Newton's method gives NaN. Can someone improve my code?

Hello everybody. I'm using Newton's method to solve a liner equation whose solution should be in [0 1]. Unfortunately, the coe I'm using gives NaN as a result for a specific combination of parameters and I would like to understand if I can improve the code I wrote for my Newton's method. In the specific case I'm considering, I reach the maximum iterations even if the tolerance is very low.

% % % % Script for solving NaN
mNAN= 16.1;
lNAN= 10^-4;
f= @(x) mNAN*x+lNAN*exp(mNAN*x)-lNAN*exp(mNAN);
Fd= @(x) mNAN*(1+lNAN*exp(mNAN*x));
tolNaN=10^-1;
nmax=10^8;
AB0 = 0.5;
[amNAN,nNAN,ierNAN]=newton(f,Fd,AB0,nmax,tolNaN);
amNAN

amNAN = NaN

Llimit=f(0)

Llimit = -982.0670

Ulimit=f(1)

Ulimit = 16.1000

fplot(@(x) mNAN*x+lNAN*exp(mNAN*x)-lNAN*exp(mNAN),[0 1.1])

function [x,n,ier] = newton(f,fd,x0,nmax,tol)
% Newton's method for non-linear equations
ier = 0;
for n = 1:nmax
    x = x0-f(x0)/fd(x0);
    if abs(x-x0) <= tol
        ier = 1;
        break
    end
    x0 = x;
end
end

Answer 1

[Torsten]

Your undamped Newton's method throws you from x = 0.5 to x = 46 appr. in the next step. Here, your function cannot be evaluated because of the large exp() term.

Conclusion: Newton's method doesn't converge in all cases.

Try a different initial point x0 in the region where the function is steeper than in 0.5 (e.g. 0.9).

Or use a damping factor 0 < d < 1:

x = x0-d*f(x0)/fd(x0);

Convergence will be slower, but safer.

Further, you should also check whether abs(f(x)) is really small when you break the iteration. abs(x-x0) < tol usually doesn't suffice for the iteration to be converged if your function is steep.

[Luca Grillo]

Thank you for your suggestions! I'll try it out and see if it gets better!