A great many functions can be expressed as a series of the form
$$ U_0(x) + U_1(x) x + U_2(x) \frac{1}{2!}x(x-1) + … $$
Where $U_r(x)$ are integrable periodic functions with period $1$. Call such functions "1 periodic normal" functions. Note that the $U_r(x)$ being periodic can be decomposed into their fourier series as:
$$ U_r(x) = \sum_{k=-\infty}^{\infty} a_{r,k} e^{2\pi i k x} $$
And so 1-periodic normal functions have a general form as:
$$ \sum_{k=-\infty}^{\infty} a_{0,k} e^{2\pi i k x} + \left( \sum_{k=-\infty}^{\infty} a_{1,k} e^{2\pi i k x} \right) x + … $$
In the event that $U_1, U_2 … $ are equal to $0$ it follows that we can use fourier analysis to determine the coefficients of $U_0$.
In particular when $U_1, U_2 … $ are equal to 0, then the operator
$$ f \rightarrow 2 \int_{0}^{1}f(x) e^{i\pi Jx} dx $$
Gives the coefficient $a_{j,0}$ of our series.
Suppose we have no guarantees about non-zero $U_r$ how could we systematically determine the $a_{j,r}$ coefficients of our series?
Best Answer
I don't see what you mean in your answer. You are supposed to start from $$f(x) = \sum_{k =0}^K (\sum_n c_{n,k} e^{2i \pi n x}) {x \choose k} = \sum_{k =0}^K (\sum_n c_{n,k} e^{2i \pi n x}) \sum_{l=0}^k s_{k,l} x^l $$
If $K$ is finite then $f$'s Fourier transform exists in the sense of (tempered) distributions and it is $$\hat{f}(y)=\sum_n \sum_{l=0}^K \delta^{(l)}(y-n) \sum_{k=l}^K c_{n,k}(-2i\pi)^{-l} s_{k,l}$$ where $\delta^{(l)}$ is $l$-th derivative of the Dirac delta, the Fourier transform of $(-2i\pi x)^l$.
Recovering the $\sum_l c_{n,k} (2i\pi)^{-l} s_{k,l}$ and the $c_{n,k}$ from $\hat{f}(y)$ is immediate.
If $K$ is infinite then you need to tell in what sense you expect convergence, there are different $c_{n,k}$ giving the same analytic functional.