We check the convergence of the secant method from the previous example.
f = @(x) x.*exp(x) - 2;
x = secant(f,1,0.5);
We don't know the exact root, so we use the built-in fzero to get a substitute.
r = fzero(f,1);
Here is the sequence of errors.
format short e
err = r - x(1:end-1)'
err =
-1.4739e-01 3.5261e-01 4.2234e-02 -1.3026e-02 4.2748e-04 4.2699e-06 -1.4055e-09 4.6629e-15
It's not so easy to see the convergence rate by looking at these numbers. But we can check the ratios of the log of successive errors.
logerr = log(abs(err));
ratios = logerr(2:end) ./ logerr(1:end-1)
ratios =
5.4444e-01 3.0358e+00 1.3717e+00 1.7871e+00 1.5938e+00 1.6486e+00 1.6190e+00
It seems to be heading toward a constant ratio of about 1.6 by the time it quits.