fsolve question

[John B. Thoo]

Hi.  I'm new to Octave and to this list, so I apologize if this is a 
FAQ.  (I did check the FAQ w/out success.  I also scanned the archive 
for 2002, but, this being Nov, there were a lot in the archive and I may 
have overlooked the topic.) :-)

Page 173 of the Octave manual gives the following example for solving a 
system of nonlinear equations (copied and pasted from my trial).


octave:1> function y = f(x)
> y(1) = -2*x(1)^2 + 3*x(1)*x(2) + 4*sin(x(2)) - 6;
> y(2) = 3*x(1)^2 - 2*x(1)*x(2)^2 + 3*cos(x(1)) + 4;
> endfunction
octave:2> [x, info] = fsolve("f", [1; 2])
x =

  0.57983
  2.54621

info = 1
octave:3>


What is the role of the condition  [1; 2]  in fsolve?  I played around 
changing them and got different solutions, but I'm not seeing how it's 
affecting the solution.  (I'm sure I'm overlooking the obvious.)  The 
manual calls them "initial conditions."  Is that like a seed value for 
Newton's method?  In particular, I'm trying to figure out what I should 
use to solve a system of 4 equations:  [1; 2; 3; 4]  or something 
cleverer?

Thanks.

---John.



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