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

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.
