What Is the Periodicity of a Function Defined by Its Taylor Series Expansion?

[jerry109]

Homework Statement

Define the periodicity of the function with the following Taylor Series expansion:

$$f=i\sum_{n=0}^{\infty} (-1)^{n} \frac {x^{3n+2}}{(2n)!}$$

Homework Equations

To find the periodicity, we take the Fourier transform of f, treated as a tempered distribution:

$$F\langle f, \varphi \rangle = \langle f, F\varphi \rangle = \langle f, \hat{\varphi}\rangle$$

The Attempt at a Solution

Following through with the integral:

$$= i\int_{-\infty}^{\infty} \sum_{n=0}^{\infty}(-1)^{n} \frac {x^{3n+2}}{(2n)!} \hat{\varphi} dx$$

$$=i\sum_{n=0}^{\infty}\int_{-\infty}^{\infty}(-1)^{n} \frac {x^{3n+2}}{(2n)!} \hat{\varphi} dx$$

I'm not sure how to finish the problem beyond this point. I was thinking of maybe taking the inverse transform and changing it into a convolution, but that didn't appear to work. I'm also unsure of how to simplify this further, or if you can insert some test function of compact support to resolve the integral.

Any help as how to finish the problem would be greatly appreciated.

Jerry109

 

[Nino MMP]

's Attempt at a SolutionWe can use the fact that the Fourier transform of $f$ is given by $$F\langle f, \varphi \rangle = \langle f, F\varphi \rangle = \langle f, \hat{\varphi}\rangle$$ where $\hat{\varphi}$ is the Fourier transform of $\varphi$. Letting $\varphi(x) = e^{i k x}$, we get $$F\langle f, e^{i k x} \rangle = \langle f, F(e^{i k x}) \rangle = \langle f, \hat{e^{i k x}}\rangle$$ where $$\hat{e^{i k x}} = \int_{-\infty}^{\infty} e^{i k x} \hat{\varphi}(x) dx$$ Substituting in our expression for $f$ and evaluating the integral, we get $$F\langle f, e^{i k x} \rangle = \langle f, F(e^{i k x}) \rangle = i \sum_{n=0}^{\infty} (-1)^{n} \frac {(ik)^{3n+2}}{(2n)!} $$ The periodicity of the function is then given by the period of the function $F(e^{i k x})$, which can be found by taking the inverse Fourier transform to get $$F^{-1}(F(e^{i k x})) = \frac{1}{2 \pi} \int_{-\infty}^{\infty} F(e^{i k x}) dk = \frac{1}{2 \pi} \int_{-\infty}^{\infty} i \sum_{n=0}^{\infty} (-1)^{n} \frac {(ik)^{3n+2}}{(2n)!} dk$$ We can now use the substitution $u = 3n + 2$ to rewrite the integral as $$\frac{1}{2 \pi} \int_{2}