What Are the Radii of Convergence for a Laurent Series Around a Singular Point?

Result is correct, and the first conclusion too. Inner radius can be greater than zero only if Laurent series don't terminate at negative direction. Now the smallest order term in the Laurent series is of order -1, so inner radius is zero.The outer radius is infinite, but you cannot conclude it from the fact that z=1/2 is a pole. Instead it is seen most easily from the fact, that the series of sine have the factorials 1/n! which make series convergent everywhere.Then for example exp(1/z) has infinite negative terms and so has an inner radius higher than zero?No.This claim is right: If Laurent series has only finite amount of non-zero terms with negative exponent, then the inner
  • #1
naggy
60
0
I'm puzzled by one thing concerning Laurent series.

If I have a series, for example f(z) =(z*sinz)/(2z-1) and I'm supposed to make a laurent series of f about the point z=1/2.

Now, what would the inner and outer radius of convergence be?

I would say that since z=1/2 is a pole, the inner radius is zero and the outer radius infinite?

If not, then how can I see the radius of convergence of a Laurent series?
 
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  • #2
naggy said:
I would say that since z=1/2 is a pole, the inner radius is zero and the outer radius infinite?

Result is correct, and the first conclusion too. Inner radius can be greater than zero only if Laurent series don't terminate at negative direction. Now the smallest order term in the Laurent series is of order -1, so inner radius is zero.

The outer radius is infinite, but you cannot conclude it from the fact that z=1/2 is a pole. Instead it is seen most easily from the fact, that the series of sine have the factorials 1/n! which make series convergent everywhere.

If not, then how can I see the radius of convergence of a Laurent series?

Most of the time it is best to try to see them by using previous knowledge of already known functions, but when all else fails, the outer and inner radius, [itex]\rho_O[/itex] and [itex]\rho_I[/itex], of a Laurent series

[tex]
\sum_{n=-\infty}^{\infty} a_n (z-z_0)^n
[/tex]

are given by the formulas

[tex]
\rho_O = \frac{1}{\underset{n\to\infty}{\textrm{lim sup}}\; \sqrt[n]{|a_n|}}
[/tex]

[tex]
\rho_I = \underset{n\to\infty}{\textrm{lim sup}}\; \sqrt[n]{|a_{-n}|}
[/tex]
 
  • #3
jostpuur said:
Result is correct, and the first conclusion too. Inner radius can be greater than zero only if Laurent series don't terminate at negative direction. Now the smallest order term in the Laurent series is of order -1, so inner radius is zero.

The outer radius is infinite, but you cannot conclude it from the fact that z=1/2 is a pole. Instead it is seen most easily from the fact, that the series of sine have the factorials 1/n! which make series convergent everywhere.
[/tex]

Then for example exp(1/z) has infinite negative terms and so has an inner radius higher than zero?


Another thing. If I wanted to find the residue of the pole(z=1/2) by writing the function as an infinite series, how would I do that? I would have to write it in terms of (z-1/2) right?

I tried doing it for sine about z=1/2 but the the denominator canceled and I did not gera a -1 order term in the series.
 
  • #4
naggy said:
Then for example exp(1/z) has infinite negative terms and so has an inner radius higher than zero?

No.

This claim is right: If Laurent series has only finite amount of non-zero terms with negative exponent, then the inner radius is zero.

This claim is wrong: If Laurent series has infinite amount of non-zero terms with negative exponent, then the inner radius is greater than zero.

These first claim does not imply the second.

(edit: and my original comment
jostpuur said:
Inner radius can be greater than zero only if Laurent series don't terminate at negative direction.
is equivalent with the first claim)

Another thing. If I wanted to find the residue of the pole(z=1/2) by writing the function as an infinite series, how would I do that? I would have to write it in terms of (z-1/2) right?

Yes. You need to solve the coefficient [itex]a_{-1}[/itex] from the series

[tex]
\frac{z\sin(z)}{2z-1} \;=\; \frac{a_{-1}}{z-\frac{1}{2}} \;+\; a_0 \;+\; a_1(z-\frac{1}{2}) \;+\; a_2(z-\frac{1}{2})^2 \;+\; \cdots
[/tex]

I tried doing it for sine about z=1/2 but the the denominator canceled and I did not gera a -1 order term in the series.

I believe you made a mistake. [itex]\frac{1}{2}\sin(\frac{1}{2})\neq 0[/itex], so the -1 order term should be there.
 
  • #5
jostpuur said:
I believe you made a mistake. [itex]\frac{1}{2}\sin(\frac{1}{2})\neq 0[/itex], so the -1 order term should be there.

I don´t understand how that leads to the a-1 term. Is it because then the function would have a rectifiable singularity, that is the limit would exist as z goes to 1/2?

These things are confusing me a bit. For example I don´t get a a-1 term in a function like f(z) = 1/(z-i) if I expand it as a geometric series. Well, I get an a-1 term but an infinite number of negative powers, and that, by definition, should not give me a pole, but something called an essential singularity.

But I KNOW that z=i is a pole, since the function goes to infinity as z->i
 
  • #6
naggy said:
I don´t understand how that leads to the a-1 term. Is it because then the function would have a rectifiable singularity, that is the limit would exist as z goes to 1/2?

I'm not sure what rectifiable singularity means. If [itex]f(z_0)=0[/itex], then [tex]\frac{f(z)}{z-z_0}[/tex] has a removable singularity at [itex]z_0[/itex], at least. It sounds like you understood this already, but merely weren't sure?

These things are confusing me a bit. For example I don´t get a a-1 term in a function like f(z) = 1/(z-i) if I expand it as a geometric series.

This function already is in the form of Laurent series. It is

[tex]
\frac{1}{z-i} = \sum_{n=-\infty}^{\infty} a_n (z-i)^n
[/tex]

with coefficients [itex]a_{-1} = 1[/itex] and [itex]a_n = 0[/itex] for [itex]n\neq -1[/itex].

If you did this:

[tex]
\frac{1}{z-i} = \frac{i}{1 + iz} = i\sum_{n=0}^{\infty} (-iz)^n = \sum_{n=0}^{\infty} (-1)^n i^{n+1} z^n
[/tex]

then you have expanded the function as Laurent series around the point [itex]z=0[/itex], and not around the point [itex]z=i[/itex]. In this case the inner radius of convergence is 0, and outer radius is 1. There does not exist [itex]z^{-1}[/itex] term with non-zero coefficient, because the original function doesn't have singularity at [itex]z=0[/itex]. The singularity is at [itex]z=i[/itex], and it has the effect that the (open) domain of convergence cannot be larger than [itex]\{z\in\mathbb{C}\;|\; |z|<1\}[/itex], because a larger ball would contain the [itex]z=i[/itex]. The singularity prevents the domain of convergence from being larger.

Well, I get an a-1 term but an infinite number of negative powers, and that, by definition, should not give me a pole, but something called an essential singularity.

Now I'm guessing, but do you mean series like this?

[tex]
\frac{1}{z-i} = \frac{1}{z} \frac{1}{1 - \frac{i}{z}} = \frac{1}{z} \sum_{n=0}^{\infty} \Big(\frac{i}{z}\Big)^n = \sum_{n=-\infty}^{-1} (-i)^{n+1} z^n
[/tex]

Here the function is written as Laurent series around the point [itex]z=0[/itex] so that the (open) domain of convergence is [itex]\{z\in\mathbb{C}\;|\; |z|>1\}[/itex]. That means the inner radius of convergence is 1, and the outer radius is [itex]\infty[/itex]. The original function doesn't have essential singularity at [itex]z=0[/itex] despite these negative powers in this expansion, because now [itex]z=0[/itex] is outside (and not even on the boundary) the domain of convergence. To see what happens at the [itex]z=0[/itex], you need to use the Laurent series which converge for [itex]|z|<1[/itex].
 
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FAQ: What Are the Radii of Convergence for a Laurent Series Around a Singular Point?

What is a Laurent series?

A Laurent series is a mathematical representation of a complex function that includes both positive and negative powers of the variable, unlike a Taylor series which only includes positive powers. It is written in the form of ∑n=-∞cn(z-a)n, where cn are complex coefficients and a is the center of the series.

What is the difference between a Laurent series and a Taylor series?

The main difference between a Laurent series and a Taylor series is that a Laurent series includes both positive and negative powers of the variable, while a Taylor series only includes positive powers. This allows Laurent series to represent functions with singularities or poles, while Taylor series can only represent analytic functions with no singularities.

How is a Laurent series used in complex analysis?

Laurent series are used in complex analysis to represent complex functions and to study their properties. They are particularly useful for functions that have singularities or poles, as they allow for a more complete understanding of the behavior of the function in the complex plane.

What is the process for finding the Laurent series of a function?

The process for finding the Laurent series of a function involves first determining the center of the series and then calculating the complex coefficients cn for both the positive and negative powers of the variable. This can be done using various techniques such as using the Cauchy integral formula or using algebraic manipulations to rewrite the function in a form suitable for a Laurent series.

Can a function have more than one Laurent series?

Yes, a function can have more than one Laurent series depending on the choice of center and the domain of the function. In general, a function will have a unique Laurent series within a specific domain, but may have different Laurent series in different domains due to the presence of singularities or poles.

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