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From: SourceForge.net <noreply@so...>  20100324 20:26:28

Bugs item #2973983, was opened at 20100321 05:28 Message generated for change (Comment added) made by crategus You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=2973983&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: Pending >Resolution: Invalid Priority: 5 Private: No Submitted By: Nobody/Anonymous (nobody) Assigned to: Nobody/Anonymous (nobody) Summary: limit givers complex infinity Initial Comment: Try (%i2) limit(a*x^3,x,inf); (%o2) infinity This is a bug. Rich  >Comment By: Dieter Kaiser (crategus) Date: 20100324 21:26 Message: For a positive a we get the limit inf and for a negative a the limit minf: (%i9) assume(a>0)$ (%i10) limit(a*x^2,x,inf); (%o10) inf (%i11) limit(a*x^2,x,inf); (%o11) minf Because the sign of a is not known in the example of this bug report, the result infinity seems to be a consistent answer. Setting the status to pending and the resolution to invalid. Dieter Kaiser  Comment By: Raymond Toy (rtoy) Date: 20100322 16:21 Message: Why is this a bug? a is unknown, so (complex) infinity seems like a reasonable answer.  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=2973983&group_id=4933 
From: SourceForge.net <noreply@so...>  20100324 20:22:32

Bugs item #2975992, was opened at 20100324 17:14 Message generated for change (Comment added) made by crategus You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=2975992&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: Nobody/Anonymous (nobody) Assigned to: Nobody/Anonymous (nobody) Summary: partfrac problem Initial Comment: Maxima cannot solve integrate(1/x(x+4),x), but it's not an integration problem. Maxima can integrate the expression 1/(x^2+4*x) that is the exact same, but before factorisation. To be clear, try those two commands: display(integrate(1/x(x+4),x)); and display(integrate(factor(1/(x^2+4*x)),x));  >Comment By: Dieter Kaiser (crategus) Date: 20100324 21:22 Message: Maxima can solve both integrals and they the results are equal: (%i4) integrate(1/(x*(x+4)),x); (%o4) log(x)/4log(x+4)/4 (%i5) integrate(1/(x^2+4*x),x); (%o5) log(x)/4log(x+4)/4 I think the multiplication operator is missing in the given example. It is 1/(x*(x+4)) and not 1/x(x+4). Setting the status to pending and the resolution to invalid. Dieter Kaiser  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=2975992&group_id=4933 
From: SourceForge.net <noreply@so...>  20100324 16:14:02

Bugs item #2975992, was opened at 20100324 16:14 Message generated for change (Tracker Item Submitted) made by nobody You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=2975992&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: partfrac problem Initial Comment: Maxima cannot solve integrate(1/x(x+4),x), but it's not an integration problem. Maxima can integrate the expression 1/(x^2+4*x) that is the exact same, but before factorisation. To be clear, try those two commands: display(integrate(1/x(x+4),x)); and display(integrate(factor(1/(x^2+4*x)),x));  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=2975992&group_id=4933 
From: SourceForge.net <noreply@so...>  20100324 15:45:54

Bugs item #2969546, was opened at 20100312 18:04 Message generated for change (Comment added) made by nobody You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=2969546&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  Solving equations Group: None Status: Closed Resolution: None Priority: 5 Private: No Submitted By: jamlatino (jamlatino) Assigned to: Nobody/Anonymous (nobody) Summary: Does not solve simple equation Initial Comment: Maxima is not able to solve this simple equation 400(800*(6x))/sqrt((6x)^2+4)=0  Comment By: Nobody/Anonymous (nobody) Date: 20100324 15:45 Message: You also can use the function find_root to numerically solve the equation. find_root(x=(sqrt(x^212*x+40)12)/2, x, 10, 10); ice1179  Comment By: Dieter Kaiser (crategus) Date: 20100316 21:47 Message: The Maxima function solve can not solve the equation of this example. I call it a missing feature of the function solve and not a bug. From the viewpoint of the algorithm of solve we have not a "simple equation". But we have the package to_poly_solver which extends the functionality of solve. to_poly_solve has the feature to solve this equation. Therefore, Maxima can solve this equation. I have added this example to a feature request to put the functionality of to_poly_solve incore. Dieter Kaiser  Comment By: Stefan Dermitzakis (stefand) Date: 20100316 13:52 Message: I have the same problem and sorry, i think you all have not understand the problem. The bug is that maxima cannot recognize the unknow x inside the sqrt(.....) in the equation. Note: the old Mcsyma have the same problem. Possibly many other bugs in maxima have the same cause.  Comment By: Dieter Kaiser (crategus) Date: 20100314 13:54 Message: The example of this bug report has been added to the open feature request ID: 2617416 "to_poly_solve in core, was: Maxima can't solve equation". Closing this bug report. Dieter Kaiser  Comment By: Dieter Kaiser (crategus) Date: 20100312 19:19 Message: Maxima is able to solve this equation with the Maxima function to_poly_solve. The package to_poly_solver has to be loaded first: (%i3) load(to_poly_solver); (%o3) "/usr/local/share/maxima/5.20post/share/contrib/to_poly_solver.mac" (%i4) eqn:400(800*(6x))/sqrt((6x)^2+4)=0$ (%i5) to_poly_solve(eqn,x); (%o5) %union([x = (2*3^(3/2)2)/sqrt(3)]) I suggest to add a comment about to_poly_solve to the documentation of solve and close this bug report. It might be arguable to extend the Maxima function solve to handle this equation too. But this would be a feature request. Dieter Kaiser  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=2969546&group_id=4933 
From: SourceForge.net <noreply@so...>  20100324 09:32:16

Bugs item #2953369, was opened at 20100217 03:56 Message generated for change (Comment added) made by dgildea You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=2953369&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: Definite Integration of 1/(ab*cos(x)) wrong Initial Comment: Maxima 5.20.1 with wxMaxima. integrate(1/(ab*cos(x)),x,0,%pi); where a>0, 0<b<a yields 0.  >Comment By: Dan Gildea (dgildea) Date: 20100324 05:32 Message: possible solution: (defun unitcir (grand var) (numden grand) (let ((result (princip (res nn* dn* #'(lambda (pt) (eq (let ((limitp nil)) ($asksign (m+ 1 (cabs pt)))) '$neg)) #'(lambda (pt) (alike1 1 (cabs pt))))))) (cond (result (m* '$%pi result)) (t nil))))  Comment By: Raymond Toy (rtoy) Date: 20100323 08:03 Message: The incorrect result comes from polelist failing to identify the locations of the poles. Since the integrand is even, we can integrate from %pi to %pi (or 0 to 2*%pi) and take half of the result. This integral is converted to the contour integral of 2/(b*yy^22*a*y+b) around the unit circle. This is evaluated by residues. We want to find the poles inside the unit circle and polelist is supposed to do that. The poles are correctly determined, but unfortunately for these poles, polelist cannot find the one pole that is in the circle. Therefore the function res thinks there are no poles in the unit circle and returns 0. When a and b are numbers, polelist does a better job and normally determines the pole that is within the unit circle. Perhaps the function that determines whether the pole is in the unit circle needs to be enhanced?  Comment By: Barton Willis (willisbl) Date: 20100226 01:45 Message: Notice how a float enters into the asksign: (%i4) integrate(1/(1a*cos(x)),x); Is a^21.0 positive or negative?neg; (%o4) (2*atan(((2*a+2)*sin(x))/(2*sqrt(1a^2)*(cos(x)+1))))/sqrt(1a^2)  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=2953369&group_id=4933 