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 A Holomorphic Function Asymptotic to Tetration JmsNxn Long Time Fellow Posts: 571 Threads: 95 Joined: Dec 2010 03/24/2021, 09:58 PM (This post was last modified: 03/25/2021, 02:54 AM by JmsNxn.) Start by taking the function $\beta_\lambda$, and define a sequence of functions $\tau_\lambda^n(s)$ such that, $ \beta_\lambda(s) + \tau_\lambda^{n+1}(s) = \log(\beta_\lambda(s+1) + \tau_\lambda^n(s+1))\\$ We aren't too concerned about where $\tau_n$ is holomorphic at the moment. We know that, $ \tau_\lambda^0(s) = 0\\ \tau_\lambda^1(s) = \sum_{k=1}^\infty \frac{(-1)^k}{k} e^{-k\lambda s}\\$ Assume that, $ \tau_\lambda^n(s) = \sum_{k=1}^\infty c_{nk}(\lambda) e^{-k\lambda s}\\$ And we are reduced to the equation, $ \tau_\lambda^{n+1}(s) = \sum_{k=1}^\infty c_{(n+1)k}(\lambda) e^{-k\lambda s} = \log(\beta_\lambda(s+1) + \sum_{k=1}^\infty c_{nk}(\lambda)e^{-k\lambda} e^{-k\lambda s}) - \beta_\lambda(s)\\$ Now we can calculate $c_{(n+1)k}(\lambda)$ inductively on $k$. Assuming we know $c_{(n+1)j}(\lambda)$ for $1 \le j \le k$; then we can find $c_{(n+1)(k+1)}(\lambda)$ by, $ \lim_{\Re(s) \to \infty} \frac{\tau_\lambda^{n+1}(s) - \sum_{j=1}^k c_{(n+1)j}(\lambda) e^{-j\lambda s}}{e^{(k+1)\lambda s}} = c_{(n+1)(k+1)}(\lambda)\\$ Of which this limit necessarily exists; and constructs a sequence $c_{(n+1)k}(\lambda)$. And now our problem is reduced to solving $\lim_{n\to\infty} c_{nk}(\lambda)$. We are also given the quick relation, $ \tau^{n+1}_\lambda(s) = \log(\beta_\lambda(s+1) + \tau^n_\lambda(s+1)) - \beta_\lambda(s)\\ = -\log(1+e^{-\lambda s}) + \log(1 + e^{-\beta_\lambda(s)}(e^{-\lambda s} + 1) \tau^n(s+1))\\ =\sum_{k=1}^\infty \frac{(-1)^k}{k}e^{-k\lambda s} + \log(1+e^{-\beta_\lambda(s)}(e^{-\lambda s} +1)\tau^{(n)}(s+1))\\$ Then since we know that $\tau^{n+1}_\lambda(s) \to 0$ we know that, $ e^{-\beta_\lambda(s)}(e^{-\lambda s} +1)\tau_\lambda^{n}(s+1) \to 0\\$ Because, $e^{-\beta_\lambda(s)}(e^{-\lambda s} +1) = \frac{1}{\beta(s+1)}$, this implies that, $ \frac{\tau^n_\lambda (s)}{\beta_\lambda(s)} \to 0\\$ This gives us hope we can devise a normality condition on $\tau^n_\lambda(s)$ as it controls how fast $\beta(s)$ can dip to zero; which is the exact problem we had with $\phi$. This can be more aptly written, $ \frac{1}{\beta(s)} = o(e^{\lambda s})\\$ But we can say something stronger. We know that $\tau^n_\lambda(s) = \mathcal{O}(e^{-\lambda s})$ and this means that, $ \frac{\tau^n_\lambda (s)}{\beta_\lambda(s)} = \mathcal{O}(e^{-\lambda s})\\$ As this is the term in the logarithm, and in order for $\tau^{n+1}_\lambda(s)$ to tend to zero exponentially the term in the logarithm has to as well. Therefore we can say that, $ \frac{1}{\beta(s)} = \mathcal{O}(h(s))\\$ Where for any $\delta > 0$ we have $e^{-\delta s} h(s) \to 0$ as $\Re(s) \to \infty$. Which tells us that, $ |\beta(s)| \ge \frac{A}{h(s)}$ Which informs us asymptotically how $\beta(s)$ approaches zero at infinity... slower than any exponential. This should help us with the limit; in the end, and making sure we stay away from zero when defining our function $F_n(s) = \log^{\circ n} \beta(s+n)$. As to the second part, when we actually go about and take this limit we want $\lambda$ to be a function of $s$. And in this sense, we want $\beta_{\lambda(s)}(s) : \{s \in \mathbb{C} \,:\, |\arg(s)| < \theta\} \to \mathbb{C}$ where  $(s,\lambda(s)) \in \mathbb{L}$ and $\lambda : \mathbb{R}^+ \to \mathbb{R}^+$. This would mean for every imaginary part of $s$ there is a large enough $N$ in which $F_n(s)$ exists for $n > N$. This is where we'll need a clever Riemann mapping on $\lambda$ over some simply connected domain $\mathbb{D} \subset \mathbb{L}$; which I'm not sure how to do; especially as this requires a two variable ideation. Forget I said anything about periodic functions, I don't think that's necessary here. I'm going to keep plugging and playing with this, but the more I fiddle the more this has none of the problems $\phi$ has. Regards, James. EDIT: Some even more evidence this is doable. Let $\lambda(s)$ be an implicitly defined function such that, $ \log \beta_{\lambda(s+1)}(s+1) - \log \beta_{\lambda(s)}(s+1) = \log(1+ e^{-\lambda(s) s})\\ \log \beta_{\lambda(s+1)}(s+1) - \beta_{\lambda(s)}(s) + \log(1+e^{-\lambda(s)s}) = \log(1+ e^{-\lambda(s) s})\\ \log \beta_{\lambda(s+1)}(s+1) = \beta_{\lambda(s)}(s)\\$ Which would require an implicit solution $\lambda(s)$; hinting that the logarithm trick might work for correctly chosen $\lambda$; as it's idempotent in this case. « Next Oldest | Next Newest »

 Messages In This Thread A Holomorphic Function Asymptotic to Tetration - by JmsNxn - 03/23/2021, 12:21 AM RE: A Holomorphic Function Asymptotic to Tetration - by JmsNxn - 03/23/2021, 08:11 AM RE: A Holomorphic Function Asymptotic to Tetration - by JmsNxn - 03/24/2021, 09:58 PM

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