       Re: How to find solutions for conditioned equations?

• To: mathgroup at smc.vnet.net
• Subject: [mg19920] Re: [mg19861] How to find solutions for conditioned equations?
• From: "Allan Hayes" <hay at haystack.demon.co.uk>
• Date: Tue, 21 Sep 1999 02:22:51 -0400
• References: <7s3pia\$cna@smc.vnet.net>
• Sender: owner-wri-mathgroup at wolfram.com

```Bob,
Your solution to this particular problem does not need InequalitySolve (but,
of course, using it in

soln[a_ /; 0 <= a <= 1] := Module[{x}, x /.
Solve[(x + 1)/2 == a || (3 - x)/2 == a, x]]

Table[soln[a], {a, 0, 1, .2}]

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

and  at a = 0 we should have 3, since x is restricted to be >-1.

We can use Inequality solve do do more of the work:

Needs["Algebra`InequalitySolve`"];

is = InequalitySolve[
((x + 1)/2 == a && -1 < x <= 1 ||
(3 - x)/2 ==  a && 1 < x <= 3), {a, x}]

a == 0 && x == 3 || 0 < a < 1 && (x == -1 + 2 a || x == 3 - 2 a) ||
a == 1 && x == 1

Now we can define the multi-valued inverse function

Apply[Set[s[a_ /; #1], x /. {ToRules[#2]}] &, sol, {1}];
s[a_/;-Infinity<a<Infinity] = {};

Table[s[a], {a, -.2, 1.2, .2}]

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

Allan
---------------------
Allan Hayes
Mathematica Training and Consulting
Leicester UK
www.haystack.demon.co.uk
hay at haystack.demon.co.uk
Voice: +44 (0)116 271 4198
Fax: +44 (0)870 164 0565

<BobHanlon at aol.com> wrote in message news:7s3pia\$cna at smc.vnet.net...
> 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}];
>
> Needs["Algebra`InequalitySolve`"];
>
> soln[a_ /; 0 <= a <= 1] := Module[{x}, x /.
>         {ToRules[InequalitySolve[(x + 1)/2 == a || (3 - x)/2 == a, x]]}];
>
> Table[soln[a], {a, 0, 1, .2}]
>
> {{-1, 3}, {-0.6, 2.6}, {-0.2, 2.2}, {0.2, 1.8}, {0.6, 1.4}, {1.}}
>
> Bob Hanlon
>
> In a message dated 9/19/1999 5:43:39 AM, d8442803 at student.nsysu.edu.tw
writes:
>
> >Suppose I have an equation
> >
> >A(x)= (x+1)/2, for -1<x<=1
> >    = (3-x)/2, for 1<x<=3
> >    = 0      , otherwise
> >
> >Now if I want to find solutions for A(x)==a, how can I otain the
> >solutions simultaneously (i.e., represented as (2a-1, 3-2a))? I always
> >
> >use the stupid method to solve them separately, but I found it's laborous
> >
> >when I have B(x), C(x) needed to be solved together. (I am trying to use
> >
> >alpha-cut to proceed the interval operations) Any suggestions?
> >
>

```

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