The sequence $ \{x_{n}\}_{n \ge 1}$ is defined by \[ x_{1} = 2, x_{n + 1} = \frac {2 + x_{n}}{1 - 2x_{n}}\;\; (n \in \mathbb{N}). \] Prove that a) $ x_{n}\not = 0$ for all $ n \in \mathbb{N}$, b) $ \{x_{n}\}_{n \ge 1}$ is not periodic.
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Tags: trigonometry, induction, Recursive Sequences
TTsphn
21.10.2007 04:23
Peter wrote: The sequence $ \{x_{n}\}_{n \ge 1}$ is defined by \[ x_{1} = 2, x_{n + 1} = \frac {2 + x_{n}}{1 - 2x_{n}}\;\; (n \in \mathbb{N}). \] Prove that $ x_{n}\not = 0$ for all $ n \in \mathbb{N}$, $ \{x_{n}\}_{n \ge 1}$ is not periodic. $ a = \arctan {2}$ $ x_1 = \tan{a}$ $ x_{n + 1} = \frac {x_n + \tan{a}}{1 - x_n\tan{a}}$ We prove by induction that: $ x_{n} = \tan{na}$ Suppose $ x_n$ is period T. $ \tan{(n + T)a} = \tan{na}$ $ \rightarrow Ta\equiv 0(\mod \pi)$ So $ a = \frac {mk\pi}{T}$ We has a result : $ k\in Q$ then $ \tan{k\pi}$ is not equal 2.
chess64
15.04.2008 04:58
As above let $ \alpha = \arctan 2$ so that $ x_n = \tan n\alpha$. For (a) suppose for the sake of contradiction that $ x_n = 0$ for some $ n$. Let $ m = n/2$ so that \[ x_n = x_{2m} = \frac{2\tan m\alpha}{1-\tan^2 m\alpha} = \frac{2x_m}{1-x_m^2}.\] If $ n$ is even, $ x_n = 0 \iff x_m = 0$. Otherwise let $ n = 2^k (2j+1)$ for some nonnegative integers $ j,k$. We can divide out powers of 2 until we are left with \[ x_{n/2^k} = x_{2j+1} = \frac{2+x_{2j}}{1-x_{2j}} = 0,\] so $ x_{2j} = -2$. But \[ x_{2j} = \frac{2x_j}{1-x_j^2} = -2.\] This equation has no rational roots so we have a contradiction (inductively, $ x_n$ must be rational since $ x_1 = 2$ is rational). Therefore there are no zero values of $ \{x_n\}$.
For (b) suppose that $ x_{k+t} = x_{k}$ for some positive integers $ k,t$ with $ t\ge 1$. Then \[ x_{k+t} - x_{k} = \tan (k+t)\alpha - \tan k\alpha = \frac{\tan t\alpha}{\cos(k+t)\alpha} = \frac{x_t}{\cos(k+t)\alpha} = 0,\] so $ x_t = 0$. However, this is impossible by (a). Thus all terms of $ \{x_n\}$ are distinct.