Timeline for Optimizing a smoothing function with the Prime Number Theorem in mind
Current License: CC BY-SA 4.0
32 events
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| S Nov 1, 2022 at 22:07 | history | bounty ended | CommunityBot | ||
| S Nov 1, 2022 at 22:07 | history | notice removed | CommunityBot | ||
| Oct 31, 2022 at 13:55 | comment | added | H A Helfgott | Yup, it's easy to prove from properties of the modified Bessel function $I_0(x)$ :). | |
| Oct 31, 2022 at 13:46 | comment | added | H A Helfgott | So, perhaps $f(x) = \frac{1}{2} (1-\textrm{erf}((\log x)/\epsilon))$ is optimal? I'll work out the optimal $\epsilon$; it is clear it should be inversely proportional to $\sqrt{\log T}$. Fortunately this $f(x)$ has a beautiful Mellin transform. | |
| Oct 31, 2022 at 13:39 | comment | added | H A Helfgott | As detailed below, the $f$ I proposed gives good results for high $f$. Büthe's function also works nicely in practice (even though Logan's function really optimizes something a bit different; Büthe needs to optimize the quantity we are discussing here, essentially). This makes sense: the $f$ I proposed converges to $(1-erf((log x)/\epsilon))/2$, and a plot suggests that Büthe's function is very close to that (should be easy to prove from properties of the modified Bessel function $I_0(x)$). | |
| Oct 30, 2022 at 10:33 | history | edited | Max Lonysa Muller | CC BY-SA 4.0 |
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| Oct 30, 2022 at 8:14 | answer | added | H A Helfgott | timeline score: 1 | |
| Oct 29, 2022 at 6:29 | answer | added | H A Helfgott | timeline score: 1 | |
| Oct 28, 2022 at 14:26 | comment | added | H A Helfgott | @2734364041 Well, I forgot a factor of $k!$, but I think that was clear! | |
| Oct 28, 2022 at 14:24 | history | edited | H A Helfgott | CC BY-SA 4.0 |
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| Oct 25, 2022 at 11:32 | comment | added | H A Helfgott | My point is that it's not that lossy; let me write that in an answer. Why is $x-n$ better than $\log(x/n)$, or a polynomial on $x$ better than a polynomial on $\log(x/n)$? | |
| Oct 25, 2022 at 11:26 | comment | added | 2734364041 | @HAHelfgott my point is that "unsmoothing" from $\sum_{n\leq x} \Lambda(n)(\log x/n)^k$ to $\sum_{n\leq x}\Lambda(n)$ is very lossy when $k\geq 2$ (unless you want to assume GRH). It's tough to detail this in the comments. But if you only care about the smoothed sum, then it is a very numerically efficient kernel, as you have observed. But even with this kernel when $k=1$ is not numerically optimal; one can do better with $\log(x/n)$ replaced by $x-n$, as Grenie and Molteni used frequently. | |
| Oct 25, 2022 at 10:11 | comment | added | H A Helfgott | Sure, or $\sum_{n\leq x} \mu(n) (\log x/n)^k$. Those sums are relatively straightforward to estimate (and have their own applications). | |
| Oct 25, 2022 at 10:08 | comment | added | 2734364041 | @HAHelfgott Is the idea behind your choice of $f$ that you would like to estimate the sum $\sum_{n\leq x}\Lambda(n) (\log x/n)^k$? Or am I missing something? | |
| Oct 25, 2022 at 9:57 | comment | added | H A Helfgott | @2734364041 Ah, forgot the factor of $(\log y)^{-k}$ (but then I'm sure you knew that). | |
| Oct 25, 2022 at 9:57 | history | edited | H A Helfgott | CC BY-SA 4.0 |
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| Oct 25, 2022 at 8:50 | comment | added | H A Helfgott | It's an interesting smoothing function, but I see no claims of optimality. | |
| Oct 25, 2022 at 8:48 | comment | added | H A Helfgott | Why would it be "very lossy"? The optimal value of $k$ seems to be considerably higher than $3$. Obviously $y$ will depend on $k$. | |
| Oct 25, 2022 at 8:42 | comment | added | 2734364041 | Your choice of $f$ is good if you are interested in a smoothed prime number theorem. If you are interested in un-smoothing, then this choice of $f$ is very lossy when $k \geq 3$. You might want to take a look at Lemma 2.2 in arxiv.org/abs/1606.09238v2 (which approximates $\mathbf{1}_{[1/2,1]}$ instead of $\mathbf{1}_{[0,1]}$). | |
| Oct 25, 2022 at 7:24 | history | edited | H A Helfgott | CC BY-SA 4.0 |
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| Oct 24, 2022 at 20:03 | history | edited | H A Helfgott | CC BY-SA 4.0 |
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| S Oct 24, 2022 at 20:03 | history | bounty started | H A Helfgott | ||
| S Oct 24, 2022 at 20:03 | history | notice added | H A Helfgott | Canonical answer required | |
| Oct 24, 2022 at 13:53 | comment | added | H A Helfgott | Sure - I was giving the general definition. | |
| Oct 24, 2022 at 13:52 | comment | added | Steven Clark | Thanks, I was a bit confused after looking at mathworld.wolfram.com/L1-Norm.html. Since $f(x)=1$ for $0\leq x\leq 1$, it seems $|f-1_{[0,1]}|_1=\int_1^\infty |f(x)| dx$. | |
| Oct 24, 2022 at 8:45 | history | edited | H A Helfgott | CC BY-SA 4.0 |
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| Oct 24, 2022 at 8:22 | comment | added | H A Helfgott | In other words: $\int_0^1 |f(x)-1| dx + \int_1^\infty |f(x)| dx$. | |
| Oct 23, 2022 at 19:28 | comment | added | H A Helfgott | It's the $L^1$ norm. | |
| Oct 22, 2022 at 19:35 | comment | added | Steven Clark | I assume the $\cdot$ operator is normal mutliplication? I assume $1_{[0,1]}$ is the indicator function (see en.wikipedia.org/wiki/Indicator_function), but what is the meaning of $|f-1_{[0,1]}|_1$ (i.e. the meaning of the subscript $1$ on the the absolute value)? | |
| Oct 21, 2022 at 21:37 | history | edited | H A Helfgott |
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| Oct 21, 2022 at 14:13 | history | edited | H A Helfgott | CC BY-SA 4.0 |
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| Oct 21, 2022 at 12:25 | history | asked | H A Helfgott | CC BY-SA 4.0 |