Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Can you help prove the functional equation: $$\sum_{n=-\infty}^\infty e^{-\pi n^2x}=x^{-1/2}\sum_{n=-\infty}^\infty e^{-\pi n^2/x}.$$

Specifically, I am looking for a solution using complex analysis, but I am interested in any solutions.

Thanks!

share|improve this question
add comment

5 Answers

up vote 8 down vote accepted
+50

Define $\theta(t) = \sum_{n \in \mathbb{Z}} e^{-\pi i n^{2} t}$. The identity that you quote is the Jacobi theta functional identity \begin{align} \theta(t) = t^{-1/2} \theta(t^{-1}). \end{align} This identity can be used to prove the Riemann zeta functional identity. To prove it, first recall that the Fourier transform of an integrable function $f \colon \mathbb{R} \to \mathbb{C}$ is simply \begin{align} \tilde{f}(s) = \int_{\mathbb{R}} f(x) e^{2 \pi i x s} dx. \end{align} and (presupposing that $f$ is also uniformly continuous) the Poisson summation formula is the identity \begin{align} \sum_{n \in \mathbb{Z}} \tilde{f}(n) = \sum_{n \in \mathbb{Z}} f(n). \end{align} Now, recall the integral \begin{align} e^{-\pi s^{2}} = \int_{\mathbb{R}} e^{-\pi x^{2}} e^{2 \pi i x s} dx \end{align} Observe that \begin{align} \sum_{n \in \mathbb{Z}} e^{- \pi n^{2} s^{2}} = \sum_{n \in \mathbb{Z}} \int_{\mathbb{R}} e^{-\pi x^{2}} e^{2 \pi i x (ns)} dx = \sum_{n \in \mathbb{Z}} s^{-1} \int_{\mathbb{R}} e^{-\pi (x^{\prime}/s)^{2}} e^{2 \pi i n x^{\prime}} dx^{\prime}, \end{align} where we have changed variables, $x^{\prime} = x s$. Now use the fourier transform, \begin{align} \sum_{n \in \mathbb{Z}} s^{-1} \int_{\mathbb{R}} e^{-\pi (x^{\prime}/s)^{2}} e^{2 \pi i n x^{\prime}} dx^{\prime} = s^{-1} \sum_{n \in \mathbb{Z}} e^{- \pi (n / s)^{2}}. \end{align} Take $s = \sqrt{t}$ and conclude the result.

share|improve this answer
1  
Your second sentence has two bad mistakes: the paper in which Riemann used this identity was not his thesis (it was several years after that) and the Jacobi theta-function identity was not "first quoted" in any sense by Riemann. –  KCd Apr 7 '11 at 7:42
    
You want $\tilde{f}(s)$, not $\tilde{f}(k)$ in your definition of the transform. –  Thomas Andrews Apr 9 '11 at 19:17
    
Corrected. Thanks. –  user02138 Apr 9 '11 at 21:14
    
I accept. Although I do still want a more complex analysis like approach. –  Eric Naslund Jun 7 '11 at 16:36
    
@Eric: Thanks. And if I find a more complex analytic proof, I'll post it. –  user02138 Jun 7 '11 at 17:24
add comment

Hint: Apply the Poisson Summation Formula to a suitable function.

share|improve this answer
    
It's also an identity of Jacobi Theta functions. –  Robert Israel Apr 6 '11 at 15:33
4  
@Robert: ... which you prove using Poisson summation. Yes? –  Qiaochu Yuan Apr 6 '11 at 22:08
add comment

The following is not a proof but it gives some physical intuition why such a formula is true.

If at time $t=0$ a unit of heat is concentrated at $x=0$ and dissipates according to the heat equation $u_t=u_{xx}$ along the $x$-axis then the temperature $x\mapsto u(x,t)$ is a Gaussian becoming flatter and flatter as $t$ increases: $$u(x,t)={1\over\sqrt{4\pi t}}e^{-x^2/(4t)}\qquad(t>0)\ .$$ Now if at time $t=0$ we have such a unit of heat at each integer point $k$ then the resulting temperature will be $$u(x,t)={1\over\sqrt{4\pi t}}\sum_{k\in{\mathbb Z}}e^{-(x-k)^2/(4t)}\ .$$ In particular the temperature at $x=0$ will be $$U(t)={1\over\sqrt{4\pi t}}\sum_{k\in{\mathbb Z}}e^{-k^2/(4t)}\qquad(t>0)\ .\qquad {\rm (a)}$$ On the other hand the process considered here is periodic in $x$ with period $1$. So the temperature $u(x,t)$ must have a description of the form $$u(x,t)=\sum_{k\in{\mathbb Z}}a_k(t)e^{2\pi i k x}$$ for certain functions $a_k(t)$. Plugging this into the heat equation gives $a_k(t)=c_k \exp(-4\pi^2 k^2 t)$ for constants $c_k$, so that we now have $$u(x,t)=\sum_{k\in{\mathbb Z}}c_k \exp(-4\pi^2 k^2 t)e^{2\pi i k x}\ .$$ The $c_k$ have to be determined by the initial condition which is a delta-function at $x=0$. Here we have to cheat a little: We replace the delta-function by a rectangle of width $2\epsilon$ and area $1$. The computation gives $c_0=1$ and $$c_k={\sin(2\pi k\epsilon)\over 2\pi k\epsilon}\qquad(k\ne 0)$$ which tends to $1$ with $\epsilon\to0$. This means that "in the limit" we have $$u(x,t)=\sum_{k\in{\mathbb Z}} \exp(-4\pi^2 k^2 t)e^{2\pi i k x}\qquad(t>0)\ .$$ Putting $x=0$ here gives $$U(t)=\sum_{k\in{\mathbb Z}} \exp(-4\pi^2 k^2 t)\qquad(t>0)\ .\qquad{\rm (b)}$$ If you believe that (a) and (b) are the same thing then you have the stated formula with $4\pi t$ instead of $x$.

share|improve this answer
add comment

Biane, Pitman and Yor describe one classical analytic approach that admits a probabilistic interpretation: http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.17.7395

(Not sure if the link is behind a paywall. Just google the title if so).

share|improve this answer
add comment

See the book Elliptic Functions by Chandrasekharan, K. at page 73 the section entitled

$\S \;8.\; The\;transformation\;formula\;connecting\;\theta_{3}(v,\tau)\;and\;\theta_{3}\left(v,-\frac{1}{\tau}\right)$

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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