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Given $x \in [0, 1]$, we denote by $e(x)$ the complex number $e^{2 \pi i x}$.

Can we characterise the set of rationals $x$ for which the sum

$$A_N(x)\, :=\, \sum_{n = 0}^N e(2^n x)$$

remains bounded as $N \to \infty$?

Remarks:

  • We allow the bound to depend on $x$, but not $N$.

  • It can be shown that the sum is unbounded for almost every (but not every!) irrational $x$.

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    $\begingroup$ Nearly everyone uses the notation $e(x)$ for $e^{2\pi i x}$ if they use any notation at all. You care about the sum $\sum_{n = 0}^N e(2^n x)$ (you have the index of the sum as $i$ by mistake). $\endgroup$ Commented Oct 13 at 17:08
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    $\begingroup$ Let $q$ be the denominator of $x$. We may assume that $q$ is odd. Note that the summands are periodic with period $\varphi(q)$ (Euler totient function). So the question actually is when $\sum_{n=1}^{\varphi(q)}\alpha_x^{2^n}=0$. $\endgroup$ Commented Oct 13 at 17:09
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    $\begingroup$ If I am not mistaken, for $x$ with denominators $p^m$ with odd prime $p$ this holds iff $\nu_p(2^{p-1}-1)<m$ $\endgroup$ Commented Oct 13 at 18:58
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    $\begingroup$ @FedorPetrov I think this is right and this shows that a necessary condition is that there exists a prime $m$ dividing the denominator with multiplicity $m$ such that $v_p ( 2^{p-1}-1)<m$. (Proof: Multiply a bunch of sums with prime power denominator together to get a sum of Galois conjugates of the original sum.) But probably that's not also sufficient. $\endgroup$
    – Will Sawin
    Commented Oct 14 at 15:52
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    $\begingroup$ @WillSawin a slightly different sufficient condition is that the order of 2 modulo $n$ is not equal to the order of 2 modulo the product of all prime divisors of $n$ (then we may partition everything onto regular $p$-gons). I expect it to be necessary. $\endgroup$ Commented Oct 14 at 16:05

2 Answers 2

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As Fedor Petrov expects in the comments, this is true for $x=a/b$ if and only if the order of $2$ modulo $b$ is greater than the order of $2$ modulo the product of prime factors of $b$.

To prove this, the first key observation is given by Peter Mueller in the comments: Since the sequence is eventually periodic, it is bounded if and only if the sum over a period is $0$. We can also just drop the first few terms by multiplying $x$ by a power of $2$ until $b$ is odd, in which case the sequence is literally periodic. So it is equivalent to say $$\sum_{n=0}^{m-1} e( 2^n a/b) =0 $$ where $m$ is the multiplicative order of $2$ modulo $b$.

The next key observation is from GH from MO in the comments, which is that $\sum_{n=0}^{m-1} e( 2^n a/b) $ is a Galois conjugate of $\sum_{n=0}^{m-1} e( 2^n a'/b) $ whenever $a'$ and $a$ are both relatively prime to $b$, so one of them is zero if and only if all of these are zero, i.e. if and only if $$\sum_{n=0}^{m-1} e( 2^n a'/b) =0 \textrm{ for all } a' \in (\mathbb Z/b\mathbb Z)^\times$$

Now finally I will make a contribution of my own: By Fourier analysis, we have $$\sum_{n=0}^{m-1} e( 2^n a'/b) =0 \textrm{ for all } a' \in (\mathbb Z/b\mathbb Z)^\times$$ if and only if $$ \sum_{c \in (\mathbb Z/b\mathbb Z)^\times} e( c/b) \chi(c) =0 \textrm{ for all } \chi\colon (\mathbb Z/b\mathbb Z)^\times \to \mathbb C^\times \textrm{ such that } \chi(2)=1.$$

Indeed the first sums depend on the equivalence class of $a'$ in $(\mathbb Z/b\mathbb Z)^\times/\langle 2\rangle$, and the second sums give the finite Fourier transform over this group.

Now the theory of Gauss sums tells us that $\sum_{c \in (\mathbb Z/b\mathbb Z)^\times} e( c/b) \chi(c) =0 $ if and only if $\chi$ factors through $(\mathbb Z/(b/p)\mathbb{Z})^\times$ for some $p$ with $p^2$ dividing $b$.

If the order of $2$ modulo $b$ is greater than the order of $2$ modulo the product of prime factors of $p$, raising $2$ to the order modulo the product of prime factors and then raising to a suitable product of prime factors, we can find a power of $2$ that is congruent to $1$ mod $b$ but not modulo $b/p$ for some such prime, showing that every $\chi$ admits such a factorization. Conversely, if these orders are equal, then every character of the kernel of $(\mathbb Z/ b \mathbb Z)^\times \to (\mathbb Z/r \mathbb Z)^\times$, with $r$ the product of prime factors, extends to a character of $(\mathbb Z/ b \mathbb Z)^\times$ trivial on $2$. The kernel is cyclic and extending a faithful character produces a nonzero Gauss sum.

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    $\begingroup$ 1. This is a nice example of virtual MO collaboration (transcending space and time)! The credit is all yours, of course. 2. Your argument reminds me of certain nonvanishing results for central $L$-values (e.g. by Rohrlich) where the values are Galois conjugates, so it suffices to show that their sum is nonzero. 3. I fixed some typos. $\endgroup$
    – GH from MO
    Commented Oct 15 at 0:44
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    $\begingroup$ @GHfromMO Collaboration transcending time, space and even language has been a big theme in my life recently... anyway really nicely done to everyone. $\endgroup$
    – Nate River
    Commented Oct 15 at 2:41
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Not a full answer, but here is a potentially useful observation - let $x$ be a real number, rational or otherwise. By writing $x$ in binary, truncating to a certain number of binary digits to the right of the decimal point, and then counting the error terms, we can show that

$$A_N (x) \leq o(1) + \inf_{k \geq 2 \log_2 N} \sum_{n = 1}^{N} e(T_{k}(2^n x)).$$

where $T_m$ denotes the truncation to $m$ binary digits operator, and the $o(1)$ denotes an error term independent of $x$ that goes to $0$ in absolute value as $N \to \infty$.

This formula allows to get very precise estimates on the sum. Indeed, since $x \to 2^n x$ is just the shift map in binary, we can use this formula to show that in fact, for the rational number $\frac{10}{99} = 0.10101010\dots$, the sum $A_N (\frac{10}{99})$ is in fact bounded as $N \to \infty$ (see here for proof!), and we may even obtain the exact explicit bound if desired.

The formula may also be useful for checking boundedness on irrationals.

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