Timeline for multiplicative functions from values at primes
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
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| Jul 28, 2017 at 3:58 | comment | added | Gerry Myerson | I'd say it's certainly about the extension to prime powers. Your three categories are $f(p^n)=(f(p))^n$, $f(p^n)=f(p)$, and $f(p^n)=0$ (for $n\ge2$). Phi-function is $f(p^n)=p^{n-1}f(p)$, divisor function is $f(p^n)=n-1+f(p)$. The relative apriori naturality of these is not clear to me. | |
| Jul 27, 2017 at 14:33 | comment | added | Igor Rivin | @GerryMyerson ... but if you notice, I am phrasing this as an extension problem, and presumably some extensions are a priori more natural than others (the Euler phi is extending $p-1,$ the divisor function $p+1,$ the number of divisors, $2,$ and if you did not know about their deeper meanings none of the extensions look particularly natural (notice that once you define the function on prime powers, there is no choice, so maybe the question is about the extension to prime powers...) | |
| Jul 27, 2017 at 13:17 | comment | added | Igor Rivin | @GerryMyerson Yes, granted. | |
| Jul 27, 2017 at 12:46 | comment | added | Gerry Myerson | The Euler phi-function is pretty natural, and falls into none of your three categories. Ditto for the number of divisors, the sum of divisors,.... | |
| Jul 27, 2017 at 6:38 | comment | added | Lior Bary-Soroker | The second is a la von Mangoldt. | |
| Jul 27, 2017 at 4:14 | comment | added | reuns | In the context of L-functions, you'll want to write $$F(s) = \prod_p (1+f(p) p^{-s}) \prod_p (1+\sum_{k \ge 2} g(p^k)p^{-sk})$$ Such that (the Dirichlet series for) $\sum_p \log (1+\sum_{k \ge 2} g(p^k)p^{-sk})$ converges (absolutely) for $\Re(s) > \sigma$ the abscissa of convergence of $\sum_{p} \log(1+f(p) p^{-s})$ or $\sum_p f(p) p^{-s}$. | |
| Jul 27, 2017 at 3:19 | answer | added | Wlod AA | timeline score: 2 | |
| Jul 27, 2017 at 2:35 | history | asked | Igor Rivin | CC BY-SA 3.0 |