Timeline for Best smoothing for the Prime Number Theorem?
Current License: CC BY-SA 4.0
14 events
| when toggle format | what | by | license | comment | |
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| Nov 17 at 18:34 | comment | added | H A Helfgott | @username Faber and Kadiri start by applying Cauchy-Schwarz, and then optimise an $L^2$ norm (see (3.1) in their paper). This is unfortunate - it would have made more sense not to Cauchy-Schwarz (the factor ((b-a) u + a) in (2.5) is almost constant in practice, as b will be taken very close to a). If one optimizes without Cauchy-Schwarzing, one gets an optimal g that is not given by a Legendre polynomial (the optimal g is rather simple to find, really) and the result improves by a constant factor (which may depend on m). CSing biases the problem. | |
| Oct 31, 2022 at 14:11 | comment | added | H A Helfgott | I now suspect that $\frac{1}{2} (1-\textrm{erf}((\log t)/\delta))$ is best (for $\delta$ proportional to $\sqrt{\log c T}/T$; I'll work out the constant). See the discussion in mathoverflow.net/questions/432914/… . Büthe's choice (from Logan) is actually very close to this, because $I_0(C \sqrt{1-x^2})/e^C \sim e^{-x^2/2}/\sqrt{2\pi C}$ for $C$ large and $x$ small. | |
| Oct 30, 2022 at 13:13 | comment | added | H A Helfgott | The reason is that one can choose $f$ such that $f-1_{[0,1]}$ support broader than $[e^-\delta,e^\delta]$, yet $|f-1_{[0,1]}|_1 = o(\delta)$ (as $T$ grows). That's not the case for Büthe's function (whose derivative is in fact discontinuous at $-\delta$ and $\delta$). | |
| Oct 30, 2022 at 13:11 | comment | added | H A Helfgott | OK, matters are now a little clearer to me. What Büthe is doing is using a result due to Logan, which, in effect, gives a function $f$ (in closed form!) with minimal $\int_T^\infty |M f(i t)| dt$ among the functions $f$ such that $f-1_{[0,1]}$ has support in $[e^{-\delta},e^\delta]$. That's impressive, and leads to good results, but it is not actually optimal - either in my phrasing above or for what Büthe is doing. It seems to me one can do asymptotically better. | |
| Oct 26, 2022 at 22:23 | comment | added | H A Helfgott | I don't see a clear claim being made in Büthe, but is it the case that Büthe's choice does lead to a (somewhat complicated) closed-form minimizer for $\int_T^\infty |Mf(1+it)| dt$ under the condition that $f(x)-1_{[0,1]}$ be supported in an interval $[1,1+\delta]$? (Short version: the Mellin transform is a Fourier transform with a change of variables, and $M(x f'(x)) = -s Mf(s)$. I'm skimming, so I may be missing something.) | |
| Oct 26, 2022 at 21:49 | comment | added | H A Helfgott | Well, if there's a closed-form formula for the minimizer (or, at least, for the minimum), then I would like to know it. | |
| Oct 26, 2022 at 21:34 | comment | added | username | You hope for closed form formula for the minimizer (happens rarely)? Or you would like to know an algorithm to compute an approximation of a minimizer? | |
| Oct 26, 2022 at 21:29 | comment | added | H A Helfgott | I should have expressed myself more carefully: the way that Faber and Kadiri set up their problem - minimizing, not an integral of the Mellin transform itself, but the $L^2$ norm of a $k$th-order derivative of the smoothing function - leads to a (Legendre) polynomial being the optimal solution. | |
| Oct 26, 2022 at 21:10 | comment | added | username | From the paper cited, the function found in Faber-Kadiri is found by calculus of variation, minimising a quotient, and imposing null derivatives at the endpoints : it happens to be that the minimum is a Legendre polynomial, this wasn't an assumption. The origin of the Bessel function appearing in Büthe's paper is explained in another paper where it refers to this paper to explain why the choice is sharp. You wish to ask a different optimisation problem by relaxing something else. In what function space? | |
| Oct 26, 2022 at 20:29 | comment | added | H A Helfgott | Hm, I can't do a brute-force computation up to just any $T$. | |
| Oct 26, 2022 at 20:28 | comment | added | username | Perhaps you could write it as $\min_f \limsup_T ...$? | |
| Oct 25, 2022 at 6:28 | comment | added | H A Helfgott | Sure, you can always assume T is not small (even $T>10^9$) to simplify. | |
| Oct 25, 2022 at 0:15 | comment | added | Zach Hunter | perhaps it is worthwhile to mention you are also given $T>10^6$ (based off the linked question) | |
| Oct 24, 2022 at 14:02 | history | asked | H A Helfgott | CC BY-SA 4.0 |