Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Both Fourier transform and Taylor series are means to represent functions in a different form.

My question:
What is the connection between these two? Is there a way to get from one to the other (and back again)? Is there an overall, connceting (geometric?) intuition?

share|improve this question
1  
Not answering your question, but a summary of the reasons that the Fourier transform tends to be favoured in engineering applications can be found in the "Motivation" section of the introduction to this paper. –  yasmar Oct 20 '10 at 23:12
3  
Thanks for asking the question; all four answers are illuminating. –  ShreevatsaR Jul 2 '11 at 4:27
add comment

4 Answers

up vote 44 down vote accepted

Assume that the Taylor expansion $f(x)=\sum_{k=0}^\infty a_k x^k$ is convergent for some $|x|>1$. Then $f$ can be extended in a natural way into the complex domain by writing $f(z)=\sum_{k=0}^\infty a_k z^k$ with $z$ complex and $|z|≤1$. So we may look at $f$ on the unit circle $|z|=1$. Consider $f$ as a function of the polar angle $\phi$ there, i.e., look at the function $F(\phi):=f(e^{i\phi})$. This function $F$ is $2\pi$-periodic, and its Fourier expansion is nothing else but $F(\phi)=\sum_{k=0}^\infty a_k e^{ik\phi}$ where the $a_k$ are the Taylor coefficients of the "real" function $x\mapsto f(x)$ we started with.

share|improve this answer
2  
Great answer. But as Qiaochu says, "it's worth mentioning that the Fourier transform is much more general than this" — the Fourier expansion exists even for functions not got in this way (as the restriction to the unit circle of some function whose Taylor series has radius of convergence greater than 1). –  ShreevatsaR Jul 2 '11 at 4:18
add comment

A holomorphic function in an annulus containing the unit circle has a Laurent series about zero which generalizes the Taylor series of a holomorphic function in a neighborhood of zero. When restricted to the unit circle, this Laurent series gives a Fourier series of the corresponding periodic function. (This explains the connection between the Cauchy integral formula and the integral defining the coefficients of a Fourier series.)

But it's worth mentioning that the Fourier transform is much more general than this and applies in a broad range of contexts. I don't know that there's a short, simple answer to this question.

Edit: I guess it's also worth talking about intuition. One intuition for the Taylor series of a function $f(x)$ at at a point is that its coefficients describe the displacement, velocity, acceleration, jerk, and so forth of a particle which is at location $f(t)$ at time $t$. And one intuition for the Fourier series of a periodic function $f(x)$ is that it describes the decomposition of $f(x)$ into pure tones of various frequencies. In other words, a periodic function is like a chord, and its Fourier series describes the notes in that chord.

(The connection between the two provided by the Cauchy integral formula is therefore quite remarkable; one takes an integral of $f$ over the unit circle and it tells you information about the behavior of $f$ at the origin. But this is more a magic property of holomorphic functions than anything else. One intuition to have here is that a holomorphic function describes, for example, the flow of some ideal fluid, and integrating over the circle gives you information about "sources" and "sinks" of that flow within the circle.)

share|improve this answer
1  
The "chord" analogy should not be taken at face value, by the way; any note you hear played on a physical instrument is not a pure sine wave but comes with a collection of overtones, and the relative strength of these overtones is what gives different instruments their different sounds. Actual pure sine waves - without overtones - can really only be generated electronically. –  Qiaochu Yuan Oct 21 '10 at 17:05
3  
displacement, velocity, acceleration, jerk, ... or for the non-physics intuition, $\mathrm{Taylor} \rightsquigarrow$ 0, mean, variance, skewness, kurtosis, ... in statistics. Or $\mathrm{Taylor} \rightsquigarrow$ height, tilt, curve, wiggle, womp, ... in terms of the graph $f(x)=x$. –  isomorphismes Mar 18 '11 at 22:11
add comment

There is a big difference between the Taylor series and Fourier transform-the Taylor series is a local approximation, while the Fourier transform uses information over a range of the variable. The theorem Qiaochu mentions is very important in complex analysis and is one indication of how restrictive having a derivative in the complex plane is on functions.

share|improve this answer
add comment

There is an analogy, more direct for fourier series. Both Fourier series and Taylor series are decompositions of a function $f(x)$, which is represented as a linear combination of a (countable) set of functions. The function is then fully specified by a sequence of coefficients, instead of by its values $f(x)$ for each $x$. In this sense, both can be called a transform ($f(x) \leftrightarrow \{ a_0, a_1, ...\}$).

For the Taylor series (around 0, for simplicity), the set of functions is $\{1, x , x^2, x^3...\}$. For the Fourier series is $\{1, \sin(\omega x), \cos(\omega x), \sin(2 \omega x), \cos(2 \omega x) ...\}$.

Actually the Fourier series is one the many transformations that uses an orthonomal basis of functions. It is shown that, in that case, the coefficients are obtained by "projecting" $f(x)$ onto each basis function, which amounts to an inner product, which (in the real scalar case) amounts to an integral. This implies that the coefficients depends on a global property of the function (over the full "period" of the function).

The Taylor series (which does not use a orthonormal basis) is conceptually very different, in that the coeffients depends only in local properties of the function, i.e., its behaviour in a neighbourhood (it's derivatives).

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.