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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 
From: SourceForge.net <noreply@so...>  20090922 10:19:18

Bugs item #2863014, was opened at 20090920 19:23 Message generated for change (Comment added) made by willisbl You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=2863014&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: Contrib: Calculate gain, phase of transfer function Initial Comment: Here's some code for calculating the gain and phase of a function in sinusoidal steady state. It works with both timedomain functions with sine/cosine terms and transient terms (By taking the Laplace transform and removing the sinuosidal term in the Laplace domain) and with transfer functions in the Laplace domain, assuming that t > infinity. It might make sense to simply roll these functions into some larger package, to avoid excessively fine grained packages.  >Comment By: Barton Willis (willisbl) Date: 20090922 05:19 Message: Suggestion: Post your code to http://sourceforge.net/apps/phpbb/maxima/viewforum.php?f=3&sid=c18d1b11a001b12002d3139e0c709503 Also, you could append your convolution function to this code.  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=2863014&group_id=4933 