TITLE: Closed-form for recursive "geometric-like" recursion
QUESTION [0 upvotes]: I asked this question of MSE, but to no avail; alas, here I am.
Let $k>0$, $C\geq 1$, $\alpha \in (0,1]$, and let $(x_n)_{n\geq 1}$, be a sequence of real numbers given by the recursion
$$
x_{n+1} = k |C|^{\alpha} + |x_{n}|^{\alpha} \qquad x_0=0.
$$
Is there a simple "non-recursive" expression for $x_{n}$, for $n>0$?  At the very worst, is there a tight-upper bound for $x_n$ only depending on $C,k,$ and $\alpha$?  I was trying to obtain an upper-bound by the geometric sums $\sum_{i=1}^n{k C}^{\alpha}$ but I'm no longer convinced this is correct.

REPLY [1 votes]: It seems extremely unlikely that a simple "non-recursive" expression for $x_n$ is possible. However, let us obtain an exact upper bound on the $x_n$'s.
Let $a:=\alpha\in(0,1]$ and $b:=k|C|^a\in(0,\infty)$. It is clear that $x_n\ge0$ for all $n$. So,
$$x_0=0,\quad x_1=b,\tag{1}$$
and
$$x_{n+1}=g(x_n)\tag{2}$$
for $n\ge1$, where $g(u):=b+u^a$. To avoid trivialities, assume that $a\ne1$, so that $a\in(0,1)$. Then $h(u):=g(u)-u$ is concave in real $u\ge0$, with $h(0)=b>0$ and $h(\infty-)=-\infty$. So, there is a unique root $u_b\in(0,\infty)$ of the equation $h(u)=0$ and, moreover, $h\ge0$ on the interval $[0,u_b]$, that is,
$$g(u)\ge u\quad \forall u\in[0,u_b],\tag{2.5}$$
whereas
$$b+u_b^a=g(u_b)=u_b.\tag{2.75}$$
In particular, it follows that
$$u_b\ge b.\tag{3}$$
Let us show, by induction on $n\ge0$, that
$$x_n\le u_b\tag{4}$$
for all $n\ge0$. Indeed, by (1) and (3), (4) holds for $n=0,1$. Supposing now that (4) holds for some $n\ge1$, we have
$$x_{n+1}=g(x_n)\le g(u_b)=u_b,$$
in view of (2) and because the function $g$ is increasing. So, (4) holds for all $n\ge0$.
Now (2), (2.5), and (4) imply
$$x_{n+1}=g(x_n)\ge x_n.$$
So, $(x_n)$ is a nondecreasing sequence in $[0,u_b]$. Therefore and in view of (2) and (2.75),
$$x_n\uparrow u_b$$
as $n\uparrow\infty$.
Thus, in view of (4), $u_b$ is the best upper bound on the $x_n$'s.

If $a$ is rational, then the unique root $u_b$ of the equation $u_b=g(u_b)$ will be algebraic in $b$. E.g., if $a=1/2$, then $u_b=(1+2 b+\sqrt{1+4 b})/2$.
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