Timeline for multiplicative functions of powers
Current License: CC BY-SA 3.0
18 events
| when toggle format | what | by | license | comment | |
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| Oct 30, 2015 at 3:02 | review | Close votes | |||
| Oct 30, 2015 at 5:03 | |||||
| Oct 25, 2015 at 22:24 | history | edited | Igor Rivin | CC BY-SA 3.0 |
toned down the language some
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| Oct 25, 2015 at 20:21 | vote | accept | Igor Rivin | ||
| Oct 25, 2015 at 20:20 | history | edited | Igor Rivin | CC BY-SA 3.0 |
added info and vitriol.
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| Oct 25, 2015 at 20:19 | answer | added | so-called friend Don | timeline score: 6 | |
| Oct 25, 2015 at 1:26 | comment | added | Alexey Ustinov | "On $d( f (n))$ and $d(d( f (n))),$ $f$ a polynomial" (Handbook of number theory. I, page 66); $\sum_{n≤x}\sigma (f(n))$, $f$ a polynomial" (page 85) "On $\sum_{n≤x}f (P(n))$, $f$ a certain arithmetic function." (page 150) | |
| Oct 24, 2015 at 23:46 | comment | added | Brad Rodgers | For $\tau(n)$ the count of divisors, $F(s) := \sum \frac{\tau(n^k)}{n^s} = \prod_p \Big(1+ \frac{\tau(p^k)}{p^s} + \frac{\tau(p^{2k})}{p^{2s}}+\cdots\Big)$. But $\tau(p^{k\ell}) = k\ell+1$, so with a little work (if I make no mistake with the algebra) one may check that $F(s) = \zeta(s)^{k+1}\prod_p (1+ (k-1)p^{-s})(1-p^{-s})^{k-1}$, where the latter Euler product converges absolutely for $\Re s > 1/2$. I think this should be enough to get an asymptotic formula by residue calculus for instance. | |
| Oct 24, 2015 at 22:17 | comment | added | Igor Rivin | I really have no idea why anyone would either downvote or vote to close this. | |
| Oct 24, 2015 at 22:16 | comment | added | Igor Rivin | @AlexeyUstinov I have the Handbook of number theory, and I could not find anything relevant (but since it is 700 pages, perhaps you have a specific place in mind). I don't know why you consider the question too broad. Going from $f$ to $f(n^k)$ is a natural transformation to consider on multiplicative functions, which may, or may not, have nice properties (perhaps under side conditions). That strikes me as a nice question - nice enough that it might have been considered by ancient sages. | |
| Oct 24, 2015 at 21:49 | answer | added | asdasdasd | timeline score: 3 | |
| Oct 24, 2015 at 9:18 | comment | added | Igor Rivin | @GerryMyerson Yes, but this shows there is no obvious connection between the asymptotic for $f$ and its composition with the $k$-th power. | |
| Oct 24, 2015 at 2:04 | comment | added | Alexey Ustinov | You can find some results of this type in Sándor, J.; Mitrinović, D. S. & Crstici, B. Handbook of number theory. I Springer, 2006. In current form your question looks too broad. | |
| Oct 24, 2015 at 0:51 | comment | added | Anthony Quas | It's not identically 0. $f(1)=1$. (Friday p.m. pedantry) | |
| Oct 23, 2015 at 23:43 | comment | added | Gerry Myerson | I would have thought that if $f(n^k)$ is identically zero, then the behavior of your sum would be exceptionally easy to understand. | |
| Oct 23, 2015 at 21:16 | comment | added | Igor Rivin | @JeremyRouse Well, as I said, I wasn't sure if it should be easy or hard, being unwise in this sort of thing (though the Moebius function really makes one wonder). By the way, the specific $f$ which brought this to mind was $f(n) = \tau(n)$ (the number of divisors). There might be a trick which works there... | |
| Oct 23, 2015 at 21:04 | comment | added | Jeremy Rouse | It's not as easy as you think - you don't just evaluate the Dirichlet series at $ks$. In the case that $f(n)$ is the coefficient of a modular form, Iwaniec says (at the very end of Topics in Classical Automorphic Forms), that the Dirichlet series does not admit an analytic continuation past $s = 0$ if $k$ is large enough. | |
| Oct 23, 2015 at 20:42 | review | Close votes | |||
| Oct 23, 2015 at 22:46 | |||||
| Oct 23, 2015 at 19:41 | history | asked | Igor Rivin | CC BY-SA 3.0 |