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I have the following iteration process: $$ p_{n+1} = \frac{{p_{n}}^3 + 3 a p_{n} }{3 {p_{n}}^2 + a } , $$ where $a > 0$.

Q1: How to prove that this iteration process converges for every number $p_0 > p > 0 $ to the fixed point $p = g(p), p > 0$ ?

Q2: How to prove that this iteration process converges for every number $p > p_0 > 0$ to the fixed point $p = g(p) , p > 0 $ ?

Perhaps the following theorem helps: Let $f \in C^2 [a,b]$ such that $f(p) = 0$ and $f'(p) \neq 0 $. Then there is a $\delta > 0 $ such that Newton's method generates a sequence $ \{ p_n \} $, which converges to $p$ for every $p_0 \in [p - \delta , p + \delta]$.

In my case, $f(x) = x^2 - a$.

I am guessing this theorem probably could help (especially for Q2), but I don't see how I can apply it.

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To apply the theorem, you may start by identifying $f$. – Did Jun 13 '13 at 12:30
@Did I'm sorry, should've mentioned that. In my case, $f(x) = x^2 - a $. – Max Muller Jun 13 '13 at 12:32
Thus, the result holds for every $p_0$ close enough to $p$, right? For other starting points, I suggest to write down $p_{n+1}-p$ as a function of $p_n-p$. – Did Jun 13 '13 at 12:35
@Did o.k. I tried. I came up with $p_{n+1} - p = \frac{ {p_{n}}^3 + a (p_n - p) - 3 p {p_n}^2 }{3 {p_{n}}^2 + a } $, but I don't really see how this could help me answering the questions. – Max Muller Jun 13 '13 at 13:50
Your expression for $p_{n+1}-p$ is incorrect. – Did Jun 13 '13 at 14:56
up vote 2 down vote accepted

Consider $p_{n+1} - p_n = \frac{-2p_n^3 + 2ap_n}{3p_n^2 + a}$. We have $$p_{n+1} - p_n =\begin{cases}\le 0, & \text{if } p_n \ge \sqrt{a}\\ = 0, & \text{if } p_n = \sqrt{a} \\ \ge 0, & \text{if } p_n \le \sqrt{a}\end{cases}$$ Moreover, let $f(x) = \frac{x^3 + 3ax}{3x^2 + a}$. Then $$f'(x) = \frac{(3x^2+3a)(3x^2 + a)-6x(x^3+3ax)}{(3x^2+a)^2}=\frac{3x^4-6ax^2+3a^2}{(3x^2+a)^2}=\frac{3(x^2-a)^2}{(3x^2+a)^2} \ge 0.$$ Thus it follows that $f$ is increasing. In particular $$p_{n+1} = f(p_n) \le f(\sqrt{a}) = \sqrt{a} \Leftrightarrow p_n \le \sqrt{a}.$$

The behaviour of the sequence is now clear: $p_n$ increases without going past $\sqrt{a}$ if $p_0 \le \sqrt{a}$ and decreases if $p_0 \ge \sqrt{a}$, again without going past $\sqrt{a}$. Thus as a bounded monotone sequence $p_n$ has a limit, which has to be $\sqrt{a}$ by taking limits on both sides of the defining equation of $p_{n+1}$.

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Thanks! Is this answer to Q1 or Q2? And how did you find that formula for $p_{n+1} - p_n$ ? – Max Muller Jun 13 '13 at 15:59
@Max Muller: This should answer both Q1 and Q2. The formula for $p_{n+1} - p_n$ is a straightforward calculation. – J. J. Jun 13 '13 at 16:37

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