1
$\begingroup$

$u_{1}=\alpha$ 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 $\alpha$?

$\endgroup$
2
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ That explains! Thanks $\endgroup$
    – LanaDR
    Aug 13 '20 at 1:03
1
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ 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 $\sqrt{15} = \sqrt{23 - 10 \sqrt{5} - 2 \sqrt{3 (85 - 38 \sqrt{5})}}$. $\endgroup$ Aug 25 '20 at 19:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.