The brackets denote the fractional part. Based on heuristic arguments of a probabilistic nature (the fact that $M(b) = E(b) \log 2$ where $E(b)$ is the expectation of the equilibrium distribution of $x(n) = \{b^n x\}$ with $x$ a normal number, and using Frullani integrals), the result seems plausible. However, computations suggest that the result is almost correct, but not exactly. Is it equal to 1/4, or not? For the context about this problem, see section 4.3 in my new article on stochastic processes, available here.

As a curiosity, $M((1+\sqrt{5})/2)= \sqrt{5}/2 \log 2$, and $M(b)$ is usually not known explicitly, except for a few rare $b$'s such as the golden ratio, supergolden ratio, and the plastic number.

  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – Aloizio Macedo Mar 28 '19 at 23:50

The answer is $(\log 2)/2$. This follows from Koksma's General Metric Theorem. This is Theorem 4.3 of 'Uniform Distribution of Sequences' by Kuipers and Niederreiter:


Let $u_n(x), n=1,2,\ldots$ be a sequence of real numbers defined for all $x$ in an interval $[a,b]$. For each $n\geq 1$, let $u_n(x)$ be continuously differentiable on $[a,b]$. Suppose that for any two positive integers $m\neq n$, the function $u_m'(x)-u_n'(x)$ is monotone with respect to $x$ and that $|u_m'(x)-u_n'(x)|\geq K>0$, where $K$ does not depend on $x, m,$ and $n$. Then $u_n(x)$ is uniformly distributed modulo $1$ for almost all $x$ in $[a,b]$.

For this problem, let $x\neq 0$ be fixed, and write $u_n(b)=b^n x$. Since $b\in [1,2]$ and $x\neq 0$, we have $u_m'(b)-u_n'(b)=(mb^{m-1}-nb^{n-1})x$ is monotone with respect to $b$. The assumptions of this theorem is satisfied. Moreover, we obtain from the proof that for any $h\in \mathbb{Z}-\{0\}$, $$ S_h(N,b)=\frac 1N \sum_{n=1}^N e^{2\pi i h u_n(b)}, \ \ b\in [1,2],$$

satisfies $$ |S_h(N,b)|^2=O_b(\frac{\log N}N) $$ for almost all $b\in [1,2]$. Let $\mathcal{E}$ be the exceptional set.

Then we apply Erdos-Turan inequality and Koksma inequality for $b\in [1,2]-\mathcal{E}$.


  1. (Koksma) Let $f$ be a function on $I=[0,1]$ of bounded variation $V(f)$, and suppose we are given $N$ points $u_1, \ldots , u_N$ in $I$ with discrepancy $$ D_N:=\sup_{0\leq a\leq b\leq 1} \left|\frac1N \#\{1\leq n\leq N: u_n \in (a,b) \} -(b-a)\right|. $$ Then $$ \left|\frac1N \sum_{n\leq N} f(u_n) - \int_I f(u)du \right|\leq V(f)D_N. $$
  2. (Erdos-Turan) Let $u_1, \ldots, u_N$ be $N$ points in $I=[0,1]$. Then there is an absolute constant $C>0$ such that for any positive integer $m$, $$ D_N\leq C \left( \frac1m+ \sum_{h=1}^m \frac1h \left| \frac1N\sum_{n=1}^N e^{2\pi i h u_n}\right|\right). $$

Taking $f(x)=\{x\}$, $u_n=u_n(b)$ and $m=N$, we obtain $$ \left|\frac1N \sum_{n\leq N}\{u_n(b)\}- \frac12\right|=O_b(\frac{(\log N)^{3/2}}{\sqrt N}). $$

By partial summation, we have $$ \sum_{k=1}^{N} \frac{\{b^k x\} }k= \frac12 + O(\frac{(\log N)^{3/2}}{\sqrt N})+\int_{1-}^N \frac{ \frac12 t + O_b(\sqrt t (\log t)^{3/2}) }{t^2} dt. $$ Applying this with $N=2n$, and $N=n$, then subtract. We have $$ \sum_{k=n+1}^{2n} \frac{\{b^k x\} }k=\frac12 \log 2 + O_b(\frac{(\log n)^{3/2}}{\sqrt n}). $$ Thus, if $b\in [1,2]-\mathcal{E}$, then $M(b) = \frac12 \log 2$.

Note that for $\mathcal{E}$ has Lebesgue measure zero.

Combining these, we finally have $$ \int_1^2 M(b) db =\frac12 \log 2. $$

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  • $\begingroup$ The exception set is the whole interval $[1, 2]$ here, where $\{b^k x\}$ is not a uniform sequence if $x$ is a good seed (the set of bad seeds $x$ has lebesgue measure zero, true.) But the value should be $D \log 2$ with $D$ a constant around 0.38, not 1/2. See my chart for $E(b) = M(b) / \log 2$, the last chart in my article at dsc.news/2HLUjTO. $E(b)$ is always below 1/2 if $b \in [1,2]$. $\endgroup$ – Vincent Granville Mar 26 '19 at 13:55
  • $\begingroup$ I was concerned about the table too. How did you obtain the table for $E(b)$? With my answer here, the table should show that $E(b)$ is almost always $1/2$. It might have happened that the values you chose for $b$ to plot are in the exceptional sets. The limitation of my argument here is that we never know whether a given $b$ is in the exceptional set or not. $\endgroup$ – Sungjin Kim Mar 26 '19 at 14:12

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