Timeline for Calculate the number theory function $\sum_{n \equiv a(\bmod q)} \frac{\phi(n)-\mu(n)}{n^{s+1}}$
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
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| Mar 28, 2023 at 22:13 | comment | added | tomos | oh, i didn't see you posted it here too. on the stackexchange one i posted an answer for you | |
| Mar 28, 2023 at 5:06 | review | Close votes | |||
| Apr 12, 2023 at 3:07 | |||||
| Mar 28, 2023 at 4:58 | comment | added | fractal | But it is still difference between the final equation. I get the thing:$$ \begin{aligned} \frac{1}{b^{s+1}} \frac{(\phi(b) L(s, \chi))}{L(s+1, \chi) \prod_{p \mid b,p \mid n'}\left(1-p^{-1}\right)} \end{aligned} $$ and \begin{aligned} \frac{1}{b^{s+1}} \frac{\mu(b)*1_{(b,n')}}{L(s+1, \chi)} \end{aligned}, not the $$ \begin{aligned} \frac{1}{b^{s+1}} \frac{(\phi(b) L(s, \chi)-\mu(b))}{L(s+1, \chi) \prod_{p \mid b}\left(1-\chi(p) p^{-s-1}\right)} \end{aligned} $$. @PeterHumphries | |
| Mar 28, 2023 at 4:49 | comment | added | Peter Humphries | Just use the fact that $$\phi(bn') = \frac{\phi(b) \phi(n') (b,n')}{\phi((b,n'))}, \qquad \mu(bn') = \mu(b) \mu(n') 1_{(b,n')=1}.$$ | |
| S Mar 28, 2023 at 4:43 | review | First questions | |||
| Mar 28, 2023 at 5:14 | |||||
| S Mar 28, 2023 at 4:43 | history | asked | fractal | CC BY-SA 4.0 |