[Cross posted from Math.SE due to lack of attention]

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?

Some Motivation:

If you were given the functions $\cos(2\pi x)2^x$ and $2^x$. You would find they agree on all integer points. So if one formed a "forward" difference taylor series centered at 0 for either you end up with $$ 1 + x + \frac{x(x-1)}{2!} + \frac{x(x-1)(x-2)}{3!} ... $$

But this is only agrees with $2^x$ globally and not for the the other function, so this leads to me believe there is a missing piece, which should combine the theory of forward differences with fourier analysis to give us the entire picture.

My motivation is in some sense mostly aesthetic but I do believe there is some interesting mathematics here.