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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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    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.2011-11-29

1 Answers 1

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So, you have:

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

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.

If, in addition to the above, you have to actually check if the sequence converges then you need additional machinery. See the wiki for a description of the proof of convergence of Newton's method which you may be able to adapt to your setting.