Decomposing functions to Taylor-Fourier series 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? 
 A: 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.
A: Consider a function $f$ defined at a set $D$ with the property that if a point $p$ is present in $D$ then $p+1, p+2,p+3 .... $ are also present in $D$. 
Thus we can form a discrete-difference expansions of $f$ around each point $p\in D$ by evaluating 
$$ f(p) + D[f](p)*(x-p) + \frac{1}{2!}D^2[f](p)*(x-p)(x-p-1) + \frac{1}{3!}D^3[f](p)*(x-p)(x-p-1)(x-p-2) + ... $$ 
We can then isolate the constant term (w.r.t powers of x) which gives us:
$$  a_0 = f(p) - pD[f](p) + \frac{p(p+1)}{2!}D^2[f](p) -\frac{p(p+1)(p+2)}{3!}D^3[f](p) .... $$
Similarly
$$ a_1 =  D[f](p) - \frac{p + (p+1)}{2!}D^2[f](p)  + \frac{p(p+2) + p(p+1) + (p+1)(p+2)}{3!}D^3[f](p) ...  $$ 
And so on and so forth...
Each of the $a_i$ can now be decomposed via fourier analysis (by varying p on an interval of length of 1), to turn them into fourier series in $p$, and once done we can swap $p$ for $x$ 
