Timeline for integral of a "sin-omial" coefficients=binomial
Current License: CC BY-SA 3.0
16 events
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| Nov 6, 2020 at 18:31 | comment | added | Aditya Guha Roy | It is just slick and so beautiful ! Bravo ! | |
| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| Nov 24, 2016 at 22:18 | comment | added | Fedor Petrov | @AlexSelby I know a similar proof that $\sum 1/n^2=\pi^2/6$: take integral $\int_{-1}^1 \log(1+z)/z dz$ (which equal $1/2$ times $\sum 1/(2k+1)^2$ as follows from series expanding) and replace the contour to an arc of a unit circle. It is borrowed from D. Russel, Another Eulerian-type proof. Math. Mag. 1991 60, p.349. | |
| Nov 22, 2016 at 0:57 | comment | added | Alex Selby | Something interesting about this proof is that it is "backwards" from the normal way of substituting. Writing $u(t)=t+is(t)$, you'd normally want to have $t$ expressed analytically in terms of $u$ to get from the sinomial expression with $\sin^{\alpha}(\alpha t)$ etc to $(e^{-i\alpha u}+e^{i(1-\alpha)u})^n$. But there is presumably no such simple expression of $t$ in terms of $u$, so you had to know to start from $(e^{-i\alpha u}+e^{i(1-\alpha)u})^n$ and work the other way, using magic to end up with the sinomial expression. (Slightly wondering if other integrals can be unlocked like this.) | |
| Nov 20, 2016 at 23:10 | vote | accept | T. Amdeberhan | ||
| Nov 20, 2016 at 9:55 | history | edited | Fedor Petrov | CC BY-SA 3.0 |
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| Nov 19, 2016 at 20:19 | comment | added | esg | Simply beautiful! | |
| Nov 19, 2016 at 19:08 | comment | added | Alex Selby | Bravo, nice proof! It works for any (incl complex) values of $n$ and $k$ too, not just integer. Consider $(1/2\pi)\int_{-\pi}^\pi e^{-ikt}(1+e^{it})^n dt$ for $\Re(k)<0$, $\Re(n)>0$, then the integrand is analytic for $-\pi<\Re(t)<\pi$. We can deform the contour to two the vertical strips $\pm\pi+iy$ for $y>0$. Then you get a Beta function and the integral becomes $-(1/\pi)\sin(k\pi)B(-k,n+1)$ which is $\binom{n}{k}$, using the reflection formula $\sin(k\pi)=\pi/(\Gamma(k)\Gamma(1-k))$. The derivation was valid for $\Re(k)<0$, but both sides are analytic, so continuation gives you all $k$. | |
| Nov 19, 2016 at 14:07 | history | edited | Fedor Petrov | CC BY-SA 3.0 |
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| Nov 19, 2016 at 13:38 | comment | added | Mark Wildon | I cannot imagine a better explanation for this beautiful identity. | |
| Nov 19, 2016 at 13:32 | comment | added | T. Amdeberhan | This is very good. | |
| Nov 19, 2016 at 13:11 | history | edited | Fedor Petrov | CC BY-SA 3.0 |
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| Nov 19, 2016 at 11:45 | history | edited | Fedor Petrov | CC BY-SA 3.0 |
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| Nov 19, 2016 at 10:17 | history | edited | Fedor Petrov | CC BY-SA 3.0 |
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| Nov 19, 2016 at 1:55 | history | edited | Fedor Petrov | CC BY-SA 3.0 |
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| Nov 19, 2016 at 1:18 | history | answered | Fedor Petrov | CC BY-SA 3.0 |