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Laurent series around infinity

  • #1
141
5
Laurent series at infinity point

I already calculated it, but my work was too long, I really wish to find a shorter route.
Calculate the Laurent series of [itex]\frac{1}{(z^{2}+1)^{2}}[/itex] around [itex]z_{0} = 0[/itex]
First, I used simple fractions and I got:
[itex]\frac{1}{(z^{2}+1)^{2}}=\frac{-i}{4} \frac{1}{z-i} + \frac{i}{4} \frac{1}{z+i} + \frac{-1}{4} \frac{1}{(z-i)^{2}} + \frac{-1}{4} \frac{1}{(z+i)^{2}}[/itex]
Then I did:
[itex]\frac{1}{z-i}=\frac{1}{-i(1+iz)}= i \displaystyle \sum_{n=0}^{\infty} (-iz)^{n}[/itex]
[itex]\frac{1}{z+i}=\frac{1}{i(1-iz)}= -i \displaystyle \sum_{n=0}^{\infty} (iz)^{n}[/itex]
[itex]\frac{1}{(z-i)^{2}} = \frac{\partial}{\partial z} ( - \frac{1}{z-i} )[/itex]
[itex]\frac{1}{(z+i)^{2}} = \frac{\partial}{\partial z} ( - \frac{1}{z+i} )[/itex]
Then we got:
[itex]\frac{1}{(z-i)^{2}} = \frac{\partial}{\partial z}( -i \displaystyle \sum_{n=0}^{\infty} (-iz)^{n} ) = -i \displaystyle \sum_{n=0}^{\infty} (n+1) (-i) (-iz)^{n}[/itex]
[itex]\frac{1}{(z+i)^{2}} = \frac{\partial}{\partial z}( i \displaystyle \sum_{n=0}^{\infty} (iz)^{n} ) = i \displaystyle \sum_{n=0}^{\infty} (n+1) (i) (iz)^{n}[/itex]
Then we have
[itex]\frac{1}{4} (-i i \displaystyle \sum_{n=0}^{\infty} (-iz)^{n} + i(-i) \displaystyle \sum_{n=0}^{\infty} (iz)^{n} + (-1) (-i) \displaystyle \sum_{n=0}^{\infty} (n+1) (-i) (-iz)^{n} + (-1) i \displaystyle \sum_{n=0}^{\infty} (n+1) (i) (iz)^{n} =[/itex]
[itex]= \frac{1}{4} (\displaystyle \sum_{n=0}^{\infty} (-iz)^{n} + \displaystyle \sum_{n=0}^{\infty} (iz)^{n} + i \displaystyle \sum_{n=0}^{\infty} (n+1) (-i) (-iz)^{n} + (-i) \displaystyle \sum_{n=0}^{\infty} (n+1) (i) (iz)^{n} ) =[/itex]
[itex]= \frac{1}{4} (\displaystyle \sum_{n=0}^{\infty} (-iz)^{n} + (iz)^{n} + i(n+1) (-i) (-iz)^{n} + (-i) (n+1) (i) (iz)^{n} ) = [/itex]
[itex]= \frac{1}{4} (\displaystyle \sum_{n=0}^{\infty} ((-iz)^{n} + (iz)^{n}) + (n+1)((-iz)^{n} + (iz)^{n}) )) =[/itex]
[itex]= \frac{1}{4} (\displaystyle \sum_{n=0}^{\infty} (n+2)((-iz)^{n} + (iz)^{n}))[/itex]
So we have [itex]a_{n}=0[/itex] [itex]n=2k+1[/itex] and if [itex]n=2k[/itex] [itex]a_{n} = (2n+4)/4 = 2k+1[/itex]
This is the Taylor series around [itex]z_{0}=0[/itex]
We want the Laurent series around [itex]z_{0} = \infty[/itex] so we do:
[itex]\frac{1}{(z^{2}+1)^{2}} = \frac{1}{z^{4}} \frac{1}{(z^{-2}+1)^{2}}[/itex]
We take [itex]w = \frac{1}{z}[/itex] and when [itex]w \to 0[/itex] we have [itex]\frac{1}{z} \to 0[/itex]
As [itex]\frac{1}{z^{4}} \frac{1}{(z^{-2}+1)^{2}} = w^{4} \frac{1}{(w^{2}+1)^{2}}[/itex], the Laurent series of this function is the same we previously calculated with a +4 in the exponent.
We revert the change of variable and we got the series.

[itex]\frac{1}{4} \displaystyle \sum_{n=0}^{\infty} (n+2)((\frac{1}{iz})^{n+4} + (\frac{1}{-iz})^{n+4})[/itex]

Finally, there are a lots of things I did I'm not really sure if were well done, I verified my work with calculator and later with wolframalpha, but I'm not sure what I this is right.
 
Last edited:

Answers and Replies

  • #2
1,796
53
That's well-done in my opinion. Suppose you could have outright made the substitution [itex]z\to 1/w[/itex] to obtain

[tex]w^4\left(-\frac{1}{2w}\frac{d}{dw} \frac{1}{1+w^2}\right)[/tex]

and just expand the [itex]\frac{1}{1+w^2}[/itex] about zero.
 
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  • #3
141
5
Thanks, it's a waaay shorter.
[itex]z = \frac{1}{w} \Rightarrow \frac{1}{(z^{2}+1)^{2}} = \frac{1}{(\frac{1}{w^{2}}+1)^{2}} = \frac{1}{\frac{1}{w^{4}} ( 1+w^{2})^{2}} = w^{4} \frac{1}{(w^{2}+1)^{2}} [/itex]
[itex]\frac{d}{d w} (\frac{1}{w^{2}+1}) = \frac{-2w}{(w^{2}+1)^{2}} \Rightarrow \frac{1}{(w^{2}+1)^{2}} = -\frac{1}{2w} \frac{d}{d w} (\frac{1}{w^{2}+1})[/itex]
[itex]w^{4} (-\frac{1}{2w} \frac{d}{d w} (\frac{1}{w^{2}+1})) = (-\frac{w^{3}}{2} \frac{d}{d w} (\frac{1}{w^{2}+1}))[/itex]
[itex]\frac{1}{1+w^{2}} = \displaystyle \sum_{n=0}^{\infty} (-w^{2})^{n}[/itex]
[itex]\frac{d}{d w} \frac{1}{1+w^{2}} = \displaystyle \sum_{n=0}^{\infty} (n+1) (-w^{2})^{n} (-2w) \Rightarrow (-\frac{w^{3}}{2} \frac{d}{d w} (\frac{1}{w^{2}+1})) = \displaystyle \sum_{n=0}^{\infty} (n+1) (-w^{2})^{n} (-2w) (-\frac{w^{3}}{2})=[/itex]
[itex]= \displaystyle \sum_{n=0}^{\infty} (n+1) (-w^{2})^{n} (w^{4}) = \displaystyle \sum_{n=0}^{\infty} (n+1) (-1)^{n} w^{2n+4} = \displaystyle \sum _{n=0}^{\infty} (n+1) (-1)^{n} \frac{1}{z^{2n+4}}[/itex]

Which is valid for all [itex]z \in \mathbb{C} / | z | > 1[/itex] (In the mess I forget to mention it my the previous post). We know it's valid here because it's where the geometric series equality is true.
 
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