Sequences and series or trigonometry problem or both?

The IITJEE Mains are to be held on April 4 2015.  So, let’s focus on some worthwhile basic problems.

Example. Let x_{0}=2003 and let x_{n+1}=\frac {1+x_{n}}{1-x_{n}} for

n \geq 1. Compute x_{2004}.

Solution:

At the first sight, you might think it’s a plain manipulation of recurring sequences. But, wait…let’s see…

With a little algebraic computation, we can show that this sequence has a period of 4; that is, x_{n+4}=x_{n} for all n \geq 1. But, why? We reveal the secret with trigonometric substitution; that is, we define

\alpha_{n} with -90 \deg < \alpha_{n} < 90 \deg such that x_{n}=\tan \alpha_{n}. It is clear that if x_{n} is a real number, such an \alpha_{n} is unique, because

\tan: (-90 \deg, 90 \deg) \rightarrow \Re is a bijection. Because 1=\tan 45 \deg, we can rewrite the given condition as

\tan \alpha_{n+1}=\frac{\tan 45\deg + \tan \alpha_{n}}{1-\tan 45\deg\tan\alpha_{n}}, which is

\tan(45\deg+\alpha_{n}) by the addition and subtraction formulae. Consequently,

\alpha_{n+1}=45\deg+\alpha{n}, or \alpha_{n+1}=45\deg+\alpha_{n}-180\deg (because tan has a period of 180\deg). In any case, it is not difficult to see that \alpha_{n+4}=\alpha_{n}+k.180\deg for some integer k. Therefore, x_{n+4}=\tan{\alpha_{n}}=x_{n}, that is, the sequence

\{x_{n}\}_{n \geq 0} has a period 4, implying that x_{2004}=x_{0}=2003.

More later,

Nalin Pithwa

 

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