How do I prove that
$$\left(\frac{1}{10^5} \sum_{n=-\infty}^{\infty} e^{\frac{-n^2}{10^{10}}}\right)^2 \approx \pi?$$
I am thinking of start with gaussian function and using the poison summation formula
Does that will work?
fourier analysispoisson-summation-formula
How do I prove that
$$\left(\frac{1}{10^5} \sum_{n=-\infty}^{\infty} e^{\frac{-n^2}{10^{10}}}\right)^2 \approx \pi?$$
I am thinking of start with gaussian function and using the poison summation formula
Does that will work?
Best Answer
We do have
\begin{align*} \int_{-\infty}^{\infty} e^{-y^2} dy &= \sqrt{\pi} \end{align*}
Now, substitute $y = \alpha x$, $dy = \alpha dx$:
\begin{align*} \alpha \int_{-\infty}^{\infty} e^{-\alpha^2 x^2} dx &= \sqrt{\pi} \\ \left( \alpha \int_{-\infty}^{\infty} e^{-\alpha^2 x^2} dx \right)^2 &= \pi \end{align*}
This suggests that we may estimate the given sums by interpreting them as a Riemann sum belonging to the integral we've just computed. However, we have to be careful.
$\sum_{n=1}^{\infty} e^{-\alpha^2 n^2} < \int_{0}^{\infty} e^{-\alpha^2 x^2} dx$
$\sum_{n=-\infty}^{\infty} e^{-\alpha^2 n^2} = 1 + 2 \sum_{n=1}^{\infty} e^{-\alpha^2 n^2} < 1 + \int_{-\infty}^{\infty} e^{-\alpha^2 x^2} \, dx$
And for the other bound use
$\int_0^{\infty} e^{-\alpha^2 x^2} \, dx < \sum_{n=0}^{\infty} e^{-\alpha^2 n^2}$
To find
$\sum_{n=-\infty}^{\infty} e^{-\alpha^2 n^2} = 2 \sum_{n=0}^{\infty} e^{-\alpha^2 n^2} - 1 > \int_{-\infty}^{\infty} e^{-\alpha^2 x^2} dx$
Collect both halves and multiply with ${\alpha}$
$\alpha \left( \int_{-\infty}^{\infty} e^{-\alpha^2 x^2} dx - 1 \right) < \alpha \sum_{n=-\infty}^{\infty} e^{-\alpha^2 n^2}$
$< \alpha \left( \int_{-\infty}^{\infty} e^{-\alpha^2 x^2} dx + 1 \right)$
$\sqrt{\pi} - \alpha < \alpha \sum_{n=-\infty}^{\infty} e^{-\alpha^2 n^2} < \sqrt{\pi} + \alpha$
If $\alpha \to 0$, both sides tend to $\sqrt{\pi}^2$, and the result follows. In fact, we may use this to find
\begin{align*} &\left| \alpha \sum_{n=-\infty}^{\infty} e^{-\alpha^2 n^2} - \sqrt{\pi} \right| < \alpha \\ &\left| \left( \alpha \int_{-\infty}^{\infty} e^{-\alpha^2 n^2} dx \right)^2 - \pi \right| < \alpha(\alpha + 2 \sqrt{\pi}) \approx 2 \alpha \sqrt{\pi}. \end{align*}
Where the approximation is valid for small $α$ similar to the given value $α=1/10^5$