Timeline for Find the tight upper bound of $\sum_{i=1}^n \frac{i}{i+x_i}$, where the $x_i$'s are distinct in $\{1,2,...,n\}$
Current License: CC BY-SA 4.0
9 events
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| Nov 3, 2018 at 18:32 | comment | added | Martin Rubey | Yes, I did it just like @js21, assuming that $k=c\cdot n$, but using $\pi_k$. In general, it's not clear to me which elements are large. For example, the permutation $[6,5,2,4,1,3]$ has larger (!) value ($106/35$) than $[6,5,4,2,1,3]$ ($127/42$). | |
| Nov 3, 2018 at 16:46 | comment | added | Wolfgang | @MartinRubey That sounds good. How did you come up with that? I suppose by solving $f(\pi_k)=f(\pi_{k+1})$ for big n, using asymptotics of the harmonic series? | |
| Nov 3, 2018 at 13:31 | comment | added | Martin Rubey | Another comment: if I made no mistake, given the conjectural form, I think that $k\simeq c\cdot n$ should be such that $c$ is the zero of $(c-2)\ln(\frac{c}{2}-1) + 2(2c-3)(c-1)$. So, $1/c=6.303250770424045$ as you write. | |
| Nov 3, 2018 at 13:25 | comment | added | Martin Rubey | OK, I just thought you had some way to check that the general form I mentioned in the comment was correct. By the way, the poset of permutations ordered by $f$ is quite interesting! | |
| Nov 3, 2018 at 13:21 | history | edited | Wolfgang | CC BY-SA 4.0 |
clarified edits
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| Nov 3, 2018 at 13:10 | comment | added | Wolfgang | @MartinRubey Of course, like for most conjectures, there is a certain amount of educated guess involved. I have not checked millions of sums, I just say "it looks like", but I don't claim... One of my assumptions is a kind of monotonicity, which seems highly plausible: if for a fixed $n$, we take $\pi_k:=[n,n-1,...,n-k+1,1,2,...,n-k]$, then there is a $k$ such that $f(\pi_1)<\cdots<f(\pi_k)>f(\pi_{k+1}>\cdots$. And the first $\pi_k$ seem to be the best candidates, so... My "indeed" referred rather to your first comment there than to js21's permutations. :) In any case, we must have $x_1=n$. | |
| Nov 2, 2018 at 22:08 | comment | added | Martin Rubey | How can you check that the maximum for such a large $n$ is of the given form? Besides, I think that @js21 considered a different permutation (which yields a large value, but not the maximum)... | |
| Nov 2, 2018 at 20:27 | history | edited | Wolfgang | CC BY-SA 4.0 |
added new conjectured best partition
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| Nov 1, 2018 at 9:35 | history | answered | Wolfgang | CC BY-SA 4.0 |