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}
(48) 
_{Sep}
(39) 
_{Oct}
(20) 
_{Nov}

_{Dec}

S  M  T  W  T  F  S 




1
(2) 
2

3
(1) 
4
(2) 
5

6
(1) 
7
(4) 
8
(1) 
9

10

11

12

13
(1) 
14

15
(2) 
16
(2) 
17
(2) 
18

19

20

21

22

23
(1) 
24

25
(1) 
26
(4) 
27
(2) 
28
(9) 
29
(1) 



From: SourceForge.net <noreply@so...>  20120207 13:13:04

Bugs item #3485344, was opened at 20120207 05:13 Message generated for change (Tracker Item Submitted) made by johnlapeyre You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=3485344&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: John Lapeyre (johnlapeyre) Assigned to: Nobody/Anonymous (nobody) Summary: $lex in sym package not maxima function Initial Comment: There is an undocumented function (defun $lex (term1 term2) ... in ./share/sym/util.lisp that clobbers any other definition when sym is loaded. But, the arguments to $lex are lisp lists, not maxima lists, so it can't be called from maxima in any case. There appear to be other functions with the same problem in the same file. eg: $degre.  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=3485344&group_id=4933 
From: SourceForge.net <noreply@so...>  20120207 11:04:26

Bugs item #3485031, was opened at 20120206 09:28 Message generated for change (Comment added) made by aleksasd You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=3485031&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: Stavros Macrakis (macrakis) Assigned to: Nobody/Anonymous (nobody) Summary: rootscontract anomaly Initial Comment: ex: (sqrt(2)+1)^(2/3)$ rootscontract(ex)=> (sqrt(2)+1)^(2/3) <<< does nothing rootscontract(2*ex) => 2*(2^(3/2)+3)^(1/3) <<< squares the inner expression Though the documentation is not explicit about what it should do in this case (it only talks about products of roots, not about powers of roots), the second behavior is more useful  if it isn't provided by rootscontract, it should be provided by *some* function. Practical result of this anomaly: rootscontract( (2^(3/2)+3)^(1/3)(sqrt(2)+1)^(2/3) ) => 0 (I don't know any other way to do this simplification in Maxima) but rootscontract( ( (2^(3/2)+3)^(1/3)(sqrt(2)+1)^(2/3) )) => remains unsimplified  Comment By: Aleksas (aleksasd) Date: 20120207 03:04 Message: Next example. see http://www.math.utexas.edu/pipermail/maxima/2012/027666.html How simplify (%i7) ex:sqrt(1x^2)/(22*x^2)$ (%i8) assume(abs(x)<1)$ We define transformation funkcion f(t) and its inverse g(t): (%i9) f(t):=(2*t)^2$ g(t):=sqrt(t)/2$ (%i11) assume(t>0)$ (%i12) f(g(t)); g(f(t)); (%o12) t (%o13) t (%i14) f(ex)=ratsimp(f(ex)); (%o14) (4*(1x^2))/(22*x^2)^2=1/(x^21) (%i15) g(%); (%o15) sqrt(1x^2)/(22*x^2)=1/(2*sqrt(1x^2)) Aleksas D  Comment By: Aleksas (aleksasd) Date: 20120206 23:38 Message: How simplify (%i1) ex:(sqrt(2)+1)^(2/3); (%o1) (sqrt(2)+1)^(2/3) (%i2) ex^3=expand(ex^3); (%o2) (sqrt(2)+1)^2=2^(3/2)+3 (%i3) %^(1/3); (%o3) (sqrt(2)+1)^(2/3)=(2^(3/2)+3)^(1/3) Other example. How simplify (%i4) ex:4*atan(1/5)atan(1/239); (%o4) 4*atan(1/5)atan(1/239) (%i5) tan(ex)=trigexpand(trigexpand(tan(ex))); (%o5) tan(4*atan(1/5)atan(1/239))=1 (%i6) atan(%); (%o6) 4*atan(1/5)atan(1/239)=%pi/4 Aleksas D  Comment By: Aleksas (aleksasd) Date: 20120206 22:38 Message: (%i1) r:(sqrt(2)+1)^(2/3)$ (%i2) r=expand(r^3)^(1/3); (%o2) (sqrt(2)+1)^(2/3)=(2^(3/2)+3)^(1/3) (%i3) 2*r=expand((2*r)^3)^(1/3); (%o3) 2*(sqrt(2)+1)^(2/3)=(2^(9/2)+24)^(1/3) Aleksas  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=3485031&group_id=4933 
From: SourceForge.net <noreply@so...>  20120207 07:38:53

Bugs item #3485031, was opened at 20120206 09:28 Message generated for change (Comment added) made by aleksasd You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=3485031&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: Stavros Macrakis (macrakis) Assigned to: Nobody/Anonymous (nobody) Summary: rootscontract anomaly Initial Comment: ex: (sqrt(2)+1)^(2/3)$ rootscontract(ex)=> (sqrt(2)+1)^(2/3) <<< does nothing rootscontract(2*ex) => 2*(2^(3/2)+3)^(1/3) <<< squares the inner expression Though the documentation is not explicit about what it should do in this case (it only talks about products of roots, not about powers of roots), the second behavior is more useful  if it isn't provided by rootscontract, it should be provided by *some* function. Practical result of this anomaly: rootscontract( (2^(3/2)+3)^(1/3)(sqrt(2)+1)^(2/3) ) => 0 (I don't know any other way to do this simplification in Maxima) but rootscontract( ( (2^(3/2)+3)^(1/3)(sqrt(2)+1)^(2/3) )) => remains unsimplified  Comment By: Aleksas (aleksasd) Date: 20120206 23:38 Message: How simplify (%i1) ex:(sqrt(2)+1)^(2/3); (%o1) (sqrt(2)+1)^(2/3) (%i2) ex^3=expand(ex^3); (%o2) (sqrt(2)+1)^2=2^(3/2)+3 (%i3) %^(1/3); (%o3) (sqrt(2)+1)^(2/3)=(2^(3/2)+3)^(1/3) Other example. How simplify (%i4) ex:4*atan(1/5)atan(1/239); (%o4) 4*atan(1/5)atan(1/239) (%i5) tan(ex)=trigexpand(trigexpand(tan(ex))); (%o5) tan(4*atan(1/5)atan(1/239))=1 (%i6) atan(%); (%o6) 4*atan(1/5)atan(1/239)=%pi/4 Aleksas D  Comment By: Aleksas (aleksasd) Date: 20120206 22:38 Message: (%i1) r:(sqrt(2)+1)^(2/3)$ (%i2) r=expand(r^3)^(1/3); (%o2) (sqrt(2)+1)^(2/3)=(2^(3/2)+3)^(1/3) (%i3) 2*r=expand((2*r)^3)^(1/3); (%o3) 2*(sqrt(2)+1)^(2/3)=(2^(9/2)+24)^(1/3) Aleksas  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=3485031&group_id=4933 
From: SourceForge.net <noreply@so...>  20120207 06:38:08

Bugs item #3485031, was opened at 20120206 09:28 Message generated for change (Comment added) made by aleksasd You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=3485031&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: Stavros Macrakis (macrakis) Assigned to: Nobody/Anonymous (nobody) Summary: rootscontract anomaly Initial Comment: ex: (sqrt(2)+1)^(2/3)$ rootscontract(ex)=> (sqrt(2)+1)^(2/3) <<< does nothing rootscontract(2*ex) => 2*(2^(3/2)+3)^(1/3) <<< squares the inner expression Though the documentation is not explicit about what it should do in this case (it only talks about products of roots, not about powers of roots), the second behavior is more useful  if it isn't provided by rootscontract, it should be provided by *some* function. Practical result of this anomaly: rootscontract( (2^(3/2)+3)^(1/3)(sqrt(2)+1)^(2/3) ) => 0 (I don't know any other way to do this simplification in Maxima) but rootscontract( ( (2^(3/2)+3)^(1/3)(sqrt(2)+1)^(2/3) )) => remains unsimplified  Comment By: Aleksas (aleksasd) Date: 20120206 22:38 Message: (%i1) r:(sqrt(2)+1)^(2/3)$ (%i2) r=expand(r^3)^(1/3); (%o2) (sqrt(2)+1)^(2/3)=(2^(3/2)+3)^(1/3) (%i3) 2*r=expand((2*r)^3)^(1/3); (%o3) 2*(sqrt(2)+1)^(2/3)=(2^(9/2)+24)^(1/3) Aleksas  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=3485031&group_id=4933 