Residues

02072015, 06:39 PM
(This post was last modified: 02072015 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/z1/(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... ∫aL√0mic (IT9CLU) :: HP Prime 50g 41CX 71b 42s 39s 35s 12C 15C  DM42, DM41X  WP34s Prime Soft. Lib 

02082015, 07:13 AM
Post: #2




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


02082015, 08:57 AM
Post: #3




RE: Residues
(02082015 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 ∫aL√0mic (IT9CLU) :: HP Prime 50g 41CX 71b 42s 39s 35s 12C 15C  DM42, DM41X  WP34s Prime Soft. Lib 

02082015, 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) 

02082015, 03:34 PM
(This post was last modified: 02082015 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... ∫aL√0mic (IT9CLU) :: HP Prime 50g 41CX 71b 42s 39s 35s 12C 15C  DM42, DM41X  WP34s Prime Soft. Lib 

02092015, 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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