Representing a Function as Laurent Series: Example

In summary: You should get1+\left(\frac{2x}{3!}+\frac{2x^2}{4!}+\cdots\right) = 1+\frac{x}{3}+\frac{x^2}{12}+\cdotsCan you take it from here?In summary, to represent the given function \frac{1}{e^x-x-1} in form of Laurent series around point 0, we can expand it as \frac{2}{x^2}(1+y)^{-1}, where y = \frac{x}{3}+\frac{x^2}{12}+... Then, by plugging in the expression for y and expanding it further, we get the Laurent
  • #1
Petar Mali
290
0

Homework Statement



How to represent function

[tex]\frac{1}{e^x-x-1}[/tex]

in form of Laurent series around point [tex]0[/tex]

Homework Equations



Laurent series

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

Here is [tex]z_0=0[/tex]



The Attempt at a Solution



Computer gives

[tex]\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[/tex]

in form of [tex]12[/tex] first members in series.

[tex]e^{x}=1+x+\frac{x^2}{2!}+...[/tex]

so I can say

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

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

But I don't know what to do with that.
 
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  • #2
Take [tex]2/x^2[/tex] common, the expression becomes

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

Now expand [itex](1+y)^{-1}[/itex] and then substitute back.
 
  • #3
praharmitra said:
Take [tex]2/x^2[/tex] common, the expression becomes

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

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

Ok. I will get

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

But I don't have this form

[tex]
\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
[/tex]

and I want to do that analyticaly from

[tex]
\frac{1}{e^x-x-1}
[/tex]
 
  • #4
So open up the brackets and expand it further. You will get that.
 
  • #5
Open up the brackets?
 
  • #6
Open up
[tex]
(\frac{x}{3}+\frac{x^2}{12}+...)^n
[/tex]
for the first few terms. then do the summation.
 
  • #7
Listen to me. I have just this[tex]\frac{1}{e^x-x-1}[/tex]
 
  • #8
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 [itex]{\cal O}(1)[/itex]). Do you agree that you got the following?

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

Now plug in the expression for y

first term is
[tex]
y = (\frac{x}{3}+\frac{x^2}{12}+...)
[/tex]
second is
[tex]y^2 = \frac{x^2}{9} + ...[/tex]

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

[tex]\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)[/tex]

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

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?
 
  • #9
I understand what you trying to tell me. But

[tex]

\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

[/tex]


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

I must take

[tex]
\frac{1}{e^x-x-1}
[/tex]
and get from that Laurent series!
 
  • #10
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.
 
  • #11
Ok just tell me

[tex]\frac{1}{e^x-x-1}=...[/tex]

what you have done before you put [tex]\frac{2}{x^2}[/tex]?
 
  • #12
Write
[tex]
e^x = 1+x+\frac{x^2}{2!}+...
[/tex]
 
Last edited:
  • #13
You figured out

[tex]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)[/tex]

so

[tex]\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)}[/tex]

Now compare this to what praharmitra wrote in post 8.
 

1. What is a Laurent series and how is it different from a Taylor series?

A Laurent series is a type of power series representation of a complex function, which includes both positive and negative powers of the independent variable. It is similar to a Taylor series, but a Taylor series only includes positive powers of the independent variable, making it a special case of a Laurent series.

2. When is it useful to represent a function as a Laurent series?

A Laurent series is useful for representing functions that have singularities or poles, as it allows us to analyze the behavior of the function near these points. It is also useful for studying functions that have a wide range of values, such as periodic functions.

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

The process for finding the Laurent series representation of a function involves first identifying any singularities or poles of the function. Then, using the properties of Laurent series, we can expand the function as a sum of terms containing positive and negative powers of the independent variable. The coefficients of these terms can be found using the Cauchy integral formula or by differentiating or integrating the function.

4. How accurate is the Laurent series representation of a function?

The accuracy of the Laurent series representation depends on the convergence of the series. If the series converges, then it can provide an exact representation of the function. However, if the series does not converge, then it may only be an approximation of the function near a certain point.

5. In what areas of science is the use of Laurent series common?

Laurent series are commonly used in mathematics, physics, and engineering, particularly in the study of complex functions and their behavior near singularities. They are also used in fields such as signal processing, control theory, and fluid dynamics.

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