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If $\left\{x_{n}\right\}\mid x_{1}=5,x_{n+1}=x_{n}^{2}-2,\forall n\geq 1$ find $\lim_{n\to\infty}\frac{x_{n+1}}{x_{1}x_{2}\cdots x_{n}}.$

If someone could help me out with tags, it'd be lovely. I think this is calculus and real-analysis, but I'm not sure--I had the problem scribbled down on a post-it, and I forget where it's from.

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    I feel like I've seen this in a putnam exam or something. But I can tell it's just hard.2012-03-18

2 Answers 2

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Since $\cos(2x)=2\cos^2x-1$, we find that (note we are viewing cosine as a complex function)

$2\cos\left(2\,\cos^{-1}\left(\frac{u}{2}\right)\right)=u^2-2.$

Repeated composition induces telescopy, which motivates the following (provable with induction):

$x_n=2\cos\left(2^{n-1}\cos^{-1}\left(\frac{x_1}{2}\right)\right).$

Using complex exponentials we see that ($\varphi=\cos^{-1}(x_1/2),~z=e^{i\varphi}$)

$2^n\cos\varphi~\cos2\varphi~\cdots~\cos 2^n\varphi=(z+z^{-1})(z^2+z^{-2})\cdots(z^{2^{n-1}}+z^{-2^{n-1}})$

$=\frac{z^{2^{n}}-z^{-2^n}}{z-z^{-1}}=\frac{\sin 2^n\varphi}{\sin\varphi}$

whence

$\gamma_n:=\frac{x_{n+1}}{x_1\cdots x_n}=2\cot\big(2^n\cos^{-1}(x_1/2)\big)\sqrt{1-(x_1/2)^2}.$

With $\lim\limits_{r\to+\infty}\cot(ir)=-i$, $x_1>2$, and $(\cos^{-1}v)/i$ a positive real for $v>1$, we find that

$\lim_{n\to\infty}\gamma_n=\sqrt{x_1^2-4}.$

In particular, the answer to our puzzle ($x_1=5$) is $\sqrt{21}$.


For numerical verification, we compare an approximation ($n=5$) to the intended answer:

$\color{Blue}{4.582575694955840006588047193728008488984456}\color{Red}{8357053937} \tag{a}$ $\color{Blue}{4.5825756949558400065880471937280084889844565767679719} \tag{b}$

a=5949389624883225721727/(5*23*527*277727*77132286527); b=Sqrt[21]

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    @PeterT.off: Not all all: if we multiply and divide by $a - a^{-1}$ we find the term to be $\frac{(a-a^{-1})(a^{2^n} + a^{-2^n})}{a^{2^n} - a^{-2^n}}$ whose limit is $a - a^{-1}$.2012-03-20
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This is not an answer, just too big for a comment. I hope it'll help someone else finish a proof.

Write $ y_n = \frac{x_n}{x_1 \dots x_{n-1}}, \quad n \ge 2. $ Then $x_{n+1} = x_n^2 - 2$ implies $y_{n+1} = y_n - \frac{2}{x_1 \dots x_n}$, hence by forming a telescopic sum, one gets $ y_n - y_2 = \sum_{i=2}^{n-1} y_{i+1} - y_i = \sum_{i=2}^{n-1} \frac{-2}{x_1 \dots x_i} $ and therefore $ y_n = y_2 - \sum_{i=2}^{n-1} \frac{2}{x_1 \dots x_i}= \frac {23}{5} - \frac 25 \sum_{i=2}^{n-1} \frac{1}{x_2 \dots x_i}. $ Computing the limit is therefore equivalent to being able to compute $ \sum_{i=2}^{n-1} \frac 1{x_2 \dots x_i}. $ This sum is quite easily shown to converge (pretty fast in top of that, since if we remove the "$-2$"'s in the recursion definition of the sequence $x_n$, we get something like $\sum \frac 1{5^{2^1}\dots 5^{2^n}}$ which converges REALLY faster than the geometric series). But I have no idea yet how to compute the series. Maybe this is not the way to go.

Hope that helps,

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    It's possible one could combine this with [Euler's continued fraction formula](http://en.wikipedia.org/wiki/Euler's_continued_fraction_formula) plus some kind of square root cf expansion to solve this.2012-03-18