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From: SourceForge.net <noreply@so...>  20090605 22:11:07

Bugs item #1119228, was opened at 20050209 12:20 Message generated for change (Comment added) made by crategus You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1119228&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  Limit Group: None >Status: Closed >Resolution: Fixed Priority: 3 Private: No Submitted By: Barton Willis (willisbl) Assigned to: Nobody/Anonymous (nobody) Summary: limit(1/zeroa) Initial Comment: Maxima knows that limit(zeroa) = 0, but it doesn't know that limit(1/zeroa) = inf. Same is true for zerob ( but limit(1/zerob) = minf). (%i1) limit(1/zeroa); Division by 0 (%i2) limit(zeroa); (%o2) 0 Barton  >Comment By: Dieter Kaiser (crategus) Date: 20090606 00:11 Message: With revision 1.71 of limit.lisp the examples work: (%i6) limit(1/zeroa); (%o6) inf (%i7) limit(1/zerob); (%o7) minf Closing this bug report as fixed. Dieter Kaiser  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1119228&group_id=4933 
From: SourceForge.net <noreply@so...>  20090605 22:09:16

Bugs item #1315837, was opened at 20051007 15:57 Message generated for change (Comment added) made by crategus You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1315837&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  Limit Group: None >Status: Closed >Resolution: Fixed Priority: 5 Private: No Submitted By: Barton Willis (willisbl) Assigned to: Nobody/Anonymous (nobody) Summary: limit(?foo) Initial Comment: (%i46) limit(?foo); (%o46) 0 < Bogus (%i47) limit(true); (%o47) true < OK (%i48) limit(false); (%o48) lim(false,FOO,0) < OK, but goofy (%i49) build_info(); Maxima version: 5.9.1 Maxima build date: 7:34 9/24/2004 host type: i686pcmingw32 lispimplementationtype: Kyoto Common Lisp lispimplementationversion: GCL 2.6.5 Barton  >Comment By: Dieter Kaiser (crategus) Date: 20090606 00:09 Message: With revision 1,71 the examples work: (%i3) limit(?foo); (%o3) ?foo (%i4) limit(true); (%o4) true (%i5) limit(false); (%o5) false Closing this bug report as fixed. Dieter Kaiser  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1315837&group_id=4933 
From: SourceForge.net <noreply@so...>  20090605 21:50:09

Bugs item #2802006, was opened at 20090605 21:34 Message generated for change (Comment added) made by nobody You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=2802006&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: Nobody/Anonymous (nobody) Assigned to: Nobody/Anonymous (nobody) Summary: integrate(1/(sqrt(x)+1), x, 0, 1); Initial Comment: Maxima can't solve this integral. I'm using maxima 5.17.1  Comment By: Nobody/Anonymous (nobody) Date: 20090605 21:49 Message: bug persists in the 5.18.1 release  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=2802006&group_id=4933 
From: SourceForge.net <noreply@so...>  20090605 21:36:09

Bugs item #2802006, was opened at 20090605 21:34 Message generated for change (Tracker Item Submitted) made by nobody You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=2802006&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: Nobody/Anonymous (nobody) Assigned to: Nobody/Anonymous (nobody) Summary: integrate(1/(sqrt(x)+1), x, 0, 1); Initial Comment: Maxima can't solve this integral. I'm using maxima 5.17.1  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=2802006&group_id=4933 
From: SourceForge.net <noreply@so...>  20090605 18:19:05

Bugs item #2801819, was opened at 20090605 16:18 Message generated for change (Comment added) made by nobody You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=2801819&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: Barton Willis (willisbl) Assigned to: Nobody/Anonymous (nobody) Summary: spurious Principal Value message Initial Comment: (%i1) assume(p > 0); (%o1) [p>0] OK: (%i4) integrate(exp(p * t^2),t,minf,inf); (%o4) sqrt(%pi)/sqrt(p) OK, but not a principle value: (%i5) integrate(exp(pp * t^2),t,minf,inf); Is pp positive, negative, or zero?pos; Principal Value (%o5) sqrt(%pi)/sqrt(pp)  Comment By: Nobody/Anonymous (nobody) Date: 20090605 18:18 Message: The difference between the two cases is in polesininterval. In the first case, there are no poles in the interval. In the second case, polesininterval thinks there are poles at minf and inf. Hence, maxima thinks we have a principal value integral.  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=2801819&group_id=4933 
From: SourceForge.net <noreply@so...>  20090605 16:31:15

Bugs item #2801821, was opened at 20090605 18:31 Message generated for change (Tracker Item Submitted) made by crategus You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=2801821&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  Limit Group: None Status: Open Resolution: None Priority: 5 Private: No Submitted By: Dieter Kaiser (crategus) Assigned to: Nobody/Anonymous (nobody) Summary: limit(x*expintegral_ei(x),x,0) > Error Initial Comment: For this example we get an error: (%i5) limit(x*expintegral_ei(x),x,0); expintegral_ei: expintegral_ei(0) is undefined.  an error. To debug this try debugmode(true); The error occurs in $gruntz, which calls $taylor. $taylor does not work for expintegral_ei(x) at the point zero and throws a Maxima error. This error is not catched by $gruntz. 1. The limit is known to be zero. Maxima should evaluate it. But hospital and $gruntz do not work. 2. $gruntz should catch the error in $taylor. The limit of this type of functions is important to get correct results for the definite integrals like, e. g. integrate(expintegral_ei(x),x,0,1). At the moment I have no idea to get the correct limit. I have not tried to improve $gruntz to catch the error. Perhaps someone who know $gruntz better can do it. Dieter Kaiser  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=2801821&group_id=4933 
From: SourceForge.net <noreply@so...>  20090605 16:18:50

Bugs item #2801819, was opened at 20090605 11:18 Message generated for change (Tracker Item Submitted) made by willisbl You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=2801819&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: Barton Willis (willisbl) Assigned to: Nobody/Anonymous (nobody) Summary: spurious Principal Value message Initial Comment: (%i1) assume(p > 0); (%o1) [p>0] OK: (%i4) integrate(exp(p * t^2),t,minf,inf); (%o4) sqrt(%pi)/sqrt(p) OK, but not a principle value: (%i5) integrate(exp(pp * t^2),t,minf,inf); Is pp positive, negative, or zero?pos; Principal Value (%o5) sqrt(%pi)/sqrt(pp)  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=2801819&group_id=4933 
From: SourceForge.net <noreply@so...>  20090605 16:15:30

Bugs item #671602, was opened at 20030121 06:37 Message generated for change (Comment added) made by crategus You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=671602&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  Limit Group: None >Status: Closed >Resolution: Fixed Priority: 5 Private: No Submitted By: Stavros Macrakis (macrakis) Assigned to: Nobody/Anonymous (nobody) Summary: limit x!/(x+a)! (a=0) Initial Comment: limit(x!/(x+a)!,x,inf)... is a pnz? zero; Divison by zero detected in SIGN: 1/a  an error Strangely enough, limit(gamma(x)/gamma(x+a),x,inf) (ask: a=0) gives the noun form.   >Comment By: Dieter Kaiser (crategus) Date: 20090605 18:15 Message: The initial problem has gone. The limit works for the given examples and is correct. (%i2) declare(a,integer); (%o2) done (%i3) limit(x!/(x+a)!,x,inf); Is a positive, negative, or zero? p; (%o3) 0 (%i4) limit(x!/(x+a)!,x,inf); Is a positive, negative, or zero? n; (%o4) inf (%i5) limit(x!/(x+a)!,x,inf); Is a positive, negative, or zero? z; (%o5) 1 (%i8) limit(gamma(x)/gamma(x+a),x,inf); Is a positive, negative, or zero? p; (%o8) 0 (%i9) limit(gamma(x)/gamma(x+a),x,inf); Is a positive, negative, or zero? n; (%o9) inf (%i10) limit(gamma(x)/gamma(x+a),x,inf); Is a positive, negative, or zero? z; (%o10) 1 There is one issue. We have declared a to be an integer to avoid in the first call to limit the question "is 4/(4+a) an integer". This question is unnecessary and it doesn't matter what is answered. Perhaps we should open a new bug report. Closing this bug report as fixed. Dieter Kaiser  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=671602&group_id=4933 
From: SourceForge.net <noreply@so...>  20090605 15:50:11

Bugs item #2795534, was opened at 20090522 20:46 Message generated for change (Comment added) made by crategus You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=2795534&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: Dieter Kaiser (crategus) Assigned to: Nobody/Anonymous (nobody) Summary: integrate(expintegral_ei(x),x,0,1) gives result with TRUE Initial Comment: When we do the above integral we get an expression with the symbol TRUE in it: (%i14) integrate(expintegral_ei(x),x,0,1); (%o14) expintegral_ei(1)true%e The same for expintegral_ci and expintegral_chi: (%i16) integrate(expintegral_ci(x),x,0,1); (%o16) sin(1)+expintegral_ci(1)true (%i18) integrate(expintegral_chi(x),x,0,1); (%o18) sinh(1)+expintegral_chi(1)true Dieter Kaiser  >Comment By: Dieter Kaiser (crategus) Date: 20090605 17:50 Message: A check to the routine easysubs in defint.lisp has been added. We no longer get a result with the symbol T in it. But the integral does not work, because of a bug in limit. (%i1) integrate(expintegral_ei(x),x,0,1); expintegral_ei: expintegral_ei(0) is undefined.  an error. To debug this try debugmode(true); Closing this bug report as fixed, because the initial problem in defint is solved. I will open a new bug report for the problem in limit. Dieter Kaiser  Comment By: Dieter Kaiser (crategus) Date: 20090604 22:20 Message: Maxima should be able to solve integrals of the type given in this bug report. Other integrals like integrate(expintegral_ei(x),x,1/2,1) work well and give correct results. There are two problems. 1. A bug in defint Because of a bug in defint, we get the symbol true in the result. The reason is a missing check of the return value of noerrsub against T in the routine easysubs. This bug is easy to cure. 2. A bug in limit The limit of x*expintegral_ei(x) > 0 has to be calculated. The limit of expintegral_ei(x) > 0 is known by Maxima. Therefore Maxima should be able do get the correct answer, but Maxima tries to calculated the function at the value 0. This gives an error and the calculation stops. I am looking into the code to find a solution to the second problem too. Dieter Kaiser  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=2795534&group_id=4933 