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Let $f(z)$ be defined by a Laurent series at z = 0 with real coefficients. In particular, $f(x) \in \mathbb{R}$ for $x \in \mathbb R$.

Compute the residue of $f(z)$ at z = 0 using just the restriction of $f$ to a real interval (-$\epsilon$, $\epsilon$) for $\epsilon$ > 0.

My goal is to be able to express the residue as a limit which can be evaluated from the real function germ at 0.

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    $\begingroup$ If your function has a simple pole this is easy, and if it doesn't then yuk. Can it have an essential singularity? $\endgroup$
    – Kevin Buzzard
    Commented Jan 12, 2017 at 20:05
  • $\begingroup$ Yes. If the principal part of the Laurent series is finite, this is easy to do. But I am particularly interested in the case where there is an essential singularity at 0. $\endgroup$
    – David Meyer
    Commented Jan 12, 2017 at 20:11
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    $\begingroup$ Using Taylor expansion, one can work out $f$ on the disks $D(\pm \epsilon/2, \epsilon/2)$ if $\epsilon$ is small enough. Using Taylor expansion a few more times, one can then continue $f$ to a punctured neighbourhood of the origin, at which point the residue can be recovered from the residue theorem. It's unlikely one can do much better than this, since there exist functions with non-zero residue that vanish extremely quickly on the real axis (e.g. $z \exp(-1/z^2)$). $\endgroup$
    – Terry Tao
    Commented Jan 12, 2017 at 20:28
  • $\begingroup$ In particular, the functions $f_n(z) := \frac{z}{n} \exp( \frac{-n}{z^2} )$ all have residue -1, but converge in just about every reasonable topology (e.g. $C^\infty$) on $(-1,1) \backslash \{0\}$ to zero as $n \to \infty$. So there is no way to recover the residue from any limiting operation that is continuous in reasonable topologies; some analytic continuation must come in at some point. $\endgroup$
    – Terry Tao
    Commented Jan 12, 2017 at 20:32
  • $\begingroup$ Terry, I have attempted to write out sequentially expanded Taylor series similar to what you just described, writing the taylor series for $f^{(k)}$ at x+t in terms of the taylor series at x, and then writing the taylor series at x+t+s in terms of the series at x+t. What I get (apparently) contains nested infinite sums that do not commute. If I assume they commute, it simplifies immediately to the taylor series at x evaluated at x+t+s. (so there is no improvement to the radius of convergence) $\endgroup$
    – David Meyer
    Commented Jan 12, 2017 at 20:52

1 Answer 1

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If $0$ is a pole, then there is a formula for the residue $$\mathrm{res}_0=\lim_{x\to 0}\frac{1}{(n-1)!}\frac{d^{n-1}}{dx^{n-1}}x^nf(x),$$ which involves only the restriction of $f$ on the real line.

If $0$ is an essential singularity, there is no formula, except the complex integral defining the residue, and it is hard to imagine any formula which will involve only the restriction on the real line and some limits. The reason is that in a neighborhood of an essential singularity, every continuous function on the real line can be approximated with any accuracy (error tending to zero arbitrarily fast as $x\to 0$) by an analytic function with any residue.

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  • $\begingroup$ But can all $n^{th}$ derivatives be approximated to arbitrary accuracy in a uniform way? Even in a situation such as $f(z) = z e^{-1/z^2}$, where $f^{(n)}(x)\to0$ as $x\to0$ for all $n$, it is still the case that $f^{(n)}(x)$ diverges as $n\to\infty$ for any $x > 0$. If possible to extract, the residue information must come from the "tail" of the sequence $f(x), f'(x), f''(x), \ldots$ $\endgroup$
    – David Meyer
    Commented Jan 13, 2017 at 15:04
  • $\begingroup$ Correction to last comment: $\lim_{n \to \infty} f^{(n)}(x)$ diverges for $-1 \lt x \lt 1$, $x \ne 0$ $\endgroup$
    – David Meyer
    Commented Jan 13, 2017 at 17:28
  • $\begingroup$ @David Meyer: yes all derivatives can be approximated. $\endgroup$
    – Alexandre Eremenko
    Commented Jan 13, 2017 at 22:54

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