Timeline for Proving $\sum_{i=1}^{n}\sum_{j=1}^{n}\left\{\frac{x_{i}}{x_{j}}\right\}\le \frac{9}{14}n^2$？
Current License: CC BY-SA 4.0
27 events
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| Jan 8 at 0:41 | vote | accept | math110 | ||
| Jan 5 at 12:18 | history | edited | Fred Hucht | CC BY-SA 4.0 |
LaTeX cosmetics in 2nd eq, limit direction in 3rd eq.
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| Jan 5 at 11:25 | comment | added | fedja | Moreover, since the LHS is invariant under the multiplication of all $x_j$ by the same constant and the RHS of your new inequality is obviously not, this new conjecture looks slightly outlandish as written and can easily be refuted by sending all $x_j$ to $0$ while preserving the ratios. I'm less demanding than @FredHucht in terms of etiquette but if you edit new questions into your posts, make sure that they make some sense at least. Also the word "conjecture" means something you have an extensive evidence for, just cannot prove, not some random thing that crossed your mind. | |
| Jan 5 at 11:18 | history | edited | YCor | CC BY-SA 4.0 |
formatting, added tags
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| Jan 5 at 8:21 | comment | added | Fred Hucht | @math110 It was me who corrected your example. You now have added another question in form of the conjecture, this is not so good. You should consider to accept the best answer to your original question (which might be the one by @fedja). Either move your conjecture question to a comment or ask a new question. | |
| Jan 5 at 1:09 | history | edited | math110 | CC BY-SA 4.0 |
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| Jan 5 at 0:26 | comment | added | math110 | @SidharthGhoshal,Thank you， It’s true that I wrote down the waiting list wrong at first.Ihave edit it | |
| Jan 5 at 0:25 | history | edited | math110 | CC BY-SA 4.0 |
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| Jan 4 at 17:22 | comment | added | Fred Hucht | I found the problem with the OP's example. Changing the definition of $x_i$ to, e.g., $x_i=\begin{cases} 1+i \epsilon & 0<i\leq2n/7\\ 2-(i+ 2n/7) \epsilon & 2n/7<i\leq5n/7\\ 4-(i+10n/7) \epsilon & 5n/7<i\leq n\\ \end{cases}$ gives the correct limit $9/14$. | |
| Jan 4 at 12:20 | answer | added | fedja | timeline score: 33 | |
| Jan 4 at 10:57 | comment | added | Fred Hucht | I used Mathematica, something like: Apart@FindSequenceFunction[Rationalize@N[Table[{n, n^-2 Total[FractionalPart[Outer[Divide, Array[x[##] &, n], Array[x[##] &, n]]], 2] /.x[i_] ->Piecewise[{{1 + i ee, 0 < i <= (2 n)/7}, {2 + (i - (2 n)/7) ee, (2 n)/ 7 < i <= (5 n)/7}, {4 + (i - (5 n)/7) ee, (5 n)/ 7 < i <= n}}]}, {n, 14, 100, 14}] /. ee -> 10^-30], n] | |
| Jan 4 at 10:41 | comment | added | Fred Hucht | @math110 I cannot verify your example from the edit 9h ago, please check. In this case I get $\frac{111}{196} < \frac {5}{8} < \frac {9}{14}$ for the limit. I also checked the other sign combinations ($ 1 \pm \ldots, 2 \pm \ldots, 4 \pm \ldots$). | |
| Jan 4 at 6:43 | comment | added | fedja | @FedorPetrov OK, reduced it to 1024 cases and checked them by brute force. The key idea is to prove the stronger inequality with $\{z\}+\{1/z\}$ replaced by $1+1/z$ for $z\ge 2$. The rest tomorrow: it is past midnight here now :-) | |
| Jan 4 at 5:08 | comment | added | fedja | @FedorPetrov Interesting idea, but, probably, not for large $n$. On the other hand, it is easy to reduce the problem to the consideration of a finite number of cases. Alas, my reduction needs to be refined quite a bit to make this observation useful with modern computers... | |
| Jan 4 at 1:01 | history | edited | math110 | CC BY-SA 4.0 |
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| Jan 3 at 22:08 | answer | added | Fedor Petrov | timeline score: 21 | |
| Jan 3 at 19:53 | answer | added | Fred Hucht | timeline score: 5 | |
| Jan 3 at 18:34 | comment | added | Fedor Petrov | is not it true that $\sum_{i,j} (\{x_i/x_j\}-9/14)u_iu_j\leqslant 0$ for all real $u_i$, that is, that the kernel $\{x/y\}-9/14$ is non-positive definite? | |
| Jan 3 at 18:25 | comment | added | Aleksei Kulikov | For the lower bound, if we take $n/2$ numbers slightly bigger than $1$ and $n/2$ numbers slightly less than $2$ then we can have a constant arbitrarily close to $\frac{5}{8}$ (which is very slightly less than $\frac{9}{14}$). | |
| Jan 3 at 17:29 | comment | added | Denis Serre | Just to make sure: what is exactly $\{y\}$ ? Is it $y-\lfloor y\rfloor$, or is it the distance to the integers ? | |
| Jan 3 at 17:05 | comment | added | David E Speyer | On the other hand, put $d_m = \text{sup} \tfrac{1}{m^2} \sum_{i,j=1}^m \left\{ \tfrac{x_i}{x_j} \right\}$. Then $d_n = c_n + O(1/n)$, so both sequences approach the same limit (and the $c$'s do approach a limit, since they are decreasing). We have $d_m \leq d_{km}$ (proof: take $k$ copies of each number). So each $d$ gives a lower bound for the limit, and each $c$ gives an upper bound. We should be able to get some pretty tight bounds this way. | |
| Jan 3 at 16:33 | comment | added | David E Speyer | If you put $c_m = \text{sup} \tfrac{1}{m(m-1)} \sum_{i,j=1}^m \left\{ \tfrac{x_i}{x_j} \right\}$, then we have $c_2 \geq c_3 \geq c_4 \geq \cdots$. The sup is over $\mathbb{R}_{>0}^m$. (Proof: Just take the sum $\sum_{i,j=1}^n \left\{ \tfrac{x_i}{x_j} \right\}$ and write it as a linear combination of the $\binom{n}{m}$ sums over $m$-element subsets of $\{ x_1, x_2, \ldots, x_n \}$.) So @TheBestMagician's observation is that $c_2 = 3/4$, and it seems like it would be worth computing the next few values of $c_m$. | |
| Jan 3 at 16:26 | comment | added | Christian Remling | @TheBestMagician: This gives the bound $(3/4)n^2 +O(n)$, which however is worse than $9/14<3/4$. | |
| Jan 3 at 16:23 | comment | added | TheBestMagician | Perhaps $\{x\}+\{1/x\}<3/2$ provides a good bound. I think it gives $\frac{3}{2}\binom{n}{2}+n$, or $\frac34$ instead of$\frac{9}{14}$. | |
| Jan 3 at 16:20 | comment | added | David E Speyer | I think it is pretty clear that the question is "Is it true that $\sum \left\{ \tfrac{x_i}{x_j} \right\} \leq \tfrac{9}{14} n^2$ for all positive real numbers $(x_1, x_2, \ldots, x_n)$?" Also with an implied question: "What is the smallest constant which can replace $9/14$ here?" | |
| Jan 3 at 15:59 | comment | added | Iosif Pinelis | "This inequality discovered by a Chinese student" -- Discovered in what sense? Is there a proof? Do you a reference to it? It is actually unclear what the question here is. | |
| Jan 3 at 15:25 | history | asked | math110 | CC BY-SA 4.0 |