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
Amad27
412
1

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##.
 

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

1. What is a Laurent expansion?

A Laurent expansion is a mathematical technique used to represent a complex function in terms of its singularities, or points where the function is not defined, such as poles or branch points.

2. Why is it important to find the residue of a function?

The residue of a function at a specific point can provide important information about the behavior of the function at that point, including whether it has a pole or a singularity. It is also essential in calculating integrals involving complex functions.

3. How do you find the residue of a function using Laurent expansion?

To find the residue of a function at a specific point, you can use the formula: Res(f,z0) = limz→z0[(z-z0)f(z)], where z0 is the point of interest. This involves finding the Laurent expansion of the function and then identifying the coefficient of the term with the highest negative power of z.

4. What is the significance of finding the residue at z=-2?

The point z=-2 is important because it is a singularity of the function. By finding the residue at this point, we can determine if it is a pole or a removable singularity, which can have different implications for the behavior of the function.

5. Can you explain the practical applications of finding the residue of a function?

The residue of a function has many practical applications in various fields, such as physics, engineering, and finance. For example, in physics, the residue is used in calculating the probability of a particle interacting with a potential, while in finance, it is used in options pricing models. Additionally, the residue can also be used to solve differential equations and in the study of complex analysis.

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