Timeline for Is the following series consisting of equally distributed $\pm 1$ bounded?
Current License: CC BY-SA 3.0
18 events
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| Apr 18, 2017 at 11:07 | history | edited | YCor | CC BY-SA 3.0 |
removed "MSE" from title. added tag.
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| Apr 14, 2017 at 2:46 | vote | accept | Vim | ||
| Apr 13, 2017 at 12:19 | history | edited | CommunityBot |
replaced http://math.stackexchange.com/ with https://math.stackexchange.com/
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| Apr 11, 2017 at 23:29 | comment | added | Vim | @alpoge thanks for your help! I'll try to understand these high-quality comments (might take me some(a considerable amount of) time to digest so there might be delay in my feedback). | |
| Apr 11, 2017 at 22:49 | comment | added | Vim | @GeraldEdgar sorry, I was just thinking that it might deserve some higher level attention here on MO, didn't give that conclusion much thought. | |
| Apr 11, 2017 at 20:55 | answer | added | coudy | timeline score: 17 | |
| Apr 11, 2017 at 20:03 | answer | added | Christian Remling | timeline score: 8 | |
| Apr 11, 2017 at 19:50 | comment | added | alpoge | (Also there's no need to take differences of exceptions --- it follows by the same argument that, for an exceptional $m$, $mF_{n_i - 1}$ lies in $O(1)$ many congruence classes modulo $F_{n_i}$ and then the same conclusion follows. Anyway...) | |
| Apr 11, 2017 at 19:44 | comment | added | alpoge | (Apparently I can't edit a comment after a few minutes, but $\{\cdot\}$ should be $||\cdot||_{\mathbb{R}/\mathbb{Z}}$, the distance to the integer nearest $\cdot$, not the distance to the largest integer below $\cdot$.) | |
| Apr 11, 2017 at 19:39 | comment | added | alpoge | In sum we lose an $O(\log{X})$ from the exceptions, and we are now left with bounding $\sum_{i\leq k}\sum_{2Y_{i-1} < m\leq 2Y_i} (-1)^{\lfloor \frac{m F_{n_i - 1}}{F_{n_i}}\rfloor}$. But the $i$-th inner sum is now over an interval of length $2F_{n_i}$, and we may pair off each $2Y_{i-1} < m\leq 2Y_{i-1} + F_{n_i}$ with $m + F_{n_i}$. Note that the corresponding signs of the two differ by a factor of $(-1)^{F_{n_i - 1}} = -1$ by construction, whence the sum vanishes, QED. (Let me know if I've made a mistake! I think this looks plausible, at least.) | |
| Apr 11, 2017 at 19:37 | comment | added | alpoge | It follows that $\lfloor mb\rfloor = \lfloor \frac{m F_{n_i -1}}{F_{n_i}}\rfloor$ except for at most $O(1)$ exceptions in this interval. Indeed, if there is an exception $m$, then any other exception $m'$ must satisfy (on taking differences) $\left\{\frac{(m'-m) F_{n_i - 1}}{F_{n_i}}\right\}\ll F_{n_i}^{-1}$, and so $(m'-m) F_{n_i - 1}$ lies in one of $O(1)$ congruence classes modulo $F_{n_i}$. Since $(F_{n_i - 1}, F_{n_i}) = 1$ it follows that $m'$ must lie in $O(1)$ many congruence classes modulo $F_{n_i}$, and now note that the interval was only of length $2F_{n_i}$ in the first place. | |
| Apr 11, 2017 at 19:37 | comment | added | alpoge | Next write $Y_j := \sum_{i\leq j} F_{n_i}$, so that $Y_0 = 0$ and $2Y_k = X$. Note that $Y_i\ll F_{n_i}\leq Y_i$. Now our sum is $\sum_{n\leq X} a_n = \sum_{i\leq k} \sum_{2Y_{i-1} < n\leq 2Y_i} a_n$, so it suffices to show that each of the inner sums is $O(1)$. Note that the $i$-th inner sum is over an interval of length $2F_{n_i}$. So we'll use the approximation $b = \frac{1}{\phi} = \frac{F_{n_i-1}}{F_{n_i}} + O(F_{n_i}^{-2})$, which holds by e.g. Binet. Hence for $2Y_{i-1} < m\leq 2Y_i$, we have that $mb = \frac{m F_{n_i -1}}{F_{n_i}} + O(Y_i^{-1})$, since again $F_{n_i}\asymp Y_i$. | |
| Apr 11, 2017 at 19:35 | comment | added | alpoge | To see the computer-suggested bound $\sum_{n\leq X} a_n\ll O(\log{X})$, first write $X$, which we may assume is even (since we only lose $O(1)$ by pulling off the last term) greedily as $2\sum_{i\leq k} F_{n_i}$, with $n_i\not\equiv 0\pmod{3}$ (i.e., all $F_{n_i - 1}$ odd) --- here the $F$'s are Fibonacci numbers in increasing order, and by 'greedily' I mean take the largest one less than $X$ and repeat. Note that in general the largest $F_m\leq X$ with $m\not\equiv 0\pmod{3}$ is $\gg X$, so that $k\ll \log{X}$. | |
| Apr 11, 2017 at 16:30 | comment | added | Gerald Edgar | You conclude "not much attention" in MSE after only 12 hours? | |
| Apr 11, 2017 at 16:15 | history | edited | Joseph O'Rourke | CC BY-SA 3.0 |
Epsilon improvement to the LaTeX.
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| Apr 11, 2017 at 15:58 | comment | added | Vim | A friend also hinted at using continued fractions, saying he proved it is not bounded some time ago but apparently have forgot and lost all the details. | |
| Apr 11, 2017 at 14:19 | review | First posts | |||
| Apr 11, 2017 at 14:31 | |||||
| Apr 11, 2017 at 14:15 | history | asked | Vim | CC BY-SA 3.0 |