Is the Poisson Sum Formula Equivalent to the Integral of a Function?

[lokofer]

If we have (Poisson sum formula) in the form:

$$\sum_{n=-\infty}^{\infty}f(n)= \int_{-\infty}^{\infty}dx f(x) \omega (x)$$

with $$\omega (x) = \sum_{n=-\infty}^{\infty}e^{2i \pi nx}$$

Then my question is if we would have that:

$$\sum_{n=-\infty}^{\infty} \frac{ f(n)}{ \omega (n)} = \int_{-\infty}^{\infty} dx f(x)$$ ??

 

[shmoe]

You should stop and ask if:

$$\omega (x) = \sum_{n=-\infty}^{\infty}e^{2i \pi nx}$$

makes any sense at all. You have an infinite sum and the terms aren't tending to zero.

 

[tacman]

The Poisson sum formula is a mathematical tool used to express a periodic function in terms of its Fourier coefficients. It states that the sum of the function over all integers is equal to the integral of the function multiplied by a weight function, which is itself a sum of complex exponentials. This formula is useful in many areas of mathematics, including number theory, signal processing, and quantum mechanics.

To answer the question, yes, the formula is correct. By substituting the weight function \omega(x) into the Poisson sum formula, we can see that the sum over all integers is equal to the integral of the function multiplied by the weight function. Rearranging the formula, we can then express the function as a sum of its Fourier coefficients divided by the weight function. This is equivalent to the integral of the function, as the weight function cancels out in the numerator and denominator.

In summary, the Poisson sum formula is a powerful tool that allows us to express a periodic function in terms of its Fourier coefficients. By manipulating the formula, we can also express the function as a sum of its Fourier coefficients divided by the weight function, which is equal to the integral of the function. This formula has many applications and is an important concept in mathematics.