Timeline for calculating Möbius function
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 11 at 17:12 | comment | added | LSpice | Thanks!........ | |
| Sep 10 at 21:14 | comment | added | Evan Ashoori | If you want to see why the sum converges to 1/3, you can write out the power series. It basically amounts to the fact that mu*1=delta, where mu is the mobius function, * represents Dirichlet convolution, and delta(n)=1 if n=1 and 0 otherwise. | |
| Sep 10 at 20:42 | comment | added | Evan Ashoori | The main idea (for the second one) is that we have that the sum over the positive integers of mu(n)/(3^(n)-1)=1/3. We think about the following range: (1/3-1/(2*3^(n)-2)), 1/3+1/(2*3^(n)-2))). I claim that for any sum over n>m of (a_n)/(3^n-1) with a_n=-1,0, or 1 to converge to x, |x| must be less than 1/(2*3^(m)-2). We can see this using geometric series (looking at the sum of the terms with n>m and assuming a_n=1 for all n). However, there is always only one choice of a_n that puts the partial sum in (1/3-1/(2*3^(n)-2)), 1/3+1/(2*3^(n)-2))). This is how it works! | |
| Sep 10 at 1:22 | comment | added | LSpice | Welcome to MO! Since this is a math forum, could you say mathematically, rather than only in Python, how your computation proceeds? | |
| Sep 9 at 22:48 | review | Late answers | |||
| Sep 10 at 4:00 | |||||
| S Sep 9 at 22:29 | review | First answers | |||
| Sep 10 at 4:00 | |||||
| S Sep 9 at 22:29 | history | answered | Evan Ashoori | CC BY-SA 4.0 |