1. ## Newton's method convergence

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?

2. Originally Posted by CarmineCortez
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

3. so i have f(x) = f'(p)(x-p) + (f'''(p)*(x-p)^3)/3!

then letting x = p_n and using the mean value thm.

f(p_n) = f(p_n)-f(p) + (f'''(p)*(p_n-p)^3)/3!

So something is not right here....

4. Originally Posted by CarmineCortez
so i have f(x) = f'(p)(x-p) + (f'''(p)*(x-p)^3)/3!

then letting x = p_n and using the mean value thm.

f(p_n) = f(p_n)-f(p) + (f'''(p)*(p_n-p)^3)/3!

So something is not right here....
First I would (without loss of generality) assume $p=0$ as it makes the algebra easier.

Then we have:

$\varepsilon_{n+1}=\varepsilon_n-\frac{f(\varepsilon_n)}{f'(\varepsilon_n)}$

Now I would expand the right hand side as a power series in $\varepsilon_n$ about $0$, remembering that $f(0)=f''(0)=0$.

CB