Negated Upper Index of Binomial Coefficient

Theorem

Let $r \in \R, k \in \Z$.

Then:

$\dbinom r k = \paren {-1}^k \dbinom {k - r - 1} k$

where $\dbinom r k$ is a binomial coefficient.

Corollary 1

Let $r \in \R, k \in \Z$.

Then:

$\dbinom {-r} k = \paren {-1}^k \dbinom {r + k - 1} k$

where $\dbinom {-r} k$ is a binomial coefficient.

Corollary 2

$\dbinom n m = \paren {-1}^{n - m} \dbinom {-\paren {m + 1} } {n - m}$

Complex Numbers

For all $z, w \in \C$ such that it is not the case that $z$ is a negative integer and $t, w$ integers:

$\dbinom z w = \dfrac {\map \sin {\pi \paren {w - z - 1} } } {\map \sin {\pi z} } \dbinom {w - z - 1} w$

where $\dbinom z w$ is a binomial coefficient.

Proof

$\ds \binom r k$

$=$

$\ds \frac {r^{\underline k} } {k!}$

Definition of Binomial Coefficient

$\ds $

$=$

$\ds \frac 1 {k!} \prod_{j \mathop = 0}^{k - 1} \paren {r - j}$

Definition of Falling Factorial

$\ds $

$=$

$\ds \frac {\paren {-1}^k} {k!} \prod_{j \mathop = 0}^{k - 1} \paren {-\paren {r - j} }$

$\ds $

$=$

$\ds \frac {\paren {-1}^k} {k!} \prod_{j \mathop = 0}^{k - 1} \paren {j - r}$

$\ds $

$=$

$\ds \frac {\paren {-1}^k} {k!} \prod_{j \mathop = 0}^{k - 1} \paren {\paren {k - 1} - j - r}$

Permutation of Indices of Product

$\ds $

$=$

$\ds \frac {\paren {-1}^k} {k!} \prod_{j \mathop = 0}^{k - 1} \paren {\paren {k - r - 1} - j}$

$\ds $

$=$

$\ds \paren {-1}^k \frac {\paren {k - r - 1}^{\underline k} } {k!}$

Definition of Falling Factorial

$\ds $

$=$

$\ds \paren {-1}^k \binom {k - r - 1} k$

Definition of Binomial Coefficient

$\blacksquare$

Sources

1964: Milton Abramowitz and Irene A. Stegun: Handbook of Mathematical Functions ... (previous) ... (next): $3$: Elementary Analytic Methods: $3.1$ Binomial Theorem etc.: Binomial Coefficients: $3.1.3$: Binomial Coefficients
1997: Donald E. Knuth: The Art of Computer Programming: Volume 1: Fundamental Algorithms (3rd ed.) ... (previous) ... (next): $\S 1.2.6$: Binomial Coefficients: $\text{G}$: $(17)$