Can anyone provide a proof of the equidistribution theorem using Weyl's criterion for the case of $c*a \,\,\, \text{(mod 1)}$ where $c=2^n: \,\,\, \forall n \in N_0$ for irrational algebraic $a$? The original equidistribution theorem is for the case that $c = n: \,\,\, \forall n \in N_0$. Much thanks in advance.
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4$\begingroup$ Theorem 4.1 of Kuipers & Niederreiter says that if $a_1,a_2,\dots$ is any sequence of distinct integers then the sequence $a_1x,a_2x,\dots$ is uniformly distributed mod 1 for almost all real $x$. Of course this says nothing about any particular $x$, nor about the set of algebraic $x$, as that set has measure zero. The Theorem is due to Weyl, and it's noted that the case $a_n=b^n$, $b\ge2$ an integer, was studied by Hardy & Littlewood in 1914. $\endgroup$– Gerry MyersonCommented Jul 17, 2014 at 23:39
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4$\begingroup$ Asking that $\{2^n \alpha\}_{n \ge 1}$ be equidistributed mod $1$ is exactly equivalent to asking that $\alpha$ be normal in base $2$. Not a single irrational algebraic number is known to be $2$-normal; for some partial results, see this paper of Bailey, Borwein, Crandall, and Pomerance: crd.lbl.gov/~dhbailey/dhbpapers/algebraic.pdf $\endgroup$– so-called friend DonCommented Jul 18, 2014 at 3:55
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2 Answers
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You are asking whether a number $a$ is a normal number in base 2 or not. This is an extremely complicated problem, and very far from being solved. Generally speaking, there exist several constructions of normal numbers, but we don't have any tool to determine whether a given number is normal or not. It is (sometimes) conjectured that all algebraic irrationals are normal, but a proof of such a statement is very, very, very, very, very far away.
You may be interested in the paper
Bailey, D.; Crandall, R.E.: On the random character of fundamental constant expansions. Experiment. Math. 10 (2001), no. 2, 175–190.
where this topic is discussed.
(By the way, the topic has a connection with ergodic theory. You are looking at the orbit of $a$ under the map $T:~x \mapsto 2x \mod 1$.)
answered Jul 18, 2014 at 8:39
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For the coefficients $2^n$ the equidistribution theorem fails. In fact it is easy to exhibit an irrational $a$ such that the sequence $(2^na)_{n=0}^\infty$ is not even dense in $(0,1)$ modulo $1$. For example, take an $a\in (0,1)$ whose binary expansion consists of increasing blocks of $1$'s seperated by $0$'s: $$ a=0.101101110111101111101111110\dots $$
Added. For algebraic irrationals such as $a=\sqrt{2}$, it is not even known if every digit occurs infinitely often in a given base, let alone $(2^n a)_{n=0}^\infty$ being dense or even equidistributed in $(0,1)$ modulo $1$.
answered Jul 17, 2014 at 21:11
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$\begingroup$ Bah, I meant irrational algebraics. My bad. I've corrected it in the question above. I've still voted for your answer since you answered my original question. That's a clever construction, though I'm guessing it's not algebraic. It reminds me of Liouville's proof of the existence of transcendentals. If it is algebraic, how about the case of $a = \sqrt(x): x \in W$ since that's what I'm particularly interested in. Thanks again for the help. $\endgroup$– jgonagleCommented Jul 17, 2014 at 21:49
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$\begingroup$ @jgonagle: The number in my response is irrational, and probably transcendental as well. Note that transcendental is a stronger property than irrational. Also, what do you denote by $W$? The natural numbers are usually denoted by $\mathbb{N}$. $\endgroup$ Commented Jul 17, 2014 at 21:50
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1$\begingroup$ @jgonagle: If you ask about $2^n$ times a quadratic irrational modulo $1$, such a sequence is probably equidistributed in $(0,1)$, but I doubt it can be proved by present technology. $\endgroup$ Commented Jul 17, 2014 at 21:52
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$\begingroup$ Right, that's why I want to exclude transcendentals and just focus on the algebraics. As for the use of $W$, that's just an old habit to disambiguate $N$ vs $N_0$. I really should have said $N_0$. Thanks for the tip. $\endgroup$– jgonagleCommented Jul 17, 2014 at 21:57
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$\begingroup$ @jgonagle: See my added section. $\endgroup$ Commented Jul 17, 2014 at 22:03