# Poisson's summation formula

1. Jun 13, 2009

### O.J.

I am trying to understand the derivation of the Poisson's sum formula. Wikipedia's article is like crosswords to me. I checked mathworld's take on it. It looked simple, but it stated that the equation is derived from a more general result without stating or proving that general result. Here's the link. Can you please tell me how that generalisation is derived? thank u.

http://mathworld.wolfram.com/PoissonSumFormula.html

2. Jun 13, 2009

### O.J.

can you at least tell me under which category in mathematics does this topic lie so that I know what kind of book I'm searching for.?

3. Jun 13, 2009

### Count Iblis

It is very easy to derive the formula using Fourier series.

4. Jun 13, 2009

### Count Iblis

Consider the functions:

$$e_{k}(x) = \exp(2 \pi i k x)$$

For functions defined on the interval from n-1/2 to n + 1/2 with n some arbitrary integer, we define the inner product:

$$\langle f,g\rangle = \int_{n - 1/2}^{n +1/2} f(x)g^{*}(x)dx$$

Then we see that:

$$\langle e_k, e_r\rangle = \delta_{k,r}$$

where $\delta_{k,r}$ is the Kronecker delta, which is zero if the indices are differet and 1 if the indices are equal. This means that the $e_k$ form an orthonormal basis of functions. It can be shown that these functions also form a complete set which then implies that we can expand any function f as:

$$f(x) = \sum_{k=-\infty}^{\infty} \langle f, e_{k}\rangle e_{k}(x)$$

[Note that this is completly analogous to how you can expand any vector in terms of basis vectors. If V is a 3d vector and ex, ey and ez are the unit vectors in the x, y, and z, directions, then V = <V,ex>ex +<V,ey> + <V,ez>ez. The inner products <V,ei> are, of course, the components of V in the ith direction.]

At x = n, we have

$$e_{k}(n)= \exp(2 \pi i k n) = 1$$,

so we have:

$$f(n) = \sum_{k=-\infty}^{\infty} \langle f, e_k\rangle = \sum_{k=-\infty}^{\infty}\int_{n-1/2}^{n+1/2} f(x) \exp(-2 \pi i k x) dx$$

Now, sum both sides over n from minus infinity to infinity:

$$\sum_{n=-\infty}^{\infty}f(n) = \sum_{n=-\infty}^{\infty}\sum_{k=-\infty}^{\infty}\int_{n-1/2}^{n+1/2} f(x) \exp(-2 \pi i k x) dx$$

Interchange the summations over n and k:

$$\sum_{n=-\infty}^{\infty}f(n) = \sum_{k=-\infty}^{\infty}\sum_{n=-\infty}^{\infty}\int_{n-1/2}^{n+1/2} f(x) \exp(-2 \pi i k x) dx$$

Summing over n yields

$$\sum_{n=-\infty}^{\infty}\int_{n-1/2}^{n+1/2}f(x)\exp(-2\pi i k x) dx=\int_{-\infty}^{\infty}f(x)\exp(-2\pi i k x)dx=\hat{f}(k)$$

So, we have:

$$\sum_{n=-\infty}^{\infty}f(n)=\sum_{k=-\infty}^{\infty}\hat{f}(k)$$

Last edited: Jun 14, 2009
5. Jun 14, 2009

### O.J.

Count thank you greatly but I cannot understand through the language of the proof can you use latex? please? :)

6. Jun 14, 2009

### Count Iblis

I've edited the posting.

7. Jun 14, 2009

### O.J.

"For functions defined on the interval from n-1/2 to n + 1/2 with n some arbitrary integer, we define the inner product:

LaTeX Code: \\langle f,g\\rangle = \\int_{n - 1/2}^{n +1/2} f(x)g^{*}(x)dx

Then we see that:

LaTeX Code: \\langle e_k, e_r\\rangle = \\delta_{k,r}

where LaTeX Code: \\delta_{k,r} is the Kronecker delta, which is zero if the indices are differet and 1 if the indices are equal. This means that the LaTeX Code: e_k form an orthonormal basis of functions. It can be shown that these functions also form a complete set which then implies that we can expand any function f as:"

I did not understand something here, what is the nature of g(x)? is it an arbitrary function or not? And how exactly did you deduce tht it is a kronecker's delta? the product of the two exponentials indeed will be 1 if the indices are the same but how will it always be zero if they arent? it could be exp (2pix (k-n)) for example..

8. Jun 16, 2009

### O.J.

I do not understand how <ek , er> results in a delta function. Isnt the integrand going to be e^(2pi i k) * e^(-2pi i r). While this is equal to 1 if k = r, it isnt necessarily 0 if r isnt equal to k. How did u interpret it as delta? please clarify.

9. Jun 16, 2009

### Redbelly98

Staff Emeritus
Equivalently, the integrand is e^[2 pi i (k-r)].

(k-r) is an integer, and the integral is done over a range of 1 (from n-½ to n+½). So the integral is done over an integer number of periods of the function e^[2 pi i (k-r)]. Therefore, the integral is zero.