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From: SourceForge.net <noreply@so...>  20070613 21:41:20

Bugs item #1635372, was opened at 20070114 16:10 Message generated for change (Comment added) made by rtoy You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1635372&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: Robert Dodier (robert_dodier) Assigned to: Nobody/Anonymous (nobody) Summary: specint returns expression containing internal Lisp variable Initial Comment: assume(p > 0, a > 0, b > 0); sin(a*t)*cosh(b*t^2)*%e^(p*t); radcan(specint(%,t)); => long expression containing failonf24p146test Not sure what's going on here, haven't looked into it.  >Comment By: Raymond Toy (rtoy) Date: 20070613 17:41 Message: Logged In: YES user_id=28849 Originator: NO The special variable is *hypreturnnounformp*, defaulting to T. The expression no longer returns failonf24p146test. Closing bug report.  Comment By: Raymond Toy (rtoy) Date: 20070118 10:48 Message: Logged In: YES user_id=28849 Originator: NO Oh, it would be a Lisp special, internal to maxima, not exposed to Maxima and not intended for the general user. The default would be to return the noun form. For debugging, I would be to turn it on to enable the old behavior so I can track down the issues more easily.  Comment By: Robert Dodier (robert_dodier) Date: 20070118 10:21 Message: Logged In: YES user_id=501686 Originator: YES About the global variable, I guess I'm not opposed to that. Maybe it can be a Lisp special or maybe it can be in a new Lisp package :specint or something (in either case so it doesn't clutter the Maxima namespace).  Comment By: Raymond Toy (rtoy) Date: 20070117 20:46 Message: Logged In: YES user_id=28849 Originator: NO Since integrate doesn't call specint, I think it should be specint. The only thing preventing me from actually fixing this is that it makes it easy to figure out why specint failed. Now, it's pretty easy to find the cause. As a compromise, perhaps I'll add a global variable (yuck) to control whether the symbol or noun form is returned.  Comment By: Robert Dodier (robert_dodier) Date: 20070116 23:07 Message: Logged In: YES user_id=501686 Originator: YES Yes, I think returning a noun form is the right thing to do here. Not sure whether it should be an integrate noun or specint noun; maybe it doesn't matter too much.  Comment By: Raymond Toy (rtoy) Date: 20070115 19:42 Message: Logged In: YES user_id=28849 Originator: NO This means that maxima thought the integrand looked like t^(v1)*exp(t^2/8/a), but it didn't satisfy the requirements that Re(a) > 0 and Re(v) > 0. At this point it doesn't know how to proceed. I have not convinced myself that this Laplace transform exists. Perhaps we could just return the noun form. If we do that, the answer contains specint(%i/4*exp(b*t^2(p+%i)*t),t) + specint(%i/4*exp(b*t^2(p%i*a)*t)) which is equivalent to specint(exp(b*t^2p*t)*sin(a*t),t)/2.  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1635372&group_id=4933 
From: SourceForge.net <noreply@so...>  20070613 21:26:32

Bugs item #1714044, was opened at 20070507 01:21 Message generated for change (Comment added) made by rtoy You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1714044&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: Robert Dodier (robert_dodier) Assigned to: Nobody/Anonymous (nobody) Summary: Maxima asks unnecessary questions in integration Initial Comment: Following example shows Maxima asking a lot of questions, although the result doesn't depend on the answers; I consider that a bug. This is from the mailing list 20070413, "integration asks to many questions". (%i1) p:(1+a*cos(kp*x1)+a*cos(kp*x2)+a*a*cos(kp*x1)*cos(kp*x2))*sin(kp*(x1x2)/4); (%i2) p2:p*(1+c*cos(kp*(x1+x2)/2)); (%i3) integrate(integrate(p2^2,x1,4*%pi/kp,4*%pi/kp),x2,5*%pi/kp/2,7*% pi/kp/2); Is kp positive or negative? p; Is a zero or nonzero? n; Is a c positive, negative, or zero? n; Is c zero or nonzero? n; Is a positive or negative? n; Is c positive or negative? p; Is cos(kp x2) positive, negative, or zero? p; Is kp positive or negative? p; <result here>  >Comment By: Raymond Toy (rtoy) Date: 20070613 17:26 Message: Logged In: YES user_id=28849 Originator: NO Oops. There is one difference. Problem 209 in rtest15 returns 2/3/sqrt(2) instead of sqrt(2)/3. But these are equivalent.  Comment By: Raymond Toy (rtoy) Date: 20070613 17:11 Message: Logged In: YES user_id=28849 Originator: NO Here is a replacement for easysubs. If the antiderivative doesn't involve the inverse of a trig function, or hyperbolic function or isn't a log, we can substitute in the limits directly (if they're finite). If the limit succeeds, we are done. This change gets rid of all the questions and doesn't introduce any additional issues in the testsuite. The change is the new line containing involve. (defun easysubs (e ll ul) (cond ((or (infinityp ll) (infinityp ul)) ()) (t (cond ((or (polyinx e var ()) (not (involve e '(%log %asin %acos %atan %asinh %acosh %atanh)))) (let ((llval (noerrsub ll e)) (ulval (noerrsub ul e))) (cond ((and llval ulval) (m ulval llval)) (t ())))) (t ())))))  Comment By: Raymond Toy (rtoy) Date: 20070613 15:01 Message: Logged In: YES user_id=28849 Originator: NO FWIW, the maxima is computing this integral via methodradicalpoly. It can find the antiderivative and is now carefully substituting the limits in via intsubs. Since the antiderivative only involves trig functions (no inverses), there shouldn't be a problem just substituting in the limits. Perhaps easysubs needs to be extended to handle this case?  Comment By: Nobody/Anonymous (nobody) Date: 20070612 23:18 Message: Logged In: NO I had the same silly question when integrating exp(ïky)*hermite(n,x)*x^j  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1714044&group_id=4933 
From: SourceForge.net <noreply@so...>  20070613 21:11:53

Bugs item #1714044, was opened at 20070507 01:21 Message generated for change (Comment added) made by rtoy You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1714044&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: Robert Dodier (robert_dodier) Assigned to: Nobody/Anonymous (nobody) Summary: Maxima asks unnecessary questions in integration Initial Comment: Following example shows Maxima asking a lot of questions, although the result doesn't depend on the answers; I consider that a bug. This is from the mailing list 20070413, "integration asks to many questions". (%i1) p:(1+a*cos(kp*x1)+a*cos(kp*x2)+a*a*cos(kp*x1)*cos(kp*x2))*sin(kp*(x1x2)/4); (%i2) p2:p*(1+c*cos(kp*(x1+x2)/2)); (%i3) integrate(integrate(p2^2,x1,4*%pi/kp,4*%pi/kp),x2,5*%pi/kp/2,7*% pi/kp/2); Is kp positive or negative? p; Is a zero or nonzero? n; Is a c positive, negative, or zero? n; Is c zero or nonzero? n; Is a positive or negative? n; Is c positive or negative? p; Is cos(kp x2) positive, negative, or zero? p; Is kp positive or negative? p; <result here>  >Comment By: Raymond Toy (rtoy) Date: 20070613 17:11 Message: Logged In: YES user_id=28849 Originator: NO Here is a replacement for easysubs. If the antiderivative doesn't involve the inverse of a trig function, or hyperbolic function or isn't a log, we can substitute in the limits directly (if they're finite). If the limit succeeds, we are done. This change gets rid of all the questions and doesn't introduce any additional issues in the testsuite. The change is the new line containing involve. (defun easysubs (e ll ul) (cond ((or (infinityp ll) (infinityp ul)) ()) (t (cond ((or (polyinx e var ()) (not (involve e '(%log %asin %acos %atan %asinh %acosh %atanh)))) (let ((llval (noerrsub ll e)) (ulval (noerrsub ul e))) (cond ((and llval ulval) (m ulval llval)) (t ())))) (t ())))))  Comment By: Raymond Toy (rtoy) Date: 20070613 15:01 Message: Logged In: YES user_id=28849 Originator: NO FWIW, the maxima is computing this integral via methodradicalpoly. It can find the antiderivative and is now carefully substituting the limits in via intsubs. Since the antiderivative only involves trig functions (no inverses), there shouldn't be a problem just substituting in the limits. Perhaps easysubs needs to be extended to handle this case?  Comment By: Nobody/Anonymous (nobody) Date: 20070612 23:18 Message: Logged In: NO I had the same silly question when integrating exp(ïky)*hermite(n,x)*x^j  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1714044&group_id=4933 
From: SourceForge.net <noreply@so...>  20070613 19:50:35

Bugs item #1736719, was opened at 20070613 19:50 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=1736719&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  Taylor Group: None Status: Open Resolution: None Priority: 5 Private: No Submitted By: Sanjoy Mahajan (sm324) Assigned to: Nobody/Anonymous (nobody) Summary: taylor() segfaults when given wrong argument order Initial Comment: I called taylor() with the wrong argument order (putting the R third instead of second), and that causes a segfault if done twice: (%i1) taylor(exp(R), 0, R, 10); Maxima encountered a Lisp error: Error in PROGN [or a callee]: Caught fatal error [memory may be damaged] Automatically continuing. To reenable the Lisp debugger set *debuggerhook* to nil. (%i2) taylor(exp(R), 0, R, 10); Segmentation fault (core dumped) This is with Maxima version: 5.12.0 Maxima build date: 3:18 5/23/2007 host type: i686pclinuxgnu lispimplementationtype: GNU Common Lisp (GCL) lispimplementationversion: GCL 2.6.7 (it's the Ubuntu 5.12.01ubuntu1 package for Ubuntu 7.10 recompiled for Ubuntu 7.04.) Sanjoy  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1736719&group_id=4933 
From: SourceForge.net <noreply@so...>  20070613 19:01:05

Bugs item #1714044, was opened at 20070507 01:21 Message generated for change (Comment added) made by rtoy You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1714044&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: Robert Dodier (robert_dodier) Assigned to: Nobody/Anonymous (nobody) Summary: Maxima asks unnecessary questions in integration Initial Comment: Following example shows Maxima asking a lot of questions, although the result doesn't depend on the answers; I consider that a bug. This is from the mailing list 20070413, "integration asks to many questions". (%i1) p:(1+a*cos(kp*x1)+a*cos(kp*x2)+a*a*cos(kp*x1)*cos(kp*x2))*sin(kp*(x1x2)/4); (%i2) p2:p*(1+c*cos(kp*(x1+x2)/2)); (%i3) integrate(integrate(p2^2,x1,4*%pi/kp,4*%pi/kp),x2,5*%pi/kp/2,7*% pi/kp/2); Is kp positive or negative? p; Is a zero or nonzero? n; Is a c positive, negative, or zero? n; Is c zero or nonzero? n; Is a positive or negative? n; Is c positive or negative? p; Is cos(kp x2) positive, negative, or zero? p; Is kp positive or negative? p; <result here>  >Comment By: Raymond Toy (rtoy) Date: 20070613 15:01 Message: Logged In: YES user_id=28849 Originator: NO FWIW, the maxima is computing this integral via methodradicalpoly. It can find the antiderivative and is now carefully substituting the limits in via intsubs. Since the antiderivative only involves trig functions (no inverses), there shouldn't be a problem just substituting in the limits. Perhaps easysubs needs to be extended to handle this case?  Comment By: Nobody/Anonymous (nobody) Date: 20070612 23:18 Message: Logged In: NO I had the same silly question when integrating exp(ïky)*hermite(n,x)*x^j  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1714044&group_id=4933 
From: SourceForge.net <noreply@so...>  20070613 18:31:02

Bugs item #657382, was opened at 20021221 20:20 Message generated for change (Comment added) made by rtoy You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=657382&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: 4 Private: No Submitted By: Stavros Macrakis (macrakis) Assigned to: Nobody/Anonymous (nobody) Summary: defint/limit infinite loop Initial Comment: integrate(1/(x^51),x,1,inf) appears to get into an infinite loop inside $limit (over 4 hours CPU).  >Comment By: Raymond Toy (rtoy) Date: 20070613 14:31 Message: Logged In: YES user_id=28849 Originator: NO FWIW, this still happens in 5.12 cvs. What's happening is that maxima has computed the antiderivative correctly and is now trying to carefully substitute in the limits of integration to make sure everything is on the right sheet. This is basically done in takeprincipal and intsubs. I don't understand why maxima does the limit essentially twice like limit(anti,x,1+eps,plus)  limit(anti,x,1eps,minus). This seems to be where maxima is getting stuck. If it were to finish, maxima would then go and take the limit as eps goes to zero from above. Perhaps if the pole is at one of the limits of integration as it is here, maxima should do something else? I think the current code assumes the pole is within the integration interval.  Comment By: Stavros Macrakis (macrakis) Date: 20021221 20:21 Message: Logged In: YES user_id=588346 Sorry,. I forgot to mention that this is under 5.5 GCL/Windows 2000.  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=657382&group_id=4933 
From: SourceForge.net <noreply@so...>  20070613 16:20:01

Bugs item #917505, was opened at 20040316 13:40 Message generated for change (Comment added) made by rtoy You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=917505&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: Stavros Macrakis (macrakis) Assigned to: Nobody/Anonymous (nobody) Summary: integral: general case handled, not special Initial Comment: assume(b>1)$ integrate(1/(cos(a*x)+b),x); works but integrate(1/(cos(4*x)+3),x); does not, even though it is just the first case with a=b=2. It turns out that 4/(cos(4*x)+3) == 1/(sin(x)^4+cos(x) ^4). The sin(x)^4 form integrates, but the cos(4*x) form doesn't....  >Comment By: Raymond Toy (rtoy) Date: 20070613 12:20 Message: Logged In: YES user_id=28849 Originator: NO I agree. Closing this bug report.  Comment By: Harald Geyer (hgeyer) Date: 20070608 07:43 Message: Logged In: YES user_id=929336 Originator: NO I have tested this and similiar cases under 5.11.0 und 5.12.0 both Clisp. I can reproduce the problem with 5.11.0 but 5.12.0 seems to work and yield correct results for this and similiar cases. I think this bug report can be closed. Regards, Harald  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=917505&group_id=4933 
From: SourceForge.net <noreply@so...>  20070613 15:36:51

Bugs item #1736216, was opened at 20070612 23:16 Message generated for change (Comment added) made by rtoy You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1736216&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: Pending Resolution: None Priority: 5 Private: No Submitted By: Nobody/Anonymous (nobody) Assigned to: Nobody/Anonymous (nobody) >Summary: integration integrate(1/((x3)^4+1/2),x,0,1);=0???? Initial Comment: In the Moses tutorial, there is the integration %i1) integrate(1/((x3)^4+1/2),x,0,1); (%o1) 0 while the true result is 0.02880...... which was obtained using romberg (%i2) romberg(1/((x3)^4+1/2),x,0,1); Is that a bug, or is there a trick. (%o2) 0.028806333924554  >Comment By: Raymond Toy (rtoy) Date: 20070613 11:36 Message: Logged In: YES user_id=28849 Originator: NO Recent CVS version doesn't return 0. It returns some really long and hairy expression that can't be printed because it's too wide. But radcan(sqrtdenest(integrate(1/((x3)^4+1/2),x,0,1))); returns something sensible, which evaluates to 0.028806... What version of maxima? Marking this as pending.  Comment By: Harald Geyer (hgeyer) Date: 20070613 09:22 Message: Logged In: YES user_id=929336 Originator: NO I don't know, what the submitter (alas he left no contact info) did wrong, but I can't reproduce this bug with 5.11.0 and 5.12.0 (both clisp). Maxima gives a very long expression, which is evaluated by float() to something almost equal to the result of romberg().  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1736216&group_id=4933 
From: SourceForge.net <noreply@so...>  20070613 13:22:20

Bugs item #1736216, was opened at 20070613 05:16 Message generated for change (Comment added) made by hgeyer You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1736216&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: Nobody/Anonymous (nobody) Assigned to: Nobody/Anonymous (nobody) Summary: integration integrate(1/((x3)^4+1/2),x,0,1);=0???? Initial Comment: In the Moses tutorial, there is the integration %i1) integrate(1/((x3)^4+1/2),x,0,1); (%o1) 0 while the true result is 0.02880...... which was obtained using romberg (%i2) romberg(1/((x3)^4+1/2),x,0,1); Is that a bug, or is there a trick. (%o2) 0.028806333924554  Comment By: Harald Geyer (hgeyer) Date: 20070613 15:22 Message: Logged In: YES user_id=929336 Originator: NO I don't know, what the submitter (alas he left no contact info) did wrong, but I can't reproduce this bug with 5.11.0 and 5.12.0 (both clisp). Maxima gives a very long expression, which is evaluated by float() to something almost equal to the result of romberg().  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1736216&group_id=4933 
From: SourceForge.net <noreply@so...>  20070613 03:18:17

Bugs item #1714044, was opened at 20070506 22:21 Message generated for change (Comment added) made by nobody You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1714044&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: Robert Dodier (robert_dodier) Assigned to: Nobody/Anonymous (nobody) Summary: Maxima asks unnecessary questions in integration Initial Comment: Following example shows Maxima asking a lot of questions, although the result doesn't depend on the answers; I consider that a bug. This is from the mailing list 20070413, "integration asks to many questions". (%i1) p:(1+a*cos(kp*x1)+a*cos(kp*x2)+a*a*cos(kp*x1)*cos(kp*x2))*sin(kp*(x1x2)/4); (%i2) p2:p*(1+c*cos(kp*(x1+x2)/2)); (%i3) integrate(integrate(p2^2,x1,4*%pi/kp,4*%pi/kp),x2,5*%pi/kp/2,7*% pi/kp/2); Is kp positive or negative? p; Is a zero or nonzero? n; Is a c positive, negative, or zero? n; Is c zero or nonzero? n; Is a positive or negative? n; Is c positive or negative? p; Is cos(kp x2) positive, negative, or zero? p; Is kp positive or negative? p; <result here>  Comment By: Nobody/Anonymous (nobody) Date: 20070612 20:18 Message: Logged In: NO I had the same silly question when integrating exp(ïky)*hermite(n,x)*x^j  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1714044&group_id=4933 
From: SourceForge.net <noreply@so...>  20070613 03:16:03

Bugs item #1736216, was opened at 20070612 20:16 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=1736216&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: Nobody/Anonymous (nobody) Assigned to: Nobody/Anonymous (nobody) Summary: integration integrate(1/((x3)^4+1/2),x,0,1);=0???? Initial Comment: In the Moses tutorial, there is the integration %i1) integrate(1/((x3)^4+1/2),x,0,1); (%o1) 0 while the true result is 0.02880...... which was obtained using romberg (%i2) romberg(1/((x3)^4+1/2),x,0,1); Is that a bug, or is there a trick. (%o2) 0.028806333924554  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1736216&group_id=4933 