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From: SourceForge.net <noreply@so...>  20090829 20:15:25

Bugs item #1797296, was opened at 20070918 22:00 Message generated for change (Comment added) 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: Open Resolution: None 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: 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...>  20090829 19:34:01

Bugs item #2846949, was opened at 20090829 20:42 Message generated for change (Comment added) made by crategus You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=2846949&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: Nobody/Anonymous (nobody) Assigned to: Nobody/Anonymous (nobody) Summary: ilt(expr,s,t) cannot calculate some difficult expressions Initial Comment: Hi! I use wxmaxima 0.8.2 in Ubuntu. I have some rational function and I need to do inverse Laplace transformation. So I use ilt(), but it can't solve this. I tryed maxima 5.10.0 and 5.13.0, the result is the same (see in the file). In 5.17.1 there is a Lisp error when I'm trying to evaluate. But when I try a slightly changed function (see in the file too) it can be solved. Mathcad 14 for Windows could even solve this with the first function. The result is in the file. But when I did a Laplace transformation with the result, mathcad gave me a very difficult solvation, and using wxmaxima I got a very simple result. Both of these results weren't the primary functon. So I don't know the right answer but I really need to!  >Comment By: Dieter Kaiser (crategus) Date: 20090829 21:33 Message: I have tried the examples with current Maxima 5.19post. Furthermore I have reformulated the integral a bit. It is equivalent, but looks simpler. You are right we get no solution for the following expression: (%i2) ilt(1/(s^2*(1 + a*s)*(1 + b*s)*(1 + c*s) + d*s),s,t); (%o2) 'ilt((a*b*c*s^3+((b+a)*c+a*b)*s^2+(c+b+a)*s+1) /(d*(a*b*c*s^4+((b+a)*c+a*b)*s^3+(c+b+a)*s^2+s+d)),s,t) +1/d As you have observed, we get a solution, when we omit the extra term +d*s: (%i3) ilt(1/(s^2*(1 + a*s)*(1 + b*s)*(1 + c*s)),s,t); (%o3) c^3*%e^(t/c)/(c^2+(ba)*c+a*b)b^3*%e^(t/b)/((ba)*cb^2+a*b) +a^3*%e^(t/a)/((ba)*ca*b+a^2)+tcba The Laplace transformation gives the original expression: (%i4) laplace(%,t,s); (%o4) c^3/((c^2+(ba)*c+a*b)*(s+1/c))b^3/(((ba)*cb^2+a*b)*(s+1/b)) +a^3/(((ba)*ca*b+a^2)*(s+1/a))c/sb/s a/s+1/s^2 We can see it more easy, when we factor the last result: (%i5) factor(%); (%o5) 1/(s^2*(a*s+1)*(b*s+1)*(c*s+1)) I can not see a bug. I have tried the first integral with Wolfram alpha, but get no solution too. Perhaps you can post the expected answer for the first integral. Dieter Kaiser  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=2846949&group_id=4933 
From: SourceForge.net <noreply@so...>  20090829 18:42:59

Bugs item #2846949, was opened at 20090829 18:42 Message generated for change (Tracker Item Submitted) made by nobody You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=2846949&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: Nobody/Anonymous (nobody) Assigned to: Nobody/Anonymous (nobody) Summary: ilt(expr,s,t) cannot calculate some difficult expressions Initial Comment: Hi! I use wxmaxima 0.8.2 in Ubuntu. I have some rational function and I need to do inverse Laplace transformation. So I use ilt(), but it can't solve this. I tryed maxima 5.10.0 and 5.13.0, the result is the same (see in the file). In 5.17.1 there is a Lisp error when I'm trying to evaluate. But when I try a slightly changed function (see in the file too) it can be solved. Mathcad 14 for Windows could even solve this with the first function. The result is in the file. But when I did a Laplace transformation with the result, mathcad gave me a very difficult solvation, and using wxmaxima I got a very simple result. Both of these results weren't the primary functon. So I don't know the right answer but I really need to!  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=2846949&group_id=4933 
From: SourceForge.net <noreply@so...>  20090829 18:40:00

Bugs item #2820202, was opened at 20090712 03:24 Message generated for change (Comment added) 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: Open Resolution: None 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: 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...>  20090829 11:06:37

Bugs item #2846665, was opened at 20090828 22:20 Message generated for change (Comment added) made by willisbl You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=2846665&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: Nobody/Anonymous (nobody) Assigned to: Nobody/Anonymous (nobody) Summary: Multiplication of block matrices Initial Comment: a: matrix( [1,2], [3,4] ); b: matrix( [a,a], [a,a] ); c:b.b; gives the output for c matrix([matrix([2,8],[18,32]),matrix([2,8],[18,32])],[matrix([2,8],[18,32]),matrix([2,8],[18,32])]) each element in 'c' is the corresponding element of 'a' squared and then multiplied by 2. I think the result of the multiplication of block matrices should be the same as if the matrices were unblocked.  Maxima version: 5.19.0 Maxima build date: 20:33 8/9/2009 host type: i686pcmingw32 lispimplementationtype: GNU Common Lisp (GCL) lispimplementationversion: GCL 2.6.8  >Comment By: Barton Willis (willisbl) Date: 20090829 06:06 Message: Maybe you want to set matrix_element_mul to "." (%i6) matrix_element_mul : "."$ (%i7) a: matrix([1,2], [3,4]); b: matrix([a,a],[a,a]); c:b.b; (%o7) matrix([1,2],[3,4]) (%o8) matrix([matrix([1,2],[3,4]),matrix([1,2],[3,4])],[matrix([1,2],[3,4]),matrix([1,2],[3,4])]) (%o9) matrix([matrix([2,8],[18,32]),matrix([2,8],[18,32])],[matrix([2,8],[18,32]),matrix([2,8],[18,32])])  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=2846665&group_id=4933 
From: SourceForge.net <noreply@so...>  20090829 03:20:57

Bugs item #2846665, was opened at 20090829 03:20 Message generated for change (Tracker Item Submitted) made by nobody You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=2846665&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: Nobody/Anonymous (nobody) Assigned to: Nobody/Anonymous (nobody) Summary: Multiplication of block matrices Initial Comment: a: matrix( [1,2], [3,4] ); b: matrix( [a,a], [a,a] ); c:b.b; gives the output for c matrix([matrix([2,8],[18,32]),matrix([2,8],[18,32])],[matrix([2,8],[18,32]),matrix([2,8],[18,32])]) each element in 'c' is the corresponding element of 'a' squared and then multiplied by 2. I think the result of the multiplication of block matrices should be the same as if the matrices were unblocked.  Maxima version: 5.19.0 Maxima build date: 20:33 8/9/2009 host type: i686pcmingw32 lispimplementationtype: GNU Common Lisp (GCL) lispimplementationversion: GCL 2.6.8  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=2846665&group_id=4933 