Timeline for Minimization problem for convolution
Current License: CC BY-SA 4.0
11 events
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| Oct 3, 2019 at 8:48 | comment | added | Kurisuto Asutora | PS: As a general remark, I am more interested in the continuous problem, not in the discrete one. So in the discretization probably one would need to obtain some asymptotics as $n \to \infty$, to say something about the continuous problem. | |
| Oct 3, 2019 at 8:46 | comment | added | Kurisuto Asutora | @DavidESpeyer: No, I do not claim that I have any "provable" numerical result, this is just what Mathematica returned after I used Mateusz' code, with the only change of replacing n=75 by n=95. I did not make any effort to control the size of errors. | |
| Oct 3, 2019 at 7:15 | comment | added | Mateusz Kwaśnicki | @DavidESpeyer: However, three out of fifty conditions $x*x(i)\geqslant 1$ are not satisfied by this solution: they are off by roughly $10^{-9}$ (this is how the numerical method is supposed to work). Still, $\tilde{x}(i) = x(i)+10^{-8}$ is a true solution with norm 1.11968 (this is the norm of $x$ plus $10^{-8} \sqrt{50}$), less than the norm of the explicit solution. (2/2) | |
| Oct 3, 2019 at 7:15 | comment | added | Mateusz Kwaśnicki | @DavidESpeyer: Interesting. I just did the same calculation for $n = 50$ ($n = 95$ takes too long on my laptop), and I am surprised to find that indeed one can improve slightly upon the explicit solution that you mentioned! Details: Explicit solution has norm 1.12556. Mathematica implementation of the Nelder–Mead method leads to a "solution" that has norm 1.11968. (1/2) | |
| Oct 3, 2019 at 0:53 | comment | added | David E Speyer | @KurisutoAsutora For $n=95$, I computed $L:=\sum_{k=0}^{94} \tfrac{1 \cdot 3 \cdots (2k-1)}{2 \cdot 4 \cdots (2k)}$ in exact arithmetic (it is a rational number) and then computed $\tfrac{L}{\sqrt{95}}$ with Mathematica using 20 digits of internal precision. I got 1.12689544 and I believe all those digits are right. Are all the digits of your 1.12594 provably achievable? (I.e. has your algorithm definitely found a point which achiees this?) If so, this is what I mean when I see you can improve a little. | |
| Oct 2, 2019 at 21:16 | comment | added | Mateusz Kwaśnicki | @DavidESpeyer: This explicit solution corresponds to the green line in the plot, I believe. I am not sure if one can really improve upon this, differences between minimizers found by different algorithms seem to be within the range of admissible error. | |
| Oct 2, 2019 at 21:07 | comment | added | David E Speyer | Note that $x_k = \tfrac{1 \cdot 3 \cdot 5 \cdots (2k-1)}{2 \cdot 4 \cdot 6 \cdots 2k}$ solves $\sum_{j=0}^i x_j x_{i-j}=1$ exactly and has the same asymptotic $2\sqrt{n}/\sqrt{\pi}$. However, your data suggests that it is possible to improve on this a little for finite $n$. | |
| Oct 1, 2019 at 12:18 | comment | added | Kurisuto Asutora | I am not quite sure if that has any real significance. The numerical values I get using the same code for n=85 are {1.12831, 1.12672, 1.1287}, and for n=95 they are {1.12594, 1.13249, 1.1269}. In comparison, $2 / \sqrt{\pi} \approx 1.1284$. | |
| Oct 1, 2019 at 11:51 | comment | added | Mateusz Kwaśnicki | I have not tried that, but apparently one can find intermediate solutions between the two that I mentioned in my comment. If I find time later today, I will see if I can figure out the details. | |
| Oct 1, 2019 at 10:42 | comment | added | Ilya Bogdanov | Have you found any other minimizers in an analytical form? | |
| Sep 30, 2019 at 11:08 | history | answered | Mateusz Kwaśnicki | CC BY-SA 4.0 |