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Let $f(a,x)=\sqrt{a(a-1)+x}$, $f^{(n)}(a,x)$ denotes the $n^{th}$ iteration of $x$, where \begin{align}f^{(1)}(a,x)=f(a,x),f^{(n)}(a,x)=f^{(n-1)}(a,f(a,x))\end{align} Find \begin{align}\lim_{n\to\infty}(2a)^{\frac{n}{2}}\sqrt{a-f^{(n)}(a,0)}\end{align}

This is a problem from a discussion in a forum, someone say that when $a=2$, the limit is equal to $\pi$, but I can't find the pattern behind this structure, Thanks for your attention.

PS: Expanding it shows \begin{align}\lim_{n\to\infty}(2a)^{\frac{n}{2}}\sqrt{a-\sqrt{a(a-1)+\sqrt{a(a-1)+\cdots}}}\end{align} where there are $n$ square root sign behind the minus sign, it is a $\infty \cdot 0$ problem.

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Why not discuss it in that forum (secret from us?) instead of here? – GEdgar Oct 23 '12 at 16:41
@GEdgar : Because none of us has found a way to find the limit – Golbez Oct 23 '12 at 17:17
But I still think it would be better to include the link of that forum. – GEdgar Oct 23 '12 at 17:28
up vote 2 down vote accepted

Steps for solution ... With $a=2$ write $f(2,x) = f(x)$. For the appropriate values of $\theta$ ...

$$\begin{align} &2\cos\theta = \sqrt{2+2\cos(2\theta)} \\ &f^{(n)}(2\cos\theta) = 2\cos\frac{\theta}{2^n} \\ &f^{(n)}(0) = 2\cos\frac{\pi}{2^{n+1}} \\ &\sqrt{2-2\cos\theta} = 2 \sin \frac{\theta}{2} \\ &\sqrt{2-f^{(n)}(0)} = 2\sin\frac{\pi}{2^{n+2}} \\ &\lim_{n\to\infty} 2^n\cdot 2\sin\frac{\pi}{2^{n+1}} = \frac{\pi}{2} \end{align}$$

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When $a$ is arbitary, is it possible to calculate the answer? – Golbez Oct 24 '12 at 13:29
The only complex numbers $a$ for which solution can be done by this method are $a=-1,0,1,2$. Function $f$ is inverse to $z^2+c$ where $c=-a(a-1)$. Iteration of $z^2+c$ in closed form is known only when $c=0$ or $c=-2$. And $-a(a-1)=0$ yields $a=0$ or $a=1$; and $-a(a-1)=-2$ yields $a=-1$ or $a=2$. – GEdgar Oct 24 '12 at 16:03

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