Residues 2

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residues 2 :)

Homework Statement



Ok, so I have this function and I need to find the residues for it.
I developed the function into Laurent series and my question is, how can I find a(-1) which is the residues?

Thanks in advance.


Homework Equations





The Attempt at a Solution

 

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Answers and Replies

  • #2
HallsofIvy
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The function is
[tex]f(z)= cos(\frac{1}{z-2})[/tex]
and you correctly used the McLaurin series for cos(u), with u= 1/(z-2), to write the Laurent series
[tex]\sum_{n= 0}^\infty \frac{(-1)^n}{(2n)!} (z-2)^{-2n}[/tex]
Since the power of z- 2 is always even, there is no (z-2)-1 term: the coefficient is 0.

HOWEVER, because that series has an infinite number of negative powers with non-zero coefficient, z= 2 is an essential singularity, not a pole. There is NO residue.
 
  • #3
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The function is
[tex]f(z)= cos(\frac{1}{z-2})[/tex]
and you correctly used the McLaurin series for cos(u), with u= 1/(z-2), to write the Laurent series
[tex]\sum_{n= 0}^\infty \frac{(-1)^n}{(2n)!} (z-2)^{-2n}[/tex]
Since the power of z- 2 is always even, there is no (z-2)-1 term: the coefficient is 0.

HOWEVER, because that series has an infinite number of negative powers with non-zero coefficient, z= 2 is an essential singularity, not a pole. There is NO residue.

Oh, so an essential singularity has no residue?

Thanks.
 
  • #4
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How about this one?

Thanks again.
 

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  • #5
HallsofIvy
Science Advisor
Homework Helper
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The function is
[tex]f(z)= cos(\frac{1}{z-2})[/tex]
and you correctly used the McLaurin series for cos(u), with u= 1/(z-2), to write the Laurent series
[tex]\sum_{n= 0}^\infty \frac{(-1)^n}{(2n)!} (z-2)^{-2n}[/tex]
Since the power of z- 2 is always even, there is no (z-2)-1 term: the coefficient is 0.

HOWEVER, because that series has an infinite number of negative powers with non-zero coefficient, z= 2 is an essential singularity, not a pole. There is NO residue.

I am informed that, according to Wikipedia, I am wrong about this. That is, that the residue is still the coefficient of the -1 power in the Laurent series. The difference is just that the formula for the residue, taking the nth derivative where n is the order of the pole, cannot be used but you still can find it by finding the Laurent series itself. In the case of your first problem, the residue is 0 and for your new problem, it is 1.
 

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