# Thread: Summation of e^(-k n^2)

1. ## Summation of e^(-k n^2)

Hi All,

This is my first thread so please forgive any mistakes regarding forum rules.

I have been trying to do some calculations for my Physics project. And I get stuck at following summations,

1. $\sum_{n=-\infty}^{\infty}\exp{\left(-\alpha n^2\right)}$

2. $\sum_{n=-\infty}^{\infty}n^2\exp{\left(-\alpha n^2\right)}$

3. $\sum_{n=-\infty}^{\infty}(-1)^n\exp{\left(-\alpha n^2\right)}$

We have already tried to approximate the summations as an integration of Gaussian. Apparently this is not good enough approximation for our purpose. Any help or suggestions will be appreciated. The first one seems to be key for other two.

Regards,
Shantanu

2. ## Re: Summation of e^(-k n^2)

Hi All,

This is my first thread so please forgive any mistakes regarding forum rules.

I have been trying to do some calculations for my Physics project. And I get stuck at following summations,

1. $\sum_{n=-\infty}^{\infty}\exp{\left(-\alpha n^2\right)}$

2. $\sum_{n=-\infty}^{\infty}n^2\exp{\left(-\alpha n^2\right)}$

3. $\sum_{n=-\infty}^{\infty}(-1)^n\exp{\left(-\alpha n^2\right)}$

We have already tried to approximate the summations as an integration of Gaussian. Apparently this is not good enough approximation for our purpose. Any help or suggestions will be appreciated. The first one seems to be key for other two.

Regards,
Shantanu
1. \displaystyle \begin{align*} \sum_{n = -\infty}^{\infty}e^{-\alpha \, n^2} &= \sum_{n = -\infty}^{-1} \left( e^{-\alpha \, n^2} \right) + 1 + \sum_{n = 1}^{\infty} \left( e^{-\alpha \, n^2} \right) \end{align*}

This entire sum can't possibly converge because if \displaystyle \begin{align*} \alpha > 0 \end{align*}, the terms in the first smaller sum don't go to 0, and if \displaystyle \begin{align*} \alpha < 0 \end{align*}, the terms in the second smaller sum don't go to 0.

2. Doesn't converge for the same reason.

3. ## Re: Summation of e^(-k n^2)

Hi Prove It,

Thank you for your reply. I don't quite understand what you mean to say. Did you perhaps miss the fact that it is [TEX]n^2[\TEX] and not n in the exponential? Your first and third terms will give exactly the same value!

-Shantanu

4. ## Re: Summation of e^(-k n^2)

Hi Prove It,

Thank you for your reply. I don't quite understand what you mean to say. Did you perhaps miss the fact that it is [TEX]n^2[\TEX] and not n in the exponential? Your first and third terms will give exactly the same value!

-Shantanu
Actually you are right, I did miss the \displaystyle \begin{align*} n^2 \end{align*} in the power. Are we assuming that \displaystyle \begin{align*} \alpha > 0 \end{align*}?

5. ## Re: Summation of e^(-k n^2)

Originally Posted by Prove It
Actually you are right, I did miss the \displaystyle \begin{align*} n^2 \end{align*} in the power. Are we assuming that \displaystyle \begin{align*} \alpha > 0 \end{align*}?
Yes. $\alpha>0$ and it is real!

-Shantanu

6. ## Re: Summation of e^(-k n^2)

Hi Shantanu,

These sums involve the Jacobi theta functions.
Jacobi Theta Functions -- from Wolfram MathWorld

7. ## Re: Summation of e^(-k n^2)

Hi JJacquelin,

That solves my problem. Thanks a lot for your help!

-Shantanu