Thread: newton method

Code:

if (x2 - x1)/x2 < err
return
end

Is this terminating condition correct for newton raphson method?

Originally Posted by ilikec

Code:

 
if (x2 - x1)/x2 < err
return
end

Is this terminating condition correct for newton raphson method?

No, it at least needs an abs() on the left of the inequality.

CB

cant i say

if feval(fx,x) < err
return
end

fx is the function
x is the value of x at current iteration

So fx(x) < err means x is very close to the root

Am i correct in this case?

Originally Posted by ilikec

cant i say

if feval(fx,x) < err
return
end

fx is the function
x is the value of x at current iteration

So fx(x) < err means x is very close to the root

Am i correct in this case?

suppose err=0.0001 and fx(x)=-200 ?

CB

oh yeah i forgot

it should be

abs(f(x) ) < err


absolute value so -200 will be 200 in this case

is it okay to use this technique?

Originally Posted by ilikec

oh yeah i forgot

it should be

abs(f(x) ) < err


absolute value so -200 will be 200 in this case

is it okay to use this technique?

Yes

CB