How is the generalization of Poisson's summation formula derived?

In summary, the Poisson's sum formula can be derived using Fourier series, where the functions e_k form an orthonormal basis of functions. By expanding any function f in terms of these basis functions and manipulating the sums and integrals, we can arrive at the formula.
  • #1
O.J.
199
0
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.
link:

http://mathworld.wolfram.com/PoissonSumFormula.html
 
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  • #2
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
It is very easy to derive the formula using Fourier series.
 
  • #4
Consider the functions:

[tex]e_{k}(x) = \exp(2 \pi i k x)[/tex]

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

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

Then we see that:

[tex]\langle e_k, e_r\rangle = \delta_{k,r}[/tex]

where [itex]\delta_{k,r}[/itex] is the Kronecker delta, which is zero if the indices are differet and 1 if the indices are equal. This means that the [itex]e_k[/itex] 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:


[tex]f(x) = \sum_{k=-\infty}^{\infty} \langle f, e_{k}\rangle e_{k}(x) [/tex]

[Note that this is completely 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

[tex]e_{k}(n)= \exp(2 \pi i k n) = 1[/tex],

so we have:


[tex]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[/tex]


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

[tex]\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[/tex]


Interchange the summations over n and k:

[tex]\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[/tex]


Summing over n yields

[tex]\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)[/tex]

So, we have:

[tex]\sum_{n=-\infty}^{\infty}f(n)=\sum_{k=-\infty}^{\infty}\hat{f}(k)[/tex]
 
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  • #5
Count thank you greatly but I cannot understand through the language of the proof can you use latex? please? :)
 
  • #6
O.J. said:
Count thank you greatly but I cannot understand through the language of the proof can you use latex? please? :)


I've edited the posting.
 
  • #7
"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
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 isn't necessarily 0 if r isn't equal to k. How did u interpret it as delta? please clarify.
 
  • #9
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.
 

What is Poisson's summation formula?

Poisson's summation formula is a mathematical formula that relates the discrete Fourier transform (DFT) of a function to the values of the function at evenly spaced points. It is used to convert a sum over integers to a sum over the real numbers, making it easier to evaluate certain types of integrals and infinite series.

What is the significance of Poisson's summation formula?

Poisson's summation formula has many applications in mathematics and physics. It is often used to solve problems in number theory, probability, signal processing, and quantum mechanics. It can also be used to evaluate integrals that would be difficult or impossible to solve using other methods.

How is Poisson's summation formula derived?

Poisson's summation formula can be derived using the properties of the Fourier transform and the Poisson summation formula for continuous functions. By taking the limit as the spacing between points approaches zero, the discrete version of the formula can be obtained.

What are the limitations of Poisson's summation formula?

Poisson's summation formula is only applicable to functions that are defined on the real numbers and have a Fourier transform. It also assumes that the function and its Fourier transform decay rapidly enough at infinity for the formula to converge. In addition, it may not always provide an exact solution, but rather an approximation.

How is Poisson's summation formula related to the Riemann zeta function?

Poisson's summation formula can be used to evaluate the Riemann zeta function at integer values. This is because the Fourier transform of the Riemann zeta function is related to the Bernoulli numbers, which can be expressed in terms of the zeta function. This connection has been used to prove important results in number theory, such as the Basel problem and the prime number theorem.

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