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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...

The residue should be undefined, 0 is an essential singularity.

(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

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)

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...

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.