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Cauchy residue theorem tells us that for a function $$f(z) = \sum_{k \in \mathbb{Z}} a(k) z^k,$$ the coefficient $a(k)$ can be extracted by an integral formula $$a(k) = \frac{1}{2\pi i}\oint f(z) z^{-k-1},$$ with a contour around zero. Now, there is nothing to prevent us from thinking of $a\colon \mathbb{Z} \to \mathbb{C}$ as a function over the complex domain, defined by the above integral. In this way, we have naturally "continued" the function $a(k)$ over the integers to one over the complex numbers. Is there any meaning to this beyond just something amusing?

For example, consider $f(z)=\exp(z)$. In this case, $a(k)=1/k!$ which one would think would be continued to $1/\Gamma(1+k)$. But at least if you attempt to plot what happens above, this is not the case. We get a function with Bessel-like oscillations (blue) which nevertheless coincides with $1/\Gamma(1+k)$ (orange) on integral points!

enter image description hereenter image description here

One can experiment with other choices of $f(z)$. When $f(z)=(z+1)^n$, it seems like the continuation actually coincides with what one would naturally think (the Beta function). Which is actually quite strange. For example for $n=5$ and $f(z) = 1 + 5 z + 10 z^2 + 10 z^3 + 5 z^4 + z^5$, how does this computation "learn" that the sequence $1,5,10,10,5,1$ actually corresponds to the binomial coefficients and thus should be continued to the Beta function? If we perturb one of the coefficients, say change $5 z^4$ to $2 z^4$, the result would look like a "perturbed" Beta function (blue below vs the Beta function in orange).

enter image description here

Any explanation (or references in the literature) for what is going on here?

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    $\begingroup$ Just a remark. Your formula for $a(k)$ with non-integer $k$ is not actually uniquely defined. Because of the branch cuts in $z^{-k-1}$ for non-integer $k$, the value of the integral depends precisely on the shape of the contour (well, more precisely on the location of your chosen branch cut and where it intersects the contour). I noticed that you didn't specify the contour in your post. So, as you change the contour, your blue curve will change, except for its intersections with the orange curve. $\endgroup$ – Igor Khavkine Feb 22 at 6:55
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Interpolating coefficients of a power series by an entire function of $n$ is a common powerful method of the study of singularities of power series. There is a close relation between the growth of this entire function and analytic continuation of the power series. See, for example,

L. Bieberbach, Analytische Fortsetzung, Springer 1955,

or V. Bernstein, Lecons sur les progres recents de la theorie de series de Dirichlet, Paris, 1933.

Sorry, I do not know any books in English, but you can look at this paper arXiv:0709.2360, Theorem C.

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The usual way to interpret the integral $$\frac{1}{2\pi i} \oint \exp(z) z^{-s} \, \mathrm{d}z$$ for nonintegral $s$ is to take a Hankel contour, and in this case the value is indeed $\frac{1}{\Gamma(s)}$ (see 5.9.2). Without knowing how you produced these plots, I'm not sure what else to say about your first example.

This is also related to Ramanujan's master theorem, which expresses this analytic continuation as a Mellin transform. Hardy gave some conditions under which this identity can be made rigorous.

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  • $\begingroup$ Interesting, thanks. I just used a circular contour around the origin for the plots. Not sure why Hankel contour produces a different result, though especially for such a "clean" function as exp(z). $\endgroup$ – MCH Feb 22 at 18:16
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    $\begingroup$ @MCH The exp(z) is fine, it's the $z^{-s}$ which gives you difficulties. The contour matters quite a bit. You can extend your circular contour to the Hankel contour by adding lines along the negative real axis so I suppose the function in your plot will be $1 / \Gamma(s)$ plus or minus some expression like $(e^{2\pi i s} - 1) \int_R^{\infty} e^{-t} t^s \, \mathrm{d}t$ if the circle contour had radius $R$ (I haven't checked the details.) $\endgroup$ – user152705 Feb 22 at 18:55
  • $\begingroup$ Interesting. I'll check it now. $\endgroup$ – MCH Feb 23 at 0:55
  • $\begingroup$ On RMT and the Mellin transform, see, e.g., mathoverflow.net/questions/79868/… $\endgroup$ – Tom Copeland Feb 23 at 21:33

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