5
$\begingroup$

How to prove the following transformation formula: $$ \theta(x)=\frac{1}{\sqrt{x}} \theta\left(\frac{1}{x}\right), $$ where $\theta$ is the Jacobi theta function $\theta(x)=\sum_{n\in \mathbb{Z}} e^{-\pi n^2 x}$?

$\endgroup$
5
$\begingroup$

One can use the Poisson summation formula: $$ \sum_{n\in\mathbb{Z}}f(n)=\sum_{k\in\mathbb{Z}}\hat{f}(k),$$ where $\hat{f}(\nu)$ denotes the Fourier transform of $f(t)$, $$ \hat{f}(\nu)=\int_{-\infty}^{\infty}f(t)e^{-2\pi i \nu t}dt.$$ Namely, setting $f(t)=e^{-\pi x t^2}$ in the above, we obtain $$\theta(x)=\sum_{n\in\mathbb{Z}}f(n)=\sum_{k\in\mathbb{Z}}\underbrace{\int_{-\infty}^{\infty}e^{-\pi x t^2-2\pi i k t}dt}_{\hat{f}(k)}= \sum_{k\in\mathbb{Z}}\frac{e^{-\pi k^2/x}}{\sqrt{x}}=\frac{\theta\left(x^{-1}\right)}{\sqrt{x}}.$$

$\endgroup$
  • $\begingroup$ How did you solve $\hat{f}(k) = \frac{e^{-\pi k^2/x}}{\sqrt{x}}$? I understand that if $g(t) = e^{-\pi t^2},$ then $\hat{g}(v) = g(v).$ The text I am referencing tells me to do a substitution, but I am having trouble. $\endgroup$ – Nick Jan 10 '15 at 6:08
  • $\begingroup$ @nick Try the substitution $t=\alpha t'+\beta$ and find out for which $\alpha$ and $\beta$ the function to integrate reduces to $e^{-\pi t'^2}$. $\endgroup$ – Start wearing purple Jan 10 '15 at 12:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.