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 Another asymptotic development, similar to 2sinh method JmsNxn Ultimate Fellow Posts: 889 Threads: 110 Joined: Dec 2010 07/05/2011, 06:34 PM (This post was last modified: 07/05/2011, 07:37 PM by JmsNxn.) http://math.eretrandre.org/tetrationforu...hp?tid=635 Thanks to mike's post here (near the bottom), we are given an asymptotic development for $e^x \cdot \ln(x)$, namely: $e^x \cdot \ln(x) \sim \sum_{n=0}^{\infty} x^n \frac{\psi_0(n+1)}{n!}$ which after solving for ln(x) gives: $\ln(x) \sim \sum_{n=0}^{\infty} x^n (\sum_{k=0}^{n} (-1)^k \cdot \frac{\psi_0(n-k)}{k!(n-k)!})$ I'm unsure about the convergence radius of this series, I've been unable to prove anything about it, however I'll continue with the conjecture leaving it open whether the series converges infinitely or not. It's a product of two infinite convergent series, so I hope it converges. Given this development, we can write: $\lambda(x) = \sum_{n=0}^{\infty} x^n (\sum_{k=0}^{n} (-1)^k \cdot \frac{\psi_0(n-k)}{k!(n-k)!})$ which I'm wondering if it's possible to perform regular iteration on lambda. If it is, we are given a formula similar to Tommy's 2sinh method. $\exp^{\circ z}(x) = \lim_{n\to\infty} \ln^{\circ n}(\lambda^{\circ -z}(\exp^{\circ n}(x)))$ I'm just posting this because it popped in my head and I thought it would be interesting. Could this method actually work (for real x, that is)? I'm unsure about performing regular iteration on lambda. I couldn't find a fixpoint :\, so maybe it's not possible. For bases other than e we are given: $\lambda_b(x) = \frac{\lambda(x)}{\ln(b)} = \sum_{n=0}^{\infty} \frac{x^n}{\ln(b)} (\sum_{k=0}^{n} (-1)^k \cdot \frac{\psi_0(n-k)}{k!(n-k)!})$ $\exp_b^{\circ z}(x) = \lim_{n\to\infty} \log_b^{\circ n}(\lambda_b^{\circ -z}(\exp_b^{\circ n}(x)))$ I think actually for other bases we're more likely to find a fixpoint, so regular iteration might be easier. For base eta the fixpoint occurs at around 1.9 « Next Oldest | Next Newest »

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