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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 