Is this Newton binomial legal?

Question

is this equation proper?

$$ \binom {9} {12} = 0 $$

so if we have a binomial $$ \binom {n} {k} $$

and when $$ n<k $$ the result is 0?

You can define $\binom{n}{k}$ for nonnegative integer $n$ and integer $k$ with the convention that it equals $0$ if $k<0$ or $k>n$. Pascal stays valid. Sometimes that is indeed handsome. — drhab, Jan 21 '14 at 08:56
$f(x)=\binom{x}{12}$ is a twelfth degree polynomial. It has $12$ distinct roots, namely the integers $0,1,2,\dots,11$. In particular, yes, $f(9)=0$. — bof, Jan 21 '14 at 09:09
There is a general scheme even for complex numbers. There is a nice lecture over here — Felix Marin, Jan 21 '14 at 12:05
Yes. And then we can state the binomial theorem as $(1+x)^9 = \sum\binom{9}{k} x^k$, where the sum is over all integers $k$. — GEdgar, Jan 21 '14 at 12:50

robjohn · Answer 1 · 2014-01-21T15:46:11.543

3

In general, for $k\in\mathbb{Z}$ and $k\ge0$, $$ \binom{n}{k}=\frac{n(n-1)(n-2)\cdots(n-k+1)}{k!} $$ If $n\in\mathbb{Z}$ and $k\gt n\ge0$, then the numerator has a factor of $0$.

As drhab mentions, if $k\lt0$, it is makes sense to define $$ \binom{n}{k}=0 $$ which is often done to simplify the limits of sums.

edited Jan 21 '14 at 15:46

answered Jan 21 '14 at 09:11

robjohn

345,667

Is there any objection to define it as $0$ if $k$ is negative? Quite handsome in for instance http://math.stackexchange.com/a/633183/75923. No bothering about where to end also where to begin. – drhab Jan 21 '14 at 09:44
@drhab: indeed. I have added some to my answer to incorporate this. – robjohn Jan 21 '14 at 15:47

score 3 · Answer 2 · edited Jan 21 '14 at 12:01

In fact, the concept of binomial coefficient can be extended to allow the upper index to be real (or even complex). (Ronald L. Graham, Donald E. Knuth, and Oren Patashnik. Concrete Mathematics: A Foundation for Computer Science, 2nd ed. Addison-Wesley, 1994, Chap. 5)

In this way we can define binomial coefficient by: $$ \binom{r}{k} = \begin{cases} \dfrac{r(r-1)(r-2)\dots(r-k+1)}{k(k-1)(k-2)\dots(1)} = \dfrac{r^{\underline{k}}}{k!}, \; k \in \mathbb{Z}_{+} \\ 0, \quad k<0. \end{cases} $$ Where $r\in \mathbb{R}$

That means we can compute:

$$ \binom{-1}{3}=\dfrac{(-1)(-2)(-3)}{3!}=-1 $$

Despite the fact there's no combinatorial interpretation here.

Also, as pointed by bof, we can also look into $\dbinom{r}{k}$ as a $k$th-degree polynomial in r.

I really recommend the book mentioned above if you want to understand more about binomial coefficients.

Edit: Just to complement, in your example: $$ \binom{9}{12}=\dfrac{9\cdot 8 \cdot 7 \cdots \cdot 0 \cdot (-1) \cdot (-2)}{12!}=\dfrac{0}{12!}=0. $$

Is this Newton binomial legal?

2 Answers2