A simple recursion for 2016 and 2018!

Algebra Level 4

$\large x_{n+1} = \dfrac{1+x_n}{1-x_n}$ Consider the recurrence relation above with $x_0 = 2016$ . Compute $\lfloor 10000 | x_{2018} | \rfloor$ .

2 solutions

Yong See Foo
Jan 6, 2016

The function $f(x)=\frac{1+x}{1-x}$ looks very suspicious, indeed quick algebra shows that

$f^2(x)=\frac{1+\frac{1+x}{1-x}}{1-\frac{1+x}{1-x}}=\frac{1-x+1+x}{1-x-(1+x)}=-\frac{1}{x}.$

From here it is then easily seen that $f^4(x)=x$ by applying the above line twice, and thus

$x_{2018}=f^{2018}(2016)=f^2(2016)=-\frac{1}{2016}$ and the required value is $\lfloor\frac{10000}{2016}\rfloor=\boxed{4}$ .

For completeness, we have to check that the values are valid. For example, $f^2 (x)$ is not defined for $x = 1, 0$ .

But otherwise, it's good to recognize the cyclicality of such transforms.

Calvin Lin Staff - 5 years, 5 months ago

Aakash Khandelwal
Mar 5, 2016

x(n+1)*x(n-1)=-1. Either a term is x(0) or -1/x(0). Accordingly we have x(2018) as -1/x(0). Please note that the term in (.) is in subscript.