## 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

This site uses Akismet to reduce spam. Learn how your comment data is processed.