prove statement about fixed point iteration

I have the following fixed point iteration: $$p_{n+1} = \frac{p_n^3 + 3ap_n}{3p_n^2 + a}$$ By defining $$g(x) = \frac{x^3 + 3ax}{3x^2 + a}$$ en some algebra I found that the fixed point is $x = \sqrt{a}$. So $g(\sqrt{a}) = \sqrt(a)$.

Now I need to show that this iteration proces converges for every $p_0$ with $0< p_0 < \sqrt{a}$ to the fixed point $\sqrt{a}$.

Maybe it isn't of any importancy but I have already proven the above statement for every $p_0$ with $0<\sqrt{a} < p_0$

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1 Answer

Note that $$g'(x)=\frac{(3x^2+3a)(3x^2+a)-(x^3+3ax)6x}{(3x^2+a)^2}=\frac{3x^4-6x^2+3a^2}{(3x^2+a)^2}=\frac{3(x^2-a)^2}{(3x^2+a)^2}$$

Note that $\frac{p_{n+1}}{p_n}=\frac{p_n^2+3a}{a+3p_n^2}$ is $>1$ if $p_n^2+3a>a+3p_n^2$, i.e. if $(0<)p_n<\sqrt a$. Hence we have the following possibiliteis if $0<p_0<\sqrt a$:

• The sequence is monotonincally increasing until for some $n$ we have $p_n\ge \sqrt a$. In this case, your (therefore important) argument shows that in converges $\to\sqrt a$ from then on
• The sequence is monotonically increasing but stays below $\sqrt a$. In that case it must converge to some limit. This limit must be a fixed point of $g$, i.e. $p_n\to \sqrt a$.
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Thanks for the answer, but why must the limit be a fixed point of $g$? – user54297 Jun 12 '13 at 12:07