Trying to use Newton's method until solution converges

So this is probably a dumb question, but I'm terrible at Matlab and am really struggling and can't seem to find any examples for my problem. I understand how Newton's method works, but not sure how to apply it in my case. My problem is I'm trying to solve this equation

for delta_sigma . All the other variables are known already.

This is the code I have right now:

% Variables
Eps = 0.011;
E = 73100;
H = 662;
n = 0.07;

maxIter = 100; % # of iterations
i = 1;
x(i) = 500; % Initial guess
while i <= maxIter
    Eps = x/E+2.*(x./(2*H)).^(1/n);
    dEps = 2./(n.*x).*(x./(2*H)).^(1/n);
    y = x(i)-(Eps/dEps);
    i = i+1;
    x(i) = y;
end

which I think would be correct if I was trying to find the delta_epsilon value. Could someone help me out? Thanks

[Torsten]

I assume delta_Eps = Eps ?

[Christina Kersten]

Yes

Answer 1

[Torsten]

% Variables
Eps = 0.011;
E = 73100;
H = 662;
n = 0.07;
maxIter = 100; % # of iterations
Tol = 1e-8;
i = 1;
error = 1.0;
X(i) = 500; % Initial guess
x = X(i);
while i <= maxIter && error > Tol
    f = x/E + 2*(x/(2*H))^(1/n) - Eps;
    df = 1/E + 2 * 1/n * (x/(2*H))^(1/n-1) * 1/(2*H);  
    x = X(i) - f/df;
    error = abs(x-X(i));
    i = i+1;
    X(i) = x;
end
i

i = 7

X

X = 1×7

500.0000 802.8180 765.4778 756.0581 755.6437 755.6430 755.6430

x/E + 2*(x/(2*H))^(1/n)- Eps

ans = 0

[Christina Kersten]

Wow thanks so much!

[Christina Kersten]

Could you explain why you have error as 1?

[Torsten]

You can choose any value > "Tol" for "error". It's only an initialization such that the while-loop is entered.

A value for error <= 1e-8 won't work because MATLAB then assumes that a solution has already been found.

[Matt J]

Wow thanks so much!

@Christina Kersten If so, you should accept-click Torsten's asnwer