# MATLAB Matlab, iteration with newton's method, noobs & errors

#### Waxterzz

Hi,

I have to describe the famous newton-rapshon algorithm for finding a root of the function
this is what i came up with, i aint not familiar with matlab and programming at all, so this is somewhat a melt from a lot of matlab tutorials, I dont master the code, but what's really a challenge is understanding the errors
--------------------------------------------------------
function x = newton(fun,dfun,x0,tol,Nmax)
x = x0;
f(x) = feval(fun,x);
df(x)= feval(dfun,x);
steps = 0;

re = tol;
myrel = 1;
while myrel > re && (steps < Nmax)
xold = x;
x = x - f(x)/df(x);
steps = steps + 1;
myrel = abs((x-xold)/x);
end;
if myrel <= re
disp( 'Zero found at' )
disp( x )
else
end;
--------------------------------------------------------

when i test it, i find 2 very strange facts

fact 1: it cannot handle non-integer, if I want to test the function in the prompt with sin and cos and x0=3 , it stops on the first iteration: 3,14255...
this is the error:
attempted to acces f(3.14255); index must be a positive inter or logical

My idea: i thought matlab had no problems at all with any kind of number, in c++(also a big noob at that) u have to declarate everything, but not so in matlab yes?)

and fact2
i only tested the function with sin, cos and 3

when i test it with a simple Polynomial, let's say x²-2 for finding the value of square root of 2 it gives an error in the second line: f(x) = feval...

how can that line work with the first example and not with a simple polynomial in x,

can someone get me on track

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bump ? anyone ??

#### CRGreathouse

Homework Helper
It sounds like f is set up as an array instead of a function.

#### eys_physics

The problem is that when you write f(x), Matlab treat this as an vector. So the solution is to write something like f=feval(...) inside the while loop (and remove the line f(x)=... and similar for df(x)

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