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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



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