Fourier Transform: All It Takes Is Wild Imagination?

傅里叶变换：只需要异想天开？ 苏剑林 · 2014-04-25 · Mathematics series integral transform complex number Fourier

When we look back at mathematics or physics with hindsight, we often stumble upon strange phenomena, and sometimes stumble into deeper, more interesting results. The Fourier transform discussed in this post is a case in point—it can be arrived at through a rather "wild" line of thought.

Laurent Expansion

As we know, a function that behaves nicely at the origin can be expanded as a Taylor series

$$f(x)=\sum_{n=0}^{\infty}a_n x^n$$

Notice that all the powers above are positive—so why shouldn't we allow negative powers of $x$? For instance, expanding $\frac{\sin z}{z^2}$ as

$$\frac{1}{z}-\frac{z}{6}+\frac{z^3}{120}\dots$$

is clearly also a reasonable thing to do. Combining this with complex analysis, we arrive at the Laurent expansion of an analytic function

$$f(z)=\sum_{n=-\infty}^{+\infty}a_n z^n$$

This is a two-sided expansion of the function, where more

$$a_n=\frac{1}{2\pi i}\int_{\gamma} \frac{f(z)}{z^{n+1}}dz$$

$\gamma$ is the contour $|z|=\rho,\rho>0$. This formula is based on the following rather obvious fact ($\alpha\in\mathbb{Z}$):

$$\int_{\gamma} z^{\alpha}dz=\left\{ {\begin{array}{\cdot {20}{c}} {2\pi i,\alpha=-1;}\\ {0,\alpha\neq -1.} \end{array}} \right.$$

Half-Integer Power Series

But here's another question: why can't we allow fractional powers? How would we expand a function like $\sqrt{z}$? This suggests that the Laurent expansion can be extended further. As an illustration, let's throw in "half-integer powers" too, and consider:

$$f(z)=\sum_{n=-\infty}^{+\infty}a_n z^{n/2}$$

How do we find the coefficients of each term? We just need to convert this into the familiar Laurent expansion. Let $z^{1/2}=\xi$, then

$$f(\xi^2)=\sum_{n=-\infty}^{+\infty}a_n \xi^n$$

We already know how to find the coefficients in a Laurent expansion, namely

$$\begin{aligned} a_n=\frac{1}{2\pi i}\int_{\gamma}\frac{f(\xi^2)}{\xi^{n+1}}d\xi &=\frac{1}{2\pi i}\int_{2\gamma}\frac{f(z)}{z^{(n+1)/2}}dz^{1/2}\\ &=\frac{1}{4\pi i}\int_{2\gamma}\frac{f(z)}{z^{n/2+1}}dz \end{aligned}$$

Here $\gamma$ is a circle centered at the origin (traversed once counterclockwise), while $n\gamma$ means going counterclockwise around the origin $n$ times. This new series is one generalization of the Laurent expansion.

Once and for All

But then another question arises: why can't we have powers that are multiples of 1/3? Why not irrational powers? This kind of "wild imagination" could go on forever. So, to preempt any further questioning, let's just throw in all real powers at once—hence the "once and for all" in the title of this section. Consider the function

$$f(z)=\int_{-\infty}^{+\infty}a(x)z^x dx$$

which encompasses both of the forms above. (The discrete case $a_n$ corresponds to $a(x)$ containing Dirac delta functions $\delta(x)$.) Next, let's derive the expression for $a(x)$. We discretize the sum, starting from the origin and summing with step size $\Delta x$:

$$f(z)=\lim_{\Delta x \to 0} \sum_{n=-\infty}^{+\infty}a(n\Delta x)\Delta x z^{n\Delta x}$$

Let $z^{\Delta x}=u$, substituting in gives

$$f(\xi^{1/\Delta x})=\sum_{n=-\infty}^{+\infty}a(n\Delta x)\Delta x \xi^{n}$$

Again using the Laurent-expansion method to find the coefficients, we get

$$\begin{aligned}a(n\Delta x)\Delta x=\frac{1}{2\pi i}\int_{\gamma}\frac{f(\xi^{1/\Delta x})}{\xi^{n+1}}d\xi &=\frac{1}{2\pi i}\int_{\left(\frac{1}{\Delta x}\right)\gamma}\frac{f(z)}{z^{(n+1)\Delta x}}dz^{\Delta x}\\ &=\frac{1}{2\pi i}\int_{\left(\frac{1}{\Delta x}\right)\gamma}\frac{\Delta x f(z)}{z^{n\Delta x+1}}dz \end{aligned}$$

that is,

$$\begin{aligned}a(x)=\frac{1}{2\pi i}\int_{\left(\frac{1}{\Delta x}\right)\gamma}\frac{f(z)}{z^{x+1}}dz \end{aligned}$$

where the integration path winds around the origin infinitely many times. Since the transformations are invertible, this becomes a mutual transform between two functions.

The Fourier Transform

Now let's switch to more familiar notation. Let $z=e^{-i\omega}$, and replace $a(x)$ with the notation $ f(x)$，对应的$f(z)$则换成$F(\omega)$. We have

$$F(\omega)=\int_{-\infty}^{+\infty}f(x)e^{-i\omega x} dx$$

as well as

$$f(x)=-\frac{1}{2\pi i}\int_{-\infty}^{+\infty} \frac{F(\omega)}{e^{-i\omega(x+1)}}de^{-i\omega}=\frac{1}{2\pi }\int_{-\infty}^{+\infty} F(\omega) e^{i\omega x}d\omega$$

The contour $\gamma$ we chose is counterclockwise, but $e^{-i\omega x}$ is clockwise, hence the extra minus sign. And this gives us the Fourier transform and its inverse.

A Brief Summary

The biggest question in readers' minds is probably this: why do the limits of integration run from negative infinity to positive infinity, rather than from 0 to positive infinity? This is genuinely hard to explain cleanly—a rigorous derivation of the inverse Fourier transform still requires the machinery found in textbooks on Methods of Mathematical Physics. Still, for what it's worth, from the perspective of this post we can offer some kind of explanation (an explanation, not a proof) for the choice of limits.

If the upper limit of integration is chosen appropriately, the lower limit can be chosen rather freely—for instance, if we take the lower limit to be 0, then the upper limit must be $2\pi N,N\in \mathbb{Z}$ with $N\to \infty$. But imposing a restriction when approaching infinity ($N\in\mathbb{Z}$ rather than $N\in \mathbb{R}$) always feels awkward, since positive infinity carries an inherent indeterminacy. So we might as well introduce the same kind of indeterminacy at the lower limit too (negative infinity)—this way, since there's no fixed starting point, there's no need to impose a restriction on the endpoint either. Hence the limits of integration can simply be written as negative and positive infinity, and whatever subtlety arises from the "infinite" regions gets naturally smoothed over in the process of integration. (PS: This explanation is admittedly rather vague, but to keep the article from getting bogged down in tedious technical detail, this is as far as I'll go. In fact, based on the rigorous theory of improper integrals in mathematical analysis, one could turn the above paragraph into an actual proof—mainly by using integration techniques such as $\int_0^{\infty} f(x)\cos(\omega x)dx$. If any readers want to dig deeper, why not give it a try?)

In truth, virtually all of the derivations above are not rigorous. The point of this post is to offer a "fairly natural" way of understanding the Fourier transform through this line of thinking, and to give readers a way of connecting two seemingly unrelated things—the Fourier transform and the Laurent expansion. Through complex analysis, many branches of mathematics can be woven together in surprisingly elegant ways. This seems to be one guiding principle of the universe's design: harmony and unity. And if that's really the case, then we have an even deeper reason to explore mathematics, to explore science—namely, for the sake of beauty!