Timeline for Binomial coefficient identity
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4 events
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| Jan 11, 2015 at 4:25 | comment | added | Todd Trimble | Yes, certainly @KConrad -- that's one nice way of looking at it. Another method is to use a finite difference analogue of the product rule. Speaking of Newtonian series and such, we also have (something like) $f(x) = \sum_{m \geq 0} (\Delta^m f)(0) \frac{x^{\underline{m}}}{m!}$ for a reasonable class of functions. All of this is further illuminated by passing to generating functions -- I recommend the book by Graham, Knuth, and Patashnik for anyone interested (can't find my copy at the moment). | |
| Jan 11, 2015 at 4:17 | comment | added | KConrad | To prove that rational function identity, the rational function $\frac{1}{x(x+1)\cdots(x+m)}$ can be expanded by partial fractions as $\sum_{k=0}^m \frac{c_k}{x+k}$ for unknown constants $c_k$. For any $i$ from $0$ to $m$, multiply both sides by $x+i$ and take a limit as $x \rightarrow -i$: $c_i = \lim_{x \rightarrow -i} \frac{1}{x(x+1)\cdots(x+i-1)(x+i+1)\cdots(x+m)}$, which is $\frac{1}{(-1)^ii!}\cdot \frac{1}{(m-i)!} = \frac{(-1)^i}{m!}\binom{m}{i}$. | |
| Jan 11, 2015 at 4:10 | comment | added | KConrad | In fact, when expressed as the rational function identity $\sum_{k=0}^m \binom{m}{k} (-1)^k/(x+k) = m!/(x(x+1)\cdots(x+m))$ this is equivalent (after changing $k$ to $m-k$ and replacing $x$ with $x-m$) to the rational function identity $\sum_{k=0}^m \binom{m}{k} (-1)^k/(x-k) = (-1)^mm!/(x(x-1)\cdots(x-m))$, which appears in the middle of the page en.wikipedia.org/wiki/Table_of_Newtonian_series with a sign error (which someone may fix after seeing this comment). | |
| Jan 11, 2015 at 4:06 | history | answered | Todd Trimble | CC BY-SA 3.0 |