Timeline for Binomial again, and again
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 17, 2016 at 22:44 | history | edited | Dan Petersen | CC BY-SA 3.0 |
edited body
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| Nov 19, 2016 at 14:36 | comment | added | Pietro Majer | Also, if we adopt the trigonometric definition of $ {n\choose x}$, the relation ${n-1\choose x}+ {n-1 \choose x-1}= {n\choose x}$ may be derived from the identity: $${1\over(n-1-x)\dots(1-x)\cdot x}-{1\over(n-x)\dots(2-x)\cdot(x-1)}={n\over(n-x)(n-1-x)\dots(1-x)\cdot x}$$ multiplying by $\sin\pi x=-\sin\pi(x-1)$ | |
| Nov 19, 2016 at 13:33 | comment | added | T. Amdeberhan | I like the simplicity here. | |
| Nov 19, 2016 at 10:46 | comment | added | Pietro Majer | The starting point for the second part, that is the identity $\int_{-\infty}^{+\infty} {n\choose x}dx={2^n}\int_{-\infty}^{+\infty} {0\choose x}dx={2^n\over \pi}\int_{-\infty}^{+\infty} {\sin\pi x\over x}dx,$ may also be derived inductively, integrating ${n\choose x}={n-1\choose x}+{n-1\choose x-1}.$ | |
| Nov 19, 2016 at 9:26 | comment | added | Pietro Majer | this is very very nice | |
| Nov 19, 2016 at 5:05 | history | answered | Noam D. Elkies | CC BY-SA 3.0 |