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}
(45) 
_{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...>  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 