# The asymptotic distribution of the scaled remainder for pseudo golden ratio
expansions of a continuous random variable

I. Herbst1, J. Møller2, A.M. Svane2

($1$ Department of Mathematics, University of Virginia, Charlottesville, Va,
22903, U.S.A.
$2$ Department of Mathematical Sciences, Aalborg University, Aalborg Ø,
Denmark

)

###### Abstract

Let $X=\sum_{k=1}^{\infty}X_{k}\beta^{-k}$ be the base-$\beta$ expansion of a
continuous random variable $X$ on the unit interval where  $\beta$ is the
positive solution to $\beta^{n}=1+\beta+\cdots+\beta^{n-1}$ for an integer
$n\geq 2$ (i.e., $\beta$ is a generalization of the golden mean for which
$n=2$). We study the asymptotic distribution and convergence rate of the
scaled remainder $\sum_{k=1}^{\infty}X_{m+k}\beta^{-k}$ when $m$ tends to
infinity.

##  1 Introduction

In an experiment whose outcome follows a certain probability  distribution and
we keep only a certain number of digits in the data, what is the distribution
of the remainder? In this paper we consider  a random variable $X$ with a
probability density function (PDF) $f$ concentrated on $[0,1)$ and with binary
digits $X_{k}\in\\{0,1\\}$ so that $X=\Sigma_{k=1}^{\infty}X_{k}\beta^{-k}$ is
its base $\beta$-expansion when $\beta$ is the positive solution to

$\beta^{n}=1+\beta+\cdots+\beta^{n-1}$ (1.1)

for some integer $n\geq 2$ (see [10] where $\beta$-expansions are discussed
for general $\beta>1$, in which case the digits are not necessarily binary).
The distribution of the scaled remainder,
$\Sigma_{k=1}^{\infty}X_{k+m}\beta^{-k}$, and the speed at which it converges
to its limit has been discussed in  first [4] when $\beta$ is an integer $q$
(and the digits then live in $\\{0,1,...,q-1\\}$) and second in [5] when
$\beta=(1+\sqrt{5})/2$ is the golden mean. Since the golden mean corresponds
to the case $n=2$ in (1.1), for $n\geq 2$ the positive solution
$\beta=\beta_{n}$ to (1.1) is called a pseudo golden mean.

Let us introduce some notation before we state the main result of this paper.
Given $x\in[0,1)$ let

$t(x)=\beta x-\lfloor\beta x\rfloor,\quad t^{0}(x)=x$

where $\lfloor\beta x\rfloor$ is the integer part of $\beta x$. Assuming only
that $1<\beta<2$ it follows that $x\in[0,1)$ implies $t(x)\in[0,1).$ In
addition one can show that with $x_{k}\coloneqq\lfloor\beta
t^{k-1}(x)\rfloor\in\\{0,1\\}$ we have

$x=\Sigma_{k=1}^{\infty}x_{k}\beta^{-k}$ (1.2)

with $0\leq\beta^{-m}t^{m}(x)=x-\Sigma_{k=1}^{m}x_{k}\beta^{-k}<\beta^{-m}$
and  $m\in\mathbb{N}$ (the natural numbers,1,2,…). Since for example
$\Sigma_{k=1}^{\infty}\beta^{-k}=(\beta-1)^{-1}>1$ there must be restrictions
on the sequence of zeros and ones which arise from $x_{k}=\lfloor\beta
t^{k-1}(x)\rfloor$. These restrictions arise automatically from the
definitions of $t$ and $x_{k}$ (see [8]). Since we restrict to $\beta$ given
by the unique positive solution to  (1.1), the restriction on the sequence of
zeros and ones is that in the sequence, $(x_{1},x_{2},...)$ in (1.2), no more
than $n-1$ $x_{j}$’s in succession can equal $1$ (see [8]). This is consistent
with the definitions above. For this and other properties of this
$\beta$-expansion and the other roots of $p(x)=x^{n}-(1+x+\cdots+x^{n-1})$,
see Appendix A and B.

Since $X$ has PDF $f$, $P(X\in B)=\int_{B}f(x)\,\mathrm{d}x$ for any
$B\in\mathcal{B}$, the Borel subsets of $[0,1)$. The main object of our
interest is the scaled remainder
$t^{m}(X)=\beta^{m}\Sigma_{k=m+1}^{\infty}X_{k}\beta^{-k}=\Sigma_{k=1}^{\infty}X_{k+m}\beta^{-k}$
for $m\in\mathbb{N}$. Note that $t$ induces a transformation $T$ on the PDF’s
concentrated on $[0,1)$, with $P(t(X)\in
B)=\int_{B}Tf(x)\,\mathrm{d}x,B\in\mathcal{B}$. Let  $f_{m}\coloneqq
T^{m}f,m\in\mathbb{N}$. With the definition $P_{m}(B)\coloneqq P(t^{m}(X)\in
B)$ we have  $P_{m}(B)=\int_{B}f_{m}(x)\,\mathrm{d}x$ where $f_{m}$ is a PDF
concentrated on $[0,1)$. Further let

$f_{\beta}(x)=\left(\sum_{j=1}^{n}F_{j}(x)\right)D(\beta)^{-1},\quad
F_{j}=(\beta^{j}-\beta^{j-1}-\beta^{j-2}-\cdots-\beta)\chi_{[\beta^{-1}+\cdots+\beta^{-(j-1)},\beta^{-1}+\cdots+\beta^{-j})}$
(1.3)

where $\chi_{I}$ is the indicator function for the interval $I$ and $D(\beta)$
is defined so that $\int_{[0,1)}f_{\beta}\,\mathrm{d}x=1$ (we also define
$D(\beta)$ in (3.16) below). Then $f_{\beta}$ is a PDF on $[0,1)$. Define
$P_{\beta}(B)\coloneqq\int_{B}f_{\beta}\,\mathrm{d}x,B\in\mathcal{B}$. In
addition let

$d_{TV}(P_{1},P_{2})=\sup_{B\in\mathcal{B}}|P_{1}(B)-P_{2}(B)|,$

the total variation distance between two probability distributions $P_{1}$ and
$P_{2}$.  Finally, we use the notation
$\|h-f_{\beta}\|_{\infty}=\sup_{x\in|0,1)\setminus{D}}|h(x)-f_{\beta}(x)|$,
where $D$ is the set of points of discontinuity of $f_{\beta}$.

Our main result is the following theorem,  where we let $\lambda_{2}$ be the
root of the polynomial $p(x)=x^{n}-(1+x+\cdots+x^{n-1})$ whose absolute value
is the second largest of those of all $n$ roots.

######  Theorem 1.1.

We have

$\lim_{m\to\infty}d_{TV}(P_{m},P_{\beta})=0.$ (1.4)

In addition, if $f$ is a Lipschitz function with $|f(x)-f(y)|\leq L_{f}|x-y|$
for all $x,y\in[0,1)$ and if $m>2n$, then

${\color[rgb]{0,0,0}\|f_{m}(x)-f_{\beta}(x)\|_{\infty}=O(|\beta^{-1}\lambda_{2}|^{m_{2}})+O(\beta^{-m_{1}})L_{f}}$
(1.5)

where $m_{1},m_{2}\in\mathbb{N}$ with $m_{2}>2n-1$ and $m_{1}+m_{2}=m$.

######  Remark 1.2.

Here, $m_{1}$ and $m_{2}$ should be chosen so that (1.5) is optimized once
additional information is known about $|\lambda_{2}|$. Note that we do know
that $|\lambda|_{2}<1$. Instead of assuming $f$ is Lipschitz, we could assume
that $f$ is only Hölder continuous with an appropriate change in the
exponential rate of convergence in (1.5) depending on the Hölder exponent.

In [10],  Rényi showed the unique existence of a PDF, $f_{\beta}$, invariant
under $T$. A formula (up to a constant factor) for $f_{\beta}$, for arbitrary
$\beta>1$ is given in [8]. His formula can be seen to give exactly equation
(4.19)  below without the factor $D(\beta)^{-1}$) if one notes that
$t^{j}(1)=\beta^{-1}+\cdots+\beta^{n-j}$ for $j\geq 1$. We remark that
$1>|\lambda_{2}|\geq\beta^{-1/(n-1)}$ and $|\lambda_{2}|=\beta^{-1/(n-1)}$ if
$n=2$ or $3$ (see Appendix B). Then under  the Lipschitz continuity assumption
in Theorem 1.1, for $n=2$

$\|f_{m}-f_{\beta}\|_{\infty}=O(\beta^{-2m/3}),$

(see [4]) and for $n=3$

$\|f_{m}-f_{\beta}\|_{\infty}=O(\beta^{-3m/5}),$

while with no further knowledge of $|\lambda_{2}|$ other than
$|\lambda_{2}|<1$ we have

$\|f_{m}-f_{\beta}\|_{\infty}=O(\beta^{-m/2}).$

We mention a very different proof of the limit (1.4) of Theorem 1.1 (see [1],
which analyzes the operator $T$ defined above). In the proof of Theorem 1.1 we
will use (see [12], Lemma 2.1)

$d_{TV}(P_{m},P_{\beta})=(1/2)\int_{[0,1)}|f_{m}-f_{\beta}|{\color[rgb]{0,0,0}\,\mathrm{d}x.}$
(1.6)

As part of the proof of Theorem 1.1 we are going to divide $[0,1)$ into
intervals. For  $m\in\mathbb{N}$ define

$\Omega_{m}=\\{(j_{1},j_{2},\cdots,j_{m})\,|\,j_{k}=\mbox{0 or 1 and no more
than $n-1$ $j$'s in succession may equal 1}\\}.$

If $J\in\Omega_{m}$ define

$L_{J,m}={\color[rgb]{0,0,0}\sum_{k=1}^{m}}j_{k}\beta^{-k}.$

Let  $r(J)$ be the number of ones in succession ending at $j_{m}$.  Then
$0\leq r(J)\leq\max(m,n-1).$ How large can
$x=\sum_{k=1}^{\infty}\beta^{-k}x_{k}$ be, given that $x_{k}=j_{k},1\leq k\leq
m$ (and the fact that the $x_{j}$’s follow the standard restrictions)? If for
example $r(J)=n-1$ we can add
$\beta^{-(m+1)}[(\beta^{-1}+\cdots+\beta^{-(n-1)})+\beta^{-n}(\beta^{-1}+\cdots+\beta^{-(n-1)})+\cdots]=\beta^{-(m+1)}$
to $L_{J,m}$. In general a calculation shows we can add a maximum of
$\beta^{-(m+1)}+\beta^{-(m+2)}+\cdots\beta^{-(m+n-r(J))}$ to $L_{J,m}$. Thus
we define the interval

$I_{J}={\color[rgb]{0,0,0}I_{J,m}=}[L_{J,m},L_{J,m}+\beta^{-(m+1)}+\beta^{-(m+2)}+\cdots\beta^{-(m+n-r(J))}).$

We give some properties of the  $J$’s in $\Omega_{m}$ as well as of the
base-$\beta$ expansion of a number $x\in[0,1)$ in the following proposition.

######  Proposition 1.3.

  1. 1)

If $J$ and $K$ are in $\Omega_{m}$ then $I_{J}\cap I_{K}=\emptyset$ unless
$J=K$.

  2. 2)

$\cup_{J\in\Omega_{m}}I_{J}=[0,1)$.

  3. 3)

Any $x\in[0,1)$ has an expansion $x=\Sigma_{k=1}^{\infty}x_{k}\beta^{-k}$ with
the $x_{j}$’s obeying the standard restrictions, unique except for
base-$\beta$ fractions of the form $L_{J,m}.$

###### Proof.

See Appendix A and B where we also discuss the roots of
$p(x)=x^{n}-(1+x+\cdots+x^{n-1})$. ∎

##  2 Formula for $f_{m}$, the PDF of $t^{m}(X)$

As we saw above

$t^{m}(X)={\color[rgb]{0,0,0}\sum_{k=1}^{\infty}}X_{m+k}\beta^{-k}.$

We have $P(t^{m}(X)\leq x)=\Sigma_{J\in\Omega_{m}}P(t^{m}(X)\leq x,X\in
I_{J,m})$, where $t^{m}(X)\leq x$ is the same as
$\Sigma_{k=m+1}^{\infty}X_{k}\beta^{-k}\leq\beta^{-m}x$, so

$P(t^{m}(X)\leq x,X\in I_{J,m})=P(L_{J,m}\leq X\leq L_{J,m}+\beta^{-m}x,X\in
I_{J,m}).$ (2.7)

For $J\in\Omega_{m}$ let

$K(J)=\beta^{-1}+\cdots+\beta^{-(n-r(J))},$

so $\beta^{-1}\leq K(J)\leq 1$. Thus
$I_{J,m}=[L_{J,m},L_{J,m}+\beta^{-m}K(J))$ and from (2.7) we have

$\displaystyle P(t^{m}(X)\leq x)=\sum_{J\in\Omega_{m}}[$
$\displaystyle\chi_{[0,K(J))}(x)P(L_{J,m}\leq X<L_{J,m}+\beta^{-m}x)$
$\displaystyle+\chi_{[K(J),1]}(x)P(L_{J,m}\leq X<L_{J,m}+K(J)\beta^{-m})].$

It follows that the PDF of $t^{m}(X)$ is

$f_{m}(x)=\beta^{-m}{\color[rgb]{0,0,0}\sum_{J\in\Omega_{m}}}\chi_{[0,K(J))}(x)f(L_{J,m}+\beta^{-m}x).$
(2.8)

We will make use of another function, $g$, an approximation to $f$, possibly
smoother, in terms of which we define $g_{m}$ as

$g_{m}(x)=\beta^{-m}{\color[rgb]{0,0,0}\sum_{J\in\Omega_{m}}}\chi_{[0,K(J))}(x)g(L_{J,m}+\beta^{-m}x).$
(2.9)

We take $g$ to be a Lipschitz continuous function on $[0,1)$ with
$|g(x)-g(y)|\leq L_{g}|x-y|$. Moreover we assume $g\geq 0$ and $\int
g(t){\color[rgb]{0,0,0}\mathrm{d}t}=1$.  With $T$ as defined in Section 1, we
have

${\color[rgb]{0,0,0}\|f_{m}-g_{m}\|_{L^{1}}=\|T^{m}f-T^{m}g\|_{L^{1}}\leq\|f-g\|_{L^{1}}},$
(2.10)

since $T$ takes PDFs to PDFs and thus has an extension to $L^{1}([0,1))$
satisfying $\|Th\|_{L^{1}}\leq\|h\|_{L^{1}}$ for all $h\in L^{1}$.

##  3 Counting the number of $J$’s in $\Omega_{m,r}$

Let

$\Omega_{m,r}=\\{J\in\Omega_{m}\,|\,r(J)=r\\}.$

In our proof of Theorem 1.1 it will be important to know the size of
$\Omega_{m,r}$. Denote this number by $N_{r}(m)$,  which is determined in
Lemma 3.2 below using first the following considerations.

We have

$N_{0}(m+1)={\color[rgb]{0,0,0}\sum_{r=0}^{\min(m,n-1)}}N_{r}(m)$

and for $1\leq r\leq\min(m+1,n-1)$

$N_{r}(m+1)=N_{r-1}(m).$

For $m\leq n-1$, $N_{0}(m)=2^{m-1}$ since we can fill up the spaces to the
left of $j_{m}=0$ with ones and zeros as we like,  and $N_{1}(m)=2^{m-2}$
since $j_{m}=1,j_{m-1}=0$ and the rest of the spaces can be filled up
arbitrarily. Similarly if $r\leq m-1$, $N_{r}(m)=2^{m-(r+1)}$.  Finally
$N_{m}(m)=1$.

If $m\geq n-1$ define $M$ by the equation

$\begin{bmatrix}N_{0}(m+1)\\\ \vdots\\\ N_{n-1}(m+1)\\\
\end{bmatrix}=\begin{bmatrix}1&1&1&\cdots 1&1\\\ 1&0&0&\cdots 0&0\\\
0&1&0&\cdots 0&0\\\ \vdots\\\ 0&0&0&\cdots 1&0\\\
\end{bmatrix}\times\begin{bmatrix}N_{0}(m)\\\ \vdots\\\ N_{n-1}(m)\\\
\end{bmatrix}=M\times\begin{bmatrix}N_{0}(m)\\\ \vdots\\\ N_{n-1}(m)\\\
\end{bmatrix}$ (3.11)

Thus

$\begin{bmatrix}N_{0}(m+1)\\\ \vdots\\\ N_{n-1}(m+1)\\\
\end{bmatrix}=M^{m+1-(n-1)}\times\begin{bmatrix}N_{0}(n-1)\\\ \vdots\\\
N_{n-1}(n-1)\\\ \end{bmatrix}=M^{m+1-(n-1)}\times\begin{bmatrix}2^{n-2}\\\
\vdots\\\ 2^{n-2-r}\\\ \vdots\\\ 1\\\ \end{bmatrix}$ (3.12)

It is not hard to show by expanding by minors of the first column  of $xI-M$
that

$\det(xI-M)=(x-1)x^{n-1}-(x^{n-2}+x^{n-3}+\cdots+x+1){\color[rgb]{0,0,0}=x^{n}-(1+x+\cdots+x^{n-1}).}$

Using (1.1), an unnormalized eigenvector corresponding to the unique positive
eigenvalue $\beta$ of $M$ is seen to be

$u=\begin{bmatrix}\beta^{n-1}\\\ \beta^{n-2}\\\ \vdots\\\ 1\\\ \end{bmatrix}$
(3.13)

and an unnormalized eigenvector with eigenvalue $\beta$ for the transposed
matrix $M^{*}$ is

$v=\begin{bmatrix}1\\\ \beta-1\\\ \beta^{2}-\beta-1\\\ \vdots\\\
\beta^{n-1}-(\beta^{n-2}+\cdots+1)\end{bmatrix}$ (3.14)

As must be the case, the components of (3.14) all have the same sign. In fact,
for $1\leq j\leq n$,

$\beta^{j-1}-(\beta^{j-2}+\beta^{j-3}+\cdots+1)=\beta^{-1}+\beta^{-2}+\cdots+\beta^{-(n-(j-1))}.$
(3.15)

To see this subtract the right hand side from the left, factor out
$\beta^{j-1}$, and use (1.1).

Define

$D(\beta)=\beta^{-(n-1)}{\color[rgb]{0,0,0}\langle v,u\rangle}$ (3.16)

where $D(\beta)$ is the same as appears in (1.3). We calculate

${\color[rgb]{0,0,0}\langle
v,u\rangle}=n\beta^{n-1}-{\color[rgb]{0,0,0}\sum_{m=0}^{n-2}}(m+1)\beta^{m}={\color[rgb]{0,0,0}\frac{\mathrm{d}}{\mathrm{d}x}}(x^{n}-x^{n-1}-x^{n-2}-\cdots-1)|_{x=\beta}{\color[rgb]{0,0,0}=\prod_{j=2}^{n}}(\beta-\lambda_{j}).$

where the $\lambda_{j}$’s are the roots of $x^{n}-(x^{n-1}+\cdots+1)$. Of
course these are the eigenvalues of $M$. We order the $\lambda_{j}$’s so that
$\beta=\lambda_{1}>|\lambda_{2}|\geq\cdots\geq|\lambda_{n}|$.

Let

$w=\begin{bmatrix}2^{n-2}\\\ \vdots\\\ 2^{n-2-r}\\\ \vdots\\\ 2^{n-n}\\\ 1\\\
\end{bmatrix}$ (3.17)

Note that this is an $n$ dimensional column vector with
$(n-1)\textsuperscript{st}$ entry  given by $2^{n-n}=1$.

######  Lemma 3.1.

${\color[rgb]{0,0,0}\langle v,w\rangle}=\beta^{n-1}.$ (3.18)

###### Proof.

It turns out that the lemma is true for $\beta$ replaced by any symbol $x$
which does not necessarily satisfy (1.1). Thus for the purposes of the proof
of this lemma, we replace $\beta$ by $x$ in (3.14). First note that the lemma
is true for $n=2$. Then assuming it is true for $n$ we have for $n+1$

$\displaystyle\langle v,w\rangle=$
$\displaystyle\,2[2^{n-2}+(x-1)2^{n-3}+\cdots+(x^{n-2}-(x^{n-3}+\cdots+1))]+x^{n-1}-$
$\displaystyle\,(x^{n-2}+\cdots+1)+x^{n}-(x^{n-1}+\cdots+1)$ $\displaystyle=$
$\displaystyle\,2[x^{n-1}-(x^{n-1}-(x^{n-2}+\cdots+1))]+x^{n-1}-$
$\displaystyle\,(x^{n-2}+\cdots+1)+x^{n}-(x^{n-1}+\cdots+1)$ $\displaystyle=$
$\displaystyle\,x^{n}.$

∎

With these notations, for $m>n-1$ we have from (3.12):

######  Lemma 3.2.

$N_{s}(m)=\beta^{m+n-1-s}\langle v,w\rangle^{-1}+O(|\lambda_{2}|^{m})$

###### Proof.

We use the fact that the eigenvalues of $M$ are simple [7]. Thus $M$ has a
basis of eigenvectors, $Mu_{j}=\lambda_{j}u_{j},1\leq j\leq n,u_{1}=u$, and
similarly $M^{*}$ has a basis of eigenvectors,
$M^{*}v_{j}=\bar{\lambda_{j}}v_{j},1\leq j\leq n,v_{1}=v$. It follows that
$\langle v_{j},u_{k}\rangle=0,j\neq k$ and $\langle v_{j},u_{j}\rangle\neq 0$.
We write $w=\Sigma_{j=1}^{n}\langle v_{j},w\rangle\langle
v_{j},u_{j}\rangle^{-1}u_{j}$. It follows that

$\displaystyle N_{s}(m)$ $\displaystyle=\sum_{j=1}^{n}(M^{m-(n-1)}\langle
v_{j},w\rangle\langle v_{j},u_{j}\rangle^{-1}u_{j})_{s}$
$\displaystyle=\beta^{m-(n-1)}\langle v,w\rangle\langle
v,u\rangle^{-1}(u)_{s}+O(|\lambda_{2}|^{m}).$

Now using $\langle v,w\rangle=\beta^{n-1}$ and $(u)_{s}=\beta^{n-1-s}$ we find
the result. ∎

##  4 Estimating $|g_{m}-f_{\beta}|$

######  Lemma 4.1.

With $f_{\beta}$ as in (1.3),

$f_{\beta}(x)={\color[rgb]{0,0,0}\sum_{s=0}^{n-1}\chi_{[0,\beta^{-1}+\cdots+\beta^{-(n-s)})}(x)\beta^{-s}D(\beta)^{-1}.}$
(4.19)

###### Proof.

We have

$\sum_{s=0}^{n-1}\chi_{[0,\beta^{-1}+\cdots+\beta^{-(n-s)})}(x)\beta^{-s}=\sum_{j=0}^{n-1}(1+\beta^{-1}+\cdots+\beta^{-(n-j)})\chi_{[\beta^{-1}+\cdots+\beta^{-(n-(j+1))},\beta^{-1}+\cdots+\beta^{-(n-j)})}$

where the term for $j=0$ is interpreted as
$(1+\beta^{-1}+\cdots+\beta^{-(n-1)})\chi_{[0,\beta^{-1})}$. From (3.15),

$1+\beta^{-1}+\cdots+{\color[rgb]{0,0,0}\beta^{{-(n-j)}}}=\beta^{j}-(\beta^{j-1}+\cdots+\beta)$

which gives the result. ∎

######  Proposition 4.2.

With $f_{\beta}$ and $g_{m}$ defined in (1.3) and (2.9) respectively,  if
$m>2n$ then

${\color[rgb]{0,0,0}\|g_{m}-f_{\beta}\|_{\infty}=O(\beta^{-m_{1}})L_{g}+O((\beta^{-1}|\lambda_{2}|)^{m_{2}})}$
(4.20)

where $m_{1},m_{2}\in\mathbb{N}$ with $m_{2}>2n-1$ and $m_{1}+m_{2}=m$.

######  Remark 4.3.

We have the estimates $\beta^{-1/(n-1)}\leq|\lambda_{2}|<1$ and for $n=2$ and
$3$ the lower estimate for $|\lambda_{2}|$ is actually an equality (see Lemma
B.1). The error is meant to be optimized by a good choice of $m_{1}$ and
$m_{2}$ once one knows more about $|\lambda_{2}|.$

###### Proof.

We localize to subintervals of the intervals $I_{K}$ with $K\in\Omega_{m}$ as
in [5]. Define for fixed $J\in\Omega_{m_{1}}$, $m>m_{1}$,
$\Omega_{m}(J)=\\{(j_{1},j_{2},..,j_{m})\in\Omega_{m}|(j_{1},j_{2},..,j_{m_{1}})=J\\}$.
Note that

$\sum_{\tilde{J}\in\Omega_{m}}h(\tilde{J})=\sum_{J\in\Omega_{m_{1}}}\sum_{J^{\prime}\in\Omega_{m}(J)}h(J^{\prime}).$

Thus from (2.9)

$g_{m}(x)=\sum_{J\in\Omega_{m_{1}}}\sum_{J^{\prime}\in\Omega_{m}(J)}\chi_{[0,K(J^{\prime}))}(x)\beta^{-m}g(L_{J^{\prime},m}+\beta^{-m}x).$
(4.21)

We have $I_{J,m_{1}}=\cup_{J^{\prime}\in\Omega_{m}(J)}I_{J^{\prime},m}$ so
that $I_{J^{\prime},m}\subset I_{J,m_{1}}$ and for
$y=L_{J^{\prime},m}+\beta^{-m}x\in I_{J^{\prime},m},t\in I_{J,m_{1}}$,
$|y-t|\leq|I_{J,m_{1}}|=K(J)\beta^{-m_{1}}$. From (2.18), the size of the set
$\Omega_{m}$ is $O(\beta^{m})$ so that

$g_{m}(x)=\left[\sum_{J\in\Omega_{m_{1}}}\sum_{J^{\prime}\in\Omega_{m}(J)}\bigg{(}\chi_{[0,K(J^{\prime}))}(x)\beta^{-m}|I_{J,m_{1}}|^{-1}\int_{I_{J,m_{1}}}g(t)\,\mathrm{d}t\bigg{)}\right]+O(\beta^{-m_{1}})L_{g}.$

Since $|I_{J,m_{1}}|=\beta^{-m_{1}}K(J)$,

$g_{m}(x)=\left[\sum_{J\in\Omega_{m_{1}}}\sum_{J^{\prime}\in\Omega_{m}(J)}\bigg{(}\chi_{[0,K(J^{\prime}))}(x)K(J)^{-1}\beta^{-m+m_{1}}\int_{I_{J,m_{1}}}g(t)\,\mathrm{d}t\bigg{)}\right]+O(\beta^{-m_{1}})L_{g}.$
(4.22)

Let $\Omega_{m_{1},r_{1}}$ be the set of $J\in\Omega_{m_{1}}$ with
$r(J)=r_{1}$. Then counting the number of $J^{\prime}s$ with $r(J)=r_{1}$ and
$r(J^{\prime})=s$,

$\displaystyle
g_{m}(x)=\bigg{[}\sum_{r_{1}=0}^{n-1}\sum_{J\in\Omega_{m_{1},r_{1}}}\Sigma_{s=0}^{n-1}\chi_{[0,\beta^{-1}+\cdots+\beta^{-(n-s)})}(x)K(J)^{-1}\times$
$\displaystyle(N_{s}(m_{2}-1)+N_{s}(m_{2}-2)+\cdots+N_{s}(m_{2}-(n-r_{1})))\beta^{-m_{2}}\int_{I_{J,m_{1}}}g(t)\,\mathrm{d}t\bigg{]}+O(\beta^{-m_{1}})L_{g}.$

From Proposition 3.2

$N_{s}(m_{2})=\beta^{m_{2}+n-1-s}\langle
v,u\rangle^{-1}+O(|\lambda_{2}|^{m_{2}}).$

Thus

$\displaystyle\sum_{s=0}^{n-1}\chi_{[0,\beta^{-1}+\cdots+\beta^{-(n-s)})}(x)K(J)^{-1}\times$
$\displaystyle\hskip
14.22636pt(N_{s}(m_{2}-1)+N_{s}(m_{2}-2)+\cdots+N_{s}(m_{2}-(n-r_{1})))\beta^{-m_{2}}$
$\displaystyle=\,$ $\displaystyle
K(J)^{-1}\sum_{s=0}^{n-1}\sum_{k=1}^{n-r_{1}}\beta^{n-1-s-k}\chi_{[0,\beta^{-1}+\cdots+\beta^{-(n-s)})}(x)\langle
v,u\rangle^{-1}+O(|\beta^{-1}\lambda_{2}|^{m_{2}}).$

We have $\sum_{k=1}^{n-r_{1}}\beta^{-k}=K(J)$. Thus

$\displaystyle g_{m}(x)=$
$\displaystyle\sum_{J\in\Omega_{m_{1}}}\Sigma_{s=0}^{n-1}\chi_{[0,\beta^{-1}+\cdots+\beta^{-(n-s)})}(x)\beta^{n-1-s}\int_{I_{J,m_{1}}}g(t)\,\mathrm{d}t\,\langle
v,u\rangle^{-1}+O(|\beta^{-1}\lambda_{2}|^{m_{2}})+O(\beta^{-m_{1}})L_{g}$
$\displaystyle=\sum_{s=0}^{n-1}\chi_{[0,\beta^{-1}+\cdots+\beta^{-(n-s)})}(x)\beta^{n-1-s}\angle
v,u\rangle^{-1}+O(|\beta^{-1}\lambda_{2}|^{m_{2}})+O(\beta^{-m_{1}})L_{g}.$

Using (3.16) and (4.19) we have

$g_{m}(x)=f_{\beta}(x)+O(|\beta^{-1}\lambda_{2}|^{m_{2}})+O(\beta^{-m_{1}})L_{g}.$

This is (4.20). ∎

Proof of Theorem 1.1. Using (2.10) and (4.20) we have

$\|f_{m}-f_{\beta}\|_{L^{1}}\leq\|f_{m}-g_{m}\|_{L^{1}}+\|g_{m}-f_{\beta}\|_{L^{1}}\leq\|f-g\|_{L^{1}}+O(\beta^{-m/2})(1+L_{g}).$

Thus

$\limsup_{m\to\infty}\|f_{m}-f_{\beta}\|_{L^{1}}\leq\|f-g\|_{L^{1}}$

which can be made as small as we like. Using (1.4), this gives

$d_{TV}(P_{m},P_{\beta})\rightarrow 0.$

Equation (1.5) follows from (4.20) setting $f=g$ and $f_{m}=g_{m}$.

## Funding

Supported by the Danish Council for Independent Research Natural Sciences,
grant DFF - 10.46540/2032-00005B.

##  Appendix A The beta expansion

In this appendix we gather some properties of the $\beta$-expansion. We use
the notation from the main text.

######  Lemma A.1.

If $J$ and $K$ are in $\Omega_{m}$ then either $J=K$ or $I_{J}\cap
I_{K}=\varnothing$.

###### Proof.

Let $J=(j_{1},\cdots,j_{m})$ and $K=(k_{1},\cdots,k_{m})$. Assume $L_{J,m}\leq
L_{K,m}$ but $J\neq K$. If $j_{1}>k_{1}$ then $L_{J,m}\geq\beta^{-1}$ while
$L_{K,m}<\beta^{-1}(\beta^{-1}+\cdots+\beta^{-(n-1)})+\beta^{-1}\beta^{-n}(\beta^{-1}+\cdots+\beta^{-(n-1)})+\cdots=\beta^{-1}(\beta^{-1}+\cdots+\beta^{-(n-1)})(1-\beta^{-n})^{-1}=\beta^{-1}$,
a contradiction. A similar calculation assuming $j_{s}=k_{s}$ for $1\leq s\leq
l_{0}-1$ and $j_{l_{0}}>k_{l_{0}}$ gives the same contradiction. We have shown
that for some $l_{0}$, $j_{s}=k_{s}$ for $1\leq s\leq l_{0}-1$ and
$j_{l_{0}}<k_{l_{0}}$. And by the way, if $L_{J,m}=L_{K,m}$ then $J=K$.

We claim that if $L_{J,m}<L_{K,m}$ then $L_{K,m,}\geq\sup I_{J}$. Assume that
$j_{1}=k_{1},\cdots,j_{l_{0}-1}=k_{l_{0}-1},j_{l_{0}}<k_{l_{0}}.$ Then $\sup
I_{J}\leq\sum_{s=1}^{(l_{0}-1)}j_{s}\beta^{-s}+\beta^{-l_{0}}(\beta^{-1}+\cdots+\beta^{-(n-1})(1-\beta^{-n})^{-1}=\sum_{s=1}^{(l_{0}-1)}+\beta^{-l_{0}}$.
While $L_{K,m}\geq\Sigma_{s=1}^{-(l_{0}-1)}j_{s}\beta^{-s}+\beta^{-l_{0}}$.
Thus $L_{K,m}\geq\sup I_{J}$. Here we have used that $\sup I_{J}$ is by
definition arrived at by using the maximum that it could be using the
restrictions that no row of $n$ adjacent $x_{k}$’s can equal $1$ if
$x=\sum_{k=1}^{\infty}x_{k}\beta^{-k}\in I_{J}$ and $(x_{1},\cdots,x_{m})=J$.
Without knowing $j_{l_{0}+1},\cdots,j_{m}$, we have (possibly) overestimated
$\sup I_{J}$ by taking $j_{l_{0}+1},\cdots,j_{m}$ to give the maximum possible
$\sup I_{J}$. This proves the result. ∎

######  Lemma A.2.

$\cup_{J\in\Omega_{m}}I_{J}=[0,1).$

###### Proof.

If $m=1$ then there are two $J$’s in  $\Omega_{1}$: $J_{1}=\\{(0)\\}$ and
$J_{2}=\\{(1)\\}$. From the definition we compute $I_{J_{1}}=[0,\beta^{-1})$
and $I_{J_{2}}=[\beta^{-1},1)$. Thus the lemma is true for $m=1$. Inductively
assume it is true for $m$ and suppose $J\in\Omega_{m}$. We have
$I_{J}=[L_{J,m},L_{J,m}+\beta^{-(m+1)}+\cdots+\beta^{-(n+l(J))})$. If
$r(J)=n-1$ then from the definition of $\beta$,
$I_{J}=[L_{J,m},L_{J,m}+\beta^{-(m+1)})$. If
$\tilde{J}=(j_{1},\cdots,j_{m},0)\in\Omega_{m+1}$, then
$L_{\tilde{J},m+1}=L_{J,m},r(\tilde{J})=0$ and
$I_{\tilde{J}}=[L_{J,m},L_{J,m}+\beta^{-(m+2)}+\cdots+\beta^{-(n+m+1)})=[L_{J,m},L_{j,m}+\beta^{-(m+1)})=I_{J}$.
On the other hand suppose $r(J)<n-1$. As before if
$\tilde{J}=(j_{1},\cdots,j_{m},0)\in\Omega_{m+1}$ we have
$I_{\tilde{J}}=[L_{J,m},L_{J,m}+\beta^{-(m+1)})$. Let
$J^{\prime}=(j_{1},\cdots,j_{m},1)\in\Omega_{m+1}$ so that
$r(J^{\prime})=r(J)+1$. Then

$\displaystyle I_{J^{\prime}}$
$\displaystyle=[L_{J^{\prime},m+1},L_{J^{\prime},m+1}+\beta^{-(m+2)}+\cdots+\beta^{-(m+n-r(J))})$
$\displaystyle=[L_{J,m}+\beta^{-(m+1)},L_{J,m}+\beta^{-(m+1)}+\cdots+\beta^{-(m+n-r(J))}).$

Since $I_{J}=[L_{J,m},L_{J,m}+\beta^{-(m+1)}+\cdots+\beta^{-(m+n-r(J))})$ we
have

$I_{\tilde{J}}\cup I_{J^{\prime}}=I_{J}.$

We have shown that every $I_{J}$ with $J\in\Omega_{m}$ is a union $I_{K}$’s
with $K\in\Omega_{m+1}$. Since our induction hypothesis was
$\cup_{J\in\Omega_{m}}I_{J}=[0,1)$ it follows that
$\cup_{J\in\Omega_{m+1}}I_{J}=[0,1)$ completing the induction. ∎

######  Corollary A.3.

Any $x\in[0,1)$ has an expansion $x=\sum_{k=1}^{\infty}x_{k}\beta^{-k}$,
unique except for  base-$\beta$ fractions of the form $L_{J,m}>0$ as long as
the $x_{k}$’s follow the standard restrictions.

###### Proof.

Given $x\in[0,1)$ and $m\geq 1$, $x\in I_{J}$ for a unique $J\in\Omega_{m}$.
We label this $J$, $J_{m}$. As we saw in the proof of Lemma A.2, $I_{J_{m}}$
is the disjoint union of at most two $I_{J}$’s with $J\in\Omega_{m+1}$, each
of which has the same $j_{1},j_{2},\cdots,j_{m}$.  Now, $x$ must be in one of
them whose $J$ we label $J_{m+1}$. Thus there is a sequence of intervals
$I_{J_{m}}$ with $x\in I_{J_{m+1}}\subset I_{J_{m}}$. If we set
$(x_{1},x_{2},\cdots,x_{m})=(j_{1},j_{2},\cdots,j_{m})=J_{m}$, then
$|x-\sum_{k=1}^{m}x_{k}\beta^{-k}|<\beta^{-m}\rightarrow 0$ as $m\to\infty.$
This shows that every $x\in[0,1)$ has a base-$\beta$ expansion.

Suppose
$x=\sum_{i=1}^{\infty}x_{i}\beta^{-i}=\sum_{i=1}^{\infty}y_{i}\beta^{-i}$. We
assume the sequences of $x_{i}$’s and $y_{i}$’s follow the usual restrictions
and that $x$ is not a base-$\beta$ fraction. Suppose that $x_{i}=y_{i}$ for
$i<k$ and $x_{k}<y_{k}$. Let $S_{k}=\sum_{i=1}^{k-1}x_{i}\beta^{-i}$. Then
$x\leq
S_{k}+(\beta^{-(k+1)}+\cdots+\beta^{-(k+n-1)})(1+\beta^{{}_{n}}+\cdots)=S_{k}+(\beta^{-(k+1)}+\cdots+\beta^{-(k+n-1)})(1-\beta^{-n})^{-1}=S_{k}+\beta^{-k}.$
On the other hand, using the representations with $y_{i}$’s,
$x>S_{k}+\beta^{-k}$, greater because $x$ is not a base-$\beta$ fraction. This
contradiction proves the uniqueness. ∎

##  Appendix B Some properties of the roots  of $x^{n}-(1+\cdots+x^{(n-1)})$

We list some properties for $n\geq 2$ in the form of a lemma:

######  Lemma B.1.

1) There is a unique positive root. This root, $\beta\in(1,2)$.
2) If $n$ is even there is one negative root. All other roots are non-real. If
$n$ is odd all roots other than $\beta$ are non-real.
3) If we label the roots $\lambda_{j}$ with
$|\lambda_{1}|>|\lambda_{2}|\geq\cdots\geq|\lambda_{n}|$, with
$\lambda_{1}=\beta$, then $|\lambda_{j}|<1$ for $j\geq 2$,
$|\lambda_{2}|\geq\beta^{-1/(n-1)}$ for all $n$. If $n$ is even then
$\lambda_{n}<0$ and in fact $|\lambda_{n}|<|\lambda_{n-1}|$. We also have
$|\lambda_{j}|\geq 3^{-1/n}$. The inequality
$|\lambda_{2}|\geq\beta^{-1/(n-1)}$ is actually an equality for $n=2$ and $3$.
3) All roots are simple.
4) $\beta$ is increasing in $n$. In fact $2-\beta\leq\beta^{-n}$ so that
$\beta$ approaches $2$ exponentially fast as $n$ increases.

######  Remark B.2.

What is missing is a better upper bound for $|\lambda_{2}|.$

###### Proof.

Much of this is well known, see below. We note first that
$p(x)=x^{n}-(1+x+\cdots+x^{n-1})$ is the determinant of the matrix $xI-M$,
cf. (3.11). $M$ has non-negative entries and $M^{n}$ has all positive entries
and therefore it is well known [6] that there is a unique positive root which
is strictly larger than the absolute value of all other roots. All roots are
simple and $|\lambda_{j}|<1$ for $j>1$ (see [7]). The latter fact is important
for the convergence rate of $f_{m}$ to $f_{\beta}$ and it is not trivial to
prove. The equation $x^{n}=1+x+\cdots+x^{n-1}$ can be rewritten
$1=x^{-1}+x^{-2}+\cdots+x^{-n}$. From this equation it is easily seen that the
positive solution, $\beta>1$, increases with $n$. We have
$1+x+\cdots+x^{n-1}=(1-x^{n})/(1-x)$ so that we can rewrite the first equation
as

$x^{n}=(x^{n}-1)/(x-1)$

or

$x^{n+1}+1=2x^{n}$

which gives

$2-x=x^{-n}.$

This shows $\beta<2$ and that since $\beta$ increases with $n$, $2-\beta$
decreases exponentially in $n$. If we factor
$p(x)=\prod_{j=1}^{n}(x-\lambda_{j})$ we see that
$\prod_{j=1}^{n}\lambda_{j}=(-1)^{n+1}$ which implies
$\beta|\lambda_{2}|^{n-1}\geq 1$ or $|\lambda_{2}|\geq\beta^{-1/(n-1)}$. The
following proof is from [1]: From the equation $x^{n+1}+1=2x^{n}$ we find
$2|x|^{n}\geq 1-|x|^{n+1}$ or $|x|^{n}\geq(2+|x|)^{-1}$. Applying this to
$x=\lambda_{j},j\geq 2$, using $|\lambda_{j}|<1$, we find
$|\lambda_{j}|>3^{-1/n}$. In particular all roots other than $\beta$ approach
the unit circle as $n$ increases.

Suppose $x$ is a negative root. If $n$ is odd we have $-2|x|^{n}=1+|x|^{n+1}$
which is impossible; so  there are no negative roots if $n$ is odd. If $n$ is
even we have $2|x|^{n}=1-|x|^{n+1}$. If $\lambda$ is another root then
$2|\lambda|^{n}=|1+\lambda^{n+1}|\geq 1-|\lambda|^{n+1}$. Subtracting we find
$2(|\lambda|^{n}-|x|^{n})\geq|x|^{n+1}-|\lambda|^{n+1}$. This implies
$|\lambda|\geq|x|$. If $|\lambda|=|x|$ then the same argument as we just gave
shows that $2|\lambda|^{n}>1-|\lambda|^{n+1}$ results in $0>0$ so we must have
$2|\lambda|^{n}=1-|\lambda|^{n+1}=|1+\lambda^{n+1}|$. This implies
$\lambda^{n+1}<0$ and since $2\lambda^{n}=1+\lambda^{n+1}$, $\lambda<0$.
Finally this means $\lambda=x$. Thus there is only one negative root (and then
only when $n$ is even) and all other roots are strictly larger in absolute
value. If $n$ is even we know there is at most one negative root and there is
one positive root. As before  $\prod_{j=1}^{n}\lambda_{j}=(-1)^{n+1}$. Since
all non-real roots come in complex conjugate pairs this shows that if $n$ is
even there is indeed one negative root.

If $n=2$ or $3$ we have $\beta\lambda_{2}=-1$ and $\beta|\lambda_{2}|^{2}=1$
respectively. So in both cases we have $|\lambda_{2}|=\beta^{-1/(n-1)}.$ ∎

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