Page 201 - Applied Numerical Methods Using MATLAB

190 NONLINEAR EQUATIONS

function [x,fx,xx] = secant(f,x0,TolX,MaxIter,varargin)
% solve f(x)=0by using the secant method.
%input : f = ftn to be given as a string ’f’ if defined in an M-file
% x0 = the initial guess of the solution
% TolX = the upper limit of |x(k) - x(k - 1)|
% MaxIter = the maximum # of iteration
%output: x = the point which the algorithm has reached
% fx = f(x(last)), xx = the history of x
h = 1e-4; h2 = 2*h; TolFun=eps;
xx(1) = x0; fx = feval(f,x0,varargin{:});
fork=1: MaxIter
if k <= 1, dfdx = (feval(f,xx(k) + h,varargin{:})-...
feval(f,xx(k) - h,varargin{:}))/h2;
else dfdx = (fx - fx0)/dx;
end
dx = -fx/dfdx;
xx(k + 1) = xx(k) + dx; %Eq.(4.5.2)
fx0 = fx;
fx = feval(f,xx(k+1));
if abs(fx) < TolFun | abs(dx) < TolX, break; end
end
x = xx(k + 1);
if k == MaxIter, fprintf(’The best in %d iterations\n’,MaxIter), end

This secant iterative formula is cast into the MATLAB routine “secant()”,
which never needs anything like the derivative as an input argument. We can
use this routine “secant()” to solve a nonlinear equation like that dealt with
in Example 4.2, by typing the following statement into the MATLAB command
window. The process is depicted in Fig. 4.5.

>>[x,err,xx] = secant(f42,2.5,1e-5,50) %with initial guess 1.8

2.5
2
1.5
1
0.5
x 5 x x x
0 3 2 0
x 1 x 4
−0.5

−1
−1.5
−2
1.8 1.9 2 2.1 2.2 2.3 2.4 2.5 2.6
Figure 4.5 Solving a nonlinear equation by the secant method.