- #1

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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.

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- Thread starter asi123
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- #1

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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.

- #2

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)

HOWEVER, because that series has an

- #3

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[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 aninfinitenumber of negative powers with non-zero coefficient, z= 2 is anessential singularity, not apole. There is NO residue.

Oh, so an essential singularity has no residue?

Thanks.

- #4

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- #5

HallsofIvy

Science Advisor

Homework Helper

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- 964

[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 aninfinitenumber of negative powers with non-zero coefficient, z= 2 is anessential singularity, not apole. 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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