# Recurrence relation with a period of 4

$$u_{1}=\alpha$$ and $$u_{n+1}=\frac{1+u_n}{1-u_n}$$. Prove that $$n=5$$ gives the first value of $$n$$ for which $$u_{n}=\alpha$$ and that this is so for all but three values of $$\alpha$$

So I have shown that $$u_2=\frac{1+\alpha}{1-\alpha}$$,$$u_3=-\frac{1}{\alpha}$$,$$u_4=\frac{\alpha-1}{\alpha+1}$$ and finally $$u_5=\alpha$$ and we need $$\alpha\ne0,-1,1$$.

I know I am also supposed to check that $$u_2,u_3,u_4$$ can never be $$\alpha$$.

But the suggested answer says "...need to consider cases where a term prior to $$u_5$$ could be equal to $$u_1$$ and this requires scrutiny of both $$u_2$$ and $$u_3$$ but not $$u_4$$..."

I don't understand why "...but not $$u_4$$..."? Is there a need to check if $$u_4$$ can be $$\alpha$$?

If $$u_4=u_1$$ then $$u_5=u_2$$, $$u_6=u_3$$ etc, so that the sequence repeats with period $$3$$ as well as period $$4$$. But in this case $$u_4=u_1$$ and we know $$u_5=u_1$$ also. As $$u_5=u_2$$ then $$u_2=u_1$$ and you have ruled out this case.
You can check periodicy also like this. Define $$\alpha _n = \arctan(u_n)$$ then we have $$\tan (\alpha _{n+1}) = {\tan {\pi \over 4} + \tan (\alpha _n)\over 1- \tan {\pi \over 4}\tan (\alpha _n) } = \tan ({\pi \over 4} + \alpha _n)$$ so $$u_{n+4} =\tan (\alpha _{n+4}) =\tan (\pi+ \alpha _{n}) = u_n$$ and we are done.
• This also shows that none of the intermediate values are equal. It also suggests looking at $a_{n+1} = \tan(a_n+\pi/m)$ for integer m to create a recurrence in disguise. Use, for example, $\tan(\pi/3) = \sqrt{3}$ or, according to Wolfy, $\tan(\pi/5) = \sqrt{5-2\sqrt{5}}$ and $\sqrt{15} = \sqrt{23 - 10 \sqrt{5} - 2 \sqrt{3 (85 - 38 \sqrt{5})}}$. Aug 25 '20 at 19:09