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Residues - salvomic - 02-07-2015 06:39 PM
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...
RE: Residues - parisse - 02-08-2015 07:13 AM
The residue should be undefined, 0 is an essential singularity.
RE: Residues - salvomic - 02-08-2015 08:57 AM
(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
RE: Residues - parisse - 02-08-2015 02:53 PM
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)
RE: Residues - salvomic - 02-08-2015 03:34 PM
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...
RE: Residues - parisse - 02-09-2015 08:01 AM
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.