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From: SourceForge.net <noreply@so...>  20070301 09:18:40

Bugs item #1671527, was opened at 20070301 01:18 Message generated for change (Tracker Item Submitted) made by Item Submitter You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1671527&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: integrate loops Initial Comment: Hello I had to interrupt maxima when it was at calculating the following integral: (%i101) m(r,th); 2 2 (%o101) SQRT(z + 2 r rM (1  COS(th)) + (rM  r) ) (%i102) F(d); 2 2 d0 4 A d0 (epsi   + 1) w 2 d (%o102)   3 d (%i103) integrate(1/m(th,r)^2*F(m(th,r))*z^2,th,0,%PI); 2 2 2 2 Is z  COS (r) rM + rM positive, negative, or zero? pos; 2 2 2 Is z + (1  COS (r)) rM positive, negative, or zero? pos; Maxima encountered a Lisp error: Console interrupt. Best wishes, Jocelyn  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1671527&group_id=4933 
From: SourceForge.net <noreply@so...>  20070301 09:19:34

Bugs item #1671527, was opened at 20070301 01:18 Message generated for change (Comment added) made by nobody You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1671527&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: integrate loops Initial Comment: Hello I had to interrupt maxima when it was at calculating the following integral: (%i101) m(r,th); 2 2 (%o101) SQRT(z + 2 r rM (1  COS(th)) + (rM  r) ) (%i102) F(d); 2 2 d0 4 A d0 (epsi   + 1) w 2 d (%o102)   3 d (%i103) integrate(1/m(th,r)^2*F(m(th,r))*z^2,th,0,%PI); 2 2 2 2 Is z  COS (r) rM + rM positive, negative, or zero? pos; 2 2 2 Is z + (1  COS (r)) rM positive, negative, or zero? pos; Maxima encountered a Lisp error: Console interrupt. Best wishes, Jocelyn  Comment By: Nobody/Anonymous (nobody) Date: 20070301 01:19 Message: Logged In: NO Version is Maxima 5.9.1  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1671527&group_id=4933 
From: SourceForge.net <noreply@so...>  20070301 09:22:08

Bugs item #1671527, was opened at 20070301 01:18 Message generated for change (Comment added) made by nobody You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1671527&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: integrate loops Initial Comment: Hello I had to interrupt maxima when it was at calculating the following integral: (%i101) m(r,th); 2 2 (%o101) SQRT(z + 2 r rM (1  COS(th)) + (rM  r) ) (%i102) F(d); 2 2 d0 4 A d0 (epsi   + 1) w 2 d (%o102)   3 d (%i103) integrate(1/m(th,r)^2*F(m(th,r))*z^2,th,0,%PI); 2 2 2 2 Is z  COS (r) rM + rM positive, negative, or zero? pos; 2 2 2 Is z + (1  COS (r)) rM positive, negative, or zero? pos; Maxima encountered a Lisp error: Console interrupt. Best wishes, Jocelyn  Comment By: Nobody/Anonymous (nobody) Date: 20070301 01:22 Message: Logged In: NO sorry about the mess... the functions are: m(r,th):=sqrt(z^2+(rMr)^2+2*rM*r*(1cos(th))); and F(d):=4*w/d*(d0/d)^2*(1+epsi(d0/d)^2)*A;  Comment By: Nobody/Anonymous (nobody) Date: 20070301 01:19 Message: Logged In: NO Version is Maxima 5.9.1  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1671527&group_id=4933 
From: SourceForge.net <noreply@so...>  20070303 02:59:46

Bugs item #1671527, was opened at 20070301 04:18 Message generated for change (Comment added) made by rtoy You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1671527&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: integrate loops Initial Comment: Hello I had to interrupt maxima when it was at calculating the following integral: (%i101) m(r,th); 2 2 (%o101) SQRT(z + 2 r rM (1  COS(th)) + (rM  r) ) (%i102) F(d); 2 2 d0 4 A d0 (epsi   + 1) w 2 d (%o102)   3 d (%i103) integrate(1/m(th,r)^2*F(m(th,r))*z^2,th,0,%PI); 2 2 2 2 Is z  COS (r) rM + rM positive, negative, or zero? pos; 2 2 2 Is z + (1  COS (r)) rM positive, negative, or zero? pos; Maxima encountered a Lisp error: Console interrupt. Best wishes, Jocelyn  >Comment By: Raymond Toy (rtoy) Date: 20070302 21:59 Message: Logged In: YES user_id=28849 Originator: NO It does seem to loop. Not sure if it really does, because I didn't wait for very long. Note, however, that all of the integrals are of the form 1/(th^2+a*th*b)^(n/2) where n is odd (5 or 7 here). Maxima can evaluate all of these integrals quite quickly. I think, but I'm not sure, that maxima is busy simplifying complicated intermediate expressions because the coefficients are not simple.  Comment By: Nobody/Anonymous (nobody) Date: 20070301 04:22 Message: Logged In: NO sorry about the mess... the functions are: m(r,th):=sqrt(z^2+(rMr)^2+2*rM*r*(1cos(th))); and F(d):=4*w/d*(d0/d)^2*(1+epsi(d0/d)^2)*A;  Comment By: Nobody/Anonymous (nobody) Date: 20070301 04:19 Message: Logged In: NO Version is Maxima 5.9.1  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1671527&group_id=4933 
From: SourceForge.net <noreply@so...>  20070307 15:30:30

Bugs item #1671527, was opened at 20070301 04:18 Message generated for change (Comment added) made by rtoy You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1671527&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: integrate loops Initial Comment: Hello I had to interrupt maxima when it was at calculating the following integral: (%i101) m(r,th); 2 2 (%o101) SQRT(z + 2 r rM (1  COS(th)) + (rM  r) ) (%i102) F(d); 2 2 d0 4 A d0 (epsi   + 1) w 2 d (%o102)   3 d (%i103) integrate(1/m(th,r)^2*F(m(th,r))*z^2,th,0,%PI); 2 2 2 2 Is z  COS (r) rM + rM positive, negative, or zero? pos; 2 2 2 Is z + (1  COS (r)) rM positive, negative, or zero? pos; Maxima encountered a Lisp error: Console interrupt. Best wishes, Jocelyn  >Comment By: Raymond Toy (rtoy) Date: 20070307 10:30 Message: Logged In: YES user_id=28849 Originator: NO What is the integrand? You have the function m(r,th), but the integral says m(th,r). Given the limits of 0 and %pi, I think your integrand is wrong. Did you really want m(r,th)?  Comment By: Raymond Toy (rtoy) Date: 20070302 21:59 Message: Logged In: YES user_id=28849 Originator: NO It does seem to loop. Not sure if it really does, because I didn't wait for very long. Note, however, that all of the integrals are of the form 1/(th^2+a*th*b)^(n/2) where n is odd (5 or 7 here). Maxima can evaluate all of these integrals quite quickly. I think, but I'm not sure, that maxima is busy simplifying complicated intermediate expressions because the coefficients are not simple.  Comment By: Nobody/Anonymous (nobody) Date: 20070301 04:22 Message: Logged In: NO sorry about the mess... the functions are: m(r,th):=sqrt(z^2+(rMr)^2+2*rM*r*(1cos(th))); and F(d):=4*w/d*(d0/d)^2*(1+epsi(d0/d)^2)*A;  Comment By: Nobody/Anonymous (nobody) Date: 20070301 04:19 Message: Logged In: NO Version is Maxima 5.9.1  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1671527&group_id=4933 
From: SourceForge.net <noreply@so...>  20070621 16:14:57

Bugs item #1671527, was opened at 20070301 04:18 Message generated for change (Comment added) made by rtoy You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1671527&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: None Priority: 5 Private: No Submitted By: Nobody/Anonymous (nobody) Assigned to: Nobody/Anonymous (nobody) Summary: integrate loops Initial Comment: Hello I had to interrupt maxima when it was at calculating the following integral: (%i101) m(r,th); 2 2 (%o101) SQRT(z + 2 r rM (1  COS(th)) + (rM  r) ) (%i102) F(d); 2 2 d0 4 A d0 (epsi   + 1) w 2 d (%o102)   3 d (%i103) integrate(1/m(th,r)^2*F(m(th,r))*z^2,th,0,%PI); 2 2 2 2 Is z  COS (r) rM + rM positive, negative, or zero? pos; 2 2 2 Is z + (1  COS (r)) rM positive, negative, or zero? pos; Maxima encountered a Lisp error: Console interrupt. Best wishes, Jocelyn  >Comment By: Raymond Toy (rtoy) Date: 20070621 12:14 Message: Logged In: YES user_id=28849 Originator: NO No additional information from poster. Marking as Pending  Comment By: Raymond Toy (rtoy) Date: 20070307 10:30 Message: Logged In: YES user_id=28849 Originator: NO What is the integrand? You have the function m(r,th), but the integral says m(th,r). Given the limits of 0 and %pi, I think your integrand is wrong. Did you really want m(r,th)?  Comment By: Raymond Toy (rtoy) Date: 20070302 21:59 Message: Logged In: YES user_id=28849 Originator: NO It does seem to loop. Not sure if it really does, because I didn't wait for very long. Note, however, that all of the integrals are of the form 1/(th^2+a*th*b)^(n/2) where n is odd (5 or 7 here). Maxima can evaluate all of these integrals quite quickly. I think, but I'm not sure, that maxima is busy simplifying complicated intermediate expressions because the coefficients are not simple.  Comment By: Nobody/Anonymous (nobody) Date: 20070301 04:22 Message: Logged In: NO sorry about the mess... the functions are: m(r,th):=sqrt(z^2+(rMr)^2+2*rM*r*(1cos(th))); and F(d):=4*w/d*(d0/d)^2*(1+epsi(d0/d)^2)*A;  Comment By: Nobody/Anonymous (nobody) Date: 20070301 04:19 Message: Logged In: NO Version is Maxima 5.9.1  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1671527&group_id=4933 
From: SourceForge.net <noreply@so...>  20070706 02:20:15

Bugs item #1671527, was opened at 20070301 01:18 Message generated for change (Comment added) made by sfrobot You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1671527&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: Nobody/Anonymous (nobody) Assigned to: Nobody/Anonymous (nobody) Summary: integrate loops Initial Comment: Hello I had to interrupt maxima when it was at calculating the following integral: (%i101) m(r,th); 2 2 (%o101) SQRT(z + 2 r rM (1  COS(th)) + (rM  r) ) (%i102) F(d); 2 2 d0 4 A d0 (epsi   + 1) w 2 d (%o102)   3 d (%i103) integrate(1/m(th,r)^2*F(m(th,r))*z^2,th,0,%PI); 2 2 2 2 Is z  COS (r) rM + rM positive, negative, or zero? pos; 2 2 2 Is z + (1  COS (r)) rM positive, negative, or zero? pos; Maxima encountered a Lisp error: Console interrupt. Best wishes, Jocelyn  >Comment By: SourceForge Robot (sfrobot) Date: 20070705 19:20 Message: Logged In: YES user_id=1312539 Originator: NO This Tracker item was closed automatically by the system. It was previously set to a Pending status, and the original submitter did not respond within 14 days (the time period specified by the administrator of this Tracker).  Comment By: Raymond Toy (rtoy) Date: 20070621 09:14 Message: Logged In: YES user_id=28849 Originator: NO No additional information from poster. Marking as Pending  Comment By: Raymond Toy (rtoy) Date: 20070307 07:30 Message: Logged In: YES user_id=28849 Originator: NO What is the integrand? You have the function m(r,th), but the integral says m(th,r). Given the limits of 0 and %pi, I think your integrand is wrong. Did you really want m(r,th)?  Comment By: Raymond Toy (rtoy) Date: 20070302 18:59 Message: Logged In: YES user_id=28849 Originator: NO It does seem to loop. Not sure if it really does, because I didn't wait for very long. Note, however, that all of the integrals are of the form 1/(th^2+a*th*b)^(n/2) where n is odd (5 or 7 here). Maxima can evaluate all of these integrals quite quickly. I think, but I'm not sure, that maxima is busy simplifying complicated intermediate expressions because the coefficients are not simple.  Comment By: Nobody/Anonymous (nobody) Date: 20070301 01:22 Message: Logged In: NO sorry about the mess... the functions are: m(r,th):=sqrt(z^2+(rMr)^2+2*rM*r*(1cos(th))); and F(d):=4*w/d*(d0/d)^2*(1+epsi(d0/d)^2)*A;  Comment By: Nobody/Anonymous (nobody) Date: 20070301 01:19 Message: Logged In: NO Version is Maxima 5.9.1  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1671527&group_id=4933 
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