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From: SourceForge.net <noreply@so...>  20120723 16:11:34

Bugs item #3547652, was opened at 20120723 08:48 Message generated for change (Comment added) made by rtoy You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=3547652&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: Open Resolution: None Priority: 5 Private: No Submitted By: https://www.google.com/accounts () Assigned to: Nobody/Anonymous (nobody) Summary: Differentiation of a function yields wrong answer Initial Comment: build_info("5.27.0","20120508 11:27:57","i686pcmingw32","GNU Common Lisp (GCL)","GCL 2.6.8") The differentiation gives erroneous answer kr2(x):=(sin(%pi*x/2)*sin(%pi*x/3))^2; kr2(x):=(sin((%pi*x)/2)*sin((%pi*x)/3))^2 diff(kr2(x),x,1); (2*%pi*cos((%pi*x)/3)*sin((%pi*x)/3)*sin((%pi*x)/2)^2)/3 + %pi*sin((%pi*x)/3)^2*cos((%pi*x)/2)*sin((%pi*x)/2) This is clearly wrong. The right answer is: (%pi*cos((%pi*x)/3)*sin((%pi*x)/3)*sin((%pi*x)/2)^2)/3 + (%pi*sin((%pi*x)/3)^2*cos((%pi*x)/2)*sin((%pi*x)/2))/2  >Comment By: Raymond Toy (rtoy) Date: 20120723 09:11 Message: Can you explain why you think maxima is wrong? From your example, kr2(x) = kr(x)^2 where kr(x) = sin(%pi*x/2)*sin(%pi*x/3). So diff(kr2(x),x) = 2*kr(x)*diff(kr(x),x) which gives maxima's answer, not yours. In fact, maxima's result is exactly 2 times your proposed correct answer.  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=3547652&group_id=4933 
From: SourceForge.net <noreply@so...>  20120723 16:04:54

Bugs item #3547315, was opened at 20120722 09:34 Message generated for change (Comment added) made by rtoy You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=3547315&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  Floating point Group: None Status: Open Resolution: None Priority: 5 Private: No Submitted By: christoph reineke (chrisrein) Assigned to: Nobody/Anonymous (nobody) Summary: Problems with bfloat Initial Comment: Enter: (%i2) fpprec:30; (%o2) 30 (%i3) bfloat((45.12*34.78923)^2); (%o3) 2.46392687692829128354787826538b6 2.46392687692829131776b6 is the correct value. build_info("5.27.0","20120508 11:27:57","i686pcmingw32","GNU Common Lisp (GCL)","GCL 2.6.8") Regards Chris  Comment By: Raymond Toy (rtoy) Date: 20120723 09:04 Message: When you write 45.12, maxima converts that to a floating point number. But 45.12 cannot be represented exactly in floating point. Same for 34.78923. (You can see what the actual value is by using rationalize(45.12), which is close t o 4512/100, but not the same.) So maxima uses floatingpoint arithmetic to compute (45.12*34.78923)^2. Then that floating point number is converted to a bfloat, giving the result that maxima produces. If you want exact results, use exact rational arithmetic: bfloat((4512/100 * 3478923/100000)^2); 2.46392687692829131776b6 Of course, even that is an approximation because the exact result is 240617859075028449/97656250000, and that can't be represented exactly as a bfloat.  Comment By: christoph reineke (chrisrein) Date: 20120723 02:12 Message: Sorry, but I don’t understand you. <Can you explain why you think the value you give is the correct value?> Which value? (45.12*34.78923)^2=2.46392687692829131776b6=the correct value. Since I expect that the correct value has more than 16 digits after the comma, I use bfloat and fpprec=30. I quote from the documentation: “bfloat (expr) Function Converts all numbers and functions of numbers in expr to bigfloat numbers. The number of significant digits in the resulting bigfloats is specified by the global variable fpprec.” …all numbers…!! Now Maxima returns 30 digits, but the result is wrong. That’s all. <Thus, the simplifier has simplified the float expression…> Which simplifier? If a CAS computes a simple multiplication each digit of the displayed floating point number should be correct (apart from the last), no matter how many digits the user wants. Even if I erroneously “simplify” something, Maxima should never display a wrong result. That’s my opinion. How do I get the correct result? Thanks for your help!  Comment By: Raymond Toy (rtoy) Date: 20120722 13:07 Message: Can you explain why you think the value you give is the correct value? Consider this transcript: (%i1) fpprec:30; (%o1) 30 (%i2) trace(bfloat); (%o2) [bfloat] (%i3) bfloat((45.12*34.78923)^2); 1 Enter bfloat [2463926.876928291] 1 Exit bfloat 2.46392687692829128354787826538b6 (%o3) 2.46392687692829128354787826538b6 Thus, the simplifier has simplified the float expression before bfloat gets a chance to look at it.  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=3547315&group_id=4933 
From: SourceForge.net <noreply@so...>  20120723 15:48:31

Bugs item #3547652, was opened at 20120723 08:48 Message generated for change (Tracker Item Submitted) made by You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=3547652&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: Open Resolution: None Priority: 5 Private: No Submitted By: https://www.google.com/accounts () Assigned to: Nobody/Anonymous (nobody) Summary: Differentiation of a function yields wrong answer Initial Comment: build_info("5.27.0","20120508 11:27:57","i686pcmingw32","GNU Common Lisp (GCL)","GCL 2.6.8") The differentiation gives erroneous answer kr2(x):=(sin(%pi*x/2)*sin(%pi*x/3))^2; kr2(x):=(sin((%pi*x)/2)*sin((%pi*x)/3))^2 diff(kr2(x),x,1); (2*%pi*cos((%pi*x)/3)*sin((%pi*x)/3)*sin((%pi*x)/2)^2)/3 + %pi*sin((%pi*x)/3)^2*cos((%pi*x)/2)*sin((%pi*x)/2) This is clearly wrong. The right answer is: (%pi*cos((%pi*x)/3)*sin((%pi*x)/3)*sin((%pi*x)/2)^2)/3 + (%pi*sin((%pi*x)/3)^2*cos((%pi*x)/2)*sin((%pi*x)/2))/2  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=3547652&group_id=4933 
From: SourceForge.net <noreply@so...>  20120723 09:12:07

Bugs item #3547315, was opened at 20120722 09:34 Message generated for change (Comment added) made by chrisrein You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=3547315&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  Floating point Group: None Status: Open Resolution: None Priority: 5 Private: No Submitted By: christoph reineke (chrisrein) Assigned to: Nobody/Anonymous (nobody) Summary: Problems with bfloat Initial Comment: Enter: (%i2) fpprec:30; (%o2) 30 (%i3) bfloat((45.12*34.78923)^2); (%o3) 2.46392687692829128354787826538b6 2.46392687692829131776b6 is the correct value. build_info("5.27.0","20120508 11:27:57","i686pcmingw32","GNU Common Lisp (GCL)","GCL 2.6.8") Regards Chris  >Comment By: christoph reineke (chrisrein) Date: 20120723 02:12 Message: Sorry, but I don’t understand you. <Can you explain why you think the value you give is the correct value?> Which value? (45.12*34.78923)^2=2.46392687692829131776b6=the correct value. Since I expect that the correct value has more than 16 digits after the comma, I use bfloat and fpprec=30. I quote from the documentation: “bfloat (expr) Function Converts all numbers and functions of numbers in expr to bigfloat numbers. The number of significant digits in the resulting bigfloats is specified by the global variable fpprec.” …all numbers…!! Now Maxima returns 30 digits, but the result is wrong. That’s all. <Thus, the simplifier has simplified the float expression…> Which simplifier? If a CAS computes a simple multiplication each digit of the displayed floating point number should be correct (apart from the last), no matter how many digits the user wants. Even if I erroneously “simplify” something, Maxima should never display a wrong result. That’s my opinion. How do I get the correct result? Thanks for your help!  Comment By: Raymond Toy (rtoy) Date: 20120722 13:07 Message: Can you explain why you think the value you give is the correct value? Consider this transcript: (%i1) fpprec:30; (%o1) 30 (%i2) trace(bfloat); (%o2) [bfloat] (%i3) bfloat((45.12*34.78923)^2); 1 Enter bfloat [2463926.876928291] 1 Exit bfloat 2.46392687692829128354787826538b6 (%o3) 2.46392687692829128354787826538b6 Thus, the simplifier has simplified the float expression before bfloat gets a chance to look at it.  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=3547315&group_id=4933 
From: SourceForge.net <noreply@so...>  20120723 00:00:51

Bugs item #3542438, was opened at 20120710 23:37 Message generated for change (Comment added) made by rtoy You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=3542438&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  Complex Group: None Status: Open Resolution: None Priority: 5 Private: No Submitted By: Aleksas (aleksasd) Assigned to: Nobody/Anonymous (nobody) Summary: wrong residue Initial Comment: Wrong: (%i1) residue(1/(x^8+1),x,exp(%i*%pi/8)); (%o1) 0 The correct result is: (%i2) exp((7*%i*%pi)/8)/8; (%o2) %e^((7*%i*%pi)/8)/8 (%i3) rectform(%); (%o3) cos((7*%pi)/8)/8(%i*sin((7*%pi)/8))/8 (%i4) float(%), numer; (%o4) 0.047835429045636*%i0.11548494156391 or (%i5) exp(%i*%pi/8)/8; (%o5) %e^((%i*%pi)/8)/8 (%i6) rectform(%); (%o6) (%i*sin(%pi/8))/8cos(%pi/8)/8 (%i7) float(%), numer; (%o7) 0.047835429045636*%i0.11548494156391 build_info("5.27.0","20120424 08:52:03","i686pcmingw32","GNU Common Lisp (GCL)","GCL 2.6.8") Aleksas D  >Comment By: Raymond Toy (rtoy) Date: 20120722 17:00 Message: This appears to be a bug in the function coeff, called from resm1 in src/residu.lisp. AFAICT, the taylor series returns the correct value, but coeff is unable to get the coefficient of 1/(xp). A workaround is the remove the call ($ratsimp ($rectform pole)). This fixes both this bug and bug 3541292, without causing a regression in bug 1504505. But coeff should really be fixed to return the correct value.  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=3542438&group_id=4933 