Timeline for An inequality improvement on AMM 11145
Current License: CC BY-SA 3.0
19 events
| when toggle format | what | by | license | comment | |
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| Jul 6, 2019 at 4:59 | answer | added | River Li | timeline score: 1 | |
| Jul 5, 2019 at 14:16 | comment | added | River Li | @Timothy Chow A earlier source: In 2013/09/19, someone asked for proof of the inequality with constant 2 in a Chinese forum and one hour later, Ji Chen gave a proof and proposed the inequality with constant $2-\dfrac{7\ln{2}}{8\ln{n}}$. | |
| Apr 13, 2017 at 12:19 | history | edited | CommunityBot |
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| Nov 27, 2015 at 22:52 | comment | added | dxiv | @r9m The intuition for $\frac{c}{\ln n}$ could come from the case where $a(k)\,=\,k$. In that case $C(n)=2 (1 - \frac{n} {(n+1) H_n})$ where $H_n$ is the $n^{th}$ harmonic number, and it's known that $H_n$ is $\mathcal{O}(\ln n)$. Of course, to move from heuristics to proof, one would have to actually show that $a(k)=k$ is the "worst case" sequence i.e. that it maximizes the ratio of the LHS vs RHS. | |
| Nov 27, 2015 at 18:31 | comment | added | r9m | @Suvrit the constant $\frac{7\ln 2}{16}$ doesn't matter, what is important is how the $\frac{1}{\log n}$ comes into play. Do you have a proof (or even a heuristics) that supports the claim? thanks! | |
| Nov 27, 2015 at 15:03 | comment | added | Suvrit | I think the constant is a red herring -- I feel the "true" bound should be of the form $2\left(1 - \frac{n}{(n+1)\log n}\right)$.... | |
| Nov 27, 2015 at 6:45 | comment | added | dxiv | @TimothyChow It could interesting to find out if the OP had a particular sequence in mind that hit that particular constant. The transIation sounds like "stronger inequality" not necessarily "the best inequality". | |
| Nov 27, 2015 at 6:34 | answer | added | dxiv | timeline score: 4 | |
| Nov 27, 2015 at 4:20 | answer | added | Brendan McKay | timeline score: 5 | |
| Nov 27, 2015 at 3:12 | comment | added | Brendan McKay | The right side is independent of the order of the $a_i$s but the left side is not. So which order makes the left side largest? | |
| Nov 26, 2015 at 20:41 | comment | added | Timothy Chow | The constant is claimed here: math.org.cn/forum.php?mod=viewthread&tid=28918 (The Chinese above the inequality simply says, "We have the stronger [inequality]".) Although that was 2 years ago, perhaps someone could try posting a followup asking for a proof. | |
| Nov 26, 2015 at 20:33 | comment | added | Steven Stadnicki | $\frac78\ln2$ is an unusual (and remarkably precise, for want of a better word) value for the constant here - can you speak to its specific genesis? | |
| Nov 26, 2015 at 20:03 | answer | added | Fedor Petrov | timeline score: 5 | |
| Nov 26, 2015 at 18:03 | comment | added | robjohn | @r9m: each of our proofs shows that we can get $2\left(1-\frac1{(n+1)^2}\right)$, but this is not as good as $2\left(1-\dfrac{7\ln{2}}{16\ln{n}}\right)$. | |
| Nov 26, 2015 at 17:09 | comment | added | r9m | @S.Zoalroshd thanks! I am aware of the proof for the constant being $2$ (see here ) but I couldn't prove this improved inequality claiming it can be replaced by: $2 - \frac{7\ln 2}{8\ln n}$ | |
| Nov 26, 2015 at 17:05 | comment | added | BigM | there is a solution here with constant equal to 2. your constant is sharper . | |
| Nov 26, 2015 at 17:01 | comment | added | r9m | @FedorPetrov robjohn verified the inequality for $n = 2$ case (mentioned in comment ), but I don't have a access to a mathematical software that might check the validity of the result in general. | |
| Nov 26, 2015 at 16:57 | comment | added | Fedor Petrov | Why are you sure that this is true? | |
| Nov 26, 2015 at 16:55 | history | asked | r9m | CC BY-SA 3.0 |