# Limit involving iterated function $f_a(x)=x^2+a^2$

I have long ago give up trying to find a nice formula for the $$n$$th iteration of functions in the form $$f_a(x)=x^2+a^2$$ However, it would be interesting to consider the asymptotic growth of the iteration of these functions. It is apparent that $$f_a^{\circ n}(x)$$ should behave like $$\text{b}^{2^n}$$, but calculating the value of the base $$b$$ in terms of $$a$$ and $$x$$ has been more difficult than I anticipated. We may solve for $$b$$ using a limit: $$b_a(x)=\lim_{n\to\infty} \big[f_a^{\circ n}(x)\big]^{2^{-n}}$$ but this is as far as I have gotten, and I can’t make this any neater. The best thing I have managed to find is the rather trivial observation $$b_a(x^2+a^2)=b_a^2(x)$$ and I have noticed that the graph of $$b_a(x)$$ typically looks like (but is not) a hyperbola.

Can anyone calculate any special values of $$b_a(x)$$ in closed form, or in terms of an infinite series or integral? Can you find any more interesting (non-trivial) functional or differential equations?

• To get some idea of how complicated this can get, see oeis.org/A000058 and the links therein, for $f(x)=x^2-x+1$. Or oeis.org/A003095 for $f(x)=x^2+1$. – Gerry Myerson Mar 2 at 4:03
• Had a look at those sites? Convinced? – Gerry Myerson Mar 3 at 5:33

## 1 Answer

From Wikipedia, it states that the $$n$$th iteration of $$ax^2+bx+\frac{b^2-2b-8}{4a}$$ is $$\frac{2c^{2^n}+2c^{-2^n}-b}{2a}$$ where $$c=\frac{2ax+b\pm\sqrt{(2ax+b)^2-16}}{4}$$

Thus, we can solve the case $$f(x)=x^2-2$$.

$$b_{\pm 2i}(x)=\frac{x+\sqrt{x^2-4}}2$$

• Yes, but that's a very special case. In the general quadratic case, there's no closed-form formula. – Gerry Myerson Mar 2 at 4:01