Re: how to find polynomial roots in OCTAVE?help...

[John B. Thoo]

Hi, Henry.  When I tried your example, the "plot" choked on me (see  
below).  Any ideas?  I'm using 2.1.73 on Mac OS X 10.4.9.

octave:1> function [y] = f(x)
> y=x^(4/3)-21.3*x^(1/3)-4.53^(4/3);
> end
octave:2> f(0)
ans = -7.4954
octave:3> x = linspace (-100, 100, 1000);
octave:4> plot (x, f(x))
error: for A^b, A must be square
error: evaluating binary operator `^' near line 2, column 4
error: evaluating binary operator `-' near line 2, column 10
error: evaluating binary operator `-' near line 2, column 23
error: evaluating assignment expression near line 2, column 2
error: called from `f'
error: evaluating argument list element number 2
octave:4> [X, INFO, MSG] = fsolve ('f', 10)
X = 23.902
INFO = 1
MSG = solution converged within specified tolerance

Thanks.
---John.


On Jun 15, 2007, at 6:39 PM, Henry F. Mollet wrote:

> fsolve was suggested and it works.
> octave:10> function [y]=f(x)
> > y=x^(4/3)-21.3*x^(1/3)-4.53^(4/3);
> > end
> octave:11> f(0)
> ans = -7.4954
>
> octave:20> x=linspace(-100,100,1000);
> octave:21> plot (x,f(x))
> Shows that good initial value is important. x=0 won't do.
>
> octave:22> [X, INFO, MSG] = fsolve ('f', 10)
> X = 23.902
> INFO = 1
> MSG = solution converged within specified tolerance
> Henry
>
>
> on 6/14/07 7:57 PM, jub at address@hidden wrote:
>
> > James Sherman Jr.-2 wrote:
> > >  then just have y^3 = x;
> >
> > you mean y^(1/3)=x is it? yeah the 'h' is the 'x', sorry :-P