Services & Resources / Wolfram Forums / MathGroup Archive
-----

MathGroup Archive 2010

[Date Index] [Thread Index] [Author Index]

Search the Archive

Re: Mathematica Collect function

  • To: mathgroup at smc.vnet.net
  • Subject: [mg110766] Re: Mathematica Collect function
  • From: Bob Hanlon <hanlonr at cox.net>
  • Date: Mon, 5 Jul 2010 06:03:10 -0400 (EDT)

expr1 =
  -(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;

expr2 = ((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);

To get the form of expr2 from expr1

expr22 = Simplify[expr1, {
    P10 + P11 == t,
    P20 + P21 == t,
    P30 + P31 == t,
    P10 - P11 == x1,
    P20 - P21 == x2,
    P30 - P31 == x3}] /. {
   t -> 1,
   x1 -> P10 - P11,
   x2 -> P20 - P21,
   x3 -> P30 - P31}

1/4 ((i-Sqrt[2]+1) (P10-P11) (P30-P31)+
(Sqrt[2] i+i-1) (P10-P11)+
((Sqrt[2]-1) i+1) (P20-P21) (P30-P31)-
(i+Sqrt[2]+1) (P20-P21))

expr2 == expr22 // Simplify

True


Bob Hanlon

---- Minh <dminhle at gmail.com> wrote: 

=============
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



  • Prev by Date: Re: Mathematica Collect function
  • Next by Date: a 4d algebraic geometry problem
  • Previous by thread: Re: Mathematica Collect function
  • Next by thread: Re: Problems running Mathematica and WordMS together in Win7