Let f(x) = x^2 and let N(x) be the associated Newton’s method iteration function for f. Show that zero is attractive but not super-attractive for f.

# Newtons method is attractive but not super-attractive

**Ed_Boy**#2

Let f(x) = x^2.

Then N(x)=x-\dfrac{x^2}{2x}= \dfrac{x}{2}.

Then N'(x) = \dfrac{1}{2}

Then N'(0) = \dfrac{1}{2}.

So 0 is attractive but not super-attractive for N(x). (But it is super-attractive for f)