Timeline for Is there a good approximation for this Gaussian-like integration?
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
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| Dec 27, 2019 at 18:49 | history | edited | user64494 | CC BY-SA 4.0 |
A typo in the title
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| Dec 27, 2019 at 11:12 | history | became hot network question | |||
| Dec 27, 2019 at 10:53 | history | edited | Carlo Beenakker | CC BY-SA 4.0 |
inline image
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| Dec 27, 2019 at 10:47 | answer | added | Carlo Beenakker | timeline score: 1 | |
| Dec 27, 2019 at 10:39 | answer | added | user64494 | timeline score: 0 | |
| Dec 27, 2019 at 7:05 | answer | added | Zurab Silagadze | timeline score: 0 | |
| Dec 27, 2019 at 3:39 | comment | added | Michael Engelhardt | @LeechLattice Ok, then the dominant contribution comes from the endpoints, $x=\pm \eta $. | |
| Dec 27, 2019 at 3:37 | comment | added | LeechLattice | @MichaelEngelhardt I understand the question as $n→+\infty$ for fixed $η$, so $x=\sqrt{2n}$ is outside the range of integration. | |
| Dec 27, 2019 at 3:34 | comment | added | Michael Engelhardt | Write $x^{2n} = e^{2n \ln x} $. When $n$ becomes large, the integrand is strongly peaked around the maximum of the exponent, i.e., at $x=\sqrt{2n} $. If that's within the range of integration, saddle point approximation should be good. | |
| Dec 27, 2019 at 3:34 | comment | added | LeechLattice | The integral is asympotically $Θ(1/n)$, so the $1.06$ is actually $1$. Are you asking for the coefficient in the $Θ$? | |
| Dec 27, 2019 at 3:15 | review | First posts | |||
| Dec 27, 2019 at 4:06 | |||||
| Dec 27, 2019 at 3:11 | history | asked | CPW | CC BY-SA 4.0 |