Timeline for A challenging (for me) limit calculation
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 23, 2023 at 22:21 | comment | added | Will Jagy | $$\alpha(z) = \frac{3}{z^2} + \frac{6 \log z }{5} + \frac{79 z^2}{1050} + \frac{29 z^4}{2625} + O(z^6)$$ | |
| Oct 23, 2023 at 22:12 | comment | added | Will Jagy | @TimothyChow turns out it was a constant plus the solution of Abel's equation, $\psi (\sin x) - \psi(x) = 1.$ I had thought, incorrectly, that something bounded was there, but $\psi(x) = \frac{3}{x^2} + \frac{6}{5} \log x + bounded. $ Anyway, what I once knew is in mathoverflow.net/questions/45608/… | |
| Oct 23, 2023 at 21:49 | comment | added | Timothy Chow | @WillJagy In your formula, what is $g(x)$? | |
| Oct 23, 2023 at 17:28 | comment | added | Michael Lugo | Fun fact: replace $\sin$ with $\tan^{-1}$ and the corresponding limit converges to $\sqrt{3/5}$. Since $\tan^{-1} x = x - x^3/3 + O(x^5)$ you have the approximation $f^\prime(x) = -f(x)^3/3$ and the rest of the proof carries through. | |
| Oct 23, 2023 at 9:43 | comment | added | C. WANG | Thank you for your solution! | |
| Oct 23, 2023 at 9:40 | vote | accept | C. WANG | ||
| Oct 23, 2023 at 9:40 | vote | accept | C. WANG | ||
| Oct 23, 2023 at 9:40 | |||||
| Oct 23, 2023 at 9:17 | history | edited | Daniel Weber | CC BY-SA 4.0 |
added 357 characters in body
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| Oct 23, 2023 at 8:42 | history | answered | Daniel Weber | CC BY-SA 4.0 |