I am currently trying to grasp spherical harmonics and try to digest that we proved that the sine and cosine functions are a basis for the $L^2$ space of the squared-integrable functions.
So as far as I have understood it, the functions that can be integrated with $\int_0^1 \mathrm dx \, f^2(x)$ are forming a vector space. Then all the $e_n = \cos(n \pi x)$ (and sine) form a basis for that space. So any (even, since I like to drop the sine terms) function $f$ can be represented as a linear combination of the basis vectors like: $f = \sum a_n e_n$
To get the coefficients $a_n$, I need to project the vector (i. e. the function) onto the basis vector (i. e. the sine) using the inner (dot) product, like so:
$ a_n = \left\langle f(x), e_n \right\rangle_F = \int_0^1 \mathrm dx \, f(x) \cos(n \pi x)$
Now I was wondering whether the Taylor series is such a representation with orthogonal functions $e_n = x^n$ as well. Is the “Taylor inner product” something like this then?
$ a_n = \left\langle f(x), x^n \right\rangle_T = \frac{1}{n!} \left. \frac{\mathrm d^n f(x)}{\mathrm d x^n} \right|_{x = 0}$
In the end, I will have a series like so :
$ f = \sum a_n e_n =\sum\limits_{n} \frac{1}{n!} \left. \frac{\mathrm d^n f(x)}{\mathrm d x^n} \right|_{x = 0} x^n$