Empty output with newtons method

Hi all

im new to matlab and im attempting to create a newtons method calculator. I keep getting an empty square bracket as an output. Can you kindly look into my code to see where the problem arises:

function approx = newton(f,nmax,tol,x0)

i = 1;

while (i<nmax)

x1 = x0 - f(x0)./diff(f(x0));

if (abs((x1 - x0)/x1) < tol)

break

end

i = i + 1;

x0 = x1;

disp(x1)

end

approx = x1;

end

Thank you

Answer 1

[Samuele Sandrini]

Hi, the problem is in this line:

  x1 = x0 - f(x0)./diff(f(x0));

because f(x0) is a number and the function diff doesn't return the derivative of f, but of a vector X it returns a vector containg the differences between adjacent elements like this (Y =diff(X)= [X(2)-X(1) X(3)-X(2) ... X(m)-X(m-1)]). So, if f(x0) is a number, diff(f(x0)) it returns nothing ([]).

If you want to apply the Newton Method you have to know the derivative of f,and pass it as an input of function newton (newton(f,df,nmax,tol,x0)), and in that point you evaluate the derivative of f in x0 (you can compute it in symbolic way and then convert it in function handle using matlabFunction(df)).

However, your idea is not wrong, but is similar (but incorrectly implemented because you would need to divide it by (x1-x0) to have the derivative but you don't know x1 yet) to the Secant Method, that is used whenever the derivative of f is unknown or diffucult to compute: you substitute the derivative of f in x0 with an approximation that is the angular coefficient of the line passing between a node () and the previous one ():

so the algorithm became something like this:

where

Note that, in this case at starting point you have to pass not only one point x0 but also another one point x1 (Here an example of implemantation).

[ahmed saheed]

Thank you :)