Suppose that f(x) has a zero at x = p and that f'(p) != 0. Show that when f''(p) = 0 Newton's methos converges to p with order at least 3.
Is this solved using a Taylor series expansion. And finding a delta and an M?
Yes it used Taylor series, but I think you need the existance of the third derivative to prove it, and its continuity at p (without these conditions you just get that the convergence is faster than quadratic).
-Opps.. f''(p)=0, sp yes, with suitable conditions on the existance and continuity of the third derivative convergence is of order 3 or better.
CB
First I would (without loss of generality) assume $\displaystyle p=0$ as it makes the algebra easier.
Then we have:
$\displaystyle \varepsilon_{n+1}=\varepsilon_n-\frac{f(\varepsilon_n)}{f'(\varepsilon_n)}$
Now I would expand the right hand side as a power series in $\displaystyle \varepsilon_n$ about $\displaystyle 0$, remembering that $\displaystyle f(0)=f''(0)=0$.
CB