TITLE: Explicit Form for $a_{n+1}=\sqrt{ka_n+l}$?
QUESTION [7 upvotes]: I found the explicit form for the case $k=1, l=2, a_1=\sqrt{2}$ (because it was on my midterm test...)
Here's how: 

First, let us consider the first few terms to find a pattern. $$a_1=\sqrt2=2\cos{\frac{\pi}{2^2}}$$$$a_2=\sqrt{2+\sqrt2}=2\cos{\frac{\pi}{2^3}}$$$$a_2=\sqrt{2+\sqrt{2+\sqrt2}}=2\cos{\frac{\pi}{2^4}}$$
  From this, I predicted that $a_n=2\cos{\frac{\pi}{2^{n+1}}}$ and using mathematical induction, it was proven to be true.

So, my question is, is this case really special, or is there some kind of a way of solving for the explicit form for $a_{n+1}=\sqrt{ka_n+l}$?
(That form is used very often as examples of Monotone Convergence Theorem, Fixed Point Iteration Method...etc, so I just got curious.)

REPLY [1 votes]: Not a full analysis, but a generalization of your observations to different initial conditions, and to another special case. If you take $a_0$ between $-2$ and $2$, you can use the fact that
$$a_0=2\cos \arccos\frac{a_0}{2}$$
to prove by mathematical induction (the proof is straightforward, so I'll omit it) that
$$a_n=2\cos\big(2^{-n}\arccos\frac{a_0}{2}\big)$$
However, if you want to use $a_0\gt 0$, you can use the hyperbolic cosine function, since it has the same half-angle formula as the cosine function. So, if $a_0\gt 2$,
$$a_n=2\cosh\big(2^{-n}\operatorname{arccosh} \frac{a_0}{2}\big)$$
If you instead let $k=l=1/2$, you end up with something similar:
$$a_n=\cos\big(2^{-n}\arccos\frac{a_0}{2}\big)$$
for $|a_n|\le 1$, and
$$a_n=2\cosh\big(2^{-n}\operatorname{arccosh} \frac{a_0}{2}\big)$$
for $a_n\gt 1$.
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