Timeline for Asymptotic limit of truncated Legendre sieve
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
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| Apr 24, 2015 at 1:55 | comment | added | Lucia | @user45947: You're right that this is insufficiently well known. For example, I think I should have known this, but I didn't! One lives and learns. | |
| Apr 23, 2015 at 16:14 | comment | added | Terry Tao | @Lucia You are in good company: en.wikipedia.org/wiki/Legendre%27s_constant Indeed, the numerical discrepancy that caused Legendre to be off by about 8% may in fact be related to the one in this question. | |
| Apr 23, 2015 at 13:22 | comment | added | user45947 | I don’t know whether you agree, but to me it appears that the reason behind the discrepancy between the probabilistic estimate given by Merten’s product theorem and the prime number theorem is often poorly explained in the literature. That just adding the constraint $d\leq x$ results in $S(x)\sim 1/\log x$ is—to me at least—very instructive, even though the prime number theorem itself was applied in your proof. Thanks again for a great contribution! | |
| Apr 23, 2015 at 12:57 | history | edited | Lucia | CC BY-SA 3.0 |
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| Apr 23, 2015 at 12:47 | history | edited | Lucia | CC BY-SA 3.0 |
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| Apr 23, 2015 at 12:43 | comment | added | Lucia | @TerryTao: Yes indeed it is $1$. I should have thought an epsilon more! | |
| Apr 23, 2015 at 8:32 | comment | added | user45947 | Excellent answer! It seems from numerical testing that @TerryTao is correct with respect to the sum $\sum_{k=1}^\infty M(k)\log(1+\frac{1}{k})=1$. A plot demonstrating this is added to the original question. I had sort of suspected $S(x)\sim 1/\log x$ from a rather hand waving heuristic (which I also have added to my posting), so it's nice to see that this actually seems to be the case. | |
| Apr 23, 2015 at 8:21 | vote | accept | user45947 | ||
| Apr 23, 2015 at 3:26 | comment | added | Terry Tao | Formally, one has $M(z) = - \sum_{n>z} \mu(n)/n$, so the sum $\sum_{k=1}^\infty M(k) \log(1+\frac{1}{k})$ appears to telescope formally to $\sum_n \frac{-\mu(n) \log n}{n} = 1$. One can presumably justify these formal computations by zeta regularization. | |
| Apr 23, 2015 at 2:50 | history | answered | Lucia | CC BY-SA 3.0 |