Re: Mathematica Collect function

*To*: mathgroup at smc.vnet.net*Subject*: [mg110748] Re: Mathematica Collect function*From*: Minh <dminhle at gmail.com>*Date*: Sun, 4 Jul 2010 06:09:56 -0400 (EDT)*References*: <i0n9v6$htp$1@smc.vnet.net>

Many thanks to all who have responded to my question. The question I posted was actually a smaller section of what I had originally planned on asking. I was hoping to use what I had learnt from your answers and apply it to a larger problem but I haven't had much success. What I want is to go from expression1: -((P10 P20 P30)/Sqrt[2]) + (i P10 P20 P30)/Sqrt[2] - i P11 P20 P30 + i P10 P21 P30 + (P11 P21 P30)/Sqrt[2] - (i P11 P21 P30)/Sqrt[2] - P10 P20 P31 - (P11 P20 P31)/Sqrt[2] - (i P11 P20 P31)/Sqrt[2] + ( P10 P21 P31)/Sqrt[2] + (i P10 P21 P31)/Sqrt[2] + P11 P21 P31 to expression2: ((1 + Sqrt[2]) i - 1)/4*(P10 - P11) - ( 1 + Sqrt[2] + i)/4*(P20 - P21) + ( 1 - Sqrt[2] + i)/4*(P10 - P11)*(P30 - P31) + ( 1 + (Sqrt[2] - 1) i)/4*(P20 - P21)*(P30 - P31) Given that P10 + P11=1,P20 + P21=1 and P30 + P31=1, expression 2 becomes expression3: ((1 + Sqrt[2]) i - 1)/4*(P10 - P11)*(P20 + P21)*(P30 + P31) - ( 1 + Sqrt[2] + i)/4*(P10 + P11)*(P20 - P21)*(P30 + P31) + ( 1 - Sqrt[2] + i)/4*(P10 - P11)*(P20 + P21)*(P30 - P31) + ( 1 + (Sqrt[2] - 1) i)/4*(P10 + P11)*(P20 - P21)*(P30 - P31) I know that they are equal because when I use Expand[expression3], I obtain expression1. I've tried forcing the simplification by introducing temporary expressions and back substituting to go from expression1 to expression 3 but I've realized then it doesn't work when introducing the extra terms (P10 + P11),(P20 + P21) and (P30 + P31). Any ideas? Thanks, Minh