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From: SourceForge.net <noreply@so...>  20070117 14:19:02

Bugs item #1635365, was opened at 20070114 16:00 Message generated for change (Comment added) made by rtoy You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1635365&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: Robert Dodier (robert_dodier) Assigned to: Nobody/Anonymous (nobody) Summary: specint calls asksign about internal variable PSEY Initial Comment: assume(p > 0, a > 0); t^(1/2)*%e^(p*ta/t); specint(%,t); => "Is psey positive, negative, or zero?" PSEY is an internal variable, should not be visible to user. See comments in src/hypgeo.lisp about PSEY. >From the comments, PSEY represents the parameter in a Laplace transform. Maybe it will resolve the problem to do assume(psey > 0) or something like that. Ideally that assumption would be attached to the result (after substituting the original parameter symbol for PSEY).  >Comment By: Raymond Toy (rtoy) Date: 20070117 09:18 Message: Logged In: YES user_id=28849 Originator: NO By not doing the substitution, maxima no longer asks questions about psey. That's good. However, problem 53 in rtest14 and problems 111, 112 in rtesthyp. I think problem 53 is ok, just a different respresntation that maxima didn't simplify. However problems 111 and 112 are more fundamental. It calls LTEP to match a form, but the Laplace variable is p+%i*a instead of a simple variable. Perhaps some changes to LTEP could fix this.  Comment By: Robert Dodier (robert_dodier) Date: 20070114 16:31 Message: Logged In: YES user_id=501686 Originator: YES Another example: assume (a > 0, p > 0); t*bessel_i(0,a*t/2)*bessel_i(1,a*t/2)*%e^(p*t); specint(%,t); => Is (psey  a) (psey + a) positive, negative, or zero?  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1635365&group_id=4933 
From: SourceForge.net <noreply@so...>  20070117 04:59:22

Bugs item #1636971, was opened at 20070116 10:42 Message generated for change (Comment added) made by robert_dodier You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1636971&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: Problem not in Maxima Group: None >Status: Pending Resolution: None Priority: 5 Private: No Submitted By: Nobody/Anonymous (nobody) Assigned to: Nobody/Anonymous (nobody) Summary: ?? Initial Comment: Maxima version: 5.9.3.99rc3Maxima build date: 12:37 9/3/2006host type: i686pcmingw32lispimplementationtype: GNU Common Lisp (GCL)lispimplementationversion: GCL 2.6.7 I wanted to know differential from ((x+k*sqrt(1x^2))^(k*asin(x))); I've written: diff((x+k*sqrt(1x^2))^(k*asin(x))); I 've seen: (k*sqrt(1x^2)+x)^(k*asin(x))*((k*log(k*sqrt(1x^2)+x))/sqrt(1x^2)+(k*(1(k*x)/sqrt(1x^2))*asin(x))/(k*sqrt(1x^2)+x))*del(x)+(k*sqrt(1x^2)+x)^(k*asin(x))*(asin(x)*log(k*sqrt(1x^2)+x)+(k*sqrt(1x^2)*asin(x))/(k*sqrt(1x^2)+x))*del(k) After simplify: (((2*k^2*x^32*k^2*x)*(k*sqrt(1x^2)+x)^(k*asin(x))*log(k*sqrt(1x^2)+x)+((kk^3)*x^3+(k^3k)*x)*(k*sqrt(1x^2)+x)^(k*asin(x))*asin(x))*del(x)+(((1k^2)*x^4+(2*k^21)*x^2k^2)*(k*sqrt(1x^2)+x)^(k*asin(x))*asin(x)*log(k*sqrt(1x^2)+x)+(k^2*x^4+2*k^2*x^2k^2)*(k*sqrt(1x^2)+x)^(k*asin(x))*asin(x))*del(k)+sqrt(1x^2)*((((k^3k)*x^2k^3)*(k*sqrt(1x^2)+x)^(k*asin(x))*log(k*sqrt(1x^2)+x)+(2*k^2*x^2k^2)*(k*sqrt(1x^2)+x)^(k*asin(x))*asin(x))*del(x)+((2*k*x^32*k*x)*(k*sqrt(1x^2)+x)^(k*asin(x))*asin(x)*log(k*sqrt(1x^2)+x)+(k*x^3k*x)*(k*sqrt(1x^2)+x)^(k*asin(x))*asin(x))*del(k)))/((1k^2)*x^4+sqrt(1x^2)*(2*k*x^32*k*x)+(2*k^21)*x^2k^2) But: diff((x+k*sqrt(1x^2))^(k*asin(x))) = (1+k^2)*%e^(k*asin(x))  >Comment By: Robert Dodier (robert_dodier) Date: 20070116 21:59 Message: Logged In: YES user_id=501686 Originator: NO A couple of comments  (1) Probably you want to specify a variable to differentiate with respect to  e.g. diff(<stuff>, x) instead of just diff(<stuff>). (2) I don't see any bug in diff here  I'm guessing the problem is that Maxima is not simplifying the result as much as you want it to. You can take up this question on the mailing list. (http://maxima.sourceforge.net/maximalist.html) (3) I guess I don't see that diff((x+k*sqrt(1x^2))^(k*asin(x))) = (1+k^2)*%e^(k*asin(x))  the righthand side seems to be somewhat different from what Maxima computes for diff((x+k*sqrt(1x^2))^(k*asin(x)), x) (which correct if I'm not mistaken). I've set the status of this report to "pending" so that it is closed automatically in 2 weeks since it doesn't appear to be a bug in diff.  Comment By: Raymond Toy (rtoy) Date: 20070116 11:36 Message: Logged In: YES user_id=28849 Originator: NO What do you want diff to do? I don't understand how you got your expected answer.  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1636971&group_id=4933 
From: SourceForge.net <noreply@so...>  20070117 04:36:11

Bugs item #1636926, was opened at 20070116 09:31 Message generated for change (Comment added) made by robert_dodier You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1636926&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: Tests Group: None Status: Open Resolution: None Priority: 5 Private: No Submitted By: Barton Willis (willisbl) Assigned to: Nobody/Anonymous (nobody) Summary: rtest16 #39 Initial Comment: Test 39 in rtest16 is: /* Bug 593344 */ limit(abs(infinity)); infinity; The correct answer is inf, not infinity. I believe.  >Comment By: Robert Dodier (robert_dodier) Date: 20070116 21:36 Message: Logged In: YES user_id=501686 Originator: NO Agreed, should be inf.  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1636926&group_id=4933 
From: SourceForge.net <noreply@so...>  20070117 04:07:55

Bugs item #1635372, was opened at 20070114 14:10 Message generated for change (Comment added) made by robert_dodier You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1635372&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: Robert Dodier (robert_dodier) Assigned to: Nobody/Anonymous (nobody) Summary: specint returns expression containing internal Lisp variable Initial Comment: assume(p > 0, a > 0, b > 0); sin(a*t)*cosh(b*t^2)*%e^(p*t); radcan(specint(%,t)); => long expression containing failonf24p146test Not sure what's going on here, haven't looked into it.  >Comment By: Robert Dodier (robert_dodier) Date: 20070116 21:07 Message: Logged In: YES user_id=501686 Originator: YES Yes, I think returning a noun form is the right thing to do here. Not sure whether it should be an integrate noun or specint noun; maybe it doesn't matter too much.  Comment By: Raymond Toy (rtoy) Date: 20070115 17:42 Message: Logged In: YES user_id=28849 Originator: NO This means that maxima thought the integrand looked like t^(v1)*exp(t^2/8/a), but it didn't satisfy the requirements that Re(a) > 0 and Re(v) > 0. At this point it doesn't know how to proceed. I have not convinced myself that this Laplace transform exists. Perhaps we could just return the noun form. If we do that, the answer contains specint(%i/4*exp(b*t^2(p+%i)*t),t) + specint(%i/4*exp(b*t^2(p%i*a)*t)) which is equivalent to specint(exp(b*t^2p*t)*sin(a*t),t)/2.  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1635372&group_id=4933 