + 3k^2 + 3k + 1 \\
=&\sum_{k=0}^{n}k^3 + 3\sum_{k=0}^{n}k^2 + 3\sum_{k=0}^{n}k + \sum_{k=0}^{n}1 = C_n + 3S_n + 3\dfrac{n(n+1)}{2} + (n+1).
\end{split}
\end{equation}
Hence
\begin{equation}
\begin{split}
S_n =& \dfrac{(n+1)^3}{3} - \dfrac{n(n+1)}{2} - \dfrac{n+1}{3} = \dfrac{(n+1)(2(n+1)^2 - 3n - 2)}{6} \\
=& \dfrac{(n+1)(2n^2+n)}{6} = \dfrac{n(n+1)(2n+1)}{6}.
\end{split}
\end{equation}
|
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"timestamp": "2023-03-29T00:00:00",
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|
Deriving the addition formula of $\sin u$ from a total differential equation How do we derive the addition formula of $\sin u$ from the following equation?
$$\frac{dx}{\sqrt{1 - x^2}} + \frac{dy}{\sqrt{1 - y^2}} = 0$$
Motivation
Let $u = \int_{0}^{x}\frac{dt}{\sqrt{1 - t^2}}$
Then $x = \sin u$
Let $v = \int_{0}^{y}\frac{dt}{\sqrt{1 - t^2}}$
Then $y = \sin v$
Let $u + v = const.$
Then $d(u + v) = \frac{dx}{\sqrt{1 - x^2}} + \frac{dy}{\sqrt{1 - y^2}} = 0$
|
Let $u = \int_{0}^{x}\frac{dt}{\sqrt{1 - t^2}}$.
Then $x = \sin u$.
Let $v = \int_{0}^{y}\frac{dt}{\sqrt{1 - t^2}}$.
Then $y = \sin v$.
Let $c$ be a constant.
$u + v = c$ is a solution of the equation:
$$\frac{dx}{\sqrt{1 - x^2}} + \frac{dy}{\sqrt{1 - y^2}} = 0$$
It suffices to prove that $\sin c = x\sqrt{1 - y^2} + y\sqrt{1 - x^2}$.
Since $v = c - u$, the right hand side is a function of $u$.
We write this function by $f(u)$.
Namely, $$f(u) = x\sqrt{1 - y^2} + y\sqrt{1 - x^2}$$
Let us calculate $\frac{df}{du}$.
$\frac{dx}{du} = 1/\frac{du}{dx} = \sqrt{1 - x^2}$
$\frac{dy}{du} = -\frac{dy}{dv} = -1/\frac{dv}{dy} = -\sqrt{1 - y^2}$
$\frac{d^2x}{du^2}= \frac{d\sqrt{1 - x^2}}{du}\cdot\frac{dx}{du} = \frac{-x}{\sqrt{1 - x^2}} \sqrt{1 - x^2} = -x$
$\frac{d^2y}{du^2}= \frac{d^2y}{dv^2} = -y$
Hence
$\frac{df}{du} = \frac{d}{du}(-x\frac{dy}{du} + y\frac{dx}{du})
= (-\frac{dx}{du}\frac{dy}{du} - x\frac{d^2y}{du^2}) + ( \frac{dy}{du}\frac{dx}{du}+y\frac{d^2x}{du^2}) = xy - yx = 0$
Hence $f(u)$ is constant.
Hence $f(u) = f(0) = y = \sin(v) = \sin(c)$ as desired.
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Proving inequality $\frac{a}{\sqrt{a+2b}}+\frac{b}{\sqrt{b+2c}}+\frac{c}{\sqrt{c+2a}}\lt \sqrt{\frac{3}{2}}$ In the pdf which you can download here I found the following inequality which I can't solve it.
Exercise 2.1.10 Let $a,b,c$ be positive real numbers such that $a+b+c=1$. Prove that
$$\displaystyle \frac{a}{\sqrt{a+2b}}+\frac{b}{\sqrt{b+2c}}+\frac{c}{\sqrt{c+2a}}\lt \sqrt{\frac{3}{2}} . $$
Thanks.
|
By C-S $$\left(\sum_{cyc}\frac{a}{\sqrt{a+2b}}\right)^2\leq\sum_{cyc}\frac{a}{(a+2b)(a+2c)}\sum_{cyc}a(a+2c)=\sum_{cyc}\frac{a}{(a+2b)(a+2c)}.$$
Thus, it remains to prove that
$$\sum_{cyc}\frac{a}{(a+2b)(a+2c)}<\frac{3}{2(a+b+c)}$$ or
$$\sum_{cyc}(4a^3b^3+4a^4bc+24a^3b^2c+24a^3c^2b+25a^2b^2c^2)>0.$$
Done!
|
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"timestamp": "2023-03-29T00:00:00",
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|
Second order homogenous linear ODE How can I solve this ODE:
$y(x)+Ay'(x)+Bxy'(x)+Cy''(x)+Dx^{2}y''(x)=0$
Can you please also show the derivation.
|
I don't think you'll find an "elementary" solution in general. Maple finds a rather complicated solution involving hypergeometric functions:
$$\displaystyle S\, := \,y \left( x \right) ={\it \_C1}\,{\mbox{$_2$F$_1$}(1/2\,{\frac {-d+B+ \sqrt{{d}^{2}+ \left( -2\,B-4 \right) d+{B}^{2}}}{d}},1/2\,{\frac {-d+B- \sqrt{{d}^{2}+ \left( -2\,B-4 \right) d+{B}^{2}}}{d}};\,-1/2\\
\mbox{}\, \left( A-B \sqrt{-{\frac {C}{d}}} \right) {d}^{-1} \left( \sqrt{-{\frac {C}{d}}} \right) ^{-1};\,1/2\, \left( C- \sqrt{-{\frac {C}{d}}}xd \right) {C}^{-1})}+{\it \_C2}\, \left( C- \sqrt{-{\frac {C}{d}}}xd \right) ^{1/2\, \left( \left( -B+2\,d \right) \sqrt{-{\frac {C}{d}}}+A \right) {d}^{-1} \left( \sqrt{-{\frac {C}{d}}} \right) ^{-1}}{\mbox{$_2$F$_1$}(1/2\, \left( d \sqrt{-{\frac {C}{d}}}- \sqrt{{d}^{2}+ \left( -2\,B-4 \right) d+{B}^{2}} \sqrt{-{\frac {C}{d}}}+A \right) {d}^{-1} \left( \sqrt{-{\frac {C}{d}}}\\
\mbox{} \right) ^{-1},1/2\, \left( d \sqrt{-{\frac {C}{d}}}+ \sqrt{{d}^{2}+ \left( -2\,B-4 \right) d+{B}^{2}} \sqrt{-{\frac {C}{d}}}+A \right) {d}^{-1\\
\mbox{}} \left( \sqrt{-{\frac {C}{d}}} \right) ^{-1};\,1/2\, \left( \left( 4\,d-B \right) \sqrt{-{\frac {C}{d}}}+A \right) {d}^{-1} \left( \sqrt{-{\frac {C}{d}}} \right) ^{-1};\,1/2\, \left( C- \sqrt{-{\frac {C}{d}}}xd \right) {C}^{-1})}
$$
(I used $d$ instead of $D$ because $D$ has a special meaning in Maple)
|
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|
Proof of $\frac{d}{dx}e^x = e^x$ I'm working through the proof of $\frac{d}{dx}e^x = e^x$, and trying to understand it, but my mind has gotten stuck at the last step.
Starting with the definition of a derivative, we can formulate it like so:
$$\frac{d}{dx} e^x = \lim_{h \to 0} \frac{e^{x+h}-e^x}{h}$$
After some algebra, we arrive at:
$$\frac{d}{dx} e^x = e^x \lim_{h \to 0} \frac{e^h-1}{h}$$
As $h\to0$, the expression approaches $\frac{0}{0}$, which makes it indeterminate. And, this is where my understanding ends. I've tried looking at wikipedia and other descriptions of the proof, but couldn't understand those explanations. It has usually been something along the lines of, "plot $\frac{e^x - 1}{x}$ and see the function's behavior at $0$," which ends up approaching $1$, which can substitute the limit to give the result of the derivative:
$$\frac{d}{dx} e^x = e^x \cdot 1 = e^x$$
I vaguely understand the concept of indeterminate forms, and why it is difficult to know what is happening with the function. But is there a better explanation of how the result of $1$ is obtained?
|
Let's define $b^x$ as
$$ b^x = a_0 + a_1x +a_2 \frac{x^2}{2!} + \ldots =\sum_{k=0}^{\infty} a_k\frac{x^k}{k!}$$
for $x=0$,
$ b^0 =1 $ Thus $a_0=1$
$$ b^{x} = 1 + a_1x +a_2 \frac{x^2}{2!} + \ldots = 1 + \sum_{k=1}^{\infty} a_k\frac{x^k}{k!} \tag 1$$
$$\frac{d}{dx} b^x = \lim_{h \to 0} \frac{b^{x+h}-b^x}{h}=b^x \lim_{h \to 0} \frac{b^{h}-1}{h}=b^x \lim_{h \to 0} \frac{(1 + a_1h +a_2 \frac{h^2}{2!} + \ldots)-1}{h}$$
$$\frac{d}{dx} b^x=b^x \lim_{h \to 0} (a_1 +a_2 \frac{h}{2!} + \ldots)=a_1b^x $$
Let's select $a_1=1$ then we define $b=e$
and then we need to find all $a_k$ values and $e$.
After selecting $a_1=1$ and $b=e$, we have :
$$\frac{d}{dx} e^x=e^x \tag 2$$
According to this defination, $a_k=1$ for $k \geq 2$
Proof:
for $a_1=1$ and $b=e$ ,
And using relation (1)
$$ e^x = 1 + x +a_2 \frac{x^2}{2!} + \ldots =1 + x+ \sum_{k=2}^{\infty} a_k\frac{x^k}{k!}$$
$$ \frac{d}{dx} e^x = 1 + a_2 x +a_3 \frac{x^2}{2!} + \ldots =$$
According to the result (2),
$$ \frac{d}{dx} e^x =e^x= 1 + x +a_2 \frac{x^2}{2!} + \ldots= 1 + a_2 x +a_3 \frac{x^2}{2!} + \ldots $$
Now We need to equal all cooeffients of $x^k$,
we find $a_{2}=1$ and $a_{k+1}=a_{k}$ for $k \geq 2$
If we solve that relation, we get: $a_{k}=1 $ for $k \geq 2$
Thus $ e^x$ can be written as power series as shown below
$$ e^x= 1 + x + \frac{x^2}{2!} + \ldots= \sum_{k=0}^{\infty} \frac{x^k}{k!} $$
and to find $e$: put $x=1$, $e^1=1 + 1 + \frac{1^2}{2!} + \ldots= \sum_{k=0}^{\infty} \frac{1}{k!}$
Note:
If we select $a_1=m$ and follow the same way as shown in the proof, we will get
$$ b^{x}= 1 + mx + \frac{(mx)^2}{2!} + \ldots= \sum_{k=0}^{\infty} \frac{(mx)^k}{k!} =e^{mx}=(e^m)^x$$
$$ b^{x}=(e^m)^x$$
Thus
$ b=e^m$
$\ln(b)=\ln(e^m)=m$
$$\frac{d}{dx} b^x=a_1 b^x =m b^x=\ln(b) b^x $$
$$\frac{d}{dx} (b^x)=\ln(b) b^x $$
|
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|
How to solve for $ x$ when $x^2 \times a = (x-10)^2 \times b$ I'm trying to figure this one out, and I'm stuck. I'm going:
$x^2 \times a = (x-10)^2 \times b $
$x^2 \times a/b = (x-10)^2 /b $
$a/b = \frac{(x-10)^2} { x^2} / x^2$
$\sqrt{a/b} = (x-10)/x$
I'm not sure how I proceed or if this is even the best way to start. (I do fine with everything up to linear algebra.)
|
$\textbf{Hint}$ : Try to put it into the form $Dx^2 + Ex + F$ and use the quadratic formula to find the solutions. Move cursor over the box for more details.
$ax^2 = b(x - 10)^2$
$ax^2 = b(x^2 - 20x + 100)$
$ax^2 = bx^2 - 20bx + 100b$
$0 = (b - a)x^2 - 20bx + 100b$
Now using the quadratic formula :
$x = \frac{20b \pm \sqrt{(20b)^2 - 4(100)b(b - a)}}{2(b - a)}$
$= \frac{20b \pm \sqrt{400ab}}{2(b-a)} = \frac{20b \pm 20\sqrt{ab}}{2(b-a)}$
$= 10 \frac{b \pm \sqrt{ab}}{b - a}$
|
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|
Prove $\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\cdot\cdot\cdot+\frac{1}{x^2}<\frac{x-1}{x}$ Please help me for proving this inequality $\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\cdot\cdot\cdot+\frac{1}{x^2}<\frac{x-1}{x}$
|
$$\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\cdot\cdot\cdot+\frac{1}{x^2}<\frac{x-1}{x}$$
Employ the inequality: $$\frac{1}{a^2}=\frac{1}{a}\cdot\frac{1}{a}<\frac{1}{a}\cdot\frac{1}{a-1}=\frac{1}{a-1}-\frac{1}{a}$$.
In
$$\displaylines{
\frac{1}{{{2^2}}} + \frac{1}{{{3^2}}} + \frac{1}{{{4^2}}} + \cdots + \frac{1}{{{x^2}}} < \cr
\left( {1 - \frac{1}{2}} \right) + \left( {\frac{1}{2} - \frac{1}{3}} \right) + \left( {\frac{1}{3} - \frac{1}{4}} \right) + \cdots + \left( {\frac{1}{{x - 1}} - \frac{1}{x}} \right) = \cr
= 1 - \frac{1}{x} = \frac{{x - 1}}{x} \cr} $$
|
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|
Sequence Sum {1/2 + 1/4 + 1/6 +...} to infinite I've been told, the following series converges:
$$\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\ldots+\frac{1}{2k}+\ldots$$
I can't get my head around, how to prove this converges; any hints?
|
Let
$$S = 1 + \frac{1}{2} + \left( \frac{1}{3} + \frac{1}{4} \right) + \left( \frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8}\right) + \;...$$
$$[Parentheses \; have \; been \; put \; for \; comparing \; the \; series \; S \; and \; T]$$
Then the given series is simply $2S$. Hence in order to investigate the convergence of the given series, we have to investigate $S$ which is a well known series called the harmonic series and is a nice example of a series whose $t_n \rightarrow 0$ as $n \rightarrow\infty$ but the series is still divergent. To show it, we will construct another series $T$ such that $T < S$ and $T$ is divergent. Let
$$T = 1+ \frac{1}{2} + \left( \frac{1}{4} + \frac{1}{4}\right) + \left( \frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8}\right) + \; ...$$
It's easy to see that $T < S$. Also, the series $T$ is the same as the series $1 + \frac{1}{2} + \frac{1}{2} + \frac{1}{2}+\; ... \; = \infty$. Hence $S$ is a divergent series as well and as a result, the given series in the question diverges.
|
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|
Calculate $\lim_{n\to\infty}[(1+x)(1+x^2)(1+x^4)\cdot\cdot\cdot(1+x^{2n})]$, $|x|<1$ Please help me solving $\lim_{n\to\infty}[(1+x)(1+x^2)(1+x^4)\cdot\cdot\cdot(1+x^{2^n})]$, in the region $|x|<1$.
|
The product expands out as
$$
(1 + x)(1 + x^2)(1 + x^4)\ldots(1 + x^{2^n}) = \sum_{k=0}^{2^{n+1} -1} x^k = \frac{1 - x^{2^{n+1}}}{1 - x}.
$$
Since $|x| < 1$, this converges to $\frac{1}{1 - x}$ as $n \to \infty$.
|
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|
Inequality. $3\left(IA^2+IB^2+IC^2\right) \geq AB^2+BC^2+CA^2$ (Korea 1998) Let $I$ be the incenter of a triangle $ABC$.Prove that:
$$3\left(IA^2+IB^2+IC^2\right) \geq AB^2+BC^2+CA^2.$$
Please help me to improve this kind of inequalities.
Thanks :)
|
First note that $IA^2=(p-a)bc/p$ and similarly for $IB, IC$. Substitute to get$$
3abc(\frac{p-a}{ap}+\frac{p-b}{bp}+\frac{p-c}{cp})-a^2-b^2-c^2>=0
$$
Replace $p$ with $(a+b+c)/2$ and get rid of the fractions to arrive at:$$
-a^3 + 2 a^2 b + 2 a b^2 - b^3 + 2 a^2 c - 9 a b c + 2 b^2 c +
2 a c^2 + 2 b c^2 - c^3>=0
$$
Substitute $a=x+y, b=x+z, c=y+z$ because they are triangle sides and simplify to get:$$
x^3 - x^2 y - x y^2 + y^3 - x^2 z + 3 x y z - y^2 z - x z^2 -
y z^2 + z^3>=0
$$
This I think you know how to prove.
|
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Finding the values of the real constants such that the limit exists Find the values of the real constants $c$ and $d$ such that
$$\lim_{x\to 0}\frac{\sqrt{c+dx}-\sqrt{3}}{x}=\sqrt{3}$$
I really have no clue how to even get started.
|
$$\lim_{x\to 0}\frac{\sqrt{c+dx}-\sqrt{3}}{x}=\sqrt{3}$$
Since the denominator goes to $0$, the limit cannot exist unless the numerator also goes to $0$. The numerator is $\sqrt{c+dx}-\sqrt{3}$, so that would have to go to $0$ as $x$ goes to $0$. But it goes to $\sqrt{c+d\cdot0} - \sqrt{3}$. Hence $c+d\cdot0$ must be $3$. That tells you $c$ is $3$, and you've got
$$\lim_{x\to 0}\frac{\sqrt{3+dx}-\sqrt{3}}{x}=\sqrt{3}.$$
Now rationalize the numerator:
$$
\frac{\sqrt{3+dx}-\sqrt{3}}{x} = \frac{\left(\sqrt{3+dx}-\sqrt{3}\right)\left(\sqrt{3+dx}+\sqrt{3}\right)}{x\left(\sqrt{3+dx}+\sqrt{3}\right)}= \frac{dx}{x\left(\sqrt{3+dx}+\sqrt{3}\right)}.
$$
The $x$s cancel and we get
$$
\frac{d}{\sqrt{3+dx}+\sqrt{3}}.
$$
The limit of that as $x\to0$ is $d/(2\sqrt{3})$. So you want $d/(2\sqrt{3}) = \sqrt{3}$. Multiply both sides by $2\sqrt{3}$ and you get $d=6$.
|
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|
Laurent series for $\sin z \sin(1/z)$? How can I find a laurent series for $\sin z \sin(1/z)$ for $z \neq 0$?
Is it possible to multiply two laurent series? I saw that in wikipedia, it's not generally possible.
|
Operate formally. Fix finite $z\not=0$. Then each expand each sine into Taylor series:
$$\begin{eqnarray}
\sin(z) \sin\left(\frac{1}{z}\right) &=& \sum_{n=0}^\infty (-1)^n \frac{z^{2n+1}}{(2n+1)!} \sum_{m=0}^\infty (-1)^n \frac{z^{-2m-1}}{(2m+1)!}
\\ &=& \sum_{n=0}^\infty \sum_{m=0}^\infty \frac{(-z^2)^{n-m}}{(2n+1)!(2m+1)!} = \Big| \text{insert identity}\Big|
\\ &=& \sum_{k=-\infty}^\infty \sum_{n=0}^\infty \sum_{m=0}^\infty \frac{(-z^2)^{n-m}}{(2n+1)!(2m+1)!} \delta_{k,n-m}
\\ &=& \sum_{k=-\infty}^\infty (-z^2)^k \sum_{m=\max(0,-k)}^\infty \frac{1}{(2(k+m)+1)!(2m+1)!}
\\ &=& \sum_{k=-\infty}^\infty (-z^2)^k \sum_{m=0}^\infty \frac{1}{(2m+2 |k| +1)!(2m+1)!}
\end{eqnarray}
$$
The latter sum can be evaluated in terms of Bessel functions, giving:
$$
\sin(z) \sin\left(\frac{1}{z}\right) = \frac{1}{2} \sum_{k=-\infty}^\infty (-z^2)^k \left(I_{2|k|}(2)- J_{2|k|}(2)\right)
$$
Indeed:
$$\begin{eqnarray}
\sum_{m=0}^\infty \frac{1}{(2m+2 |k| +1)!(2m+1)!} &=& \sum_{m=0}^\infty \frac{1+(-1)^m}{2} \frac{1}{(m+2 |k| +1)!(m+1)!} \\
&=& \sum_{m=0}^\infty \frac{1-(-1)^m}{2} \frac{1}{(m+2 |k|)!(m)!} \\
&=& \frac{1}{2} I_{2|k|}(2) - \frac{1}{2} J_{2|k|}(2)
\end{eqnarray}
$$
|
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|
What That Mean "In Base 18"? i am a programmer who interest in math , lately in palindromic numbers , so if it's stupid question i am sorry !
i was reading about palindromic numbers in wikipedia , at some point it says
In base 18, some powers of seven are palindromic:
- 7^3 = 111
- 7^4 = 777
- 7^6 = 12321
- 7^9 = 1367631
What that mean , In Base 18 , Thanks in advance
|
For example, $12321_{18}=1\cdot 18^4+2\cdot 18^3+3\cdot 18^2+2\cdot 18+1=117649_{10} \\ =1\cdot 10^5+1\cdot 10^4+7\cdot 10^3+6\cdot 10^2+4\cdot 10+9=7^6$
|
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|
Proof : $ \sum_{n=1}^{\infty}\frac{1}{2nx^{2n}}=-\frac{1}{2}\ln (1-\frac{1}{x^2})$ How to prove that :
$$ \sum_{n=1}^{\infty}\frac{1}{2nx^{2n}}=-\frac{1}{2}\ln \left(1-\frac{1}{x^2}\right)$$
|
We factor out $\frac{1}{2}$ as it is a constant
$$
\sum_{n=1}^{\infty}\frac{1}{2nx^{2n}}=
\frac{1}{2}\sum_{n=1}^{\infty}\frac{1}{nx^{2n}}
$$
Now, the Taylor's series for $-\log (1-x)$ is, for $-1\le x < 1$
$$-\log (1-x)=\sum^{\infty}_{n=1} \frac{x^n}n$$
Thus
$$-\log \left(1-\frac{1}{x}\right)=\sum^{\infty}_{n=1} \frac{1}{n x^n}$$
$$-\log \left(1-\frac{1}{x^2}\right)=\sum^{\infty}_{n=1} \frac{1}{n x^{2n}}$$
and, by multiplying both side by $\frac{1}{2}$, we get
$$\frac{1}{2}\sum_{n=1}^{\infty}\frac{1}{nx^{2n}}=-\log \left(1-\frac{1}{x^2}\right)$$
|
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|
The exact value of $\sum_{n=1}^{\infty}\frac{n^2+n+1}{3^n}$ What is the value of :
$$\sum_{n=1}^{\infty}\frac{n^2+n+1}{3^n}$$
|
In most practical applications, you can just ask Mathematica, and it will tell you it's $\frac{11}{4}$.
If you want to arrive at the formula in a more rigorous way, you can do the following:
Consider the function $f$: $$f(x) = \frac{1}{1 - x} = \sum_{n=0}^\infty x^n$$
An initial observation is that $f(\frac{1}{3}) = \sum_{n=0}^\infty \frac{1}{3^n}$ so it looks a little like your sum. Now, consider $f'$:
$$f'(x) = \frac{1}{(1 - x)^2} = \sum_{n=1}^\infty n x^{n-1} = \sum_{n=0}^\infty (n+1) x^{n} $$
When you plug in $\frac{1}{3}$ again, you find $f'(\frac{1}{3}) = \sum_{n=0}^\infty \frac{n+1}{3^n}$. Finally, consider $f''$:
$$f''(x) = \frac{2}{(1 - x)^3} = \sum_{n=0}^\infty (n+2)(n+1) x^{n} = \sum_{n=0}^\infty (n^2 + 3n + 2) x^{n}$$
When you plug in $\frac{1}{3}$ once more, you find $f''(\frac{1}{3}) = \sum_{n=0}^\infty \frac{n^2 + 3n + 2}{3^n}$. Now, all you have to do is to express $n^2 + n +1$ as:
$$ n^2 + n + 1 = (n^2 + 3n + 1) - 2 (n+1) + 1 $$
so you get that the sought sum $S$ is:
$$ S = f''\left(\frac{1}{3}\right) - 2 f'\left(\frac{1}{3}\right) - f\left(\frac{1}{3}\right)$$
The remaining computation is not pleasant, but it is definitely doable, and does not involve any new creative ideas.
I was a little too careless: the computation I did was for a sum raging from $n=0$ rather than $n=1$ as in the problem. It is however easy to mend - just subtract the initial term, which is just $\frac{1}{3}$.
|
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|
Show that the following is indeed a mass function for R.V. $Y$ which can take values $2^n$ and $-2^n$ with probability $\frac{1}{2^{n+2}}$ So I think I have the pieces, just having trouble putting the puzzle together.
$P(Y=2^{n})=P(Y=-2^{n}) = \frac{1}{2^{n+2}}$
\begin{align*}
\sum_{n=0}^{\infty} p_y(y)
&=\sum_{n=0}^{\infty} \frac{1}{2^{n+2}}\\
&=\frac{1}{4} \sum_{n=0}^{\infty} \frac{1}{2^n}\\
&=1/2
\end{align*}
Is this on the right track? Is it $\sum_{n=0}^{\infty} p_y(y) =2*\sum_{n=0}^{\infty} \frac{1}{2^{n+2}} =\frac{1}{2} \sum_{n=0}^{\infty} \frac{1}{2^n}=1$ instead.
I also need to show that the expectation does not exist. So I have
So far I have that $E(Y) = \sum_i y_ip_y(y_i)$, thus
\begin{align*}
E(Y)= \sum_{n=0}^{\infty} 2^n \frac{1}{2^{n+2}} = \infty\\
E(Y)= \sum_{n=0}^{\infty} -2^n \frac{1}{2^{n+2}} = -\infty
\end{align*}
and thus the expectation does not exist
As always, thanks for any help.
|
Overall you are doing the problem correctly. The probability mass function has been given to you explicitly. Perhaps you were asked to show that it is indeed a pmf, though that is not clear from the question.
If that was asked for, you need to sum the probabilities ("weights") and show that the sum is $1$. The sum of the weights at numbers of the form $2^n$ is $\sum_{n=0}^\infty \frac{1}{2^{n+2}}$, which you showed is $\frac{1}{2}$. The sum of the weights at numbers of the form $-2^n$ is also $\sum_{n=0}^\infty \frac{1}{2^{n+2}}$, another $\frac{1}{2}$, for a total of $1$. So we were indeed given a pmf.
For the expectation, we need to find
$$\sum_{n=0}^\infty 2^n \cdot\frac{1}{2^{n+2}} + \sum_{n=0}^\infty (-2^n) \cdot\frac{1}{2^{n+2}}.$$
Both of the above sums diverge, since $2^n\cdot\frac{1}{2^{n+2}}=\frac{1}{4}$.
Thus the expectation does not exist.
This is an interesting example of a pmf which is symmetrical about $x=0$, but such that the mean is not $0$. In all symmetrical situations, if the mean exists, then it is at the centre of symmetry.
|
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|
Relations between the roots of a cubic polynomial How do I solve the last two of these problems?
The roots of the equation $x^3+4x-1=0$ are $\alpha$, $\beta$, and $\gamma$. Use the substitution $y=\dfrac{1}{1+x}$ to show that the equation $6y^3-7y^2+3y-1=0$ has roots $\dfrac{1}{\alpha+1}$, $\dfrac{1}{\beta+1}$, and $\dfrac{1}{\gamma+1} $. $\quad \quad \quad (2)$
For the cases $n=1$ and $n=2$, find the value of
$$\dfrac{1}{(\alpha+1)^n}+\dfrac{1}{(\beta+1)^n}+\dfrac{1}{(\gamma+1)^n}. \tag{2}$$
Deduce the value of $\dfrac{1}{(\alpha+1)^3}+\dfrac{1}{(\beta+1)^3}+\dfrac{1}{(\gamma+1)^3}. \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \, \, \,(2)$
Hence show that $\dfrac{(\beta+1)(\gamma+1)}{(\alpha+1)^2}+\dfrac{(\gamma+1)(\alpha+1)}{(\beta+1)^2}+\dfrac{(\alpha+1)(\beta+1)}{(\gamma+1)^2}=\dfrac{73}{36} \quad \quad \quad \quad \quad \quad \quad (3)$
|
Part 1,
Since , $\frac{1}{\alpha +1}$ is the root of the equation $6y^3-7y^2+3y-1=0\implies $ $$\frac{1}{(\alpha +1)^3}=\frac{7}{(\alpha +1)^2}-\frac{3}{\alpha +1}+1$$
Similar equality follows for $\beta, \gamma$
Part 2,
Product of roots= $\frac{1}{6}\implies \frac{1}{1+\alpha}\frac{1}{1+\beta}\frac{1}{1+\gamma}=\frac{1}{6}\implies (1+\beta)(1+\gamma)=\frac{6}{1+\alpha}\implies \frac{(1+\beta)(1+\gamma)}{(1+\alpha)^2}=\frac{6}{(1+\alpha)^3}$
Therefore, $\frac{(1+\beta)(1+\gamma)}{(1+\alpha)^2}+\frac{(1+\beta)(1+\gamma)}{(1+\alpha)^2}+\frac{(1+\beta)(1+\gamma)}{(1+\alpha)^2}=6$(result of second part)
Now, you can easily calculate the last sum.
|
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Solving $ \sin x + \sqrt 3 \cos x = 1 $ - is my solution correct? I have an equation that I'm trying to solve:
$$ \sin x + \sqrt 3 \cos x = 1 $$
After pondering for a while and trying different things out, this chain of steps is what I ended up with:
$$ \sin x + \sqrt 3 \cos x = 1 $$
$$ \sin x = 1 - \sqrt 3 \cos x $$
$$ \left(\sin x \right)^2 = \left(1- \sqrt 3 \cos x\right)^2 $$
$$ \sin^2 x = 1 - 2 \sqrt 3 \cos x + 3 \cos^2 x $$
$$ 2 \sqrt 3 \cos x - 3 \cos^2 x = 1 - \sin^2 x $$
$$ 2 \sqrt 3 \cos x - 3 \cos^2 x = \cos^2 x $$
$$ 2 \sqrt 3 \cos x = \cos^2 x + 3 \cos^2 x $$
$$ 4 \cos^2 x = 2 \sqrt 3 \cos x $$
$$ \frac{4 \cos^2 x}{\cos x} = 2 \sqrt 3 $$
$$ 4 \cos x = 2 \sqrt 3 $$
$$ \cos x = \frac{2 \sqrt 3}{4} $$
$$ \cos x = \frac{\sqrt 3}{2} $$
The fraction $ \frac{\sqrt 3}{2} $ can be rewritten as $ \cos \left(\pm \frac{\pi}{6}\right) $, so my solutions are:
$$ \cos x = \cos \left(\frac{\pi}{6}\right) \quad \text{or} \quad \cos x = \cos \left(-\frac{\pi}{6}\right) $$
$$ x = \frac{\pi}{6} + 2\pi n \quad \text{or} \quad x = -\frac{\pi}{6} + 2\pi n $$
Since I earlier on exponentiated both sides I have to check my solutions:
$$ x = \frac{\pi}{6} + 2\pi \Rightarrow \text{LHS} = \sin \left(\frac{\pi}{6} + 2\pi\right) + \sqrt 3 \cos \left(\frac{\pi}{6} + 2\pi\right) = 2 \not = \text{RHS} $$
$$ x = -\frac{\pi}{6} + 2\pi \Rightarrow \text{LHS} = \sin \left(-\frac{\pi}{6} + 2\pi\right) + \sqrt 3 \cos \left(-\frac{\pi}{6} + 2\pi\right) = 1 = \text{RHS} $$
Leaving $ x = -\frac{\pi}{6} + 2\pi n $ as the answer since its positive counterpart was not equal to $ 1 $.
$$ \text{Answer:} \: x = -\frac{\pi}{6} + 2\pi n $$
Have I done anything wrong or does this look good? I haven't really done this before so I feel uncertain not just about the solution, but also my steps and notation...
|
You went a little bit astray after $4 \cos^2 x = 2 \sqrt 3 \cos x$, when you divided by $\cos x$: what if $\cos x=0$?
It’s better at that point to bring everything to one side and factor: $4\cos^2x-2\sqrt3\cos x=0$, so $2\cos x(2\cos x-\sqrt3)=0$. Now appeal to the fact that if a product is $0$, at least one of the factors must be $0$. Obviously $2\ne 0$, so either $\cos x=0$, or $2\cos x-\sqrt3=0$. As it happens, both of these possibilities give you solutions. You found the second set, but not the first set.
If $\cos x=0$, we need $\sin x=1$ to have a solution. If $\sin x=1$, $\cos x$ is automatically $0$, so you just need to find the solutions to $\sin x=1$ to complete your solution.
|
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|
Solve $(\sqrt{5+2\sqrt{6}})^{x}+(\sqrt{5-2\sqrt{6}})^{x}=10$.
Solve $(\sqrt{5+2\sqrt{6}})^{x}+(\sqrt{5-2\sqrt{6}})^{x}=10$
I square the both sides and get $(5+2\sqrt{6})^{x}+(5-2\sqrt{6})^{x}=98$. But I don't know how to carry on. Please help. Thank you.
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Let $t_1=(\sqrt{5+2\sqrt6})^x$, and $t_2=(\sqrt{5-2\sqrt6})^x$.
Now the given equation is:
$$\tag 1 t_1+t_2=10$$
But
$$\displaylines{
{t_1}\cdot{t_2}&=& {\left( {\sqrt {5 + 2\sqrt 6 } } \right)^x}{\left( {\sqrt {5 - 2\sqrt 6 } } \right)^x} \cr
&=& {\left[ {\sqrt {\left( {5 + 2\sqrt 6 } \right)\left( {5 - 2\sqrt 6 } \right)} } \right]^x} \cr
&=& {\left[ {\sqrt {{5^2} - {{\left( {2\sqrt 6 } \right)}^2}} } \right]^x} \cr
&=& {\left( {\sqrt {25 - 24} } \right)^x} \cr &=& {1^x} = 1 \cr} $$
Thus
$$\begin{cases} t_1+t_2=10 \\t_1\cdot t_2=1\end{cases}$$
$\Rightarrow$ $t^2-10t+1=0$
Since $$\frac{1}{{{t_1}}} = {t_2}$$
we
have $${t_1} + \frac{1}{{{t_1}}} = 10$$ or
$$t_1^2 - 10{t_1} + 1 = 0$$
Note the system is symmetric on the unknowns.
For this quadratic equation we have:
$t_{1,2}=\frac{10\pm\sqrt{100-4}}{2}$
$t_{1,2}=\frac{10\pm 4\sqrt{6}}{2}$
$t_1=5+2\sqrt 6$, $t_2=5-2\sqrt 6$
Now return the inital substition:
$t_1=(\sqrt{5+2\sqrt6})^x$
$5+2\sqrt 6=(\sqrt{5+2\sqrt6})^x$
$(\sqrt{5+2\sqrt 6})^2=(\sqrt{5+2\sqrt6})^x$ $\Rightarrow$ $x=2$, and
$(\sqrt{5+2\sqrt6})^x=5-2\sqrt 6$
$(\sqrt{5+2\sqrt6})^x=(5+2\sqrt 6)^{-2}$ $\Rightarrow$ $x=-2$
Definitly $x=2$, and $x=-2$ is solve.
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"timestamp": "2023-03-29T00:00:00",
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|
Evaluating limit $\lim_{x\to\infty}(x^2-\sqrt{x^4 + 7x^2 + 1})$ The problem is evaluate
$$\lim_{x\to\infty}(x^2-\sqrt{x^4 + 7x^2 + 1})$$
I understand all of the calculus involved, but am having trouble figuring out how to get started with the algebra. I have tried factoring and using conjugates, but the only answer I am able to get is $-7$, which is incorrect. Any help would be appreciated.
What I have done so far:
$$\frac{(x^2-\sqrt{x^4+7x^2+1})(x^2+\sqrt{x^4+7x^2+1})}{ x^2+\sqrt{x^4+7x^2+1}}$$
results in
$$\frac{-7x^2-1}{x^2+\sqrt{x^4+7x^2+1}}$$
factor out the $x^4$ under the radical, then divide numerator and denominator by $x^2$ to get
$$\frac{-7-1/x^2}{1 + \sqrt{1+7/x^2+1/x^4}}$$
at this point the limit as x approaches infinity would be -7/2 or -3.5.
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Let $y = \sqrt{x^4+7x^2+1}$ and $L=\lim (x^2-y)$. Throughout this answer $\lim$ means $\lim_{x\rightarrow \infty}$.
I'll first work on simplifying $y$ as follows. We can take an $x^2$ out of the square root , so that $y= x^2 \sqrt{1+7/x^2 + 1/x^4}$. We can expand the right side out by using the binomial expansion $\sqrt{1+\epsilon} = 1 + \epsilon/2 + H(\epsilon)$, where $H$ stands for the higher order terms in the expansion. Thus, $y = x^2(1+ \frac{7}{2x^2} + \frac{1}{2x^4} + H(x))$. Note that $H(x)$ contains large negative powers of $x$, from $x^{-4}$ onwards.
Now we can find $L$. We know that
\begin{equation}
L = \lim (x^2 - y) \\
=\lim x^2 - x^2(1+ \frac{7}{2x^2} + \frac{1}{2x^4} + H(x)) \\
= \lim x^2 - (x^2 + \frac{7}{2} + \frac{1}{2x^2} + H(x)x^2) \\
= \lim 0 - 7/2 - H(x)x^2
\end{equation}
where $H(x)x^2$ contains only negative powers of $x$. As such the limit evaluates to $L = -7/2$.
This is very similar to what you have done, except that I simplified the algebra and swept all unwanted terms into $H$. The answer matches yours.
|
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|
How to integrate $1/(16+x^2)^2$ using trig substitution My answer: $1/128 (\tan^{-1}(x/4) + 4x/(16+x^2)^4)$
|
$$
\int\frac{dx}{(16+x^2)^2} = \int\frac{4\sec^2\theta\,d\theta}{(16\sec^2\theta)^2} = \frac{1}{64} \int\frac{d\theta}{\sec^2\theta} = \frac{1}{64}\int \cos^2\theta\,d\theta = \cdots\cdots
$$
\begin{align}
x & = 4\tan\theta \\[6pt]
dx & = 4\sec^2\theta\,d\theta \\[6pt]
16+x^2 & = 16 + 16\tan^2\theta = 16\sec^2\theta
\end{align}
After getting a function of $\theta$, it may be useful to know that
$$
\text{If }x=4\tan\theta\text{ then }\sin\theta = \frac{x}{\sqrt{x^2+16}}\text{ and }\cos\theta = \frac{4}{\sqrt{x^2+16}}.
$$
I.e., remember trigonometry.
Appendix on trigonometry in response to comments:
If $\tan\theta=\dfrac x4$, then you can construct a triangle in which $\text{opposite}=x$ and $\text{adjacent}=4$. Consequently $\text{hypotenuse}=\sqrt{x^2+4^2}$. So $\sin\theta=\dfrac{\text{opposite}}{\text{hypotenus}}=\dfrac{x}{\sqrt{x^2+4^2}}$ and $\cos\theta=\dfrac{\text{adjacent}}{\text{hypotenus}}=\dfrac{4}{\sqrt{x^2+4^2}}$.
I'm not sure whether this actually addresses the concerns raised in the comments.
Later edit in order to compare answers:
\begin{align}
\frac{1}{64}\int\cos^2\theta\,d\theta & = \frac{1}{64}\int \frac12 + \frac12 \cos(2\theta)\,d\theta = \frac{1}{64} \left(\frac\theta2 + \frac14\sin(2\theta)\right)+C \\[12pt]
& = \frac{1}{64}\left(\frac12\arctan\left(\frac x4\right)+\frac12\sin\theta\cos\theta\right)+C \\[12pt]
& = \frac{1}{128}\arctan\left(\frac x4\right)+\frac{1}{128}\cdot\frac{4x}{x^2+16} + C \\[12pt]
& = \frac{1}{128}\arctan\left(\frac x4\right)+\frac{1}{32}\cdot\frac{x}{x^2+16} + C.
\end{align}
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|
Integral calculus What is the integral
$$\int \arccos(z/\sqrt{R^2-x^2})dx$$
This is one of 4 equations for integrating the area of a major sector of a circle within a sphere between limits to find its volume. Two of the functions are easily integrated (first and last), but the above and $\int\arccos(x^2(z/\sqrt{R^2-x^2}))dx$ are difficult to do. I also need the integration of this equation too.
The full equation to be integrated is:
$$
Area = \pi(R^2-x^2)-(R^2-x^2)\arccos(z/\sqrt{R^2-x^2}+z*\sqrt{R^2-x^2-z^2}.
$$
The function $z$ is the distance to the major segment chord from the center.
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To me, it looks like that $\arccos$ part is the source of all ugliness for this problem. I'm not sure using the substitution $x=R\sin\theta$ will simplify this $\arccos$ into something manageable. Therefore, I myself would try to eliminate the $\arccos$ through integration by parts.
$$u=\arccos[z(R^2-x^2)^{-\frac12}],du=\frac{d[z(R^2-x^2)^{-\frac12}]}{\sqrt{1-z(R^2-x^2)^{-\frac12}}}=$$
$$\frac{-\frac12z(R^2-x^2)^{-\frac32}(-2x)dx}{\sqrt{1-z(R^2-x^2)^{-\frac12}}}$$
Multiplying top and bottom by $(R^2-x^2)^\frac32$
$$du=\frac{xzdx}{\sqrt{(R^2-x^2)^3-z(R^2-x^2)^\frac52}}$$
$$\int\arccos(\frac{z}{\sqrt{R^2-x^2}})dx=x\arccos(\frac{z}{\sqrt{R^2-x^2}})-\int\frac{zx^2dx}{\sqrt{(R^2-x^2)^3-z(R^2-x^2)^\frac52}}$$
Ugly, I know, but at least we have some directions we can try. Let's try
$$x=(R^2-u^2)^\frac12,dx=\frac{du}{2\sqrt{R^2-u^2}}$$
$$\int\frac{zx^2dx}{\sqrt{(R^2-x^2)^3-z(R^2-x^2)^\frac52}}=\int\frac{z(R^2-u^2)du}{2\sqrt{(R^2-u^2)(u^6-zu^5)}}=\frac z2\int\sqrt{\frac{R^2-u^2}{u^6-zu^5}}du$$
Well, I gave it a valiant effort, but I appear to be stuck at this point. Maybe someone else can pick up where I left off (or more likely, someone will link to Wolfram making all of the above completely irrelevant...).
|
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|
Problems regarding integrals involving Legendre polynomials I am finding difficulty doing this integral involving Legendre polynomials.
$$\int_{-1}^1 x^2 P_{n-1}(x)P_{n+1}(x)dx = \frac{2n(n+1)}{(2n-1)(2n+1)(2n+3)}$$ I have two strategies in my mind both of them have failed to produce results. One is that I could somehow use the orthogonality of Legendre polynomials after using Bonnet's recursion formula to get Legendre polynomials, to simplify the integrals, or I could use the Rodrigues formula by doing integration by parts. The first approach fails because of the $x^2$ in the integral. The second approach is not giving me integrated parts, where limits can be applied easily.How do I do this?
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By Bonnet's formula, we have
\[ (2n-1)xP_{n-1} = nP_n + (n-1)P_{n-2} \]
and
\[ (2n+3)xP_{n+1} = (n+2)P_{n+2} + (n+1)P_n \]
so
\begin{align*}
\int_{-1}^1 x^2 P_{n-1}P_{n+1}\; dx
&= \frac 1{(2n-1)(2n+3)}\int_{-1}^1 \bigl( nP_n + (n-1)P_{n-2}\bigr)\bigl((n+2)P_{n+2} + (n+1)P_n\bigr)\; dx\\
&= \frac 1{(2n-1)(2n+3)}\int_{-1}^1 n(n+1)P_n^2\; dx\\
&= \frac {2n(n+1)}{(2n-1)(2n+1)(2n+3)}
\end{align*}
I have another denominator, I saw. But I will show also by example mine is right ;-): We have ($n=1$) that $P_0 = 1$, $P_2(x) = \frac 12(3x^2 - 1)$. It holds
\begin{align*}
\frac 12 \int_{-1}^1 (3x^4 - x^2) \; dx &= \frac 22\left(\frac 35 - \frac 13\right)\\
&= \frac 4{15}\\
&= \frac{2\cdot 1 \cdot (1+1)}{(2\cdot 1 - 1)(2 \cdot 1 \mathbin{{\color{red}+}} 1)(2\cdot 1 + 3)}
\end{align*}
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|
Find a $4\times 4$ matrix $A$ where $A\neq I$ and $A^2 \neq I$, but $A^3 = I$. Give an example of a $4 \times 4$ matrix where $A \neq I$, $A^2 \neq I$, and $A^3 = I$.
I found a $2 \times 2$ matrix where $A \neq I$ and $A^2 = I$, but this problem is more complex and has me completely stumped.
|
Here is a $2\times2$ example
$$
\begin{bmatrix}
-\frac12&\frac{\sqrt{3}}{2}\\
-\frac{\sqrt{3}}{2}&-\frac12
\end{bmatrix}\tag{1}
$$
This works because it is a matrix representation of $e^{2\pi i/3}=-\frac12+i\frac{\sqrt{3}}{2}$; that is, a rotation by $\frac{2\pi}{3}$. Thus, squaring it gives another $2\times2$ example
$$
\begin{bmatrix}
-\frac12&-\frac{\sqrt{3}}{2}\\
\frac{\sqrt{3}}{2}&-\frac12
\end{bmatrix}\tag{2}
$$
$(1)$ and $(2)$ can easily be extended to $4\times4$ examples in many ways. Here is one using $(1)$:
$$
\begin{bmatrix}
-\frac12&\frac{\sqrt{3}}{2}&0&0\\
-\frac{\sqrt{3}}{2}&-\frac12&0&0\\
0&0&\hspace{7pt}1\hspace{7pt}\vphantom{-\frac12}&0\\
0&0&0&\hspace{5pt}1\hspace{5pt}\vphantom{-\frac12}
\end{bmatrix}\tag{3}
$$
Yet another $2\times2$ matrix:
As Hagen von Eitzen points out, we can consider the unit vectors $u,v,w$, which are separated by $\frac{2\pi}{3}$
$\hspace{5cm}$
and note that $u+v+w=0$ to get that $w=-u-v$. Let $\begin{bmatrix}a\\b\end{bmatrix}$ be a coordinate vector using the basis vectors $\{u,v\}$. Then rotation by $\frac{2\pi}{3}$ has the following action:
$$
\begin{bmatrix}u&v\end{bmatrix}\begin{bmatrix}a\\b\end{bmatrix}\mapsto\begin{bmatrix}v&w\end{bmatrix}\begin{bmatrix}a\\b\end{bmatrix}=\begin{bmatrix}u&v\end{bmatrix}\begin{bmatrix}0&-1\\1&-1\end{bmatrix}\begin{bmatrix}a\\b\end{bmatrix}\tag{4}
$$
Thus, the matrix for rotating by $\frac{2\pi}{3}$ under the basis $\{u,v\}$ is
$$
\begin{bmatrix}0&-1\\1&-1\end{bmatrix}\tag{5}
$$
which, as Gerry Myerson points out, is the companion matrix for $x^2+x+1$. A companion matrix is annihilated by its polynomial, so $(5)$ is annihilated by $x^3-1=(x-1)(x^2+x+1)$.
The square of $(5)$ also satisfies the specified conditions:
$$
\begin{bmatrix}-1&1\\-1&0\end{bmatrix}\tag{6}
$$
The answer given by Marc van Leeuwen uses $(6)$.
|
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|
Using generating functions find the sum $1^3 + 2^3 + 3^3 +\dotsb+ n^3$ I am quite new to generating functions concept and I am really finding it difficult to know how to approach problems like this. I need to find the sum of $1^3 + 2^3 + 3^3 +\dotsb+ n^3$ using generating functions. How do I proceed about it?
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If you really want to get blown away, consider the following, taken from Aigner's "A Course in Enumeration" (Springer, 2007). Define:
$$
s_m(n) = \sum_{0 \le k < n} k^m
$$
and it's exponential generating function:
$
\begin{align}
\widehat{S}_n(z)
&= \sum_{m \ge 0} s_m(n) \frac{z^m}{m!} \\
&= \sum_{1 \le k \le n - 1} \sum_{m \ge 0} \frac{k^m z^m}{m!} \\
&= \sum_{1 \le k \le n - 1} \mathrm{e}^{k z} \\
&= \frac{\mathrm{e}^{n z} - 1}{\mathrm{e}^z - 1}
\end{align}
$
This is almost the exponential generating function for the powers:
$
\begin{align}
\widehat{P}(z)
&= \sum_{m \ge 0} n^m \frac{z^m}{m!} \\
&= \mathrm{e}^{n z}
\end{align}
$
Sadly, the series $\mathrm{e}^z - 1$ has no reciprocal, as it has no constant term. But we can write:
$$
(\widehat{P}(z) - 1) \widehat{B}(z)
= z \widehat{S}(z)
$$
where:
$$
\widehat{B}(z)
= \frac{z}{\mathrm{e}^z - 1}
$$
whose coefficients are the Bernoulli numbers:
$\begin{align}
\widehat{B}(z)
&= \sum_{n \ge 0} B_n \frac{z^n}{n!} \\
&= 1 - \frac{1}{2} z + \frac{1}{6} \frac{z^2}{2!}
- \frac{1}{30} \frac{z^4}{4!} + \frac{1}{42} \frac{z^6}{6!}
- \frac{1}{30} \frac{z^8}{8!} + \frac{5}{66} \frac{z^{10}}{10!}
- \frac{691}{2130} \frac{z^{12}}{12!}
+ \frac{7}{6} \frac{z^{14}}{14!}
- \dotsb
\end{align}$
and finally:
$$
\sum_{m \ge 0} s_m(n) \frac{z^{m + 1}}{m!}
= \sum_{m \ge 0} z^m
\sum_{0 \le k \le m}
\binom{m}{k} \frac{(n z)^{m - k}}{(m - k)!} B_k
$$
Comparing coefficients of $z^{m + 1}$ and simplifying:
$$
s_m(n)
= \frac{1}{m + 1}
\sum_{0 \le k \le m} (-1)^k \binom{m + 1}{k} B_k n^{m + 1 - k}
$$
|
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|
how to find the series $x + x^{1 + \frac{1}{2}} + x^{1 + \frac{1}{2}+ \frac{1}{3}} +...$ is convergent. The series $x + x^{1 + \frac{1}{2}} + x^{1 + \frac{1}{2}+ \frac{1}{3}} +...$
is convergent if
(A) $x>e$
(B) $x<e $
(C) $x<1/e$
(D) $x>1/e$
I think the answer is C, but I could not determine the condition... how to solve it. Plz help.
|
Nth term of series is given by $u_n=x^{\sum_{i=1}^n \frac{1}{i}}$
Logarithmic test says
Suppose $\sum_{ n \geq 1} a_n $ is a series of positive terms. Suppose
that
$$ \lim_{n \to \infty} n \ln \dfrac{ a_n }{a_{n+1} } = g $$
and if $g>1$ series is convergent and divergent if $g<1$.
now firstly take $x\ge0$, at $x=0$ series converges, and at $x=1$ series diverges(option B discarded).
Now $1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4} \cdots$ is divergent series, so $\lim( 1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4} \cdots)$ tends to infinity so if $x>1$, $lim(x^{1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4} \cdots}) \ne 0$, so not convergent for $x>1$(option A discarded).
applying log test for $0<x<1$, $$\lim_{n \to \infty} n \log \dfrac{ a_n }{a_{n+1} }=n \log(x^{\frac{-1}{n}})=-\log x$$
so if $-\log x > 1$ or $x<\frac{1}{e}$ , series converges and if $x>\frac{1}{e}$ series diverges.
For negative values of $x$ , series is absolutely convergent for $x<\frac{1}{e}$ so it is convergent in $x<\frac{1}{e}$ region.
|
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|
Monotonicity of $\ln n\over \sqrt{n}$ Is it possible to prove the monotonicity of $(a_n)_n =\frac{\ln n}{\sqrt n}$ without using derivatives?
|
For $a_n=\dfrac{\ln{n}}{\sqrt{n}}$ we denote $f_n=\dfrac{1}{a_n^2}=\dfrac{n}{\ln^2{n}}.$ In order to prove that $\left\lbrace a_n \right\rbrace $ is decreasing, we consider
\begin{gather*}
f_{n+1}-f_n=\dfrac{n+1}{\ln^2(n+1)}- \dfrac{n}{\ln^2{n}}=
\dfrac{(n+1)\ln^2{n}-n\ln^2(n+1)}{\ln^2{n} \ \ln^2(n+1)}=\dfrac{A}{B}
\end{gather*}
and show that sequence $\left\lbrace f_n \right\rbrace$ increases: $f_{n+1}-f_n >0 \quad \left(\forall n \geqslant n_0 \right).$
Numerator $A$ may be rewritten as
\begin{gather*}
A=(n+1)\ln^2{n}-n\ln^2(n+1)=(n+1)\ln^2{n}-n\ln^2\left(n\left(1+\dfrac{1}{n}\right)\right)=\\
(n+1)\ln^2{n}-n\left(\ln{n}+\ln\left(1+\dfrac{1}{n}\right)\right)^2=\\
n \ln^2{n}+\ln^2{n}-n \ln^2{n}-2n \ln{n}\ln\left(1+\dfrac{1}{n}\right)-n \ln^2\left(1+\dfrac{1}{n}\right)=\\
\ln^2{n}-n \ln{n}\ln\left(1+\dfrac{1}{n}\right)-n \ln{n}\ln\left(1+\dfrac{1}{n}\right)-n \ln^2\left(1+\dfrac{1}{n}\right)=\\
\ln{n}\left(\ln{n}-n\ln\left(1+\dfrac{1}{n}\right)\right)-n\ln\left(1+\dfrac{1}{n}\right)\left( \ln{n}+\ln\left(1+\dfrac{1}{n}\right)\right)=\\
\ln{n}\left(\ln{n}-\ln\left(1+\dfrac{1}{n}\right)^n\right)-\ln\left(1+\dfrac{1}{n}\right)^n\left( \ln{n}+\ln\left(1+\dfrac{1}{n}\right)\right).
\end{gather*}
Next, applying the estimate
$${2} \leqslant \left(1+\dfrac{1 }{n}\right)^n \leqslant {3} \quad \left(\forall n \in \mathbb{N}\right),$$
we obtain
\begin{gather}
\ln{n}\left(\ln{n}-\ln\left(1+\dfrac{1}{n}\right)^n\right) \geqslant
\ln{n}\left(\ln{n}-\ln{3} \right);\\
-\ln\left(1+\dfrac{1}{n}\right)^n\left( \ln{n}+\ln\left(1+\dfrac{1}{n}\right)\right) \geqslant -\ln{3}(\ln{n}+\ln{2}).
\end{gather}
Thus, by adding these inequalities,
$$
A \geqslant {\ln{n}\left(\ln{n}-\ln{3} \right) -\ln{3}(\ln{n}+\ln{2})}=
\ln^2{n}-2\ln{3} \ \ln{n}-\ln{2} \ \ln{3}.$$
Quadratic polynomial
$
x^2-2\ln{3}\cdot x-\ln{2}\ln{3}>0 $ for $x\in\left(-\infty,\, x_1\right)\cup{\left(x_2, \,+\infty\right)},
$ where
\begin{gather*}x_1=\ln{3}- \sqrt{\ln^2{3}+\ln{2}\ln{3}}, \\
x_2=\ln{3}+ \sqrt{\ln^2{3}+\ln{2}\ln{3}}.
\end{gather*}
Thus, $A>0$ for $\ln{n}>\ln{3}+ \sqrt{\ln^2{3}+\ln{2}\ln{3}}\approx 2.501626534,$ respectively $n\geqslant n_0= \left \lfloor{e^{2.501626534}}\right \rfloor+1=\left\lfloor {12.20232533}\right\rfloor+1=13.$
This value is less precise than exact value $e^2,$ which is point of minimum for $f(x)=\dfrac{x}{\ln^2{x}}$.
|
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|
Show that $n^3-n$ is divisible by $6$ using induction As homework, I have to prove that
$\forall n \in \mathbb{N}: n^3-n$ is divisible by 6
I used induction
1) basis: $A(0): 0^3-0 = 6x$ , $x \in \mathbb{N}_0$ // the 6x states that the result is a multiple of 6, right?
2) requirement: $A(n):n^3-n=6x$
3) statement: $A(n+1): (n+1)^3-(n+1)=6x$
4) step: $n^3-n+(n+1)^3-n=6x+(n+1)^3-n$
So when I resolve that I do get the equation: $n^3-n=6x$ so the statement is true for $\forall n \in \mathbb{N}$
Did I do something wrong or is it that simple?
|
No need for induction.
$$n^3-n=n(n^2-1)=n(n-1)(n+1)$$
which are three consecutive integers. So one must be divisible by 3.
Check for $n=1$: $1^3-1=0=3\cdot 0$
Assume it's true for $n=k$.
If you let $n=k+1$ you get
$$\begin{align*}
(k+1)^3-(k+1)&=k^3+3k^2+2 \\
&=k^3+3k^2+2k\\
&=3\cdot (k^2+k)+(k^3-k)\end{align*}$$
which is divisible by 3
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"timestamp": "2023-03-29T00:00:00",
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|
When is $1^5 + 2^5 + \ldots + n^5$ a square? When is $1^5 + 2^5 + \ldots + n^5$ a square? I found that this happens sometimes: $n=13$ gives $1001^2$, $n=133$ gives $9712992^2$ and $n=1321$ gives $942162299^2$.
I feel that the identity$$\displaystyle\sum_{i=1}^n i^5 = \tfrac{1}{12}[2n^6+6n^5+5n^4-n^2]$$ will be useful, since it's all square powers except one.. but I see no way to connect that.
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$$1^5 + 2^5 + ... + n^5 = \frac{1}{12} n^2 (n+1)^2 (2 n^2+2 n-1) $$
Now we need to solve $$\frac{1}{12} n^2 (n+1)^2 (2 n^2+2 n-1) $$ is a perfect square
$$\iff\frac{2 n^2+2 n-1}{12} = k^2$$
[do some thing here] I don't know how to solve more
|
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"timestamp": "2023-03-29T00:00:00",
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|
For what values of the variable x does the following inequality hold: $\ \frac{4x^2}{\Bigl(1-\sqrt{\ 1\ +2x}\Bigr)^2} <
2x+9$
... IMO-1960
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Maybe you wanna write everything as
$$\ \frac{((2x+1)-2)(2x+1)+1}{\Bigl(1-\sqrt{2x+1}\Bigr)^2} <
2x+1+8$$
Then, let's denote $2x+1=y$ that yields
$$\ \frac{(y-2)y+1}{\Bigl(1-\sqrt{y}\Bigr)^2} <
y+8$$
$$\ \frac{(1-y)^2}{\Bigl(1-\sqrt{y}\Bigr)^2} <
y+8$$
$$\ \left(\frac{(1-\sqrt{y})(1+\sqrt{y})}{1-\sqrt{y}}\right)^2 <
y+8$$
$$\ (1+\sqrt{y})^2 <
y+8$$
$$y<\frac{49}{4}$$
Or
$$2x+1<\frac{49}{4}$$
$$x<\frac{45}{8} \tag1$$
At the same time we know that
$$x\ge -\frac{1}{2} \tag2$$
and pay attention at $x=0$.
From $(1)$ and $(2)$ we conclude that
$$x\in \left[-\frac{1}{2}, 0\right)\cup \left(0,\frac{45}{8}\right).$$
Chris.
|
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|
What are the steps to simplify the following modulus expression? I have no clue how to do this exactly. Is there a systematic way of doing this or you just have to do it by trial and error?
$n^2 \equiv 9 \pmod {72}$
to $n \equiv a \pmod b$?
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Note that $72=8\cdot 9$ and that $n^2\equiv 9\pmod{72}$ implies (and is in fact equivalent to) $n^2\equiv 9\pmod 9$ and $n^2\equiv 9\pmod 8$.
The first simplifies to $n^2\equiv 0\pmod 9$, i.e. $3^2|n^2$. This is equivalent to $3|n$. The other equation simplifieds to $n^2 \equiv 1\pmod 8$. You may try these few cass by hand, but one should know this special property of the prime $2$:
*
*The square of an even number is always $\equiv 0\pmod 4$
*The square of an odd number is always $\equiv 1\pmod 8$
Thus $n^2\equiv 9\pmod{72}$ is the same as $n\equiv 0\pmod 3$ and $n\equiv 1\pmod 2$, i.e. $$n\equiv 3\pmod 6.$$
Check: If $n=6k+3$, then $n^2=36 k^2+36 k + 9=36\cdot k(k+1)+9$. Since one od the numbers $k, k+1$ is even, $k(k+1)$ is even and $36k(k+1)$ is a multiple of $72$.
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{
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"timestamp": "2023-03-29T00:00:00",
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|
Proving that $|CA|+|CB|=2|AB|$ in a general $ABC$ triangle How in this situation (presented in image) can I prove that $|CA|+|CB|=2|AB|$?
|
Here's a calculatory approach:
Call $AE=x$ and $BD=y$ and let the angles at $A$, $B$, $C$ be $\alpha$, $\beta$, $\gamma$, respectively.
Now, the area of the triangle $ABC$ equals $$\frac 12 (a+x) (b+y) \sin \gamma.$$
But it is also the sum of the three smaller regions:
$$ab\sin\gamma +\frac 12 ay\sin\beta +\frac 12 xb \sin \alpha.$$
Now, equate the two expressions, divide everything by $\sin \gamma$ and use $\dfrac{\sin \beta}{\sin \gamma} = \dfrac{a+x}{a+b}$ and $\dfrac{\sin \alpha}{\sin \gamma} = \dfrac{b+y}{a+b}$ to obtain
$$ab+ \frac 12 ay \frac {a+x}{a+b} + \frac 12 xb \frac{y+b}{a+b} = \frac 12 (a+x)(b+y).$$
Multiply everything by $2(a+b)$ and develop to get:
$$a^2b+ab^2=aby+xab.$$
Divide by $ab$ to get
$$a+b=x+y$$
which is equivalent to the required equation: $$2|AB| =2a+2b = (a+x)+(b+y)=|CA|+|CB|.$$
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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.