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From: SourceForge.net <noreply@so...>  20090930 12:36:17

Bugs item #2870695, was opened at 20090930 08:24 Message generated for change (Settings changed) made by rtoy You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=2870695&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: Pending >Resolution: Out of Date Priority: 5 Private: No Submitted By: Nobody/Anonymous (nobody) Assigned to: Nobody/Anonymous (nobody) Summary: integration mistery Initial Comment: integrate(exp(sqrt(x)),x,0,2) return "Is yx positive or negative?" It's very strange because integrate(exp(sqrt(x)),x) return the good result and integrate(2^(sqrt(x)/(log(2)),x,0,2) return the good result too....  >Comment By: Raymond Toy (rtoy) Date: 20090930 08:36 Message: This no longer happens with the 5.19post version of maxima. Marking as pending  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=2870695&group_id=4933 
From: SourceForge.net <noreply@so...>  20090930 12:24:39

Bugs item #2870695, was opened at 20090930 12:24 Message generated for change (Tracker Item Submitted) made by nobody You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=2870695&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: Nobody/Anonymous (nobody) Assigned to: Nobody/Anonymous (nobody) Summary: integration mistery Initial Comment: integrate(exp(sqrt(x)),x,0,2) return "Is yx positive or negative?" It's very strange because integrate(exp(sqrt(x)),x) return the good result and integrate(2^(sqrt(x)/(log(2)),x,0,2) return the good result too....  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=2870695&group_id=4933 
From: SourceForge.net <noreply@so...>  20090929 23:56:33

Bugs item #2869181, was opened at 20090928 15:30 Message generated for change (Settings changed) made by willisbl You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=2869181&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: None Group: None >Status: Closed >Resolution: Fixed Priority: 4 Private: No Submitted By: Barton Willis (willisbl) Assigned to: Barton Willis (willisbl) Summary: is(equal(0, exp(%i * x))) > unknown Initial Comment: OK: (%i3) is(equal(exp(%i * x),0)); (%o3) false OK, but should be better: (%i4) is(equal(0, exp(%i * x))); (%o4) unknown I'm testing a fix for meqp and meqpbycsign.  >Comment By: Barton Willis (willisbl) Date: 20090929 18:56 Message: Fixed by compar.lisp CVS revision 1.55; appended tests to rtestequal  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=2869181&group_id=4933 
From: SourceForge.net <noreply@so...>  20090929 19:01:11

Bugs item #2870018, was opened at 20090929 19:01 Message generated for change (Tracker Item Submitted) made by nobody You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=2870018&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: Nobody/Anonymous (nobody) Assigned to: Nobody/Anonymous (nobody) Summary: %i should never be an integer, even in integration Initial Comment: integrate(exp(x^%i),x,0,1); yields a question about whether %i is an integer. But integer(%i) => false.  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=2870018&group_id=4933 
From: SourceForge.net <noreply@so...>  20090929 17:35:24

Bugs item #2869955, was opened at 20090929 17:35 Message generated for change (Tracker Item Submitted) made by nobody You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=2869955&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: limit incorrect for x/sqrt(1x^2) Initial Comment: Maxima 5.19.1 <snip> (%i1) limit(x/sqrt(1x^2),x,1,minus); (%o1) infinity (%i2) domain(x); (%o2) real(x) The answer should be minf.  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=2869955&group_id=4933 
From: SourceForge.net <noreply@so...>  20090929 17:02:04

Bugs item #2869933, was opened at 20090929 13:02 Message generated for change (Tracker Item Submitted) made by dsimcha You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=2869933&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: None Group: None Status: Open Resolution: None Priority: 5 Private: No Submitted By: dsimcha (dsimcha) Assigned to: Nobody/Anonymous (nobody) Summary: atvalue() doesn't allow functions with subscripts Initial Comment: (%i1) atvalue(x[1](t), t=0, 0); Improper argument to atvalue: x (t) 1  an error. To debug this try debugmode(true); (%i2) atvalue(x(t), t=0, 0); (%o2) 0 (%i3)  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=2869933&group_id=4933 
From: SourceForge.net <noreply@so...>  20090928 21:41:20

Bugs item #2867727, was opened at 20090927 01:30 Message generated for change (Settings changed) made by crategus You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=2867727&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: Closed >Resolution: Fixed Priority: 5 Private: No Submitted By: Dieter Kaiser (crategus) Assigned to: Nobody/Anonymous (nobody) Summary: specint: wrong result for Parabolic Cylinder D function Initial Comment: A look at the implemented algorithm for the Laplace transform of a hypergeometric function shows that the integral for the Parabolic Cylinder D function should work only for an argument t^k with an exponent k < 1, e.g with an argument sqrt(t) (I have already introduced a new symbol in my sandbox for the Parabolic Cylinder D function): (%i2) factor(ratsimp(specint(exp(s*t)*parabolic_cylinder_d(0,sqrt(t)),t))); (%o2) 4/(4*s+1) This is a correct result. We can verify the result using a specific expansion: parabolic_cylinder_d(0,z) = exp(z^2/4) parabolic_cylinder_d(1,z) = z*exp(z^2/4) parabolic_cylinder_d(2,z) = (z^21)*exp(z^2/4) Now some more results with the first expansion for v=0: (%i3) d0(z):=exp(z^2/4)$ We can verify the result for the Parabolic Cylinder D function with an argument sqrt(t): (%i6) factor(ratsimp(specint(exp(s*t)*d0(sqrt(t)),t))); (%o6) 4/(4*s+1) Now we do the integration with an argument t using the expansion. We get a result which contains the Error function: (%i7) factor(ratsimp(specint(exp(s*t)*d0(t),t))); (%o7) sqrt(%pi)*%e^s^2*(erf(s)1) The following integral should not work, but it does and the answer is wrong: (%i8) factor(ratsimp(specint(exp(s*t)*parabolic_cylinder_d(0,t),t))); (%o8) 2^(3/2)*gamma(3/4)/(4*s+1)^(3/4) The routine f16p217test does not fail as expected and produces the wrong result. Remarks: 1. $specint uses the Parabolic Cylinder D function and its Laplace transform for the Hermite and the Erfc function. For the Hermite function the algorithm does not work as expected. For the erfc function it is more simple to integrate (1erf(z)). 2. The Parabolic Cylinder D function is transformed in terms of the Whittaker W function. In a second step the Whittaker W function is transformed to a hypergeometric representation. This representation is integrated. But we already have a routine simpdtf which does the transformation to a hypergeometric representation in one step. Dieter Kaiser  >Comment By: Dieter Kaiser (crategus) Date: 20090928 23:41 Message: Fixed in hypgeo.lisp revision 1.67, Closing this bug report. Dieter Kaiser  Comment By: Dieter Kaiser (crategus) Date: 20090928 00:14 Message: I had again a look at the algorithm of whittest. It is not necessary to return in all cases a noun form. We only have to make sure that 1/2uk and 1/2+uk are not zero or a negative integer. Only for these cases the transformation will fail. This is the extension to check the indices and to call the transformation to the Whittaker M function: (defun whittest (r a i1 i2) (let (n m) (cond ((f16p217test r a i1 i2)) ((and (not (and (maximaintegerp (setq n (sub (sub '((rat simp) 1 2) i2) i1))) (member ($sign n) '($zero $neg $nz)))) (not (and (maximaintegerp (setq m (sub (add '((rat simp) 1 2) i2) i1))) (member ($sign m) '($zero $neg $nz))))) ;; 1/2uk and 1/2+uk are not zero or a negative integer ;; Transform to Whittaker M and try again. (distrexecinit ($expand (mul (init r) (wtm a i1 i2))))) (t ;; Both conditions fails, return a noun form. (setq *hypreturnnounflag* 'whittestfailed))))) Dieter Kaiser  Comment By: Dieter Kaiser (crategus) Date: 20090927 22:37 Message: The error is in the routine f16p217test. The Laplace transform of the Whittaker W function is only valid for an argument a*t and not a*t^v. There is no check for v=1. This is a possible correction of the involved part of the code in the routine f16p217test: ;; Ok, we satisfy the conditions. Now extract the arg. ;; The transformation is only valid for an argument a*t. We have ;; to specialize the pattern to make sure that we satisfy the condition. (let ((l (m2a*t a) ; more special pattern for the argument ; (m2 a ; '((mplus) ; ((coeffpt) (f hasvar) (a freevar)) ; ((coeffpp) (c zerp))) ; nil) )) If we correct this part of code the Laplace transform for the Parabolic Cylinder D function for an argument sqrt(t) will work, but will fail for an argument t as expected. Furthermore, we have to change the algorithm of whittest. When f16p217test fails a transformation to a Whittaker M function is done. But the algorithm for the Laplace transform of the Whittaker W function is already the general Laplace transform of the corresponding representation in terms of Whittaker M. The problem is, that we get errors when trying this transformation. So it is the best to return a noun form at this point. This could be the change to the routine whittest: (defun whittest (r a i1 i2) (cond ((f16p217test r a i1 i2)) (t ; The formula used in f16p217test already is already the general Laplace ; transformation. The transformation to a Whittaker M representation ; does not give any new information, but will fail with errors like undefined gamma ; functions or divisions by zero. Therefore, we return a noun form at this point. ; ;; Convert to M function and try again. ; (distrexecinit ($expand (mul (init r) ; (wtm a i1 i2)))) (setq *hypreturnnounflag* 'whittestfailed) ))) With these changes the Laplace transform of the Parabolic Cylinder D function will work as expected. There are no problems with the testsuite. Dieter Kaiser  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=2867727&group_id=4933 
From: SourceForge.net <noreply@so...>  20090928 20:30:40

Bugs item #2869181, was opened at 20090928 15:30 Message generated for change (Tracker Item Submitted) made by willisbl You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=2869181&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: None Group: None Status: Open Resolution: None Priority: 4 Private: No Submitted By: Barton Willis (willisbl) Assigned to: Barton Willis (willisbl) Summary: is(equal(0, exp(%i * x))) > unknown Initial Comment: OK: (%i3) is(equal(exp(%i * x),0)); (%o3) false OK, but should be better: (%i4) is(equal(0, exp(%i * x))); (%o4) unknown I'm testing a fix for meqp and meqpbycsign.  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=2869181&group_id=4933 
From: SourceForge.net <noreply@so...>  20090927 22:14:08

Bugs item #2867727, was opened at 20090927 01:30 Message generated for change (Comment added) made by crategus You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=2867727&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: Dieter Kaiser (crategus) Assigned to: Nobody/Anonymous (nobody) Summary: specint: wrong result for Parabolic Cylinder D function Initial Comment: A look at the implemented algorithm for the Laplace transform of a hypergeometric function shows that the integral for the Parabolic Cylinder D function should work only for an argument t^k with an exponent k < 1, e.g with an argument sqrt(t) (I have already introduced a new symbol in my sandbox for the Parabolic Cylinder D function): (%i2) factor(ratsimp(specint(exp(s*t)*parabolic_cylinder_d(0,sqrt(t)),t))); (%o2) 4/(4*s+1) This is a correct result. We can verify the result using a specific expansion: parabolic_cylinder_d(0,z) = exp(z^2/4) parabolic_cylinder_d(1,z) = z*exp(z^2/4) parabolic_cylinder_d(2,z) = (z^21)*exp(z^2/4) Now some more results with the first expansion for v=0: (%i3) d0(z):=exp(z^2/4)$ We can verify the result for the Parabolic Cylinder D function with an argument sqrt(t): (%i6) factor(ratsimp(specint(exp(s*t)*d0(sqrt(t)),t))); (%o6) 4/(4*s+1) Now we do the integration with an argument t using the expansion. We get a result which contains the Error function: (%i7) factor(ratsimp(specint(exp(s*t)*d0(t),t))); (%o7) sqrt(%pi)*%e^s^2*(erf(s)1) The following integral should not work, but it does and the answer is wrong: (%i8) factor(ratsimp(specint(exp(s*t)*parabolic_cylinder_d(0,t),t))); (%o8) 2^(3/2)*gamma(3/4)/(4*s+1)^(3/4) The routine f16p217test does not fail as expected and produces the wrong result. Remarks: 1. $specint uses the Parabolic Cylinder D function and its Laplace transform for the Hermite and the Erfc function. For the Hermite function the algorithm does not work as expected. For the erfc function it is more simple to integrate (1erf(z)). 2. The Parabolic Cylinder D function is transformed in terms of the Whittaker W function. In a second step the Whittaker W function is transformed to a hypergeometric representation. This representation is integrated. But we already have a routine simpdtf which does the transformation to a hypergeometric representation in one step. Dieter Kaiser  >Comment By: Dieter Kaiser (crategus) Date: 20090928 00:14 Message: I had again a look at the algorithm of whittest. It is not necessary to return in all cases a noun form. We only have to make sure that 1/2uk and 1/2+uk are not zero or a negative integer. Only for these cases the transformation will fail. This is the extension to check the indices and to call the transformation to the Whittaker M function: (defun whittest (r a i1 i2) (let (n m) (cond ((f16p217test r a i1 i2)) ((and (not (and (maximaintegerp (setq n (sub (sub '((rat simp) 1 2) i2) i1))) (member ($sign n) '($zero $neg $nz)))) (not (and (maximaintegerp (setq m (sub (add '((rat simp) 1 2) i2) i1))) (member ($sign m) '($zero $neg $nz))))) ;; 1/2uk and 1/2+uk are not zero or a negative integer ;; Transform to Whittaker M and try again. (distrexecinit ($expand (mul (init r) (wtm a i1 i2))))) (t ;; Both conditions fails, return a noun form. (setq *hypreturnnounflag* 'whittestfailed))))) Dieter Kaiser  Comment By: Dieter Kaiser (crategus) Date: 20090927 22:37 Message: The error is in the routine f16p217test. The Laplace transform of the Whittaker W function is only valid for an argument a*t and not a*t^v. There is no check for v=1. This is a possible correction of the involved part of the code in the routine f16p217test: ;; Ok, we satisfy the conditions. Now extract the arg. ;; The transformation is only valid for an argument a*t. We have ;; to specialize the pattern to make sure that we satisfy the condition. (let ((l (m2a*t a) ; more special pattern for the argument ; (m2 a ; '((mplus) ; ((coeffpt) (f hasvar) (a freevar)) ; ((coeffpp) (c zerp))) ; nil) )) If we correct this part of code the Laplace transform for the Parabolic Cylinder D function for an argument sqrt(t) will work, but will fail for an argument t as expected. Furthermore, we have to change the algorithm of whittest. When f16p217test fails a transformation to a Whittaker M function is done. But the algorithm for the Laplace transform of the Whittaker W function is already the general Laplace transform of the corresponding representation in terms of Whittaker M. The problem is, that we get errors when trying this transformation. So it is the best to return a noun form at this point. This could be the change to the routine whittest: (defun whittest (r a i1 i2) (cond ((f16p217test r a i1 i2)) (t ; The formula used in f16p217test already is already the general Laplace ; transformation. The transformation to a Whittaker M representation ; does not give any new information, but will fail with errors like undefined gamma ; functions or divisions by zero. Therefore, we return a noun form at this point. ; ;; Convert to M function and try again. ; (distrexecinit ($expand (mul (init r) ; (wtm a i1 i2)))) (setq *hypreturnnounflag* 'whittestfailed) ))) With these changes the Laplace transform of the Parabolic Cylinder D function will work as expected. There are no problems with the testsuite. Dieter Kaiser  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=2867727&group_id=4933 
From: SourceForge.net <noreply@so...>  20090927 20:37:42

Bugs item #2867727, was opened at 20090927 01:30 Message generated for change (Comment added) made by crategus You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=2867727&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: Dieter Kaiser (crategus) Assigned to: Nobody/Anonymous (nobody) Summary: specint: wrong result for Parabolic Cylinder D function Initial Comment: A look at the implemented algorithm for the Laplace transform of a hypergeometric function shows that the integral for the Parabolic Cylinder D function should work only for an argument t^k with an exponent k < 1, e.g with an argument sqrt(t) (I have already introduced a new symbol in my sandbox for the Parabolic Cylinder D function): (%i2) factor(ratsimp(specint(exp(s*t)*parabolic_cylinder_d(0,sqrt(t)),t))); (%o2) 4/(4*s+1) This is a correct result. We can verify the result using a specific expansion: parabolic_cylinder_d(0,z) = exp(z^2/4) parabolic_cylinder_d(1,z) = z*exp(z^2/4) parabolic_cylinder_d(2,z) = (z^21)*exp(z^2/4) Now some more results with the first expansion for v=0: (%i3) d0(z):=exp(z^2/4)$ We can verify the result for the Parabolic Cylinder D function with an argument sqrt(t): (%i6) factor(ratsimp(specint(exp(s*t)*d0(sqrt(t)),t))); (%o6) 4/(4*s+1) Now we do the integration with an argument t using the expansion. We get a result which contains the Error function: (%i7) factor(ratsimp(specint(exp(s*t)*d0(t),t))); (%o7) sqrt(%pi)*%e^s^2*(erf(s)1) The following integral should not work, but it does and the answer is wrong: (%i8) factor(ratsimp(specint(exp(s*t)*parabolic_cylinder_d(0,t),t))); (%o8) 2^(3/2)*gamma(3/4)/(4*s+1)^(3/4) The routine f16p217test does not fail as expected and produces the wrong result. Remarks: 1. $specint uses the Parabolic Cylinder D function and its Laplace transform for the Hermite and the Erfc function. For the Hermite function the algorithm does not work as expected. For the erfc function it is more simple to integrate (1erf(z)). 2. The Parabolic Cylinder D function is transformed in terms of the Whittaker W function. In a second step the Whittaker W function is transformed to a hypergeometric representation. This representation is integrated. But we already have a routine simpdtf which does the transformation to a hypergeometric representation in one step. Dieter Kaiser  >Comment By: Dieter Kaiser (crategus) Date: 20090927 22:37 Message: The error is in the routine f16p217test. The Laplace transform of the Whittaker W function is only valid for an argument a*t and not a*t^v. There is no check for v=1. This is a possible correction of the involved part of the code in the routine f16p217test: ;; Ok, we satisfy the conditions. Now extract the arg. ;; The transformation is only valid for an argument a*t. We have ;; to specialize the pattern to make sure that we satisfy the condition. (let ((l (m2a*t a) ; more special pattern for the argument ; (m2 a ; '((mplus) ; ((coeffpt) (f hasvar) (a freevar)) ; ((coeffpp) (c zerp))) ; nil) )) If we correct this part of code the Laplace transform for the Parabolic Cylinder D function for an argument sqrt(t) will work, but will fail for an argument t as expected. Furthermore, we have to change the algorithm of whittest. When f16p217test fails a transformation to a Whittaker M function is done. But the algorithm for the Laplace transform of the Whittaker W function is already the general Laplace transform of the corresponding representation in terms of Whittaker M. The problem is, that we get errors when trying this transformation. So it is the best to return a noun form at this point. This could be the change to the routine whittest: (defun whittest (r a i1 i2) (cond ((f16p217test r a i1 i2)) (t ; The formula used in f16p217test already is already the general Laplace ; transformation. The transformation to a Whittaker M representation ; does not give any new information, but will fail with errors like undefined gamma ; functions or divisions by zero. Therefore, we return a noun form at this point. ; ;; Convert to M function and try again. ; (distrexecinit ($expand (mul (init r) ; (wtm a i1 i2)))) (setq *hypreturnnounflag* 'whittestfailed) ))) With these changes the Laplace transform of the Parabolic Cylinder D function will work as expected. There are no problems with the testsuite. Dieter Kaiser  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=2867727&group_id=4933 
From: SourceForge.net <noreply@so...>  20090926 23:30:08

Bugs item #2867727, was opened at 20090927 01:30 Message generated for change (Tracker Item Submitted) made by crategus You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=2867727&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: Dieter Kaiser (crategus) Assigned to: Nobody/Anonymous (nobody) Summary: specint: wrong result for Parabolic Cylinder D function Initial Comment: A look at the implemented algorithm for the Laplace transform of a hypergeometric function shows that the integral for the Parabolic Cylinder D function should work only for an argument t^k with an exponent k < 1, e.g with an argument sqrt(t) (I have already introduced a new symbol in my sandbox for the Parabolic Cylinder D function): (%i2) factor(ratsimp(specint(exp(s*t)*parabolic_cylinder_d(0,sqrt(t)),t))); (%o2) 4/(4*s+1) This is a correct result. We can verify the result using a specific expansion: parabolic_cylinder_d(0,z) = exp(z^2/4) parabolic_cylinder_d(1,z) = z*exp(z^2/4) parabolic_cylinder_d(2,z) = (z^21)*exp(z^2/4) Now some more results with the first expansion for v=0: (%i3) d0(z):=exp(z^2/4)$ We can verify the result for the Parabolic Cylinder D function with an argument sqrt(t): (%i6) factor(ratsimp(specint(exp(s*t)*d0(sqrt(t)),t))); (%o6) 4/(4*s+1) Now we do the integration with an argument t using the expansion. We get a result which contains the Error function: (%i7) factor(ratsimp(specint(exp(s*t)*d0(t),t))); (%o7) sqrt(%pi)*%e^s^2*(erf(s)1) The following integral should not work, but it does and the answer is wrong: (%i8) factor(ratsimp(specint(exp(s*t)*parabolic_cylinder_d(0,t),t))); (%o8) 2^(3/2)*gamma(3/4)/(4*s+1)^(3/4) The routine f16p217test does not fail as expected and produces the wrong result. Remarks: 1. $specint uses the Parabolic Cylinder D function and its Laplace transform for the Hermite and the Erfc function. For the Hermite function the algorithm does not work as expected. For the erfc function it is more simple to integrate (1erf(z)). 2. The Parabolic Cylinder D function is transformed in terms of the Whittaker W function. In a second step the Whittaker W function is transformed to a hypergeometric representation. This representation is integrated. But we already have a routine simpdtf which does the transformation to a hypergeometric representation in one step. Dieter Kaiser  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=2867727&group_id=4933 
From: SourceForge.net <noreply@so...>  20090926 18:53:03

Bugs item #2867499, was opened at 20090926 18:02 Message generated for change (Settings changed) made by crategus You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=2867499&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: Closed >Resolution: Fixed Priority: 5 Private: No Submitted By: Dieter Kaiser (crategus) Assigned to: Nobody/Anonymous (nobody) Summary: specint(exp(s*t)*CONST*t^2*bessel_y(1,t),t) wrong Initial Comment: Maxima does not take into a account a constant part of an integrand when doing the Laplace transform of bessel_y(v,a*t) where v is an integer order. The symbol CONST is missing in the following result: (%i5) ratsimp(specint(exp(s*t)*CONST*t^2*bessel_y(1,t),t)); (%o5) (3*s*log((sqrt(s^2+1)+s)/(sqrt(s^2+1)s))+sqrt(s^2+1)*(2*s^24)) /(%pi*sqrt(s^2+1)*(s^4+2*s^2+1)) The error is in the routine f2p105v2cond. The constant part is extracted from the variable l but not multiplied to the result of the Laplace transform. Dieter Kaiser  >Comment By: Dieter Kaiser (crategus) Date: 20090926 20:53 Message: Fixed in hypgeo.lisp revision 1.66. Closing this bug report. Dieter Kaiser  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=2867499&group_id=4933 
From: SourceForge.net <noreply@so...>  20090926 18:52:20

Bugs item #2867434, was opened at 20090926 15:35 Message generated for change (Settings changed) made by crategus You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=2867434&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: Closed >Resolution: Fixed Priority: 5 Private: No Submitted By: Dieter Kaiser (crategus) Assigned to: Nobody/Anonymous (nobody) Summary: specint(exp(s*t)*t^2*%h[1,1](t),t) does not work Initial Comment: We know hankel_1(v,z) = bessel_j(v,z) + %i * bessel_y(v,z) and hankel_2(v,z) = bessel_j(v,z)  %i * bessel_y(v,z). If we try to get the Laplace transform of %h[v,sort](z), that is the Hankel function known to specint, we get: (%i2) specint(exp(s*t)*t^2*%h[1,1](t),t); Division by 0  an error. To debug this try debugmode(true); (%i3) specint(exp(s*t)*t^2*%h[1,2](t),t); Division by 0  an error. To debug this try debugmode(true); But both Laplace transforms Maxima can calculate: (%i7) factor(ratsimp(specint(exp(s*t)*t^2*(bessel_j(1,t)+%i*bessel_y(1,t)),t))); (%o7) (3*s*sqrt(s^2+1)*log((sqrt(s^2+1)+s)/(sqrt(s^2+1)s)) +3*%pi*s*sqrt(s^2+1)+2*s^42*s^24) /(%pi*(s^2+1)^3) (%i8) factor(ratsimp(specint(exp(s*t)*t^2*(bessel_j(1,t)%i*bessel_y(1,t)),t))); (%o8) (3*s*sqrt(s^2+1)*log((sqrt(s^2+1)+s)/(sqrt(s^2+1)s)) +3*%pi*s*sqrt(s^2+1)+2*s^42*s^24) /(%pi*(s^2+1)^3) The error for the hankel functions %h[v,sort](t) occurs, because the transformation to Bessel J and Bessel Y functions is implemented wrongly in the routine htjory. Dieter Kaiser  >Comment By: Dieter Kaiser (crategus) Date: 20090926 20:52 Message: Fixed in hypgeo.lisp revision 1.66. Closing this bug report. Dieter Kaiser  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=2867434&group_id=4933 
From: SourceForge.net <noreply@so...>  20090926 16:02:31

Bugs item #2867499, was opened at 20090926 18:02 Message generated for change (Tracker Item Submitted) made by crategus You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=2867499&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: Dieter Kaiser (crategus) Assigned to: Nobody/Anonymous (nobody) Summary: specint(exp(s*t)*CONST*t^2*bessel_y(1,t),t) wrong Initial Comment: Maxima does not take into a account a constant part of an integrand when doing the Laplace transform of bessel_y(v,a*t) where v is an integer order. The symbol CONST is missing in the following result: (%i5) ratsimp(specint(exp(s*t)*CONST*t^2*bessel_y(1,t),t)); (%o5) (3*s*log((sqrt(s^2+1)+s)/(sqrt(s^2+1)s))+sqrt(s^2+1)*(2*s^24)) /(%pi*sqrt(s^2+1)*(s^4+2*s^2+1)) The error is in the routine f2p105v2cond. The constant part is extracted from the variable l but not multiplied to the result of the Laplace transform. Dieter Kaiser  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=2867499&group_id=4933 
From: SourceForge.net <noreply@so...>  20090926 13:38:01

Bugs item #2866802, was opened at 20090925 20:23 Message generated for change (Settings changed) made by crategus You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=2866802&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: Closed >Resolution: Fixed Priority: 5 Private: No Submitted By: Dieter Kaiser (crategus) Assigned to: Nobody/Anonymous (nobody) Summary: specint(exp(s*t)*t^(5/2)*bessel_j(1/2,sqrt(t))^2,t) wrong Initial Comment: We have the following known failure in rtest14.mac: ********************** Problem 57 *************** Input: specint(t^(5/2)*bessel_y(1/2,t^(1/2))^2*%e^(p*t),t) In $specint the expression with the bessel_y function is transformed to the square of the bessel_j function. So we get the following integrand: (1) t^(5/2)*(bessel_j(1/2,sqrt(t))^2 Furthermore, this expression is equivalent to: (2) 2/%pi*cos(sqrt(t))^2 Maxima can do this transformation: (%i17) t^(5/2)*bessel_j(1/2,sqrt(t))^2,besselexpand:true; (%o17) 2*cos(sqrt(t))^2*t^2/%pi The problem is that the integrands (1) and (2) give different Laplace transforms: First the result for bessel_j(1/2,sqrt(t))^2: (%i14) res1:factor(ratsimp(specint(exp(s*t)*t^(5/2)*bessel_j(1/2,sqrt(t))^2,t))); (%o14) %e^(1/s)*(8*s^3*%e^(1/s)18*s^2*%e^(1/s)+4*s*%e^(1/s) +15*sqrt(%pi)*%i*erf(%i/sqrt(s))*s^(5/2) 20*sqrt(%pi)*%i*erf(%i/sqrt(s))*s^(3/2) +4*sqrt(%pi)*%i*erf(%i/sqrt(s))*sqrt(s)) /(2*%pi*s^6) Next, the result for cos(sqrt(t))^2 (we use the flag besselexpand): (%i15) res2 : factor(ratsimp(specint(exp(s*t) * t^(5/2) * bessel_j(1/2, sqrt(t))^2, t))), besselexpand:true; (%o15) %e^(1/s)*(16*s^3*%e^(1/s)18*s^2*%e^(1/s)+4*s*%e^(1/s) +15*sqrt(%pi)*%i*erf(%i/sqrt(s))*s^(5/2) 20*sqrt(%pi)*%i*erf(%i/sqrt(s))*s^(3/2) +4*sqrt(%pi)*%i*erf(%i/sqrt(s))*sqrt(s)) /(4*%pi*s^6) The results differ by a factor 2 in most, but not in all terms. I had a long search for the bug and I have found the problem in the algorithm for the product of hypergeometric functions. Maxima does the following transformation for our case of two bessel_j(1/2,sqrt(t)) functions: bessel_j(1/2,sqrt(t))^2 > 2/%pi*2F3([0,1/2], [1/2,1/2,0], t) Next the hypergeometric function is reduced in two steps: 2F3([0,1/2], [1/2,1/2,0], t) > 1F2([1/2], [1/2,1/2], t) > 0F1([], [1/2], t) But, 0F1([],[1/2],t) represents cos(2*sqrt(t)) and not cos(sqrt(t))^2 as expected. Therefore, we get the Laplace transform of cos(2*sqrt(t)) and not of cos(sqrt(t))^2, when we use the hypergeometric algorithm. We can check this by doing the Laplace transform of cos(2*sqrt(t)) directly. A correct hypergeometric representation of cos(sqrt(t))^2 is 1/2*(0F1([],[1/2],t) + 1). The error is, that we do the following transformation for a parameter a=0: 2F3([a,1/2],[1/2,1/2,a],t) > 1F2([1/2], [1/2,1/2], t). I think this transformation is not valid for a=0, because the hypergeometric function 2F3 is not well defined for this case. The parameter a is zero for v+u=1, where v and u are the order of the two Bessel J functions involved in the Laplace transformation. For the square of bessel functions this is the case for v=1/2. Remark: There are a lot of more possibilities for wrong results. The example in rtest14.mac is not the simplest one. Dieter Kaiser  >Comment By: Dieter Kaiser (crategus) Date: 20090926 15:38 Message: The handling of the special case besse_j(1/2,t)^2 has been implemented. Now the Laplace transform of the example works as expected. Closing this bug report as fixed. Dieter Kaiser  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=2866802&group_id=4933 
From: SourceForge.net <noreply@so...>  20090926 13:36:00

Bugs item #2867434, was opened at 20090926 15:35 Message generated for change (Tracker Item Submitted) made by crategus You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=2867434&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: Dieter Kaiser (crategus) Assigned to: Nobody/Anonymous (nobody) Summary: specint(exp(s*t)*t^2*%h[1,1](t),t) does not work Initial Comment: We know hankel_1(v,z) = bessel_j(v,z) + %i * bessel_y(v,z) and hankel_2(v,z) = bessel_j(v,z)  %i * bessel_y(v,z). If we try to get the Laplace transform of %h[v,sort](z), that is the Hankel function known to specint, we get: (%i2) specint(exp(s*t)*t^2*%h[1,1](t),t); Division by 0  an error. To debug this try debugmode(true); (%i3) specint(exp(s*t)*t^2*%h[1,2](t),t); Division by 0  an error. To debug this try debugmode(true); But both Laplace transforms Maxima can calculate: (%i7) factor(ratsimp(specint(exp(s*t)*t^2*(bessel_j(1,t)+%i*bessel_y(1,t)),t))); (%o7) (3*s*sqrt(s^2+1)*log((sqrt(s^2+1)+s)/(sqrt(s^2+1)s)) +3*%pi*s*sqrt(s^2+1)+2*s^42*s^24) /(%pi*(s^2+1)^3) (%i8) factor(ratsimp(specint(exp(s*t)*t^2*(bessel_j(1,t)%i*bessel_y(1,t)),t))); (%o8) (3*s*sqrt(s^2+1)*log((sqrt(s^2+1)+s)/(sqrt(s^2+1)s)) +3*%pi*s*sqrt(s^2+1)+2*s^42*s^24) /(%pi*(s^2+1)^3) The error for the hankel functions %h[v,sort](t) occurs, because the transformation to Bessel J and Bessel Y functions is implemented wrongly in the routine htjory. Dieter Kaiser  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=2867434&group_id=4933 
From: SourceForge.net <noreply@so...>  20090925 19:06:23

Bugs item #2865951, was opened at 20090924 21:01 Message generated for change (Comment added) made by crategus You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=2865951&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: Closed >Resolution: Fixed Priority: 5 Private: No Submitted By: Dieter Kaiser (crategus) Assigned to: Nobody/Anonymous (nobody) Summary: specint(exp(s*t)*bessel_y(0,a*t),t) not simplifed Initial Comment: We have a special algorithm in $specint to express the Laplace transform of Bessel Y for an integer order in terms of the Associated Legendre Q function. Maxima knows this function and can simplify it. Unfortunately, the result of $specint is not fully simplified: (%i2) assume(s>0)$ (%i3) specint(exp(s*t)*bessel_y(0,a*t),t); (%o3) 2*legendre_q(0,s/sqrt(s^2+a^2))/(%pi*sqrt(s^2+a^2)) Because the Associated Legendre Q function is not implemented as a simplifying function we need an additional eval to get a simplified result: (%i4) ev(%); (%o4) log((sqrt(s^2+a^2)+s)/(sqrt(s^2+a^2)s))/(%pi*sqrt(s^2+a^2)) It is easy to change this. We do not call legen in the routine leg2fsimp, which returns an unsimplifed noun form (legen m v z '$q) but do a Maxima function call (the function is not in core and has to be autoloaded) (mfuncall '$assoc_legendre_q m v z)) When we change the code, we get immediately the simplifed result: (%i6) specint(exp(s*t)*bessel_y(0,a*t),t); (%o6) log((sqrt(s^2+a^2)+s)/(sqrt(s^2+a^2)s))/(%pi*sqrt(s^2+a^2)) This result is equivalent to atanh(s/sqrt(s^2+a^2)/(%pi*sqrt(s^2+a^2)) The tabels I know give the result: 2*asinh(s/a)/(%pi*sqrt(s^2+a^2)) But this is equivalent to the atanh expression. We can show it with the help of the relation: atan(z)=1/2 * asinh(2*z/(z^2+1)) All expressions of the form t^n*bessel_y(v,t) with v an positive integer and n>=v will simplify accordingly. Dieter Kaiser  >Comment By: Dieter Kaiser (crategus) Date: 20090925 21:06 Message: Fixed in hypgeo.lisp revision 1.64. Closing this bug report. Dieter Kaiser  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=2865951&group_id=4933 
From: SourceForge.net <noreply@so...>  20090925 18:23:57

Bugs item #2866802, was opened at 20090925 20:23 Message generated for change (Tracker Item Submitted) made by crategus You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=2866802&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: Dieter Kaiser (crategus) Assigned to: Nobody/Anonymous (nobody) Summary: specint(exp(s*t)*t^(5/2)*bessel_j(1/2,sqrt(t))^2,t) wrong Initial Comment: We have the following known failure in rtest14.mac: ********************** Problem 57 *************** Input: specint(t^(5/2)*bessel_y(1/2,t^(1/2))^2*%e^(p*t),t) In $specint the expression with the bessel_y function is transformed to the square of the bessel_j function. So we get the following integrand: (1) t^(5/2)*(bessel_j(1/2,sqrt(t))^2 Furthermore, this expression is equivalent to: (2) 2/%pi*cos(sqrt(t))^2 Maxima can do this transformation: (%i17) t^(5/2)*bessel_j(1/2,sqrt(t))^2,besselexpand:true; (%o17) 2*cos(sqrt(t))^2*t^2/%pi The problem is that the integrands (1) and (2) give different Laplace transforms: First the result for bessel_j(1/2,sqrt(t))^2: (%i14) res1:factor(ratsimp(specint(exp(s*t)*t^(5/2)*bessel_j(1/2,sqrt(t))^2,t))); (%o14) %e^(1/s)*(8*s^3*%e^(1/s)18*s^2*%e^(1/s)+4*s*%e^(1/s) +15*sqrt(%pi)*%i*erf(%i/sqrt(s))*s^(5/2) 20*sqrt(%pi)*%i*erf(%i/sqrt(s))*s^(3/2) +4*sqrt(%pi)*%i*erf(%i/sqrt(s))*sqrt(s)) /(2*%pi*s^6) Next, the result for cos(sqrt(t))^2 (we use the flag besselexpand): (%i15) res2 : factor(ratsimp(specint(exp(s*t) * t^(5/2) * bessel_j(1/2, sqrt(t))^2, t))), besselexpand:true; (%o15) %e^(1/s)*(16*s^3*%e^(1/s)18*s^2*%e^(1/s)+4*s*%e^(1/s) +15*sqrt(%pi)*%i*erf(%i/sqrt(s))*s^(5/2) 20*sqrt(%pi)*%i*erf(%i/sqrt(s))*s^(3/2) +4*sqrt(%pi)*%i*erf(%i/sqrt(s))*sqrt(s)) /(4*%pi*s^6) The results differ by a factor 2 in most, but not in all terms. I had a long search for the bug and I have found the problem in the algorithm for the product of hypergeometric functions. Maxima does the following transformation for our case of two bessel_j(1/2,sqrt(t)) functions: bessel_j(1/2,sqrt(t))^2 > 2/%pi*2F3([0,1/2], [1/2,1/2,0], t) Next the hypergeometric function is reduced in two steps: 2F3([0,1/2], [1/2,1/2,0], t) > 1F2([1/2], [1/2,1/2], t) > 0F1([], [1/2], t) But, 0F1([],[1/2],t) represents cos(2*sqrt(t)) and not cos(sqrt(t))^2 as expected. Therefore, we get the Laplace transform of cos(2*sqrt(t)) and not of cos(sqrt(t))^2, when we use the hypergeometric algorithm. We can check this by doing the Laplace transform of cos(2*sqrt(t)) directly. A correct hypergeometric representation of cos(sqrt(t))^2 is 1/2*(0F1([],[1/2],t) + 1). The error is, that we do the following transformation for a parameter a=0: 2F3([a,1/2],[1/2,1/2,a],t) > 1F2([1/2], [1/2,1/2], t). I think this transformation is not valid for a=0, because the hypergeometric function 2F3 is not well defined for this case. The parameter a is zero for v+u=1, where v and u are the order of the two Bessel J functions involved in the Laplace transformation. For the square of bessel functions this is the case for v=1/2. Remark: There are a lot of more possibilities for wrong results. The example in rtest14.mac is not the simplest one. Dieter Kaiser  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=2866802&group_id=4933 
From: SourceForge.net <noreply@so...>  20090924 19:02:08

Bugs item #2865951, was opened at 20090924 21:01 Message generated for change (Settings changed) made by crategus You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=2865951&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: Dieter Kaiser (crategus) Assigned to: Nobody/Anonymous (nobody) Summary: specint(exp(s*t)*bessel_y(0,a*t),t) not simplifed Initial Comment: We have a special algorithm in $specint to express the Laplace transform of Bessel Y for an integer order in terms of the Associated Legendre Q function. Maxima knows this function and can simplify it. Unfortunately, the result of $specint is not fully simplified: (%i2) assume(s>0)$ (%i3) specint(exp(s*t)*bessel_y(0,a*t),t); (%o3) 2*legendre_q(0,s/sqrt(s^2+a^2))/(%pi*sqrt(s^2+a^2)) Because the Associated Legendre Q function is not implemented as a simplifying function we need an additional eval to get a simplified result: (%i4) ev(%); (%o4) log((sqrt(s^2+a^2)+s)/(sqrt(s^2+a^2)s))/(%pi*sqrt(s^2+a^2)) It is easy to change this. We do not call legen in the routine leg2fsimp, which returns an unsimplifed noun form (legen m v z '$q) but do a Maxima function call (the function is not in core and has to be autoloaded) (mfuncall '$assoc_legendre_q m v z)) When we change the code, we get immediately the simplifed result: (%i6) specint(exp(s*t)*bessel_y(0,a*t),t); (%o6) log((sqrt(s^2+a^2)+s)/(sqrt(s^2+a^2)s))/(%pi*sqrt(s^2+a^2)) This result is equivalent to atanh(s/sqrt(s^2+a^2)/(%pi*sqrt(s^2+a^2)) The tabels I know give the result: 2*asinh(s/a)/(%pi*sqrt(s^2+a^2)) But this is equivalent to the atanh expression. We can show it with the help of the relation: atan(z)=1/2 * asinh(2*z/(z^2+1)) All expressions of the form t^n*bessel_y(v,t) with v an positive integer and n>=v will simplify accordingly. Dieter Kaiser  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=2865951&group_id=4933 
From: SourceForge.net <noreply@so...>  20090924 19:01:09

Bugs item #2865951, was opened at 20090924 21:01 Message generated for change (Tracker Item Submitted) made by crategus You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=2865951&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: None Group: None Status: Open Resolution: None Priority: 5 Private: No Submitted By: Dieter Kaiser (crategus) Assigned to: Nobody/Anonymous (nobody) Summary: specint(exp(s*t)*bessel_y(0,a*t),t) not simplifed Initial Comment: We have a special algorithm in $specint to express the Laplace transform of Bessel Y for an integer order in terms of the Associated Legendre Q function. Maxima knows this function and can simplify it. Unfortunately, the result of $specint is not fully simplified: (%i2) assume(s>0)$ (%i3) specint(exp(s*t)*bessel_y(0,a*t),t); (%o3) 2*legendre_q(0,s/sqrt(s^2+a^2))/(%pi*sqrt(s^2+a^2)) Because the Associated Legendre Q function is not implemented as a simplifying function we need an additional eval to get a simplified result: (%i4) ev(%); (%o4) log((sqrt(s^2+a^2)+s)/(sqrt(s^2+a^2)s))/(%pi*sqrt(s^2+a^2)) It is easy to change this. We do not call legen in the routine leg2fsimp, which returns an unsimplifed noun form (legen m v z '$q) but do a Maxima function call (the function is not in core and has to be autoloaded) (mfuncall '$assoc_legendre_q m v z)) When we change the code, we get immediately the simplifed result: (%i6) specint(exp(s*t)*bessel_y(0,a*t),t); (%o6) log((sqrt(s^2+a^2)+s)/(sqrt(s^2+a^2)s))/(%pi*sqrt(s^2+a^2)) This result is equivalent to atanh(s/sqrt(s^2+a^2)/(%pi*sqrt(s^2+a^2)) The tabels I know give the result: 2*asinh(s/a)/(%pi*sqrt(s^2+a^2)) But this is equivalent to the atanh expression. We can show it with the help of the relation: atan(z)=1/2 * asinh(2*z/(z^2+1)) All expressions of the form t^n*bessel_y(v,t) with v an positive integer and n>=v will simplify accordingly. Dieter Kaiser  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=2865951&group_id=4933 
From: SourceForge.net <noreply@so...>  20090923 21:56:39

Bugs item #2864588, was opened at 20090922 23:24 Message generated for change (Settings changed) made by crategus You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=2864588&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: Closed >Resolution: Fixed Priority: 5 Private: No Submitted By: Dieter Kaiser (crategus) Assigned to: Nobody/Anonymous (nobody) Summary: specint(exp(s*t)*bessel_i(1/2,t)^2,t) not correct Initial Comment: I think we have some problems with the Laplace transform of Bessel functions with a half integral index. This is an example for the square of the bessel_i(1/2,t) function: (%i1) assume(s>0)$ We do the Laplace transform for the square of the expanded bessel_i(1/2,t) function and use the algorithm for the sinh function to get the Laplace transform: (%i2) expr1:bessel_i(1/2,t)^2,besselexpand:true; (%o2) 2*sinh(t)^2/(%pi*t) (%i3) specint(exp(s*t)*expr1,t); (%o3) log(14/s^2)/(2*%pi) Aagain, but we do not expand the function. Now, the hypergeometric algorithm is used. (%i4) expr2:bessel_i(1/2,t)^2; (%o4) bessel_i(1/2,t)^2 (%i5) specint(exp(s*t)*expr2,t); (%o5) (%i1)*%i*log(14/s^2)/(2^(3/2)*%pi) The answers differ by a factor sqrt(%i). First the ratio of the two answers: (%i17) specint(exp(s*t)*expr1,t)/specint(exp(s*t)*expr2,t); (%o17) sqrt(2)*%i/(%i1) The ratio differs by a factor sqrt(%i): (%i18) rectform(sqrt(%i)*%); (%o18) 1 I think we have more of such problems with the Laplace transform of Bessel functions. There is one expected failure for the Laplace transform of bessel_y(1/2,sqrt(t))^2 in the testfile rtest14.mac which might be related to this problem. I had already a long search to find the bug, but had no success. Perhaps it is a mathematical problem related to the hypergeometric transformation and integration for a half integral index. Dieter Kaiser  >Comment By: Dieter Kaiser (crategus) Date: 20090923 23:56 Message: The problem for bessel_i(1/2,t)^2 is fixed in hypgeo.lisp revsion 1.62. The problem for bessel_y(1/2,sqrt(t)) is still open. But it is another bug. Closing this bug report. Dieter Kaiser  Comment By: Dieter Kaiser (crategus) Date: 20090923 00:37 Message: I have found the bug for the square of the Bessel I function. We transform to two Bessel J functions. In the transformation is missing a factor %i^v. (In the following code I have already replaced the function 1fact and have inserted the powers of %i). ;; Laplace transform of square of Bessel I function (cond ((setq l (onei^2 u)) (setq index1 (cdras 'v l) arg1 (mul '$%i (cdras 'w l)) rest (mul (power '$%i (neg index1)) (power '$%i (neg index1)) ; the missing factor (cdras 'u l))) (return (lt1j^2 rest arg1 index1)))) Now we get: (%i3) assume(s>0)$ We expand the Bessel function: (%i4) specint(exp(s*t)*bessel_i(1/2,t)^2,t),besselexpand:true; (%o4) log(14/s^2)/(2*%pi) Now we use the hypergeometric code and get the same result: (%i5) specint(exp(s*t)*bessel_i(1/2,t)^2,t); (%o5) log(14/s^2)/(2*%pi) Dieter Kaiser  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=2864588&group_id=4933 
From: SourceForge.net <noreply@so...>  20090923 20:44:41

Bugs item #2865255, was opened at 20090923 16:44 Message generated for change (Tracker Item Submitted) made by dsimcha You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=2865255&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: None Group: None Status: Open Resolution: None Priority: 5 Private: No Submitted By: dsimcha (dsimcha) Assigned to: Nobody/Anonymous (nobody) Summary: Can't do ilt(1 / (s + a)^n, s, t) In Proper Environment Initial Comment: What actually happens: (%i1) assume(n > 0); (%o1) [n > 0] (%i2) declare(n, integer); (%o2) done (%i3) ilt( 1 / (s + a)^n, s, t); 1 (%o3) ilt(, s, t) n (s + a) Expected answer (Not necessarily in this form) : t^(n  1) exp(a * t) / (t  1)!  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=2865255&group_id=4933 
From: SourceForge.net <noreply@so...>  20090922 22:37:48

Bugs item #2864588, was opened at 20090922 23:24 Message generated for change (Comment added) made by crategus You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=2864588&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: Dieter Kaiser (crategus) Assigned to: Nobody/Anonymous (nobody) Summary: specint(exp(s*t)*bessel_i(1/2,t)^2,t) not correct Initial Comment: I think we have some problems with the Laplace transform of Bessel functions with a half integral index. This is an example for the square of the bessel_i(1/2,t) function: (%i1) assume(s>0)$ We do the Laplace transform for the square of the expanded bessel_i(1/2,t) function and use the algorithm for the sinh function to get the Laplace transform: (%i2) expr1:bessel_i(1/2,t)^2,besselexpand:true; (%o2) 2*sinh(t)^2/(%pi*t) (%i3) specint(exp(s*t)*expr1,t); (%o3) log(14/s^2)/(2*%pi) Aagain, but we do not expand the function. Now, the hypergeometric algorithm is used. (%i4) expr2:bessel_i(1/2,t)^2; (%o4) bessel_i(1/2,t)^2 (%i5) specint(exp(s*t)*expr2,t); (%o5) (%i1)*%i*log(14/s^2)/(2^(3/2)*%pi) The answers differ by a factor sqrt(%i). First the ratio of the two answers: (%i17) specint(exp(s*t)*expr1,t)/specint(exp(s*t)*expr2,t); (%o17) sqrt(2)*%i/(%i1) The ratio differs by a factor sqrt(%i): (%i18) rectform(sqrt(%i)*%); (%o18) 1 I think we have more of such problems with the Laplace transform of Bessel functions. There is one expected failure for the Laplace transform of bessel_y(1/2,sqrt(t))^2 in the testfile rtest14.mac which might be related to this problem. I had already a long search to find the bug, but had no success. Perhaps it is a mathematical problem related to the hypergeometric transformation and integration for a half integral index. Dieter Kaiser  >Comment By: Dieter Kaiser (crategus) Date: 20090923 00:37 Message: I have found the bug for the square of the Bessel I function. We transform to two Bessel J functions. In the transformation is missing a factor %i^v. (In the following code I have already replaced the function 1fact and have inserted the powers of %i). ;; Laplace transform of square of Bessel I function (cond ((setq l (onei^2 u)) (setq index1 (cdras 'v l) arg1 (mul '$%i (cdras 'w l)) rest (mul (power '$%i (neg index1)) (power '$%i (neg index1)) ; the missing factor (cdras 'u l))) (return (lt1j^2 rest arg1 index1)))) Now we get: (%i3) assume(s>0)$ We expand the Bessel function: (%i4) specint(exp(s*t)*bessel_i(1/2,t)^2,t),besselexpand:true; (%o4) log(14/s^2)/(2*%pi) Now we use the hypergeometric code and get the same result: (%i5) specint(exp(s*t)*bessel_i(1/2,t)^2,t); (%o5) log(14/s^2)/(2*%pi) Dieter Kaiser  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=2864588&group_id=4933 
From: SourceForge.net <noreply@so...>  20090922 21:24:44

Bugs item #2864588, was opened at 20090922 23:24 Message generated for change (Tracker Item Submitted) made by crategus You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=2864588&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: Dieter Kaiser (crategus) Assigned to: Nobody/Anonymous (nobody) Summary: specint(exp(s*t)*bessel_i(1/2,t)^2,t) not correct Initial Comment: I think we have some problems with the Laplace transform of Bessel functions with a half integral index. This is an example for the square of the bessel_i(1/2,t) function: (%i1) assume(s>0)$ We do the Laplace transform for the square of the expanded bessel_i(1/2,t) function and use the algorithm for the sinh function to get the Laplace transform: (%i2) expr1:bessel_i(1/2,t)^2,besselexpand:true; (%o2) 2*sinh(t)^2/(%pi*t) (%i3) specint(exp(s*t)*expr1,t); (%o3) log(14/s^2)/(2*%pi) Aagain, but we do not expand the function. Now, the hypergeometric algorithm is used. (%i4) expr2:bessel_i(1/2,t)^2; (%o4) bessel_i(1/2,t)^2 (%i5) specint(exp(s*t)*expr2,t); (%o5) (%i1)*%i*log(14/s^2)/(2^(3/2)*%pi) The answers differ by a factor sqrt(%i). First the ratio of the two answers: (%i17) specint(exp(s*t)*expr1,t)/specint(exp(s*t)*expr2,t); (%o17) sqrt(2)*%i/(%i1) The ratio differs by a factor sqrt(%i): (%i18) rectform(sqrt(%i)*%); (%o18) 1 I think we have more of such problems with the Laplace transform of Bessel functions. There is one expected failure for the Laplace transform of bessel_y(1/2,sqrt(t))^2 in the testfile rtest14.mac which might be related to this problem. I had already a long search to find the bug, but had no success. Perhaps it is a mathematical problem related to the hypergeometric transformation and integration for a half integral index. Dieter Kaiser  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=2864588&group_id=4933 
From: SourceForge.net <noreply@so...>  20090922 10:52:59

Bugs item #2864197, was opened at 20090922 05:52 Message generated for change (Tracker Item Submitted) made by willisbl You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=2864197&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: None Group: None Status: Open Resolution: None Priority: 5 Private: No Submitted By: Barton Willis (willisbl) Assigned to: Nobody/Anonymous (nobody) Summary: algsys with domain : complex Initial Comment: Wrong: (%i1) algsys([y  (1)^(1/3), x = y^3],[x,y]), domain : complex; (%o1) [] OK: (%i2) algsys([y  (1)^(1/3), x = y^3],[x,y]), domain : real; (%o2) [[x=1,y=1]]  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=2864197&group_id=4933 