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Series Contest (SC3)

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Post Fri Oct 11, 2013 7:25 pm
sos440 User avatar
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galactus wrote:
Here is one that appears to be a little tough, but it has an interesting result. Maybe not, though, you guys are quite clever.

Problem 5

\(\displaystyle \displaystyle \lim_{n\to \infty}\left(\frac{\sum_{k=1}^{\infty}\sin(\frac{2k}{2n})}{\sum_{k=1}^{\infty}\sin(\frac{2k-1}{2n})}\right)^{n}=e^{-2\tan(1/2)}\)

You may want to refer to this.
Every integral harbors a story as fascinating as a comic book!

Post Sat Oct 12, 2013 4:33 am
Shobhit Site Admin
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Posts: 850
Location: Jaipur, India

Here is another one:

Problem 6

Prove that \(\displaystyle \sum_{n=-\infty}^\infty e^{-5\pi n^2}= \frac{\sqrt[4]{\pi}}{5\Gamma \left(\frac{3}{4} \right)}\sqrt{10+5\sqrt{5}}\)

You are allowed to use the already established result: \(\displaystyle \sum_{n=-\infty}^\infty e^{-\pi n^2}=\frac{\sqrt[4]{\pi}}{\Gamma \left(\frac{3}{4} \right)}\)

[HINT] This problem is a little tough. I would suggest to start by proving that

\(\displaystyle \sqrt{\alpha} \sum_{n=-\infty}^{\infty}e^{-\alpha^2 n^2} = \sqrt{\beta} \sum_{n=-\infty}^{\infty}e^{-\beta^2 n^2}\)

where \(\alpha \beta = \pi\) and \(\Re \alpha,\Re \beta >0\).

Post Sat Oct 12, 2013 5:29 am
Shobhit Site Admin
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Post Sat Oct 12, 2013 1:11 pm
galactus User avatar
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Thanks, SOS. Is there any problem you can't eventually solve? :):)

Anyway, my attempt was to use the exponential for sin. I chipped away at this off over the last several days. I think I made some headway.

Using what is inside the parentheses:

\(\displaystyle \sum_{k=1}^{n}\sin(k/n)=\frac{1}{2i}\sum_{k=1}^{n}\left(e^{\frac{ki}{n}}-e^{\frac{-ki}{n}}\right)\)

\(\displaystyle \sum_{k=1}^{n}\sin(\frac{2k-1}{2n})=\frac{1}{2i}\sum_{k=1}^{n}\left(e^{\frac{(2k-1)i}{2n}}-e^{\frac{-(2k-1)i}{2n}}\right)\)

So, we get:

\(\displaystyle \frac{\frac{1}{2i}\sum_{k=1}^{n}\left(e^{\frac{ki}{n}}-e^{\frac{-ki}{n}}\right)}{\frac{1}{2i}\sum_{k=1}^{n}\left(e^{\frac{(2k-1)i}{2n}}-e^{\frac{-(2k-1)i}{2n}}\right)}\)

Now, factor a little:

\(\displaystyle \frac{\sum_{k=1}^{n}\left(e^{i/n}\right)^{k}-\sum_{k=1}^{n}\left(e^{-i/n}\right)^{k}}{e^{-i/2n}\sum_{k=1}^{n}\left(e^{i/n}\right)^{k}-e^{i/2n}\sum_{k=1}^{n}\left(e^{-i/2n}\right)^{k}}\)

These are partial geometric series. They should simplify down.

The top left one evaluates to \(\displaystyle \frac{(e^{i}-1)e^{i/n}}{e^{i/n}-1}\)

The top right one: \(\displaystyle \frac{(e^{i}-1)e^{-i}}{e^{i/n}-1}\)

The bottom left: \(\displaystyle \frac{e^{-i/2n}(e^{i}-1)e^{i/n}}{e^{i/n}-1}\)

The bottom right: \(\displaystyle \frac{e^{i/2n}(e^{i}-1)e^{-i}}{e^{i/n}-1}\)

Putting these altogether, it should whittle down to some trig functions involving sin and/or cos. But, I have not finished yet.

It looks encouraging though.

EDIT:

Well, I spent some time trying to hammer down the results of the sums above.

They become:

\(\displaystyle \frac{\sin(1/n+1)-\sin(1/n)-\sin(1)}{\sin(1/2n+1)+\sin(1/2n-1)-2\sin(1/2n)}\)

this is equivalent to:

\(\displaystyle \frac{\cos(1/2n)\sin(1/2)+\sin(1/2n)\cos(1/2)}{\sin(1/2)}\)

Using the product-to-sum formulas on the numerator, it whittles down to:

\(\displaystyle =\frac{\sin(\frac{n+1}{2n})}{\sin(1/2)}\)

But, I admit I left tech do most of the work and played around with some trial and error.

So, we finally get:

\(\displaystyle \lim_{n\to \infty}\left(\frac{\sin(\frac{n+1}{2n})}{\sin(1/2)}\right)^{n}\)

Now, since this is a limit and there is an 'e' in the required solution, I figured I would make the sub \(\displaystyle n=1/k\) in order to get something that resembles the 'e' limit.

\(\displaystyle \lim_{k\to 0}\left(\frac{\sin(\frac{k+1}{2})}{\sin(1/2)}\right)^{1/k}\)

So, take logs:

\(\displaystyle \lim_{k\to 0}\frac{1}{k}[log(\sin(\frac{k+1}{2})-log(\sin(1/2))]\)

Using L'Hopital and taking this limit:

\(\displaystyle \lim_{k\to 0}\frac{1}{2}\cot(\frac{k+1}{2})\)

results in \(\displaystyle 1/2\cot(1/2)\)

Now, e:

\(\displaystyle e^{1/2\cot(1/2)}=e^{\frac{1}{2\tan(1/2)}}\)

WHEW!!!!. I messed around with this hoping it would lead somewhere eventually. The worst part, for me anyway, was finding a workable trig identity out of all of that.

Post Sun Oct 13, 2013 3:44 pm
Random Variable Integration Guru
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Posts: 381
I'd liked to see a somewhat self-contained proof of the relationship between the complete elliptic integral of the first kind and the theta function \(\displaystyle \theta(q) = \sum_{n=-\infty}^{\infty} q^{n^{2}}\).

Specifically that \(\displaystyle K(k) = \int_{0}^{\frac{\pi}{2}} \frac{d \phi}{\sqrt{1-k^{2} \sin^{2} \phi}} \ d \phi = \frac{\pi}{2} \ _{2}F_{1} \left( \frac{1}{2}, \frac{1}{2}; 1; k^{2} \right) = \frac{\pi}{2} \theta^{2} \Bigg( \exp \Big(- \frac{K(\sqrt{1-k^{2}})}{K(k)} \pi \Big) \Bigg)\)

The one given by Bruce C Berndt in his books on Ramanujan's notebooks is scattered among multiple books.

Post Mon Oct 14, 2013 6:35 am
Shobhit Site Admin
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Posts: 850
Location: Jaipur, India

Random Variable wrote:
I'd liked to see a somewhat self-contained proof of the relationship between the complete elliptic integral of the first kind and the theta function \(\displaystyle \theta(q) = \sum_{n=-\infty}^{\infty} q^{n^{2}}\).

Specifically that \(\displaystyle K(k) = \int_{0}^{\frac{\pi}{2}} \frac{d \phi}{\sqrt{1-k^{2} \sin^{2} \phi}} \ d \phi = \frac{\pi}{2} \ _{2}F_{1} \left( \frac{1}{2}, \frac{1}{2}; 1; k^{2} \right) = \frac{\pi}{2} \theta^{2} \Bigg( \exp \Big(- \frac{K(\sqrt{1-k^{2}})}{K(k)} \pi \Big) \Bigg)\)

The one given by Bruce C Berndt in his books on Ramanujan's notebooks is scattered among multiple books.


Let \(\displaystyle y=\text{sn}(u,k) = \frac{\vartheta_3 }{\vartheta_2}\frac{\vartheta_1(u \vartheta_3^{-2} )}{\vartheta_4(u \vartheta_3^{-2} )}\) and \(\displaystyle k=\frac{\vartheta_3^2}{\vartheta_2^2}\)

Using formula for derivatives of ratios of theta functions, we get

\(\displaystyle \begin{align*}
\left( \frac{dy}{du} \right)^2 &= \left( \frac{1}{\vartheta_3\vartheta_2} \frac{\vartheta_4^2 \vartheta_2(\vartheta_3^{-2}u)\vartheta_3( \vartheta_3^{-2}u)}{\vartheta_4^2( \vartheta_3^{-2}u)} \right)^2 \\
&= (1-y^2)(1-k^2 y^2) \tag{*}
\end{align*}\)

Here we used that

\(\displaystyle 1-y^2 = \text{cn}(u,k)=\frac{\vartheta_2^2(u\vartheta_3^{-2})\vartheta_4^2}{\vartheta_2^2 \vartheta_4^2( \vartheta_3^{-2}u)}\tag{1}\)

and

\(\displaystyle 1-k^2 y^2 = \text{dn}(u,k)= \frac{\vartheta_3^2 (\vartheta_3^{-2}u) \vartheta_4^2}{\vartheta_3^2 \vartheta_4^2(\vartheta_3^{-2}u)}\tag{2}\)

The Differential Equation, (*) may now be written as

\(\displaystyle u = \int_0^{y} \frac{1}{\sqrt{(1-t^2)(1-k^2 t^2)}}dt \tag{3}\)

When \(y=1\), (1) will become

\[\vartheta_2^2(u \vartheta_3^{-2})=0\]

This is possible only if

\[u \vartheta_3^{-2} = \frac{\pi}{2} \implies u = \frac{\pi}{2}\vartheta_3^2\]
Putting this in equation (3), we get

\[ {\frac{\pi}{2}\vartheta_3^2 = K} \tag{4}\]

Note \(\displaystyle \vartheta_i \equiv \vartheta_i(0,q)\) and \(\displaystyle \vartheta_i(z) \equiv \vartheta_i(z,q)\).

Post Mon Oct 14, 2013 11:55 am
Shobhit Site Admin
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Posts: 850
Location: Jaipur, India

Should I post the solution to Problem 6? :geek:

Post Mon Oct 14, 2013 7:47 pm
Random Variable Integration Guru
Integration Guru

Posts: 381
I don't know enough about the properties of the Jacobi theta functions to follow that proof.


The identity \(\displaystyle \sqrt{\alpha} \sum_{n=-\infty}^{\infty}e^{-\alpha^2 n^2} = \sqrt{\beta} \sum_{n=-\infty}^{\infty}e^{-\beta^2 n^2}\) (where \(\displaystyle \alpha \beta = \pi\)) follows from the identity \(\displaystyle \sum_{n=-\infty}^{\infty} e^{-\pi \tau n^2} = \frac{1}{\sqrt{\tau}} \sum_{n=-\infty}^{\infty} e^{-\pi n^2 / \tau}\).

post204.html#p204


Let \(\displaystyle \tau = \frac{\alpha}{\beta}\).

Then \(\displaystyle \sqrt{\alpha} \sum_{n=-\infty}^{\infty} e^{- \pi \frac{\alpha}{\beta} n^{2}} = \sqrt{\beta} \sum_{n=-\infty}^{\infty} e^{- \pi \frac{\beta}{\alpha} n^{2}}\)

\(\displaystyle \implies \sqrt{\alpha} \sum_{n=-\infty}^{\infty} e^{- \pi \alpha \frac{\alpha}{\pi} n^{2}} = \sqrt{\beta} \sum_{n=-\infty}^{\infty} e^{- \pi \beta \frac{\beta}{\pi} n^{2}}\)

\(\displaystyle \implies \sqrt{\alpha} \sum_{n=-\infty}^{\infty} e^{- \alpha^{2} n^{2}} = \sqrt{\beta} \sum_{n=-\infty}^{\infty} e^{- \beta^{2} n^{2}}\)

Post Mon Oct 14, 2013 11:28 pm
Random Variable Integration Guru
Integration Guru

Posts: 381
Are you sure that's the answer to problem 6?

According to that formula we need the singular value \(\displaystyle k_{25}\) to evaluate \(\displaystyle \sum_{n=-\infty}^\infty e^{-5\pi n^2}\).

According to Wolfram Mathworld, \(\displaystyle K(k_{25}) = \frac{\sqrt{5}+2}{20} \frac{\Gamma^{2} (\frac{1}{4})}{\sqrt{\pi}}\)

http://mathworld.wolfram.com/EllipticIn ... Value.html


So \(\displaystyle K(k_{25}) = \frac{\sqrt{5}+2}{20} \frac{\Gamma^{2} (\frac{1}{4})}{\sqrt{\pi}} = \frac{\pi}{2} \theta^{2} (e^{- 5 \pi} ) = \frac{\pi}{2} \Big( \sum_{k=-\infty}^{\infty} e^{- 5 \pi n^{2}} \Big)^{2}\)

\(\displaystyle \implies \sum_{k=-\infty}^{\infty} e^{- 5 \pi n^{2}} = \sqrt{\frac{\sqrt{5}+2}{10}} \frac{\Gamma (\frac{1}{4})}{\pi^{3/4}} = \sqrt{\frac{\sqrt{5}+2}{10}} \frac{\pi \csc (\frac{\pi}{4})}{\Gamma( \frac{3}{4}) \pi^{3/4}}\)

\(\displaystyle = \sqrt{\frac{\sqrt{5}+2}{5}} \frac{\pi^{\frac{1}{4}}}{\Gamma (\frac{3}{4})}\)

Post Tue Oct 15, 2013 9:18 am
Shobhit Site Admin
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Posts: 850
Location: Jaipur, India

Random Variable wrote:
I don't know enough about the properties of the Jacobi theta functions to follow that proof.


The identity \(\displaystyle \sqrt{\alpha} \sum_{n=-\infty}^{\infty}e^{-\alpha^2 n^2} = \sqrt{\beta} \sum_{n=-\infty}^{\infty}e^{-\beta^2 n^2}\) (where \(\displaystyle \alpha \beta = \pi\)) follows from the identity \(\displaystyle \sum_{n=-\infty}^{\infty} e^{-\pi \tau n^2} = \frac{1}{\sqrt{\tau}} \sum_{n=-\infty}^{\infty} e^{-\pi n^2 / \tau}\).

post204.html#p204


Let \(\displaystyle \tau = \frac{\alpha}{\beta}\).

Then \(\displaystyle \sqrt{\alpha} \sum_{n=-\infty}^{\infty} e^{- \pi \frac{\alpha}{\beta} n^{2}} = \sqrt{\beta} \sum_{n=-\infty}^{\infty} e^{- \pi \frac{\beta}{\alpha} n^{2}}\)

\(\displaystyle \implies \sqrt{\alpha} \sum_{n=-\infty}^{\infty} e^{- \pi \alpha \frac{\alpha}{\pi} n^{2}} = \sqrt{\beta} \sum_{n=-\infty}^{\infty} e^{- \pi \beta \frac{\beta}{\pi} n^{2}}\)

\(\displaystyle \implies \sqrt{\alpha} \sum_{n=-\infty}^{\infty} e^{- \alpha^{2} n^{2}} = \sqrt{\beta} \sum_{n=-\infty}^{\infty} e^{- \beta^{2} n^{2}}\)


I am continuing on from RV.

... For simplicity let \(\displaystyle \varphi(q)=\sum_{n=-\infty}^\infty q^{n^2}\).

\(\displaystyle \sqrt{\alpha}\varphi(e^{-\alpha^2})=\sqrt{\beta}\varphi(e^{-\beta^2})\)

If we set \(\displaystyle \alpha^2=5\pi \quad \text{and} \quad \beta^2=\frac{\pi}{5}\), we find that

\(\displaystyle \sqrt{5}\varphi(e^{-5\pi})=\varphi(e^{-\pi/5})\tag{1}\)

Now put \(\displaystyle \alpha^2 = \pi(1+2i) , \beta^2 = \frac{\pi(i-2i)}{5}\) and separate real parts to deduce that

\(\displaystyle \begin{align*}
\sqrt{1+\sqrt{5}}\varphi(e^{-\pi}) &= 2\sqrt{2} \left( \frac{1}{2}+\sum_{n=1}^\infty e^{-n^2 \pi/5}\cos \frac{2\pi n^2}{5}\right)\\
&= 2\sqrt{2}\left( \frac{1}{2}+\sum_{n=1}^\infty e^{-(5n)^2 \pi/5} +\cos \frac{2\pi}{5}\sum_{\begin{matrix}n=1 \\ n \not\equiv 0 \; (\text{mod }5)\end{matrix}}^\infty e^{-n^2 \pi/5}\right) \\
&= \sqrt{2} \left(\left( 1-\cos\frac{2\pi}{5}\right)\varphi(e^{-5\pi})+\cos \frac{2\pi }{5}\varphi(e^{-\pi/5}) \right) \\
&= \frac{5-\sqrt{5}}{\sqrt{2}}\varphi(e^{-5\pi}) \quad \text{by (1)}
\end{align*}\)

This upon simplification should give the required result.

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