Is Fourier series an “inverse” of Taylor series?

Question

I've understood Taylor series as being the representation of a "transcendental" function, using power functions with coefficents represented by appropriate derivatives. (Or maybe it is the MacLauren series, where $\cos x= 1-\frac{x^2}{2!}+\frac{x^4}{4!}+...$)

I've understood Fourier series as being the representation of periodic linear functions using integral coefficients multiplied by transcendental functions.

Is my understanding correct in either or both cases? And if one consists of derivatives and the other of integrals, does that mean that Fourier series are the converse (or "inverse") of Taylor series?

Answer 1

Just a brief comparison:

Fourier series are:

• Global in nature. Fourier series are computed using an integral over one period (they represent the entire function over one period even if it is discontinuous, piece-wise continuous, etc.... Gibbs phenomenon not withstanding)
• Fourier series decompose a function by representing it as a linear combination of basis functions (sine and cosine). These basis functions are orthogonal.
• Fourier series are invertable. That is once you have your Fourier coefficients you can reconstruct the entire function from the coefficients (up to a point, i.e. Gibbs phenomenon).

On the other hand:

Taylor series are:

• Local in nature. Taylor series are computed using an infinite number of derivatives at one point (therefore they cannot represent functions which are discontinuous, piece-wise continuous, etc).
• Taylor series decompose a function by representing it as a fixed combination of derivatives. These "basis" functions are not orthogonal.
• Taylor series are invertable only in the neighborhood of a point. You cannot, in general, recover the entire function from a Taylor series.

Answer 2

Your understanding is incorrect in both cases.

A Taylor series is able to represent an "arbitrary" function $f$ in the neigbourhood of a given point $a$ in the domain of $f$ as a power series: $$f(x)=\sum_{k=0}^\infty c_k (x-a)^k\ ,$$ i.e., in the form of an "infinite polynomial". Neither the series nor its finite partial sums are "linear functions". The coefficients $c_k$ of this power series are connected to the represented function $f$ by the formula $c_k=f^{(k)}(a)/k!\ $. So we see here the values $f^{(k)}(a)$ entering in a linear way, but the derivatives $f^{(k)}$ as functions do not appear in the representation.

A Fourier series is able to represent an "arbitrary" periodic function $f$ of period $2\pi$ as an "infinite linear combination" of the basic periodic functions $t\mapsto \sin(k t)$, $t\mapsto \cos(k t)$; so it has the form $$f(t)={a_0\over 2}+\sum_{k=1}^\infty (a_k\cos(kt)+b_k\sin(kt))\ .$$ The coefficients $a_k$, $b_k$ in this representation are connected to the given $f$ via certain integrals (which I won't write down here).

In fact there is a certain connection between these two paradigms. It works in the realm of functions of a complex variable $z$, but there is no question of the same function $f$ of a real variable $x$ or $t$ being represented with more or less the same coefficients $c_k$, $a_k$, $b_k$ first as a Taylor series and then as a Fourier series.

Answer 3

Both Fourier series and Taylor series are decompositions of a function, the difference is that Taylor series are inherently local, while Fourier series are inherently global.

You can find out more here:
Connection between Fourier transform and Taylor series