# what do we know about the product $\prod_{n\in \mathbb{N}}(1-\frac{x^m}{n^m})?$

My question is what do we know about the product $$\prod_{n\in \mathbb{N}}(1-\frac{x^m}{n^m})?$$ which is a slightly modified product from the eulerian product.

• For $m=2$, it's a well-known expression for sinc; Mathematica will give you similar generalizations for fixed even $m$, and reciprocal $\Gamma$ functions for fixed odd $m$ ... not sure about a nice general expression for variable $m$ ... – Michael Engelhardt Aug 4 '19 at 18:36
• Also Euler product is for products over primes, Euler's formula for $\sin(x)$ is called Weierstrass product – reuns Aug 5 '19 at 3:43

Well, for $$m=2$$, we have the standard product $$\prod_{n\in \mathbb N} (1-\frac {x^2} {n^2})=\frac {\sin(\pi x)} {\pi x}$$.
If we denote your product by $$P_m(x)$$, then we can evaluate eg. $$P_4(x)$$ like this:

$$(1-\frac {x^4} {n^4}) = (1-\frac {x^2} {n^2})(1+\frac {x^2} {n^2})=(1-\frac {x^2} {n^2})(1-\frac {(ix)^2} {n^2}).$$ Thus, $$P_4(x)=\frac {\sin(\pi x)} {\pi x} \frac {\sin(i \pi x)} {i\pi x}=\frac {\sin(\pi x) \sinh(\pi x)} {\pi^2 x^2}$$

We can write an interesting identity:

$$\log P_m(x)=\sum_{n\in\mathbb N} \log (1-\frac {x^m} {n^m})=-\sum_{n=1}^\infty\sum_{k=1}^\infty \frac 1 k (\frac x m)^{mk}=-\sum_{k=1}^\infty \frac 1 k \zeta (mk) x^{mk}$$

By differentiating, we get

$$\frac {P'_m(x)} {P_m(x)}=-\frac mx\sum_{k=1}^\infty \zeta(mk)x^{mk}$$

EDIT: Also, by using $$1/\Gamma(z)=ze^{\gamma z}\Pi_{n\in \mathbb N} (1+\frac zn)e^{-z/n}$$ and expanding $$(1-z^3/n^3)$$ as $$(1-z/n)(\omega-z/n)(\omega^2-z/n)$$, we get:

$$P_3(z)=\frac {1/z^3} {\Gamma(-z)\Gamma(-\omega z)\Gamma(-\omega^2 z)}$$

This is similar to

$$P_2(z) = \frac {-1/z^2} {\Gamma(-z)\Gamma(z)} = \frac {\sin(\pi z)} {\pi z}$$ (by the reflection formula)

I think it is easy to see now that those formulas generalize to all $$m$$. We just take $$1/z^m$$ (or $$-1/z^m$$ for $$m$$ even) and divide by the product of $$\Gamma(-\rho z)$$, where $$\rho$$ goes over all $$m$$-th roots of unity.

$$P_m(z)=(-1)^{m+1}\frac 1 {z^m} \prod_\rho \frac 1 {\Gamma(-\rho z)}$$

For even $$m$$, this can be further reduced using the reflection formula, because the roots of unity form pairs (both $$\rho$$ and $$-\rho$$ are roots of unity).

The even case $$\prod_{n \ge 1} (1-(x/n)^{2m}) =\prod_{a=1}^m \prod_{n \ge 1} (1- e^{2i \pi a/m}(x/n)^2)= \prod_{a=1}^m \frac{\sin(\pi e^{i \pi a/ m} x)}{\pi e^{i \pi a/ m} x}$$

Otherwise $$\prod_{n \ge 1} (1-(x/n)^m) =\lim_{N \to \infty }\prod_{a=1}^m \prod_{n=1}^N (1- e^{2i \pi a/m}x/n)= \prod_{a=1}^m \frac{-e^{2i \pi a/ m} x}{\Gamma(-e^{2i \pi a/ m} x)}$$