Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $m = 2$

share|improve this question
1  
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
add comment

1 Answer

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.