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From: SourceForge.net <noreply@so...>  20090904 17:15:37

Bugs item #2849942, was opened at 20090903 09:26 Message generated for change (Settings changed) made by rtoy You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=2849942&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: Harry Litzroth (hlitzroth) Assigned to: Nobody/Anonymous (nobody) Summary: log(x) with assume(x<0) and similar problems Initial Comment: The problem I think boils down to the fact, that integrates 1/x as log(x), even if x < 0. The attached file gives more instructions suffering from this problem.  >Comment By: Raymond Toy (rtoy) Date: 20090904 13:15 Message: See, for example, http://www.math.utexas.edu/pipermail/maxima/2009/017190.html. I think this is the expected behavior. Changing category to integration.  Comment By: Harry Litzroth (hlitzroth) Date: 20090903 11:57 Message: Integrating 1/x gives log x, instead of log x. If assume(x<0) Maxima nevertheless has no problem with plotting log(x) between [x^, 10^6, 10], and simply calls the assume inconistent. I will post ab htmlfile with more details.  Comment By: Raymond Toy (rtoy) Date: 20090903 09:37 Message: Can you please describe the problem here for those who do not use wxmaxima?  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=2849942&group_id=4933 
From: SourceForge.net <noreply@so...>  20090904 12:43:09

Bugs item #1797296, was opened at 20070918 22:00 Message generated for change (Settings changed) made by crategus You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1797296&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: Closed >Resolution: Fixed Priority: 5 Private: No Submitted By: Nobody/Anonymous (nobody) Assigned to: Nobody/Anonymous (nobody) Summary: Crazy results when doing limit of 'diff Initial Comment: Maxima version: 5.13.0Maxima build date: 15:45 9/16/2007host type: i586pclinuxgnulispimplementationtype: CLISPlispimplementationversion: 2.41 (20061013) (built 3380066971) (memory 3398964343) Maxima returns crazy results when evaluating the limit of an unevaluated derivative: Examples: limit('diff((x+1)/(x^21),x),x,1); limit('diff((x+1),x),x,1); limit('diff((x+n),x),x,1); Not only is the "with respect to" variable in the demoninator of the result wrong, i.e., d/d(x+1), but the limiting value of the variable is wrong. The limit was supposed to as x > 1, but the output shows the limit as x>0  reporter's email: joe.vender AT owensboro.net  >Comment By: Dieter Kaiser (crategus) Date: 20090904 14:41 Message: Applied suggested fix in limit.lisp revision 1.87. Closing this bug report as fixed. Dieter Kaiser  Comment By: Dieter Kaiser (crategus) Date: 20090829 22:15 Message: Limit does not try to simplify noun forms of derivatives, but replaces the noun forms by a gensym. This is done in $limit with a call to hide in the following line of code: (setq exp (resimplify (factosimp (tansc (lfibtophi (limitsimp ($expand (hide exp) 1 0) var)))))) I think the problem is, that hide is called to late. At this point the limit values have already been transformed. This is a piece of the corrected code: ;; Hide expressions with limit, derivative, integrate, sum ;; before any transformations of the limit values (setq exp (hide exp)) ;; Transform the limit value. (unless (infinityp val) (unless (zerop2 val) (setq exp (subin (m+ var val) exp))) (setq val (cond ((eq dr '$plus) '$zeroa) ((eq dr '$minus) '$zerob) (t 0))) (setq origval 0)) (setq exp (resimplify (factosimp (tansc (lfibtophi (limitsimp ($expand exp 1 0) var)))))) These are the results for the reported examples: (%i11) limit('diff((x+1)/(x^21),x),x,1); (%o11) 'limit('diff((x+1)/(x^21),x,1),x,1) (%i12) limit('diff((x+1),x),x,1); (%o12) 'limit('diff(x+1,x,1),x,1) (%i13) limit('diff((x+n),x),x,1); (%o13) 'limit('diff(x+n,x,1),x,1) This change solves similar problems with the limit of 'integrate too. Dieter Kaiser  Comment By: Stavros Macrakis (macrakis) Date: 20071008 16:03 Message: Logged In: YES user_id=588346 Originator: NO Dear "nobody" (20071007 22:49), the quotation mark (') in the original bug report is critical. There is no problem with limit(diff((x+1),x),x,1); there *is* a problem with limit('diff((x+1),x),x,1).  Comment By: Nobody/Anonymous (nobody) Date: 20071008 04:49 Message: Logged In: NO limit(diff((x+1)/(x^21),x),x,1); works fine . and also the others: limit(diff((x+1),x),x,1); limit(diff((x+n),x),x,1);  Comment By: Stavros Macrakis (macrakis) Date: 20070920 19:34 Message: Logged In: YES user_id=588346 Originator: NO The original bug is valid. A simple case: limit('diff(y,x),x,1) => 'limit('diff(y,x+1,1),x,0) The followup comment is confused. The syntax (..., ..., ...) in Maxima evaluates each of the elements of the list and returns the last value. This is the correct behavior.  Comment By: Nobody/Anonymous (nobody) Date: 20070918 22:45 Message: Logged In: NO also; limit(('diff(x^n),x),x,1); returns 1. Notice the mismatch of the parentheses. The problem lies in that adding ",x" after 'diff(x^n) and putting parentheses around the whole expression returns whatever is put after the comma instead of (del(x^n),x). Ex. 'diff(x^n) returns del(x^n) ('diff(x^n),x) returns x ('diff(x^n),abc) returns abc which is then evaluated by the limit function. It appears that when entering something like (f(x),f(y)) maxima always outputs f(y)  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1797296&group_id=4933 
From: SourceForge.net <noreply@so...>  20090903 16:45:49

Bugs item #2847387, was opened at 20090830 14:53 Message generated for change (Comment added) made by rtoy You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=2847387&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: Dieter Kaiser (crategus) Assigned to: Nobody/Anonymous (nobody) Summary: hgfred([3/2,b],[5/2],1) bogus Initial Comment: The general result for the definite integral integrate(sqrt(t)*(t+1)^b,t,0,1) is 2/3*hypergeometric([3/2,b],[5/2],1). The new hypergeometric code gives correct answers for b a positive integer: (%i128) 2/3*hypergeometric([3/2,1],[5/2],1); (%o128) 16/15 (%i129) 2/3*hypergeometric([3/2,2],[5/2],1); (%o129) 184/105 (%i130) 2/3*hypergeometric([3/2,3],[5/2],1); (%o130) 928/315 For negative integers I have tried hgfred. This is an example for b=3: (%i112) 2/3*hgfred([3/2,3],[5/2],1); (%o112) (12*((1/(2*(1("*"()))^3*(1))+5/(8*(1("*"()))^2*(1)^2) 15/(16*(1("*"()))*(1)^3) +15*atanh(sqrt(1))/(16*(1)^(7/2)) 3/(1("*"()))^4) *sqrt(1("*"())) 2*(1/(4*(1("*"()))^2*(1))+3/(8*(1("*"()))*(1)^2) 3*atanh(sqrt(1))/(8*(1)^(5/2)) 1/(1("*"()))^3) /sqrt(1("*"())) 3*(1/(4*(1("*"()))*(1))+atanh(sqrt(1))/(4*(2*sqrt(1))) 1/(2*(1("*"()))^2)) /(2*(2*sqrt(1("*"())))) 3*(atanh(sqrt(1))/(2*sqrt(1))1/(2*(1("*"())))) /(2*(1("*"()))^(5/2)) 15*(1atanh(sqrt(1))*sqrt(1))/(16*(1("*"()))^(7/2))) *(2*sqrt(1("*"())))*(1)^2 36*((1/(4*(1("*"()))^2*(1))+3/(8*(1("*"()))*(1)^2) 3*atanh(sqrt(1))/(8*(1)^(5/2)) 1/(1("*"()))^3) *sqrt(1("*"())) 3*(1/(4*(1("*"()))*(1))+atanh(sqrt(1))/(4*(2*sqrt(1))) 1/(2*(1("*"()))^2)) /(2*sqrt(1("*"()))) 3*(atanh(sqrt(1))/(2*sqrt(1))1/(2*(1("*"())))) /(4*(2*sqrt(1("*"())))) 3*(1atanh(sqrt(1))*sqrt(1))/(8*(1("*"()))^(5/2))) *sqrt(1("*"()))*(1)^2 +9*((1/(4*(1("*"()))*(1))+atanh(sqrt(1))/(4*(2*sqrt(1))) 1/(2*(1("*"()))^2)) *sqrt(1("*"())) (atanh(sqrt(1))/(2*sqrt(1))1/(2*(1("*"())))) /sqrt(1("*"())) (1atanh(sqrt(1))*sqrt(1))/(4*(2*sqrt(1("*"())))))*(1)^2 /sqrt(1("*"())) 48*((1/(4*(1("*"()))^2*(1))+3/(8*(1("*"()))*(1)^2) 3*atanh(sqrt(1))/(8*(1)^(5/2)) 1/(1("*"()))^3) *sqrt(1("*"())) 3*(1/(4*(1("*"()))*(1))+atanh(sqrt(1))/(4*(2*sqrt(1))) 1/(2*(1("*"()))^2)) /(2*sqrt(1("*"()))) 3*(atanh(sqrt(1))/(2*sqrt(1))1/(2*(1("*"())))) /(4*(2*sqrt(1("*"())))) 3*(1atanh(sqrt(1))*sqrt(1))/(8*(1("*"()))^(5/2))) *(2*sqrt(1("*"())))*1 +72*((1/(4*(1("*"()))*(1))+atanh(sqrt(1))/(4*(2*sqrt(1))) 1/(2*(1("*"()))^2)) *sqrt(1("*"())) (atanh(sqrt(1))/(2*sqrt(1))1/(2*(1("*"())))) /sqrt(1("*"())) (1atanh(sqrt(1))*sqrt(1))/(4*(2*sqrt(1("*"()))))) *sqrt(1("*"()))*1 +24*((1/(4*(1("*"()))*(1))+atanh(sqrt(1))/(4*(2*sqrt(1))) 1/(2*(1("*"()))^2)) *sqrt(1("*"())) (atanh(sqrt(1))/(2*sqrt(1))1/(2*(1("*"())))) /sqrt(1("*"())) (1atanh(sqrt(1))*sqrt(1))/(4*(2*sqrt(1("*"()))))) *(2*sqrt(1("*"())))) /3 The answer is not simplified and contains bad subexpressions. But if we do an extra expand we get the correct solution: (%i113) expand(%); (%o113) %pi/16 We have the same problem with other negative integers for b too. For a positive integer we get an answer in terms of the jacobi_p function which does not simplify to a rational number. There is a problem with b=2: (%i124) 2/3*hgfred([3/2,2],[5/2],1); (%o124) 16*jacobi_p(2,3/2,2*false5/2,3)/105 The answer contains the boolean value false. Dieter Kaiser  >Comment By: Raymond Toy (rtoy) Date: 20090903 12:45 Message: I think the fundamental issue is that hgfred is meant for symbolic work with symbolic argument. Now, for the first problem, hgfred([3/2,2],[5/2],1) does something bad probably from the call to subst at the end of hypatanh. Perhaps it should use $subst. The second problem with jacobi_p is another example of where maxima is assuming the argument is symbolic, not numeric. Since people use hgfred all the time with numeric argument, perhaps hgfred should either warn/error about that, or it should replace the argument with a gensym, do the simplification, and then subst/limit the result with the numeric argument.  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=2847387&group_id=4933 
From: SourceForge.net <noreply@so...>  20090903 16:02:36

Bugs item #2850079, was opened at 20090903 18:02 Message generated for change (Tracker Item Submitted) made by hlitzroth You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=2850079&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: Xmaxima or other UI Group: None Status: Open Resolution: None Priority: 5 Private: No Submitted By: Harry Litzroth (hlitzroth) Assigned to: Nobody/Anonymous (nobody) Summary: log(x), assume, integrate(1/x,x) in maxima. Initial Comment: This is only meant to send an illustrating HTMLfile for nr. 2849942.  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=2850079&group_id=4933 
From: SourceForge.net <noreply@so...>  20090903 15:57:45

Bugs item #2849942, was opened at 20090903 15:26 Message generated for change (Comment added) made by hlitzroth You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=2849942&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: Xmaxima or other UI Group: None Status: Open Resolution: None Priority: 5 Private: No Submitted By: Harry Litzroth (hlitzroth) Assigned to: Nobody/Anonymous (nobody) Summary: log(x) with assume(x<0) and similar problems Initial Comment: The problem I think boils down to the fact, that integrates 1/x as log(x), even if x < 0. The attached file gives more instructions suffering from this problem.  >Comment By: Harry Litzroth (hlitzroth) Date: 20090903 17:57 Message: Integrating 1/x gives log x, instead of log x. If assume(x<0) Maxima nevertheless has no problem with plotting log(x) between [x^, 10^6, 10], and simply calls the assume inconistent. I will post ab htmlfile with more details.  Comment By: Raymond Toy (rtoy) Date: 20090903 15:37 Message: Can you please describe the problem here for those who do not use wxmaxima?  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=2849942&group_id=4933 
From: SourceForge.net <noreply@so...>  20090903 13:37:17

Bugs item #2849942, was opened at 20090903 09:26 Message generated for change (Comment added) made by rtoy You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=2849942&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: Xmaxima or other UI Group: None Status: Open Resolution: None Priority: 5 Private: No Submitted By: Harry Litzroth (hlitzroth) Assigned to: Nobody/Anonymous (nobody) Summary: log(x) with assume(x<0) and similar problems Initial Comment: The problem I think boils down to the fact, that integrates 1/x as log(x), even if x < 0. The attached file gives more instructions suffering from this problem.  >Comment By: Raymond Toy (rtoy) Date: 20090903 09:37 Message: Can you please describe the problem here for those who do not use wxmaxima?  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=2849942&group_id=4933 
From: SourceForge.net <noreply@so...>  20090903 13:26:43

Bugs item #2849942, was opened at 20090903 15:26 Message generated for change (Tracker Item Submitted) made by hlitzroth You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=2849942&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: Xmaxima or other UI Group: None Status: Open Resolution: None Priority: 5 Private: No Submitted By: Harry Litzroth (hlitzroth) Assigned to: Nobody/Anonymous (nobody) Summary: log(x) with assume(x<0) and similar problems Initial Comment: The problem I think boils down to the fact, that integrates 1/x as log(x), even if x < 0. The attached file gives more instructions suffering from this problem.  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=2849942&group_id=4933 
From: SourceForge.net <noreply@so...>  20090902 16:13:57

Bugs item #2820202, was opened at 20090712 03:24 Message generated for change (Settings changed) made by crategus You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=2820202&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: 6 Private: No Submitted By: Barton Willis (willisbl) Assigned to: Nobody/Anonymous (nobody) Summary: rootscontract(%i/2); Initial Comment: (%i1) rootscontract(%i/2); (%o1) %i/2 (%i2) 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 (%o2)  >Comment By: Dieter Kaiser (crategus) Date: 20090902 18:13 Message: The code which causes the wrong results has been commented out. Closing this bug report as fixed. Dieter Kaiser  Comment By: Dieter Kaiser (crategus) Date: 20090829 20:39 Message: The following line in the routine rtcon in comm2.lisp is responsible for the bug: (setq e (list* (car e) 1 '((mexpt) 1 ((rat simp) 1 2)) (delete '$%i (copylist (cdr e)) :count 1 :test #'eq))) This code replaces the symbol %i with (1)*(1)^(1/2). The expression (1)^(1/2) is put on a list of roots. This is mathematically correct, but does not work. The routine rtcfixitup, which constructs the result does not handle inverse roots. In rtcfixitup (1)^(1/2) is replaced by %i but it has to be %i. I have no idea, why %i is not replaced simply by (1)^(1/2). I have tried this change. %i is replaced by (1)^(1/2): (setq e (list* (car e) '((mexpt) 1 ((rat simp) 1 2)) (delete '$%i (copylist (cdr e)) :count 1 :test #'eq)))) I have got no problems with the testsuite and the reported bug will vanish. Dieter Kaiser  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=2820202&group_id=4933 
From: SourceForge.net <noreply@so...>  20090901 21:00:02

Bugs item #2848682, was opened at 20090901 21:51 Message generated for change (Comment added) made by crategus You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=2848682&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  Assume Group: None Status: Open Resolution: None Priority: 5 Private: No Submitted By: Dieter Kaiser (crategus) Assigned to: Nobody/Anonymous (nobody) Summary: abs(log(1)) > log(1) Initial Comment: Maxima always simplifies abs(log(x)) to log(x) if x<1: (%i38) assume(x<0)$ (%i39) abs(log(x)); (%o39) log(x) This happens for negative numbers too: (%i40) abs(log(1)); (%o40) log(1) As a consequence the absolute value of e.g. log(1) is imaginary: (%i45) rectform(abs(log(1))); (%o45) %i*%pi This is the correct result: (%i46) abs(rectform(log(1))); (%o46) %pi The reason is that for every argument x<1 the function sign returns neg. (%i42) sign(log(x)); (%o42) neg But this is true only for 0 < x < 1. For x<0 sign should return pnz. This could be a correction to the routine sign in compar.lisp: ((eq (caar x) '%log) ;; Return a negative or positive sign only when the argument ;; is a real positive value in all other cases '$pnz. (cond ((eq (setq sign (compare (cadr x) 0)) '$neg) ;; Negative argument. (setq sign '$pnz)) (t ;; Positive argument to the Log function. ;; Check argument < 1 or > 1. (compare (cadr x) 1)))) Dieter Kaiser  >Comment By: Dieter Kaiser (crategus) Date: 20090901 22:59 Message: Remark: When we improve the routine sign for the log function, we get one error with the testsuite: Running tests in rtest16: ********************** Problem 112 *************** Input: limit(abs(log(x)), x, 0) Result: 'limit(abs(log(x)),x,0) This differed from the expected result: inf We get again the expected limit, when we improve the limit for the absolute value too. Dieter Kaiser  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=2848682&group_id=4933 
From: SourceForge.net <noreply@so...>  20090901 19:51:19

Bugs item #2848682, was opened at 20090901 21:51 Message generated for change (Tracker Item Submitted) made by crategus You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=2848682&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  Assume Group: None Status: Open Resolution: None Priority: 5 Private: No Submitted By: Dieter Kaiser (crategus) Assigned to: Nobody/Anonymous (nobody) Summary: abs(log(1)) > log(1) Initial Comment: Maxima always simplifies abs(log(x)) to log(x) if x<1: (%i38) assume(x<0)$ (%i39) abs(log(x)); (%o39) log(x) This happens for negative numbers too: (%i40) abs(log(1)); (%o40) log(1) As a consequence the absolute value of e.g. log(1) is imaginary: (%i45) rectform(abs(log(1))); (%o45) %i*%pi This is the correct result: (%i46) abs(rectform(log(1))); (%o46) %pi The reason is that for every argument x<1 the function sign returns neg. (%i42) sign(log(x)); (%o42) neg But this is true only for 0 < x < 1. For x<0 sign should return pnz. This could be a correction to the routine sign in compar.lisp: ((eq (caar x) '%log) ;; Return a negative or positive sign only when the argument ;; is a real positive value in all other cases '$pnz. (cond ((eq (setq sign (compare (cadr x) 0)) '$neg) ;; Negative argument. (setq sign '$pnz)) (t ;; Positive argument to the Log function. ;; Check argument < 1 or > 1. (compare (cadr x) 1)))) Dieter Kaiser  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=2848682&group_id=4933 