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

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

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

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.