Trying to find a fast converging series of normalization constants; plus a recap JmsNxn Ultimate Fellow     Posts: 1,214 Threads: 126 Joined: Dec 2010 10/26/2021, 02:12 AM (This post was last modified: 10/26/2021, 07:03 AM by JmsNxn.) So, I've been asking myself, how do we normalize the beta method for each iteration. If we start with the naive idea that, \begin{align} \widetilde{F}_{\lambda,b}^1(z) &= \beta_{\lambda,b}(z) - \log_b(1+\exp(-\lambda z))\\ \widetilde{F}_{\lambda,b}^n(z) &= \log^{\circ n}_b \beta_{\lambda,b}(z+n) = \beta_{\lambda,b} (z) + \tau_{\lambda,b}(z)\\ F^n_{\lambda,b}(z) &= \widetilde{F}_{\lambda,b}^n(z+k_n)\\ F^n_{\lambda,b}(0) &= \widetilde{F}_{\lambda,b}^n(k_n) = 1\\ \end{align} Where, \begin{align} \beta_{\lambda,b}(z) &= \Omega_{j=1}^\infty \dfrac{e^{bt}}{e^{\lambda(j- z)} + 1} \bullet t\\ \beta_{\lambda,b}(z+1) &= \dfrac{e^{b\beta_{\lambda,b}(z)}}{e^{-\lambda z}+1}\\ \end{align} We start to approach a more effective programming. First of all, I want to work through $\beta_{\lambda,b}$. This function is holomorphic everywhere that: $\sum_{j=1}^\infty \dfrac{e^{bt}}{1+e^{\lambda(j-z)}}$ Converges compactly normally for $(t,z,b,\lambda)$--as we throw $t$ away, this just needs to work for $t\approx 0$. All in all, this beast converges where ever: $\sum_{j=1}^\infty \left| \left|\dfrac{e^{bt}}{1+e^{\lambda(j-z)}}\right| \right|_{\mathcal{K},\mathcal{S},\mathcal{B}, \mathcal{L}} < \infty\\$ Where $t \in \mathcal{K}$ is a compact set of $\mathbb{C}$. Where $z \in \mathcal{S}$ is a compact set of $\mathbb{C}$. And where $b \in \mathcal{B}$ is a compact set of $\mathbb{C}$. Finally $\lambda \in \mathcal{L}$, a compact set of $\mathbb{C}$. Checking where the sum has this summative property is elementary. And we get that, \begin{align} \beta_{\lambda,b}(z)\,\,&\text{is holomorphic on}\,\,\mathbb{S}\\ \mathbb{S} &= \{(z,b,\lambda) \in \mathbb{C}^3\,|\,\Re \lambda>0,\, \lambda(j-z)\neq (2k+1)\pi i,\,\forall j,k \in \mathbb{Z},\,j\ge 1,\,b \neq 0\}\\ \end{align} Because this is exactly where the sum converges compactly normally. So, for all compact $\mathcal{A} \subset \mathbb{S}$ we necessarily get that, \[ \sum_{j=1}^\infty \left| \left|\dfrac{e^{bt}}{1+e^{\lambda(j-z)}}\right| \right|_{\mathcal{A}, |t|

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