Fourier transform on a distribution

[jerry109]

Homework Statement

Determine the Fourier transform on the tempered distribution:

[tex]
\langle f, \varphi \rangle
[/tex]

Where [tex] f[/tex] can be given by they taylor series representation:

[tex]
f = i\sum_{n=0}^{\infty} \frac {x^{3n+2}}{(2n)!}
[/tex]

The Attempt at a Solution

Fourier transform on tempered distribution is:

[tex]
F\langle f, \varphi \rangle = \langle F f, \varphi \rangle = \langle f, F \varphi \rangle

= i\int \sum_{n=0}^{\infty} \frac {x^{3n+2}}{(2n)!} \hat{\varphi} dx

= i\sum_{n=0}^{\infty}\int \frac {x^{3n+2}}{(2n)!} \hat{\varphi} dx
[/tex]

I'm stuck as how to resolve the infinite sum involving the Fourier transform of the test function [tex]\varphi[/tex].

Perhaps:

[tex]
i\sum_{n=0}^{\infty}\int \frac {x^{3n+2}}{(2n)!} \hat{\varphi} dx = i\sum_{n=0}^{\infty}\int \frac {x^{3n+2}}{(2n)!} \hat{\delta}(x-x_0) dx
= i\sum_{n=0}^{\infty}\int \frac {x^{3n+2}}{(2n)!} e^{2{\pi}ikx_0}dx= ie^{2{\pi}ik{x_0}}\sum_{n=0}^{\infty}\int \frac {x^{3n+2}}{(2n)!} dx

= ie^{2{\pi}ik{x_0}}\sum_{n=0}^{\infty}\frac {x^{3n+3}}{(2n)! * (3n+3)!}

[/tex]Any help would be fantastically appreciated,
Jerry109

 

[mighty2000]

Dear Jerry109,

Thank you for sharing your attempt at solving this problem. I can see that you are on the right track, but there are a few things that need to be clarified.

Firstly, when considering the Fourier transform on tempered distributions, it is important to remember that it is defined as an integral, not a sum. So instead of using a summation notation, we should be using an integral notation. Also, the tempered distribution f is not given by a Taylor series representation, but rather by an integral representation. So we should be using an integral representation for f as well.

With that being said, here is how I would approach this problem:

We know that the Fourier transform of a tempered distribution f is given by:

F\langle f, \varphi \rangle = \langle F f, \varphi \rangle = \langle f, F \varphi \rangle

= \int f(x) \hat{\varphi}(x) dx

= \int i\sum_{n=0}^{\infty} \frac {x^{3n+2}}{(2n)!} \hat{\varphi}(x) dxNow, we need to use the definition of the Fourier transform for a function g(x) given by:

\hat{g}(k) = \int_{-\infty}^{\infty} g(x) e^{-2\pi i kx} dx

So, substituting g(x) = i\sum_{n=0}^{\infty} \frac {x^{3n+2}}{(2n)!} \hat{\varphi}(x) in the above equation, we get:

F\langle f, \varphi \rangle = \int i\sum_{n=0}^{\infty} \frac {x^{3n+2}}{(2n)!} \hat{\varphi}(x) dx

= \int i\sum_{n=0}^{\infty} \frac {x^{3n+2}}{(2n)!} \int_{-\infty}^{\infty} \hat{\varphi}(y) e^{-2\pi i yx} dy dx

= i\sum_{n=0}^{\infty} \int_{-\infty}^{\infty} \hat{\varphi}(y) \left(\int_{