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

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

and I want to do that analyticaly from

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

 

[praharmitra]

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

 

[Petar Mali]

Open up the brackets?

 

[praharmitra]

Open up
$$(\frac{x}{3}+\frac{x^2}{12}+...)^n$$
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
$$y = (\frac{x}{3}+\frac{x^2}{12}+...)$$
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

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


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

I must take

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

 

[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.