Timeline for Maximize $f(0)+\cdots+f(n-1)$ subject to $f(x)f(y) + f(x+y) \leq 1$
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
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| Dec 30, 2018 at 23:03 | vote | accept | Sean Eberhard | ||
| Dec 30, 2018 at 16:42 | comment | added | Gerhard Paseman | If you index from 0, your vector e does not satisfy the inequality. $e_0 + e_j \leq 0$ holds for every $j$. If $e_1$ and $e_n$ are positive, we have $e_{n-1} + e_n$ is negative. Now either $e_1$ is non positive, or else we can start with $k=n$ and build up the sum(and decrement $k$ by one or by two) to show it is negative, adding either nonpositive $e_k$ or negative $(e_{k-1} + e_k)$ to get the whole sum is non positive. Gerhard "Nothing Up My Sleeve, Presto!" Paseman, 2018.12.30. | |
| Dec 30, 2018 at 8:39 | comment | added | Sean Eberhard | I don't follow how your "either $e_n \leq 0$ or $e_{n-j} + e_n < 0$" condition implies $e_0 + \cdots + e_n \leq 0$. E.g., if $j = 1$ and $e = (.1, -.11, .1)$? I think I must be missing something... | |
| Dec 28, 2018 at 18:29 | history | edited | Gerhard Paseman | CC BY-SA 4.0 |
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| Dec 28, 2018 at 18:21 | history | edited | Gerhard Paseman | CC BY-SA 4.0 |
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| Dec 28, 2018 at 13:44 | comment | added | Gerhard Paseman | I think I have it now. The first variable that is greater than 1/phi affects the other variables, and allows one to partition the other variables into subsums which average less than 1/phi each. Gerhard "Will Fill In Details Later" Paseman, 2018.12.28. | |
| Dec 23, 2018 at 16:41 | comment | added | Gerhard Paseman | I am trying to show that the average is at most 1/phi. This is clear for n=1 and n=2. For n=3 at most one of the x's can be greater than 1/phi, and use the n=2 case to show the average is below. For larger n, the post above shows that if only the last x is greater than 1/phi, then the average is below 1/phi. It remains to handle the case when two or more x are above 1/phi, which I believe can be done. Gerhard "Still Doing Calculations With Epsilon" Paseman, 2018.12.23. | |
| Dec 23, 2018 at 10:34 | comment | added | Sean Eberhard | Thanks for your comments. Indeed I put the system through a generic optimizer before posting the question (viewable here: colab.research.google.com/drive/…). But I'm afraid I don't quite follow the details of your strategy for making this into a proof. | |
| Dec 23, 2018 at 4:32 | history | edited | Gerhard Paseman | CC BY-SA 4.0 |
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| Dec 22, 2018 at 21:51 | history | edited | Gerhard Paseman | CC BY-SA 4.0 |
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| Dec 22, 2018 at 21:45 | history | answered | Gerhard Paseman | CC BY-SA 4.0 |