Timeline for Average involving the Euler phi function
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
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| Jul 12, 2013 at 22:43 | comment | added | Eric Naslund | @i707107 Doing some rearrangements, we can rewrite the double sum as a sum with the Von Mangoldt function: $$\frac{1}{N^{2}}\sum_{ab\leq N}\frac{\Lambda(a)b}{\log ab}.$$ From here, we can easily bound the sum over $ab\leq( N/\log(N))^2$, and then show that $$\frac{1}{N^{2}}\sum_{\frac{N}{\left(\log N\right)^{2}}\leq ab\leq N}\frac{\Lambda(a)b}{\log ab}\sim\frac{1}{N^{2}\log N}\sum_{\frac{N}{\left(\log N\right)^{2}}\leq ab\leq N}\Lambda(a)b, $$ and by the prime number theorem, the right hand side above is asymptotic to $$-\frac{\zeta'(2)}{2\zeta(2)}\frac{1}{\log N}.$$ | |
| Jun 18, 2013 at 17:41 | vote | accept | wongpin101 | ||
| May 7, 2013 at 19:52 | comment | added | Sungjin Kim | I think so too. | |
| May 7, 2013 at 17:50 | comment | added | Greg Martin | Agreed. I think the constant is $|\zeta'(2)|/2\zeta(2)$, though I might have missed something. | |
| May 7, 2013 at 2:53 | comment | added | Sungjin Kim | It is also possible to obtain an asymptotic formula with main term $C\frac{1}{\log N}$ for some positive constant $C$. | |
| May 7, 2013 at 2:44 | history | answered | Greg Martin | CC BY-SA 3.0 |