# Decomposing functions to Taylor-Fourier series

[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.

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:

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