If $a$ is a sequence defined recursively by $a_{n+1} = \frac{a_n-1}{a_n+1}$ and $a_1=1389$ then can you find what $a_{2000}$ and $a_{2001}$ are?
it would be really appreciated if you could give me some pointers on how to solve this mathematically.
(Obviously using the formula to find the values one by one is possible but is extremely tedious, actually i wrote a simple piece of software to calculate it and the answers were a little odd, $a_{2000}=-1.0014409$ and $a_{2001}=1389.0381$)
By the way here is the code i used for calculating the answers (it's in java) :
float answer = 1389;
for (int i = 2; i <= 2001; i++) {
answer = (answer - 1) / (answer + 1);
}
(i ran it two times the first time i set the loop condition to 2000.)
The fixed points of the recursion $$a\to\frac{a-1}{a+1},$$ are $\mathrm i$ and $-\mathrm i$. Even if the whole sequence $(a_n)$ is real valued, this suggests to look at the dynamics of the modified (complex) variable $$b_n=\frac{a_n-\mathrm i}{a_n+\mathrm i}.$$ Behold! It happens that $$b_{n+1}=-\mathrm i\,b_n,$$ hence $b_{n+1}=(-\mathrm i)^nb_1$ for every $n$. Furthermore, $(-\mathrm i)^4=1$ hence the sequences $(a_n)$ and $(b_n)$ have period $4$.
In particular, $a_{2001}=a_1=1389$. It remains to note that $$\frac{a_4-1}{a_4+1}=a_5=a_1,$$ to deduce $$a_{2000}=a_4=-\frac{a_1+1}{a_1-1}=-\frac{695}{694}.$$
Edit: Likewise, the recursion $$a\to\frac{a-u}{a+v},$$ is periodic with period $n$ for every starting point if and only if there exists some integer $k$ such that $$\cos^2\left(\frac{k\pi}n\right)=\frac14\frac{(1+v)^2}{u+v}.$$ If $u=v=1$, $n=4$ and $k=1$ solve this.