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Find the residue at $z=-2$ for

$$g(z) = \frac{\psi(-z)}{(z+1)(z+2)^3}$$

I know that:

$$\psi(z+1) = -\gamma - \sum_{k=1}^{\infty} (-1)^k\zeta(k+1)z^k$$

Let $z \to -1 - z$ to get:

$$\psi(-z) = -\gamma - \sum_{k=1}^{\infty} \zeta(k+1)(z+1)^k$$

therefore we divide by the other part to get:

$$\frac{\psi(-z)}{(z+1)(z+2)^3} = -\frac{\gamma}{(z+1)(z+2)^3} - \sum_{k=1}^{\infty} \frac{\zeta(k+1)(z+1)^{k-1}}{(z+2)^3}$$

I have to somehow get the coefficient of $\frac{1}{z+2}$ because I want to evaluate the residue of

$g(z)$ at $z=-2$

The problem is I cant ever get a factor of $\frac{1}{z+2}$ what should I do?

Jack D'Aurizio
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Amad27
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1 Answers1

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$\psi(z)$ is regular over $\Re(z)>0$, hence $z=-2$ is a triple pole for $g(z)$. This gives: $$\operatorname{Res}\left(g(z),z=-2\right)= \frac{1}{2}\left.\frac{d^2}{dz^2} (z+2)^3 g(z)\right|_{z=-2}=\frac{1}{2}\left.\frac{d^2}{dz^2} \frac{\psi(-z)}{z+1}\right|_{z=-2}\tag{1}$$ so: $$\operatorname{Res}\left(g(z),z=-2\right)=\left.\frac{\psi(-z)}{(1+z)^3}+\frac{ \psi'(-z)}{(1+z)^2}+\frac{\psi''(-z)}{2(1+z)}\right|_{z=-2} = -\psi(2)+\psi'(2)-\frac{\psi''(2)}{2}$$ leading to: $$\begin{eqnarray*}\operatorname{Res}\left(g(z),z=-2\right)&=& -(1-\gamma)+\left(1-\zeta(2)\right)-\frac{\psi''(2)}{2}\\&=&\gamma-\zeta(2)+\sum_{n\geq 0}\frac{1}{(n+2)^3}\\&=&\color{red}{\gamma-\zeta(2)+\zeta(3)-1}.\tag{2}\end{eqnarray*}$$

Jack D'Aurizio
  • 353,855
  • I am trying to avoid the limit approach. Thanks anyway!This derivative method is too cumbersome without a calculator. How did you calculate $\psi"(2)$ without a calculator? – Amad27 Jan 13 '15 at 13:57
  • @Amad27: Since for any $z$ in the right halfplane $$\psi'(z)=\sum_{n\geq 0}\frac{1}{(n+z)^2},$$ it follows that $$\psi''(z)=-2\sum_{n\geq 0}\frac{1}{(n+z)^3},$$ hence $\psi''$ at the positive integers simply depends on $\zeta(3)$. – Jack D'Aurizio Jan 13 '15 at 14:00
  • @Amad27: Anyway, when dealing with a triple pole, to avoid derivatives is a mere dream. I think this method is way faster then rearranging the series for $\psi(z)$ through geometric series expansions. – Jack D'Aurizio Jan 13 '15 at 14:01
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    I will accept the answer because this is a nice method but otherwise, we could use the taylor series representation for $\psi(z + 1)$ – Amad27 Jan 13 '15 at 14:16
  • With this series:

    $$\psi(-z) = -\gamma - \sum_{k=1}^{\infty} \zeta(k+1)(z+1)^k$$

    First we divide by our part:

    $$\frac{\psi(-z)}{(z+1)(z+2)^3} = -\frac{\gamma}{(z+1)(z+2)^3} - \sum_{k=1}^{\infty} \frac{\zeta(k+1)(z+1)^{k-1}}{(z+2)^3}$$

    We see (z+1)=((z+2) - 1) And then take this to the third power we get a factor of $(z+2)^2$ somewhere.

    We have to let $k=3$ to get:

    $$... + \frac{\zeta(4)}{(z+2)} + ...$$

    But this gives the incorrect coefficient?

     – Amad27 Jan 13 '15 at 14:35
  • @Amad27: Nope, you have to multiply $$\psi(-z)=(1-\gamma)-\sum_{k=1}^{+\infty}(1-\zeta(k+1))z^k$$ and $$\frac{1}{z+1}=-\sum_{k\geq 0}(z+2)^k $$ then compute the coefficient of $(z+2)^2$. – Jack D'Aurizio Jan 13 '15 at 14:36
  • Why is my approach incorrect? – Amad27 Jan 13 '15 at 14:38
  • @Amad27: because your right hand sides are not power series in the variable $(z+2)$, they have to be rearranged to reach that form. – Jack D'Aurizio Jan 13 '15 at 14:38
  • It follows from definition though: http://en.wikipedia.org/wiki/Digamma_function#Taylor_series – Amad27 Jan 13 '15 at 14:39
  • @Amad27: I was talking about the expansion you gave for $g(z)$. It is not a power series, in that form. – Jack D'Aurizio Jan 13 '15 at 14:40
  • I thought: $$\frac{1}{a}\sum_{n=1}^{\infty} b_n = \sum_{n=1}^{infty} \frac{b_n}{a}$$ ?? – Amad27 Jan 13 '15 at 14:42
  • @Amad27: but in this particular case, both $a$ and $b_n$ depend on $z$, and we can extract a residue only from a power series, not from a generic series. – Jack D'Aurizio Jan 13 '15 at 14:46
  • $b_n$ depends on $n$ not $z$ though? – Amad27 Jan 13 '15 at 14:47
  • If $\sum b_n$ is the Taylor expansion for some function $h(z)$, $b_n$ does depend on $z$. Think about it, I have no time for digging into details you should have studied by yourself before approaching these questions. – Jack D'Aurizio Jan 13 '15 at 14:49
  • Consider the function $\frac{1}{1-x}$ we have:

    $$\frac{1}{1-x} = \sum_{n \ge 1} x^{n-1}$$ Yes okay, for a certain value of $x$ I see what you mean. But at the same time:

    $$\frac{x}{1-x} = \sum_{n \ge 1} x^{n}$$

    We took $x$ inside the sum, we can do the same with $$\frac{1}{(n+1)(n+2)^3}$$ by this logic cant we?

     – Amad27 Jan 13 '15 at 15:19
  • How about using the traditional definition of the constant $a_n$ given here: http://en.wikipedia.org/wiki/Laurent_series – Amad27 Jan 13 '15 at 15:25
  • So, if there are many series representing a function, such a function has many residues in $z=-2$? Nope, the issue is that $$\sum_{n\geq 0}\zeta(n) (z+2)^n (z+1)^{19} $$ is not a power series in $(z+2)$, end story. – Jack D'Aurizio Jan 13 '15 at 15:30
  • Okay. How about the approached used here: http://math.stackexchange.com/questions/1101277/sum-n-1-infty-frac1n1n23-using-complex-analysis/1101765#1101765 – Amad27 Jan 13 '15 at 15:33
  • @Amad27: I see nothing wrong in that approach. By Cauchy integral formula, we can compute derivatives (like I did) by integrating along a path. – Jack D'Aurizio Jan 13 '15 at 15:35
  • But he did not consider $f(z)$ completely there. He left out $\frac{1}{(z+1)}$ and what about the other power of (z+2)? – Amad27 Jan 13 '15 at 15:36
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    He was integrating $g(z)$ along a circle centered in the right point. Nothing wrong. By the way, why do you have two accounts? – Jack D'Aurizio Jan 13 '15 at 15:38
  • So he does not need to consider $\frac{1}{(z+1)(z+2)^3}$ I am using this account because I cannot access my account, this is not a personal computer, I do not want to use my credentials on this machine. – Amad27 Jan 13 '15 at 15:39
  • Jack D' Aurizio, how would you find the residue at a general $n$?

    http://math.stackexchange.com/questions/1103957/finding-the-residue-at-z-n

     – Amad27 Jan 14 '15 at 13:27
  • @Amad27: where are the singularities of $\phi(-z)$? can you prove they are all simple poles? – Jack D'Aurizio Jan 14 '15 at 13:54
  • Jack, yes because they are only undefined for one part, $\psi(z)$ – Amad27 Jan 14 '15 at 13:57
  • Then, since $\frac{1}{(z+1)(z+2)^3}$ is a regular function over $\Re(z)>0$, the residue of $g(z)$ in $z=n$ simply equals the residue of $\psi(-z)$ at $z=n$ times $\frac{1}{(n+1)(n+2)^3}$. This is the reason for using such function to encode the series Marko Riedel solved. – Jack D'Aurizio Jan 14 '15 at 14:02
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    You are amazing, upvotes * 1 million! but which law or property states this? Ive never seen this definition/ – Amad27 Jan 14 '15 at 14:05
  • Assume that $f(z)=g(z)h(z)$, where $g(z)$ is holomorphic in a neighbourhood of $z_0$ while $h(z)$ has a simple pole in $z_0$. Then $f(z)$ has a simple pole in $z_0$ too, and: $$\operatorname{Res}(f(z),z=z_0)=\lim_{z\to z_0}f(z)(z-z_0) = g(z_0)\lim_{z\to z_0}h(z)(z-z_0) = g(z_0)\operatorname{Res}(h(z),z=z_0)$$ by the definition of residue in a simple pole. – Jack D'Aurizio Jan 14 '15 at 14:19
  • Obviously, that does not hold anymore if $h(z)$ has a double pole or worse in $z_0$. – Jack D'Aurizio Jan 14 '15 at 14:21
  • Okay. So how do we calculate residue of digamma?

    $$Res = \lim_{z \to n} \psi(-z)(z-n)$$

    Somehow, it equals $1$ but I am not sure how. Please only hints, do not give a direct answer! Thanks!

     – Amad27 Jan 14 '15 at 14:23
  • @Amad27: I just gave you a complete answer in your other question, look at there. It is one, by the way, you're right. – Jack D'Aurizio Jan 14 '15 at 14:28
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    Ok nice! I used the series representation, the coefficient of $\frac{1}{z-n}$ is $1$ The question remains, what is the residue of $$\frac{(\psi(-z) + \gamma)^2}{(z+1)(z+2)^3}$$ at $z=-2$ I will try it, and post a question later. – Amad27 Jan 14 '15 at 14:32
  • Jack, can you help here: http://math.stackexchange.com/questions/1104046/residue-of-frac-psi-z-gamma2z1z23-at-z-n I did some work but I am not sure how to proceed with the other part? – Amad27 Jan 14 '15 at 14:41
  • I cant figure out the residue at $z=-2$ can you help? I tried the limit way but the square creates an issue. What should I do? – Amad27 Jan 14 '15 at 14:58
  • I just remarked that my lemma works for simple poles, but obviously when introducing a square such a function has a double pole in $z=n\geq 0$. Do you want also the residue in $z=-2$? In such a case the numerator is a holomorphic function and the remaining part has a triple pole. Multiply by $(z+2)^3$ and differentiate twice. It should be clear now. – Jack D'Aurizio Jan 14 '15 at 15:01
  • Differentiation is insanely hard. – Amad27 Jan 14 '15 at 15:02
  • In such a case I cannot help you. – Jack D'Aurizio Jan 14 '15 at 15:04
  • Actually, the lemma you stated should work for all. Robjohn did the same: http://math.stackexchange.com/questions/1081800/explanation-for-summation-complex-analysis-method – Amad27 Jan 14 '15 at 15:04