On arithmetical nature of Tichy-Uitz's function

TL;DR

This paper investigates the arithmetical properties of a specific singular function from a family introduced by Tichy and Uitz, focusing on the case where the parameter is related to the golden ratio.

Contribution

It provides a detailed analysis of the arithmetical nature of the Tichy-Uitz function for a particular parameter value, extending understanding of its properties.

Findings

Identifies the arithmetical characteristics of g_λ for λ=(3−√5)/2

Connects the function's properties to number theory and algebraic structures

Enhances understanding of singular functions related to continued fractions

Abstract

R.F.Tichy and J.Uitz introduced a one parameter family $g_{\lambda}$, $\lambda \in (0,1)$, of singular functions. When $\lambda=1/2$ the function $g_{\lambda}$ coincides with the famous Minkowski question mark function. In this paper we describe the arithmetical nature of the function $g_{\lambda}$ when $\lambda = \frac{3-\sqrt{5}}{2}$.

Figures

Equations

$${\mathcal{F}}_n=\left\{0=x_{0,n}<x_{1,n}<\dots <x_{N(n),n}=1\right\},$$

$${\mathcal{F}}_{n+1}={\mathcal{F}}_n\cup Q_{n+1}$$

$$Q_{n+1}=\left\{x_{j-1,n}\oplus x_{j,n},j=1,\dots ,N(n)\right\}.$$

$$\frac{a}{b}\oplus \frac{c}{d}=\frac{a+b}{c+d}$$

$$\xi =[0;a_1,a_2,\dots ,a_m]=0+\frac{1}{a_1+\frac{1}{a_2+\dots +\frac{1}{a_m}}},a_j\in \mathbb{N},a_m⩾2.$$

$$S(\xi ):=a_1+...+a_m=n+1.$$

$$g_{\lambda }(0)=g_{\lambda }(0/1)=0,g_{\lambda }(1)=g_{\lambda }(1/1)=1.$$

$$g_{\lambda }(x_{j-1,n}\oplus x_{j,n})=g_{\lambda }(x_{j-1,n})+\left(g_{\lambda }(x_{j,n})-g_{\lambda }(x_{j-1,n})\right)\lambda .$$

$$?(0)=?(0/1)=0,?(1)=?(1/1)=1.$$

$$?(x_{j-1,n}\oplus x_{j,n})=\frac{?(x_{j-1,n})+?(x_{j,n})}{2}.$$

$$x=[0;a_1,a_2,\dots ,a_m,\dots ]=\frac{1}{a_1+\frac{1}{a_2+\dots +\frac{1}{a_m+\frac{1}{\dots }}}},$$

$$?(x)=\frac{1}{2^{a_1-1}}-\frac{1}{2^{a_1+a_2-1}}+\frac{1}{2^{a_1+a_2+a_3-1}}-\dots$$

$$?(x)=\underset{n\to \infty }{\lim }\frac{♯\left\{\xi \in {\mathcal{F}}_n:\xi ⩽x\right\}}{♯{\mathcal{F}}_n}=\underset{n\to \infty }{\lim }\frac{♯\left\{\xi \in {\mathcal{F}}_n:\xi ⩽x\right\}}{2^n+1}.$$

$$\begin{gathered} g_{\lambda }(x)={\lambda }^{a_1-1}-{\lambda }^{a_1-1}(1-\lambda {)}^{a_2}+{\lambda }^{a_1-1}(1-\lambda {)}^{a_2}{\lambda }^{a_3}-\dots + \\ +(-1{)}^{m+1}{\lambda }^{\sum _{\left(1⩽i⩽m,i\equiv 1\mathrm{mod}2\right)}{a}_i-1}(1-\lambda {)}^{\sum _{\left(1⩽i⩽m,i\equiv 0\mathrm{mod}2\right)}{a}_i}+\dots . \end{gathered}$$

$$g_{\lambda }(0)=0,g_{\lambda }(1)=1$$

$$g_{\lambda }(x_{j-1,n}\oplus x_{j,n})=g_{\lambda }(x_{j-1,n})+\left(g_{\lambda }(x_{j,n})-g_{\lambda }(x_{j-1,n})\right)\lambda ,$$

$$g_{\lambda }(x_{j-1,n}\oplus x_{j,n})=g_{\lambda }(x_{j,n})-\left(g_{\lambda }(x_{j,n})-g_{\lambda }(x_{j-1,n})\right)(1-\lambda ).$$

$$g_{\lambda }(1/a_1)={\lambda }^{a_1-1}$$

$$\begin{gathered} g_{\lambda }(y)=g_{\lambda }([0;a_1,\dots ,a_{m-1}]\oplus [0;a_1,\dots ,a_m])= \\ =g_{\lambda }([0;a_1,\dots ,a_{m-1}])+\lambda (g_{\lambda }([0;a_1,\dots ,a_m])-g_{\lambda }([0;a_1,\dots ,a_{m-1}]))= \\ =g_{\lambda }([0;a_1,\dots ,a_{m-1}])+{\lambda }^{\sum _{\left(1⩽i⩽m,i\equiv 1(\mathrm{mod}2)\right)}{a}_i-1}(1-\lambda {)}^{\sum _{\left(1⩽i⩽m,i\equiv 1\mathrm{mod}2\right)}{a}_i}\lambda , \end{gathered}$$

$$\begin{gathered} g_{\lambda }(z)=g_{\lambda }([0;a_1,\dots ,a_m+1]\oplus [0;a_1,\dots ,a_m])= \\ =g_{\lambda }([0;a_1,\dots ,a_m])-(1-\lambda )(g_{\lambda }([0;a_1,\dots ,a_m])-g_{\lambda }([0;a_1,\dots ,a_m+1])= \\ =g_{\lambda }([0;a_1,\dots ,a_m])-{\lambda }^{\sum _{\left(1⩽i⩽m,i\equiv 1\mathrm{mod}2\right)}{a}_i-1}(1-\lambda {)}^{\sum _{\left(1⩽i⩽m,i\equiv 1\mathrm{mod}2\right)}{a}_i}(1-\lambda {)}^2. \end{gathered}$$

$$x=[[b_0;b_1,b_2,\dots ,b_l,...]]=b_0-\frac{1}{b_1-\frac{1}{b_2-\dots -\frac{1}{b_l-\frac{1}{\dots }}}},b_i⩾2,$$

$$x=[[1;b_1,\dots ,b_l]].$$

$${\Xi }_n:=\{0,1\}\cup \left(\underset{1⩽k⩽n}{\cup }{\Theta }_k\right),$$

$${\Xi }_k=\{0={\xi }_{1,n}<{\xi }_{2,n}<\dots <{\xi }_{♯{\Xi }_n,n}=1\}.$$

$$\begin{gathered} g_{{\tau }^2}(x)={\tau }^{2a_1-2}-{\tau }^{2a_1+a_2-2}+{\tau }^{2a_1+a_2+2a_3-2}-\dots + \\ +(-1{)}^{m+1}{\tau }^{{\sum }_{i=1}^m{\alpha }_i{a}_i-2}+\dots , \end{gathered}$$

$$\alpha_{m}=\left\{\begin{array}[]{ll}1,&\text{if }m\text{ is even},\\ 2,&\text{if }m\text{ is odd}.\end{array}\right.$$

$$g_{{\tau }^2}(x)=\underset{n\to \infty }{\lim }\frac{♯\{\xi \in {\Xi }_n:\xi ⩽x\}}{♯{\Xi }_n},x\in (0,1).$$

$$\mathcal{M}(x):=\underset{n\to \infty }{\lim }\frac{♯\{\xi \in {\Xi }_n:\xi ⩽x\}}{♯{\Xi }_n},x\in (0,1).$$

$$\mathcal{M}(x\oplus y)=\mathcal{M}(x)+\left(\mathcal{M}(y)-\mathcal{M}(x)\right){\tau }^2.$$

$$♯{\Theta }_1=1,♯{\Theta }_2=1,♯{\Theta }_{n+1}=♯{\Theta }_n+♯{\Theta }_{n-1},$$

Full text

On arithmetical nature of Tichy-Uitz’s function

Jabitskaya E. N. 111 Research is supported by RFBR grant No 09-01-00371-a

In [1] R. F. Tichy and J. Uitz introduced a one parameter family $g_{\lambda}$, $\lambda\in(0,1)$, of singular functions. When $\lambda=1/2$ the function $g_{\lambda}$ coincides with the famous Minkowski question mark function. In this paper we describe the arithmetical nature of the function $g_{\lambda}$ when $\lambda=\frac{3-\sqrt{5}}{2}$.

Key words: continued fractions, Minkowski’s function.

2000 Mathematical Subject Classification: 11J70.

1 Stern-Brocot sequences

We remind the definition of Stern-Brocot sequences $\mathcal{F}_{n},n=0,1,2,\ldots$.

Consider the two-point set $\mathcal{F}_{0}=\left\{\frac{0}{1},\frac{1}{1}\right\}$. Let $n\geqslant 0$ and

[TABLE]

where $x_{j,n}=p_{j,n}/q_{j,n}$, $(p_{j,n},q_{j,n})=1$, $j=0,\ldots,N(n)$ and $N(n)=2^{n}$. Then

[TABLE]

with

[TABLE]

Here

[TABLE]

is the mediant of the fractions $\frac{a}{b}$ and $\frac{c}{d}$.

The elements of $Q_{n}$ can be characterized in the following way. A rational number $\xi\in[0,1]$ belongs to $Q_{n}$ if and only if in the continued fraction expansion

[TABLE]

the sum of partial quotients is exactly $n+1$:

[TABLE]

So $\mathcal{F}_{n}$ consists of all rational $\xi\in[0,1]$ such that $S(\xi)\leqslant n+1$.

2 Tichy-Uitz’s singular functions

In [1] R. F. Tichy and J. Uitz considered a one parameter family $g_{\lambda}$, $\lambda\in(0,1)$, of singular functions. In this section we describe the construction of $g_{\lambda}$ from [1]. This construction is an inductive one.

Given $\lambda\in(0,1)$ put

[TABLE]

Suppose that $g_{\lambda}(x)$ is defined for all elements $x\in\mathcal{F}_{n}$. Then we define $g_{\lambda}(x)$ for $x\in Q_{n+1}$. Each $x\in Q_{n+1}$ is of the form $x=x_{j-1,n}\oplus x_{j,n}$ where $x_{j-1,n}$ and $x_{j,n}$ are consecutive elements from $\mathcal{F}_{n}$. Then

[TABLE]

So we have defined $g_{\lambda}$ for all rational numbers from $[0,1]$. One can see that the function $g_{\lambda}(x)$ is a continuous function from $\mathbb{Q}\cap[0,1]$ to $[0,1]$. So it can be extended to a continuous function from the whole segment $[0,1]$ to $[0,1]$.

For every $\lambda$ the function $g_{\lambda}(x)$ increases in $x\in[0,1]$. By the Lebesgue theorem $g_{\lambda}(x)$ is a differentiable function almost everywhere. Moreower, it is easy to see that $g_{\lambda}^{\prime}(x)=0$ almost everywhere (is the scence of Lebesgue mesure). Certain properties of functions $g_{\lambda}(x)$ were investigated in [1]. Some related topics one can find in [6] and [7]. Here we should note that in the case $\lambda=1/2$ the function $g_{1/2}(x)$ coincides with the famous Minkowski question mark function $?(x)$.This function may be considered as the limit distribution function for Stern-Brocot sequences $\mathcal{F}_{n}$. The aim of the present paper is to explain the arithmetical nature of the function $g_{\lambda}(x)$ when $\lambda=\frac{3-\sqrt{5}}{2}$.

3 Minkowski’s function $?(x)$

Let us consider the function $g_{1/2}(x)=?(x)$. This function was introduced by Minkowski. As it follows from the definition of $g_{\lambda}$ for $\lambda=1/2$:

[TABLE]

and for $x_{j-1,n},x_{j,n}\in\mathcal{F}_{n}$

[TABLE]

The definition of $?(x)$ for irrational $x$ follows by continuity.

R. Salem in [2] found a new presentation for $?(x)$. If $x\in(0,1)$ is represented in the form of regular continued fraction

[TABLE]

then

[TABLE]

For rational $x$ the representation (2) and consequently (3) is finite.

Minkowski’s question mark function may be treated as the limit distribution function for Stern-Brocot sequences in the following sense:

[TABLE]

A finite formula for the right hand side of (4) was given by T. Rivoal in the preprint [5]. Various properties of Minkowski question mark function were investigated in papers [8] by A. Denjoy, [3] by P. Viader, J. Paradis, L. Bibiloni and in [4] by A. A. Dushistova, I. D. Kan and N. G. Moshchevitin.

4 General form of formula (3)

The formula (3) can be generalized on the whole family of functions $g_{\lambda}$ in the following way.

Proposition Let $x,\lambda\in(0,1)$ and $x=[0;a_{1},\ldots,a_{m},\ldots]$ is the regular continued fraction expansion of $x$, then

[TABLE]

Proof: By definition of $g_{\lambda}$

[TABLE]

and

[TABLE]

where $x_{j-1,n}$ and $x_{j,n}$ are consecutive elements from $\mathcal{F}_{n}$. We can also rewrite the formula (6) in the form

[TABLE]

The equality

[TABLE]

follows from the formula (6) immediately since $1/a_{1}=\underbrace{0\oplus\ldots\oplus 0\oplus}_{(a_{1}-1)\,\,\text{times}}1$. Suppose that the formula (5) is proved for $x=[0;a_{1},\ldots,a_{m}]$, then it is enough to prove it for $y=[0;a_{1},\ldots,a_{m}+1]$ and for $z=[0;a_{1},\ldots,a_{m},2]$.

Let $m$ is odd, then by applying formula (6) we get

[TABLE]

and by applying formula (7) we get

[TABLE]

For even $m$ the proof is analogously.

5 Regular reduced continued fractions and the main result

Any real number $x$ can be expressed uniquely in the form

[TABLE]

which is known as regular reduced continued fraction (eine reduziert-regelmassige Kettenbrouch [10], [9]).

For a rational number $x\in(0,1)$ the representation (10) takes the form:

[TABLE]

For such $x$ we denote $L(x)=b_{1}+\ldots+b_{l}$.

Analogously to the sequence $\mathcal{F}_{n}$ we define the sequence $\Xi_{n}$:

[TABLE]

where $\Theta_{k}=\{x\in Q:L(x)=k+1\}$, $k\geqslant 1$.

We arrange the elements of $\Xi_{n}$ in the increasing order:

[TABLE]

We would like to note that in the special case $\lambda=\tau^{2}$ formula (5) gives:

[TABLE]

where

[TABLE]

For rational $x$ the representation (12) is finite.

The Theorem 1 below is the main results of the present paper. It generalizes the formula (4) on the regular reduced continued fractions.

Theorem 1

Function $g_{\lambda}$, where $\lambda=\tau^{2}=\frac{3-\sqrt{5}}{2}$, $\tau=\frac{\sqrt{5}-1}{2}$ coincides with the distributional function of the sequence $\Xi_{n}$, that is

[TABLE]

Now we consider the function

[TABLE]

Our purpose is to prove that $\mathcal{M}(x)=g_{\lambda}$. Function $\mathcal{M}(x)$ is increasing as a distribution function, so it is enough to prove that $\mathcal{M}(x)$ coincides with $g_{\tau^{2}}(x)$ for rational $x$, that is

[TABLE]

for any two consecutive elements of $\Xi_{n}$ for any $n$.

6 Auxiliary results

The following result is well known. We present it without a proof.

Lemma 1

Let $x$ is represented in the form (1) and in the form (11). To get the set $(b_{1},\ldots,b_{l})$ from $(a_{1},\ldots,a_{m})$ we should replace $a_{i}$ by

$\underbrace{2\ldots 2}_{a_{i}-1}$* if $i$ is odd (empty string if $a_{i}=1$).* 2. 2.

$a_{i}+2$* if $i$ is even and $i\neq m$.* 3. 3.

$a_{i}+1$* if $i$ is even and $i=m$.*

Lemma 2

For the number of elements in $\Theta_{n}$ one has

[TABLE]

so $\sharp\Theta_{n}=F_{n},$ where

[TABLE]

is the nth Fibonacci number.

Proof. We prove the lemma by induction. Since $\Theta_{1}=\{1/2\}$, $\Theta_{2}=\{2/3\}$, then the base of induction is true. Let us suppose that the lemma is true for $k\leqslant n$ and $x=[[1;b_{1},\ldots,b_{l}]]\in\Theta_{n+1}$, then $b_{1}+\ldots+b_{l}=n+2$. There are two cases: either $b_{l}=2$ or $b_{l}\geqslant 2$. In the first case $b_{1}+\ldots+b_{l-1}=n$, so $[[1;b_{1},\ldots,b_{l-1}]]\in\Theta_{n-1}$, in the second case $b_{1}+\ldots+b_{l}-1=n+1$, so $[[1;b_{1},\ldots,b_{l}-1]]\in\Theta_{n}$. Thus we have one-one correspondence between $\Theta_{n-1}\cup\Theta_{n}$ and $\Theta_{n+1}$, and so $\sharp\Theta_{n+1}=\sharp\Theta_{n}+\sharp\Theta_{n-1}$.

Defenition 1

Let $x,y,z$ be consecutive elements of $\Xi_{n}$, $y\in\Theta_{n}$. We denote the mediant $x\oplus y$ by $y^{l}$, the mediant $y\oplus z$ we denote by $y^{r}$.

Lemma 3

Let $x,y,z$ be consecutive elements of $\Xi_{n}$, $y\in\Theta_{n}$, then $y^{l}\in\Theta_{n+2}$, $y^{r}\in\Theta_{n+1}$.

Proof. Let $y=[[1;b_{1},\ldots,b_{s}]]$. Then $y^{l}=[[1;b_{1},\ldots,b_{s},2]]$, $y^{r}=[[1;b_{1},\ldots,b_{s}+1]]$.

Now let us construct an infinite tree $D$ whose nodes are labeled by rationals in $(0,1)$. We identify the nodes with the rationals they labeled by. The root is labeled by $1/2$. From node $x$ come two arrows: the left arrow goes to $x^{l}$ and the right arrow goes to $x^{r}$. The nodes of the tree $D$ are partitioned into levels. $1/2$ belongs to the level 1. If $x$ belongs to the level $n$, then $x^{r}$ belongs to the level $n+1$, and $x^{l}$ belongs to the level $n+2$ (figure 1).

It follows from the construction of the tree that nodes from level $n$ of $D$ are marked by numbers from $\Theta_{n}$. So $x$ belongs to the level $n$ if and only if $x\in\Theta_{n}$.

The subtree of $D$ with root in the node $x$ we denote by $D^{(x)}$. The set of nodes of $D$ from level 1 to level $n$ we denote by $D_{n}$. The set of nodes of $D^{(x)}\cap D_{n}$ we denote by $D^{(x)}_{n}$. Note that there exist a levels preserving isomorphism between $D$ and $D^{(x)}$. If $x$ belongs to the level $n$, then

[TABLE]

Besides

[TABLE]

7 Proof of Theorem 1

We remind that it is enough to prove (13) for any consecutive elements of $\Xi_{n}$ $x$ and $y$.

To prove the equality (13) we consider the subtree $D^{(x\oplus y)}$ of $D$. Note that

[TABLE]

Consequently

[TABLE]

On the other hand

[TABLE]

Let $x\oplus y\in\Theta_{k}$, then $(x\oplus y)^{l}\in\Theta_{k+2}$. Therefore

[TABLE]