I was working on an exploratory activity for one of my graduate courses, and we were asked to see what we may notice about functions when they are composed with themselves an infinite number of times.
The examples given were $f(x) = ax^2$, $g(x) = a\sqrt{x}$, and $h(x) = ax^3$, where $a$ would be a parameter we change and our $x$ would be pulled from a sequence of seeds $\{x_0, x_1, \dots\}$ such that $x_i \in \mathbb{R}$ and $x_i = f(x_{i-1})$ for all $i \geq 1$.
As an example (so that my question may be more easily understood), we first calculated the infinite composition of $f(x)$ when $a = 1, 2, 3$ and $x_0 = 0.5$. We found that the compositions behaved differently based on their seed value and after plugging in various other values for $a$ and trying some different seed values, we found the following relationship to hold true:
$$ f(x) \begin{cases} -\infty & \text{ if } |ax|>1 \text{ and } a<0 \\ -x & \text{ if } ax=-1 \\ 0 & \text{ if } |ax|<1 \\ x & \text{ if } ax=1 \\ \infty & \text{ if } |ax|>1 \text{ and } a>0 \\ \end{cases}$$ We then had to prove Algebraically this relationship was true and found that the following series represented our infinite composition and held our relationship true. $$x_i = a^{2^{i+1}-1} x^{2^{i+1}}$$
Now, we were asked to choose functions and explore them and one of the functions I thought might be interesting to explore was the one in the title, $F(x)=ae^x$.
After starting my work, however, I realized how erratic its behavior was regardless of the seed and parameters I chose. So far, I have come up with the hypothesis that the relationship is as follows:
- If $a < -e$, then $F(x)$ oscillates between two values: one, (we will denote $\alpha$) which $ \alpha \approx a$ (very close in larger cases, less so as $a \rightarrow e$) and another (we will call $\beta$) which is on a point such that there exists some line with slope $m =1$ connecting the two points $(\alpha, F(\alpha)$ and $\beta, F(\beta)$.
- If $a = -e$, $F(x)$ will converge to $-1$.
- If $0 > a > -e$, then $F(x)$ converges to some specific decimal(i.e., when $a = -\frac{1}{10}$, $F(x)$ would always converge to $-0.091277$, regardless of the $x_0$ I initially chose.)
- If $a=0$, then $F(x) = 0$.
- If $\frac{1}{e} > a > 0$, then $F(x)$ converges to some specific decimal(I was unable in the time I spent working on this to discover exactly what values these related to, but I did find that regardless of the seed, each parameter in this scenario would converge to the exact same one every time. i.e., when $a = \frac{1}{10}$, $F(x)$ would always converge to $0.111833$, regardless of the $x_0$ I initially chose.)
- If $a = \frac{1}{e}$, then $F(x)$ would always converge to $1$.
- If $a > \frac{1}{e}$, then $F(x)$ would always converge to $\infty$
Note how for $0 < |a| < |e|$ how the values that $a$ and $-a$ converge to almost seem to add up to $|a|$. Like how $|-0.091277| + |0.111833| \approx 0.2$. (I do not know what this means but I was taking anything at this point!)
This whole thing has brought me down a rabbit hole I want to explore more (I've already read up on Infinite Compositions of Analytic functions, tons of stuff on Taylor Series, and I know this has convergence and limits tied into it, along with who knows what else) but I really wanted to ask the community whether or not this function composition could be represented AS a series of some sort, whether geometric, Taylor, analytic, etc.
TL:DR - Can the infinite function compositions of $F(x) = ae^x$, where $a \in \mathbb{R}$ is a parameter and $x$ is in the series $\{x_0, x_1, \dots\}$ such that $x_i \in \mathbb{R}$ and $x_i = F(x_{i-1})$ for all $i \geq 1$, be represented algebraically as some series? If so, what would it be, what kind of series would it be, and how would one go about finding it?
NOTE: I already turned in my assignment and what I had done was more than sufficient for the activity. Now it's just my curiosity that's getting the better of me.
Any help is appreciated. Thank you!