# Tag Info

169

Note that if $\pi$ were rational (with even numerator), then $\sin(n)$ would equal $1$ periodically, so the series would diverge. Similarly if $\pi$ were a sufficiently strong Liouville number. Thus, to establish convergence, one must use some quantitative measure of the irrationality of $\pi$. It is known that the irrationality measure $\mu$ of $\pi$ is ...

37

On the OP request, here is the plot of first 10000 partial sums. Following Terry Tao's suggestion, here is the plot of ($n$th partial sum) $+2^{\frac32}/\sqrt{\pi n}$ for $n$ up to one million: The thick line in the beginning actually consists of high frequency oscillations - in the range up to 2000 it looks like this: (I hope there are no rounding ...

35

Let $a_n$ be an increasing sequence of positive integers which grows really fast, say $a_{n+1} > \exp(a_n)$. Take $A = \{10^{-1}, \sum 10^{-a_n}\}$. Then $d_A(a_n) \leq 2\cdot 10^{- a_{n+1}} \leq 2\cdot 10^{-\exp a_n}$, so $A$ cannot be tame. EDIT. One could replace $1/10$ by some transcendental $0<x<1$ such that $\sum x^{-a_n}$ is transcendental ...

31

See Georges Rhin: Approximations de Padé et mesures effectives d'irrationalité. (French) [Padé approximants and effective measures of irrationality] Séminaire de Théorie des Nombres, Paris 1985–86, 155–164, Progr. Math., 71, Birkhäuser Boston, Boston, MA, 1987. Inequality (8) there shows that if $u_0$, $u_1$ and $u_2$ are integers with $H= \max(|u_1|,|u_2|)$...

27

It's indeed unbounded for every irrational $x$. Let me identify points of $\mathbb{R}/\mathbb{Z}$ with their representatives on $[0,1)$, and order it by the usual order $<$ of $\mathbb{R}$ applied to the representatives. Replace $x$ by $y = x/2$, and the question becomes whether for all $m$ there exists $k$ such that the orbit of $0$ in the irrational ...

26

$$\sqrt{2+\sqrt{3}}-\sqrt{2- \sqrt{3}}=\sqrt{2}$$

26

Mahler proved in 1957 (see here) that if $q$ is a positive rational number which is not an integer, then the distance of $q^n$ to the nearest integer is $(1-o(1))^n$. In particular, taking $q=3/2$, we have for $n$ sufficiently large that $$\lfloor q^n\rfloor+1-q^n>(q/2)^n>\lfloor q^n\rfloor/2^n.$$ Rearranging the two sides, we get for $n$ sufficiently ...

25

Here's an argument that for $A$ a finite set of algebraic numbers, $d_A(n)$ decays at most exponentially. Suppose $A$ is contained in some number field $K$. For $x\in K^\times$, there's a product formula $$\prod_v |x|_v=1,$$ where $v$ runs over the valuations on $K$ and $|\cdot|_v$ is a suitably normalized absolute value. In order for some absolute value ...

25

As Terry mentions in the comments, the reason for the $\sqrt{5}$ is that the limiting case, the golden ratio, forces it. There is a very neat explanation of all of this in the classic number theory book by Hardy and Wright, pages 209 to 212. I give a brief sketch of the ideas. Why $\phi$ is the worst case. As Hardy and Wright put it, "from the point of ...

24

The product does not tend to the limit zero. For any irrational number $\alpha$ one can show that $$\limsup_{N\to \infty} \prod_{n=1}^{N} |1- e(n\alpha)| = \infty. \tag{1}$$ (Here I use the usual notation $e(\alpha) =e^{2\pi i\alpha}$, and the product in the question is stated for $\alpha/2$ rather than $\alpha$). I'll prove a slightly weaker result; ...

20

Embarrassingly, it turns out that Greg Kuperberg previously answered my question: see Theorem 2.11 of this paper. In particular, he proved that all sets of algebraic numbers of tame, and he also observed that there exist sets including non-algebraic numbers that are non-tame (though he didn't include that observation in his paper). Moreover, he did this ...

20

Delone (1930) and Nagell (1928) showed for any nonzero integer $d$ that the equation $x^3 - dy^3 = 1$ has at most one solution in integers $(x,y)$ besides $(1,0)$, with no constraint on the signs of $x$ and $y$. In particular, since $x^3 - 2y^3 = 1$ has the integral solution $(-1,-1)$, there is no integral solution $(x,y)$ in positive integers. This ...

20

Semilog plot building on მამუკა ჯიბლაძე's picture, this time to $10^7$

19

This integral has a series counterpart $$\sum_{k=0}^\infty \frac{240}{(4k+5)(4k+6)(4k+7)(4k+9)(4k+10)(4k+11)}=\frac{22}{7}-\pi$$ https://math.stackexchange.com/a/1657416/134791 (UPDATE Peter Bala New series for old functions https://oeis.org/A002117/a002117.pdf, 2009, formula 5.1) Equivalently, $$\sum_{k=1}^\infty \frac{240}{(4k+1)(4k+2)(4k+3)(4k+5)(4k+6)... 19 You can do this without much number theory. View your field as a finite-dimensional vector space over \mathbb Q. Then every element acts linearly on the field, so it acts as some matrix with rational entries. The actual element you want to bound is an eigenvalue of this matrix. We can lower bound it by lower bounding the determinant and upper bounding ... 19 Robert Israel's guess that this is true for t an algebraic irrational does actually follow from Schanuel's conjecture. Apply Schanuel's conjecture to \log p and t \log p for n different primes p. These are linearly independent unless t is the ratio of the logarithms of two integers, which it isn't because t is an algebraic irrational. So we ... 19 Setting a_0=A^7, b_0=B^7 and then a_{n+1}=[b_n^{-1},a_n], b_{n+1}=[a_n,b_n] seems very efficient. The length grows like C \alpha^n with \alpha= \frac{3 + \sqrt{17}}2 and the operator norm distance to the identity matrix gets squared in each step, since$$\|1 - [a,b]\| \leq 2 \|1-a\| \cdot \|1-b\|.$$I used a similar construction in On the length of ... 18 Any number field K which has a real place is generated by a root of a Pisot polynomial. So the answer is no. For example, we can take p a prime \geq 11, and use the real subfield of \mathbb{Q}(\zeta_p), to get a Pisot polynomial whose splitting field is cyclic of degree (p-1)/2. Suppose K has r real places, and 2s complex places. Let V_i ... 16 It seems that such a sequence exists. 1. Firstly, we take an auxiliary sequence a(n)=\{(n+1)\sqrt2\} for n\geq 0. For every m>n the standard estimate yields, say,$$ |a(m)-a(n)|=|(m-n)\sqrt2-p|=\frac{|2(m-n)^2-p^2|}{(m-n)\sqrt2+p}>\frac1{10|m-n|}; $$here p=[(m+1)\sqrt2]-[(n+1)\sqrt2]\leq 8(m-n). 2. Now we construct the sequence of points ... 16 Squares of elements of the 3-adic integers \mathbb{Z}_3 are congruent to 0 or 1 modulo 3, thus they are all at 3-adic distance 1 from 2. Squares of elements of \mathbb{Q}_3 \setminus \mathbb{Z}_3 are 3-adically even further away from the target. Thus in \mathbb{Q}_3, you cannot approximate a square root of 2 at all well - just like ... 16 What is known is that every real irrational has a 0 in its g-ary expansion for infinitely many g. WLOG take 0 < x < 1. Taking an even-numbered convergent of the continued fraction of x gives us a rational p/q such that$$\frac{p}{q} < x < \frac{p}{q} + \frac{1}{q^2}$$so that the first two digits in the base-q expansion of x ... 15 I have no idea how to approach the general problem, but here is a quick observation: A. Let \alpha = 2\beta so that \beta is irrational if and only if \alpha is so. Define f by f(x) = \log|2\sin\pi x|. Then$$ \log \left| \prod_{k=1}^{n} (1 - e^{\pi k i \alpha} ) \right| = \sum_{k=1}^{n} f(k\beta). $$Now by the Riemann-integrable criterion for ... 14 David Speyer has beaten me to it, but for what it's worth, a simple explicit example is$$p(X)=X^4-2X^3-5X^2-4X-1$$which is a Pisot polynomial, and has Galois group the dihedral group of order 8. How did I find it? I started from observing that if \vartheta is the positive root of X^2-X-1, then \alpha=1+\sqrt{\vartheta} is an algebraic unit with only ... 13 You cannot do much better than what you observed, because the fractional parts \{2^n x\} are essentially the tails in the binary expansion of x, so they can be bounded away from zero (even for rational numbers x). However, Furstenberg (1967) proved that the fractional parts \{2^m3^n x\} are dense in (0,1) for any irrational x, and in fact the ... 13 We have Theorem. Let \psi(x) and \varphi(x) be positive increasing functions such that$$\int_1^\infty \frac{dx}{\psi(x)}=+\infty,\qquad \int_1^\infty \frac{dx}{\varphi(x)}<+\infty.$$Then for almost all \alpha\in(0,1) we have$$\Omega(\log N\cdot \psi(\log\log N))\le\sup_{n\le N}\sum_{j=1}^n(-1)^{\lfloor j\alpha\rfloor}=O(\log N\cdot \varphi(\...

12

For a fixed $\alpha$, the number $N_{\alpha}(\epsilon)$ is bounded by a polynomial function of $1/\epsilon$. The proof of this requires either Faltings's product theorem, or Esnault and Viehweg's multidimensional Dyson lemma. See section 6.5 of Bombieri and Gubler's book (Heights in Diophantine Geometry), with particular attention to point 6.5.8. Roth's ...

12

Here is a full solution for the modified problem, inspired by Gro-Tsen's valuable comment. 1. There are infinitely many rational numbers $a/b\in\mathbb{Q}$ in lowest terms such that $$\left|\frac{a^2}{b^2}-2\right|_\infty\ll\frac{1}{b}\qquad\text{and}\qquad \left|\frac{a^2}{b^2}-2\right|_7\ll\frac{1}{b}.$$ To see this, we work in $\mathbb{Z}[\sqrt{2}]$, ...

11

Fix $\varepsilon>0$. We have to prove (for large enough $n$) that there exist non-negative integers $a,b$ such that $\log_2 n\leqslant a+b\log_2 3<\log_2 n+\varepsilon$. This follows from the fact that fractional parts of $b\alpha$, $\alpha:=\log_2 3$, are dense in $(0,1)$. Indeed, choose non-negative integers $b_1,\dots,b_k$ so that fractional parts ...

11

I believe that there is a completely explicit upper bound for $N_\alpha(\epsilon)$ (more generally counting in relative number fields and using more then one, possibly non-archimedean, absolute value) in the following paper: Robert Gross, A note on Roth's theorem. J. Number Theory 36 (1990), no. 1, 127–132. MR1068678 Of course, it's not going to be pretty....

11

The answer is no. Start with $\alpha$ between 3.25 and 3.75; take squares of those 2 to see that one can restrict $\alpha^2$ to being between 11.25 and 11.75; take 3/2 powers of those to see that one can restrict $\alpha^3$ to being between 38.25 and 38.75; take 4/3 powers of those etc. One gets the idea: since $\alpha>3$ a 0.5 wide interval is magnified ...

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