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MathGroup Archive 1999

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Re: Re: How to find solutions for conditioned equations?

  • To: mathgroup at smc.vnet.net
  • Subject: [mg19958] Re: [mg19919] Re: How to find solutions for conditioned equations?
  • From: BobHanlon at aol.com
  • Date: Wed, 22 Sep 1999 04:11:29 -0400
  • Sender: owner-wri-mathgroup at wolfram.com

A numerical approach comes to mind:

Clear[A];
A[x_ /; -1 < x <= 1] := (x + 1)/2;
A[x_ /; 1 < x <= 3] := (3 - x)/2;
A[x_] := 0;

Plot[A[x], {x, -1.2, 3.2}]; 

Rewrite the definition of A using UnitStep

Clear[A]
A[x_] := UnitStep[x + 1]*(x + 1)/2 + UnitStep[x - 1]*(-x + 1) + 
    UnitStep[x - 3]*(x - 3)/2

Plot[A[x], {x, -1.2, 3.2}]; 

Look for a root in each region

x /. FindRoot[A[x] == .4, {x, #}] & /@ {.5, 1.5}

{-0.2, 2.2}

x /. Table[FindRoot[A[x] == a, {x, #}] & /@ {.5, 1.5}, {a, 0, 1, .2}]

{{-1., 3.}, {-0.6, 2.6}, {-0.2, 2.2}, {0.2, 1.8}, {0.6, 1.4}, {1., 1.}}

Bob Hanlon

In a message dated 9/21/1999 8:50:13 AM, d8442803 at student.nsysu.edu.tw writes:

>Now my question is whether I can use A[x]==a directly in the solving 
>equation? Say, for example,
>
>soln[a_]=InequalitySolve[A[x]== a, x]]}];  // substituted by A[x]==a
>


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