Timeline for Maximal (minimal) value of $S=x_1^2x_2+x_2^2x_3+\cdots+x_{n-1}^2x_n+x_n^2x_1$ on condition that $x_1^2+x_2^2+\cdots+x_n^2=1$
Current License: CC BY-SA 4.0
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| Feb 13, 2021 at 1:58 | comment | added | fedja | @WillSawin Partly yes, but not in full yet. BTW, the quick proof you have gives at most $O(n^{-3/2})$ for the error just because the minimal term is of that order and you can break the chain anywhere. Of course, that is rather short of what I claimed. I'll think more of it and (I hope) post my observations later. Now I'll just note that the Lagrange multiplier theorem gives $\lambda x_i=x_{i-1}^2+2x_ix_{i+1}$ and, since the maximum is $>0.45$, we can multiply by $x_i$, add up, and get $\lambda>1.35$. Since $x_{i+1}$ is less than $0.15$ for most $i$, you get $x_i\le x_{i-1}^2$ most of the time. | |
| Feb 13, 2021 at 1:35 | comment | added | Will Sawin | @fedja This all sounds right to me, but can you prove this rigorously? Presumably there is a good way to turn the computer solution for finite $n$ into a rigorous proof for all $n$ but I don't know it. The only thing I have a quick proof of is that $\operatorname{max}_n \leq \max_{\infty} + o(1)$ | |
| Feb 13, 2021 at 1:30 | comment | added | fedja | @WillSawin It certainly does and incredibly fast, but there is no nice answer for the infinite problem either. What Yaakov pointed out is essentially the maximum: the maximizing sequence decays doubly exponentially, so beyond $x_{12}$ it is invisible to the computer and the trivial iteration like $x_i'\sim x_i+x_{i-1}^2+2x_ix_{i+1}$ converges quickly to the maximizer too. So, for all practical purposes, Yaakov's sequence is optimal (with more precision, of course, which changes only the last digit in his comment to $8$). | |
| Feb 10, 2021 at 20:02 | comment | added | Will Sawin | It seems likely that the maximum converges as $n\to \infty$ to the maximum to the analogous optimization problem in $\mathbb Z$, i.e. maximize $\sum_{i=-\infty}^{\infty} x_i^2 x_{i+1}$ subject to $\sum_{i=-\infty}^{\infty} x_i^2 =1$. Proving this (maybe with some estimate on the rate of convergence) would be a good solution in my opinion, even if the formulas for any individual $n$ are not so nice. | |
| Feb 10, 2021 at 1:01 | comment | added | Yaakov Baruch | @user64494. Indeed rounding your solution to x[1]=0.5873, x[2]=0.6771, x[3]=0.4224, x[4]=0.1344, x[5]=0.0133, x[6]=0.0001 and x[>=7]=0, gives $0.451389406$ for any $n\ge 7$. | |
| Feb 9, 2021 at 12:06 | comment | added | user64494 | @FedorPetrov: MattF already reported that, but he deleted his comment. The result of Mathematica shows the method of Lagrange multipliers is applied. | |
| Feb 9, 2021 at 11:20 | comment | added | Fedor Petrov | this is just undersimplified: for example, for $n=3$ we have $x(2)\to \frac{164555}{1007 \sqrt{3}}-\frac{54516 \sqrt{3}}{1007}=\frac1{\sqrt{3}}$ | |
| Feb 8, 2021 at 16:44 | comment | added | Henry | That revised $n=10$ is not far from my suggestion | |
| Feb 8, 2021 at 16:38 | comment | added | user64494 |
@Henry: There was a typo in my code. In fact, it should be {0.451389, {x[1] -> 0.000131406, x[2] -> 1.5108*10^-8, x[3] -> -4.01598*10^-8, x[4] -> -1.91736*10^-8, x[5] -> 7.39988*10^-10, x[6] -> 0.587294, x[7] -> 0.677083, x[8] -> 0.42238, x[9] -> 0.134393, x[10] -> 0.0133402}}. Small negative numbers due to round-off errrors. NMaximize may find a loval minimum in higher dimensions. Other methods give 0.451389 too.
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| Feb 8, 2021 at 16:23 | comment | added | Henry | Your $n=10$ case looks strange: you should not have any negative values for maximal $S$ - just take the positive versions to increase $S$ | |
| Feb 8, 2021 at 16:21 | comment | added | Henry | For example for $n\ge 5$, you can have (not quite maximal) $x_1=0.592, x_2=0.677, x_3=0.418, x_4=0.128, x_5 \approx 0.00995$ and all the rest $0$ giving $S\approx 0.45137$ | |
| Feb 8, 2021 at 16:19 | comment | added | user64494 |
@Henry: For $n=10$ NMaximize command of Mathematica results in {0.451389, {x[1] -> 0.677084, x[2] -> 0.422378, x[3] -> 0.134391, x[4] -> 0.0133399, x[5] -> 0.000131404, x[6] -> 5.61579*10^-9, x[7] -> -8.43825*10^-9, x[8] -> -7.99726*10^-9, x[9] -> -0.557805, x[10] -> -0.183765}}
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| Feb 8, 2021 at 15:59 | comment | added | Henry | That $0.452532$ for $n=5$ makes sense. I think you can exceed $0.451$ for any $n$. | |
| Feb 8, 2021 at 15:53 | comment | added | user64494 |
@MattF. Thank you. No, it is not so. E.g. Maximize[{x[1]^2*x[2] + x[2]^2*x[3] + x[3]^2*x[4] + x[4]^2*x[5] + x[5]^2*x[1], x[1]^2 + x[2]^2 + x[3]^2 + x[4]^2 + x[5]^2 == 1}, {x[1], x[2], x[3], x[4], x[5]}] // Simplify still produces a long output and NMaximize[{x[1]^2*x[2] + x[2]^2*x[3] + x[3]^2*x[4] + x[4]^2*x[5] + x[5]^2*x[1], x[1]^2 + x[2]^2 + x[3]^2 + x[4]^2 + x[5]^2 == 1}, {x[1], x[2], x[3], x[4], x[5]}, Method -> "DifferentialEvolution"] performs {0.452532, {x[1] -> 0.661899, x[2] -> 0.434777, x[3] -> 0.174964, x[4] -> 0.138598, x[5] -> 0.568363}}.
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| Feb 8, 2021 at 15:51 | comment | added | Henry | What is your maximum for $S$ when $n=5$? | |
| Feb 8, 2021 at 15:22 | history | answered | user64494 | CC BY-SA 4.0 |