[Math] Negative falling Factorial

Question

Please can someone tell me what is the value of $1^\underline{-2}$? I know that $1^\underline{2}=0$.

Thanks.

Answer 1

Notice that $$x^{\underline k}=\frac{x^{\underline{k+1}}}{x-k}$$ for $k\ge 0$. If we generalize this to negative $k$, we have

$$\begin{align*} x^{\underline{-1}}&=\frac{x^{\underline0}}{x-(-1)}=\frac1{x+1}\\\\ x^{\underline{-2}}&=\frac1{(x+1)(x+2)}\\\\ &\;\vdots\\\\ x^{\underline{-k}}&=\frac1{(x+1)^{\overline{k}}}=\frac1{(x+k)^{\underline{k}}} \end{align*}$$

for $k\in\Bbb Z^+$. From this definition you can prove that for arbitrary $m,n\in\Bbb Z$,

$$x^{\underline{m+n}}=x^{\underline m}(x-m)^{\underline n}\;,$$

the falling power analogue of $x^{m+n}=x^mx^n$ for ordinary powers.