Asymptotic behavior of the quadratic recurrence $x_n=x_{n-1}^2+c$.

Question

Let $c\in\mathbb{R}^+$. I am looking for a sequence $\{y_n\}$ that asymptotically (and if possible tightly) upper bounds the recurrence $$x_n=x^2_{n-1}+c.$$ I would to like to write $y_n$ as a function of $n$, $x_0$, and $c$.

Answer 1

The question is about the asymptotic behavior of $\,x_n = x_{n-1}^2+c\,$ given any $\,x_0\ge0\,$ and $\,c\ge0.\,$ A simple example is OEIS sequence A003095 where $\,x_0=0,\;c=1.\,$ The general answer depends on $$ f_c(x) := \frac1{2x} \!-\! \sum_{n=0}^\infty a_n(c) x^{2n+1} = \frac1{2x}\\ - (c)x - (2c\!+\!c^2) x^3 - (4c^2\!+\!2c^3)x^5 - (16c \!+\! 12c^2 \!+\! 12c^3 \!+\! 5c^4)x^7 - \dots$$ where $\,a_n(1)\,$ is OEIS sequence A088674. This function satisfies the equations $$ f_c(x)^2 + c = f_c(2x^2) \quad\text{ and }\quad f_c(x) = \sqrt{f_c(2x^2)-c}. $$ The second equation iterated determines the coefficients $\,a_n(c)\,$ as polynomials in $\,c.\,$ Thus, $$ x_n \simeq y_n := f_c\big(d^{-2^n}/2\big) \;\; \text{ where }\;\; d := \lim_{n\to\infty} x_n^{(2^{-n})} $$ is the asymptotic expansion of $\,x_n.\,$