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From: SourceForge.net <noreply@so...>  20090115 18:40:36

Bugs item #2501765, was opened at 20090112 11:36 Message generated for change (Settings changed) made by dgildea You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=2501765&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  Integration Group: None >Status: Closed >Resolution: Fixed Priority: 5 Private: No Submitted By: Raymond Toy (rtoy) Assigned to: Nobody/Anonymous (nobody) Summary: integrate((14*x^232)/(x^4+3*x^2+1)^2,x,0,inf); Initial Comment: integrate((14*x^232)/(x^4+3*x^2+1)^2,x,0,inf); produces a division by zero error. Maxima ought to be able to do this integral. And maxima can do the indefinite integral, producing a rational expression and another integral. Maxima can compute that definite integral.  >Comment By: Dan Gildea (dgildea) Date: 20090115 13:40 Message: Fix in solve.lisp rev 1.22 gives correct multiplicity of roots. (%i6) integrate((14*x^232)/(x^4+3*x^2+1)^2,x,0,inf); (%o6) (sqrt(3sqrt(5))*(139*sqrt(2)*sqrt(5)+345*sqrt(2))*%pi +sqrt(sqrt(5)+3)*(345*sqrt(2)139*sqrt(2)*sqrt(5))*%pi) /200 (%i7) float(%); (%o7) 14.47111834594389 (%i8) quad_qags((14*x^232)/(x^4+3*x^2+1)^2,x,0,1000); (%o8) [14.47111834594389,1.179685737489548e9,441,0]  Comment By: Raymond Toy (rtoy) Date: 20090114 08:46 Message: I think this fails because maxima computes the multiplicities of the roots of the denominator incorrectly. The roots are double roots, but maxima thinks they're single and hence gets a division by zero error when trying to compute the residue by differentiating the denominator once instead of twice. See also https://sourceforge.net/tracker2/?func=detail&aid=2505241&group_id=4933&atid=104933  Comment By: Raymond Toy (rtoy) Date: 20090112 12:01 Message: Maxima is using a keyhole contour and is computing the residues of the integrand. I think Maxima doesn't notice that the poles are multiple poles, and just uses the simple method of computing the residue by differentiating the denominator. This produces a division by zero.  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=2501765&group_id=4933 
From: SourceForge.net <noreply@so...>  20090115 18:35:30

Bugs item #2505241, was opened at 20090113 14:20 Message generated for change (Settings changed) made by dgildea You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=2505241&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: None Group: None >Status: Closed >Resolution: Fixed Priority: 5 Private: No Submitted By: Raymond Toy (rtoy) Assigned to: Nobody/Anonymous (nobody) Summary: solve multiplicities wrong Initial Comment: This is right: (%i543) solve((x^2+2*x+1)^7,x); (%o543) [x =  1] (%i544) multiplicities; (%o544) [14] But this is wrong: (%i545) solve((y^4+3*y^2+1)^7,y)$ (%i546) multiplicities; (%o546) [1, 1, 1, 1] Each root has multiplicity 7, not 1. In solve.lisp, after easycases, there is a comment from Barton about commenting out some code dealing with multiplicities. Not sure if this would fix the problem, or what the actual problem is in the comment.  >Comment By: Dan Gildea (dgildea) Date: 20090115 13:35 Message: In solve.lisp rev 1.22, pass multiplicity parameter through easycases to solve. (%i2) solve((x^2+2*x+1)^7,x); (%o2) [x = 1] (%i3) multiplicities; (%o3) [14] (%i4) solve((y^4+3*y^2+1)^7,y)$ (%i5) multiplicities; (%o5) [7,7,7,7] I didn't touch the commented code.  Comment By: Barton Willis (willisbl) Date: 20090115 06:26 Message: Also after uncommenting, the code in solve.lisp I commented out in 2004, the testsuite reports two problems: Error found in c:\maximacvs3\maxima\tests\rtest15.mac, problem: (137) Error found in c:\maximacvs3\maxima\tests\rtestint.mac, problem: (45)  Comment By: Barton Willis (willisbl) Date: 20090115 06:11 Message: I wish that my comment had more details. I tried uncommenting the codethen (%i1) solve((x^2+2*x+1)^7,x)$ Wrong, but previously correct. (%i2) multiplicities; (%o2) [2] Wrong: (%i3) solve((y^4+3*y^2+1)^7,y)$ (%i4) multiplicities; (%o4) [1, 1, 1, 1]  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=2505241&group_id=4933 
From: SourceForge.net <noreply@so...>  20090115 11:26:55

Bugs item #2505241, was opened at 20090113 13:20 Message generated for change (Comment added) made by willisbl You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=2505241&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: None Group: None Status: Open Resolution: None Priority: 5 Private: No Submitted By: Raymond Toy (rtoy) Assigned to: Nobody/Anonymous (nobody) Summary: solve multiplicities wrong Initial Comment: This is right: (%i543) solve((x^2+2*x+1)^7,x); (%o543) [x =  1] (%i544) multiplicities; (%o544) [14] But this is wrong: (%i545) solve((y^4+3*y^2+1)^7,y)$ (%i546) multiplicities; (%o546) [1, 1, 1, 1] Each root has multiplicity 7, not 1. In solve.lisp, after easycases, there is a comment from Barton about commenting out some code dealing with multiplicities. Not sure if this would fix the problem, or what the actual problem is in the comment.  >Comment By: Barton Willis (willisbl) Date: 20090115 05:26 Message: Also after uncommenting, the code in solve.lisp I commented out in 2004, the testsuite reports two problems: Error found in c:\maximacvs3\maxima\tests\rtest15.mac, problem: (137) Error found in c:\maximacvs3\maxima\tests\rtestint.mac, problem: (45)  Comment By: Barton Willis (willisbl) Date: 20090115 05:11 Message: I wish that my comment had more details. I tried uncommenting the codethen (%i1) solve((x^2+2*x+1)^7,x)$ Wrong, but previously correct. (%i2) multiplicities; (%o2) [2] Wrong: (%i3) solve((y^4+3*y^2+1)^7,y)$ (%i4) multiplicities; (%o4) [1, 1, 1, 1]  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=2505241&group_id=4933 
From: SourceForge.net <noreply@so...>  20090115 11:11:44

Bugs item #2505241, was opened at 20090113 13:20 Message generated for change (Comment added) made by willisbl You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=2505241&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: None Group: None Status: Open Resolution: None Priority: 5 Private: No Submitted By: Raymond Toy (rtoy) Assigned to: Nobody/Anonymous (nobody) Summary: solve multiplicities wrong Initial Comment: This is right: (%i543) solve((x^2+2*x+1)^7,x); (%o543) [x =  1] (%i544) multiplicities; (%o544) [14] But this is wrong: (%i545) solve((y^4+3*y^2+1)^7,y)$ (%i546) multiplicities; (%o546) [1, 1, 1, 1] Each root has multiplicity 7, not 1. In solve.lisp, after easycases, there is a comment from Barton about commenting out some code dealing with multiplicities. Not sure if this would fix the problem, or what the actual problem is in the comment.  >Comment By: Barton Willis (willisbl) Date: 20090115 05:11 Message: I wish that my comment had more details. I tried uncommenting the codethen (%i1) solve((x^2+2*x+1)^7,x)$ Wrong, but previously correct. (%i2) multiplicities; (%o2) [2] Wrong: (%i3) solve((y^4+3*y^2+1)^7,y)$ (%i4) multiplicities; (%o4) [1, 1, 1, 1]  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=2505241&group_id=4933 
From: SourceForge.net <noreply@so...>  20090115 05:34:42

Bugs item #2508928, was opened at 20090115 00:34 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=2508928&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  Simplification Group: None Status: Open Resolution: None Priority: 5 Private: No Submitted By: Raymond Toy (rtoy) Assigned to: Nobody/Anonymous (nobody) Summary: hgfred([a+5/6,a+7/3],[2*a+23/3],4*z*(1z)) fails Initial Comment: This returns a result with false in the result. The error comes from HYPCOS. The arguments, a+11/6, a+7/3, 2*(a+11/6)+1, are correct for HYPCOS but HYPCOS fails to find the correct expression and returns NIL.  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=2508928&group_id=4933 
From: SourceForge.net <noreply@so...>  20090115 04:48:40

Bugs item #2508877, was opened at 20090114 23:48 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=2508877&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  Simplification Group: None Status: Open Resolution: None Priority: 2 Private: No Submitted By: Raymond Toy (rtoy) Assigned to: Nobody/Anonymous (nobody) Summary: simp2f1willcontinuein Initial Comment: There are many cases where Maxima will print SIMP2F1WILLCONTINUEIN. hgfred([a,a+1/2],[3/2],x) is one such expression. This is a reminder that Maxima doesn't know how to reduce this to some other form. But A&S 15.2.4 can reduce 3/2 to 1/2, and Maxima knows what hgfred([a,a+1/2],[1/2],x) is when a is a positive integer.  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=2508877&group_id=4933 
From: SourceForge.net <noreply@so...>  20090115 01:06:38

Bugs item #2505945, was opened at 20090113 21:24 Message generated for change (Comment added) made by rtoy You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=2505945&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  Simplification Group: None >Status: Closed >Resolution: Fixed Priority: 5 Private: No Submitted By: Barton Willis (willisbl) >Assigned to: Raymond Toy (rtoy) Summary: hgfred([2,1/2],[3],x^2); Initial Comment: (%i45) hgfred([2,1/2],[3],x^2); 1 Enter hgfred[[2,1/2],[3],x^2] Nonvariable 2nd argument to diff: x^2 But hgfred([2,1/2],[3],x) doesn't give an error.  >Comment By: Raymond Toy (rtoy) Date: 20090114 20:06 Message: Fixed in hyp.lisp, rev 1.93. We don't try differentiating wrt to a nonvariable anymore.  Comment By: Raymond Toy (rtoy) Date: 20090114 08:51 Message: FWIW, this is a bug in the code. It's using A&S 15.2.4 to transform F(a,b;c;z) to F(a,b;cn;z): F(a,b;cn;z) = 1/poch(cn,n)/z^(cn1)*diff(z^(c1)*F(a,b;c;z), ,z,n)  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=2505945&group_id=4933 