Laurent series throwing away terms

[Meggle]

Laurent series "throwing away" terms

Homework Statement

Veeeery similar to https://www.physicsforums.com/showthread.php?p=1868354#post1868354":

Determine the Laurent series and residue for f(z) = \frac{1}{(e^{z} - 1)^{z}}

Homework Equations

We know that the Taylor series expansion of e^{z} is = 1 + z + z^{2}/2! + ...

The Attempt at a Solution

Dick responded with the advice "You expand around z=0. exp(z)-1=z+z^2/2!+...=z(1+z/2!+...). Write that as z(1+a) where a=z/2!+z^2/3!+... So 1 over that squared is (1/z^2)*(1/(1+a)^2). The series expansion for 1/(1+a)^2 is 1-2a+3a^2+... As you said, you only need the first few terms. Now start throwing away terms that you know won't contribute to the terms you need. E.g. a^2 starts with a z^2 term."

So I have:
\frac{1}{(e^{z} - 1)^{z}} = \frac{1}{z^{2}} [1 - 2a + 3a^{2} ...]

= \frac{1}{z^{2}} [1 - \frac{2z}{2!} + \frac{3z^{2}}{2!} ...
-\frac{2z^{2}}{3!} + \frac{3z^{4}}{3!}...]

But now I'm stuck again. How do I know what to throw away? It doesn't looks like terms cancel. How do I know terms won't contribute to the term I need?

 

[jackmell]

I would say there isn't a Laurent series about any point 2n pi because the function is multi-valued and neither are there any (isolated) poles with residues. You can however compute a Taylor series for a particular analytic "determination" away from and not including any branch-points.

Also, the function:

f(z)=\frac{1}{(e^z-1)^z}

is qualitatively different than:

g(z)=\frac{1}{(e^z-1)^2}

The function g(z) is analytic except for poles at 2npi. However, the function

f(z)=e^{-z\log(e^z-1)}

is multi-valued because of the multivalued log term and the points at 2n\pi are non-isolated branch-point singularities.

 

[vela]

Homework Statement

Veeeery similar to https://www.physicsforums.com/showthread.php?p=1868354#post1868354":

Determine the Laurent series and residue for f(z) = \frac{1}{(e^{z} - 1)^{z}}

Homework Equations

We know that the Taylor series expansion of e^{z} is = 1 + z + z^{2}/2! + ...

The Attempt at a Solution

Dick responded with the advice "You expand around z=0. exp(z)-1=z+z^2/2!+...=z(1+z/2!+...). Write that as z(1+a) where a=z/2!+z^2/3!+... So 1 over that squared is (1/z^2)*(1/(1+a)^2). The series expansion for 1/(1+a)^2 is 1-2a+3a^2+... As you said, you only need the first few terms. Now start throwing away terms that you know won't contribute to the terms you need. E.g. a^2 starts with a z^2 term."

So I have:
\frac{1}{(e^{z} - 1)^{z}} = \frac{1}{z^{2}} [1 - 2a + 3a^{2} ...]

= \frac{1}{z^{2}} [1 - \frac{2z}{2!} + \frac{3z^{2}}{2!} ...
-\frac{2z^{2}}{3!} + \frac{3z^{4}}{3!}...]

But now I'm stuck again. How do I know what to throw away? It doesn't looks like terms cancel. How do I know terms won't contribute to the term I need?

I'll assume the exponent in the denominator is supposed to be 2, not z. You have

\frac{1}{(e^z-1)^2} = \frac{1}{(z+z^2/2!+z^3/3!+\cdots)^2} = \frac{1}{z^2(1+z/2!+z^2/3!+\cdots)^2}

In your work, you use a=z, but it's actually equal to a=z/2!+z²/3!+... When you do the expansion, you need to find powers of that series.

 

[Meggle]

I'll assume the exponent in the denominator is supposed to be 2, not z. You have

\frac{1}{(e^z-1)^2} = \frac{1}{(z+z^2/2!+z^3/3!+\cdots)^2} = \frac{1}{z^2(1+z/2!+z^2/3!+\cdots)^2}

In your work, you use a=z, but it's actually equal to a=z/2!+z²/3!+... When you do the expansion, you need to find powers of that series.

Yes, sorry, the exponent in the denominator is supposed to be 2.

I was trying to show using a=z/2!+z^2/3!+... in the last two lines, i.e. the first line was for z/2!, the second line was for z^2/3!...
So what I have is:

\frac{1}{(e^{z}-1)^{2}} = \frac{(1 - 2(\frac{z}{2!}) + 3(\frac{z}{2!})^{2} -2(\frac{z^{2}}{3!}) - 2(\frac{z^{3}}{4!}) +...)}{z^{2}}

(Because I think my expansion of \frac{1}{(1+a)^{2}} is in positive powers of z? Using \frac{1}{(1+z)^{2}} = \sum^{\infty}_{n=0} (-1)^{n}(n+1)z^{n} for modulus z <1 ) )

\frac{1}{(e^{z}-1)^{2}} = \frac{1}{z^{2}}(1 - z + (\frac{3}{2!} - \frac{2}{3!})z^{2} -\frac{z^{3}}{2} + (\frac{3}{3!} - \frac{2}{4!})z^{2}...)

\frac{1}{(e^{z}-1)^{2}} = \frac{1}{z^{2}} + \frac{1}{z} + (\frac{3}{2!} - \frac{2}{3!}) -\frac{z}{2} + (\frac{3}{3!} - \frac{2}{4!})z^{2} ...

I get that the first two terms of the last one are negative powers of z and then there's a term where z^{0} followed by positive powers of z. I've been trying to make that into a power series (either as one piece or with a_{n} with positive powers and b_{n} overnight but I can't seem to work it out.

(And the preview function completely isn't working, so hopefully the equations will come out ok!)

 

[vela]

The preview thing is a bug. Just refresh the page, and the images will update to the correct rendering.

I don't think you're going to find a simple expression for the coefficients. Usually in this type of problem, you just work out the first few terms to demonstrate you know what you're doing. If you need to find the residue, you need to get b₁.

It helps to realize you can factor z out of a, so the aⁿ term will only contribute to terms in the series of order zⁿ and higher.

Just so you know what you're shooting for, this is what Mathematica gave for the series expansion:

\frac{1}{e^z-1} = \frac{1}{z^2}-\frac{1}{z}+\frac{5}{12}-\frac{z}{12}+\frac{z^2}{240}+\cdots