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Find the residue of:

$$f(z) = \frac{(\psi(-z) + \gamma)^2}{(z+1)(z+2)^3} \space \text{at} \space z=n \space \text{for a general} \space n$$

How can I start? Using advice from Jack D Aurizio, I get:

$$\mathrm{Res}_{z=n}\space f(z) = \frac{1}{(n+1)(n+2)^3} \cdot \mathrm{Res}_{z=n} (\psi(-z) + \gamma)^2 $$

The question remains, how do I find:

$$\mathrm{Res}_{z=n} (\psi(-z) + \gamma)^2$$

??

Amad27
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2 Answers2

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Wait, my advice does not work in such a case, because $(\psi(-z)+\gamma)^2$ has a double pole in $z=n$, due to: $$ g(z)=\gamma+\psi(-z)= \frac{1}{z}+\sum_{n\geq 1}\left(\frac{1}{n}+\frac{1}{z-n}\right).$$

Anyway, given that $z=n$ is a double pole for $(\psi(-z)+\gamma)^2$ and $h(z)=\frac{1}{(z+1)(z+2)^3}$ is holomorphic over $\Re(z)>-1$, then $z=n$ is a double pole for $f(z)$, so:

$$\operatorname{Res}\left(f(z),z=n\right) = \frac{d}{dz}\left.(z-n)^2 f(z)\right|_{z=n} = \left.2(z-n)f(z)\right|_{z=n}+\left.(z-n)^2 f'(z)\right|_{z=n}\tag{1}$$ where: $$ \left. 2(z-n)f(z)\right|_{z=n} = 2\operatorname{Res}(f(z),z=n)=\frac{2}{(n+1)(n+2)^3}$$ by the correct version of my advice (simple poles), and: $$ \left.(z-n)^2 f'(z)\right|_{z=n} = \left. (z-n)^2\left(g'(z) h(z)+g(z) h'(z)\right)\right|_{z=n}=h(n)\cdot\left.(z-n)^2g'(z)\right|_{z=n}.\tag{2} $$ Since: $$ g'(z) = -\sum_{n\geq 0}\frac{1}{(z-n)^2} $$ we have that $(2)$ just equals $\frac{1}{(n+1)(n+2)^3}$, so: $$ \operatorname{Res}\left( f(z), z=n \right) = \color{red}{\frac{1}{(n+1)(n+2)^3}} \tag{3} $$ like in the case when $g(z)$ was not squared.

Jack D'Aurizio
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  • And what about $z=-2$? – Amad27 Jan 14 '15 at 15:01
  • I dont think this is correct. There should be a $H_n$ in the numerator – Amad27 Jan 14 '15 at 16:30
  • I see, you didnt square it? Why no $H_n$ there? Also, When you said:

    $2\mathrm{Res}(f(z), z =n)$$ Do you mean the SINGLE ORDER residue? How? The function $f(z)$ is still of order two poles.?

     – Amad27 Jan 15 '15 at 14:05
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Use the series for $\psi(-z)+\gamma$, then square it.

We only need the first few terms to find the residue because it is the coefficient of the 1/z term.

$$\psi(-z)+\gamma=\frac{1}{z-n}+2H_{n}+\cdot\cdot\cdot$$

$$(\psi(-z)+\gamma)^{2}=\left(\frac{1}{z-n}+2H_{n}+\cdot\cdot\cdot \right)\left(\frac{1}{z-n}+2H_{n}+\cdot\cdot\cdot \right)$$

$$=\frac{1}{(z-n)^{2}}+\frac{2H_{n}}{z-n}+\cdot\cdot\cdot $$

Thus, the residue at $z=n$ is $$2H_{n}$$

Cody
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  • then how about $$\frac{(\psi(-z) + \gamma)^2}{z^2}$$? Will it be $\frac{2H_n}{n^2}$ or something else? – Amad27 Jan 15 '15 at 13:29
  • The series for $\psi(-z)+\gamma$ is: $$\frac{1}{z-n}+H_{n}+\sum_{k=1}^{\infty}(-1)^{k}H_{n}^{(k+1)}-\zeta(k+1)^{k}, ;\ n\geq 0$$. Just manipulate accordingly, then look at the coefficient of the 1/z term to find the residue. Here is a paper you will like if you do not already have it. http://algo.inria.fr/flajolet/Publications/FlSa98.pdf – Cody Jan 15 '15 at 22:13