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In the paper Transformations of infinite series, Bryden Cais gives the following transformations of infinite products

Theorem 4. If $$ f(t) = \frac{\cosh(\pi t)-1}{\sinh(\pi t)}\frac{\cosh(2\pi t)+1}{\sinh(2\pi t)}\frac{\cosh(3\pi t)-1}{\sinh(3\pi t)}\cdots, $$ then $$ f(t) = t^{\frac12}f(\tfrac1t). \tag{8} $$ Proposition 26. Let $$ f(\alpha) = \prod_{n=1}^\infty \left(1 - \frac{2\sqrt{5}}{1+\sqrt{5}+4\cosh(n\alpha)}\right)^{\left(\frac n5\right)}. $$ If $\alpha\beta=\frac{4\pi^2}{25}$, then $$ f(\alpha)=f(\beta). $$

With some modification of Cais's method using contour integration one can obtain the following generalizations of these infinite product transformations $$ \prod_{n=1}^\infty\left(\frac{1-e^{-\pi\alpha\sqrt{n^2+\beta^2}}}{1+e^{-\pi\alpha\sqrt{n^2+\beta^2}}}\right)^{(-1)^n}=\sqrt{\frac{\tanh\frac{\pi\beta}{2}}{\tanh\frac{\pi\alpha\beta}{2}}}\prod_{n=1}^\infty\left(\frac{1-e^{-\pi\sqrt{n^2/\alpha^2+\beta^2}}}{1+e^{-\pi\sqrt{n^2/\alpha^2+\beta^2}}}\right)^{(-1)^n},\tag{1} $$ $$ \prod_{n=1}^\infty\left(1-\tfrac{2\sqrt{5}}{1+\sqrt{5}+4\cosh{\frac{2\pi\alpha\sqrt{n^2+\beta^2}}{5}}}\right)^{\left(\frac{n}{5}\right)}=\prod_{n=1}^\infty\left(1-\tfrac{2\sqrt{5}}{1+\sqrt{5}+4\cosh{\frac{2\pi\sqrt{n^2/\alpha^2+\beta^2}}{5}}}\right)^{\left(\frac{n}{5}\right)}.\tag{2} $$ It is clear that $(1)$ and $(2)$ reduce to theorem 4 and proposition 26 respectively, when $\beta=0$.

Note that infinite products in theorem 4 and proposition 26 are modular forms, however the infinite products in $(1)$ and $(2)$ are not.

Q1: What are the most general modular forms that admit generalized transformation formulas like $(1)$ and $(2)$?

In chapter 5 of his paper Cais gives a general methodology to construct modular forms which then can be generalized as above, thus giving an infinite family of formulas like $(1)$ and $(2)$. Then one can take linear combination of arbitrary number of these functions. Will this be the most general function of this kind or there are others?

This question is related to the previous question. The formula $$ \prod_{n=0}^\infty\frac{1+e^{-\pi\alpha\sqrt{(2n+1)^2+\beta^2}}}{1+e^{-\pi\sqrt{(2n+1)^2/\alpha^2+\beta^2}}}=\exp\left\{\frac{1}{2}\int_0^\infty\ln\frac{1+e^{-\pi\alpha\sqrt{x^2+\beta^2}}}{1+e^{-\pi\sqrt{x^2/\alpha^2+\beta^2}}}\ dx\right\}.\tag{1} $$ is a limiting case ($m,n\to\infty$, with $m/n$ fixed) of the following proposition

If $\cos\frac{\pi (j-\frac{1}{2})}{n}+\cosh\alpha_j= \cos\frac{\pi (k-\frac{1}{2})}{m}+\cosh\beta_k=x$ for all integers $1\le j\le n,\ 1\le k\le m$ then

$$ \prod_{j=1}^n2\cosh m\alpha_j=\prod_{k=1}^m2\cosh n\beta_k.\tag{1a} $$

The formulas defining $\alpha_j$ and $\beta_k$ arise during solution of Helmholtz equation on a finite rectangular lattice with suitable boundary conditions (see e.g. Phillips, B.; Wiener, N. (1923). Nets and the Dirichlet problem. Journal of Math. and Physics, Massachusetts Institute, 105–124).

Q2: Is there a finite analog of $(1)$ similar to $(1a)$?

Any references regarding these infinite products are welcomed. If the question is not clear please ask in the comments and I will clarify it.

Note. Q2 has been answered, however Q1 is still open.

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EDIT (Feb 2024): The question has been generally answered in the article:

  • Martin Nicholson, Finite and infinite product transformations, arXiv:1712.06097.

Question $2$ has a surprisingly simple answer:

If $$\cosh\alpha_j+\cos\frac{\pi j}{2n}=\cosh\beta_k+\cos\frac{\pi k}{2m}=x\tag{3}$$ for all $1\le j\le 2n-1,\ 1\le k\le 2m-1$, then $$ \prod_{j=1}^{2n-1}\left(\frac{\tanh m\alpha_j}{\sinh\alpha_j}\right)^{(-1)^j}=\prod_{k=1}^{2m-1}\left(\frac{\tanh n\beta_k}{\sinh\beta_k}\right)^{(-1)^k}. \tag{4} $$ Proof. We use two well known formulas $$ 2^{m-1} \prod _{j=1}^{m-1} \left(\cosh \alpha-\cos \frac{\pi j}{m}\right)=\frac{\sinh m \alpha}{\sinh \alpha}, $$ $$ 2^{m-1} \prod _{j=1}^m \left(\cosh \alpha-\cos \frac{\pi (j-1/2)}{m}\right)=\cosh m \alpha, $$ to write the products in $(4)$ in symmetric form: \begin{align} &\prod_{j=1}^{2n-1}\left(\frac{\tanh m\alpha_j}{\sinh\alpha_j}\right)^{(-1)^j}=\frac{\sinh\alpha_{2n-1}}{\tanh m\alpha_{2n-1}}\prod_{j=1}^{n-1}\frac{\tanh m\alpha_{2j}}{\sinh\alpha_{2j}}\frac{\sinh\alpha_{2j-1}}{\tanh m\alpha_{2j-1}}\\ &=\frac{\sinh\alpha_{2n-1}}{\tanh m\alpha_{2n-1}}\prod_{j=1}^{n-1}\frac{\cosh\alpha_{2j-1}-\cos\frac{\pi (m-1/2)}{m}}{\cosh\alpha_{2j}-\cos\frac{\pi(m-1/2)}{m}}\\ &\times \prod_{j=1}^{n-1}\prod_{k=1}^{m-1}\frac{\cosh\alpha_{2j-1}-\cos\frac{\pi (k-1/2)}{m}}{\cosh\alpha_{2j}-\cos\frac{\pi(k-1/2)}{m}}\frac{\cosh\alpha_{2j}-\cos\frac{\pi k}{m}}{\cosh\alpha_{2j-1}-\cos\frac{\pi k}{m}}\\ &=\frac{1}{\cosh\alpha_{2n-1}-\cos\frac{\pi (m-1/2)}{m}}\frac{\sinh\alpha_{2n-1}}{\tanh m\alpha_{2n-1}}\frac{\sinh\beta_{2m-1}}{\tanh n\beta_{2m-1}}\\ &\times \prod_{j=1}^{n-1}\prod_{k=1}^{m-1}\frac{x-\cos\frac{\pi (j-1/2)}{n}-\cos\frac{\pi (k-1/2)}{m}}{x-\cos\frac{\pi j}{n}-\cos\frac{\pi(k-1/2)}{m}}\frac{x-\cos\frac{\pi j}{n}-\cos\frac{\pi k}{m}}{x-\cos\frac{\pi (j-1/2)}{n}-\cos\frac{\pi k}{m}}. \end{align} Now, in this last formula the factor $$ \frac{1}{\cosh\alpha_{2n-1}-\cos\frac{\pi (m-1/2)}{m}}=\frac{1}{x+\cos\frac{\pi}{2n}+\cos\frac{\pi }{2m}} $$ is obviously symmetric when $m$ and $n$ are interchanged, and so is the factor $\frac{\sinh\alpha_{2n-1}}{\tanh m\alpha_{2n-1}}\frac{\sinh\beta_{2m-1}}{\tanh n\beta_{2m-1}}$ and the double product. So both sides in $(4)$ are symmetric when $m$ and $n$ are interchanged, and hence they are equal. $~\Box$

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  • $\begingroup$ @SamHopkins , who cares? $\endgroup$ Feb 26 at 20:32
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    $\begingroup$ @CaveJohnson please don't self-delete this answer, even if it has been answered elsewhere. Someone might stumble on this exact posting wanting to know the answer, and having the answer here is more helpful than if it is not here. $\endgroup$
    – David Roberts
    Feb 26 at 23:35
  • $\begingroup$ even better than removing a self-contained answer and citing a 12-page paper, is leaving your answer, and also citing the paper, meaning people can read the short solution here, and consult the longer paper if they want. $\endgroup$
    – David Roberts
    Feb 27 at 6:49
  • $\begingroup$ Please don't delete the content of this answer again. $\endgroup$
    – David Roberts
    Feb 27 at 6:51
  • $\begingroup$ Not a threat, a polite request to keep the useful information here on the page. Blanking one's own posts comes under the technical name "self-vandalism" (a term of art in the SE software), and is frowned on network-wide The first edit done here would count as that, to my mind, because it removed all useful content. The second edit with the arXiv link was marginally better, but is counter to normal MO practice to update answers with new info, not replace them entirely, with a more opaque reference. $\endgroup$
    – David Roberts
    Feb 27 at 8:54

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