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From: <noreply@so...>  20020708 18:28:12

Bugs item #578806, was opened at 20020708 13:28 You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=578806&group_id=4933 Category: None Group: None Status: Open Resolution: None Priority: 5 Submitted By: Barton Willis (willisb) Assigned to: Nobody/Anonymous (nobody) Summary: two bugs in antid.mac Initial Comment: 1. nonzeroandfreeof, defined in antid.mac, has a bug: /* nonzeroandfreeof(x,0) should evaluate to false because the second argument is zero. */ (C1) nonzeroandfreeof(x,0); (D1) TRUE To fix this, change (a typo in second argument) NONZEROANDFREEOF(XARGUMENTX, BARGUMENTE) := IS(EARGUMENTE # 0 AND FREEOF(XARGUMENTX, EARGUMENTE))$ to NONZEROANDFREEOF(XARGUMENTX, EARGUMENTE) := IS(EARGUMENTE # 0 AND FREEOF(XARGUMENTX, EARGUMENTE))$ (What's the story with these lengthy identifiers?) 2. antidiff, also defined in antid.mac, assumes that listarith is true; setting it to false, we get sillyness: (C2) listarith : false; (D2) FALSE (C3) antidiff(x + diff(f(x),x),x,f(x)); (D3) 'INTEGRATE([x^2/2,0],x)+[f(x),0] /* setting listarith to true, we get the correct value. */ (C4) listarith : true; (D4) TRUE (C5) antidiff(x + diff(f(x),x),x,f(x)); (D5) f(x)+x^2/2 To fix this, in the function antid locally bind listarith to true. Thus change ANTID(FN, VAR, UNK) := BLOCK( [LCLVARIABLELCL, DVARIABLED, AARGUMENTA, BARGUMENTB, TVARIABLET, UVARIABLEU, KILLER], DVARIABLED : DERIVDEGREE(FN, UNK, VAR), ... to ANTID(FN, VAR, UNK) := BLOCK( [LCLVARIABLELCL, DVARIABLED, AARGUMENTA, BARGUMENTB, TVARIABLET, UVARIABLEU, KILLER, listarith : true], ... Barton  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=578806&group_id=4933 
From: <noreply@so...>  20020708 17:35:51

Bugs item #541030, was opened at 20020408 14:21 You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=541030&group_id=4933 Category: None Group: None Status: Open Resolution: None Priority: 9 Submitted By: David Billinghurst (billingd) Assigned to: Nobody/Anonymous (nobody) Summary: integrate(sqrt(x+1/x2),x,0,1) wrong Initial Comment: The definite integral integrate(sqrt(x+1/x2),x,0,1) => 4/3 with cvs maxima/gcl under windows. The answer should be 4/3 according to Michael Wester, "A Review of CAS Mathematical Capabilities", 15 April 1995. (Problem 84). It definitely should be positive.  Comment By: Pedro Fortuny Ayuso (pfortuny) Date: 20020708 17:35 Message: Logged In: YES user_id=519681 FWIW, mupad 2.5 linux gives: int(sqrt(x+1/x2),x); > int(sqrt(x+1/x2),x); (i.e. not evaluated). while int(abs(x1)/sqrt(x),x); > sign(x1)(2*x^(3/2)/32*x^(1/2)) which, when evaluated at 1 gives: 4/3*sign(0). ... ... ...  Comment By: Juan Hierro (buscaideas) Date: 20020612 00:03 Message: Logged In: YES user_id=528795 I am not sure whether this should be a real bug or just a funny consequence of the indetermination in the sign of a square root. Looking at the comment by toy@..., one may realize that the result is correct provided that the sign of the square root in the integrand is shared with that of (x1). This problems appears in much simpler funcions too. For instance, let us say one has f(x):=sqrt(x*x2*x+1); Then, the sin algorithm performs the integration with sqrt always positive, while the risch algorithm works with sqrt sharing the sign of (x1). factor(diff(integrate(f(x),x),x)); ==> sqrt(x*x2*x+1) factor(diff(risch(f(x),x),x)); ==> (x1) Is it wrong this behaviour? Should there be any way to specify which branch to employ when handling multivaluated functions? f(x) in this last situation may be either abs(x1), or (x1), or abs(x1), or (1x). sin seems to work with abs(x1) and risch with (x1). In the same way, sqrt(x+1/x2) is either abs(x1)/sqrt(x) or (x1)/sqrt(x) where both signs are admisible in sqrt(x). In this case, both sin and risch algorithms seem to work with (x1)/sqrt(x).  Comment By: David Billinghurst (billingd) Date: 20020410 03:23 Message: Logged In: YES user_id=365569 toy@... (toy@...) wrote: I guess maxima gets this wrong because it says: (C1) integrate(sqrt(x+1/x2),x); 3/2 2 x  6 SQRT(x) (D1)  3 which is only true if x1 is positive. For some reason it has assumed x1 is positive somewhere during integration. Yet another integration bug. Ray  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=541030&group_id=4933 
From: <noreply@so...>  20020708 16:42:55

Bugs item #578757, was opened at 20020708 16:42 You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=578757&group_id=4933 Category: None Group: None Status: Open Resolution: None Priority: 5 Submitted By: Pedro Fortuny Ayuso (pfortuny) Assigned to: Nobody/Anonymous (nobody) Summary: Emaxima update all GUI does not work Initial Comment: When I try to update all or tex update all cells in an Emaxima file, the following happens: If I call that process from the X menus (pop down menu, submenu, etc...), it does not work: after asking "interactive update", any of the answers (y or n) makes emacs stop (recoverable by pressing Cg Cg, though). If I do it by keyboard (either Mx maximaupdateall or Cc Cu a), then it works properly. Thanks, Pedro Fortuny.  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=578757&group_id=4933 