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From: SourceForge.net <noreply@so...>  20061105 02:44:57

Bugs item #1504505, was opened at 20060611 18:24 Message generated for change (Comment added) made by rtoy You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1504505&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: Stavros Macrakis (macrakis) Assigned to: Nobody/Anonymous (nobody) Summary: integrate( 1/(x^81),x,0,1/2) => internal error Initial Comment: integrate( 1/(x^81), x, 0, 1/2 ) => Divison by 0 Smaller exponents don't cause problem. Same problem with gcd: spmod. Using the indefiniteintegral at the limits works fine. 5.9.3 gcl windows  >Comment By: Raymond Toy (rtoy) Date: 20061104 21:44 Message: Logged In: YES user_id=28849 The bug is really in $TAYLOR. A workaround is to call $RATSIMP in RESM1 to simplify the pole somewhat. Closing this bug.  Comment By: Raymond Toy (rtoy) Date: 20061104 11:54 Message: Logged In: YES user_id=28849 Not sure if this is right, but the maxima calculates the roots of 17*x^4+64*x^3+96*x^2+16 in a rather messy form. This seems to confuse $residue (used in RES). If we modify the calls to $residue to call $rectform for the pole to simplify it, this integral works. Not exactly sure whether $residue should do this itself or if RES should do it.  Comment By: Raymond Toy (rtoy) Date: 20061104 11:03 Message: Logged In: YES user_id=28849 For the record, here's is what is happening. The integral is converted to an integral from 0 to inf. Since it's still rational, it uses contour integration over the poles of the rational inside the keyhole contour. To compute this, a partial fraction expansion is done, and all parts are integrated separately. This works until the rational (4*x^2+8*x+4)/(17*x^4+64*x^3+96*x^2+64*x+16) is integrated. At this point the function RES is called and gets a division by zero. Not sure why yet. Perhaps some confusion on computing the residues for the complicated roots of the denominator above?  Comment By: Raymond Toy (rtoy) Date: 20060831 14:55 Message: Logged In: YES user_id=28849 What is happening, partially, is that maxima wants to convert the given integral to an equivalent integral from 0 to infinity. I don't know why we get an internal error for 1/(x^81). However, we can work around the problem by changing the order in methodradicalpoly. The third cond clause could be moved up one more, just before the second clause containing the calls to ratp and ratfnt (which converts the finite integral to an infinite integral). What this does is try to compute the indefinite integral, and if we succeed, we just substitute the limits in. Then all of the examples here work, and the test suite passes.  Comment By: Stavros Macrakis (macrakis) Date: 20060621 01:24 Message: Logged In: YES user_id=588346 integrate(1/(x^4+1),x,0,1/2) same problem. This is a subproblem (by partfrac) of previous problem.  Comment By: Robert Dodier (robert_dodier) Date: 20060620 23:38 Message: Logged In: YES user_id=501686 Assign category (Lisp Core  Integration). I looked at this a little bit but I can't tell what's going on. trace output from :lisp (trace polyform methodbylimits initialanalysis) suggests something interesting is happening but I can't tell what.  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1504505&group_id=4933 
From: SourceForge.net <noreply@so...>  20061104 16:54:28

Bugs item #1504505, was opened at 20060611 18:24 Message generated for change (Comment added) made by rtoy You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1504505&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: Stavros Macrakis (macrakis) Assigned to: Nobody/Anonymous (nobody) Summary: integrate( 1/(x^81),x,0,1/2) => internal error Initial Comment: integrate( 1/(x^81), x, 0, 1/2 ) => Divison by 0 Smaller exponents don't cause problem. Same problem with gcd: spmod. Using the indefiniteintegral at the limits works fine. 5.9.3 gcl windows  >Comment By: Raymond Toy (rtoy) Date: 20061104 11:54 Message: Logged In: YES user_id=28849 Not sure if this is right, but the maxima calculates the roots of 17*x^4+64*x^3+96*x^2+16 in a rather messy form. This seems to confuse $residue (used in RES). If we modify the calls to $residue to call $rectform for the pole to simplify it, this integral works. Not exactly sure whether $residue should do this itself or if RES should do it.  Comment By: Raymond Toy (rtoy) Date: 20061104 11:03 Message: Logged In: YES user_id=28849 For the record, here's is what is happening. The integral is converted to an integral from 0 to inf. Since it's still rational, it uses contour integration over the poles of the rational inside the keyhole contour. To compute this, a partial fraction expansion is done, and all parts are integrated separately. This works until the rational (4*x^2+8*x+4)/(17*x^4+64*x^3+96*x^2+64*x+16) is integrated. At this point the function RES is called and gets a division by zero. Not sure why yet. Perhaps some confusion on computing the residues for the complicated roots of the denominator above?  Comment By: Raymond Toy (rtoy) Date: 20060831 14:55 Message: Logged In: YES user_id=28849 What is happening, partially, is that maxima wants to convert the given integral to an equivalent integral from 0 to infinity. I don't know why we get an internal error for 1/(x^81). However, we can work around the problem by changing the order in methodradicalpoly. The third cond clause could be moved up one more, just before the second clause containing the calls to ratp and ratfnt (which converts the finite integral to an infinite integral). What this does is try to compute the indefinite integral, and if we succeed, we just substitute the limits in. Then all of the examples here work, and the test suite passes.  Comment By: Stavros Macrakis (macrakis) Date: 20060621 01:24 Message: Logged In: YES user_id=588346 integrate(1/(x^4+1),x,0,1/2) same problem. This is a subproblem (by partfrac) of previous problem.  Comment By: Robert Dodier (robert_dodier) Date: 20060620 23:38 Message: Logged In: YES user_id=501686 Assign category (Lisp Core  Integration). I looked at this a little bit but I can't tell what's going on. trace output from :lisp (trace polyform methodbylimits initialanalysis) suggests something interesting is happening but I can't tell what.  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1504505&group_id=4933 
From: SourceForge.net <noreply@so...>  20061104 16:27:48

Bugs item #1590528, was opened at 20061104 11:27 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=1590528&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: Open Resolution: None Priority: 1 Private: No Submitted By: Raymond Toy (rtoy) Assigned to: Nobody/Anonymous (nobody) Summary: Does debugmode(true) actually do anything? Initial Comment: debugmode(true)$ integrate((4*x^2+8*x+4)/(17*x^4+64*x^3+96*x^2+64*x+16),x,0,inf); This should give a division by zero error, and we enter debug mode: (dbm:1) :bt (dbm:1) I tried this with gcl, clisp, and cmucl. Nothing really seems to happen. Does debugmode work for anything? At least with CMUCL, it doesn't seem to matter too much, because I can press Cc to get to CMUCL's debugger which can then produce backtraces and such.  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1590528&group_id=4933 
From: SourceForge.net <noreply@so...>  20061104 16:03:03

Bugs item #1504505, was opened at 20060611 18:24 Message generated for change (Comment added) made by rtoy You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1504505&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: Stavros Macrakis (macrakis) Assigned to: Nobody/Anonymous (nobody) Summary: integrate( 1/(x^81),x,0,1/2) => internal error Initial Comment: integrate( 1/(x^81), x, 0, 1/2 ) => Divison by 0 Smaller exponents don't cause problem. Same problem with gcd: spmod. Using the indefiniteintegral at the limits works fine. 5.9.3 gcl windows  >Comment By: Raymond Toy (rtoy) Date: 20061104 11:03 Message: Logged In: YES user_id=28849 For the record, here's is what is happening. The integral is converted to an integral from 0 to inf. Since it's still rational, it uses contour integration over the poles of the rational inside the keyhole contour. To compute this, a partial fraction expansion is done, and all parts are integrated separately. This works until the rational (4*x^2+8*x+4)/(17*x^4+64*x^3+96*x^2+64*x+16) is integrated. At this point the function RES is called and gets a division by zero. Not sure why yet. Perhaps some confusion on computing the residues for the complicated roots of the denominator above?  Comment By: Raymond Toy (rtoy) Date: 20060831 14:55 Message: Logged In: YES user_id=28849 What is happening, partially, is that maxima wants to convert the given integral to an equivalent integral from 0 to infinity. I don't know why we get an internal error for 1/(x^81). However, we can work around the problem by changing the order in methodradicalpoly. The third cond clause could be moved up one more, just before the second clause containing the calls to ratp and ratfnt (which converts the finite integral to an infinite integral). What this does is try to compute the indefinite integral, and if we succeed, we just substitute the limits in. Then all of the examples here work, and the test suite passes.  Comment By: Stavros Macrakis (macrakis) Date: 20060621 01:24 Message: Logged In: YES user_id=588346 integrate(1/(x^4+1),x,0,1/2) same problem. This is a subproblem (by partfrac) of previous problem.  Comment By: Robert Dodier (robert_dodier) Date: 20060620 23:38 Message: Logged In: YES user_id=501686 Assign category (Lisp Core  Integration). I looked at this a little bit but I can't tell what's going on. trace output from :lisp (trace polyform methodbylimits initialanalysis) suggests something interesting is happening but I can't tell what.  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1504505&group_id=4933 
From: SourceForge.net <noreply@so...>  20061103 21:40:21

Bugs item #1582625, was opened at 20061023 00:22 Message generated for change (Comment added) made by rtoy You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1582625&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(t^2*log(t)/((t^21)*(t^4+1)), t, 0, 1) wrong? Initial Comment: Symbolic integration seems to return an incorrect result: (%i1) integrate(t^2*log(t)/((t^21)*(t^4+1)), t, 0, 1); (sqrt(2) + 1) %pi (%o1)  16 sqrt(2) (%i2) float(%); (%o2) 1.053029287545515 (%i3) romberg(t^2*log(t)/((t^21)*(t^4+1)), t, 0.0000001, 0.9999999); (%o3) 0.1806718095951 (%i4) float(%pi^2/(16*(2+sqrt(2)))); (%o4) 0.18067126259065  >Comment By: Raymond Toy (rtoy) Date: 20061103 16:40 Message: Logged In: YES user_id=28849 The issue appears to be in logimag02%pi. Some of the poles are of the form (1)^(1/4) or sqrt(%i). The call to simplify %plog(pole) doesn't actually simplify and the noun form is returned (I think). If we replace (defun logimag02%pi (x) (let ((plog (simplify ((%plog) ,x)))) with (defun logimag02%pi (x) (let ((plog (simplify ($rectform `((%plog) ,x))))) maxima returns (sqrt(2)2)*%pi^2/32 which is .1806712625906549, which corresponds pretty well with the numerical result from romberg and quad_qags. The test suite runs fine with this change. I think the real problem is in the simplifier for plog, but I'm not too motivated in fixing that.  Comment By: Raymond Toy (rtoy) Date: 20061023 13:57 Message: Logged In: YES user_id=28849 Maxima uses the substitution t = exp(y) to change the integral from 0 to 1 to 0 to inf. Then it uses its routine to handle this infinite integral by converting it to an integral from minf to inf, because the integrand is even. Finally, it uses rectzto%pi2 to integrate this final integrand. rectzto%pi2 needs to find the poles of the denominator. I'm guessing it's getting that wrong.  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1582625&group_id=4933 
From: SourceForge.net <noreply@so...>  20061101 14:42:47

Bugs item #1588623, was opened at 20061101 08:42 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=1588623&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: romberg with % or %o argument Initial Comment: (%i1) sqrt(1+x^3); (%o1) sqrt(x^3+1) (%i2) romberg(%, x,0,1); Maxima encountered a Lisp error: Also, the command romberg(%o1,x,0,1) gives a Lisp error. This works OK: (%i4) romberg(sqrt(1+x^3), x,0,1); (%o4) 1.111447999508385 (%i1) build_info(); Maxima version: 5.10.0 Maxima build date: 19:9 9/21/2006 host type: i686pcmingw32 lispimplementationtype: GNU Common Lisp (GCL) lispimplementationversion: GCL 2.6.7  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1588623&group_id=4933 