Product of $(1-m)(1-2m)\cdots(1-nm)$

Question

Given $0\le nm<1$, what is $(1-m)(1-2m)\cdots(1-nm)$? AM–GM suggests an upper bound of $(1-\frac{m}{2}(1+n))^n$. If $m\ll1$, I believe a decent approximation is $(\sqrt{1-m(1+n)})^n$.

Answer 1

This is a multifactorial, specifically with

$$(1-m)(1-2m)\dots(1-nm)=\frac{(1-m)!^{(m)}}{(1-(n+1)m)!^{(m)}}$$

As the Wikipedia shows,

$$z!^{(\alpha)}=\alpha^{(z-1)/\alpha}\frac{\Gamma\left(\frac z\alpha+1\right)}{\Gamma\left(\frac1\alpha+1\right)}$$

Thus, we get

$$(1-m)!^{(m)}=m^{-1}\frac{\Gamma\left(\frac1m\right)}{\Gamma\left(\frac1m+1\right)}=1$$

$$(1-(n+1)m)!^{(m)}=m^{-(n+1)}\frac{\Gamma\left(\frac1m-n\right)}{\Gamma\left(\frac1m+1\right)}$$

Using the Gamma function's reflection formula,

$$\Gamma(z+1)=z\Gamma(z)$$

We may deduce that

$$(1-m)(1-2m)\dots(1-nm)=m^{n+1}\frac{\Gamma(\frac1m+1)}{\Gamma\left(\frac1m-n\right)}$$

Where $\Gamma$ is the Gamma function, which may be evaluated using the recursive definition/formula.

For example, at $n=2$, we have

$$(1-m)(1-2m)=m^3\frac{\Gamma\left(\frac1m+1\right)}{\Gamma\left(\frac1m-2\right)}$$

We know that:

$$\Gamma\left(\frac1m+1\right)=\frac1m\Gamma\left(\frac1m\right)=\frac1m\left(\frac1m-1\right)\Gamma\left(\frac1m-1\right)\\=\frac1m\left(\frac1m-1\right)\left(\frac1m-2\right)\Gamma\left(\frac1m-2\right)$$

Thus,

$$(1-m)(1-2m)=m^3\frac1m\left(\frac1m-1\right)\left(\frac1m-2\right)$$

which checks out.

If you want it in terms of seemingly more elementary functions, we may use the factorial to get

$$(1-m)(1-2m)\dots(1-nm)=m^{n+1}\frac{\left(\frac1m\right)!}{\left(\frac1m-n-1\right)!}$$

Or in terms of the binomial coefficients:

$$(1-m)(1-2m)\dots(1-nm)=m^{n+1}(n+1)!\binom{1/m}{n+1}$$

Answer 2

The product is: $$P=\prod_{k=1}^N(1-km)$$ The result is: $$P=(-m)^N\dfrac{\Gamma\dfrac{(Nm+m-1)}{m}}{\Gamma\dfrac{(m-1)}{m}}$$ So you can get the exact value of $P$