Dear Nils, Thanks for the careful explanation. There's a good reason I don't teach this stuff (even if I enjoy it!).
Rob On Thursday, November 5, 2015 at 7:35:47 PM UTC-8, Nils Bruin wrote: > > On Thursday, November 5, 2015 at 5:11:42 PM UTC-8, Rob Beezer wrote: >> >> LIke I said, my complex analysis is rusty. And I am really asking for >> somebody else. Is the chunk of code that produces 2 (above) not following >> the definition of the residue? >> > > No it's not. If t=1/z then you don't get a truncated taylor expansion > around t=0 by substituting t=1/z in the taylor expansion around z=0. > > In general, if you define the residue by a contour integral, then the > residue of exp(1/z) around z=0 is 0: Any contour around z=0 is a boundary > of a simply connected domain of the Riemann sphere on which the function is > regular, so integrating along the contour gets you 0. > > exp(1/z) has an essential singularity at z=0, so there is no laurent > expansion to read off the residue from. > -- You received this message because you are subscribed to the Google Groups "sage-support" group. To unsubscribe from this group and stop receiving emails from it, send an email to sage-support+unsubscr...@googlegroups.com. To post to this group, send email to sage-support@googlegroups.com. Visit this group at http://groups.google.com/group/sage-support. For more options, visit https://groups.google.com/d/optout.
