Work at SourceForge, help us to make it a better place! We have an immediate need for a Support Technician in our San Francisco or Denver office.
Close
From: SourceForge.net <noreply@so...>  20030516 16:15:53

Bugs item #738848, was opened at 20030516 09:15 Message generated for change (Tracker Item Submitted) made by Item Submitter You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=738848&group_id=4933 Category: Share Libraries Group: Fix for 5.9.0 Status: Open Resolution: None Priority: 5 Submitted By: Nobody/Anonymous (nobody) Assigned to: Nobody/Anonymous (nobody) Summary: Poverseries error for 1/(1t)^2 Initial Comment: It seems powerseries(1(ab*t)^n, t, 0) produces an error for n=2, but is correct for n=1,3,4: (C1) POWERSERIES(1/(ab*t)^2, t, 0); INF ==== \ I1  I1  2 I1 (D1) > a b (I1 + 1) t / ==== I1 = 0 [I believe that powers of a and b above have opposite signs].  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=738848&group_id=4933 
From: SourceForge.net <noreply@so...>  20030517 18:32:39

Bugs item #738848, was opened at 20030516 09:15 Message generated for change (Comment added) made by nobody You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=738848&group_id=4933 Category: Share Libraries Group: Fix for 5.9.0 Status: Open Resolution: None Priority: 5 Submitted By: Nobody/Anonymous (nobody) Assigned to: Nobody/Anonymous (nobody) Summary: Poverseries error for 1/(1t)^2 Initial Comment: It seems powerseries(1(ab*t)^n, t, 0) produces an error for n=2, but is correct for n=1,3,4: (C1) POWERSERIES(1/(ab*t)^2, t, 0); INF ==== \ I1  I1  2 I1 (D1) > a b (I1 + 1) t / ==== I1 = 0 [I believe that powers of a and b above have opposite signs].  Comment By: Nobody/Anonymous (nobody) Date: 20030517 11:32 Message: Logged In: NO (C1) verbose:true; (D1) TRUE (C2) POWERSERIES(1/(ab*t)^2, t, 0); In the first simplification we have returned: 1  2 2 2 b t  2 a b t + a trying to do a rational function expansion of 1  2 2 2 b t  2 a b t + a Using a special rule for expressions of form M  N (A + C VAR ) Here we have [N = 2, A =  a, C = b, M = 1] INF ==== \ I1  I1  2 I1 (D2) > a b (I1 + 1) t / ==== I1 = 0 (C3)  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=738848&group_id=4933 
From: SourceForge.net <noreply@so...>  20030519 08:32:06

Bugs item #738848, was opened at 20030516 16:15 Message generated for change (Comment added) made by kratt5 You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=738848&group_id=4933 Category: Share Libraries Group: Fix for 5.9.0 Status: Open Resolution: None Priority: 5 Submitted By: Nobody/Anonymous (nobody) Assigned to: Nobody/Anonymous (nobody) Summary: Poverseries error for 1/(1t)^2 Initial Comment: It seems powerseries(1(ab*t)^n, t, 0) produces an error for n=2, but is correct for n=1,3,4: (C1) POWERSERIES(1/(ab*t)^2, t, 0); INF ==== \ I1  I1  2 I1 (D1) > a b (I1 + 1) t / ==== I1 = 0 [I believe that powers of a and b above have opposite signs].  Comment By: Martin Rubey (kratt5) Date: 20030519 08:32 Message: Logged In: YES user_id=651552 This is fixed with the fix for [727542] powerseries wrong/fix Martin (please, somebody merge it into cvs, it is IMPORTANT!)  Comment By: Nobody/Anonymous (nobody) Date: 20030517 18:32 Message: Logged In: NO (C1) verbose:true; (D1) TRUE (C2) POWERSERIES(1/(ab*t)^2, t, 0); In the first simplification we have returned: 1  2 2 2 b t  2 a b t + a trying to do a rational function expansion of 1  2 2 2 b t  2 a b t + a Using a special rule for expressions of form M  N (A + C VAR ) Here we have [N = 2, A =  a, C = b, M = 1] INF ==== \ I1  I1  2 I1 (D2) > a b (I1 + 1) t / ==== I1 = 0 (C3)  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=738848&group_id=4933 
From: SourceForge.net <noreply@so...>  20030528 16:41:10

Bugs item #738848, was opened at 20030516 12:15 Message generated for change (Comment added) made by rtoy You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=738848&group_id=4933 Category: Share Libraries Group: Fix for 5.9.0 >Status: Closed >Resolution: Duplicate Priority: 5 Submitted By: Nobody/Anonymous (nobody) Assigned to: Nobody/Anonymous (nobody) Summary: Poverseries error for 1/(1t)^2 Initial Comment: It seems powerseries(1(ab*t)^n, t, 0) produces an error for n=2, but is correct for n=1,3,4: (C1) POWERSERIES(1/(ab*t)^2, t, 0); INF ==== \ I1  I1  2 I1 (D1) > a b (I1 + 1) t / ==== I1 = 0 [I believe that powers of a and b above have opposite signs].  >Comment By: Raymond Toy (rtoy) Date: 20030528 12:41 Message: Logged In: YES user_id=28849 Bug marked as duplicate. I'll apply your patch for bug 727542.  Comment By: Martin Rubey (kratt5) Date: 20030519 04:32 Message: Logged In: YES user_id=651552 This is fixed with the fix for [727542] powerseries wrong/fix Martin (please, somebody merge it into cvs, it is IMPORTANT!)  Comment By: Nobody/Anonymous (nobody) Date: 20030517 14:32 Message: Logged In: NO (C1) verbose:true; (D1) TRUE (C2) POWERSERIES(1/(ab*t)^2, t, 0); In the first simplification we have returned: 1  2 2 2 b t  2 a b t + a trying to do a rational function expansion of 1  2 2 2 b t  2 a b t + a Using a special rule for expressions of form M  N (A + C VAR ) Here we have [N = 2, A =  a, C = b, M = 1] INF ==== \ I1  I1  2 I1 (D2) > a b (I1 + 1) t / ==== I1 = 0 (C3)  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=738848&group_id=4933 