# Convergence of Newton's Method

Suppose $m \geq 2$ be an integer such that $m \geq 2$, let $a, x_0 >0$. The sequence is then

$x_{n+1} = \frac{1}{m}\left( (m-1)x_n + \frac{a}{x^2 _{m+1}} \right )$

Show that the sequence $(x_n)_{n \in \mathbb{N}}$ converges to a unique positive solution of $x^m = a$

Attempt

Honestly, I thought I would just go and prove this without the whole m getting in my way first. So I would try to prove by m = 2 first and then use the fact that $m \geq 2$ to cheat my way through it.

But when I got to the limit part to converging to $x^m = a$, I couldn't get it. I got nothing like it.

EDIT

Sorry, it should have been

$x_{n+1} = \frac{1}{m}\left( (m-1)x_n + \frac{a}{x^{m-1}_n} \right )$

@ Gerry

When $m = 2$

$x_{n+1} = \frac{1}{2}\left( (2-1)x_n + \frac{a}{x^{2-1}_n} \right ) = \frac{1}{2}\left( (x_n + \frac{a}{x_n}) \right )$

EDIT2: yes it worked for $m = 2$

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Lol, read again your first sentence, it's funny. You said something twice. – Patrick Da Silva Nov 29 '11 at 4:34
Your idea of proving for $m=2$ first is a good one, but I didn't follow your subsequent remarks: did you succeed for $m=2$ or not? – Srivatsan Nov 29 '11 at 4:35
What is the point of using this algorithm? If this is a fixed point method (Newton's method is), then you should have that if x_{n+1} = x_n then x_n^m = a, which is obviously not the case here. Is this homework? Where did this come from? I can't find the mistake by myself in this question, but I don't feel like it'll converge towards a root of $x^m$ because the sequence is not related in anyway to $x^m$. Welcome to MSE by the way! =D – Patrick Da Silva Nov 29 '11 at 4:41
I think the last bit of your formula is wrong. For $m=2$, Newton says $x_{n+1}=x_n-((x_n^2-a)/(2x_n))$, which rewrites as $x_{n+1}=(1/2)(x_n+(a/x_n))$, which isn't what you have. – Gerry Myerson Nov 29 '11 at 4:53

$$x_{n+1} = \frac{1}{m}\left( (m-1)x_n + \frac{a}{x^{m-1}_n} \right )$$
Supposing that the sequence does in fact converge. Then, we can set $x_{n+1}=x_n = x$ in the above equation. If you simplify the resulting expression you should get the desired result.