Timeline for A number theoretical identity of exponential sum
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
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| Dec 26, 2020 at 18:59 | vote | accept | user267839 | ||
| Dec 24, 2020 at 2:07 | comment | added | Alapan Das | Because the Mangoldt function $\Lambda(n)=log(p),$ when $n=p^{a} \rightarrow \text{perfect power of some primes}$. Now, $n$ can't be any power of the primes $q_i$ as these divides $q$, but doesn't divide $a$ as $(a,q)=1$. For $n$ which isn't a perfect power of prime Mangoldt function is zero. | |
| Dec 24, 2020 at 1:41 | comment | added | user267839 | could you elaborate a bit more extensive why the identity $$\left(\sum_{(a,q)=1 ,a<q} e(a\frac{p}{q})\sum_{\substack{n \leq N \\ n \equiv a (\text{modulo q})}} \Lambda(n) \right)=\sum_{\substack{p \leq N \\ p \nmid q}} \log(p) \sum_{r=0}^{a(p): p^{a(p)} \leq N} e(p^r\alpha)$$. is true. I not understand. | |
| Dec 23, 2020 at 8:19 | history | edited | Alapan Das | CC BY-SA 4.0 |
added 61 characters in body
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| Dec 23, 2020 at 8:18 | history | undeleted | Alapan Das | ||
| Dec 23, 2020 at 8:13 | history | deleted | Alapan Das | via Vote | |
| Dec 23, 2020 at 8:09 | history | answered | Alapan Das | CC BY-SA 4.0 |