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Residues
02-07-2015, 06:39 PM (This post was last modified: 02-07-2015 11:20 PM by salvomic.)
Post: #1
Residues
hi all,
calculating residue of \( sin(z)^{-1} \) in z=0 I get 1, ok.
Calculating residue of \( sin(z^{-1}) \) in z=0 I get 0, but I think I should get 1, instead...
(The Laurent series is 1/z-1/(6 z^3)+1/(120 z^5)-1/(5040 z^7)+1/(362880 z^9)-1/(39916800 z^11)+o((1/z)^13) ...)
Am I wrong?

Thank you
Salvo

EDIT: here (and in a book of mine) they say 1...

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02-08-2015, 07:13 AM
Post: #2
RE: Residues
The residue should be undefined, 0 is an essential singularity.
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02-08-2015, 08:57 AM
Post: #3
RE: Residues
(02-08-2015 07:13 AM)parisse Wrote:  The residue should be undefined, 0 is an essential singularity.

yes, 0 is a singularity, in fact, but some books consider the value of this residue 1: I would understand the real reason...

Parisse, please, there is a way in Prime (and emulator) to get coefficients ak and bk of the Laurent Series (or the whole series), to confront result with residues in some case?

Thanks a lot for kindness

Salvo

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02-08-2015, 02:53 PM
Post: #4
RE: Residues
What's ak and bk?
You can get Laurent series with series, for example
series(1/sin(x)^3,x=0,1)
residue(1/sin(x)^3,x=0)
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02-08-2015, 03:34 PM (This post was last modified: 02-08-2015 03:54 PM by salvomic.)
Post: #5
RE: Residues
I meant this, with ak, bk coefficient of Laurent Series...
If we know b1, we have also residue, generally speaking...

With series(sin(x^-1),x,0,7) I get always sin(1/x), so the unique coefficient should be 1, if I'm not wrong...

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02-09-2015, 08:01 AM
Post: #6
RE: Residues
series is more general than Laurent series, it does series expansion with respect to the most rapidly varying subexpression. For sin(1/x) you can't expand more than sin(1/x), that's why it is left unchanged.
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