Defining "$a\text{ in base }b$" as
$$a_b=\sum^{\infty}_{k=0}\frac{a_{[k]}}{b^k}$$
with $a_{[k]}$ being the digits of $a\in (1,10)$ and $b\gt 1$. For example, if $a=3.1416$ and $b=4.6$, then
$$a_b=\frac{3}{4.6^0}+\frac{1}{4.6^1}+\frac{4}{4.6^2}+\frac{1}{4.6^3}+\frac{6}{4.6^4}\approx 3.43010138$$
Now we define a sequence, placing brackets to indicate order of operations:
$$S(a)=(a, a_a, a_{(a_a)}, a_{(a_{(a_a)})}, ...)$$
The most general (and seemingly quite difficult) question that comes to mind is: Given an $a\in (1,10)$, what is the long term behavior of $S(a)$?
A (maybe) simpler question, that I also pondered is: If $a_{[k]}$ is finite, does that imply, that $S(a)$ doesn't diverge?
Behavior of $S(a)$ for numbers $a$ between $1$ and $2$. For each $a\in\left\{1.000, 1.001, 1.002, ..., 1.998, 1.999 \right\}$ the entries of $S(a)$ in positions $10^6$ and $10^6-1$ are computed and drawn as circles in the picture, they are then connected with a line.
More info
If $a=2$, it is easy to see, that $S(a)$ stays fixed at $2$. Indeed for any integer $a\in(1,10)$, it holds, that $S(a)$ stays fixed at $a$.
Other curious examples are
- For $a = 1.1$, the sequence converges to $\frac{1+\sqrt{5}}{2}=\phi$
- For $a = 1.04$, the sequence converges to $2$
A few other things can happen, for particular values of $a$. For example if $a=1.05$, then $S(a)$ doesn't converge to a single value, but to a two-term limit cycle. (and interestingly, the sum of the two terms is $5$).
It has been found in [1], that only one of five things can happen:
- $a$ is a fixed point (as with $a=2$)
- $S(a)$ converges to a single value (as with $a=1.1$)
- $S(a)$ falls into a two-term cycle (as with $a=1.\overline{1}$)
- $S(a)$ converges to a two-term limit cycle (as with $a=1.05$)
- $S(a)$ diverges, alternately approaching $1$ and $\infty$ (as with $a=1.\overline{01}$)
Background
Watched a video on YouTube called "Dungeon Numbers (extra) - Numberphile". I got obsessed with it for a while and tried doing something similar for complex numbers some years back. I came back to the problem again and felt it was solvable not using too crazy techniques.
My attempt
(EDIT: This doesn't seem to work quite right as it breaks for $a=1.\overline{1}$ as will be shown below)
After many days of trying things, I finally came up with the following: Given $a\in(1,2)$ define the function:
$$\mathcal{F_a}(x)=\sum_{k=0}^{\infty}\frac{a_{[k]}}{x^k}$$ , consider the roots of $$\mathcal{D_a}(x)=\mathcal{F_a}(x) - \mathcal{F_a}^3(x)$$
(EDIT: After some more fiddling, I realized $\mathcal{D_a}(x)$ can be replaced with $\mathcal{F_a}^2(x)-x$, which makes things somewhat easier)
Where $\mathcal{F_a}^3(x)$ stands for $\mathcal{F_a}\left(\mathcal{F_a}(\mathcal{F_a}(x))\right)$. The number $3$ may be exchanged for any odd number greater than it and the following seems to hold:
Except for the trivial case, when $a$ is an integer, the roots of $\mathcal{D_a}(x)$ describe the long term behavior of $S(a)$ in this way:
- If $S(a)$ converges to a single number $x_0$, then $\mathcal{D_a}(x)$ has a single root in $\mathbb{R}^+$. This root is $x_0$
- If $S(a)$ converges to the two-term limit cycle $(x_0, x_2)$, then $\mathcal{D_a}(x)$ has three roots in $\mathbb{R}^+$. These roots are $x_0, x_1, x_2$, with $x_0\lt x_1 \lt x_2$
I'm not sure, if this holds outside the interval $(1,2)$, or if it even holds inside that interval, but everything I tested seemed to agree with it. I don't understand at all why this works though, so I'm struggling to prove things. Any help would be much appreciated.
I came up with this method when playing around with iterating the function $\mathcal{F_a}(x)$ and plotting the results. I noticed, that all the plots seemed to intersect and certain points and found that those were exactly the values, that $S(a)$ would converge to
Example Calculations:
$a=1.1$
$$\mathcal{F_a}(x)=1+\frac{1}{x}=\frac{x+1}{x}$$ $$\mathcal{F}_a^3(x)= 1+\frac{1}{1+\frac{1}{1+\frac{1}{x}}} = \frac{3x+2}{2x+1}$$ Finding positive roots of $\mathcal{D_a}(x)$ (intersection of $\mathcal{F_a}$ and $\mathcal{F}_a^3$) $$\frac{3x+2}{2x+1}=\frac{x+1}{x}$$ $$x^2-x-1$$ $$x=\frac{1}{2}+\frac{\sqrt{5}}{2}$$
$a=1.28$
$$\mathcal{F}_a(x)=1+\frac{2}{x}+\frac{8}{x^2}$$
$$\mathcal{F}_a^3(x)=1+\frac{2}{\left(1+\frac{2}{\left(1+\frac{2}{x}+\frac{8}{x^2}\right)}+\frac{8}{\left(1+\frac{2}{x}+\frac{8}{x^2}\right)^2}\right)}+\frac{8}{\left(1+\frac{2}{\left(1+\frac{2}{x}+\frac{8}{x^2}\right)}+\frac{8}{\left(1+\frac{2}{x}+\frac{8}{x^2}\right)^2}\right)^2}$$
Using WolframAlpha, it is found, that the roots of $\mathcal{D_a}(x)$ in $\mathbb{R}^+$ are:
- $x_0 = 2$
- $x_1 = \frac{1}{3}\left(1+\sqrt[3]{118+3\sqrt{1509}}+\sqrt[3]{118-3\sqrt{1509}}\right)\approx 2.7673$
- $x_2 = 4$
and indeed, when doing the iteration, the sequence does converge to the limit cycle $(2,4)$
$a=1.\overline{1}$
This doesn't work, because
$$\mathcal{F_a}(x)=\sum_{k=0}^{\infty}\frac{1}{x^k}=\frac{x}{x-1}$$ $$\mathcal{F}_a^3(x)=\frac{\frac{\frac{x}{x-1}}{\frac{x}{x-1}-1}}{\frac{\frac{x}{x-1}}{\frac{x}{x-1}-1}-1} = \frac{x}{x-1} = \mathcal{F_a}(x)$$ $$\implies \mathcal{D_a}(x) \equiv 0$$ But $$S(a)=(1.\overline{1}, 10, 1.\overline{1}, 10, ...)$$
$a=1.1110000099$
This case is also an exception, but it doesn't break completely. Since The formulas get a little unwieldy, the roots of (using the new method) $\mathcal{D_a}(x)=\mathcal{F_a}^2(x)-x$ were found using maple. In this case there are five of them.
When iterating $a$, it converges to a two-cyle: $(1.1077.., 10.2682..)$. These match the first and the fifth root of $\mathcal{D_a}(x)$.
I'm not quite sure how much this adds, but here are three cobweb plots for $a=1.03, a=1.04, a=1.05$. The curves are the functions $\mathcal{F_a}$. The origin-line is $y=x$. In the plot for $1.03$ we see a nice and quick convergence. In the plot for $1.04$ we see much slower convergence - this is reflected by the fact, that $\mathcal{F^2_a}(x)-x$ has a higher multiplicity root, than in the case for $1.03$. Finally, in the third plot, we see a convergence to a two term limit cycle.
Any and all comments, tips or critiques are very much appreciated.
