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From: SourceForge.net <noreply@so...>  20071019 21:58:54

Bugs item #1816801, was opened at 20071019 14:58 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=1816801&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: Nobody/Anonymous (nobody) Assigned to: Nobody/Anonymous (nobody) Summary: Equations scaled incorrectly Initial Comment: Now, this could be me, but I think that Maxima is not meant to actually change your expression when you ask it to factor it. I think this session best describes the problem: (%i48) (10*y11*%i+2)*(10*z+11*%i+2); (%o48) (10*y11*%i+2)*(10*z+11*%i+2) (%i49) factor(%); (%o49) (10*y11*%i+2)*(10*z+11*%i+2) (%i50) expand(%); (%o50) 100*y*z110*%i*z+20*z+110*%i*y+20*y+125 (%i51) factor(%); (%o51) 5*(10*y11*%i+2)*(10*z+11*%i+2) (%i52) expand(%); (%o52) 500*y*z550*%i*z+100*z+550*%i*y+100*y+625 (%i53) factor(%); (%o53) 25*(10*y11*%i+2)*(10*z+11*%i+2) (%i54) expand(%); (%o54) 2500*y*z2750*%i*z+500*z+2750*%i*y+500*y+3125 (%i55) factor(%); (%o55) 125*(10*y11*%i+2)*(10*z+11*%i+2) (%i56) expand(%); (%o56) 12500*y*z13750*%i*z+2500*z+13750*%i*y+2500*y+15625 (%i57) factor(%); (%o57) 625*(10*y11*%i+2)*(10*z+11*%i+2) (%i58) expand(%); (%o58) 62500*y*z68750*%i*z+12500*z+68750*%i*y+12500*y+78125 Maxima version: 5.10.0 Maxima build date: 14:57 10/18/2006 host type: i686pclinuxgnu lispimplementationtype: GNU Common Lisp (GCL) lispimplementationversion: GCL 2.6.7  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1816801&group_id=4933 
From: SourceForge.net <noreply@so...>  20071019 14:23:35

Bugs item #1778796, was opened at 20070821 12:36 Message generated for change (Comment added) made by rtoy You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1778796&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( (x^3+1)/(x^4 + 4*x + 1), 0, 1); Initial Comment: Maxima: 5.11.0 Lisp: SBCL 1.0.3 On the mailing list, I asked about a definite integral that seems to make Maxima hang. 8< (%i1) integrate( (x^3+1)/(x^4 + 4*x + 1), x); 4 log(x + 4 x + 1) (%o1)  4 (%i2) integrate( (x^3+1)/(x^4 + 4*x + 1), x, 0, 1); [... no output, sblc cpu usage at 87% ...]  Barton Willis replied and suggested to file a bug report. He said: > > To see in part what is going on, try this: > > (%i4) trace(?csign); > (%o4) [csign] > (%i5) integrate( (x^3+1)/(x^4 + 4*x + 1), x, 0, 1); > > > Yikes! all kinds of junk! > > Also try integrate((x^3+1)/(x^4 + 4*x + 1), x,a,b). > > I suppose that Maxima is struggling to show that the antiderivative > is continuous on [0,1]. But Maxima goes about it in just about the > worst of all possible ways. Maxima does know that x^4 + 4*x + 1 is > positive > for x in [0,1], so it should be able to determine that > log(x^4 + 4*x + 1) is continuous on [0,1]. > > (%i1) assume(x >= 0, x<=1); > (%o1) [x>=0,x<=1] > (%i2) sign(x^4 + 4*x + 1); > (%o2) pos  >Comment By: Raymond Toy (rtoy) Date: 20071019 10:23 Message: Logged In: YES user_id=28849 Originator: NO A summary of some emails. Maxima has converted the integral to a contour integral around a keyhole contour (circle with a slit on the positive real axis). The denominator is a quartic with 4 complex (and messy) roots. Maxima is using residues to evaluate the integral. If you wait long enough (and have enough memory), maxima does finally finish. (It took some hour or two). The result is too long to display, but if I bfloat the resullt, I get the expected numerical answer. (Which is a little surprising because there are some very large integers of 50+ digits or more in the result.) Since this is a rational function, I tried an experiment with ratfnt trying to do the antiderivative first before trying the contour integral. This works, and doesn't cause the test suite to fail. Two tests do fail, but that is because the form of the answer has changed, not the value.  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1778796&group_id=4933 
From: SourceForge.net <noreply@so...>  20071019 10:08:13

Bugs item #1812167, was opened at 20071012 12:53 Message generated for change (Comment added) made by hgeyer You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1812167&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: Installation Group: None >Status: Closed >Resolution: Fixed Priority: 5 Private: No Submitted By: Harald Geyer (hgeyer) Assigned to: Nobody/Anonymous (nobody) Summary: current cvs still installs as 5.12.99rc1 Initial Comment: Hi, it seems nobody has updated the version string in cvs since the last release. Thus current cvs installs with an lower version string than 5.13. Sorry, I can't fix it myself right now, because I don't have my cvs ssh key ready. best, Harald  >Comment By: Harald Geyer (hgeyer) Date: 20071019 12:08 Message: Logged In: YES user_id=929336 Originator: YES Fixed by configure.in revision 1.94, Thu Oct 18 13:39:45 2007 UTC  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1812167&group_id=4933 
From: SourceForge.net <noreply@so...>  20071019 02:20:14

Bugs item #1605716, was opened at 20061129 15:53 Message generated for change (Comment added) made by sfrobot You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1605716&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: Closed Resolution: Works For Me Priority: 3 Private: No Submitted By: Barton Willis (willisbl) Assigned to: Nobody/Anonymous (nobody) Summary: taylor at infinity for non algebraic Initial Comment: Consider: (%i1) log(x) + exp(x^2)$ (%i2) e : diff(%,x)/%; (%o2) (2*x*%e^x^2+1/x)/(log(x)+%e^x^2) Towards infinity, a good approximation to e is 2 x. But (%i3) taylor(e,x,inf,2); (%o3) 4/x+... And (%i4) taylor(e,x,inf,5); (%o4) 240/x^5+... And (%i5) taylor(ratsimp(e),x,inf,5); 1/(x*log(x)+x*%e^x^2) Assumed to be zero in `taylor' (%o5) 0+... It would be better if taylor just gave up. All the results are not right. Barton  >Comment By: SourceForge Robot (sfrobot) Date: 20071018 19:20 Message: Logged In: YES user_id=1312539 Originator: NO This Tracker item was closed automatically by the system. It was previously set to a Pending status, and the original submitter did not respond within 14 days (the time period specified by the administrator of this Tracker).  Comment By: Dan Gildea (dgildea) Date: 20071004 03:46 Message: Logged In: YES user_id=1797506 Originator: NO These seem ok in current cvs. (%i7) log(x) + exp(x^2)$ (%i8) e : diff(%,x)/%; (%o8) (2*x*%e^x^2+1/x)/(log(x)+%e^x^2) (%i9) taylor(e,x,inf,2); (%o9) +2*x+((+(2*log(x)))*x+1/x)*%e^x^2 +((+2*log(x)^2)*x+(+(log(x)))/x)*(%e^x^2)^2 (%i10) taylor(e,x,inf,5); (%o10) +2*x+((+(2*log(x)))*x+1/x)*%e^x^2 +((+2*log(x)^2)*x+(+(log(x)))/x)*(%e^x^2)^2 +((+(2*log(x)^3))*x+(+log(x)^2)/x)*(%e^x^2)^3 +((+2*log(x)^4)*x+(+(log(x)^3))/x)*(%e^x^2)^4 +((+(2*log(x)^5))*x+(+log(x)^4)/x)*(%e^x^2)^5 (%i11) taylor(ratsimp(e),x,inf,5); (%o11) +2*x+((+(2*log(x)))*x+1/x)*%e^x^2 +((+2*log(x)^2)*x+(+(log(x)))/x)*(%e^x^2)^2 +((+(2*log(x)^3))*x+(+log(x)^2)/x)*(%e^x^2)^3 +((+2*log(x)^4)*x+(+(log(x)^3))/x)*(%e^x^2)^4 +((+(2*log(x)^5))*x+(+log(x)^4)/x)*(%e^x^2)^5 (%i12) log(log(x)) + exp(x)$ (%i13) diff(%,x)/%; (%o13) (1/(x*log(x))%e^x)/(log(log(x))+%e^x) (%i14) taylor(%,x,inf,2); (%o14) +(+(+1/log(log(x)))/log(x))/x+(+(1/log(log(x))) +(+(+(1/log(log(x))^2))/log(x))/x) *%e^x+(+1/log(log(x))^2)*(%e^x)^2 (%i15) log(log(x)) + x$ (%i16) diff(%,x)/%; (%o16) (1/(x*log(x))+1)/(log(log(x))+x) (%i17) taylor(%,x,inf,2); (%o17) 1/x+(+(log(log(x)))+1/log(x))/x^2  Comment By: Barton Willis (willisbl) Date: 20061201 02:54 Message: Logged In: YES user_id=895922 Originator: YES Two related problems: (%i1) log(log(x)) + exp(x)$ (%i2) diff(%,x)/%$ (%i3) taylor(%,x,inf,2); Invalid call to varexpand (%i4) log(log(x)) + x$ (%i5) diff(%,x)/%$ (%i6) taylor(%,x,inf,2); (%o6) 1/x+(1/log(x)+(log(log(x))+zeroa+...)+...)/x^2+... Isn't this unsimplified? I don't know what taylor is trying to do with expressions similar to %o5. Again, maybe it would be better if taylor gave up.  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1605716&group_id=4933 