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\begin{document}

\title{Multiplicative Iteration On Infinite Products}



\author[1]{Benedict Irwin}%
\affil[1]{University of Cambridge}%


\vspace{-1em}



  
  \date{May 17, 2024}


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We have the product
\begin{equation}
\prod_{n=2}^\infty \frac{n^3-1}{n^3+1} = \frac{2}{3}
\end{equation}
if we take the prodegrand and apply a multiplicative derivative defined by 
\begin{equation}
f^*(x) = \exp\left(\frac{f'(x)}{f(x)}\right)
\end{equation}
then we get
\begin{equation}
\prod_{n=2}^\infty \exp\left(\frac{6n^2}{n^6-1}\right) \sim 1.64872
\end{equation}
This converges to \begin{equation}
\lim_{m \to \infty} \exp\left(\frac{m^4 + 2 m^3 + 2 m^2 - 3 m - 2}{2 m (m + 1) (m^2 + m + 1)}\right) = \sqrt{e}
\end{equation}
likewise for 
\begin{equation}
\prod_{n=2}^\infty \frac{n^3-1}{n^3+1} = \pi \mathrm{csch} \pi
\end{equation}
we get
\begin{equation}
\prod_{n=2}^\infty \exp\left(\frac{4n}{n^4-1}\right) = \exp(-3/2 + 2 \gamma + \psi_0(2-i) + \psi_0(2+i))
\end{equation}

we can take the product
\begin{equation}
\prod_{k=1}^\infty \frac{(1+k^{-1})^2}{(1+2k^{-1})} =2
\end{equation}
and find
\begin{equation}
\prod_{k=1}^\infty \exp\left(\frac{-2}{k(k+1)(k+2)}\right) = \frac{1}{\sqrt{e}}
\end{equation}



in the same line of thought we have \begin{equation}
\prod_{n=2}^\infty \exp\left(\frac{8n^3}{n^8-1}\right) = A
\end{equation}
\begin{equation}
A = \exp\left(\sum_{\pm} -\frac{3}{2} + 2 \gamma - \psi_0(2 \pm i) + \psi_0(2 \pm (-1)^{1/4}) + \psi_0(2 \pm (-1)^{3/4}) \right)
\end{equation}
it seems that
\begin{equation}
Q_k=\prod_{n=2}^\infty \exp\left(\frac{(2k)n^{k-1}}{n^{2k}-1}\right) =\exp\left(\frac{3}{2} + \sum_{\pm}\sum_{l=1}^{k-1} \psi_0(2\pm(-1)^{l/k})  \right)
\end{equation}

\begin{tabular}{|l|l|l|l|}
\hline
\multicolumn{4}{|c|}{List of Products} \\
\hline
$\prod_{n=2}^\infty \frac{n^3-1}{n^3+1}$ & $\frac{2}{3}$ & $\prod_{n=2}^\infty \exp\left(\frac{6n^2}{n^6-1}\right)$ & $\sqrt{e}$ \\
$\prod_{n=1}^\infty \left(1+\frac{1}{n}\right)^n e^{1/(2n)-1}$ & $\frac{e^{1+\gamma/2}}{\sqrt{2\pi}}$ & $\prod_{n=1}^\infty $ & $e^{1-\gamma - \pi^2/12}$ \\
\hline
\end{tabular}

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