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How to get a simple system of recursion equations to give output?

Iuval Clejan

Posted 14 days ago

This seems to not converge, no output after a day. Is there another way to do it, that will give output? It's just a system of recursion equations, with boundary conditions at 1 and N (here N=3).

POSTED BY: Iuval Clejan

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Iuval Clejan

Posted 7 days ago

I am trying a similar system, but.a bit different. Why is NSolve not able to substitute numerical values into the functions that are being minimized? I also tried FindRoot, with same issue.

POSTED BY: Iuval Clejan

Iuval Clejan

Posted 13 days ago

Solutions are such that the coefficients are not independent of each other.

POSTED BY: Iuval Clejan

Iuval Clejan

Posted 13 days ago

OK, great, thanks!. That means either x[1]=y[1]=0 if the expressions are not equal, or if they are, an infinite number of solutions. The real problem is with xE .ne. 0, and using your idea I get something useful for some relations between the coefficients that give some unique solutions. Will try for N>3.

POSTED BY: Iuval Clejan

Bill Nelson

Posted 14 days ago

This may or may not help you. If you try FullSimplify[Collect[Eliminate[{ x[1]==f11[x[2],y[2]], y[1]==f12[x[2],y[2]], x[2]==fx[x[1],y[1],x[3],y[3]], y[2]==fy[x[1],y[1],x[3],y[3]], x[3]==f31[x[2],y[2],0], y[3]==f32[x[2],y[2],0]}, {x[2],y[2],x[3],y[3]}], {x[1],y[1]}]] then it will quickly show you expr1x[1]+expr2y[1]==0 && expr3x[1]+expr4y[1]==0 && expr5x[1]+expr6y[1]==0 Each of those six new expressions are large, but this has removed x[2],y[2],x[3],y[3] and perhaps you can see something in those expressions that will tell you something about the solution you are looking for.

This may or may not help you. If you try

FullSimplify[Collect[Eliminate[{
    x[1]==f11[x[2],y[2]],
    y[1]==f12[x[2],y[2]],
    x[2]==fx[x[1],y[1],x[3],y[3]],
    y[2]==fy[x[1],y[1],x[3],y[3]],
    x[3]==f31[x[2],y[2],0],
    y[3]==f32[x[2],y[2],0]},
    {x[2],y[2],x[3],y[3]}],
  {x[1],y[1]}]]

then it will quickly show you

expr1*x[1]+expr2*y[1]==0 &&
expr3*x[1]+expr4*y[1]==0 &&
expr5*x[1]+expr6*y[1]==0

Each of those six new expressions are large, but this has removed x[2],y[2],x[3],y[3] and perhaps you can see something in those expressions that will tell you something about the solution you are looking for.

POSTED BY: Bill Nelson