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Here from the fraction set we have a really hard question to be answered...suppose that a sequence is defined as $a_{n} = a_{n-1} - \dfrac 1{a_{n-1}}$, where $a_0$ is given. already know what I'm asking you ... find the $n^{\text{th}}$ element of the sequence in terms of $n$ and $a_0$! :)

P.S I've been trying for a lot of time ... but until now I haven't managed to find a solution; only some very weird graphs which can puzzle you...if you wish I can send you some of them!

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It's very unlikely that a closed-form solution can be found. $a(n)$ is a rational function of $a(0)$ with numerator of degree $2^n$ and denominator of degree $2^n-1$.

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Is it provably non-conjugate to something like $x^{2^n}$ or $\cos 2^nx$ ? – zyx Nov 4 '13 at 16:27
Well, there are some constraints on conjugacy. The function $f(z) = z - 1/z$ on the Riemann sphere is two-to-one except for $f^{-1}(2i) = \{i\}$, $f^{-1}(-2i) = \{-i\}$. The only fixed point is $\infty$, and there is just one $2$-cycle $\{\pm 1/\sqrt{2}\}$. – Robert Israel Nov 5 '13 at 3:44
Counting small cycles would probably disprove conjugacy to any specific closed form by a finite computation, but ideally there would be a dynamical/ergodic invariant that distinguishes it from anything simple. – zyx Nov 5 '13 at 4:40

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