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From: SourceForge.net <noreply@so...>  20061106 18:14:03

Bugs item #1073338, was opened at 20041125 13:21 Message generated for change (Comment added) made by rtoy You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1073338&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: Robert Dodier (robert_dodier) Assigned to: Nobody/Anonymous (nobody) Summary: integrate yields incorrect result on rational function Initial Comment: "integrate" yields incorrect results on some rational functions. "Division by 0" is strange. The definite integral below is certainly greater than 0 as the integrand is positive over [0, 1]. "integrate (1/((x3)^4+1/2), x)" returns the noun form, so maybe (maybe) what happens is that the noun form is evaluated at the limits of integration and it's the same, hence 0 is the result. (Just guessing there.) Note that the difference between the two integrands is that one is 1/(something + 1), while the other is 1/(same something + 1/2).  (%i1) integrate (1/((x3)^4+1), x, 0, 1); Division by 0  an error. Quitting. To debug this try DEBUGMODE(TRUE); (%i2) integrate (1/((x3)^4+1/2), x, 0, 1); (%o2) 0 (%i3) build_info (); Maxima version: 5.9.1 Maxima build date: 21:24 9/23/2004 host type: i686pclinuxgnu lispimplementationtype: CMU Common Lisp lispimplementationversion: 19a  Same behavior observed in CVS build of 2004/11/24.  >Comment By: Raymond Toy (rtoy) Date: 20061106 13:13 Message: Logged In: YES user_id=28849 Recent changes in defint.lisp and residue.lisp has fixed the issue with the integrate(1/((x3)^4+1),x,0,1). For integrate(1/((x3)^4+1/2),x,0,1), it does seem to be either a gcd or residue problem. We can work around this issue by making 2 changes in residu.lisp. In the function RES, for the case of simple poles, we can use RES1 instead of $RESIDUE. In the function RES1, we need to call $RECTFORM for each pole to make is simpler. With these changes applied, this integral can be evaluated. However, the answer takes some time and is quite messy. We can get the simpler result by using factor(expand(sqrtdenest(<foo>))). The answer is: (2*atan((sqrt(2)4*2^(1/4)+8)/(49*2^(3/4)+sqrt(2)8)) log((2^(3/4)+12*sqrt(2)+73*2^(1/4)2)/(33*2^(1/4))) +log((2^(3/4)12*sqrt(2)+73*2^(1/4)+2)/(33*2^(1/4))) 2*atan((sqrt(2)+4*2^(1/4)+8)/(49*2^(3/4)+sqrt(2)8))) /(2*2^(3/4)) Numerical evaluation of this compares favorably with the numerical result from quad_qags.  Comment By: Stavros Macrakis (macrakis) Date: 20050130 18:04 Message: Logged In: YES user_id=588346 This is apparently another GCD problem. Fixed by setting the 'algebraic' flag: integrate (1/((x3)^4+1), x, 0, 1),algebraic:true For the indefinite integral, you can factor over the Gaussians then use partfrac: integrate ( partfrac ( gfactor( 1/((x3)^4+1) ), x ), x ) You can simplify the result using ratsimp(rectform(...)) s  Comment By: Nobody/Anonymous (nobody) Date: 20050118 16:17 Message: Logged In: NO Two remarks: The method that Maxima uses to solve such integrals is integration by residues. Trace (residue) shows that this function is called with the correct arguments but answers zeroes for integrate (1/((x3)^4+1/2), x, 0, 1); and runs into a division by zero for the other integral. The indefinite integrals can be solved with changevar: 'Integrate(1/((x3)^4+1/2), x); changevar (%, x  3  y ,y ,x); ev (%, Integrate);  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1073338&group_id=4933 
From: SourceForge.net <noreply@so...>  20061106 16:28:49

Bugs item #733071, was opened at 20030506 00:19 Message generated for change (Comment added) made by rtoy You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=733071&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: Defint unsimplified result Initial Comment: integrate(exp(x^2%i*2*%pi*x*s),x,minf,inf) => SQRT(%PI)*%E^(%I^2*%PI^2*s^2) Note the unsimplified %i^2.  >Comment By: Raymond Toy (rtoy) Date: 20061106 11:28 Message: Logged In: YES user_id=28849 FWIW, this is caused by the call to $ratsimp in linpower in defint.lisp. Something about it prevents simplifying %i^2 to 1.  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=733071&group_id=4933 
From: SourceForge.net <noreply@so...>  20061106 14:37:28

Bugs item #1590528, was opened at 20061104 11:27 Message generated for change (Comment added) made by rtoy 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.  >Comment By: Raymond Toy (rtoy) Date: 20061106 09:37 Message: Logged In: YES user_id=28849 After reading your comments and the examples in the user manual, I see that debugmode is useful. I also agree that we should change the "Quitting" part. However, I propose to leave the debugmode(true) part in. Even though it can't debug Lisp functions and such, it's quite useful because execution stops at the error and I can press Cc (in CMUCL) and use CMUCL's debugger to figure out where the problem is. I suppose I could use (setf *breakonsignals* t), but this is still useful.  Comment By: Robert Dodier (robert_dodier) Date: 20061105 11:11 Message: Logged In: YES user_id=501686 Maxima debugmode can print a backtrace of functions defined in Maxima by := . So far as I can tell, debugmode doesn't know anything about functions defined by defun or defmfun or defmspec. Given that most functions in Maxima are defined by defun, defmfun, and defmspec, it is not very informative to use debugmode. Since debugmode is useful in a certain context (namely debugging userdefined functions) I think we should keep it, but let's change "  an error. Quitting. To debug this try debugmode(true);" to just "  an error." (i.e. cut out the recommendation to use debugmode, which is usually not helpful, and also the "Quitting." because, in fact, Maxima is not executing (quit) nor quit()).  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1590528&group_id=4933 