Falling factorials count injective functions. There are no injective functions $ A \to B $ if $|A| \gt |B|$ . Is there a way to extend the definition of the falling factorial, ideally to $\mathbb{C} \times \mathbb{C} \to \mathbb{C}$, so that the zeroes are preserved? The most obvious way of informally generalizing the falling factorial involves taking ratios of the gamma function, but I'm not convinced that that works out.

When the right argument of the rising factorial is a natural number, it is defined in the following way:

$$ (x)^{\overline{k}} \stackrel{df}{=} \prod_{i = 0}^{k-1} (x + i) $$

$$ (x)^{\overline{k}} = (x)(x+1)\cdots(x+k-2)(x+k-1) $$

And the falling factorial is defined in the following way

$$ (x)^{\underline{k}} \stackrel{df}{=} \prod_{i = 0}^{k-1} (x - i) $$

$$ (x)^{\underline{k}} = (x)(x-1)\cdots (x-k+2)(x-k+1) $$

The falling factorial can be defined in terms of the rising factorial in two ways.

$$ (x)^{\underline{k}} = (-1)^k (-x)^{\overline{k}} $$


$$ (x)^{\underline{k}} = (x-k+1) ^ {\overline{k}} $$

The rising factorial satisfies the following identity:

$$ (x)^{\overline{k}} = \frac{(x+k-1)!}{(x-1)!} $$

Which suggests the generalization

$$ (x)^{\overline{k}} = \frac{\Gamma(x+k)}{\Gamma(x)} $$


$$ (x)^{\underline{k}} = \frac{\Gamma(x + 1)}{\Gamma(x-k + 1)} $$

This makes me nervous since $0 \notin \text{Im}(\Gamma)$ and $\Gamma$ has a pole when $ x \le -1 $ .

Does a nice way of extending the rising/falling factorial to complex arguments exist?

  • $\begingroup$ I do not know a better way than through the gamma ratio you show. Of course, the poles in the numerator will cancel with the denominator if both are negative integers $\endgroup$ – G Cab Nov 17 '18 at 23:18
  • 1
    $\begingroup$ by the way, it is actually $$ x^{\,\underline {\,k\,} } = \left( {x - k + 1} \right)^{\,\overline {\,k\,} } = {{\Gamma (x + 1)} \over {\Gamma (x - k + 1)}} $$ $\endgroup$ – G Cab Nov 17 '18 at 23:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.