I've been experimenting with a lot of ideas to resolve:
Decomposing functions to Taylor-Fourier series
That is given a complex function $f: \mathbb{C} \rightarrow \mathbb{C}$ expressed as
$$f(x)= \begin{pmatrix} \vdots \\ a_{0,2}e^{4i\pi x} \ + \\ a_{0,1}e^{2i\pi x} \ + \\ a_{0,0} \ + \\ a_{0,-1}e^{-2i\pi x} \ + \\ a_{0,-2}e^{-4i\pi x} \ + \\ \vdots \end{pmatrix} + \begin{pmatrix} \vdots \\ a_{1,2}e^{4i\pi x} \ + \\ a_{1,1}e^{2i\pi x} \ + \\ a_{1,0} \ + \\ a_{1,-1}e^{-2i\pi x} \ + \\ a_{1,-2}e^{-4i\pi x} \ + \\ \vdots \end{pmatrix} x +\frac{1}{2!} \begin{pmatrix} \vdots \\ a_{1,2}e^{4i\pi x} \ + \\ a_{1,1}e^{2i\pi x} \ + \\ a_{1,0} \ + \\ a_{1,-1}e^{-2i\pi x} \ + \\ a_{1,-2}e^{-4i\pi x} \ + \\ \vdots \end{pmatrix} x(x-1) + … $$
How do we determine the coefficients $a_{i,j}$?
So one sort of outlandish idea i had was "why not fourier transform the thing and see what happens?"
So I considered a concrete function
$$ f(x) = x \cos (2\pi x) $$
To explore what would happen.
Now the convolution theorem tell us that $\mathcal{F}[f\cdot g] = \mathcal{F}[f]*\mathcal{F}[g]$
So we conclude that: $$\mathcal{F}[x \cos (2\pi x)] = \int_{-\infty}^{\infty} \frac{\delta(\tau )}{\tau } \left( \delta(s – 1 – \tau) + \delta(s + 1 – \tau )\right) d\tau $$
Where $\delta$ is the "dirac delta" distribution function.
Is there anyway this can be evaluated? Also how does one reason about weird functions like these $\delta$ functions.
Best Answer
The Fourier transform of $2i\pi x$ is $\delta'$, the Fourier transform of $2i\pi x e^{2i \pi a x}$ is $(\delta' \ast \delta(-.a))(\xi) = \delta'(\xi-a)$.
If the $f_n(x) = \sum_k c_{n,k}e^{2i \pi kx}$ are $1$-periodic distributions and $$f(x)=\sum_{n=0}^\infty f_n(x)x^n$$ converges in the sense of distributions then its Fourier transform is the infinite order functional $$\hat{f}(\xi)=\sum_{n=0}^\infty \sum_k c_{n,k} (2i\pi)^{-n}\delta^{(n)}(\xi-k) $$
which is well-defined when applied to Fourier transforms of functions in $C^\infty_c$ which are entire.
If $f$ converges in the sense of tempered distributions then so does $\hat{f}$, so it has locally finite order, and it will have another expression not involving all the derivatives of $\delta(\xi-k)$. Looking at the regularized $f(x)e^{-\pi x^2/b^2}$ may give that expression as
$$\hat{f}(\xi)=\lim_{B \to \infty} \sum_{n=0}^\infty \sum_k c_{n,k} (2i\pi)^{-n}(\delta^{(n)}(\xi-k) \ast B e^{-\pi B^2 \xi^2})$$