Trying to find a fast converging series of normalization constants; plus a recap JmsNxn Ultimate Fellow     Posts: 1,064 Threads: 121 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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