Can we approximate $\prod_{n=1}^\infty (1-\frac{(2n+1)x^2}{n^2\pi^2})$?

Question

It seems like from the graph $\prod_{n=1}^\infty (1-\frac{(2n+1)x^2}{n^2\pi^2})$ is somehow alike to the graph $e^{-x^2}$, the main problem is that the limitations of the software makes it hard to graph for large numbers. Is it possible to do so or it diverges?

Answer 1

The partial product $$P_m=\prod _{n=1}^{m } \left(1-\frac{(2 n+1) }{n^2\pi ^2 }x^2\right)$$ effectively looks like a gaussian (tested for $m=10^5$). Using Pochhammer symbols $$P_m=\frac{\left(\frac{\pi^2-x^2-\sqrt{x^2 \left(x^2+\pi ^2\right)}}{\pi ^2}\right){}_m \left(\frac{\pi^2-x^2+\sqrt{x^2 \left(x^2+\pi ^2\right)} }{\pi ^2}\right){}_m}{(m!)^2}$$

Taking logarithms and expanding as series for large values of $m$, we have $$\log(P_m)\sim-\frac{2 x^2 \log (m)}{\pi ^2}\implies P_m\sim \exp\Bigg[-\frac{2 x^2 }{\pi ^2} \log (m)\Bigg]=m^{-\frac{2 }{\pi ^2}x^2}$$

Try it and let us know.

Edit

We can even do much better using the next term of the expansion which gives $$\log(P_m)\sim-\frac{2 x^2 \log (m)}{\pi ^2}-$$ $$\log \left(\Gamma \left(\frac{\pi^2-x^2+\sqrt{x^2 \left(x^2+\pi ^2\right)}}{\pi ^2}\right) \Gamma \left(\frac{\pi^2-x^2-\sqrt{x^2 \left(x^2+\pi ^2\right)}}{\pi ^2}\right)\right)$$ Expanding for small values of $x$, this gives $$ P_m\sim \exp\Bigg[-\frac{\left(12 \log (m)+\pi ^2+12 \gamma \right)}{6 \pi ^2}\,x^2\Bigg]$$ which is much better.

We could even go further introducing terms in $x^4$ but they will include some polygamma terms.

Answer 2

Well I am not able to proceed from here but you try, Let $x=1$, $$A=\prod_{n\ge1}(1-\frac{2n+1}{(\pi n)^{2}})$$ $\ln$ on both sides, $$\ln A=\sum_{n\ge1}\ln\left(1-\frac{2n+1}{(\pi n)^{2}}\right)$$ Note that $2n+1=o(n^{2})$, $$\ln A=\sum_{n\ge1}\sum_{k\ge 1}\frac{1}{k}\frac{(2n+1)^{k}}{(n\pi)^{2}}$$ Interchanging the summation signs, $$\ln A=\sum_{k\ge 1}\frac{\pi^{-2k}}{k}\sum_{n\ge 1}\frac{(2n+1)^{k}}{n^{2k}}$$ $$\ln A=\sum_{k\ge 1}\frac{\pi^{-2k}}{k}\sum_{n\ge 1}\sum_{u\le k}\binom{k}{u}n^{-2u}2^{k-u}n^{u-k}$$ $$\ln A=\sum_{k\ge 1}\frac{\pi^{-2k}}{k}\sum_{u=0}^{k}\binom{k}{u}2^{k-u}\zeta(u+k)$$ I am unable to determine in this series diverges or converges but I am sure that it simplifies the original product to some extent.

Answer 3

Calling $\Pi$ the partial product up to $N$, for $0<|x|<\pi/\sqrt{3}$ we have

$$ \begin{array}{ll} -\ln\Pi & =\displaystyle \sum_{n=1}^N \ln\Bigg(\frac{1}{1-\frac{(2n+1)x^2}{n^2\pi^2}}\Bigg) \\[5pt] & > \displaystyle \sum_{n=1}^N \ln\Big(1+\frac{(2n+1)x^2}{n^2\pi^2}\Big) \\[5pt] & >\displaystyle \sum_{n=1}^N \left[\frac{(2n+1)x^2}{n^2\pi^2}-\frac{1}{2}\left(\frac{(2n+1)x^2}{n^2\pi^2}\right)^2 \right] \\[5pt] & >\displaystyle \left[\frac{2x^2}{\pi^2}\sum_{n=1}^N \frac{1}{n}\right] - \mathcal{O}(1) ~~~\to\infty \end{array} $$

Thus, $\ln \Pi\to-\infty$ so $\Pi\to0^+$ as $N\to\infty$.