Find the residue of g(z) at z=-2 using Laurent Expansion

In summary: Once you have that information, you can then use standard techniques of analytic continuation to extend the series to other regions.
  • #1
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Homework Statement


Find the residue at z=-2 for

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



Homework Equations



$$\psi(-z)$$ represents the digamma function, $$\zeta(z)$$ represents the Riemann-Zeta-Function.

The Attempt at a Solution


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 at z=-2

The problem is I can't ever get a factor of $$\frac{1}{z+2}$$ because of the cubed (z+2) in the bottom. What should I do?
 
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  • #2
The residue for a pole of order m at k is given by:
##lim _{z->k} [\frac{1}{(m-1)!} \frac{ d^{m-1}}{dz^{m-1}} ((z-k)^m f(z))]##
For this case, what is m? k = -2. Do you know the derivatives of ##\psi(z)##?
 
  • #3
Amad27 said:

Homework Statement


Find the residue at z=-2 for

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



Homework Equations



$$\psi(-z)$$ represents the digamma function, $$\zeta(z)$$ represents the Riemann-Zeta-Function.

The Attempt at a Solution


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 at z=-2

The problem is I can't ever get a factor of $$\frac{1}{z+2}$$ because of the cubed (z+2) in the bottom. What should I do?

First: be very careful when using a series expansion such as
[tex] \psi(z+1) = -\gamma - \sum_{k=1}^{\infty} (-1)^k\zeta(k+1)z^k [/tex]
This may have limited applicability (that is, may be valid only in a subset of ##\mathbb{C}##) and may need to be extended to other regions by some standard techniques of "analytic continuation". Basically, you need to know whether or not ##\psi(-z)## has singularities, and where they are. For example, can it have a pole at ##z = -2##? If so, that would change the residue.

So, your first order of business should be to understand fully the nature of the singularities of ##g(z)##, at least near ##z = -2##.
 

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