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Residue theorem can be stated informally as $$\oint_C f(z)dz=2\pi i\sum a_{-1}$$ A contour integral sums up all the $-1$ coefficients inside.

Then, one would naturally ask:

Is there something like $$\text{something of }f(z)=\sum a_{-2}$$ where $\text{something}$ is a sort of operator?

I encounter difficulties as the residue theorem makes use of the fact that $\frac1{z-c}$ has no antiderivative, but $\frac1{(z-c)^2}$ does have an antiderivative, so the same trick can’t be used.

Any idea?

EDIT:

@David C. Ullrich's answer does rule out some possibilities. However, I am not desperate: I still have the hope that there might exist some operators $\operatorname{P}$ that:

  1. only requires information of $f(z)$ along a contour $\gamma$
  2. $$\operatorname{P}_\gamma[f(z)]=\sum_{\text{all poles included}} a_{-2}$$ where $a_{-2}$ is the coefficient of $z^{-2}$ of the Laurent expansion of $f(z)$ around the pole.

($f(z)$ can be assumed to be meromorphic on $\mathbb C$.)

Therefore, I started a bounty to draw more attention.

Szeto
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  • Doesn't $\oint_C z f(z)dz$ work? – pregunton Jul 27 '18 at 13:38
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    @pregunton You can try it out. It equals $$\sum(a_{-2}+ca_{-1})$$ where $c$ is location of pole. – Szeto Jul 27 '18 at 13:41
  • You could try $\oint (z-p) f(z) dz$ then, with $p$ chosen such that $\sum (c-p) a_{-1}=0$. That should be possible when $\oint f(z) dz \neq 0$. – Kusma Jul 27 '18 at 13:56
  • @Kusma But there are multiple poles inside the contour. We can’t choose a $p$ such that $c-p=0$ for all poles $c$. – Szeto Jul 27 '18 at 14:00
  • @Szeto I think what Kusma means is to take $p$ as the weighted average of all poles in the contour, with weights the residues. That way, even if $c-p\neq0$ for any pole, the sum $\sum (c-p)a_{-1}$ does equal zero. – pregunton Jul 27 '18 at 14:02
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    @pregunton Yes, that is what I mean. Of course you will need locations of all singularities and all residues to do this calculation. If this works, the calculation will also similarly work if $\oint f(z) dz=0$, you will just need to consider $\tilde f(z)=f(z)+\frac1{z-A}$ for some point $A$ inside $C$. – Kusma Jul 27 '18 at 14:12
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    If you have a formula for $a_{-2}$, $a_{-1}$ and $a_{-2}+c a_{-1}$, then you have one for the pole $c$. That is, you would have a formula to compute poles (or zeroes) of meromorphic functions. That seems too much to be true. If you ever find such an operator, I suggest you try if you can prove the Riemann Hypothesis with it. – Bart Michels Aug 04 '18 at 17:04
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    Well of course with no restrictions, not even linearity, on that operator there exists such an operator $P$. For example, define $P_\gamma[f]$ to be the sum of $a_{-2}$ over poles inside $\gamma$. (That requires only information about $f$ on $\gamma$ by uniqueness: If $f=g$ on $\gamma$ then $f=g$ inside $\gamma$.) – David C. Ullrich Aug 06 '18 at 21:56
  • In fact come to think of it defining $P[f]=\sum a_{-2}$ does give a linear operator. But such an operator cannot be continuous, in the sense that $f_n\to f$ uniformly on $\gamma$ implies that $P[f_n]\to P[f]$. The Riesz Representation Theorem shows that if $P$ is continuous in that sense then it's given by integration against a complex measure, which I showed can't happen. – David C. Ullrich Aug 06 '18 at 22:13

1 Answers1

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Various positive suggestions in comments depend on for example knowing where all the poles are. This is not a residue-theoremish thing; an analog of RT would give, for the disk, a complex measure $\mu$ on $|z|=1$ such that $\int_{|z|=1}f\,d\mu$ equals the sum of the $a_{-2}$ at all the poles. With no information except the values of $f$ on the boundary, is the point.

There is no such measure. Taking $f(z)=z^n$ for $n\in\Bbb Z$ tells you what the Fourier coefficients of $\mu$ would be, and it follows that the only $\mu$ that could possibly work is the one defined by $$\int_{|z|=1}f(z)\,d\mu=\frac1{2\pi i}\int_{|z|=1}zf(z)\,dz.$$

But that gives the wrong answer for other $f$, for example $f(z)=1/(z-1/2)$.

In more detail: First, it's clear that $$\frac1{2\pi i}\int_{|z|=1}z\cdot z^n\,dz=\begin{cases}1,&(n=-2), \\0,&(n\ne-2).\end{cases}$$

Now assume that $\mu$ "works". Then we have $\int_{|z|=1}f(z)\,d\mu=\frac1{2\pi i}\int_{|z|=1}zf(z)\,dz$ if $f(z)=z^n$. Since trigonometric polynomials are uniformly dense in the continuous functions on the circle it follows that $$\int_{|z|=1}f(z)\,d\mu=\frac1{2\pi i}\int_{|z|=1}zf(z)\,dz$$for every $f$ continuous on the unit circle.

But now let $f(z)=1/(z-1/2)$. It follows that $$\int_{|z|=1}f(z)\,d\mu=\frac1{2\pi i}\int_{|z|=1}zf(z)\,dz\ne0,$$by the actual Residue Theorem. But $\sum a_{-2}=0$, so $\mu$ doesn't work for this $f$.

David C. Ullrich
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