You can subscribe to this list here.
2008 
_{Jan}
(4) 
_{Feb}
(2) 
_{Mar}

_{Apr}

_{May}

_{Jun}

_{Jul}

_{Aug}
(2) 
_{Sep}

_{Oct}
(27) 
_{Nov}

_{Dec}
(6) 

2009 
_{Jan}
(3) 
_{Feb}

_{Mar}

_{Apr}
(1) 
_{May}

_{Jun}
(27) 
_{Jul}

_{Aug}

_{Sep}
(3) 
_{Oct}
(22) 
_{Nov}
(12) 
_{Dec}

2010 
_{Jan}
(11) 
_{Feb}
(3) 
_{Mar}
(2) 
_{Apr}
(11) 
_{May}
(29) 
_{Jun}
(59) 
_{Jul}

_{Aug}
(3) 
_{Sep}
(2) 
_{Oct}
(8) 
_{Nov}
(3) 
_{Dec}

2011 
_{Jan}
(5) 
_{Feb}

_{Mar}

_{Apr}
(6) 
_{May}

_{Jun}

_{Jul}
(13) 
_{Aug}
(4) 
_{Sep}
(4) 
_{Oct}
(19) 
_{Nov}

_{Dec}
(2) 
2012 
_{Jan}
(2) 
_{Feb}
(1) 
_{Mar}
(3) 
_{Apr}

_{May}

_{Jun}

_{Jul}
(1) 
_{Aug}

_{Sep}

_{Oct}

_{Nov}

_{Dec}

2013 
_{Jan}

_{Feb}

_{Mar}

_{Apr}

_{May}
(78) 
_{Jun}
(10) 
_{Jul}

_{Aug}

_{Sep}

_{Oct}

_{Nov}

_{Dec}
(2) 
2014 
_{Jan}
(2) 
_{Feb}

_{Mar}

_{Apr}

_{May}

_{Jun}

_{Jul}

_{Aug}

_{Sep}
(1) 
_{Oct}
(1) 
_{Nov}

_{Dec}

S  M  T  W  T  F  S 

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

18

19

20

21

22

23
(3) 
24

25

26

27

28

29

30

31





From: Игорь Пашев <pashev.igor@gm...>  20100823 19:26:39

Why not do it in mind ? ;) 
From: Yue Li <yue.li@gm...>  20100823 15:03:01

On Aug 23, 2010, at 9:41 AM, Urban Simoncic wrote: > Hi, > > I am wondering if I could simplify the expression with new variables. > Let say I have b:=x+y+z and a:=b+exp(b). Then I calculate derivative > D(a, x) and I get the exp(x+y+z) + 1, which is correct. But, how could I > get the expression exp(b)+1 instead of exp(x+y+z)+1? Is there a way to > exchange every occurrence of x+y+z with b? > Hi Urban You could utilize OpenAxiom's pattern matching facility to manually do so. One needs to use function ``rule" to define his/her own rewriting rule, then apply it to an expression. For instance: (8) > b := x+y+z (8) z + y + x Type: Polynomial Integer (9) > a := b + exp(b) z + y + x (9) %e + z + y + x Type: Expression Integer (10) > d := D(a, x) z + y + x (10) %e + 1 Type: Expression Integer (11) > myRule := rule(x+y+z == b) (11) z + y + x == b Type: RewriteRule(Integer,Integer,Expression Integer) (12) > myRule(d) b (12) %e + 1 Type: Expression Integer Best, Yue 
From: Urban Simoncic <urban.simoncic@gm...>  20100823 14:41:45

Hi, I am wondering if I could simplify the expression with new variables. Let say I have b:=x+y+z and a:=b+exp(b). Then I calculate derivative D(a, x) and I get the exp(x+y+z) + 1, which is correct. But, how could I get the expression exp(b)+1 instead of exp(x+y+z)+1? Is there a way to exchange every occurrence of x+y+z with b? That is an illustrative example, but I actually have much more complicated case, where the result will go to an algorithm and a new variable would significantly reduce computation time. Thanks, Urban 