Timeline for Optimal constant to compare $L^2$ norm of smooth function on $[0, 1]$ to a grid
Current License: CC BY-SA 4.0
18 events
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| Feb 27, 2023 at 16:31 | vote | accept | Drew Brady | ||
| Feb 27, 2023 at 8:40 | comment | added | Iosif Pinelis | @DrewBrady : It was not easy for me to construct this example. Your extra periodicity condition actually gave me the idea, to (indeed) expand $f$ into a sine series, then choosing the coefficients of the series carefully. Indeed, the example is sensitive to the choice of the coefficients. | |
| Feb 27, 2023 at 8:34 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Feb 27, 2023 at 7:19 | comment | added | Drew Brady | By the way, this example is very sensitive to constants. Is it necessary? What I mean by this is that if we redefine $d_n(f)$ by the equation $$d_n(f) = \int_0^1 f^2 - 2 \frac{1}{n} \sum_{j=1}^n f^2(j/n),$$ then all of a sudden $d_n(f_n) = O(1/n^4)$. Is it possible to avoid this; i.e., have that it is also multiplicatively worse? I understand that this is not the original question, and I can certainly create a new question if you feel it is necessary. | |
| Feb 27, 2023 at 7:08 | comment | added | Drew Brady | @IosifPinelis, thanks. I checked the calculations, and this looks right to me. Would you mind explaining somewhat your construction? Is the appearance of sinusoids related to some Fourier interpretation of the (incorrectly) conjectured inequality? | |
| Feb 27, 2023 at 6:51 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Feb 27, 2023 at 6:45 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Feb 27, 2023 at 6:40 | comment | added | Iosif Pinelis | The consideration is now quite complete: One cannot get a constant factor $O(n^{-4})$ good for all $f$ at once. | |
| Feb 27, 2023 at 6:38 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Feb 27, 2023 at 6:25 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Feb 26, 2023 at 21:32 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Feb 26, 2023 at 21:32 | comment | added | Iosif Pinelis | @DrewBrady : According to en.wikipedia.org/wiki/… , here for the remainder $R_p$ in the Euler–Maclaurin formula we have $|R_p|=O(n^{-p})$, so that I can actually take $p=4$ (as is now done) to get $|R_p|=O(n^{-4})$. | |
| Feb 26, 2023 at 21:22 | comment | added | Iosif Pinelis | @fedja : I used the condition that $f$ is $C^\infty$. Per this argument, the constant factor in $O(n^{-4})$ may depend on $f$. As noted in the answer, getting a constant factor in $O(n^{-4})$ good for all $f$ at once would require more care. | |
| Feb 26, 2023 at 20:28 | comment | added | fedja | I'm afraid you are using the control over more derivatives than are given to you (all you know is that $f''\in L^2$) in this argument. The truth for the original question seems to be $n^{-3}$, not $n^{-4}$ as well as the precision in the Euler-Maclaurin without the control over the third derivative. But you are absolutely right that some version of Euler-Maclaurin is the key. | |
| Feb 26, 2023 at 20:11 | vote | accept | Drew Brady | ||
| Feb 26, 2023 at 20:29 | |||||
| Feb 26, 2023 at 19:51 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Feb 26, 2023 at 19:29 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Feb 26, 2023 at 19:24 | history | answered | Iosif Pinelis | CC BY-SA 4.0 |