We again look at finding a solution of 
near 
. To apply Newton's method, we need to calculate values of both the residual function f and its derivative.
near . To apply Newton's method, we need to calculate values of both the residual function f and its derivative. f = @(x) x.*exp(x) - 2;
dfdx = @(x) exp(x).*(x+1);
We don't know the exact root, so we use the built-in fzero to determine the ``true'' value.
format long, r = fzero(f,1)
We use 
as a starting guess and apply the iteration in a loop, storing the sequence of iterates in a vector.
as a starting guess and apply the iteration in a loop, storing the sequence of iterates in a vector.x = 1;
for k = 1:6
x(k+1) = x(k) - f(x(k)) / dfdx(x(k));
end
Here is the sequence of errors (leaving out the last value, now serving as our exact root).
format short e
err = x' - r
Glancing at the exponents of the errors, they roughly form a neat doubling sequence until the error is comparable to machine precision. We can see this more precisely by taking logs.
format short
logerr = log(abs(err))
Quadratic convergence isn't as graphically distinctive as linear convergence.
semilogy(abs(err),'.-')
xlabel('k'), ylabel('|x_k-r|') % ignore this line
title('Quadratic convergence') % ignore this line
This looks faster than linear convergence, but it's not easy to say more. If we could use infinite precision, the curve would continue to steepen forever.