MaplePrimes - answers and comments on Question, nonlinear identity / matching

nonlinear identity / matching

Robert Israel — Mon, 16 Jul 2007 09:28:34 Z

You could do this:

> match(op(2,K) = -op(1,K), x, 's');
  s;

true {b = 1, c = 0, a = 0}

now this

edgar — Tue, 17 Jul 2007 18:47:11 Z

That works. Any suggestion on this? The S1 is generated in a program by solve(), I want integer solutions...There do exist integer solutions.

> S1 := {kk[1] = -sqrt(2)*kk[4]-kk[5]*sqrt(2), kk[4] = kk[4], kk[5] = kk[5], kk[2] = -kk[5]*sqrt(2)+2*sqrt(2), kk[3] = 0};

> isolve(S1);

> subs({kk[1]=0,kk[2]=0,kk[3]=0,kk[4]=-2,kk[5]=2},S1);

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integer solutions

Robert Israel — Tue, 17 Jul 2007 21:08:37 Z

Since your equations are linear and sqrt(2) is irrational, you can separate the terms that involve sqrt(2) and those that don't. It seems isolve isn't clever enough to do this itself.

> split:= proc(eq) 
    local v;
    v:= lhs(eq)-rhs(eq);
    coeff(v,sqrt(2),1), coeff(v,sqrt(2),0);
  end proc:
  S2:= map(split, S1);

S2 := {0, kk[3], kk[1], kk[2], kk[4]+kk[5], kk[5]-2}

> isolve(S2);

{kk[3] = 0, kk[1] = 0, kk[2] = 0, kk[4] = -2, kk[5] = 2}

similar ... but

edgar — Tue, 17 Jul 2007 21:13:57 Z

This will also work... subs(sqrt(2)=z,S1);

eliminate(%,z); But I want to put this inside a program, and I don't know in advance what irrationals will appear. So now what?

convert, RootOf, subsindets, etc

JacquesC — Tue, 17 Jul 2007 23:16:15 Z

Quick response now, I can turn this suggestion into code later, if someone else doesn't beat me to it. Basically, you need to first make sure you convert your expression into something linear in a set of independent bases. So if you will have a lot of irrationals, you need to convert to RootOf and normalize, to get rid of linear relations. Then you can safely replace every RootOf that's left with a fresh symbol (I would use subsindets for the latter task).