# Let $M(b)=\lim_{n\rightarrow\infty}\sum_{k=n+1}^{2n} \{b^k x\}/k$. Do we have $\int_{1}^{2}M(b)db = 1/4$?

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.

• Comments are not for extended discussion; this conversation has been moved to chat. – 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:

Theorem

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 $$b\in [1,2]-\mathcal{E}$$.

Theorem

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 $$\int_1^2 M(b) db =\frac12 \log 2.$$

• 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 $b \in [1,2]$. – Vincent Granville Mar 26 '19 at 13:55
• 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. – Sungjin Kim Mar 26 '19 at 14:12