finding the number of interations of newtons method

I'm having a problem trying to find how many interations the following function has to go through to get the root

function [root,count]=NewtonMethod(c0,c,x)

count=1;

fx1=poly_val(c0,c,x)

fx2=poly_val(c0,c,x +0.001)

m=(fx2-fx1)/(0.001)

newGuess= x-((fx1/m))

if fx1 < (1*10^-8)

fx1=poly_val(c0,c,newGuess)

fx2=poly_val(c0,c,newGuess+0.001)

m=(fx2-fx1)/(0.001)

newGuess=newGuess-((fx1)/(m))

fx1=poly_val(c0,c,newGuess)

end

root=newGuess

count=count+1

fprintf('number of times is = %i \n' , count)

the input is NewtonMethod(0,[5 1 -6 0 1],2)

and the output is 2.0943 and the count is 2 but should be 6

Answer 1

[Harsha Priya Daggubati]

Hi,

I guess the count should be incremented in the place where fx1 is being compared and it should be conditioned in a loop to get the number of iterations.

function [root,count]=NewtonMethod(c0,c,x)
count=0;
fx1 = poly_val(c0,c,x);
fx2 = poly_val(c0,c,x +0.001);
m =(fx2-fx1)/(0.001);
newGuess = x-((fx1/m));
while fx1 < (1*10^-8)
    fx1 = poly_val(c0,c,newGuess);
    fx2 = poly_val(c0,c,newGuess+0.001);
    m = (fx2-fx1)/(0.001);
    newGuess = newGuess-((fx1)/(m));
    fx1 = poly_val(c0,c,newGuess);
    count = count+1;
end
root = newGuess;

fprintf('number of times is = %i \n' , count);
end

Hope this works!