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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 