Timeline for Lower bound of exponential sum
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
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| S Nov 7, 2020 at 14:13 | history | suggested | RobPratt |
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| Nov 7, 2020 at 13:54 | review | Suggested edits | |||
| S Nov 7, 2020 at 14:13 | |||||
| Nov 7, 2020 at 0:40 | comment | added | Thomas Yang | @GerryMyerson Ah that is my bad. I forgot to multiply $N^{-1}$. | |
| Nov 7, 2020 at 0:39 | history | edited | Thomas Yang | CC BY-SA 4.0 |
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| Nov 6, 2020 at 22:32 | comment | added | Gerry Myerson | OK. But you still have $\phi$ defined on $[0,1]$ but evaluated at $2,3,4,\dots,N$. Something's gotta give. | |
| Nov 6, 2020 at 22:06 | comment | added | Thomas Yang | @GerryMyerson Sorry for the confusion. By a function $\phi\sim1$ I meant that $\phi$ is both bounded above and below by absolute constants. | |
| Nov 6, 2020 at 22:05 | history | edited | Thomas Yang | CC BY-SA 4.0 |
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| Nov 6, 2020 at 11:51 | comment | added | Gerry Myerson | I want to interpret $\phi'\sim1$ to mean $\phi'$ is asymptotic to one, but since $\phi$ is only defined on a finite interval, that can't be right, can it? But then, the summation doesn't make sense if $\phi$ is only defined on $[0,1]$, since it involves $\phi(n)$ for $n=1,2,\dots,N$. So, what do you really mean? | |
| Nov 6, 2020 at 1:40 | history | asked | Thomas Yang | CC BY-SA 4.0 |