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Since the coefficients

$$a_k = \frac1{2\pi i}\oint_C\frac{f(z)}{(z-c)^{k+1}}\,dz$$

for the Laurent series

$$f(z)\Big|_{r\le|z|\le R} = \sum_{k=-\infty}^{\infty}a_k\cdot(z-c)^k $$

of a function $f\in\mathcal H(B(r,R))$ (i.e. a function that is holomorphic on the Annulus of radii $r\le R$), evaluated on a circle of radius $\rho\in[r,R]$,

$$\tilde a_k := \rho^k a_k = \frac1{2\pi}\int\limits_{\phi=0}^{2\pi} f(c+\rho e^{i\phi}) e^{-ik\phi}\,d\phi \\\Rightarrow f(c+\rho e^{i\phi}) = \sum_{k=-\infty}^\infty \tilde a_k\,e^{ik\phi},$$

are (up to the factor $\rho^k$) equivalent to the Fourier coefficients of a function

$$\tilde f(x):=f\Big(c+\rho e^{i\tfrac x{2\pi\rho}}\Big),\quad x\in[-\pi\rho,+\pi\rho],$$

I was wondering if there is a meaningful limit $\rho\to\infty$ for the Laurent series. For the Fourier series, the limit becomes the Fourier transform with continous "coefficients", $\tilde a_k\to \tilde a(k)\equiv \mathcal F\{\tilde f(x)\}(k)$, such that

$$\tilde f(x) = \int_{-\infty}^\infty \tilde a(k) e^{ikx}\, dk, \\\tilde a(k) = \frac1{2\pi}\int_{-\infty}^\infty \tilde f(x) e^{-ikx}\, dx.$$

So in a similar way, this limit would turn the Laurent series into a "Laurent transform",

$$f(z)\Big|_{r\le|z|} = \int_{-\infty}^\infty a(k)\cdot (z-c)^k, \\ a(k) = \lim_{R\to\infty} \frac1{2\pi}\oint\limits_{|z|=R}\frac{f(z)}{(z-c)^{k+1}}\,dz$$

(up to some factors that I probably forgot, and assuming that $f$ remains holomorphic for $R\to\infty$), describing $f(z)$ by its values at $\mathbb C$-infinity.

So, does this make sense, has it been investigated upon before and/or any applications?

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  • $\begingroup$ Are you sure about the $\tilde{a}_k \to \tilde{a}(k)$ argument? Formally $\phi \to x / 2\pi \rho$ so it appears you should have $\tilde{a}_k \to \tilde{a}(k / 2\pi \rho)$, no? This makes sense as every "finite" frequency on the circle, taking your change of coordinates, should limit to "0"-frequency. (There should also be a factor of $2\pi \rho$ coming from when the change of variables hit the volume form.) $\endgroup$ – Willie Wong Jul 16 '13 at 10:49
  • $\begingroup$ @WillieWong It's sloppy notation, sorry. It's indeed more $\tilde a_k \to \tilde a(k/2\pi\rho)$ followed by $\rho\to\infty, \sum\to\int$ and a substitution to $\tilde a(\tilde k)$. I should probably expand on that, but in principle it should just be the usual transition from Fourier series to Fourier transform $\endgroup$ – Tobias Kienzler Jul 16 '13 at 10:57
  • $\begingroup$ You must "expand" on that. The way I see it, depending on normalisation, you may just have a bunch of indeterminant forms of the type $\infty \cdot 0$ equal to each other. Consider the case $f(z) = z^n$ and $c = 0$. Your $a(k) = 0$ for all integer $k$ not equal to $n-1$, and blows up with equality. Essentially it looks to me that what you are doing is doing the Laurent series expansion of $f(1/z)$ about $0$. $\endgroup$ – Willie Wong Jul 16 '13 at 11:27
  • $\begingroup$ Note that $(z-c)^k$ for $k$ non-integral is not well defined on $\mathbb{C}$; but you can define it on a Riemann surface (think of the one corresponding to $\log (z-c)$). This better captures the periodic/non-periodic divide between the Fourier series and Fourier transform. But for the integrals to converge you will need meromorphic functions which decays to zero as you move up and down the sheets. $\endgroup$ – Willie Wong Jul 16 '13 at 11:38
  • $\begingroup$ BTW, convergence is also an issue with your original formulation. Your "fourier coefficients" $\tilde{a}(k)$ will not converge if $\lim_{r \to \infty} f(c - r) \neq 0$. $\endgroup$ – Willie Wong Jul 16 '13 at 11:57
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In the same way that there is a discrete $\rightarrow$ continuous transition as one goes from Fourier series to transforms, one can make a continuous analogue to Taylor series in the form of the Laplace transform. This is beautifully explained by Arthur Mattuck in this video and this one, which I heartily recommend, but the gist is this:

Start with a Taylor series of the form $\sum_{n=0}^\infty a_nx^n=:A(x)$, and change the discrete variable $n=0,1,2,\ldots$ for a continuous variable $t\in(0,\infty)$, to get $\int_0^\infty a(t) x^t dt$. From this it is only a cosmetic change to define the variable $s=-\ln(x)$, which is positive and real in the interesting range for $x$, $0<x<1$, and this makes the integral look like $$\int_0^\infty a(t)e^{-st}dt=:A(s),$$ which is just a standard Laplace transform. Typically $A(s)$ will be defined in a half-plane $\text{Re}(s)>s_0$, as long as $a(t)$ is of exponential class for $t\rightarrow\infty$.

If you now want a continuous analogue of Laurent series, then all you need to do is extend your integration to minus infinity as well: $$\int_{-\infty}^\infty a(t)e^{-st}dt=:A(s).$$ This will make the typical domain for $s$ a strip of the form $s_0<\text{Re}(s)<s_1$. This corresponds to $x=e^{-s}$ being in an annulus around zero, which is also the typical domain of convergence of Laurent series.

You are also much more constrained in your choice of $a(t)$, in the same way that coefficient sequences are more restricted for Laurent series. This is in a way more serious here, because the class of 'interesting' functions which are now available is indeed noticeably smaller than the equivalent restriction in Laurent series. (In particular, things like 'folding' a sequence so that e.g. $a_n=1/(|n|!)$ become more problematic, and can cause $a(t)$ to not be analytic anymore.) This means that the transform is slightly less useful, and indeed it is much less common in the literature than the standard Laplace transform.

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You may want to look into the Mellin transform. The basic idea is similar to what you seem to have in mind, but the integration contours are different from what you guessed. Incidentally, one side of the Mellin transform applies even outside the context of analytic functions. Of course, the Fourier, Laplace and Mellin transforms are fundamentally related, through changes of integration variables.

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