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Series Contest (SC3)

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Post Tue Oct 15, 2013 3:24 pm
Shobhit Site Admin
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Location: Jaipur, India

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

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

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


Thanks for pointing out the typo. I will fix it right now.

Post Wed Oct 16, 2013 2:38 am
Shobhit Site Admin
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Posts: 852
Location: Jaipur, India

Problem 7

Prove that

\(\displaystyle \sum_{n=-\infty}^{\infty} \frac{\Gamma(a+n)\Gamma(b+n)}{\Gamma(c+n)\Gamma(d+n)}=\frac{\pi^2}{\sin(\pi a)\sin(\pi b)}\frac{\Gamma(c+d-a-b-1)}{\Gamma(c-a)\Gamma(c-b)\Gamma(d-a)\Gamma(d-b)}\)

where $\text{Re}(c+d-a-b)>1$.

This is known as 'Dougall's Bilateral Sum'.

Post Fri Oct 25, 2013 7:31 pm
sos440 User avatar
Integration Guru
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Location: California, US

Shobhit wrote:
Problem 7

Prove that

\(\displaystyle \sum_{n=-\infty}^{\infty} \frac{\Gamma(a+n)\Gamma(b+n)}{\Gamma(c+n)\Gamma(d+n)}=\frac{\pi^2}{\sin(\pi a)\sin(\pi b)}\frac{\Gamma(c+d-a-b-1)}{\Gamma(c-a)\Gamma(c-b)\Gamma(d-a)\Gamma(d-b)}\)

This is known as 'Dougall's Bilateral Sum'.

It seems somehow related to the Ramanujan's beta integral. Interesting...
Every integral harbors a story as fascinating as a comic book!

Post Sat Oct 26, 2013 5:30 pm
Shobhit Site Admin
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Posts: 852
Location: Jaipur, India

sos440 wrote:
It seems somehow related to the Ramanujan's beta integral. Interesting...


Yes, that is what came to my mind when I saw it. :)

The idea is to integrate \(\displaystyle f(z) = \frac{\Gamma(a+z)\Gamma(b+z)}{\Gamma(c+z)\Gamma(d+z)}\pi \cot(\pi z)\) around a large circle centered at the origin and then use the Residue Theorem.

Have a look at page no 128 of

http://journals.cambridge.org/download.php?file=%2FPEM%2FPEM25%2FS0013091500033642a.pdf&code=e68e80076d6d8b4e968140bc4db80943

Note - Here, \(\Pi(z)\) has been used to denote gamma function instead of \(\Gamma(z)\).

Post Mon Nov 04, 2013 11:12 pm

Posts: 138
Location: North Londinium, UK
Shobhit wrote:
Problem 7

Prove that

\(\displaystyle \sum_{n=-\infty}^{\infty} \frac{\Gamma(a+n)\Gamma(b+n)}{\Gamma(c+n)\Gamma(d+n)}=\frac{\pi^2}{\sin(\pi a)\sin(\pi b)}\frac{\Gamma(c+d-a-b-1)}{\Gamma(c-a)\Gamma(c-b)\Gamma(d-a)\Gamma(d-b)}\)

This is known as 'Dougall's Bilateral Sum'.



I've a proof of this in one of my books, but haven't tried it myself... Yet. It's worth noting, however, that the case \(\displaystyle d=1\) is equivalent to the hypergeometric function, since \(\displaystyle 1/\Gamma(n)=0\) for negative integers. This explains why the term \(\displaystyle 1/n!\) is introduced into hypergeometric series.

Post Fri Nov 08, 2013 1:04 pm
Shobhit Site Admin
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Posts: 852
Location: Jaipur, India

Problem 8

Prove that

\(\displaystyle \frac{4K^2}{\pi^2}=\sum_{j=0}^\infty \frac{\left(\frac{1}{2} \right)_j^3}{(j!)^3}(2kk')^{2j}={\;}_3F_2 \left( \frac{1}{2},\frac{1}{2},\frac{1}{2};1,1;(2kk')^2\right)\)

All symbols have their standard meaning in the theory of Elliptic Functions.

[BONUS]

\(\displaystyle \frac{4}{\pi}=\sum_{n=0}^\infty \frac{\left(\frac{1}{2} \right)_n^3}{(n!)^3}\frac{6n+1}{4^n}\)

[NOTE] \(\displaystyle (a)_n = \frac{\Gamma(a+n)}{\Gamma(a)}\)

Post Wed Dec 11, 2013 9:29 am

Posts: 38
Location: India, West Bengal
Shobhit wrote:
Problem 8\(\displaystyle \frac{4K^2}{\pi^2}=\sum_{j=0}^\infty \frac{\left(\frac{1}{2} \right)_j^3}{(j!)^3}(2kk')^{2j}={\;}_3F_2 \left( \frac{1}{2},\frac{1}{2},\frac{1}{2};1,1;(2kk')^2\right)\)


I had my eye on that problem for some time.
\(\displaystyle K(k) = \frac{\pi}{2} {\;}_2F_1 \left (\frac{1}{2}, \frac{1}{2}; 1; k^2 \right )\)

Using Kummer's transformation, we have

\(\displaystyle K(k) = \frac{\pi}{2} {\;}_2F_1 \left ( \frac{1}{4}, \frac{1}{4}; 1; 4k^2k'^2 \right )\)

By the Clausen Transform,

\(\displaystyle \left [ {\;}_2F_1 \left ( \frac{1}{4}, \frac{1}{4}; 1; 4k^2k'^2 \right ) \right ]^2 = {\;}_3F_2 \left( \frac{1}{2},\frac{1}{2},\frac{1}{2};1,1;4k^2k'^2\right)\)

Hence, the desired result follows.

Post Fri Dec 13, 2013 6:42 pm

Posts: 38
Location: India, West Bengal
Okay, I pass it to anyone who wants to post the next problem.

Post Thu Dec 19, 2013 2:01 pm
Shobhit Site Admin
Site Admin

Posts: 852
Location: Jaipur, India

Here is another problem for anybody interested:

Problem 9

\(\displaystyle \sum_{n=0}^\infty \frac{1}{F_{2n+1}+1}=\frac{1}{2}\sqrt5\)

where \(F_n\) is a Fibonacci Number.

Post Sat Dec 21, 2013 5:20 pm

Posts: 38
Location: India, West Bengal
Am I correct in thinking that this has a whole lot to do with thetas?

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