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From: SourceForge.net <noreply@so...>  20121108 23:09:22

Bugs item #3585486, was opened at 20121108 14:36 Message generated for change (Comment added) made by rtoy You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=3585486&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: 7 Private: No Submitted By: Stavros Macrakis (macrakis) Assigned to: Nobody/Anonymous (nobody) Summary: radexpand:all/true broken Initial Comment: block([radexpand:all], sqrt(x^2) ) => sqrt(x^2), should be x per doc block([radexpand:true,domain:'real], sqrt(x^2) ) => sqrt(x^2), should be abs(x) per doc Tested in Maxima 5.27.0 20120430 11:59:06 i686appledarwin11.3.0 SBCL 1.0.55.0abb03f9  >Comment By: Raymond Toy (rtoy) Date: 20121108 15:09 Message: Maxima version too old? I get this with the current sources (and cmucl): block([radexpand:all], sqrt(x^2) ) => x block([radexpand:true,domain:'real], sqrt(x^2) ) => abs(x)  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=3585486&group_id=4933 
From: SourceForge.net <noreply@so...>  20121108 22:36:31

Bugs item #3585486, was opened at 20121108 14:36 Message generated for change (Tracker Item Submitted) made by macrakis You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=3585486&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: 7 Private: No Submitted By: Stavros Macrakis (macrakis) Assigned to: Nobody/Anonymous (nobody) Summary: radexpand:all/true broken Initial Comment: block([radexpand:all], sqrt(x^2) ) => sqrt(x^2), should be x per doc block([radexpand:true,domain:'real], sqrt(x^2) ) => sqrt(x^2), should be abs(x) per doc Tested in Maxima 5.27.0 20120430 11:59:06 i686appledarwin11.3.0 SBCL 1.0.55.0abb03f9  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=3585486&group_id=4933 
From: SourceForge.net <noreply@so...>  20121108 17:52:25

Bugs item #3577666, was opened at 20121016 08:03 Message generated for change (Comment added) made by rtoy You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=3577666&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: Invalid Priority: 5 Private: No Submitted By: JeanYves (jyoberle) Assigned to: Nobody/Anonymous (nobody) Summary: radcan seems to give nonequvalent function Initial Comment: Hi, When doing radcan((2*x)/(sqrt(1(x^21)^2)*sqrt(1asin(x^21)^2))), the result is 2/(sqrt(2x^2)*sqrt(1asin(x^21))*sqrt(asin(x^21)+1)). But these two expressions are not equivalent. For x = 0.4, (2*x)/(sqrt(1(x^21)^2)*sqrt(1asin(x^21)^2)) gives 20.01585798944382 wheras 2/(sqrt(2x^2)*sqrt(1asin(x^21))*sqrt(asin(x^21)+1)) gives 20.01585798944383. (2*x)/(sqrt(1(x^21)^2)*sqrt(1asin(x^21)^2)) is the result of diff(acos(asin(x^21)),x) if this can help. Build info is build_info("5.28.02","20120827 23:16:48","i686pcmingw32","GNU Common Lisp (GCL)","GCL 2.6.8"). Best regards, JeanYves  >Comment By: Raymond Toy (rtoy) Date: 20121108 09:52 Message: First, let's make the expression simpler, removing the part with asin: e: 2*x/sqrt(1(x^21)^2) ratsimp(e) > 2*x/sqrt(x^2x^4) No problem with that. radcan(e) > 2/sqrt(2x^2) This is very different. From the ratsimp result, we can change the expression (manually) to 2*x/abs(x)/sqrt(2x^2). This is basically the same as what radcan has produced, but radcan has "assumed" that x is very large and positive, which makes abs(x) = x. This is what I meant by saying that radcan can do "unexpected" changes. If this is not clear, you should bring this up on the mailing list where Richard Fateman (author of radcan) can explain what's going on. This really needs to be documented better.  Comment By: JeanYves (jyoberle) Date: 20121108 08:38 Message: @rtoy: I'm answering late but I tend to agree with tomasriker. When you look at the limit for x=sqrt(sin(1)+1) which is the largest negative possible value you get: limit(f(x),x,sqrt(1+sin(1)),plus) is inf limit(radcan(f(x)),x,sqrt(1+sin(1)),plus) is minf f(x) being diff(acos(asin(x^21)),x) However, I followed Ivch advice and replaced radcan by factor in my code (with a small trick for integers).  Comment By: David Scherfgen (tomasriker) Date: 20121108 07:55 Message: @rtoy: You missed the minus sign! The one result is negative, the other is positive. So this is definitely a bug!  Comment By: Raymond Toy (rtoy) Date: 20121030 18:21 Message: Although the documentation of radcan isn't very clear on this, radcan is expected to produce results like this. I think the idea is if x is very large, then both expressions are equivalent. I think that's true for your expressions. If this is not what you want, use ratsimp or some other combination of expand and factor. Marking as pending/invalid.  Comment By: Valery Lovchikov (lvch) Date: 20121018 23:19 Message: use function factor %i1 factor(sqrt(x^2u*x^4)); %o1 sqrt(1u*x^2)*abs(x) but %i1 radcan(sqrt(x^2u*x^4)); %o1 x*sqrt(1u*x^2)  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=3577666&group_id=4933 
From: SourceForge.net <noreply@so...>  20121108 16:38:11

Bugs item #3577666, was opened at 20121016 08:03 Message generated for change (Comment added) made by jyoberle You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=3577666&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: Invalid Priority: 5 Private: No Submitted By: JeanYves (jyoberle) Assigned to: Nobody/Anonymous (nobody) Summary: radcan seems to give nonequvalent function Initial Comment: Hi, When doing radcan((2*x)/(sqrt(1(x^21)^2)*sqrt(1asin(x^21)^2))), the result is 2/(sqrt(2x^2)*sqrt(1asin(x^21))*sqrt(asin(x^21)+1)). But these two expressions are not equivalent. For x = 0.4, (2*x)/(sqrt(1(x^21)^2)*sqrt(1asin(x^21)^2)) gives 20.01585798944382 wheras 2/(sqrt(2x^2)*sqrt(1asin(x^21))*sqrt(asin(x^21)+1)) gives 20.01585798944383. (2*x)/(sqrt(1(x^21)^2)*sqrt(1asin(x^21)^2)) is the result of diff(acos(asin(x^21)),x) if this can help. Build info is build_info("5.28.02","20120827 23:16:48","i686pcmingw32","GNU Common Lisp (GCL)","GCL 2.6.8"). Best regards, JeanYves  >Comment By: JeanYves (jyoberle) Date: 20121108 08:38 Message: @rtoy: I'm answering late but I tend to agree with tomasriker. When you look at the limit for x=sqrt(sin(1)+1) which is the largest negative possible value you get: limit(f(x),x,sqrt(1+sin(1)),plus) is inf limit(radcan(f(x)),x,sqrt(1+sin(1)),plus) is minf f(x) being diff(acos(asin(x^21)),x) However, I followed Ivch advice and replaced radcan by factor in my code (with a small trick for integers).  Comment By: David Scherfgen (tomasriker) Date: 20121108 07:55 Message: @rtoy: You missed the minus sign! The one result is negative, the other is positive. So this is definitely a bug!  Comment By: Raymond Toy (rtoy) Date: 20121030 18:21 Message: Although the documentation of radcan isn't very clear on this, radcan is expected to produce results like this. I think the idea is if x is very large, then both expressions are equivalent. I think that's true for your expressions. If this is not what you want, use ratsimp or some other combination of expand and factor. Marking as pending/invalid.  Comment By: Valery Lovchikov (lvch) Date: 20121018 23:19 Message: use function factor %i1 factor(sqrt(x^2u*x^4)); %o1 sqrt(1u*x^2)*abs(x) but %i1 radcan(sqrt(x^2u*x^4)); %o1 x*sqrt(1u*x^2)  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=3577666&group_id=4933 
From: SourceForge.net <noreply@so...>  20121108 16:04:25

Bugs item #3585415, was opened at 20121108 07:45 Message generated for change (Comment added) made by rtoy You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=3585415&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: Open Resolution: None Priority: 7 Private: No Submitted By: David Scherfgen (tomasriker) Assigned to: Nobody/Anonymous (nobody) Summary: integrate loops forever with simple expression Initial Comment: if you calculate this: integrate(x^(1/3)/(x^(2/3)+1), x, 0, 8); Maxima will compute forever and never return. However, the indefinite integral works: integrate(x^(1/3)/(x^(2/3)+1), x); gives you the antiderivative: F(x) := (3*(x^(2/3)+1))/2(3*log(x^(2/3)+1))/2; Now the original definite integral can be computed as F(8)  F(0). This is OK, because F(x) doesn't change sign or shows any other notable behavior between 0 and 8. The question is: why doesn't Maxima do this? Instead it loops forever ...  >Comment By: Raymond Toy (rtoy) Date: 20121108 08:04 Message: Definite integration in maxima often does not compute the indefinite integral first, but in this case it does compute the indefinite integral. Maxima appears to be stuck computing the limits. Don't know why that should be.  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=3585415&group_id=4933 
From: SourceForge.net <noreply@so...>  20121108 15:55:59

Bugs item #3577666, was opened at 20121016 08:03 Message generated for change (Comment added) made by tomasriker You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=3577666&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: Pending Resolution: Invalid Priority: 5 Private: No Submitted By: JeanYves (jyoberle) Assigned to: Nobody/Anonymous (nobody) Summary: radcan seems to give nonequvalent function Initial Comment: Hi, When doing radcan((2*x)/(sqrt(1(x^21)^2)*sqrt(1asin(x^21)^2))), the result is 2/(sqrt(2x^2)*sqrt(1asin(x^21))*sqrt(asin(x^21)+1)). But these two expressions are not equivalent. For x = 0.4, (2*x)/(sqrt(1(x^21)^2)*sqrt(1asin(x^21)^2)) gives 20.01585798944382 wheras 2/(sqrt(2x^2)*sqrt(1asin(x^21))*sqrt(asin(x^21)+1)) gives 20.01585798944383. (2*x)/(sqrt(1(x^21)^2)*sqrt(1asin(x^21)^2)) is the result of diff(acos(asin(x^21)),x) if this can help. Build info is build_info("5.28.02","20120827 23:16:48","i686pcmingw32","GNU Common Lisp (GCL)","GCL 2.6.8"). Best regards, JeanYves  Comment By: David Scherfgen (tomasriker) Date: 20121108 07:55 Message: @rtoy: You missed the minus sign! The one result is negative, the other is positive. So this is definitely a bug!  Comment By: Raymond Toy (rtoy) Date: 20121030 18:21 Message: Although the documentation of radcan isn't very clear on this, radcan is expected to produce results like this. I think the idea is if x is very large, then both expressions are equivalent. I think that's true for your expressions. If this is not what you want, use ratsimp or some other combination of expand and factor. Marking as pending/invalid.  Comment By: Valery Lovchikov (lvch) Date: 20121018 23:19 Message: use function factor %i1 factor(sqrt(x^2u*x^4)); %o1 sqrt(1u*x^2)*abs(x) but %i1 radcan(sqrt(x^2u*x^4)); %o1 x*sqrt(1u*x^2)  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=3577666&group_id=4933 
From: SourceForge.net <noreply@so...>  20121108 15:50:36

Bugs item #3585415, was opened at 20121108 07:45 Message generated for change (Settings changed) made by tomasriker You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=3585415&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: Open Resolution: None >Priority: 7 Private: No Submitted By: David Scherfgen (tomasriker) Assigned to: Nobody/Anonymous (nobody) Summary: integrate loops forever with simple expression Initial Comment: if you calculate this: integrate(x^(1/3)/(x^(2/3)+1), x, 0, 8); Maxima will compute forever and never return. However, the indefinite integral works: integrate(x^(1/3)/(x^(2/3)+1), x); gives you the antiderivative: F(x) := (3*(x^(2/3)+1))/2(3*log(x^(2/3)+1))/2; Now the original definite integral can be computed as F(8)  F(0). This is OK, because F(x) doesn't change sign or shows any other notable behavior between 0 and 8. The question is: why doesn't Maxima do this? Instead it loops forever ...  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=3585415&group_id=4933 
From: SourceForge.net <noreply@so...>  20121108 15:45:51

Bugs item #3585415, was opened at 20121108 07:45 Message generated for change (Tracker Item Submitted) made by tomasriker You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=3585415&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: Open Resolution: None Priority: 5 Private: No Submitted By: David Scherfgen (tomasriker) Assigned to: Nobody/Anonymous (nobody) Summary: integrate loops forever with simple expression Initial Comment: if you calculate this: integrate(x^(1/3)/(x^(2/3)+1), x, 0, 8); Maxima will compute forever and never return. However, the indefinite integral works: integrate(x^(1/3)/(x^(2/3)+1), x); gives you the antiderivative: F(x) := (3*(x^(2/3)+1))/2(3*log(x^(2/3)+1))/2; Now the original definite integral can be computed as F(8)  F(0). This is OK, because F(x) doesn't change sign or shows any other notable behavior between 0 and 8. The question is: why doesn't Maxima do this? Instead it loops forever ...  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=3585415&group_id=4933 