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[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 $2^x$ (for $x>0$) 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.

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The approach for determining the Fourier series coefficient $a_{j,r}$ for $U_r(x)$ is similar to the approach for determining the Fourier series coefficient $a_{j,0}$ for $U_0(x)$. The Fourier series coefficient for $U_r(x)$ is determined from $U_r(x)$, not the entire sum $f(x)$.

It can be determined from the entire sum $f(x)$ if $U_s(x)=0$ for all $s\ne r$ in which case:

$$U_r(x)=\frac{f(x)\,r!}{\prod_{n=0}^{r-1}(x-n)}$$

It can also be determined from the entire sum $f(x)$ if there are a finite number of non-zero terms

$$U_{s_1}(x), U_{s_2}(x), ... U_{s_K}(x)$$

where $s_k\ne r$ in which case

$$f(x)=U_r(x)\frac{1}{r!}\prod_{n=0}^{r-1}(x-n)+\sum_{k=1}^K b_{s_k}$$

where

$$b_{s_k}=U_{s_k}(x)\frac{1}{s_k!}\prod_{n=0}^{s_k-1}(x-n)$$.

In this latter case:

$$U_r(x)=\frac{\left(f(x)-\sum_{k=1}^K b_{s_k}\right)\,r!}{\prod_{n=0}^{r-1}(x-n)}$$

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