When I am doing Newton's Method with a tolerance, how do I only get the values up until it reaches the tolerance?

Using Newton's method, I was wondering if I was using the while loop correctly because it keeps running and doesn't stop even though I have a tolerance put into the code.

function [R,E] = myNewton(f, df, x0, tol)

R=myNewton(f,df,x0-f(x0)/df(x0),tol);

E=myNewton(f,df,(x0-R)*df(x0),tol);

while abs(f®)>tol

    R=R+(x0-f(x0)/df(x0));

    E=E+((x0-R)*df(x0));

end

end

[Carlos]

Sorry but what does abs(f®)mean? I cannot distinguish the symbols inside abs. If you edit your code, perhaps we could see more clearly.

Regards

Answer 1

[Mischa Kim]

Christopher, I am not quite sure I understand your code.

• What's the reason for executing myNewton twice, once for R once for E?
• Why are you calling myNewton from whithin myNewton?

The code below shows a simple example on how to implement Newton's method.

 function myMain()
 tol  = 1e-4;

 f    = @(x) x^2 - 2;
 x0   = 1;
 xsol = myNewton(f, x0, tol);
 end

 function x = myNewton(f, x0, tol)
 dx = 1e-8;
 df = @(x,dx) (f(x + dx/2) - f(x - dx/2))/dx;

 x  = x0;
 while abs(f(x))>tol
    x = x - f(x)/df(x,dx);
 end
 end

[Christopher]

R(i) is the Newton estimation of the root of f for the i-th iteration.

E(i) is the value of abs(f(R(i))) for the i-th iteration of the Newton method.

I understand how to get get a value for Newton's Method, but I am uncertain how to iterate the code until I obtain a value that is less than the tolerance.

Answer 2

[Jammi]

function myMain()

tol = 1e-4;

f = @(x) x^2 - 2;

x0 = 1;

xsol = myNewton(f, x0, tol);

end

function x = myNewton(f, x0, tol)

dx = 1e-8;

df = @(x,dx) (f(x + dx/2) - f(x - dx/2))/dx;

x = x0;

while abs(f(x))>tol

x = x - f(x)/df(x,dx);

end

end