Decomposing functions to Taylor-Fourier series

Question

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?

Answer 1

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

$$ f(p) + D[f](p)*(x-p) + \frac{1}{2!}D^2[f](p)*(x-p)(x-p-1) + \frac{1}{3!}D^3[f](p)*(x-p)(x-p-1)(x-p-2) + ... $$

Now suppose $f$ is smooth then $f(x) = a_0 + a_1x + a_2x^2 + ... $ so we might wish to isolate the individual $a_i$.

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 $x$

So our function looks like:

$$ f(x) = a_0(x) + a_1(x)*x + a_2(x)*x^2 + ... $$

Where each of the $a_i$ is a periodic function in $x$. These can now be recombined into the original form

Now we know for example that $\frac{1}{2}x(x-1) = 1/2x^2 + 1/2x$ and thus we can rewrite the partial sum $a_0(x) + a_1(x)*x + a_2(x)*x^2$ as

$$ a_0(x) + \left( a_1 (x) + a_2(x) \right) x + 2 \left( a_2(x) \right) \frac{x(x-1)}{2!} $$

And in general arbitrary large numbers of terms can be truncated accordingly. This process might lead to divergences so another different angle is to express the function as

$$ f(x) = Q_0(p) + Q_1(p)x + Q_2(p)\frac{1}{2!}x(x-1) + ... $$

I.E. shifting all the $p$ dependence to the coefficients. Then again we perform fourier analysis on the individual coefficient expressions and extract directly the wave terms without further reshuffling.

Answer 2

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.