"Roger Stafford" <ellieandrogerxyzzy@mindspring.com.invalid>
wrote in message <g0gmrt$bej$1@fred.mathworks.com>...
> "Ashwini Deshpande" <vd.ashwini@mathworks.com> wrote in
message
> <g0gi9l$gtu$1@fred.mathworks.com>...
> > "Ashwini Deshpande" <vd.ashwini@mathworks.com> wrote in
> > message <g0gg50$hb4$1@fred.mathworks.com>...
> > > I have 3 unknowns and 3 nonlinear equations, as follows:
> > >
> > > x^3  3x = y;
> > > y^3  3y = z;
> > > z^3  3z = x;
> > >
> > > How do i solve this using matlab.
> > >
> > > Thanks !
> > > Ashwini
> >
> > I tried to solve these equations using 'fsolve' command, i
> > am getting confused with how to set the starting values for
> > x, y and z. The procedure which i tried to solve is as
follows:
> > function F = myfun(x)
> > F = [x(1)^3  3*x(1)  x(2);
> > x(2)^3  3*x(2)  x(3);
> > x(3)^3  3*x(3)  x(1)];
> >
> > then from command prompt:
> > x0 = [1;1;1];
> > [x,fval] = fsolve(@myfun,x0)
> >
> > The problem is for all value of x0, i am getting different
> > answers for x, y and.
> >
> > And one more doubt is i am getting one root value for x, y
> > and z. But i suppose to get 3 values for each variable as it
> > is a 3rd order equation.
> >
> > Is there anything wrong in my approach, plz suggest me the
> > correct one.
> >
> > Thanks!
> > Ashwini
> 
> This is a problem for 'roots'. By substitution, you can
obtain a 27th degree
> polynomial equation and that will have 27 roots
altogether. First substitute
> x^33x in for y in the second equation getting:
>
> (x^33x)^33(x^33x) = z
>
> Then substitute the left hand expression in this equation
for z in the third
> equation:
>
> ((x^33x)^33(x^33x))^33((x^33x)^33(x^33x)) = x
>
> This is a polynomial equation of the 27th degree in x.
For each of the 27
> roots for x, the values of y and z can be determined
uniquely from the first
> two equations, so that makes 27 possible triplets x, y,
and z which can satisfy
> the three equations simultaneously.
>
> It is easy to see that there are three triplet solutions
with x, y, and z all
> equal: (0,0,0), (2,2,2), and (2,2,2). I can only guess
at the remaining 24
> solutions, but because of the symmetry of the equations,
very likely they
> consist of four basically different solutions with six
permutations among x, y,
> and z possible for each one. I wouldn't be surprised if
many of these were
> complexvalued.
>
> Finding all such multiple solutions is a task well
beyond the capabilities of
> 'fsolve' unless you provide it with a very large quantity
of initial guesses for
> its 'x0' argument. The 'roots' function seems the only
reasonable way to do
> the problem.
>
> Roger Stafford
>
>
Thank u very much for ur reply,
I got the result, i tried as follows:
F = ((x^33*x)^33*(x^33*x))^33*((x^33*x)^33*(x^33*x))x;
F1 = expand(F);
F2 = sym2poly(F1);
F3 = roots(F2);
Thanks!
Ashwini
