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From: SourceForge.net <noreply@so...>  20080806 22:53:23

Bugs item #734851, was opened at 20030508 16:50 Message generated for change (Comment added) made by dgildea You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=734851&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  Taylor Group: None >Status: Closed >Resolution: Fixed Priority: 5 Private: No Submitted By: Stavros Macrakis (macrakis) >Assigned to: Dan Gildea (dgildea) Summary: pade interfered with by taylor Initial Comment: (C1) taylor(sin(x),x,0,3); (D1) xx^3/6 (C2) pade(d1,2,2); (D2) [6*x/(x^2+6)] Fine so far. (C3) taylor(x,x,0,3); (D3) +x (C4) pade(d1,2,2); (D4) [6/7] What's this?! Pade is being affected by the Taylor calculation in C3?! Presumably there is some global flag being set but not reset.... (C5) taylor(exp(x),x,0,3); (D5) 1+x+x^2/2+x^3/6 (C6) pade(d1,2,2); (D6) [6*x/(x^2+6)] Somehow that restored the thing that was causing the problem.  >Comment By: Dan Gildea (dgildea) Date: 20080806 18:53 Message: Logged In: YES user_id=1797506 Originator: NO Fixed in pade.lisp rev 1.8: declare varlist and genvar as special (%i4) taylor(sin(x),x,0,3); (%o4) xx^3/6 (%i5) pade(%,2,2); (%o5) [6*x/(x^2+6)] (%i6) taylor(x,x,0,3); (%o6) +x (%i7) pade(%o4,2,2); (%o7) [6*x/(x^2+6)]  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=734851&group_id=4933 
From: SourceForge.net <noreply@so...>  20080806 22:26:04

Bugs item #1996354, was opened at 20080617 14:12 Message generated for change (Comment added) made by dgildea You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1996354&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  Simplification Group: None Status: Open Resolution: None Priority: 5 Private: No Submitted By: Barton Willis (willisbl) Assigned to: Nobody/Anonymous (nobody) Summary: unsimplifed result from expand Initial Comment: (%i52) (%e^(2*sqrt(2))*(%e^(2*sqrt(2))+2*%e^sqrt(2)+1)^2)/16+(%e^(2*sqrt(2))*(%e^(2*sqrt(2)) 2*%e^sqrt(2)+1)^2)/16(%e^(2*sqrt(2))*(%e^(2*sqrt(2))1)^2)/8$ /* should be 1 */ (%i53) expand(%); (%o53) %e^(2^(3/2)2*sqrt(2))/2+1/2 (%i54) expand(%,0,0); (%o54) 1  >Comment By: Dan Gildea (dgildea) Date: 20080806 18:26 Message: Logged In: YES user_id=1797506 Originator: NO seems to be a basic simplification problem. (%i8) 2^(1/2)*2  2*2^(1/2); (%o8) 2^(3/2)2*sqrt(2) (%i9) 2^(3/2)2*sqrt(2); (%o9) 0  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1996354&group_id=4933 
From: SourceForge.net <noreply@so...>  20080806 11:11:37

Bugs item #1376860, was opened at 20051209 04:53 Message generated for change (Comment added) made by crategus You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1376860&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: None Priority: 5 Private: No Submitted By: Raymond Toy (rtoy) Assigned to: Nobody/Anonymous (nobody) Summary: specint(gammaincomplete(v,a*t)*exp(p*t),t) seems wrong Initial Comment: (%i12) assume(v>0,p>0); (%o12) [v>0,p>0] (%i13) specint(gammaincomplete(v,a*t)*exp(p*t),t); SIMP2F1WILLCONTINUEIN (%o13) a^v*(p+a)^(v1)*gamma(v+1)*%f[2,1]([v+1,3/2],[2],p/(p+a)) Compare this with (%i14) specint(gammagreek(v,a*t)*exp(p*t),t); (%o14) a^v*p^(v1)*gamma(v+1)/((a/p+1)^v*v) This matches formula 34, p 179 in Tables of Transforms. Considering gammaincomplete = gamma(n)gammagreek, the expression for gammaincomplete seems wrong. It might still be right if the hypergeometric function simplifies, but maxima can't, and I can't think of any way to simplify it either.  >Comment By: Dieter Kaiser (crategus) Date: 20080806 13:11 Message: Logged In: YES user_id=2039760 Originator: NO The code is changed as suggested. Closing the bug report. Dieter Kaiser  Comment By: Dieter Kaiser (crategus) Date: 20080719 00:17 Message: Logged In: YES user_id=2039760 Originator: NO I have suggested to use a representation as a Hypergeometric 1F1 for a<>0 for the Gammaincomplete function to get more simple results for the Laplace transform of the Gammaincomplete function. Because $specint has no code to do the Laplace transform for the Hypergeometric function itself ($specint knows only an intern representation for the Hypergeometric function) it is more easy to transform to the Gammagreek function. The Gammagreek function will be transformed to the Hypergeometric 1F1 function. The following changes produce the correct results shown in this thread: =================================================================== RCS file: /cvsroot/maxima/maxima/src/hypgeo.lisp,v retrieving revision 1.40 diff u r1.40 hypgeo.lisp  hypgeo.lisp 4 Jul 2008 14:12:52 0000 1.40 +++ hypgeo.lisp 18 Jul 2008 22:03:29 0000 @@ 2301,7 +2302,7 @@ ((eq flg 'kbateman) (kbatemantw i1 a1)) ((eq flg 'gammaincomplete)  (gammaincompletetw a1 i1)) + (gammaincompletetogammagreek a1 i1)) ((eq flg 'kti) (kti i1 a1)) ((eq flg 'erfc) @@ 2582,6 +2583,28 @@ (power '$%e (div x 2)) (wwhit x (div (sub a 1) 2)(div a 2)))) +;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;; +;;; Only for a=0 we use the general representation as a Whittaker W function. +;;; In all other cases we transform to a Gammagreek function. The Gammagreek +;;; function will be further transformed to a Hypergeometric 1F1 representation. +;;; With this change we get more simple and correct results for the Laplace +;;; transform of the Gammaincomplete function. +;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;; + +(defun gammaincompletetogammagreek (a x) + (if (eq (asksign a) '$zero) + ;; The representation as a Whittaker W function for a=0 + (mul + (power x (div (sub a 1) 2)) + (power '$%e (div x 2)) + (wwhit x (div (sub a 1) 2) (div a 2))) + ;; In all other cases the representation as a Gammagreek function + (sub + (list '(%gamma) a) + (list '($gammagreek) a x)))) + +;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;; + (defun distrexecinit (fun) (cond ((and (consp fun) (consp (car fun)) =================================================================== The testsuite has no problems with this algorithm. With this changes this bug report could be closed. Should I commit this change of the code? Dieter Kaiser  Comment By: Dieter Kaiser (crategus) Date: 20080615 18:11 Message: Logged In: YES user_id=2039760 Originator: NO I would like to suggest the following algorithm to get an evaluated integral for the Gammaincomplete function: (defun gammaincompletetohypergeometric (a x) (if (eq (asksign a) '$zero) ;; The representation as Whittaker W function for a=0 (mul* (power x (div (sub a 1) 2)) (power '$%e (div x 2)) ($whittaker_w (div (sub a 1) 2) (div a 2) x)) ; sign of (div a 2) correct? ;; In all other cases the representation as Hypergeometric 1F1 function (sub (list '(%gamma) a) (mul (power x a) (inv a) ($hypergeometric (list '(mlist) a) (list '(mlist) (add a 1)) (mul 1 x)))))) This routine has to be replaced for GAMMAINCOMPLETETW and called in the routine FRACTEST2. The idea is to use the Hypergeometric Function 1F1 when the parameter A is not equal to zero. That is equivalent to use gamma(a)  gammagreek(a,x) for the evaluation. Now we use the transformation to Whittaker W only for a=0, but for this case we get a nice result too. Here some examples with the above routine: (%i1) assume(s>0,a>0); (%o1) [s > 0,a > 0] We get the expected simple answer for gammaincomplete: (%i2) radcan(specint(%e^(s*t)*gammaincomplete(a,t),t)); (%o2) (a*gamma(a)*(s+1)^agamma(a+1))/(a*s*(s+1)^a) If we take a=0 we get a nice and correct result too: (%i3) radcan(specint(%e^(s*t)*gammaincomplete(0,t),t)); (%o3) log(s+1)/s The function %ei uses the code of gammaincomplete with a=0: (%i4) radcan(specint(%e^(s*t)*%ei(t),t)); (%o4) log(1s)/s For special values like a=1/2 or a=1 both algorithm get simple expressions. The results calulated with 1F1 are identical to the results of the orginal code which use the transformation to Whittaker W: (%i5) radcan(specint(%e^(s*t)*gammaincomplete(1/2,t),t)); (%o5) (sqrt(%pi)*sqrt(s+1)sqrt(%pi))/(s*sqrt(s+1)) (%i6) radcan(specint(%e^(s*t)*gammaincomplete(1,t),t)); (%o6) 1/(s+1) In the above code I have used the routines $hypergeometric and $whittaker_w. I have introduced this functions to make the code more expressive. It is easy to use the alternavtives %f and %w. Furthermore $specint has to be extended to integrate %f. The testsuite has no problems with this alogorithm. Dieter Kaiser  Comment By: Raymond Toy (rtoy) Date: 20060212 20:15 Message: Logged In: YES user_id=28849 The fix for transforming %w causes this return the result in terms of the associated Legendre function Q. Somewhat better, but I do not know if this is equivalent.  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1376860&group_id=4933 
From: SourceForge.net <noreply@so...>  20080806 11:07:12

Bugs item #1162505, was opened at 20050313 18:13 Message generated for change (Comment added) made by crategus You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1162505&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 Group: None >Status: Closed Resolution: None Priority: 5 Private: No Submitted By: Robert Dodier (robert_dodier) Assigned to: Nobody/Anonymous (nobody) Summary: specint returns Lisp symbol otherdefinttofollownegtest Initial Comment: specint returns a Lisp symbol, otherdefinttofollownegtest, when it fails. Example: specint (1/(1+exp(u)),u); yields otherdefinttofollownegtest specint (1/(1+exp(u))*exp(s*u),u); yields otherdefinttofollownegtest The return value should be a noun in these cases, although it's not clear what the noun should be  integrate or specint ?? My suggestion is that the return value should be an integrate noun (with the limits of integration specified as zero and infinity).  >Comment By: Dieter Kaiser (crategus) Date: 20080806 13:07 Message: Logged In: YES user_id=2039760 Originator: NO The code is changed as suggested. Closing the bug report. Dieter Kaiser  Comment By: Dieter Kaiser (crategus) Date: 20080718 22:49 Message: Logged In: YES user_id=2039760 Originator: NO The changed code of $specint (Rev. 1.40) gives for the second integral of this bug report a noun form, but the symbol 'otherdefinittofollownegtest for the first integral. The reason is, that the routine NEGTEST don't signal the return of a noun form. With the introduced global flag *hypreturnnounformflag* this can be easy corrected: =================================================================== RCS file: /cvsroot/maxima/maxima/src/hypgeo.lisp,v retrieving revision 1.40 diff u r1.40 hypgeo.lisp  hypgeo.lisp 4 Jul 2008 14:12:52 0000 1.40 +++ hypgeo.lisp 18 Jul 2008 20:34:55 0000 @@ 1108,7 +1108,8 @@ ;; We forget the rule after finishing the calculation. (mfuncall '$forget `((mgreaterp) psey 0))))))  (return 'otherdefinttofollownegtest))) + (return + (setq *hypreturnnounformflag* 'otherdefinttofollownegtest)))) =================================================================== The results after the change are: (%i6) specint(1/(1+exp(u)),u); (%o6) specint(%e^u/(%e^u+1),u) (%i7) specint(1/(1+exp(u))*exp(u),u); (%o7) specint(1/(%e^u+1),u) There a lot of other places this mechanism can be used to create a nice noun form. Should I commit this change? After this change the bug report could be closed. Dieter Kaiser  Comment By: Raymond Toy (rtoy) Date: 20050314 22:55 Message: Logged In: YES user_id=28849 Since integrate doesn't know about Laplace transforms, I would prefer specint be used since that's what I typed. Having said that, I think specint was never really finished and needs more work, and I rather like the symbols because they tell me at least where to look to find out why it failed. specint is complicated enough that this little bit of info is rather nice. Eventually, though, a noun form should be returned.  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1162505&group_id=4933 