# 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:
$B = \sum_{n=0}^{\infty} (-1)^{n}a_{n}$
$C = \sum_{n=0}^{\infty} e^{2\pi i n/v}a_{n}$
$D = \prod_{n=0}^{\infty} \frac{a_{n}}{b_{n}}$
$E = \prod_{n=0}^{\infty} \left[\frac{a_{n}}{b_{n}}\right]^{(-1)^{n}}$
$F = \prod_{n=0}^{\infty} \left[\frac{a_{n}}{b_{n}}\right]^{e^{2\pi i n/v}}$