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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 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...>  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 