Re: Mathematica Collect function
- To: mathgroup at smc.vnet.net
- Subject: [mg110731] Re: Mathematica Collect function
- From: Bob Hanlon <hanlonr at cox.net>
- Date: Sat, 3 Jul 2010 08:20:30 -0400 (EDT)
expr = Expand[((1 + Sqrt[2]) i - 1)/4*(P10 - P11) -
(1 + Sqrt[2] + i)/4*(P20 - P21)];
Your initial form is not what Mathematica considers the simplest form
expr // Simplify
(1/4)*((Sqrt[2]*i + i - 1)*P10 - (Sqrt[2]*i + i - 1)*P11 -
(i + Sqrt[2] + 1)*(P20 - P21))
You need to force it
(expr /. {P10 -> x1 + P11, P20 -> x2 + P21} //
Simplify) /.
{x1 -> (P10 - P11), x2 -> (P20 - P21)}
(1/4)*((Sqrt[2]*i + i - 1)*(P10 - P11) - (i + Sqrt[2] + 1)*
(P20 - P21))
This is almost your initial form
Bob Hanlon
---- Minh <dminhle at gmail.com> wrote:
=============
Given that:
Expand[((1 + Sqrt[2]) i - 1)/4*(P10 - P11) - (1 + Sqrt[2] + i)/
4*(P20 - P21)]
will output
-(P10/4) + (i P10)/4 + (i P10)/(2 Sqrt[2]) + P11/4 - (i P11)/4 - (
i P11)/(2 Sqrt[2]) - P20/4 - P20/(2 Sqrt[2]) - (
i P20)/4 + P21/4 + P21/(2 Sqrt[2]) + (i P21)/4
How do I get from:
-(P10/4) + (i P10)/4 + (i P10)/(2 Sqrt[2]) + P11/4 - (i P11)/4 - (
i P11)/(2 Sqrt[2]) - P20/4 - P20/(2 Sqrt[2]) - (
i P20)/4 + P21/4 + P21/(2 Sqrt[2]) + (i P21)/4
back to
((1 + Sqrt[2]) i - 1)/4*(P10 - P11) - (1 + Sqrt[2] + i)/
4*(P20 - P21)
I've tried using the Collect function as follows:
Collect[-(P10/4) + (i P10)/4 + (i P10)/(2 Sqrt[2]) + P11/4 - (i P11)/
4 - (i P11)/(2 Sqrt[2]) - P20/4 - P20/(2 Sqrt[2]) - (i P20)/4 + P21/
4 + P21/(2 Sqrt[2]) + (i P21)/4, {(P10 - P11), (P20 - P21)}]
but it doesn't seem to collect the terms {(P10 - P11), (P20 - P21)}.
Got any suggestions?