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
14 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 27, 2023 at 16:31 | vote | accept | Drew Brady | ||
| Feb 27, 2023 at 2:36 | history | edited | Drew Brady | CC BY-SA 4.0 |
added 603 characters in body
|
| Feb 26, 2023 at 20:11 | vote | accept | Drew Brady | ||
| Feb 26, 2023 at 20:29 | |||||
| Feb 26, 2023 at 19:24 | answer | added | Iosif Pinelis | timeline score: 2 | |
| Feb 26, 2023 at 17:31 | comment | added | Christian Remling | (6) this suggests that two pieces of suitably chosen shapes will give $D\gtrsim 1/n^2$; (7) since we need a connecting piece in the middle that I didn't discuss, this is not quite a proof yet, but it feels close to one. | |
| Feb 26, 2023 at 17:30 | comment | added | Christian Remling | My procedure is as follows: (1) If we were allowed to take a decreasing $g=f^2$, then your difference (call it $D$ perhaps) would be $\gtrsim 1/n$ already; (2) your assumptions prevent this, so the obvious next attempt is to make $g$ increasing on $[0,1/2]$, say, followed by a decreasing piece on $[1/2, 1]$; (3) the first piece will have a negative $D$, so there are competing effects; (4) if we look more closely, we find that the $1/n$ contributions from the two pieces will cancel each other out, independently of shape; (5) however, the $1/n^2$ terms do depend on shape; (continued below) | |
| Feb 25, 2023 at 23:38 | comment | added | Christian Remling | What I'm saying is that the discrepancy is $(1/2)f(1)^2(1/n) + C(f)/n^2+O(n^{-3})$ for a monotone function. If we follow an increasing piece by a decreasing one, the $1/n$ terms will cancel since they only depend on the final value, but the next order will depend on the shape of $f$, so can be given a non-zero constant (and it's not a proof yet since I ignored the intermediate piece). | |
| Feb 25, 2023 at 23:35 | comment | added | Christian Remling | @DrewBrady: No, your formula was $-(12c^2/42)1/n + O(n^{-2})$. | |
| Feb 25, 2023 at 22:11 | comment | added | Christian Remling | I still think, on closer reflection, that this is implausible. For example, if $g(x)=x^N$, then the discrepancy between the integral $\int_0^1 g(x)\, dx$ and its upper Riemann sum is $-1/(2n)+N/(12n^2)+O(n^{-3})$. So if we follow up such a function with its decreasing counterpart on $[1,2]$, but with a different $N$, then the $\simeq 1/n^2$ won't cancel and we expect an error $\gtrsim 1/n^2$. Of course, we need an intermediate piece for a smooth transition, but I wouldn't expect this to somehow cancel out the $1/n^2$ contribution we already have, independently of its shape. | |
| Feb 25, 2023 at 20:28 | comment | added | fedja | Have you tried to begin with $x^2$ and then to glue the upside down parabola (with the leading coefficient $-1$, say) to it extended until it hits $0$ so that an extra $1/2f(1)^2$ would be useless? | |
| Feb 25, 2023 at 18:21 | comment | added | Christian Remling | Also, we don't really need your calculation since $0$ is an obvious upper bound for my function, if I had noticed the absence of an absolute value. | |
| Feb 25, 2023 at 17:38 | comment | added | Christian Remling | Yes, with no absolute values you'd need a decreasing function $f^2$ (to make the sum the lower Riemann sum) for a $1/n$ error, which however is prevented by your assumptions. | |
| Feb 25, 2023 at 1:12 | comment | added | Christian Remling | I don't think this sounds plausible. If you just take any increasing function, $f(x)=cx^3$, say, then you are comparing the integral with its upper Riemann sum, and on each interval of size $1/n$ not close to zero, you are making an error $\simeq 1/n^2$ for an overall error of $1/n$. | |
| Feb 24, 2023 at 23:58 | history | asked | Drew Brady | CC BY-SA 4.0 |