modular function produced via infinite product of weight-k modular forms

Suppose I have a modular form $f_{k}(z)$ of weight $k$:

$f_{k}\left(\frac{az+b}{cz+d}\right) = (cz+d)^{k} f_{k}(z)$

Then consider the following infinite product:

$u(z) = \prod_{n=1}^{\infty} \frac{f_{4n-3}(z)f_{4n}(z)}{f_{4n-2}(z)f_{4n-1}(z)}$

Ignoring convergence problems for the moment, formally $u(z)$ must be a modular function, since:

$\prod_{n=1}^{\infty} \frac{f_{4n-3}(z)f_{4n}(z)}{f_{4n-2}(z)f_{4n-1}(z)} = \prod_{n=1}^{\infty} \frac{(cz+d)^{4n-3}f_{4n-3}(z)(cz+d)^{4n}f_{4n}(z)}{(cz+d)^{4n-2}f_{4n-2}(z)(cz+d)^{4n-1}f_{4n-1}(z)} = \prod_{n=1}^{\infty} \frac{f_{4n-3}\left(\frac{az+b}{cz+d}\right)f_{4n}\left(\frac{az+b}{cz+d}\right)}{f_{4n-2}\left(\frac{az+b}{cz+d}\right)f_{4n-1}\left(\frac{az+b}{cz+d}\right)}$

And
$\frac{(cz+d)^{4n-3}(cz+d)^{4n}}{(cz+d)^{4n-2}(cz+d)^{4n-1}} = (cz+d)^{4n-3+4n-4n+2-4n+1)} = (cz+d)^{0} =1$

infinite products of rational functions of Klein’s j-invariant

As a Riemann surface, the fundamental region has genus 0, and every (level one) modular function is a rational function in j; and, conversely, every rational function in j is a modular function

Well then, doesn’t that mean that:

$\Lambda(\tau) = \prod_{n=1}^{\infty} \frac{j(\tau)^{2n}-3j(\tau) +1}{j(\tau)^{3n}-i j(\tau)^{n} +1}$ is also a modular function, each of its factors being a rational function of $j(\tau)$. I’ll attempt to make a picture of $\Lambda(\tau)$ later, my computers are busy doing other things

Series representation of Bell numbers in terms of Bessel functions

The generating function of the Bell numbers $B_{n}$ is
$B(x) = \sum_{n\,=\,0}^{\infty} \frac{B_{n}x^{n}}{n!} = e^{e^{x}-1}$

Set $x=it$:
$B(it) = \sum_{n\,=\,0}^{\infty} \frac{B_{n}(it)^{n}}{n!} = e^{e^{(it)}-1}$

Factor out $e^{-1}$ and split exponential:

$e^{e^{(it)}-1} = e^{-1}e^{e^{it}} = e^{-1}e^{\cos t}e^{i \sin t}$

Replace with series from Jacobi-Anger expansions for each:

$e^{-1}e^{\cos t}e^{i \sin t} = e^{-1} \left[\sum_{m\in\mathbb{Z}} i^{m}J_{m}(-i)e^{imt} \right] \left[\sum_{n\in\mathbb{Z}} J_{n}(1)e^{int} \right]$

Collect and reorganize:

$e^{-1} \left[\sum_{m\in\mathbb{Z}} i^{m}J_{m}(-i)e^{imt} \right] \left[\sum_{n\in\mathbb{Z}} J_{n}(1)e^{int} \right] = e^{-1} \sum_{m,n\in\mathbb{Z}} i^{m}J_{m}(-i)e^{imt}J_{n}(1)e^{int}$

Let’s multiply those exponentials:

$e^{-1} \sum_{m,n\in\mathbb{Z}} i^{m}J_{m}(-i)e^{imt}J_{n}(1)e^{int} = e^{-1} \sum_{m,n\in\mathbb{Z}} i^{m}J_{m}(-i)J_{n}(1)e^{it(m+n)}$

Let’s expand that exponential as a power series:

$e^{-1} \sum_{m,n\in\mathbb{Z}} i^{m}J_{m}(-i)J_{n}(1)e^{it(m+n)} = e^{-1} \sum_{m,n\in\mathbb{Z}} i^{m}J_{m}(-i)J_{n}(1) \sum_{v\,=\,0}^{\infty} \frac{(it(m+n))^{v}}{v!}$

Shuffling things around a bit:
$e^{-1} \sum_{m,n\in\mathbb{Z}} i^{m}J_{m}(-i)J_{n}(1) \sum_{v\,=\,0}^{\infty} \frac{(it(m+n))^{v}}{v!} = e^{-1}\sum_{v\,=\,0}^{\infty} \frac{(it)^{v}}{v!} \sum_{m,n\in\mathbb{Z}} i^{m}J_{m}(-i)J_{n}(1) (m+n)^{v}$

Therefore:
$B_{a} = \frac{1}{e} \sum_{m,n\in\mathbb{Z}} i^{m}J_{m}(-i)J_{n}(1) (m+n)^{a}$

oft-worn U(1) modification paths of series and products

So I’ve been playing around with products of the gamma function like so:

$\prod_{n=1}^{\infty}\frac{\Gamma(\frac{4n-3}{4n-2})}{\Gamma(\frac{4n-1}{4n})}=\frac{\Gamma(1/2)}{\Gamma(2/3)}\frac{\Gamma(5/6)}{\Gamma(7/8)}\frac{\Gamma(9/10)}{\Gamma(11/12)}\frac{\Gamma(13/14)}\ldots$

When I notice that swapping half of the terms with their reciprocals is akin to taking the general term to the power $(-1)^{n}$

$\prod_{n=1}^{\infty}\left[\frac{\Gamma(\frac{4n-3}{4n-2})}{\Gamma(\frac{4n-1}{4n})}\right]^{(-1)^{n+1}}=\frac{\Gamma(1/2)}{\Gamma(5/6)}\frac{\Gamma(2/3)}{\Gamma(7/8)}\frac{\Gamma(9/10)}{\Gamma(15/16)}\frac{\Gamma(11/12)}\ldots$

And then it occurs to me: well, what would happen if I just changed that $(-1)^{n}$ term with some power of a root of unity: $e^{2\pi i n/v$, and one little bit of lore about infinite series that I suppose I’ve been curious about for a long time and haven’t actually seen written up anywhere comes to fore:

$A = \sum_{n=0}^{\infty} a_{n}$
well, we can multiply each term by $(-1)^{n}$ yielding:
$B = \sum_{n=0}^{\infty} (-1)^{n}a_{n}$
and then by a root of unity:
$C = \sum_{n=0}^{\infty} e^{2\pi i n/v}a_{n}$

Analogously, if we have a product:
$D = \prod_{n=0}^{\infty} \frac{a_{n}}{b_{n}}$
we can do the above trick as well, giving us:
$E = \prod_{n=0}^{\infty} \left[\frac{a_{n}}{b_{n}}\right]^{(-1)^{n}}$
$F = \prod_{n=0}^{\infty} \left[\frac{a_{n}}{b_{n}}\right]^{e^{2\pi i n/v}}$

The idea here is that there are predictable things that occur when we make the above modifications to series and products, and it would be nice to know (and I’m sure I’ll look into it more closely later, this post is just so that I get my ideas written down) what happens as a result.