Identity for an infinite product

Question

Here is an experimental "result" exhibiting the difference of two (formal) infinite products that "almost factorizes".

QUESTION. Is this true? $$\prod_{n\geq1}(1+x^{2n-1})^{24} - \prod_{n\geq1}(1-x^{2n-1})^{24} =48x+4096x^3\prod_{n\geq1}(1+x^{2n})^{24}.$$

Answer 1

The identity is true. To see this, we consider $$f(q):=q\prod_{k\geq 1}(1-q^k)^{24},\qquad |q|<1.$$ With this notation, $$\frac{qf(q)}{f(q^2)}=\prod_{\substack{k\geq 1\\\text{$k$ odd}}}(1-q^k)^{24}\qquad\text{and}\qquad\frac{f(q^4)}{q^2f(q^2)}=\prod_{\substack{k\geq 1\\\text{$k$ even}}}(1+q^k)^{24},$$ hence the identity in the original post is equivalent to $$\frac{-qf(-q)-qf(q)}{f(q^2)}=48q+4096q^3\frac{f(q^4)}{q^2f(q^2)}.$$ In other words, we need to prove that $$f(q)+f(-q)+48f(q^2)+4096f(q^4)=0.\tag{1}$$ Now we introduce the familiar notation $q=e^{2\pi iz}$, so that $$\Delta(z):=f(e^{2\pi iz})=f(q)\tag{2}$$ is the modular discriminant function, a holomorphic cusp form of weight $12$ and level $1$ on the upper half-plane. The point is that $\Delta(z)$ is an eigenfunction of all the Hecke operators $T_n$ (because the space of holomorphic cusp forms of weight $12$ and level $1$ is one-dimensional). The eigenvalues are the coefficients of $f(q)$ as a power series, also known as the values of the Ramanujan $\tau$ function. In particular, $$(T_2\Delta)(2z)=\tau(2)\Delta(2z)=-24\Delta(2z),$$ where we have evaluated $T_2\Delta$ and $\Delta$ at $2z$ (instead of $z$) for convenience. By the definition of $T_2$, the left-hand side equals $$2^{11}\sum_{ad=2}\frac{1}{d^{12}}\sum_{b\bmod d}\Delta\left(\frac{2az+b}{d}\right)=\frac{1}{2} \Delta\left(\frac{2z}{2}\right)+\frac{1}{2}\Delta\left(\frac{2z+1}{2}\right)+2^{11}\Delta(4z).$$ So this expression equals $-24\Delta(2z)$, whence $$\Delta(z)+\Delta(z+1/2)+48\Delta(2z)+2^{12}\Delta(4z)=0.$$ This is equivalent to $(1)$ in the light of $(2)$, hence we are done.

Answer 2

The question is to prove

$$ \prod_{n\geq1}(1+x^{2n-1})^{24} - \prod_{n\geq1}(1-x^{2n-1})^{24} \\ = 48x + 4096x^3\prod_{n\geq1}(1+x^{2n})^{24}. $$

A quick search through the OEIS shows that

$$ u(x) := \sum_{n\ge0} A010815(n)x^n = \prod_{n>0} (1-x^n),$$

$$ \frac{u(x^2)}{u(x)} = \sum_{n\ge0} A000009(n)x^n = \prod_{n>0} (1+x^n),$$

$$ \frac{u(x^2)^2}{u(x)u(x^4)}=\sum_{n\ge0} A000700(n)x^n=\prod_{n>0} (1+x^{2n-1}), $$

$$ \frac{u(x)}{u(x^2)} = \sum_{n\ge0} A081362(n)x^n = \prod_{n>0} (1-x^{2n-1}). $$

For brevity, use the notation $u_1 := u(x), u_2 := u(x^2), u_4 := u(x^4).$ The result to prove can now be written as

$$ S := 48x + 4096x^3\frac{u_4^{24}}{u_2^{24}} - \frac{u_2^{48}}{u_1^{24}u_4^{24}} + \frac{u_1^{24}}{u_2^{24}} = 0. $$

The sum on the left factorizes using Mathematica as

$$ S = \frac{F_1 F_2}{u_1^{24}u_2^{24}u_4^{24}} = \frac{(u_1^{16}u_4^8 + 16xu_1^8u_4^{16} -u_2^{24})F_2}{u_1^{24}u_2^{24}u_4^{24}}. $$

where $F_2$ has six terms of degree $48$. Note that $F_1 = 0$ is the very first identity $\texttt{t4_24_48}$ in my "Dedekind eta function product identities" collection. It has many proofs. Jacobi called it "aequatio identica satis abstrusa".

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Your identity is related to "Monstrous Moonshine". Specifically,

$$ x^{-1}\prod_{n\geq1}(1+x^{2n-1})^{24} $$

is the generating function of OEIS A097340 "McKay-Thompson series of class 4A for the Monster group with a(0) = 24."

$$ x^{-1}\prod_{n\geq1}(1-x^{2n-1})^{24} $$

is the generating function of OEIS A007191 "McKay-Thompson series of class 2B for the Monster group with a(0) = -24."

Their difference is $\,48 + 4096A(x^2)\,$ where $\,A(x)\,$ is the generating function of OEIS A014103. This generating function also satisfies the functional equations

\begin{align} A(x)^2 - A(x^2) &= 48A(x)A(x^2) + 4096A(x)A(x^2)^2, \\ A(x) + A(-x) &= -48A(x)A(-x) + 4096A(x)^2A(-x)^2 \end{align}

both of which are multiples of your identity.

Answer 3

This identity is exactly level $\ell=1$ case of the identity for 10 special levels $\ell=1,2,3,5,6,7,11,14,15,23$ (where $\sigma(\ell):=(\sum_{d|\ell}d)| 24)$, valid for the level $\ell$ cusp form

$\Delta_\ell(z):=\prod_{d\mid\ell} \eta(dz)^{24/\sigma(\ell)},$

and the proof follows from equations (3.4) and (3.6) in the paper

Jacobi identities, modular lattices, and modular towers by Chua and Solé (https://core.ac.uk/download/pdf/82374204.pdf)

using the fact that they are Hecke eigenform and looking at the eigen equation for $T(2)$ or $U(2)$ (when $\ell$ is even).

The relations is equivalent to some cubic and quartic units given by special values of the ratio of the cusp forms. For example for level $\ell=23$, it gives an infinite product for the plastic ratio $\rho=\sqrt[3]{108+12\sqrt{69}},$ the positive root of $x^3-x-1=0$,

$$\rho=q\prod_{n=1}^\infty(1+q^n)(1+q^{23n}),\quad q=\exp(-\pi/\sqrt{23}).$$

Answer 4

This is essentially Ramanujan's class invariants in disguise. We have $$\prod_{n\geq 1}(1+q^{2n-1})^{24}=64qG^{24}(q),\qquad\prod_{n\geq 1}(1-q^{2n-1})^{24}=64qg^{24}(q),\tag{1}$$ and in terms of elliptic moduli $k, k'$ we have $$G^{24}(q)=1/(4k^2k'^2),\qquad g^{24}(q)=k'^4/4k^2,\tag{2}$$ and hence the LHS of identity in question is $$64q\cdot\frac{1-k'^6}{4k^2k'^2}=16q\cdot\frac{1+k'^2+k'^4}{k'^2}.\tag{3}$$ Next, let us consider the Dedekind eta function $\eta(q) $ defined by $$\eta(q) =q^{1/24}\prod_{n\geq 1}(1-q^n)\tag{4}$$ and its representation in terms of elliptic integrals and moduli as $$\eta(q) =2^{-1/6}\sqrt{\frac{2K} {\pi}} k^{1/12}k'^{1/3},\qquad\eta(q^2)=2^{-1/3}\sqrt{\frac{2K} {\pi}}(kk')^{1/6}.\tag{5}$$ We have $$\prod_{n\geq 1}(1+q^{2n})=\prod_{n\geq 1}\frac{1-q^{2n}}{(1-q^n)(1+q^{2n-1})}=\frac{q^{-1/12}\eta(q^2)}{q^{-1/24}\eta(q)\cdot 2^{1/4}q^{1/24}G(q)},$$ so that $$q^2\prod_{n\geq 1}(1+q^{2n})^{24}=\frac{\eta^{24}(q^2)}{64\eta^{24}(q)G^{24}(q)}=\frac{k^4}{256k'^2}.\tag{6}$$ Note that expression in $(3)$ can be written as $$16q\left (3+\frac{k^4}{k'^2}\right)=16q(3+256q^2\prod_{n\geq 1}(1+q^{2n})^{24})$$ via $(6)$ so that identity in question holds.