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From: SourceForge.net <noreply@so...>  20090819 22:55:54

Bugs item #2836164, was opened at 20090812 13:38 Message generated for change (Comment added) made by crategus You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=2836164&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: Duplicate Priority: 4 Private: No Submitted By: Barton Willis (willisbl) Assigned to: Nobody/Anonymous (nobody) Summary: rtest_boolean #100 & 101 / Fix Initial Comment: The function predreverse sometimes returns an unsimplified expression; an example from rtest_boolean #100: (%i14) implies (a%, b%) := (not a%) or b%$ (%i15) implies(aa and bb, cc or dd); 1> (PREDREVERSE ((MAND SIMP) $AA $BB)) <1 (PREDREVERSE ((MNOT) ((MAND SIMP) $AA $BB))) (%o15) notaa or notbb or cc or dd (%i16) ?print(%); ((MOR SIMP) ((MOR SIMP) ((MNOT SIMP) $AA) ((MNOT SIMP) $BB)) $CC $DD) (%o16) notaa or notbb or cc or dd A putative fix: (defmfun predreverse (pred) (take '(mnot) pred))  >Comment By: Dieter Kaiser (crategus) Date: 20090820 00:55 Message: Closing this bug report.  Comment By: Barton Willis (willisbl) Date: 20090812 13:53 Message: Sorrythis bug was reported as bug 1725951.  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=2836164&group_id=4933 
From: SourceForge.net <noreply@so...>  20090819 22:53:40

Bugs item #2761756, was opened at 20090414 13:56 Message generated for change (Comment added) made by crategus You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=2761756&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: Invalid Priority: 5 Private: No Submitted By: Nobody/Anonymous (nobody) Assigned to: Nobody/Anonymous (nobody) Summary: possible integration bug Initial Comment: I tried below. (%i2) declare(n,integer) ; (%o2) done (%i3) integrate(sin(n*%pi*x/a)*(sin(%pi*x/a))^3,x,a,a) ; (%o3) 0 Now, (sin(sin(%pi*x/a))^3 can be converted to (3*sin(%pi*x/a)sin(3*%pi*x/a))/4 and if I do above integration manually, it yields that the result is 0 for any n except 1 or 3. If n=1 result should be 3/4*a and if n=3 result should be a/4 but not zero for both cases. Is this a bug? Maxima version: 5.17.1 Maxima build date: 20:10 3/10/2009 host type: powerpcappledarwin8.11.0 lispimplementationtype: CMU Common Lisp lispimplementationversion: Stage 3 20071108T014921 (19D)  >Comment By: Dieter Kaiser (crategus) Date: 20090820 00:53 Message: When n is not declared to be an integer the answer is: (%i6) integrate(sin(n*%pi*x/a)*(sin(%pi*x/a))^3,x,a,a); (%o6) 12*a*sin(%pi*n)/(%pi*n^410*%pi*n^2+9*%pi) This answer simplifies correctly to zero when n is an integer. Closing this bug report as invalid. Dieter Kaiser  Comment By: Nobody/Anonymous (nobody) Date: 20090414 16:02 Message: Ok, it seems it an orthogonal problem. (%i7) integrate(sin(n*%pi*x)*sin(%pi*x),x,1,1) ; (%o7) 0  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=2761756&group_id=4933 
From: SourceForge.net <noreply@so...>  20090819 22:10:49

Bugs item #2827474, was opened at 20090726 23:52 Message generated for change (Comment added) made by crategus You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=2827474&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: Works For Me Priority: 5 Private: No Submitted By: Barton Willis (willisbl) Assigned to: Nobody/Anonymous (nobody) Summary: integrate(log(sec(x)),x) > lisp error Initial Comment: (%i4) integrate(log(sec(x)),x); Maxima encountered a Lisp error: (%i5) build_info(); Maxima version: 5.18.1 Maxima build date: 20:57 4/19/2009 host type: i686pcmingw32 lispimplementationtype: GNU Common Lisp (GCL) lispimplementationversion: GCL 2.6.8 (%o5)  >Comment By: Dieter Kaiser (crategus) Date: 20090820 00:10 Message: As reported in the last posting the integral of this bug report works. The answer does not look nice but can be shown to be equivalent to the result (%i*li[2](%e^(2*%i*x)))/2(%i*x^2)/2+x*log(1+%e^(2*%i*x))+x*log(sec(x)) from wolfram alpha. Closing this bug report as "works for me". Remark: Related integrals with integrands like log(sin(x)) or log(cos(x)) give an answer too. Dieter Kaiser  Comment By: Dieter Kaiser (crategus) Date: 20090727 00:21 Message: With the current CVS version and in a fresh Maxima this problem is not present: Maxima version: 5.18post Maxima build date: 0:14 7/27/2009 host type: i686pclinuxgnu lispimplementationtype: CLISP lispimplementationversion: 2.44.1 (20080223) (built 3436700604) (memory 3457635278) (%i4) integrate(log(sec(x)),x); (%o4) (x*log(sin(2*x)^2+cos(2*x)^2+2*cos(2*x)+1) +2*%i*x*atan2(sin(2*x),cos(2*x)+1)%i*li[2](%e^(2*%i*x))%i*x^2) /2 +x*log(sec(x)) I have not checked the result. Dieter Kaiser  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=2827474&group_id=4933 
From: SourceForge.net <noreply@so...>  20090819 22:05:47

Bugs item #2840634, was opened at 20090819 22:05 Message generated for change (Tracker Item Submitted) made by nobody You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=2840634&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: CLIENT: Lost socket connection Initial Comment: I have the following file (see attached). Whenever I open wxMaxima and open the file, I get the following error after the line algsys(...) : CLIENT: Lost socket connection ...Restart maxima with 'Maxima>Restart maxima'. Sometimes, after restarting maxima, it manages to automatically continue. This also happens if I introduce the algsys(...) line within maxima (not wxmaxima). I am running ubuntu 9.04  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=2840634&group_id=4933 
From: SourceForge.net <noreply@so...>  20090819 21:42:56

Bugs item #2840566, was opened at 20090819 22:34 Message generated for change (Comment added) made by crategus You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=2840566&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: Stefano Ferri (stefano_ferri) Assigned to: Nobody/Anonymous (nobody) Summary: defint fails to determine if one of its limit is real Initial Comment: In some cases defints exits with an error, saying that one limit of integration is not real, even if it's surely real. This is an example: an integration over a line, between 0 and a value that is a length computed with Pitagora's theorem (therefore, it is obviously real): (%i2) assume(a>0,b>0); (%o2) [a > 0,b > 0] (%i3) l:sqrt((ab)^2+b^2); (%o3) sqrt(b^2+(ab)^2) (%i4) defint(x,x,0,l); defint: upper limit of integration must be real; found sqrt(b^2+(ab)^2)  an error. To debug this try debugmode(true); defint somehow thinks l is not real, but it is. In fact the sum of the square of two terms both > 0, following the given assumptions a>0 and b>0, is >0, so sqrt must be real. Maxima correctly evaluates that: (%i6) is((ab)^2+b^2>0); (%o6) true (%i7) realpart(l); (%o7) sqrt(b^2+(ab)^2) (%i8) imagpart(l); (%o8) 0 Maxima correctly says that l is real, but defint does not. Somehow defint lost sign informations about the argument of the square root, therefore an imaginary part comes out and the integral is not evaluated.  >Comment By: Dieter Kaiser (crategus) Date: 20090819 23:42 Message: Fixed in defint.lisp revision 1.65. The example of this bug report works and has been added to rtest16.mac. Dieter Kaiser  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=2840566&group_id=4933 
From: SourceForge.net <noreply@so...>  20090819 21:40:53

Bugs item #2835098, was opened at 20090810 21:15 Message generated for change (Comment added) made by crategus You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=2835098&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 Group: None >Status: Closed >Resolution: Fixed Priority: 5 Private: No Submitted By: Richard Hennessy (richardhennessy) Assigned to: Nobody/Anonymous (nobody) Summary: SIGNPREP strangeness Initial Comment: (%i1) integrate(sqrt(2*m*(E[n]U(x))),x,x[0],x[0])=(n1/2)*%pi*hbar; Maxima encountered a Lisp error: Error in PROGN [or a callee]: Bind stack overflow. "Looks like a bug all right  please file a bug report if you haven't already. Part of a stack trace is shown below. The calls from SRATSIMP through SIGNPREP keep repeating. Why does it jump from SIGNPREP to SRATSIMP again? Incidentally (SRATSIMP '((MTIMES SIMP) 1 ((%IMAGPART) (($X SIMP ARRAY) 0)))) by itself doesn't cause any trouble. FWIW" "Robert Dodier" Rich  >Comment By: Dieter Kaiser (crategus) Date: 20090819 23:40 Message: Fixed in compar.lisp revision 1.52 and in rpart.lisp revision 1.27. All examples of this bug report work and have been added to rtest16.mac. Dieter Kaiser  Comment By: Robert Dodier (robert_dodier) Date: 20090811 18:12 Message: Here is a smaller example: block ([?limitp : true], sign (foo (x))); => stack overflow Looks like the problem is strange interaction among SIGNPREP, SRATSIMP, and TRISPLIT: SIGNPREP calls TRISPLIT to get real and imaginary parts, then calls SRATSIMP on the result, which calls $CSIGN (iirc) which eventually calls SIGNPREP, and the cycle repeats. See SIGNPREP in src/compar.lisp. Changing title to reflect problem more precisely.  Comment By: Barton Willis (willisbl) Date: 20090810 22:39 Message: Another example of (I'd guess) the same bug: integrate(f(x),x,x[0],x[1]);  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=2835098&group_id=4933 
From: SourceForge.net <noreply@so...>  20090819 20:37:02

Bugs item #2840566, was opened at 20090819 22:34 Message generated for change (Settings changed) made by stefano_ferri You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=2840566&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: Stefano Ferri (stefano_ferri) Assigned to: Nobody/Anonymous (nobody) Summary: defint fails to determine if one of its limit is real Initial Comment: In some cases defints exits with an error, saying that one limit of integration is not real, even if it's surely real. This is an example: an integration over a line, between 0 and a value that is a length computed with Pitagora's theorem (therefore, it is obviously real): (%i2) assume(a>0,b>0); (%o2) [a > 0,b > 0] (%i3) l:sqrt((ab)^2+b^2); (%o3) sqrt(b^2+(ab)^2) (%i4) defint(x,x,0,l); defint: upper limit of integration must be real; found sqrt(b^2+(ab)^2)  an error. To debug this try debugmode(true); defint somehow thinks l is not real, but it is. In fact the sum of the square of two terms both > 0, following the given assumptions a>0 and b>0, is >0, so sqrt must be real. Maxima correctly evaluates that: (%i6) is((ab)^2+b^2>0); (%o6) true (%i7) realpart(l); (%o7) sqrt(b^2+(ab)^2) (%i8) imagpart(l); (%o8) 0 Maxima correctly says that l is real, but defint does not. Somehow defint lost sign informations about the argument of the square root, therefore an imaginary part comes out and the integral is not evaluated.  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=2840566&group_id=4933 
From: SourceForge.net <noreply@so...>  20090819 20:35:01

Bugs item #2840566, was opened at 20090819 22:34 Message generated for change (Tracker Item Submitted) made by stefano_ferri You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=2840566&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: Stefano Ferri (stefano_ferri) Assigned to: Nobody/Anonymous (nobody) Summary: defint fails to determine if one of its limit is real Initial Comment: In some cases defints exits with an error, saying that one limit of integration is not real, even if it's surely real. This is an example: an integration over a line, between 0 and a value that is a length computed with Pitagora's theorem (therefore, it is obviously real): (%i2) assume(a>0,b>0); (%o2) [a > 0,b > 0] (%i3) l:sqrt((ab)^2+b^2); (%o3) sqrt(b^2+(ab)^2) (%i4) defint(x,x,0,l); defint: upper limit of integration must be real; found sqrt(b^2+(ab)^2)  an error. To debug this try debugmode(true); defint somehow thinks l is not real, but it is. In fact the sum of the square of two terms both > 0, following the given assumptions a>0 and b>0, is >0, so sqrt must be real. Maxima correctly evaluates that: (%i6) is((ab)^2+b^2>0); (%o6) true (%i7) realpart(l); (%o7) sqrt(b^2+(ab)^2) (%i8) imagpart(l); (%o8) 0 Maxima correctly says that l is real, but defint does not. Somehow defint lost sign informations about the argument of the square root, therefore an imaginary part comes out and the integral is not evaluated.  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=2840566&group_id=4933 