Re: subsidiary conditions in Solve?
- To: mathgroup at smc.vnet.net
- Subject: [mg17349] Re: [mg17324] subsidiary conditions in Solve?
- From: BobHanlon at aol.com
- Date: Mon, 3 May 1999 01:45:53 -0400
- Sender: owner-wri-mathgroup at wolfram.com
In a message dated 5/1/99 5:29:03 AM, sandra at rgmiller.Stanford.EDU writes:
>I'm trying to solve simultaneous equations, and I'd like to include some
>subsidiary conditions which place restrictions on quantities (like the
>variance) to be positive. Examples of these conditions are:
>
> 2 2
>E (M) > (E(M))
>
>and
> 2
>sigma > 0
> M
>
>Here's a simple example -- in this case, I'd like to rule out the solution
>x=-3. I'm following p. 825, "3.4.9 Solving Equations with Subsidiary
>Conditions" of the 1999 Mathematica Book.
>
>In[92]:= subsid1 = Greater[x,0]
>
>Out[92]= x > 0
>
>In[93]:= Solve[x^2 + x - 6 == 0, subsid1]
>
>General::ivar: x > 0 is not a valid variable.
>
> 2
>Out[93]= Solve[-6 + x + x == 0, x > 0]
>
>
>What am I doing wrong?
>
The subsidiary conditions given in section 3.4.9 are themselves equations
(equalities) not inequalities. Here are some alternatives for inequaliities.
Select[Solve[x^2 + x - 6 == 0, x], (x /. #)>0&]
{{x -> 2}}
Cases[Solve[x^2 + x - 6 == 0, x], x_?((x /. #)>0&)]
{{x -> 2}}
Cases[Solve[x^2 + x - 6 == 0, x], x_?(Positive[x /. #]&)]
{{x -> 2}}
Needs["Algebra`InequalitySolve`"]
InequalitySolve[x^2 + x - 6 == 0 && x > 0, x]
x == 2
Bob Hanlon