From: SourceForge.net <noreply@so...>  20070629 23:18:22

Bugs item #1692729, was opened at 20070402 02:46 Message generated for change (Settings changed) made by robert_dodier You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1692729&group_id=4933 Please note that this message will contain a full copy of the comment thread, including the initial issue submission, for this request, not just the latest update. >Category: Lisp Core  Polynomials Group: None Status: Open Resolution: None Priority: 5 Private: No Submitted By: Nobody/Anonymous (nobody) Assigned to: Nobody/Anonymous (nobody) Summary: coeff, ratcoef on Taylor series around inf Initial Comment: It seems that coeff() and ratcoef() have trouble extracting coefficients from Taylor expansions around infinity: (%i119) taylor(1/x, x, inf, 1); 1 (%o119)/T/  + . . . x (%i120) ratcoef(taylor(1/x, x, inf, 1), x, 1); (%o120)/R/ 0 (%i121) coeff(taylor(1/x, x, inf, 1), x, 1); (%o121)/R/ 0 A workaround is to get rid of the Taylor property, e.g., by wrapping the expression in subst(0, 0, ...).  Comment By: Stavros Macrakis (macrakis) Date: 20070404 23:45 Message: Logged In: YES user_id=588346 Originator: NO Yes, ratcoeff is a bit confused by taylor expansions at infinity: qq: taylor(sum((i3+100)*x^(i3),i,0,5),x,inf,10) => 102*x^2+101*x+100+99/x+98/x^2+97/x^3 + ... makelist(ratcoeff(qq,x,i),i,3,4) => => [0, 102, 101, 100, 99, 98, 97, 0] In other words, it treats the 1/x^2 as the ratcoeff(...,x,2) term. So the workaround is to negate the power... but of course that will become exactly wrong when the bug is fixed....  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1692729&group_id=4933 