About generalized continued fractions

Question

Let us consider the sequences $(x_n), (a_n)$, starting with $n=0$ and $x_0\in ]0,1[$, defined by the following generalized Gaussian map: $$x_{n+1}=\frac{\lambda_n}{x_n^{\alpha_n}}-\Big\lfloor \frac{\lambda_n}{x_n^{\alpha_n}}\Big\rfloor, \mbox{ } \mbox{ } \mbox{ } a_n=\Big\lfloor\frac{\lambda_n}{ {x_n^{\alpha_n}}}\Big\rfloor$$

where $\lambda_0,\lambda_1\dots$ and $\alpha_0,\alpha_1\dots$ are positive real numbers. For convergence and to make sense, they must satisfy some conditions not discussed here. Continued fractions correspond to $\lambda_0=\lambda_1\dots=1$ and $\alpha_0=\alpha_1\dots=1$. Generalized continued fractions correspond to $\alpha_0=\alpha_1\dots=1$.

The number $x_0$ can be reconstructed using the following infinite backward recursion, starting with a large $n=N$ and replacing $x_{N+1}$ by $0$, and then iteratively going down to $n=0$. $$x_{n}=\Big(\frac{\lambda_n}{a_n+x_{n+1}}\Big)^{1/\alpha_n}$$ Convergence is very fast, so even $N=30$ will give you more than 15 digits of accuracy for $x_0$, depending on the parameters.

This is the same scheme used to compute $x_0$'s continued fraction, and it will produce the same traditional continued fraction if all $\alpha_n$'s are equal to $1$. Here I assume $\alpha_n=\alpha, \lambda_n=\lambda$: they do not depend on $n$.

My question

The behavior of $a_n$ (these are the $[a_0,a_1,\dots]$ in the continued fraction expansion of $x_0$) is very chaotic for most numbers, but if you replace $\alpha$ by a value lower than $1$, it seems that the coefficients $a_n$ are bounded, or at least their expected value is not infinite anymore. The equilibrium distribution for $(x_n)$ if $\alpha=\lambda=1$ is well known and equal to $$F_X(x)=\log_2(1+x), \mbox{ } \mbox{ } \mbox{ } \mbox{ } 0\leq x \leq 1.$$ All this relate to ergodic, chaotic dynamical systems associated with numeration systems, the base-$10$ system being based on the tenfold map, while the system described here is based on a generalization of the Gauss-Kuzmin-Wirsing operator (see here and here). The coefficients $a_n$'s are called the digits.

I am wondering is this could lead to interesting results about approaching irrational numbers by other rational or irrational numbers. For instance, if $\lambda=1,\alpha=9$ and $x_0=\pi-3$, you get, with just one iteration,

$$\pi-3\approx (43715304)^\frac{1}{9},$$

correct to $9$ digits. Also, what are the conditions for convergence? What are the conditions on $\alpha,\lambda$ so that the $a_n$'s are bounded or have finite expectation? Is there a simple expression for $F_X(x)$ or its derivative (the density) $f_X(x)$ when $\lambda\neq 1$ or $\alpha\neq 1$?

Convergents

To simplify the notation, let $\mu_k=(\lambda_k/a_k)^{1/\alpha_k}$. Here is how the three first convergents of $x_0$ are computed if you only know the digits $a_0,a_1\dots$ and try to reconstruct $x_0$:

$$\mu_0$$

$$(\lambda_0/(a_0+\mu_1))^{1/\alpha_0}$$

$$(\lambda_0/(a_0+(\lambda_1/(a_1+\mu_2))^{1/\alpha_1}))^{1/\alpha_0}$$

Update on 2/17/2021: I provided a way to iteratively solve the functional equation to find the equilibrium density, see the last section in my answer.

Answer 1

My answer applies to standard continued fractions $x_0=[a_0,a_1,\dots]$ corresponding to $\alpha=1$, as well as to the full general case. As stated in my original question, the $a_n$'s are called the digits of $x_0$ in the continued fraction system (actually, continued power if $\alpha\neq 1$). The following is true for all normal numbers $x_0$ in that system, with the definition of normality being a generalization of the concept as we know it in base $10$ or base $2$. All numbers $x_0\in[0,1]$ but a set of Lebesgue measure zero, are normal in the continued power numeration system. Yet the set of non-normal numbers is dense.

That said, let's now answer my question. The theoretical distribution $D$ of any digit $a_n$ of any normal number $x_0$ in the power system, satisfies the following:

1. It is unbounded, that is, $P(D=k)>0$ for any strictly positive integer $k$
2. $P(D=k) \sim O(k^{-(1+1/\alpha)})$ as $k\rightarrow\infty$
3. $E(D)$ is infinite if and only if $\alpha\geq 1$
4. $Var(D)$ is infinite if and only if $\alpha\geq \frac{1}{2}$

The correlation between successive digits thus exists if $\alpha<\frac{1}{2}$.

Proof of the main result (sketch)

Let $X$ be the random variable associated with the equilibrium distribution of the sequence $(x_n)$. We have

$$P(D=k)=P(m_{k+1}<X\leq m_k)=\int_{m_{k+1}}^{m_k} f_X(x)dx \sim O(m_k-m_{k+1})$$

where $m_k=(\lambda/k)^{1/\alpha}$. Since $f_X(x)$ is bounded, smooth, and away from zero in $[0,1]$, results 1 and 2 follow easily. The asymptotic result is based on the fact that $f_X$ is quite close to a uniform density on $[0,1]$.

We also have

$$E(D)=\int_0^1 \frac{\lambda}{x^\alpha}f_X(x)dx - \int_0^1 \Big\{\frac{\lambda}{x^\alpha}\Big\}f_X(x)dx$$ where the curly brackets represent the fractional part function. The rightmost integral is finite, and the left one is infinite only if $\alpha\geq 1$. Now we proved the third result. A similar argument can be used to prove the last result. Note that for continued fractions ($\alpha=\lambda=1$) the distribution of $D$ is known exactly, and verifies all the properties 1-4 listed above.

Functional equation for $f_X$

Let $x'$ be a back image of $x$, in the sense that if $x_{n+1}=x$, then $x_{n}=x'$. For any $x\in [0,1]$, there are infinitely many back images $x_k'$ indexed by $k=1,2,\dots$ and denoted as

$$x_k'=g_k(x) = \Big(\frac{\lambda}{x+k}\Big)^{1/\alpha}.$$

Now the back image of the whole interval $[x,x+\epsilon]$ (with $\epsilon$ very close to $0$) is $$\bigcup_{k=1}^\infty \Big[g_k(x+\epsilon),g_k(x)\Big].$$ The length of the $k$-th interval in the above formula is approximately $\epsilon g'_k(x)$, where $g'$ denotes the derivative with respect to $x$. Thus, we have the following fundamental functional equation: $$f_X(x)=\sum_{k=1}^\infty |g'_k(x)|\cdot f_X(g_k(x)) .$$

In particular, if $\alpha=\lambda=1$ (the standard continued fraction), it is easy to verify that $f_X(x)=\frac{1+x}{\log 2}$ satisfies the above formula and is thus the density associated with the equilibrium distribution. This is a well known result, and the proof provided here is probably one of the shortest ones.

Solving the functional equation

We can use the above formula to implement a fixed point algorithm that converges to the density $f_X$, starting with $n=0$ and a uniform density $f_0$ on $[0, 1]$, and using the following recursion:

$$f_{n+1}(x)=\sum_{k=1}^\infty |g'(x)|\cdot f_n(g_k(x), \mbox{ } \mbox{ } \mbox{ } 0\leq x \leq 1. $$

In practice, using the first few hundreds terms in the infinite sum, and using a few thousand values of $x$ evenly spread in $[0, 1]$, provides a good approximation to $f_X(x)$ in only $3$ or $4$ iterations. In the cases tested, convergence occurred when $\lambda\leq 1$. The chart below shows the first $4$ iterations with $\lambda=\alpha=1$. In that case, the theoretical $f_X$ is known and represented by the red curve in the figure below. The black line is the initial uniform density $f_0$ on $[0, 1]$. The green curve corresponds to $f_1$, the grey one to $f_2$, and the orange one to $f_3$. The orange one is very close to $f_X=f_\infty$: one can barely distinguish it from the red curve.

A Perl script for the fixed point iterations is available here. Note that there are infinitely many solutions to the functional equation (multiples of each other), but there is only one that is a density, that is, integrating to $1$.