Timeline for Growth rate of elementary sequences
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 15 at 5:58 | vote | accept | António Borges Santos | ||
| Jan 15 at 2:21 | answer | added | Christian Remling | timeline score: 1 | |
| Jan 15 at 1:15 | comment | added | Christian Remling | I of course fell into the same little trap as before, this time with $y_n=0$. | |
| Jan 15 at 0:15 | comment | added | António Borges Santos | Can you now help me with a concrete choice of $x_n$ and $y_n$? | |
| Jan 15 at 0:15 | comment | added | António Borges Santos | Okay, let's say $z_n=\log(n)^2$ then $z_n$ satisfies the assumptions since $\log(\int_0^{\infty} e^{-\varepsilon \log(x)^2} dx) $ behaves like $\frac{1}{4\varepsilon}.$ Thus $\lim_{\varepsilon \downarrow 0}\varepsilon \log( \sum_n e^{-\varepsilon z_n}) = \frac{1}{4}$ if I am not mistaken. | |
| Jan 14 at 23:52 | comment | added | António Borges Santos | @ChristianRemling yes, but remember also that $x_n \to 0$ and $y_n \to \infty$ are necessary. So you can rearrange but the tail behaviour is somehow determined... | |
| Jan 14 at 23:26 | comment | added | António Borges Santos | well, $y$ also has to have the property that $\lim_{\varepsilon \downarrow 0} \varepsilon \log(\sum_{n} e^{-\varepsilon y_n})=0$, right? | |
| Jan 14 at 23:18 | comment | added | António Borges Santos | @ChristianRemling Thanks, I added as a constraint that $(z_{n})$ is strictly increasing. Notice that $y$ cannot vanish too often, since otherwise the sum is infinite right away. Does this address your concern? | |
| Jan 14 at 23:17 | history | edited | António Borges Santos | CC BY-SA 4.0 |
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| Jan 14 at 23:12 | history | edited | António Borges Santos | CC BY-SA 4.0 |
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| Jan 14 at 22:38 | history | asked | António Borges Santos | CC BY-SA 4.0 |