You can subscribe to this list here.
2002 
_{Jan}

_{Feb}

_{Mar}

_{Apr}

_{May}

_{Jun}
(67) 
_{Jul}
(61) 
_{Aug}
(49) 
_{Sep}
(43) 
_{Oct}
(59) 
_{Nov}
(24) 
_{Dec}
(18) 

2003 
_{Jan}
(34) 
_{Feb}
(35) 
_{Mar}
(72) 
_{Apr}
(42) 
_{May}
(46) 
_{Jun}
(15) 
_{Jul}
(64) 
_{Aug}
(62) 
_{Sep}
(22) 
_{Oct}
(41) 
_{Nov}
(57) 
_{Dec}
(56) 
2004 
_{Jan}
(48) 
_{Feb}
(47) 
_{Mar}
(33) 
_{Apr}
(39) 
_{May}
(6) 
_{Jun}
(17) 
_{Jul}
(19) 
_{Aug}
(10) 
_{Sep}
(14) 
_{Oct}
(74) 
_{Nov}
(80) 
_{Dec}
(22) 
2005 
_{Jan}
(43) 
_{Feb}
(33) 
_{Mar}
(52) 
_{Apr}
(74) 
_{May}
(32) 
_{Jun}
(58) 
_{Jul}
(18) 
_{Aug}
(41) 
_{Sep}
(71) 
_{Oct}
(28) 
_{Nov}
(65) 
_{Dec}
(68) 
2006 
_{Jan}
(54) 
_{Feb}
(37) 
_{Mar}
(82) 
_{Apr}
(211) 
_{May}
(69) 
_{Jun}
(75) 
_{Jul}
(279) 
_{Aug}
(139) 
_{Sep}
(135) 
_{Oct}
(58) 
_{Nov}
(81) 
_{Dec}
(78) 
2007 
_{Jan}
(141) 
_{Feb}
(134) 
_{Mar}
(65) 
_{Apr}
(49) 
_{May}
(61) 
_{Jun}
(90) 
_{Jul}
(72) 
_{Aug}
(53) 
_{Sep}
(86) 
_{Oct}
(61) 
_{Nov}
(62) 
_{Dec}
(101) 
2008 
_{Jan}
(100) 
_{Feb}
(66) 
_{Mar}
(76) 
_{Apr}
(95) 
_{May}
(77) 
_{Jun}
(93) 
_{Jul}
(103) 
_{Aug}
(76) 
_{Sep}
(42) 
_{Oct}
(55) 
_{Nov}
(44) 
_{Dec}
(75) 
2009 
_{Jan}
(103) 
_{Feb}
(105) 
_{Mar}
(121) 
_{Apr}
(59) 
_{May}
(103) 
_{Jun}
(82) 
_{Jul}
(67) 
_{Aug}
(76) 
_{Sep}
(85) 
_{Oct}
(75) 
_{Nov}
(181) 
_{Dec}
(133) 
2010 
_{Jan}
(107) 
_{Feb}
(116) 
_{Mar}
(145) 
_{Apr}
(89) 
_{May}
(138) 
_{Jun}
(85) 
_{Jul}
(82) 
_{Aug}
(111) 
_{Sep}
(70) 
_{Oct}
(83) 
_{Nov}
(60) 
_{Dec}
(16) 
2011 
_{Jan}
(61) 
_{Feb}
(16) 
_{Mar}
(52) 
_{Apr}
(41) 
_{May}
(34) 
_{Jun}
(41) 
_{Jul}
(57) 
_{Aug}
(73) 
_{Sep}
(21) 
_{Oct}
(45) 
_{Nov}
(50) 
_{Dec}
(28) 
2012 
_{Jan}
(70) 
_{Feb}
(36) 
_{Mar}
(71) 
_{Apr}
(29) 
_{May}
(48) 
_{Jun}
(61) 
_{Jul}
(44) 
_{Aug}
(54) 
_{Sep}
(20) 
_{Oct}
(28) 
_{Nov}
(41) 
_{Dec}
(137) 
2013 
_{Jan}
(62) 
_{Feb}
(55) 
_{Mar}
(31) 
_{Apr}
(23) 
_{May}
(54) 
_{Jun}
(54) 
_{Jul}
(90) 
_{Aug}
(46) 
_{Sep}
(38) 
_{Oct}
(60) 
_{Nov}
(92) 
_{Dec}
(17) 
2014 
_{Jan}
(62) 
_{Feb}
(35) 
_{Mar}
(72) 
_{Apr}
(30) 
_{May}
(97) 
_{Jun}
(81) 
_{Jul}
(63) 
_{Aug}
(64) 
_{Sep}
(28) 
_{Oct}
(45) 
_{Nov}
(48) 
_{Dec}
(109) 
2015 
_{Jan}
(106) 
_{Feb}
(36) 
_{Mar}
(65) 
_{Apr}
(63) 
_{May}
(95) 
_{Jun}
(56) 
_{Jul}
(48) 
_{Aug}
(55) 
_{Sep}
(100) 
_{Oct}
(57) 
_{Nov}
(33) 
_{Dec}
(46) 
2016 
_{Jan}
(76) 
_{Feb}
(53) 
_{Mar}
(88) 
_{Apr}
(79) 
_{May}
(62) 
_{Jun}
(65) 
_{Jul}
(37) 
_{Aug}
(23) 
_{Sep}
(108) 
_{Oct}
(68) 
_{Nov}
(66) 
_{Dec}
(47) 
2017 
_{Jan}
(55) 
_{Feb}
(11) 
_{Mar}
(30) 
_{Apr}
(19) 
_{May}
(14) 
_{Jun}
(21) 
_{Jul}
(30) 
_{Aug}
(38) 
_{Sep}

_{Oct}

_{Nov}

_{Dec}

S  M  T  W  T  F  S 







1

2
(4) 
3

4
(1) 
5
(1) 
6
(2) 
7
(4) 
8
(9) 
9
(4) 
10
(1) 
11
(1) 
12
(6) 
13
(5) 
14
(1) 
15

16

17
(8) 
18

19
(1) 
20
(1) 
21

22
(2) 
23
(1) 
24
(1) 
25
(3) 
26
(1) 
27
(4) 
28
(4) 
29
(6) 
30
(3) 
31
(2) 





From: SourceForge.net <noreply@so...>  20080322 17:09:59

Bugs item #1923119, was opened at 20080322 12:09 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=1923119&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: 4 Private: No Submitted By: Barton Willis (willisbl) Assigned to: Nobody/Anonymous (nobody) Summary: 1/sqrt(8)  sqrt(8) / 8 /> 0 Initial Comment: (%i17) 1/sqrt(8)  sqrt(8) / 8; (%o17) 1/(2*sqrt(2))1/2^(3/2) (%i18) expand(%,0,0); (%o18) 0 Here I think (%o17) is unsimplifiedthe expand(%,0,0) shouldn't be required.  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1923119&group_id=4933 
From: SourceForge.net <noreply@so...>  20080322 12:52:48

Bugs item #1920177, was opened at 20080319 22:07 Message generated for change (Settings changed) made by crategus You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1920177&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: Includes proposed fix Status: Open Resolution: None Priority: 5 >Private: No Submitted By: Crategus (crategus) Assigned to: Nobody/Anonymous (nobody) Summary: Problems with the bessel functions Initial Comment: There are several problems with the bessel functions. The problems I have found can be divided in the following categories: 1. Inconsistent use of the internal arrays $besselarray, $yarray and $iarray. Example: Restart Maxima and Enter bessel_y(2,1.0). You get an Lisp error. Try first bessel_y(2,1.0) and then repeat bessel_y(2,1.0). Now you get the correct result 1.65...  %i * 3,30... Try bessel_j(2,1.0), you get 0.1149...  %i*2.8142... Next enter bessel_y(2,1.0), you get a Lisp error. The reason for the problems is that the routine bessely uses the global array $YARRAY to store values, but uses also the array $BESSELARRAY to calculate the answer. This is an error. I think the best is to eliminate the use of the global arrays completly. 2. Problematic roundoff errors: Try bessel_j(2,1.0). The result is 0,114903...  %i* 2.8142... e17. The correct result is pure real. The problem is the use of the transformation j[n]=exp(n*%pi*%i)*j[n](x) in the code which produce a small imaginary part. This is a roundoff error for an integer order. In the case of non integer values the imaginary part is correct. So you can not cut off the imaginary part in general. Here is a piece of code which will return an answer which is pure real when the order is an integer (I started to rewrite the code, introduced a function besselj and eliminated the use of the global arrays): (if (integerp order) (if (evenp order) (aref jvals n) ( (aref jvals n)) ) (let ((v (* (cis (* order pi)) (aref jvals n)))) (simplifya `((mplus) ,(realpart v) ((mtimes) $%i ,(imagpart v))) nil ) ) ) 3. Wrong mathematic: Try bessel_j(2.5,1,0). You get 0.04949... Correct is the result 2.8763... * %i Or bessel_j(3,2.0). You get 0,128943... Correct is the result 0.128943... The calculation of the bessel function j[n](x) for real argument x and negativ order n as (realpart (hankel1 order arg)) is not correct. The correct result can be obtained with the formula j[n](x) = 1/2 * (H1[n](x) + H2[n](x)). I tried the following code: (let ((result (* 0.5d0 (+ (hankel1 order arg) (hankel2 order arg))))) (complexify result) ; Problem: you get a small imaginary part ;in the case of a real result (like Problem 2) ; An alternative with a function fpround which round the result ; so that a small imaginary part will vanish ; ; (simplifya ; `((mplus) ; ,(fpround (realpart result) 14) ; ((mtimes) ; ,(fpround (imagpart result) 14) ; $%i ; ) ; ) ; nil ; ) ) This is the definition of the function fpround: (defun fpround (x &optional (digits 1)) (let ((fac (expt 10 digits))) (/ (round x (/ 1 fac)) (float fac)) ) ) I have redesigned a lot of the code for the bessel functions but the work is not finished. Is the work interesting for the project? I use GCL 2.6.6 on Windows XP for programing. Crategus  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1920177&group_id=4933 
From: SourceForge.net <noreply@so...>  20080320 14:20:50

Bugs item #1921102, was opened at 20080320 15:20 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=1921102&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: Includes proposed fix Status: Open Resolution: None Priority: 5 Private: Yes Submitted By: Crategus (crategus) Assigned to: Nobody/Anonymous (nobody) Summary: Test files dont work Initial Comment: I installed now Maxima 5.14.0 with GCL 2.6.6 ANSI on my MS Windows XP system. After installation the test files dont work. I get the error "unexpected end of #(input stream ..." reading the files in test mode. On my system the following extra EOF test in the routine TYIRAW works: (defun tyiraw (&optional (stream *standardinput*) eofoption) ;; added this extra EOF test, because test files generate ;; unexpected end of inputstream on my system (Windows XP, GCL 2.6.6) (when (eq (peekchar nil stream nil eofoption) eofoption) (returnfrom tyiraw eofoption) ) ;; this is the original unchanged code (let ((ch (readcharnohang stream nil eofoption))) (if ch ch (progn (when (and *promptonreadhang* *readhangprompt*) (princ *readhangprompt*) (forceoutput *standardoutput*) ) (readchar stream nil eofoption) ) ) ) ) I had the same problem with the versions 5.13.0 und 5.12.0 of Maxima on my system. Perhaps the bug is in the routine TESTBATCH. A reason may be the different EOF values used in Maxima at different places. Crategus  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1921102&group_id=4933 
From: SourceForge.net <noreply@so...>  20080319 21:07:22

Bugs item #1920177, was opened at 20080319 22:07 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=1920177&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: Includes proposed fix Status: Open Resolution: None Priority: 5 Private: Yes Submitted By: Crategus (crategus) Assigned to: Nobody/Anonymous (nobody) Summary: Problems with the bessel functions Initial Comment: There are several problems with the bessel functions. The problems I have found can be divided in the following categories: 1. Inconsistent use of the internal arrays $besselarray, $yarray and $iarray. Example: Restart Maxima and Enter bessel_y(2,1.0). You get an Lisp error. Try first bessel_y(2,1.0) and then repeat bessel_y(2,1.0). Now you get the correct result 1.65...  %i * 3,30... Try bessel_j(2,1.0), you get 0.1149...  %i*2.8142... Next enter bessel_y(2,1.0), you get a Lisp error. The reason for the problems is that the routine bessely uses the global array $YARRAY to store values, but uses also the array $BESSELARRAY to calculate the answer. This is an error. I think the best is to eliminate the use of the global arrays completly. 2. Problematic roundoff errors: Try bessel_j(2,1.0). The result is 0,114903...  %i* 2.8142... e17. The correct result is pure real. The problem is the use of the transformation j[n]=exp(n*%pi*%i)*j[n](x) in the code which produce a small imaginary part. This is a roundoff error for an integer order. In the case of non integer values the imaginary part is correct. So you can not cut off the imaginary part in general. Here is a piece of code which will return an answer which is pure real when the order is an integer (I started to rewrite the code, introduced a function besselj and eliminated the use of the global arrays): (if (integerp order) (if (evenp order) (aref jvals n) ( (aref jvals n)) ) (let ((v (* (cis (* order pi)) (aref jvals n)))) (simplifya `((mplus) ,(realpart v) ((mtimes) $%i ,(imagpart v))) nil ) ) ) 3. Wrong mathematic: Try bessel_j(2.5,1,0). You get 0.04949... Correct is the result 2.8763... * %i Or bessel_j(3,2.0). You get 0,128943... Correct is the result 0.128943... The calculation of the bessel function j[n](x) for real argument x and negativ order n as (realpart (hankel1 order arg)) is not correct. The correct result can be obtained with the formula j[n](x) = 1/2 * (H1[n](x) + H2[n](x)). I tried the following code: (let ((result (* 0.5d0 (+ (hankel1 order arg) (hankel2 order arg))))) (complexify result) ; Problem: you get a small imaginary part ;in the case of a real result (like Problem 2) ; An alternative with a function fpround which round the result ; so that a small imaginary part will vanish ; ; (simplifya ; `((mplus) ; ,(fpround (realpart result) 14) ; ((mtimes) ; ,(fpround (imagpart result) 14) ; $%i ; ) ; ) ; nil ; ) ) This is the definition of the function fpround: (defun fpround (x &optional (digits 1)) (let ((fac (expt 10 digits))) (/ (round x (/ 1 fac)) (float fac)) ) ) I have redesigned a lot of the code for the bessel functions but the work is not finished. Is the work interesting for the project? I use GCL 2.6.6 on Windows XP for programing. Crategus  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1920177&group_id=4933 
From: SourceForge.net <noreply@so...>  20080317 15:19:59

Bugs item #1913067, was opened at 20080312 17:06 Message generated for change (Comment added) made by rtoy You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1913067&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: Ximin Luo (infinity0x) Assigned to: Nobody/Anonymous (nobody) Summary: Cannot integrate 1/(1+x^n) Initial Comment: Maxima cannot integrate functions of the form 1/(1+x^n) for n >= 7. Mathematica is capable of this. Maxima attempts the integration via an algorithm which seems to involve taking partial fractions. For example, if you try to integrate 1/(1+x^7) for example, Maxima gives: log(1+x)/7  integral(x^52*x^4+3*x^34*x^2+5*x6)/(x^6x^5+x^4x^3+x^2x+1) / 7 I'm guessing a different algorithm (such as that employed by Mathematica) is required to give a fully symbolic answer. For the record, the integral(1/(1+x^7)) is: Log[1 + x]/7  (Cos[Pi/7]*Log[1 + x^2  2*x*Cos[Pi/7]])/7  (Cos[(3*Pi)/7]*Log[1 + x^2  2*x*Cos[(3*Pi)/7]])/7  (Cos[(5*Pi)/7]*Log[1 + x^2  2*x*Cos[(5*Pi)/7]])/7 + (2*ArcTan[(x  Cos[Pi/7])*Csc[Pi/7]]* Sin[Pi/7])/7 + (2*ArcTan[(x  Cos[(3*Pi)/7])* Csc[(3*Pi)/7]]*Sin[(3*Pi)/7])/7 + (2*ArcTan[(x  Cos[(5*Pi)/7])* Csc[(5*Pi)/7]]*Sin[(5*Pi)/7])/7  Additionally, this also means that we can calculate the integral(that nasty rational function), by calculating 7 * ( log(1+x)/7  integral(1/(1+x^7)) ) As a side note, Mathematica also cannot give integral(that nasty rational function) fully symbolically. Instead, it gives: RootSum[1  #1 + #1^2  #1^3 + #1^4  #1^5 + #1^6 & , (6*Log[x  #1] + 5*Log[x  #1]*#1  4*Log[x  #1]*#1^2 + 3*Log[x  #1]* #1^3  2*Log[x  #1]*#1^4 + Log[x  #1]*#1^5)/(1 + 2*#1  3*#1^2 + 4*#1^3  5*#1^4 + 6*#1^5) & ] where RootSum[ f, form ] represents the sum of form[x] for all x that satisfy the polynomial equation f[x] == 0.  >Comment By: Raymond Toy (rtoy) Date: 20080317 11:19 Message: Logged In: YES user_id=28849 Originator: NO By handcomputed, I mean someone wrote out the expansion of x^6+1 as (x^2+1)*(x^2+sqrt(3)*x+1)*(x^2sqrt(3)*x+1). I'm not sure maxima can figure that out itself. For the general case x^n+c, we would need to write that as a product of quadratics (and maybe a linear term). Easy to do since we know the roots are, more or less, the n roots of unity. I think the results will be pretty ugly for n >= 7.  Comment By: Raymond Toy (rtoy) Date: 20080317 11:19 Message: Logged In: YES user_id=28849 Originator: NO By handcomputed, I mean someone wrote out the expansion of x^6+1 as (x^2+1)*(x^2+sqrt(3)*x+1)*(x^2sqrt(3)*x+1). I'm not sure maxima can figure that out itself. For the general case x^n+c, we would need to write that as a product of quadratics (and maybe a linear term). Easy to do since we know the roots are, more or less, the n roots of unity. I think the results will be pretty ugly for n >= 7.  Comment By: Ximin Luo (infinity0x) Date: 20080317 10:27 Message: Logged In: YES user_id=2003896 Originator: YES i don't know how to integrate these without knowing the roots of the equation; i thought mathematica was, because it could integrate 1/(1+x^n) without being able to integrate (that nasty rational function). (and i thought algorithms do to this would be available somewhere else.) also, i assumed maxima was already factorising the equation for arbitrarily large values of n. but your approach seems feasible. what do you mean by handcomputed though? as in, numerically instead of algebraically? wouldn't that give imprecise answers?  Comment By: Raymond Toy (rtoy) Date: 20080317 09:37 Message: Logged In: YES user_id=28849 Originator: NO If you can, please describe how to integrate these without knowing the roots of the equation. FWIW, maxima has a special function to integrate functions of the form f(x)/(a*x^n+b). (See enprog and eprog in sinint.lisp.) It only handles the cases of n = 4, 5, 6. And it does this by essentially doing a partial fraction expansion with handcomputed quadratic factors. It seems feasible to extend this to any n.  Comment By: Raymond Toy (rtoy) Date: 20080317 09:37 Message: Logged In: YES user_id=28849 Originator: NO If you can, please describe how to integrate these without knowing the roots of the equation. FWIW, maxima has a special function to integrate functions of the form f(x)/(a*x^n+b). (See enprog and eprog in sinint.lisp.) It only handles the cases of n = 4, 5, 6. And it does this by essentially doing a partial fraction expansion with handcomputed quadratic factors. It seems feasible to extend this to any n.  Comment By: Raymond Toy (rtoy) Date: 20080317 09:37 Message: Logged In: YES user_id=28849 Originator: NO If you can, please describe how to integrate these without knowing the roots of the equation. FWIW, maxima has a special function to integrate functions of the form f(x)/(a*x^n+b). (See enprog and eprog in sinint.lisp.) It only handles the cases of n = 4, 5, 6. And it does this by essentially doing a partial fraction expansion with handcomputed quadratic factors. It seems feasible to extend this to any n.  Comment By: Raymond Toy (rtoy) Date: 20080312 20:30 Message: Logged In: YES user_id=28849 Originator: NO Actually, the roots are obviously the seven roots of 1, which maxima does know, and it could have done the partial fraction expansion to find the value of the integral. Not sure how or where to teach maxima about this, though. Some thought needed.  Comment By: Ximin Luo (infinity0x) Date: 20080312 18:32 Message: Logged In: YES user_id=2003896 Originator: YES The point is that Maxima does not NEED to know the roots of that equation. Sorry for making this unclear. By doing the integral using a different algorithm which doesn't involve taking partial fractions, you can avoid the above, and get a purely symbolic integral, like Mathematic does.  Comment By: Raymond Toy (rtoy) Date: 20080312 18:16 Message: Logged In: YES user_id=28849 Originator: NO Maxima can't give an answer because it doesn't know the roots of (x^7+1)/(x+1). If you set integrate_use_rootsof:true, then maxima can return an answer, but unless you can compute the roots of the above equation, it's probably not very helpful.  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1913067&group_id=4933 
From: SourceForge.net <noreply@so...>  20080317 15:19:36

Bugs item #1913067, was opened at 20080312 17:06 Message generated for change (Comment added) made by rtoy You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1913067&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: Ximin Luo (infinity0x) Assigned to: Nobody/Anonymous (nobody) Summary: Cannot integrate 1/(1+x^n) Initial Comment: Maxima cannot integrate functions of the form 1/(1+x^n) for n >= 7. Mathematica is capable of this. Maxima attempts the integration via an algorithm which seems to involve taking partial fractions. For example, if you try to integrate 1/(1+x^7) for example, Maxima gives: log(1+x)/7  integral(x^52*x^4+3*x^34*x^2+5*x6)/(x^6x^5+x^4x^3+x^2x+1) / 7 I'm guessing a different algorithm (such as that employed by Mathematica) is required to give a fully symbolic answer. For the record, the integral(1/(1+x^7)) is: Log[1 + x]/7  (Cos[Pi/7]*Log[1 + x^2  2*x*Cos[Pi/7]])/7  (Cos[(3*Pi)/7]*Log[1 + x^2  2*x*Cos[(3*Pi)/7]])/7  (Cos[(5*Pi)/7]*Log[1 + x^2  2*x*Cos[(5*Pi)/7]])/7 + (2*ArcTan[(x  Cos[Pi/7])*Csc[Pi/7]]* Sin[Pi/7])/7 + (2*ArcTan[(x  Cos[(3*Pi)/7])* Csc[(3*Pi)/7]]*Sin[(3*Pi)/7])/7 + (2*ArcTan[(x  Cos[(5*Pi)/7])* Csc[(5*Pi)/7]]*Sin[(5*Pi)/7])/7  Additionally, this also means that we can calculate the integral(that nasty rational function), by calculating 7 * ( log(1+x)/7  integral(1/(1+x^7)) ) As a side note, Mathematica also cannot give integral(that nasty rational function) fully symbolically. Instead, it gives: RootSum[1  #1 + #1^2  #1^3 + #1^4  #1^5 + #1^6 & , (6*Log[x  #1] + 5*Log[x  #1]*#1  4*Log[x  #1]*#1^2 + 3*Log[x  #1]* #1^3  2*Log[x  #1]*#1^4 + Log[x  #1]*#1^5)/(1 + 2*#1  3*#1^2 + 4*#1^3  5*#1^4 + 6*#1^5) & ] where RootSum[ f, form ] represents the sum of form[x] for all x that satisfy the polynomial equation f[x] == 0.  >Comment By: Raymond Toy (rtoy) Date: 20080317 11:19 Message: Logged In: YES user_id=28849 Originator: NO By handcomputed, I mean someone wrote out the expansion of x^6+1 as (x^2+1)*(x^2+sqrt(3)*x+1)*(x^2sqrt(3)*x+1). I'm not sure maxima can figure that out itself. For the general case x^n+c, we would need to write that as a product of quadratics (and maybe a linear term). Easy to do since we know the roots are, more or less, the n roots of unity. I think the results will be pretty ugly for n >= 7.  Comment By: Ximin Luo (infinity0x) Date: 20080317 10:27 Message: Logged In: YES user_id=2003896 Originator: YES i don't know how to integrate these without knowing the roots of the equation; i thought mathematica was, because it could integrate 1/(1+x^n) without being able to integrate (that nasty rational function). (and i thought algorithms do to this would be available somewhere else.) also, i assumed maxima was already factorising the equation for arbitrarily large values of n. but your approach seems feasible. what do you mean by handcomputed though? as in, numerically instead of algebraically? wouldn't that give imprecise answers?  Comment By: Raymond Toy (rtoy) Date: 20080317 09:37 Message: Logged In: YES user_id=28849 Originator: NO If you can, please describe how to integrate these without knowing the roots of the equation. FWIW, maxima has a special function to integrate functions of the form f(x)/(a*x^n+b). (See enprog and eprog in sinint.lisp.) It only handles the cases of n = 4, 5, 6. And it does this by essentially doing a partial fraction expansion with handcomputed quadratic factors. It seems feasible to extend this to any n.  Comment By: Raymond Toy (rtoy) Date: 20080317 09:37 Message: Logged In: YES user_id=28849 Originator: NO If you can, please describe how to integrate these without knowing the roots of the equation. FWIW, maxima has a special function to integrate functions of the form f(x)/(a*x^n+b). (See enprog and eprog in sinint.lisp.) It only handles the cases of n = 4, 5, 6. And it does this by essentially doing a partial fraction expansion with handcomputed quadratic factors. It seems feasible to extend this to any n.  Comment By: Raymond Toy (rtoy) Date: 20080317 09:37 Message: Logged In: YES user_id=28849 Originator: NO If you can, please describe how to integrate these without knowing the roots of the equation. FWIW, maxima has a special function to integrate functions of the form f(x)/(a*x^n+b). (See enprog and eprog in sinint.lisp.) It only handles the cases of n = 4, 5, 6. And it does this by essentially doing a partial fraction expansion with handcomputed quadratic factors. It seems feasible to extend this to any n.  Comment By: Raymond Toy (rtoy) Date: 20080312 20:30 Message: Logged In: YES user_id=28849 Originator: NO Actually, the roots are obviously the seven roots of 1, which maxima does know, and it could have done the partial fraction expansion to find the value of the integral. Not sure how or where to teach maxima about this, though. Some thought needed.  Comment By: Ximin Luo (infinity0x) Date: 20080312 18:32 Message: Logged In: YES user_id=2003896 Originator: YES The point is that Maxima does not NEED to know the roots of that equation. Sorry for making this unclear. By doing the integral using a different algorithm which doesn't involve taking partial fractions, you can avoid the above, and get a purely symbolic integral, like Mathematic does.  Comment By: Raymond Toy (rtoy) Date: 20080312 18:16 Message: Logged In: YES user_id=28849 Originator: NO Maxima can't give an answer because it doesn't know the roots of (x^7+1)/(x+1). If you set integrate_use_rootsof:true, then maxima can return an answer, but unless you can compute the roots of the above equation, it's probably not very helpful.  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1913067&group_id=4933 
From: SourceForge.net <noreply@so...>  20080317 14:27:20

Bugs item #1913067, was opened at 20080312 21:06 Message generated for change (Comment added) made by infinity0x You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1913067&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: Ximin Luo (infinity0x) Assigned to: Nobody/Anonymous (nobody) Summary: Cannot integrate 1/(1+x^n) Initial Comment: Maxima cannot integrate functions of the form 1/(1+x^n) for n >= 7. Mathematica is capable of this. Maxima attempts the integration via an algorithm which seems to involve taking partial fractions. For example, if you try to integrate 1/(1+x^7) for example, Maxima gives: log(1+x)/7  integral(x^52*x^4+3*x^34*x^2+5*x6)/(x^6x^5+x^4x^3+x^2x+1) / 7 I'm guessing a different algorithm (such as that employed by Mathematica) is required to give a fully symbolic answer. For the record, the integral(1/(1+x^7)) is: Log[1 + x]/7  (Cos[Pi/7]*Log[1 + x^2  2*x*Cos[Pi/7]])/7  (Cos[(3*Pi)/7]*Log[1 + x^2  2*x*Cos[(3*Pi)/7]])/7  (Cos[(5*Pi)/7]*Log[1 + x^2  2*x*Cos[(5*Pi)/7]])/7 + (2*ArcTan[(x  Cos[Pi/7])*Csc[Pi/7]]* Sin[Pi/7])/7 + (2*ArcTan[(x  Cos[(3*Pi)/7])* Csc[(3*Pi)/7]]*Sin[(3*Pi)/7])/7 + (2*ArcTan[(x  Cos[(5*Pi)/7])* Csc[(5*Pi)/7]]*Sin[(5*Pi)/7])/7  Additionally, this also means that we can calculate the integral(that nasty rational function), by calculating 7 * ( log(1+x)/7  integral(1/(1+x^7)) ) As a side note, Mathematica also cannot give integral(that nasty rational function) fully symbolically. Instead, it gives: RootSum[1  #1 + #1^2  #1^3 + #1^4  #1^5 + #1^6 & , (6*Log[x  #1] + 5*Log[x  #1]*#1  4*Log[x  #1]*#1^2 + 3*Log[x  #1]* #1^3  2*Log[x  #1]*#1^4 + Log[x  #1]*#1^5)/(1 + 2*#1  3*#1^2 + 4*#1^3  5*#1^4 + 6*#1^5) & ] where RootSum[ f, form ] represents the sum of form[x] for all x that satisfy the polynomial equation f[x] == 0.  >Comment By: Ximin Luo (infinity0x) Date: 20080317 14:27 Message: Logged In: YES user_id=2003896 Originator: YES i don't know how to integrate these without knowing the roots of the equation; i thought mathematica was, because it could integrate 1/(1+x^n) without being able to integrate (that nasty rational function). (and i thought algorithms do to this would be available somewhere else.) also, i assumed maxima was already factorising the equation for arbitrarily large values of n. but your approach seems feasible. what do you mean by handcomputed though? as in, numerically instead of algebraically? wouldn't that give imprecise answers?  Comment By: Raymond Toy (rtoy) Date: 20080317 13:37 Message: Logged In: YES user_id=28849 Originator: NO If you can, please describe how to integrate these without knowing the roots of the equation. FWIW, maxima has a special function to integrate functions of the form f(x)/(a*x^n+b). (See enprog and eprog in sinint.lisp.) It only handles the cases of n = 4, 5, 6. And it does this by essentially doing a partial fraction expansion with handcomputed quadratic factors. It seems feasible to extend this to any n.  Comment By: Raymond Toy (rtoy) Date: 20080317 13:37 Message: Logged In: YES user_id=28849 Originator: NO If you can, please describe how to integrate these without knowing the roots of the equation. FWIW, maxima has a special function to integrate functions of the form f(x)/(a*x^n+b). (See enprog and eprog in sinint.lisp.) It only handles the cases of n = 4, 5, 6. And it does this by essentially doing a partial fraction expansion with handcomputed quadratic factors. It seems feasible to extend this to any n.  Comment By: Raymond Toy (rtoy) Date: 20080317 13:37 Message: Logged In: YES user_id=28849 Originator: NO If you can, please describe how to integrate these without knowing the roots of the equation. FWIW, maxima has a special function to integrate functions of the form f(x)/(a*x^n+b). (See enprog and eprog in sinint.lisp.) It only handles the cases of n = 4, 5, 6. And it does this by essentially doing a partial fraction expansion with handcomputed quadratic factors. It seems feasible to extend this to any n.  Comment By: Raymond Toy (rtoy) Date: 20080313 00:30 Message: Logged In: YES user_id=28849 Originator: NO Actually, the roots are obviously the seven roots of 1, which maxima does know, and it could have done the partial fraction expansion to find the value of the integral. Not sure how or where to teach maxima about this, though. Some thought needed.  Comment By: Ximin Luo (infinity0x) Date: 20080312 22:32 Message: Logged In: YES user_id=2003896 Originator: YES The point is that Maxima does not NEED to know the roots of that equation. Sorry for making this unclear. By doing the integral using a different algorithm which doesn't involve taking partial fractions, you can avoid the above, and get a purely symbolic integral, like Mathematic does.  Comment By: Raymond Toy (rtoy) Date: 20080312 22:16 Message: Logged In: YES user_id=28849 Originator: NO Maxima can't give an answer because it doesn't know the roots of (x^7+1)/(x+1). If you set integrate_use_rootsof:true, then maxima can return an answer, but unless you can compute the roots of the above equation, it's probably not very helpful.  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1913067&group_id=4933 
From: SourceForge.net <noreply@so...>  20080317 13:37:58

Bugs item #1913067, was opened at 20080312 17:06 Message generated for change (Comment added) made by rtoy You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1913067&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: Ximin Luo (infinity0x) Assigned to: Nobody/Anonymous (nobody) Summary: Cannot integrate 1/(1+x^n) Initial Comment: Maxima cannot integrate functions of the form 1/(1+x^n) for n >= 7. Mathematica is capable of this. Maxima attempts the integration via an algorithm which seems to involve taking partial fractions. For example, if you try to integrate 1/(1+x^7) for example, Maxima gives: log(1+x)/7  integral(x^52*x^4+3*x^34*x^2+5*x6)/(x^6x^5+x^4x^3+x^2x+1) / 7 I'm guessing a different algorithm (such as that employed by Mathematica) is required to give a fully symbolic answer. For the record, the integral(1/(1+x^7)) is: Log[1 + x]/7  (Cos[Pi/7]*Log[1 + x^2  2*x*Cos[Pi/7]])/7  (Cos[(3*Pi)/7]*Log[1 + x^2  2*x*Cos[(3*Pi)/7]])/7  (Cos[(5*Pi)/7]*Log[1 + x^2  2*x*Cos[(5*Pi)/7]])/7 + (2*ArcTan[(x  Cos[Pi/7])*Csc[Pi/7]]* Sin[Pi/7])/7 + (2*ArcTan[(x  Cos[(3*Pi)/7])* Csc[(3*Pi)/7]]*Sin[(3*Pi)/7])/7 + (2*ArcTan[(x  Cos[(5*Pi)/7])* Csc[(5*Pi)/7]]*Sin[(5*Pi)/7])/7  Additionally, this also means that we can calculate the integral(that nasty rational function), by calculating 7 * ( log(1+x)/7  integral(1/(1+x^7)) ) As a side note, Mathematica also cannot give integral(that nasty rational function) fully symbolically. Instead, it gives: RootSum[1  #1 + #1^2  #1^3 + #1^4  #1^5 + #1^6 & , (6*Log[x  #1] + 5*Log[x  #1]*#1  4*Log[x  #1]*#1^2 + 3*Log[x  #1]* #1^3  2*Log[x  #1]*#1^4 + Log[x  #1]*#1^5)/(1 + 2*#1  3*#1^2 + 4*#1^3  5*#1^4 + 6*#1^5) & ] where RootSum[ f, form ] represents the sum of form[x] for all x that satisfy the polynomial equation f[x] == 0.  >Comment By: Raymond Toy (rtoy) Date: 20080317 09:37 Message: Logged In: YES user_id=28849 Originator: NO If you can, please describe how to integrate these without knowing the roots of the equation. FWIW, maxima has a special function to integrate functions of the form f(x)/(a*x^n+b). (See enprog and eprog in sinint.lisp.) It only handles the cases of n = 4, 5, 6. And it does this by essentially doing a partial fraction expansion with handcomputed quadratic factors. It seems feasible to extend this to any n.  Comment By: Raymond Toy (rtoy) Date: 20080317 09:37 Message: Logged In: YES user_id=28849 Originator: NO If you can, please describe how to integrate these without knowing the roots of the equation. FWIW, maxima has a special function to integrate functions of the form f(x)/(a*x^n+b). (See enprog and eprog in sinint.lisp.) It only handles the cases of n = 4, 5, 6. And it does this by essentially doing a partial fraction expansion with handcomputed quadratic factors. It seems feasible to extend this to any n.  Comment By: Raymond Toy (rtoy) Date: 20080317 09:37 Message: Logged In: YES user_id=28849 Originator: NO If you can, please describe how to integrate these without knowing the roots of the equation. FWIW, maxima has a special function to integrate functions of the form f(x)/(a*x^n+b). (See enprog and eprog in sinint.lisp.) It only handles the cases of n = 4, 5, 6. And it does this by essentially doing a partial fraction expansion with handcomputed quadratic factors. It seems feasible to extend this to any n.  Comment By: Raymond Toy (rtoy) Date: 20080312 20:30 Message: Logged In: YES user_id=28849 Originator: NO Actually, the roots are obviously the seven roots of 1, which maxima does know, and it could have done the partial fraction expansion to find the value of the integral. Not sure how or where to teach maxima about this, though. Some thought needed.  Comment By: Ximin Luo (infinity0x) Date: 20080312 18:32 Message: Logged In: YES user_id=2003896 Originator: YES The point is that Maxima does not NEED to know the roots of that equation. Sorry for making this unclear. By doing the integral using a different algorithm which doesn't involve taking partial fractions, you can avoid the above, and get a purely symbolic integral, like Mathematic does.  Comment By: Raymond Toy (rtoy) Date: 20080312 18:16 Message: Logged In: YES user_id=28849 Originator: NO Maxima can't give an answer because it doesn't know the roots of (x^7+1)/(x+1). If you set integrate_use_rootsof:true, then maxima can return an answer, but unless you can compute the roots of the above equation, it's probably not very helpful.  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1913067&group_id=4933 
From: SourceForge.net <noreply@so...>  20080317 13:37:54

Bugs item #1913067, was opened at 20080312 17:06 Message generated for change (Comment added) made by rtoy You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1913067&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: Ximin Luo (infinity0x) Assigned to: Nobody/Anonymous (nobody) Summary: Cannot integrate 1/(1+x^n) Initial Comment: Maxima cannot integrate functions of the form 1/(1+x^n) for n >= 7. Mathematica is capable of this. Maxima attempts the integration via an algorithm which seems to involve taking partial fractions. For example, if you try to integrate 1/(1+x^7) for example, Maxima gives: log(1+x)/7  integral(x^52*x^4+3*x^34*x^2+5*x6)/(x^6x^5+x^4x^3+x^2x+1) / 7 I'm guessing a different algorithm (such as that employed by Mathematica) is required to give a fully symbolic answer. For the record, the integral(1/(1+x^7)) is: Log[1 + x]/7  (Cos[Pi/7]*Log[1 + x^2  2*x*Cos[Pi/7]])/7  (Cos[(3*Pi)/7]*Log[1 + x^2  2*x*Cos[(3*Pi)/7]])/7  (Cos[(5*Pi)/7]*Log[1 + x^2  2*x*Cos[(5*Pi)/7]])/7 + (2*ArcTan[(x  Cos[Pi/7])*Csc[Pi/7]]* Sin[Pi/7])/7 + (2*ArcTan[(x  Cos[(3*Pi)/7])* Csc[(3*Pi)/7]]*Sin[(3*Pi)/7])/7 + (2*ArcTan[(x  Cos[(5*Pi)/7])* Csc[(5*Pi)/7]]*Sin[(5*Pi)/7])/7  Additionally, this also means that we can calculate the integral(that nasty rational function), by calculating 7 * ( log(1+x)/7  integral(1/(1+x^7)) ) As a side note, Mathematica also cannot give integral(that nasty rational function) fully symbolically. Instead, it gives: RootSum[1  #1 + #1^2  #1^3 + #1^4  #1^5 + #1^6 & , (6*Log[x  #1] + 5*Log[x  #1]*#1  4*Log[x  #1]*#1^2 + 3*Log[x  #1]* #1^3  2*Log[x  #1]*#1^4 + Log[x  #1]*#1^5)/(1 + 2*#1  3*#1^2 + 4*#1^3  5*#1^4 + 6*#1^5) & ] where RootSum[ f, form ] represents the sum of form[x] for all x that satisfy the polynomial equation f[x] == 0.  >Comment By: Raymond Toy (rtoy) Date: 20080317 09:37 Message: Logged In: YES user_id=28849 Originator: NO If you can, please describe how to integrate these without knowing the roots of the equation. FWIW, maxima has a special function to integrate functions of the form f(x)/(a*x^n+b). (See enprog and eprog in sinint.lisp.) It only handles the cases of n = 4, 5, 6. And it does this by essentially doing a partial fraction expansion with handcomputed quadratic factors. It seems feasible to extend this to any n.  Comment By: Raymond Toy (rtoy) Date: 20080317 09:37 Message: Logged In: YES user_id=28849 Originator: NO If you can, please describe how to integrate these without knowing the roots of the equation. FWIW, maxima has a special function to integrate functions of the form f(x)/(a*x^n+b). (See enprog and eprog in sinint.lisp.) It only handles the cases of n = 4, 5, 6. And it does this by essentially doing a partial fraction expansion with handcomputed quadratic factors. It seems feasible to extend this to any n.  Comment By: Raymond Toy (rtoy) Date: 20080312 20:30 Message: Logged In: YES user_id=28849 Originator: NO Actually, the roots are obviously the seven roots of 1, which maxima does know, and it could have done the partial fraction expansion to find the value of the integral. Not sure how or where to teach maxima about this, though. Some thought needed.  Comment By: Ximin Luo (infinity0x) Date: 20080312 18:32 Message: Logged In: YES user_id=2003896 Originator: YES The point is that Maxima does not NEED to know the roots of that equation. Sorry for making this unclear. By doing the integral using a different algorithm which doesn't involve taking partial fractions, you can avoid the above, and get a purely symbolic integral, like Mathematic does.  Comment By: Raymond Toy (rtoy) Date: 20080312 18:16 Message: Logged In: YES user_id=28849 Originator: NO Maxima can't give an answer because it doesn't know the roots of (x^7+1)/(x+1). If you set integrate_use_rootsof:true, then maxima can return an answer, but unless you can compute the roots of the above equation, it's probably not very helpful.  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1913067&group_id=4933 
From: SourceForge.net <noreply@so...>  20080317 13:37:41

Bugs item #1913067, was opened at 20080312 17:06 Message generated for change (Comment added) made by rtoy You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1913067&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: Ximin Luo (infinity0x) Assigned to: Nobody/Anonymous (nobody) Summary: Cannot integrate 1/(1+x^n) Initial Comment: Maxima cannot integrate functions of the form 1/(1+x^n) for n >= 7. Mathematica is capable of this. Maxima attempts the integration via an algorithm which seems to involve taking partial fractions. For example, if you try to integrate 1/(1+x^7) for example, Maxima gives: log(1+x)/7  integral(x^52*x^4+3*x^34*x^2+5*x6)/(x^6x^5+x^4x^3+x^2x+1) / 7 I'm guessing a different algorithm (such as that employed by Mathematica) is required to give a fully symbolic answer. For the record, the integral(1/(1+x^7)) is: Log[1 + x]/7  (Cos[Pi/7]*Log[1 + x^2  2*x*Cos[Pi/7]])/7  (Cos[(3*Pi)/7]*Log[1 + x^2  2*x*Cos[(3*Pi)/7]])/7  (Cos[(5*Pi)/7]*Log[1 + x^2  2*x*Cos[(5*Pi)/7]])/7 + (2*ArcTan[(x  Cos[Pi/7])*Csc[Pi/7]]* Sin[Pi/7])/7 + (2*ArcTan[(x  Cos[(3*Pi)/7])* Csc[(3*Pi)/7]]*Sin[(3*Pi)/7])/7 + (2*ArcTan[(x  Cos[(5*Pi)/7])* Csc[(5*Pi)/7]]*Sin[(5*Pi)/7])/7  Additionally, this also means that we can calculate the integral(that nasty rational function), by calculating 7 * ( log(1+x)/7  integral(1/(1+x^7)) ) As a side note, Mathematica also cannot give integral(that nasty rational function) fully symbolically. Instead, it gives: RootSum[1  #1 + #1^2  #1^3 + #1^4  #1^5 + #1^6 & , (6*Log[x  #1] + 5*Log[x  #1]*#1  4*Log[x  #1]*#1^2 + 3*Log[x  #1]* #1^3  2*Log[x  #1]*#1^4 + Log[x  #1]*#1^5)/(1 + 2*#1  3*#1^2 + 4*#1^3  5*#1^4 + 6*#1^5) & ] where RootSum[ f, form ] represents the sum of form[x] for all x that satisfy the polynomial equation f[x] == 0.  >Comment By: Raymond Toy (rtoy) Date: 20080317 09:37 Message: Logged In: YES user_id=28849 Originator: NO If you can, please describe how to integrate these without knowing the roots of the equation. FWIW, maxima has a special function to integrate functions of the form f(x)/(a*x^n+b). (See enprog and eprog in sinint.lisp.) It only handles the cases of n = 4, 5, 6. And it does this by essentially doing a partial fraction expansion with handcomputed quadratic factors. It seems feasible to extend this to any n.  Comment By: Raymond Toy (rtoy) Date: 20080312 20:30 Message: Logged In: YES user_id=28849 Originator: NO Actually, the roots are obviously the seven roots of 1, which maxima does know, and it could have done the partial fraction expansion to find the value of the integral. Not sure how or where to teach maxima about this, though. Some thought needed.  Comment By: Ximin Luo (infinity0x) Date: 20080312 18:32 Message: Logged In: YES user_id=2003896 Originator: YES The point is that Maxima does not NEED to know the roots of that equation. Sorry for making this unclear. By doing the integral using a different algorithm which doesn't involve taking partial fractions, you can avoid the above, and get a purely symbolic integral, like Mathematic does.  Comment By: Raymond Toy (rtoy) Date: 20080312 18:16 Message: Logged In: YES user_id=28849 Originator: NO Maxima can't give an answer because it doesn't know the roots of (x^7+1)/(x+1). If you set integrate_use_rootsof:true, then maxima can return an answer, but unless you can compute the roots of the above equation, it's probably not very helpful.  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1913067&group_id=4933 
From: SourceForge.net <noreply@so...>  20080317 12:05:37

Bugs item #1916517, was opened at 20080317 04:40 Message generated for change (Comment added) made by willisbl You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1916517&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: Satoshi Adachi (satoshi_adachi) Assigned to: Nobody/Anonymous (nobody) Summary: (F=0 and G=0) and (F=0 and FG=0) have different solutions. Initial Comment: Dear Developers of Maxima, I met a set of two albebraic equations F = 0 and G = 0 of two variables x and y such that (i) solve([F=0,G=0],[x,y]) returns no solution [] (this is wrong), whereas (ii) solve([F=0,FG=0],[x,y]) returns true solutions. This is strange, since the first set of equations F=0 and G=0 should have the same solutions of the second set of equations F=0 and FG=0. (1) The Maxima program to solve F=0 and FG=0 is  /* * solve_system_of_algebraic_equations_OK.maxima: * * F is a polynomial of x and y, which is degree 3 in x and 2 in y. * G is a polynomial of x and y, which is degree 3 in x and 2 in y. * * solve() and algsys() can solve the system equations F=0 and FG=0. */ display2d:false; F:2*x^3*y^2+2*a*x^2*y^2+3*x^2*y+2*a*x*y+x+1; G:x^3*y^2+2*a*x^2*y^2+a^2*x*y^2+x^2*y+2*a*x*y+a^2*y+1; solutions:solve([F=0,FG=0],[x,y]); /* OK: solutions are found. */  The result is  Maxima 5.14.0cvs http://maxima.sourceforge.net Using Lisp GNU Common Lisp (GCL) GCL 2.6.7 (aka GCL) Distributed under the GNU Public License. See the file COPYING. Dedicated to the memory of William Schelter. The function bug_report() provides bug reporting information. (%i1) batch(solve_system_of_algebraic_equations_OK.maxima) batching #p/Volumes/HFS+2/home/adachi/work/285/solve_system_of_algebraic_equations_OK.maxima (%i2) display2d : false (%o2) false (%i3) F:1+x+2*a*x*y+3*x^2*y+2*a*x^2*y^2+2*x^3*y^2 (%o3) 2*x^3*y^2+2*a*x^2*y^2+3*x^2*y+2*a*x*y+x+1 (%i4) G:1+a^2*y+2*a*x*y+x^2*y+a^2*x*y^2+2*a*x^2*y^2+x^3*y^2 (%o4) x^3*y^2+2*a*x^2*y^2+a^2*x*y^2+x^2*y+2*a*x*y+a^2*y+1 (%i5) solutions:solve([F = 0,FG = 0],[x,y]) (%o5) [[x = (sqrt(a)*(a^24*a)+sqrt(a4)*(2*aa^2))/(2*sqrt(a4)), y = 1/(a*sqrt(a^24*a))], [x = (sqrt(a4)*(a^22*a)+sqrt(a)*(a^24*a))/(2*sqrt(a4)), y = 1/(a*sqrt(a^24*a))]] (%o6) "solve_system_of_algebraic_equations_OK.maxima"  =============================================================================== (2) The Maxima program to solve F=0 and G=0 is  /* * solve_system_of_algebraic_equations_NG.maxima: * * F is a polynomial of x and y, which is degree 3 in x and 2 in y. * G is a polynomial of x and y, which is degree 3 in x and 2 in y. * * solve() and algsys() can NOT solve the system equations F=0 and G=0. */ display2d:false; F:2*x^3*y^2+2*a*x^2*y^2+3*x^2*y+2*a*x*y+x+1; G:x^3*y^2+2*a*x^2*y^2+a^2*x*y^2+x^2*y+2*a*x*y+a^2*y+1; solutions:solve([F=0,G=0],[x,y]); /* NG: solutions are NOT found! */  The result is  Maxima 5.14.0cvs http://maxima.sourceforge.net Using Lisp GNU Common Lisp (GCL) GCL 2.6.7 (aka GCL) Distributed under the GNU Public License. See the file COPYING. Dedicated to the memory of William Schelter. The function bug_report() provides bug reporting information. (%i1) batch(solve_system_of_algebraic_equations_NG.maxima) batching #p/Volumes/HFS+2/home/adachi/work/285/solve_system_of_algebraic_equations_NG.maxima (%i2) display2d : false (%o2) false (%i3) F:1+x+2*a*x*y+3*x^2*y+2*a*x^2*y^2+2*x^3*y^2 (%o3) 2*x^3*y^2+2*a*x^2*y^2+3*x^2*y+2*a*x*y+x+1 (%i4) G:1+a^2*y+2*a*x*y+x^2*y+a^2*x*y^2+2*a*x^2*y^2+x^3*y^2 (%o4) x^3*y^2+2*a*x^2*y^2+a^2*x*y^2+x^2*y+2*a*x*y+a^2*y+1 (%i5) solutions:solve([F = 0,G = 0],[x,y]) (%o5) [] (%o6) "solve_system_of_algebraic_equations_NG.maxima"  =============================================================================== (3) I also tried algebraic:true to solve F=0 and G=0. The program is  /* * solve_system_of_algebraic_equations_with_algebraic_NG.maxima: * * F is a polynomial of x and y, which is degree 3 in x and 2 in y. * G is a polynomial of x and y, which is degree 3 in x and 2 in y. * * solve() and algsys() can NOT solve the system equations F=0 and G=0. */ display2d:false; algebraic:true; F:2*x^3*y^2+2*a*x^2*y^2+3*x^2*y+2*a*x*y+x+1; G:x^3*y^2+2*a*x^2*y^2+a^2*x*y^2+x^2*y+2*a*x*y+a^2*y+1; solutions:solve([F=0,G=0],[x,y]); /* NG: internal error! */  The result is further strange for me as  Maxima 5.14.0cvs http://maxima.sourceforge.net Using Lisp GNU Common Lisp (GCL) GCL 2.6.7 (aka GCL) Distributed under the GNU Public License. See the file COPYING. Dedicated to the memory of William Schelter. The function bug_report() provides bug reporting information. (%i1) batch(solve_system_of_algebraic_equations_with_algebraic.maxima) batching #p/Volumes/HFS+2/home/adachi/work/285/solve_system_of_algebraic_equations_with_algebraic.maxima (%i2) display2d : false (%o2) false (%i3) algebraic:true (%o3) true (%i4) F:1+x+2*a*x*y+3*x^2*y+2*a*x^2*y^2+2*x^3*y^2 (%o4) 2*x^3*y^2+2*a*x^2*y^2+3*x^2*y+2*a*x*y+x+1 (%i5) G:1+a^2*y+2*a*x*y+x^2*y+a^2*x*y^2+2*a*x^2*y^2+x^3*y^2 (%o5) x^3*y^2+2*a*x^2*y^2+a^2*x*y^2+x^2*y+2*a*x*y+a^2*y+1 (%i6) solutions:solve([F = 0,G = 0],[x,y]) Polynomial quotient is not exact  an error. To debug this try debugmode(true);  Sincerely yours, Satoshi Adachi  >Comment By: Barton Willis (willisbl) Date: 20080317 07:05 Message: Logged In: YES user_id=895922 Originator: NO Thanks for reporting this bug. The function algsys isn't nearly as good as we would like. The best I can offer is my favorite workaround (that might work); it's something like: (%i92) load(topoly)$ (%i93) algebraic : true$ Use 'elim' to triangularize the equationsthis can create spurious solutions: (%i94) elim([F,G],[x,y]); (%o94) [[],[x^4+a^2*x^3a*x^3+a^3*x^2a^2*x^2+a^3*x,x^3*y^2+2*a*x^2*y^2+a^2*x*y^2+x^2*y+2*a*x*y+a^2*y+1]] Use algsys to solve these equations: (%i95) sol : algsys(first(rest(%)),[x,y])$ Try to reject the spurious solutions: (%i96) true_sol : set()$ (%i97) for si in sol do if every(lambda([a], is(equal(a,0))), ratsimp(subst(si,[F,G]))) then true_sol : adjoin(si, true_sol); (%o97) done (%i98) true_sol; (%o98) {[x=(a*sqrt(a^24*a)a^2+2*a)/2,y=sqrt(a^24*a)/(a^34*a^2)], [x=(a*sqrt(a^24*a)+a^22*a)/2,y=sqrt(a^24*a)/(a^34*a^2)]} Two solutions *might* be spurious and two checked out OK. You'll need to inspect the two rejected solutions to see if they are really spurious. And a = 0 and a = 4 might be special cases that need to be checked. Of course, all this should be automatic; unfortunately, algsys isn't there yet. Barton  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1916517&group_id=4933 
From: SourceForge.net <noreply@so...>  20080317 09:42:30

Bugs item #1916517, was opened at 20080317 18:40 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=1916517&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: Satoshi Adachi (satoshi_adachi) Assigned to: Nobody/Anonymous (nobody) Summary: (F=0 and G=0) and (F=0 and FG=0) have different solutions. Initial Comment: Dear Developers of Maxima, I met a set of two albebraic equations F = 0 and G = 0 of two variables x and y such that (i) solve([F=0,G=0],[x,y]) returns no solution [] (this is wrong), whereas (ii) solve([F=0,FG=0],[x,y]) returns true solutions. This is strange, since the first set of equations F=0 and G=0 should have the same solutions of the second set of equations F=0 and FG=0. (1) The Maxima program to solve F=0 and FG=0 is  /* * solve_system_of_algebraic_equations_OK.maxima: * * F is a polynomial of x and y, which is degree 3 in x and 2 in y. * G is a polynomial of x and y, which is degree 3 in x and 2 in y. * * solve() and algsys() can solve the system equations F=0 and FG=0. */ display2d:false; F:2*x^3*y^2+2*a*x^2*y^2+3*x^2*y+2*a*x*y+x+1; G:x^3*y^2+2*a*x^2*y^2+a^2*x*y^2+x^2*y+2*a*x*y+a^2*y+1; solutions:solve([F=0,FG=0],[x,y]); /* OK: solutions are found. */  The result is  Maxima 5.14.0cvs http://maxima.sourceforge.net Using Lisp GNU Common Lisp (GCL) GCL 2.6.7 (aka GCL) Distributed under the GNU Public License. See the file COPYING. Dedicated to the memory of William Schelter. The function bug_report() provides bug reporting information. (%i1) batch(solve_system_of_algebraic_equations_OK.maxima) batching #p/Volumes/HFS+2/home/adachi/work/285/solve_system_of_algebraic_equations_OK.maxima (%i2) display2d : false (%o2) false (%i3) F:1+x+2*a*x*y+3*x^2*y+2*a*x^2*y^2+2*x^3*y^2 (%o3) 2*x^3*y^2+2*a*x^2*y^2+3*x^2*y+2*a*x*y+x+1 (%i4) G:1+a^2*y+2*a*x*y+x^2*y+a^2*x*y^2+2*a*x^2*y^2+x^3*y^2 (%o4) x^3*y^2+2*a*x^2*y^2+a^2*x*y^2+x^2*y+2*a*x*y+a^2*y+1 (%i5) solutions:solve([F = 0,FG = 0],[x,y]) (%o5) [[x = (sqrt(a)*(a^24*a)+sqrt(a4)*(2*aa^2))/(2*sqrt(a4)), y = 1/(a*sqrt(a^24*a))], [x = (sqrt(a4)*(a^22*a)+sqrt(a)*(a^24*a))/(2*sqrt(a4)), y = 1/(a*sqrt(a^24*a))]] (%o6) "solve_system_of_algebraic_equations_OK.maxima"  =============================================================================== (2) The Maxima program to solve F=0 and G=0 is  /* * solve_system_of_algebraic_equations_NG.maxima: * * F is a polynomial of x and y, which is degree 3 in x and 2 in y. * G is a polynomial of x and y, which is degree 3 in x and 2 in y. * * solve() and algsys() can NOT solve the system equations F=0 and G=0. */ display2d:false; F:2*x^3*y^2+2*a*x^2*y^2+3*x^2*y+2*a*x*y+x+1; G:x^3*y^2+2*a*x^2*y^2+a^2*x*y^2+x^2*y+2*a*x*y+a^2*y+1; solutions:solve([F=0,G=0],[x,y]); /* NG: solutions are NOT found! */  The result is  Maxima 5.14.0cvs http://maxima.sourceforge.net Using Lisp GNU Common Lisp (GCL) GCL 2.6.7 (aka GCL) Distributed under the GNU Public License. See the file COPYING. Dedicated to the memory of William Schelter. The function bug_report() provides bug reporting information. (%i1) batch(solve_system_of_algebraic_equations_NG.maxima) batching #p/Volumes/HFS+2/home/adachi/work/285/solve_system_of_algebraic_equations_NG.maxima (%i2) display2d : false (%o2) false (%i3) F:1+x+2*a*x*y+3*x^2*y+2*a*x^2*y^2+2*x^3*y^2 (%o3) 2*x^3*y^2+2*a*x^2*y^2+3*x^2*y+2*a*x*y+x+1 (%i4) G:1+a^2*y+2*a*x*y+x^2*y+a^2*x*y^2+2*a*x^2*y^2+x^3*y^2 (%o4) x^3*y^2+2*a*x^2*y^2+a^2*x*y^2+x^2*y+2*a*x*y+a^2*y+1 (%i5) solutions:solve([F = 0,G = 0],[x,y]) (%o5) [] (%o6) "solve_system_of_algebraic_equations_NG.maxima"  =============================================================================== (3) I also tried algebraic:true to solve F=0 and G=0. The program is  /* * solve_system_of_algebraic_equations_with_algebraic_NG.maxima: * * F is a polynomial of x and y, which is degree 3 in x and 2 in y. * G is a polynomial of x and y, which is degree 3 in x and 2 in y. * * solve() and algsys() can NOT solve the system equations F=0 and G=0. */ display2d:false; algebraic:true; F:2*x^3*y^2+2*a*x^2*y^2+3*x^2*y+2*a*x*y+x+1; G:x^3*y^2+2*a*x^2*y^2+a^2*x*y^2+x^2*y+2*a*x*y+a^2*y+1; solutions:solve([F=0,G=0],[x,y]); /* NG: internal error! */  The result is further strange for me as  Maxima 5.14.0cvs http://maxima.sourceforge.net Using Lisp GNU Common Lisp (GCL) GCL 2.6.7 (aka GCL) Distributed under the GNU Public License. See the file COPYING. Dedicated to the memory of William Schelter. The function bug_report() provides bug reporting information. (%i1) batch(solve_system_of_algebraic_equations_with_algebraic.maxima) batching #p/Volumes/HFS+2/home/adachi/work/285/solve_system_of_algebraic_equations_with_algebraic.maxima (%i2) display2d : false (%o2) false (%i3) algebraic:true (%o3) true (%i4) F:1+x+2*a*x*y+3*x^2*y+2*a*x^2*y^2+2*x^3*y^2 (%o4) 2*x^3*y^2+2*a*x^2*y^2+3*x^2*y+2*a*x*y+x+1 (%i5) G:1+a^2*y+2*a*x*y+x^2*y+a^2*x*y^2+2*a*x^2*y^2+x^3*y^2 (%o5) x^3*y^2+2*a*x^2*y^2+a^2*x*y^2+x^2*y+2*a*x*y+a^2*y+1 (%i6) solutions:solve([F = 0,G = 0],[x,y]) Polynomial quotient is not exact  an error. To debug this try debugmode(true);  Sincerely yours, Satoshi Adachi  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1916517&group_id=4933 
From: SourceForge.net <noreply@so...>  20080314 10:09:23

Bugs item #1913588, was opened at 20080313 10:06 Message generated for change (Comment added) made by willisbl You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1913588&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  Polynomials Group: None Status: Open Resolution: None Priority: 5 Private: No Submitted By: Nobody/Anonymous (nobody) Assigned to: Nobody/Anonymous (nobody) Summary: algsys hangs Initial Comment: algsys hangs on the following input: eq1 : a*x + c*y + d*y^2/2 = 0; eq2 : b*x + e*x^2 + f*y  g*y^3 = h; algsys([eq1,eq2],[x,y]); System information: Maxima version: 5.12.0 Maxima build date: 15:52 7/20/2007 host type: i686pclinuxgnu lispimplementationtype: GNU Common Lisp (GCL) lispimplementationversion: GCL 2.6.7  >Comment By: Barton Willis (willisbl) Date: 20080314 05:09 Message: Logged In: YES user_id=895922 Originator: NO A workaround is to eliminate x and solve the quartic for y; something like (%i14) eliminate([eq1,eq2],[x]); (%o14) [d^2*e*y^4+(4*c*d*e4*a^2*g)*y^3+(4*c^2*e2*a*b*d)*y^2+(4*a^2*f4*a*b*c)*y4*a^2*h] (%i15) solve(%,y) Of course, Maxima should not need any help.  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1913588&group_id=4933 
From: SourceForge.net <noreply@so...>  20080313 15:06:58

Bugs item #1913588, was opened at 20080313 08:06 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=1913588&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  Polynomials Group: None Status: Open Resolution: None Priority: 5 Private: No Submitted By: Nobody/Anonymous (nobody) Assigned to: Nobody/Anonymous (nobody) Summary: algsys hangs Initial Comment: algsys hangs on the following input: eq1 : a*x + c*y + d*y^2/2 = 0; eq2 : b*x + e*x^2 + f*y  g*y^3 = h; algsys([eq1,eq2],[x,y]); System information: Maxima version: 5.12.0 Maxima build date: 15:52 7/20/2007 host type: i686pclinuxgnu lispimplementationtype: GNU Common Lisp (GCL) lispimplementationversion: GCL 2.6.7  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1913588&group_id=4933 
From: SourceForge.net <noreply@so...>  20080313 12:05:52

Bugs item #1569644, was opened at 20061002 20:15 Message generated for change (Settings changed) made by dgildea You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1569644&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: Closed >Resolution: Works For Me Priority: 5 Private: No Submitted By: Stavros Macrakis (macrakis) Assigned to: Nobody/Anonymous (nobody) Summary: expand(ceiling(...)) doesn't look inside Initial Comment: expand(ceiling(x*(x1))) doesn't expand the product.  Comment By: Harald Geyer (hgeyer) Date: 20070618 14:26 Message: Logged In: YES user_id=929336 Originator: NO I can't reproduce this bug with maxima 5.12.0 CLISP. I guess this problem has been fixed in the meanwhile. I think this bug report should be closed. Regards, Harald  Comment By: Barton Willis (willisbl) Date: 20061002 20:30 Message: Logged In: YES user_id=895922 Deleting the simp flag check in simpceiling fixes this bug. I assume that all the other functions in nummod.lisp have similar bugs. I'll fix these bugs.  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1569644&group_id=4933 
From: SourceForge.net <noreply@so...>  20080313 09:33:12

Bugs item #1909488, was opened at 20080307 06:08 Message generated for change (Settings changed) made by willisbl You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1909488&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: Closed >Resolution: Invalid Priority: 5 Private: No Submitted By: Nobody/Anonymous (nobody) Assigned to: Nobody/Anonymous (nobody) Summary: mnewton bug Initial Comment: The 3rd example given does not work. Reference: http://maxima.sourceforge.net/docs/manual/en/ maxima_63.html#Item_003aIntroductiontomnewton (%i1) load("mnewton")$ (%i2) mnewton([x1+3*log(x1)x2^2, 2*x1^2x1*x25*x1+1], [x1, x2], [5, 5]); (%o2) [[x1=3.822890025575447,x2=2.807544757033248]] (%i3) mnewton([2*a^a5],[a],[1]); (%o3) [[a=1.70927556786144]] (%i4) mnewton([2*3^uv/u5, u+2^v4], [u, v], [2, 2]); Polynomial quotient is not exact#0: mnewton(funclist=[v/u+2*3^u5,2^v+u4],varlist=[u,v],guesslist=[2,2])(mnewton.mac line 89)  an error. To debug this try debugmode(true);(%i5)  Comment By: Nobody/Anonymous (nobody) Date: 20080308 23:13 Message: Logged In: NO That must be it, this is my build info (%i1) build_info()$ Maxima version: 5.12.0 Maxima build date: 15:52 7/20/2007 host type: i686pclinuxgnu lispimplementationtype: GNU Common Lisp (GCL)lispimplementationversion: GCL 2.6.7  Comment By: Barton Willis (willisbl) Date: 20080307 17:21 Message: Logged In: YES user_id=895922 Originator: NO It works for me; maybe you have an old version. (%i2) mnewton([2*3^uv/u5, u+2^v4], [u, v], [2, 2]); (%o2) [[u=1.066618389595407,v=1.552564766841786]] (%i8) build_info(); Maxima version: 5.14.0 Maxima build date: 21:46 12/27/2007 host type: i686pcmingw32 lispimplementationtype: GNU Common Lisp (GCL) lispimplementationversion: GCL 2.6.8  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1909488&group_id=4933 
From: SourceForge.net <noreply@so...>  20080313 09:32:31

Bugs item #1910043, was opened at 20080308 06:12 Message generated for change (Settings changed) made by willisbl You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1910043&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  Complex Group: None >Status: Closed Resolution: Invalid Priority: 5 Private: No Submitted By: Nobody/Anonymous (nobody) Assigned to: Nobody/Anonymous (nobody) Summary: Wrong calculation with nested matrices Initial Comment: If I am not mistaking, these two calculations should yield the same result, but do not in Maxima: input: matrix([0,0,0,1],[0,0,1,0],[0,1,0,0],[1,0,0,0]).matrix([0,0,0,%i],[0,0,%i,0],[0,%i,0,0],[ %i,0,0,0]); output: matrix([%i,0,0,0],[0,%i,0,0],[0,0,%i,0],[0,0,0,%i]) input: matrix([0,matrix([0,1],[1,0])],[matrix([0,1],[1,0]),0]).matrix([0,matrix([0,%i],[%i,0])],[matrix([0,%i],[%i,0]),0]); output: matrix([matrix([0,%i],[%i,0]),matrix([0,0],[0,0])],[matrix([0,0],[0,0]),matrix([0,%i],[%i,0])]) thus, there is something wrong, or am I wrong with that? Maxima version: 5.12.0Maxima build date: 15:52 7/20/2007host type: i686pclinuxgnulispimplementationtype: GNU Common Lisp (GCL)lispimplementationversion: GCL 2.6.7  Comment By: Barton Willis (willisbl) Date: 20080308 10:11 Message: Logged In: YES user_id=895922 Originator: NO I don't think we want to default matrix_element_mult to "."; suppose a, b, c, and d are identifiers for real numbers. Then (%i23) m1 : matrix([a,b],[c,d])$ (%i25) m2 : matrix([d,b],[c,a])$ (%i26) matrix_element_mult : "."$ (%i27) m1 . m2; (%o27) matrix([a.db.c,b.aa.b],[c.dd.c,d.ac.b]) With noncommutative multiplication, the offdiagonal terms are nonzero, but with commutative multiplication, the offdiagonal terms are zero. That is correct for a,b,c,d real numbers. (%i28) matrix_element_mult : "*"$ (%i29) m1 . m2; (%o29) matrix([a*db*c,0],[0,a*db*c])  Comment By: fabus (fgebert) Date: 20080308 09:22 Message: Logged In: YES user_id=1126735 Originator: NO I did not know matrix_element_mult. Is there anything against defaulting matrix_element_mult to "."?  Comment By: Barton Willis (willisbl) Date: 20080308 08:25 Message: Logged In: YES user_id=895922 Originator: NO To work with nested (or block) matrices, you'll need to set matrix_element_mult to ".". Try this (%i17) matrix_element_mult : "."$ (%i18) matrix([0,matrix([0,1],[1,0])],[matrix([0,1],[1,0]),0]).matrix([0,matrix([0,%i],[%i,0])],[matrix([0,%i],[ %i,0]),0])$ (%i20) mat_unblocker(%); (%o20) matrix([%i,0,0,0],[0,%i,0,0],[0,0,%i,0],[0,0,0,%i])  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1910043&group_id=4933 
From: SourceForge.net <noreply@so...>  20080313 00:30:31

Bugs item #1913067, was opened at 20080312 17:06 Message generated for change (Comment added) made by rtoy You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1913067&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: Ximin Luo (infinity0x) Assigned to: Nobody/Anonymous (nobody) Summary: Cannot integrate 1/(1+x^n) Initial Comment: Maxima cannot integrate functions of the form 1/(1+x^n) for n >= 7. Mathematica is capable of this. Maxima attempts the integration via an algorithm which seems to involve taking partial fractions. For example, if you try to integrate 1/(1+x^7) for example, Maxima gives: log(1+x)/7  integral(x^52*x^4+3*x^34*x^2+5*x6)/(x^6x^5+x^4x^3+x^2x+1) / 7 I'm guessing a different algorithm (such as that employed by Mathematica) is required to give a fully symbolic answer. For the record, the integral(1/(1+x^7)) is: Log[1 + x]/7  (Cos[Pi/7]*Log[1 + x^2  2*x*Cos[Pi/7]])/7  (Cos[(3*Pi)/7]*Log[1 + x^2  2*x*Cos[(3*Pi)/7]])/7  (Cos[(5*Pi)/7]*Log[1 + x^2  2*x*Cos[(5*Pi)/7]])/7 + (2*ArcTan[(x  Cos[Pi/7])*Csc[Pi/7]]* Sin[Pi/7])/7 + (2*ArcTan[(x  Cos[(3*Pi)/7])* Csc[(3*Pi)/7]]*Sin[(3*Pi)/7])/7 + (2*ArcTan[(x  Cos[(5*Pi)/7])* Csc[(5*Pi)/7]]*Sin[(5*Pi)/7])/7  Additionally, this also means that we can calculate the integral(that nasty rational function), by calculating 7 * ( log(1+x)/7  integral(1/(1+x^7)) ) As a side note, Mathematica also cannot give integral(that nasty rational function) fully symbolically. Instead, it gives: RootSum[1  #1 + #1^2  #1^3 + #1^4  #1^5 + #1^6 & , (6*Log[x  #1] + 5*Log[x  #1]*#1  4*Log[x  #1]*#1^2 + 3*Log[x  #1]* #1^3  2*Log[x  #1]*#1^4 + Log[x  #1]*#1^5)/(1 + 2*#1  3*#1^2 + 4*#1^3  5*#1^4 + 6*#1^5) & ] where RootSum[ f, form ] represents the sum of form[x] for all x that satisfy the polynomial equation f[x] == 0.  >Comment By: Raymond Toy (rtoy) Date: 20080312 20:30 Message: Logged In: YES user_id=28849 Originator: NO Actually, the roots are obviously the seven roots of 1, which maxima does know, and it could have done the partial fraction expansion to find the value of the integral. Not sure how or where to teach maxima about this, though. Some thought needed.  Comment By: Ximin Luo (infinity0x) Date: 20080312 18:32 Message: Logged In: YES user_id=2003896 Originator: YES The point is that Maxima does not NEED to know the roots of that equation. Sorry for making this unclear. By doing the integral using a different algorithm which doesn't involve taking partial fractions, you can avoid the above, and get a purely symbolic integral, like Mathematic does.  Comment By: Raymond Toy (rtoy) Date: 20080312 18:16 Message: Logged In: YES user_id=28849 Originator: NO Maxima can't give an answer because it doesn't know the roots of (x^7+1)/(x+1). If you set integrate_use_rootsof:true, then maxima can return an answer, but unless you can compute the roots of the above equation, it's probably not very helpful.  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1913067&group_id=4933 
From: SourceForge.net <noreply@so...>  20080312 22:32:46

Bugs item #1913067, was opened at 20080312 21:06 Message generated for change (Comment added) made by infinity0x You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1913067&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: Ximin Luo (infinity0x) Assigned to: Nobody/Anonymous (nobody) Summary: Cannot integrate 1/(1+x^n) Initial Comment: Maxima cannot integrate functions of the form 1/(1+x^n) for n >= 7. Mathematica is capable of this. Maxima attempts the integration via an algorithm which seems to involve taking partial fractions. For example, if you try to integrate 1/(1+x^7) for example, Maxima gives: log(1+x)/7  integral(x^52*x^4+3*x^34*x^2+5*x6)/(x^6x^5+x^4x^3+x^2x+1) / 7 I'm guessing a different algorithm (such as that employed by Mathematica) is required to give a fully symbolic answer. For the record, the integral(1/(1+x^7)) is: Log[1 + x]/7  (Cos[Pi/7]*Log[1 + x^2  2*x*Cos[Pi/7]])/7  (Cos[(3*Pi)/7]*Log[1 + x^2  2*x*Cos[(3*Pi)/7]])/7  (Cos[(5*Pi)/7]*Log[1 + x^2  2*x*Cos[(5*Pi)/7]])/7 + (2*ArcTan[(x  Cos[Pi/7])*Csc[Pi/7]]* Sin[Pi/7])/7 + (2*ArcTan[(x  Cos[(3*Pi)/7])* Csc[(3*Pi)/7]]*Sin[(3*Pi)/7])/7 + (2*ArcTan[(x  Cos[(5*Pi)/7])* Csc[(5*Pi)/7]]*Sin[(5*Pi)/7])/7  Additionally, this also means that we can calculate the integral(that nasty rational function), by calculating 7 * ( log(1+x)/7  integral(1/(1+x^7)) ) As a side note, Mathematica also cannot give integral(that nasty rational function) fully symbolically. Instead, it gives: RootSum[1  #1 + #1^2  #1^3 + #1^4  #1^5 + #1^6 & , (6*Log[x  #1] + 5*Log[x  #1]*#1  4*Log[x  #1]*#1^2 + 3*Log[x  #1]* #1^3  2*Log[x  #1]*#1^4 + Log[x  #1]*#1^5)/(1 + 2*#1  3*#1^2 + 4*#1^3  5*#1^4 + 6*#1^5) & ] where RootSum[ f, form ] represents the sum of form[x] for all x that satisfy the polynomial equation f[x] == 0.  >Comment By: Ximin Luo (infinity0x) Date: 20080312 22:32 Message: Logged In: YES user_id=2003896 Originator: YES The point is that Maxima does not NEED to know the roots of that equation. Sorry for making this unclear. By doing the integral using a different algorithm which doesn't involve taking partial fractions, you can avoid the above, and get a purely symbolic integral, like Mathematic does.  Comment By: Raymond Toy (rtoy) Date: 20080312 22:16 Message: Logged In: YES user_id=28849 Originator: NO Maxima can't give an answer because it doesn't know the roots of (x^7+1)/(x+1). If you set integrate_use_rootsof:true, then maxima can return an answer, but unless you can compute the roots of the above equation, it's probably not very helpful.  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1913067&group_id=4933 
From: SourceForge.net <noreply@so...>  20080312 22:16:25

Bugs item #1913067, was opened at 20080312 17:06 Message generated for change (Comment added) made by rtoy You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1913067&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: Ximin Luo (infinity0x) Assigned to: Nobody/Anonymous (nobody) Summary: Cannot integrate 1/(1+x^n) Initial Comment: Maxima cannot integrate functions of the form 1/(1+x^n) for n >= 7. Mathematica is capable of this. Maxima attempts the integration via an algorithm which seems to involve taking partial fractions. For example, if you try to integrate 1/(1+x^7) for example, Maxima gives: log(1+x)/7  integral(x^52*x^4+3*x^34*x^2+5*x6)/(x^6x^5+x^4x^3+x^2x+1) / 7 I'm guessing a different algorithm (such as that employed by Mathematica) is required to give a fully symbolic answer. For the record, the integral(1/(1+x^7)) is: Log[1 + x]/7  (Cos[Pi/7]*Log[1 + x^2  2*x*Cos[Pi/7]])/7  (Cos[(3*Pi)/7]*Log[1 + x^2  2*x*Cos[(3*Pi)/7]])/7  (Cos[(5*Pi)/7]*Log[1 + x^2  2*x*Cos[(5*Pi)/7]])/7 + (2*ArcTan[(x  Cos[Pi/7])*Csc[Pi/7]]* Sin[Pi/7])/7 + (2*ArcTan[(x  Cos[(3*Pi)/7])* Csc[(3*Pi)/7]]*Sin[(3*Pi)/7])/7 + (2*ArcTan[(x  Cos[(5*Pi)/7])* Csc[(5*Pi)/7]]*Sin[(5*Pi)/7])/7  Additionally, this also means that we can calculate the integral(that nasty rational function), by calculating 7 * ( log(1+x)/7  integral(1/(1+x^7)) ) As a side note, Mathematica also cannot give integral(that nasty rational function) fully symbolically. Instead, it gives: RootSum[1  #1 + #1^2  #1^3 + #1^4  #1^5 + #1^6 & , (6*Log[x  #1] + 5*Log[x  #1]*#1  4*Log[x  #1]*#1^2 + 3*Log[x  #1]* #1^3  2*Log[x  #1]*#1^4 + Log[x  #1]*#1^5)/(1 + 2*#1  3*#1^2 + 4*#1^3  5*#1^4 + 6*#1^5) & ] where RootSum[ f, form ] represents the sum of form[x] for all x that satisfy the polynomial equation f[x] == 0.  >Comment By: Raymond Toy (rtoy) Date: 20080312 18:16 Message: Logged In: YES user_id=28849 Originator: NO Maxima can't give an answer because it doesn't know the roots of (x^7+1)/(x+1). If you set integrate_use_rootsof:true, then maxima can return an answer, but unless you can compute the roots of the above equation, it's probably not very helpful.  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1913067&group_id=4933 
From: SourceForge.net <noreply@so...>  20080312 21:07:11

Bugs item #1913067, was opened at 20080312 21:06 Message generated for change (Settings changed) made by infinity0x You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1913067&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: Ximin Luo (infinity0x) Assigned to: Nobody/Anonymous (nobody) Summary: Cannot integrate 1/(1+x^n) Initial Comment: Maxima cannot integrate functions of the form 1/(1+x^n) for n >= 7. Mathematica is capable of this. Maxima attempts the integration via an algorithm which seems to involve taking partial fractions. For example, if you try to integrate 1/(1+x^7) for example, Maxima gives: log(1+x)/7  integral(x^52*x^4+3*x^34*x^2+5*x6)/(x^6x^5+x^4x^3+x^2x+1) / 7 I'm guessing a different algorithm (such as that employed by Mathematica) is required to give a fully symbolic answer. For the record, the integral(1/(1+x^7)) is: Log[1 + x]/7  (Cos[Pi/7]*Log[1 + x^2  2*x*Cos[Pi/7]])/7  (Cos[(3*Pi)/7]*Log[1 + x^2  2*x*Cos[(3*Pi)/7]])/7  (Cos[(5*Pi)/7]*Log[1 + x^2  2*x*Cos[(5*Pi)/7]])/7 + (2*ArcTan[(x  Cos[Pi/7])*Csc[Pi/7]]* Sin[Pi/7])/7 + (2*ArcTan[(x  Cos[(3*Pi)/7])* Csc[(3*Pi)/7]]*Sin[(3*Pi)/7])/7 + (2*ArcTan[(x  Cos[(5*Pi)/7])* Csc[(5*Pi)/7]]*Sin[(5*Pi)/7])/7  Additionally, this also means that we can calculate the integral(that nasty rational function), by calculating 7 * ( log(1+x)/7  integral(1/(1+x^7)) ) As a side note, Mathematica also cannot give integral(that nasty rational function) fully symbolically. Instead, it gives: RootSum[1  #1 + #1^2  #1^3 + #1^4  #1^5 + #1^6 & , (6*Log[x  #1] + 5*Log[x  #1]*#1  4*Log[x  #1]*#1^2 + 3*Log[x  #1]* #1^3  2*Log[x  #1]*#1^4 + Log[x  #1]*#1^5)/(1 + 2*#1  3*#1^2 + 4*#1^3  5*#1^4 + 6*#1^5) & ] where RootSum[ f, form ] represents the sum of form[x] for all x that satisfy the polynomial equation f[x] == 0.  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1913067&group_id=4933 
From: SourceForge.net <noreply@so...>  20080312 21:06:39

Bugs item #1913067, was opened at 20080312 21:06 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=1913067&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: Ximin Luo (infinity0x) Assigned to: Nobody/Anonymous (nobody) Summary: Cannot integrate 1/(1+x^n) Initial Comment: Maxima cannot integrate functions of the form 1/(1+x^n) for n >= 7. Mathematica is capable of this. Maxima attempts the integration via an algorithm which seems to involve taking partial fractions. For example, if you try to integrate 1/(1+x^7) for example, Maxima gives: log(1+x)/7  integral(x^52*x^4+3*x^34*x^2+5*x6)/(x^6x^5+x^4x^3+x^2x+1) / 7 I'm guessing a different algorithm (such as that employed by Mathematica) is required to give a fully symbolic answer. For the record, the integral(1/(1+x^7)) is: Log[1 + x]/7  (Cos[Pi/7]*Log[1 + x^2  2*x*Cos[Pi/7]])/7  (Cos[(3*Pi)/7]*Log[1 + x^2  2*x*Cos[(3*Pi)/7]])/7  (Cos[(5*Pi)/7]*Log[1 + x^2  2*x*Cos[(5*Pi)/7]])/7 + (2*ArcTan[(x  Cos[Pi/7])*Csc[Pi/7]]* Sin[Pi/7])/7 + (2*ArcTan[(x  Cos[(3*Pi)/7])* Csc[(3*Pi)/7]]*Sin[(3*Pi)/7])/7 + (2*ArcTan[(x  Cos[(5*Pi)/7])* Csc[(5*Pi)/7]]*Sin[(5*Pi)/7])/7  Additionally, this also means that we can calculate the integral(that nasty rational function), by calculating 7 * ( log(1+x)/7  integral(1/(1+x^7)) ) As a side note, Mathematica also cannot give integral(that nasty rational function) fully symbolically. Instead, it gives: RootSum[1  #1 + #1^2  #1^3 + #1^4  #1^5 + #1^6 & , (6*Log[x  #1] + 5*Log[x  #1]*#1  4*Log[x  #1]*#1^2 + 3*Log[x  #1]* #1^3  2*Log[x  #1]*#1^4 + Log[x  #1]*#1^5)/(1 + 2*#1  3*#1^2 + 4*#1^3  5*#1^4 + 6*#1^5) & ] where RootSum[ f, form ] represents the sum of form[x] for all x that satisfy the polynomial equation f[x] == 0.  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1913067&group_id=4933 
From: SourceForge.net <noreply@so...>  20080312 20:44:33

Bugs item #1913047, was opened at 20080312 20:44 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=1913047&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: Share Libraries Group: None Status: Open Resolution: None Priority: 5 Private: No Submitted By: Matthew Gwynne (proteumus) Assigned to: Nobody/Anonymous (nobody) Summary: gf_set in gf package doesn't terminate on some inputs. Initial Comment: Summary : Calling gf_set(2,1,[x]); results in nontermination. It seems natural to want to generate GF(2), and even if the format of the modulus used isn't correct, it doesn't seem to matter as the problem lies in the combination of b and e, rather than p (as described in Other Information). Version of Maxima : 5.14 Test Case : $ maxima Maxima 5.14.0 http://maxima.sourceforge.net Using Lisp SBCL 1.0.9gentoo Distributed under the GNU Public License. See the file COPYING. Dedicated to the memory of William Schelter. The function bug_report() provides bug reporting information. (%i1) load("gf"); (%o1) /usr/share/maxima/5.14.0/share/contrib/gf/gf.mac (%i2) gf_set(2,1,[x]); doesn't terminate. Expected result : $ maxima Maxima 5.14.0 http://maxima.sourceforge.net Using Lisp SBCL 1.0.9gentoo Distributed under the GNU Public License. See the file COPYING. Dedicated to the memory of William Schelter. The function bug_report() provides bug reporting information. (%i1) load("gf"); (%o1) /usr/share/maxima/5.14.0/share/contrib/gf/gf.mac (%i2) gf_set(2,1,[x]); (%o2) true Other Information : It seems the problem occurs when gf_findprim() (line 365) is called within gf_set. fastf and slowf both evaluate to the empty list given the arguments (ifactors(1) = [] in both cases) and this leads to f being [], and therefore lf being 0. This then leads to nontermination of the while statement, beginning on line 414, due to the fact that "i" is set to 1 but the inner while statement, beginning on line 430, only executes if i <= lf and of course 1 <= 0 is false. This inner while statement is the only place "found" is set to true and this must occur for the outer while statement to terminate. Reproducible : Always  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1913047&group_id=4933 
From: SourceForge.net <noreply@so...>  20080312 05:06:42

Bugs item #1904814, was opened at 20080229 08:47 Message generated for change (Comment added) made by nobody You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1904814&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  Plotting Group: None Status: Open Resolution: None Priority: 5 Private: No Submitted By: Nobody/Anonymous (nobody) Assigned to: Nobody/Anonymous (nobody) Summary: strings in plot2doption `legend' Initial Comment: In my opinion the following example lines f(x):=1/(1+x^2); g(x):=exp(x^2); plot2d([f(x),g(x)],[x,4,4], [legend,"$\\frac{1}{1+x^2}$","$e^{x^2}$"]); plot2d([f(x),g(x)],[x,4,4], [legend,"$\\frac{1}{1+x^2}$","$e^{x^2}$"], [gnuplot_term,"eepic"], [gnuplot_out_file,"bug.tex"]); don't give the expected results: In the displayed graph, and in the eepicoutput, the backslash \ and the beginning dollar $ is missing. For instance, the output in bug.tex contains the line \put(2548,1636){\makebox(0,0)[r]{FRAC{1}{1+X^2}$}} Another problem is that frac and x get capitalized. I have been trying out maxima only for a few hours now, however my understanding of `legend' and the handling of strings let me expect a different result, the above output should be \put(2548,1636){\makebox(0,0)[r]{$\frac{1}{1+x^2}$}} It does not seem to be a gnuplot problem, because gnuplot handles the analogous construct correctly. Maxima version: 5.14.0 Maxima build date: 11:54 2/29/2008 host type: i686pclinuxgnu lispimplementationtype: CLISP lispimplementationversion: 2.39 (20060716) (built 3373656864) (memory 3413271278) Best wishes, Peter Mueller (peter.mueller@...)  Comment By: Nobody/Anonymous (nobody) Date: 20080311 22:06 Message: Logged In: NO Problem is due to src/plot.lisp assuming that Maxima strings are implemented as symbols. I have a patch to fix that which I'll commit in a few days. Robert Dodier (not logged in at the moment)  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1904814&group_id=4933 
From: SourceForge.net <noreply@so...>  20080311 00:37:29

Bugs item #1911030, was opened at 20080310 07:18 Message generated for change (Comment added) made by willisbl You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1911030&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: 4 Private: No Submitted By: Barton Willis (willisbl) Assigned to: Nobody/Anonymous (nobody) Summary: matrix_element_add with ratmx : true Initial Comment: Using a user defined value for matrix_element_add and ratmx bound to true, we get a Lisp error. Example: (%i1) matrix_element_add : lambda([a,b], max(a,b))$ (%i2) m : matrix([a,b],[c,d])$ This is OK (ratmx : false) (%i3) m.m; (%o3) matrix([max(a^2,b*c),max(a*b,b*d)], ...) This is a bug: (%i4) m.m, ratmx : true; Maxima encountered a Lisp error:  >Comment By: Barton Willis (willisbl) Date: 20080310 19:37 Message: Logged In: YES user_id=895922 Originator: YES By the time matrix multiplication makes it to multl, it's too late. Maybe we should do something like ratmx : true$ matrix_element_mult : lambda([a,b], ...) > error  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1911030&group_id=4933 