Timeline for For $x$ irrational, is $a_{n} =\sum_{k=1}^{n}(-1)^{⌊kx⌋}$ unbounded?
Current License: CC BY-SA 4.0
17 events
| when toggle format | what | by | license | comment | |
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| S Apr 29, 2020 at 2:38 | history | suggested | Hans | CC BY-SA 4.0 |
There is no need for the center dots.
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| Apr 29, 2020 at 2:22 | review | Suggested edits | |||
| S Apr 29, 2020 at 2:38 | |||||
| Apr 28, 2020 at 21:42 | history | edited | Wlod AA | CC BY-SA 4.0 |
Easier on eyes, lesz confusing at first glance.
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| Apr 28, 2020 at 18:24 | answer | added | Kurisuto Asutora | timeline score: 4 | |
| Apr 26, 2020 at 18:38 | answer | added | Paata Ivanishvili | timeline score: 11 | |
| Apr 26, 2020 at 9:01 | answer | added | juan | timeline score: 13 | |
| Apr 26, 2020 at 8:35 | comment | added | YCor | Here's a related reference: pdfs.semanticscholar.org/9cb4/… The case they mention as settled ($g(x)=\{2x\}$ sounds at least as hard as the current case. So I believe the result is true and known. | |
| Apr 26, 2020 at 8:21 | comment | added | YCor | In particular, if $T$ is considered as operator on $L^2([0,2])$, then estimating $\|(\mathcal{A}_nT)g\|_2$ should be an exercise using Fourier transform. | |
| Apr 26, 2020 at 8:14 | comment | added | YCor | It's a very natural question from the point of view of ergodic theory. One has the dynamical system $u:t\mapsto t+x$ on the circle 𝐑/2𝐙. In turn 𝑢 induces an operator 𝑇 on functions or measures by $𝑇(𝑓(𝑡))=𝑓(𝑢(𝑡))=𝑓(𝑡+𝑥)$. In ergodic theory it's natural to estimate the average $\mathcal{A}_nT=\frac1n\sum_{i=1}^nT^n$ in various ways. Here the question is just evaluating $(\mathcal{A}_nT)g$ at $0$, for $g=\mathbb{1}_{[0,1]}-\mathbb{1}_{[1,2]}$ (namely asking if it's 𝑂(1/𝑛) or not). | |
| Apr 26, 2020 at 7:48 | history | edited | YCor |
edited tags
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| Apr 26, 2020 at 7:39 | history | edited | YCor |
edited tags; edited tags
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| Apr 26, 2020 at 7:30 | history | edited | YCor | CC BY-SA 4.0 |
moved question to text
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| Apr 26, 2020 at 6:10 | answer | added | Ville Salo | timeline score: 27 | |
| Apr 26, 2020 at 5:50 | comment | added | dhy | @ToddTrimble If the infimum of $q^2|x-\frac{p}{q}|$ over all rational approximations $\frac{p}{q}$ with $q$ odd is zero, then this sum is unbounded. (For each fixed integer $k$, consider $\displaystyle\sum_{i=m}^{m+kq-1}(-1)^{\lfloor ix\rfloor}$ for large $q$ and "generic" $m$.) Almost every $x$ (in the sense of Lebesgue measure) satisfies this - this follows from the Duffin-Schaeffer conjecture (now the Koukoulopoulos-Maynard theorem), but I'm sure that's massive overkill. | |
| Apr 26, 2020 at 5:04 | comment | added | Paata Ivanishvili | I am a little bit curious where is this question coming from? | |
| Apr 26, 2020 at 1:38 | comment | added | Todd Trimble | I must confess that it's not obvious to me that it's unbounded for any irrational $x$. Can someone set me straight? | |
| Apr 26, 2020 at 1:29 | history | asked | Chennes | CC BY-SA 4.0 |