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

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Luca Grillo 2024-8-9

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编辑： Torsten 2024-8-9

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

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Answer 1

Torsten 2024-8-9

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https://ww2.mathworks.cn/matlabcentral/answers/2144209-newton-s-method-gives-nan-can-someone-improve-my-code#answer_1497189

编辑：Torsten 2024-8-9

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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.