Timeline for How can $\binom{n}{k}=\frac{n\cdot (n-1)\cdot ... \cdot (n-k+1)}{k!}$, with $k<0$ or $k>n$ be equal to $0$?
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| Jun 21 at 23:46 | comment | added | Semiclassical | One possible definition one could adopt is that $\binom{n}{k}$ is the coefficient of $z^k$ in the series expansion of $(1+z)^n$. Since the series expansion contains no negative powers and no powers larger than $z^n$, the corresponding binomial coefficients are zero as well. (Of course, the challenge with this definition is to recover the usual formula for $\binom{n}{k}$ for $k$ in the appropriate range.) | |
| Jun 21 at 16:59 | vote | accept | Benjamin Letelier Lazo | ||
| Jun 21 at 16:58 | comment | added | Andreas Blass | The formula for $n\choose k$ in the title of your question (in contrast to the formula in the main text) solves your problem when $k>n$, because in this case one of the factors in the numerator is $n-(n+1)+1=0$. As for the problem with $k<0$, the usual convention is that negative integers $m$ have $m!=\infty$. This convention makes sense both in the context of ordinary properties of factorials (consider $(m+1)!=m!\cdot (m+1)$ when $m=-1$) and in the fancier context of the Gamma function (a rather natural complex-analytic extension of the function $z\mapsto(z-1)!$). | |
| Jun 21 at 16:50 | answer | added | Masd | timeline score: 2 | |
| Jun 21 at 16:44 | comment | added | JMoravitz | If that is truly all you have to go on as a definition, then I can not prove this to you. The definition is bad and incomplete. I suggest taking the standard definition instead or extending your definition to include the extreme cases for $k$. | |
| Jun 21 at 16:22 | history | edited | Benjamin Letelier Lazo | CC BY-SA 4.0 |
I added a definition of the binomial coefficient, in the comment section there is more info on how I ended with the doubt.
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| Jun 21 at 16:08 | comment | added | Benjamin Letelier Lazo | The first slide of my binomial coefficient class was defining $\binom{n}{k} = \frac{n\cdot (n-1)\cdot ... \cdot (n-k+1)}{k!}$, so I assume that is the definition of $\binom{n}{k}$ that I should use. | |
| Jun 21 at 16:05 | comment | added | Divide1918 | In that case, you might as well adopt the convention that it is 0. Just define it as 0 and proceed with your induction. | |
| Jun 21 at 16:04 | comment | added | jjagmath | That doesn't address the important point JMoravitz is doing in the comments: How are defining $\binom{n}{k}$? | |
| Jun 21 at 16:02 | comment | added | Benjamin Letelier Lazo | The question is raised because I was trying to do a proof by induction over $n$ of $\sum_{k=0}^{n} (-1)^{k}\cdot \binom{n}{k} = 0$, with $n > 0$. I know that it can be easily proof by using the binomial theorem, but I wanted to practice the proof by induction. In the induction step I ended with $\sum_{k=0}^{n} (-1)^{k}\cdot \binom{n}{k-1} + \sum_{k=0}^{n} (-1)^{k}\cdot \binom{n}{k} + ...$, the second term is part of my induction hyphotesis, and ended with the first term where the doubt raised when $k=0$. | |
| Jun 21 at 15:51 | comment | added | JMoravitz | The most typical way to define it is combinatorially... at which $\binom{n}{k}$ is defined to be equal to the number of $k$-element subsets of an $n$-element set. There are no "negative-sized subsets" to any set... just as there are not subsets of size larger than the size of the parent set they came from... hence it equals zero. Any of that stuff about factorials comes after the fact and is only claimed to equal the binomial coefficient in those situations where it makes sense to talk about. | |
| Jun 21 at 15:49 | comment | added | JMoravitz | "based on what the binomial coefficient means" Here we get into a chicken and egg sort of scenario. To be able to communicate with one another... we must first decide what the symbols $\binom{n}{k}$ should represent. How are we defining it? Algebraically? Combinatorially? Recursively? Some other way? | |
| S Jun 21 at 15:42 | review | First questions | |||
| Jun 21 at 15:45 | |||||
| S Jun 21 at 15:42 | history | asked | Benjamin Letelier Lazo | CC BY-SA 4.0 |