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Let $x_0 = 2003$, and let $x_{n+1} = \frac{1 + x_n}{1 - x_n}$ for $n >= $1. Compute $x_{2019}$

I tried expressing $x_2$, $x_3$, $x_4$.. in terms of previous terms to see if there is a pattern but $n >= 1$ is troubling me. Please help.

Yash Gabra
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3 Answers3

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HINT: Note that

$$\begin{align*} x_{n+2}&=\frac{1+x_{n+1}}{1-x_{n+1}}\\ &=\frac{1+\frac{1+x_n}{1-x_n}}{1-\frac{1+x_n}{1-x_n}}\\ &=\frac{1-x_n+1+x_n}{1-x_n-1-x_n}\\ &=-\frac1{x_n}\,. \end{align*}$$

Brian M. Scott
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  • From this relation, $x_{2019} = -\frac{1}{x_1}$ but how would we get $x_2$ or $x_1$ seeing as to how we can't substitute $n=0$? – user1729 Nov 20 '20 at 05:36
  • Professor, then we have $x_{n+2} = x_{n-2}$ so that $x_{2019} = x_{1}$? – Derek Luna Nov 20 '20 at 05:37
  • @user 1729 when you substitute, the negative sign and reciprocation disappear I think, so that we get $x_{n} = x_{n-4}$ which is the same as what I wrote. – Derek Luna Nov 20 '20 at 05:40
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    @DerekLuna: We have $x_{n+4}=x_n$, so in particular $x_{4n}=2003$ for each $n\in\Bbb N$. That gives you $x_{2016}$, from which it’s easy to get $x_{2018}$ and then $x_{2019}$. – Brian M. Scott Nov 20 '20 at 05:40
  • @Brian M Scott, although I wonder if for some people seeing that $x_{2020} = 2003 $ and using the given recurrence might be quicker. Thank you for the help. – Derek Luna Nov 20 '20 at 05:55
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    @DerekLuna: You’re welcome. – Brian M. Scott Nov 20 '20 at 05:57
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Hint:

Let $x_r=\tan(y_r)$

$$x_{n+1}=\tan(y_{n+1})=\tan\left(\dfrac\pi4+y_n\right)$$

$$\implies\tan(y_{n+r})=\tan(y_n)\text{ if }4\mid r$$

and $$\tan(y_{n+r})=-\dfrac1{\tan(y_n)}\text{ if }r\equiv2\pmod4$$

lab bhattacharjee
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If you look here, ou will find how to solve the first-order rational difference equation in the more general case.

For $x_0=a$, this would give after simplifications

$$x_n=\frac{a \cos \left(\frac{\pi n}{4}\right)+\sin \left(\frac{\pi n}{4}\right)}{\cos \left(\frac{\pi n}{4}\right)-a \sin \left(\frac{\pi n}{4}\right)}$$

Claude Leibovici
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