Representing a Function as Laurent Series: Example

[Petar Mali]

Homework Statement

How to represent function

\frac{1}{e^x-x-1}

in form of Laurent series around point 0

Homework Equations

Laurent series

f(z)=\sum^{\infty}_{n=-\infty}a_n(z-z_0)^n

Here is z_0=0

The Attempt at a Solution

Computer gives

\frac{2}{x^2}-\frac{2}{3 x}+\frac{1}{18}+\frac{x}{270}-\frac{x^2}{3240}-\frac{x^3}{13608}-\frac{x^4}{2041200}+\frac{x^5}{874800}+\frac{13 x^6}{146966400}-\frac{307 x^7}{24249456000}-\frac{479 x^8}{203695430400}+O[x]^9

in form of 12 first members in series.

e^{x}=1+x+\frac{x^2}{2!}+...

so I can say

e^x-x-1=\sum^{\infty}_{n=2}\frac{x^n}{n!}

\frac{1}{e^x-x-1}=\frac{1}{\sum^{\infty}_{n=2}\frac{x^n}{n!}}

But I don't know what to do with that.

 

[praharmitra]

Take 2/x^2 common, the expression becomes

\frac{2}{x^2}\left(1+\frac{x}{3}+\frac{x^2}{12}+...\right)^{-1} = \frac{2}{x^2}(1+y)^{-1}
where
y = \frac{x}{3}+\frac{x^2}{12}+...

Now expand (1+y)^{-1} and then substitute back.

 

[Petar Mali]

Take 2/x^2 common, the expression becomes

\frac{2}{x^2}\left(1+\frac{x}{3}+\frac{x^2}{12}+...\right)^{-1} = \frac{2}{x^2}(1+y)^{-1}
where
y = \frac{x}{3}+\frac{x^2}{12}+...

Now expand (1+y)^{-1} and then substitute back.

Ok. I will get

\frac{2}{x^2}\sum^{\infty}_{n=0}(-1)^n(\frac{x}{3}+\frac{x^2}{12}+...)^n

But I don't have this form

<br /> \frac{2}{x^2}-\frac{2}{3 x}+\frac{1}{18}+\frac{x}{270}-\frac{x^2}{3240}-\frac{x^3}{13608}-\frac{x^4}{2041200}+\frac{x^5}{874800}+\frac{13 x^6}{146966400}-\frac{307 x^7}{24249456000}-\frac{479 x^8}{203695430400}+O[x]^9<br />

and I want to do that analyticaly from

<br /> \frac{1}{e^x-x-1}<br />

 

[praharmitra]

So open up the brackets and expand it further. You will get that.

 

[Petar Mali]

Open up the brackets?

 

[praharmitra]

Open up
<br /> (\frac{x}{3}+\frac{x^2}{12}+...)^n<br />
for the first few terms. then do the summation.

 

[Petar Mali]

Listen to me. I have just this\frac{1}{e^x-x-1}

 

[praharmitra]

I understand. What I am asking you to do is simple. Lots of irritating calculations, but simple.

Let me work out the first two terms for you (upto {\cal O}(1)). Do you agree that you got the following?

\frac{2}{x^2}(1+y)^{-1} = \frac{2}{x^2}(1-y+y^2-y^3+...)

Now plug in the expression for y

first term is
<br /> y = (\frac{x}{3}+\frac{x^2}{12}+...)<br />
second is
y^2 = \frac{x^2}{9} + ...

Note that you don't require to calculate any higher terms since those will just contribute to x^3 and higher. Now plug it in

\frac{2}{x^2}(1-y+y^2-y^3+...) = \frac{2}{x^2}\left(1 - \frac{x}{3} - \frac{x^2}{12} + \frac{x^2}{9}\right)

\Rightarrow = \frac{1}{e^x-x-1} = \frac{2}{x^2} - \frac{2}{3x}+\frac{1}{18} +...

And to go up one higher order you will have to go till y^3, and to next order in the expansion of y. Got it?

 

[Petar Mali]

I understand what you trying to tell me. But

<br /> <br /> \frac{2}{x^2}-\frac{2}{3 x}+\frac{1}{18}+\frac{x}{270}-\frac{x^2}{3240}-\frac{x^3}{13608}-\frac{x^4}{2041200}+\frac{x^5}{874800}+\frac{13 x^6}{146966400}-\frac{307 x^7}{24249456000}-\frac{479 x^8}{203695430400}+O[x]^9<br /> <br />


this relation is given by the computer and I can't use it.

I must take

<br /> \frac{1}{e^x-x-1} <br />
and get from that Laurent series!

 

[praharmitra]

I am not asking you to use it! I am asking you to calculate by series expanding by hand! And then VERIFY your answer by matching it with what the computer has given you. The method I have written is an analytic series expansion that I have done by hand. I haven't used any computer to do any part of my calculations.

This calculation is exactly that of getting the Laurent Series.

 

[Petar Mali]

Ok just tell me

\frac{1}{e^x-x-1}=...

what you have done before you put \frac{2}{x^2}?

 

[praharmitra]

Write
<br /> e^x = 1+x+\frac{x^2}{2!}+...<br />

 

[vela]

You figured out

e^x-x-1 = \frac{x^2}{2!}+\frac{x^3}{3!}+\cdots = \frac{x^2}{2}\left(1+\frac{2x}{3!}+\frac{2x^2}{4!}+\cdots\right)

so

\frac{1}{e^x-x-1} = \frac{1}{\frac{x^2}{2}\left(1+\frac{2x}{3!}+\frac{2x^2}{4!}+\cdots\right)} = \frac{2}{x^2}\,\frac{1}{1+\left(\frac{2x}{3!}+\frac{2x^2}{4!}+\cdots\right)}

Now compare this to what praharmitra wrote in post 8.