How to find a power series representation for a divergent product?

Question

Euler used the identity $$ \frac{ \sin(x) }{x} = \prod_{n=1}^{\infty} \left(1 - \frac{x^2}{n^2 \pi^2 } \right) = \sum_{n=0}^{\infty} \frac{ (-1)^n }{(2n + 1)! } x^{2n} $$ to solve the Basel problem. The product is obtained by noting that the sine function is 'just' an infinite polynomial, which can be rewritten as the product of its zeroes. The sum is found by writing down the taylor series expansion of the sine function and dividing by $x$.

Now, I am interested in finding the sum representation of the following product: $$ \prod_{n=1}^{\infty} \left(1 - \frac{x}{n \pi} \right) ,$$ which is divergent (see this article).

The infinite sum representation of this product is not as easily found (at least not by me) because it does not have an obvious formal representation like $\frac{\sin(x)}{x}$ above.

Questions: what is the infinite sum representation of the second product I mentioned? How does one obtain this sum? And is there any 'formal' represenation for these formulae (like $\frac{\sin(x)}{x}$ above).

Answer 1

Not sure if there is a precise sense in which this is meaningful, as the product is divergent for all $x\neq 0$, but the function $$ f(x):=\frac{1}{\Gamma\left(1-\frac{x}{\pi}\right)} $$ has simple zeroes at precisely the positive multiples of $\pi$, and satisfies $f(0)=1$. The reflection formula for $\Gamma$ shows that $$ f(x)f(-x)=\frac{\sin(x)}{x}=\prod_{n=1}^\infty \left(1-\frac{x}{n\pi}\right)\left(1+\frac{x}{n\pi}\right). $$

Answer 2

I never gave the full answer :)

$$\prod_{n=1}^{\infty} (1-x/n)$$ When analysing a product it's often easiest to consider the form $\prod_{n=1}^{\infty} (1+f(n))$ given that $\sum_{n=1}^{\infty}f(n)^m=G(m)$; $f(n)=-x/n$ Then $G(m)=(-x)^m\zeta(m)$.

$$\prod_{n=1}^{\infty} (1-x/n)=e^{\sum_{m=1}^{\infty}\frac{(-1)^{m+1}x^m\sum_{n=1}^{\infty}f(n)^m}{m}}$$ Because $\zeta(1)$ is the only part of the product that goes to 0 (e.g. e^-infty), the regularized product will tend to 0 just as $\zeta(1)$ goes to infinity. However this is easy fixable:

$$\prod_{n=1}^{\infty} (1-x/n)e^{x/n}=e^{\sum_{m=2}^{\infty}\frac{(-1)^{m+1}x^m\sum_{n=1}^{\infty}f(n)^m}{m}}=e^{\sum_{m=1}^{\infty}\frac{- x^{m+1}\zeta(m+1)}{m+1}}$$ which does converge. To show the above more well known representation,

$$\sum_{n=1}^{\infty}-\frac{x^{n+1}\zeta(n+1)}{n+1}=\int_{0}^{-x}\sum_{n=1}^{\infty}(-1)^nz^n\zeta(n+1)dz=\int_{0}^{-x}-H_zdz=-\ln((-x)!)+x\gamma$$ $$\prod_{n=1}^{\infty} (1-x/n)e^{x/n}=\frac{e^ {x\gamma}}{(-x)!}$$ And now with the answer I posted 6 years ago, and using the refined stirling numbers: $$\prod_{n=1}^{\infty} (1-x/n)e^{x/n}=\frac{e^ {x\gamma}}{(-x)!}=\bigg(1-x^2\zeta(2)/2-x^3 2\zeta(3)/6 +x^4 \big(3\zeta(2)^2-6\zeta(4)\big)/24+... \bigg)$$

So we can also say with p goes to inf: $$p^x\prod_{n=1}^{p} (1-x/n)=\frac{1}{(-x)!}=$$

We can rewrite our G(m) with $G(1)=-x\gamma$ and use this in the formula given above to write a polynomal representation, which is easier then multiply our previous polynome by $e^{-x\gamma}$.

$$p^x\prod_{n=1}^{p} (1-x/n)= \bigg(1+-x\gamma+x^2 \frac{\gamma^2-\zeta(2)}{2}+x^3 \frac{-\gamma^3+3\gamma\zeta(2)-2\zeta(3)}{3!}+...\big)=\frac{1}{(-x)!}$$

I find it hard to clearly explain these refined stirling numbers but if you are throwing in a bit of time, consider the following ways to represent the found product representation. It's going to be vague but i clarify it if you want.

The "nice" form is in knowing g(m) for all m. There are a lot of way to represent the refined stirling numbers, especially within this context. Lots of it is related to partitions and ways to "write" a number. e.g. for the x^4 term, $4=1+1+1+1$ so we get $g(1)^4*4!/4!/1^4$ and the next term is $1+1+2; G(1)^2 G(2) * 4!/1^2/2^1/2$ as coefficient. So we divide by the a*s^a if you have a G(s)^a. another more intunitive way is to see it as the "unique" combination of all outcomes. Another way is to write it as these unique combination, but use sum of sums (particular cool! And very easy to image things). You can also achieve it algebraric with writing the e powers out, but that's a real hassle. And if you want another way to represent these refined stirling numbers, you can construct them by using previous found stirling numbers and binominals, which is the most efficient.

I always wondered why there was no wikipedia page about them.

Answer 3

Your series is easily seen as divergent (if it tends to zero i call it divergent aswell), and it's actualy a nasty one since it's not simply sumable, see the harmonic series.

however also a divergent product can be formulate as a divergent series, but agian this is a tricky one and i've no quick answer.

A general answer to find the power series of a product is the following. It's also works if you wish to find a power series for the gamma function at integer points k since past your kth point it will only have zero's. (stirling numbers ) These is also a perfectly fine methode to extrapolate the gamma function to non integer points but you always will have an error.

$$\prod_{i=b}^c 1+a_i$$= $$1+\sum_{i=b}^{c} (a_i)+$$ $$1/2!((\sum_{i=b}^c (a_i))^2-\sum_{i=b}^c (a_i)^2$$ $$1/3!((\sum_{i=b}^c (a_i))^3-3\sum_{i=b}^c (a_i)^2\sum_{i=b}^c (a_i)+2\sum_{i=b}^c (a_i)^3)$$ $$1/4!((\sum_{i=b}^c (a_i))^4-6(\sum_{i=b}^c (a_i))^2\sum_{i=b}^c (a_i)^2+3(\sum_{i=b}^c (a_i)^2)^2+8(\sum_{i=b}^c (a_i)^3)\sum_{i=b}^c (a_i)-6\sum_{i=b}^c (a_i)^4)$$

These are the refined strirling numbers, but i guess you get the pattern, i dont master the skills yet to write this more efficient down.