# Closed-form for recursive “geometric-like” recursion

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.

• What is the significance of $f$? You don't seem to use it again. – Christian Remling Dec 6 '20 at 19:57
• Please include a link to the question on m.se, and include a link there to the question here. – Gerry Myerson Dec 6 '20 at 21:20
• @GerryMyerson I look down the MSE question. When I migrted it here – Wasserstein's Apprentice Dec 7 '20 at 7:08

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 $$u_b=(1+2 b+\sqrt{1+4 b})/2$$.