Timeline for Proportionality constant in Montgomery-Vaughan Theorem 7.20
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 3, 2021 at 5:42 | vote | accept | kodlu | ||
| Aug 2, 2021 at 21:58 | answer | added | kodlu | timeline score: 0 | |
| Jul 31, 2021 at 21:44 | comment | added | kodlu | @GregMartin, I will accept that as an answer if you would like to enter it as such. | |
| Jul 31, 2021 at 16:35 | history | edited | YCor | CC BY-SA 4.0 |
removed capitals from title
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| Jul 31, 2021 at 7:28 | comment | added | Greg Martin | Certainly $B(x,r)$ is a decreasing function of $r$, but that doesn't imply that the bound on $B(x,r)$ continues to hold for $r\ge2$. The reason that $R<2$ is required in the given proof is that it proceeds via upper bounds for $\sum_{n\le x} r^{\Omega(n)}$; this sum genuinely changes character when $r>2$, since the integers $n$ that are powers of $2$ (or nearly so) are then increased by the map $n\mapsto r^{\Omega(n)}$; for example, the largest power of $2$ less than $x$ already gives a contribution of $x^{(\log r)/\log 2}$ to the sum. | |
| Jul 31, 2021 at 6:21 | history | asked | kodlu | CC BY-SA 4.0 |