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 How it looks (i.θ)ₐ JmsNxn Long Time Fellow Posts: 291 Threads: 67 Joined: Dec 2010 04/23/2015, 11:15 PM (This post was last modified: 04/23/2015, 11:19 PM by JmsNxn.) (04/23/2015, 04:52 PM)marraco Wrote: Now, if the bases$\vspace{15}{1really turn into ellipses, then it should be easy to find an algebraic expression for tetration to real exponents, or at least an important insight for $\vspace{20}{^{\frac{1}{x}}a}$ The base $\vspace{15}{a=e^{e^{-1}} }$ seems to match an ellipse with center near c=2.65599203615835 (Don't take that precision as accurate. I got it from Excel), relation of axis b=a, and radius $\vspace{20}{r=\left(e^{\pi}+W_{(1)}\right)^{\,e^{-\pi}}}$, or $\left(imag(^{i.x}a)\right)^2 \,+\, \left({\frac{real(^{i.x}a)-c}{b}}\right)^2\,=\, r^2\\ \\ a=e^{e^{-1}} \\ b= e^{e^{-1}}\\ c= 2.655992036 \\ r=\left(e^{\pi}+W_{(1)}\right)^{\,e^{-\pi}}\\ \\ W_{(x)}.e^{W_{(x)}}=x \,\Rightarrow \, W_{(1)}=0,56714329 $ I've been trying to follow this thread, and finally I have something to contribute. For the bases $1 < a \le e^{1/e}$ it might be easier to use my expansion of this function. See http://arxiv.org/pdf/1503.07555v1.pdf I came up with a holomorphic expression for $^^z a$ for these bases. It's a fast converging expression as well. It is not as messy as a Taylor series expansion of this same function. It's also a single holomorphic expression for all $\Re(z) > 0$, greatly reducing computational time. This is for the periodic/pseudoperiodic extension of tetration (regular koenigs iteration) for bases $1 < a \le e^{1/e}$ The expression isn't so easy to write out: $^^z a= \frac{1}{\G(1-z)} (\sum_{n=0}^\infty (^^{n+1} a)\frac{(-1)^n}{n!(n+1-z)} + \int_1^\infty (\sum_{n=0}^\infty (^^{n+1}a) \frac{(-w)^n}{n!}) w^{-z}\,dw)$ It is a periodic solution, except at eta. « Next Oldest | Next Newest »

 Messages In This Thread How it looks (i.θ)ₐ - by marraco - 04/18/2015, 11:20 PM RE: How it looks (i.θ)ₐ - by sheldonison - 04/19/2015, 02:40 PM RE: How it looks (i.θ)ₐ - by marraco - 04/19/2015, 08:40 PM RE: How it looks (i.θ)ₐ - by sheldonison - 04/19/2015, 11:19 PM RE: How it looks (i.θ)ₐ - by marraco - 04/20/2015, 02:35 AM RE: How it looks (i.θ)ₐ - by Gottfried - 04/20/2015, 07:53 AM RE: How it looks (i.θ)ₐ - by marraco - 04/21/2015, 02:34 AM RE: How it looks (i.θ)ₐ - by marraco - 04/21/2015, 03:20 AM RE: How it looks (i.θ)ₐ - by sheldonison - 04/21/2015, 08:23 AM RE: How it looks (i.θ)ₐ - by marraco - 04/23/2015, 04:52 PM RE: How it looks (i.θ)ₐ - by JmsNxn - 04/23/2015, 11:15 PM RE: How it looks (i.θ)ₐ - by sheldonison - 04/23/2015, 11:20 PM RE: How it looks (i.θ)ₐ - by marraco - 04/26/2015, 12:50 AM RE: How it looks (i.θ)ₐ - by sheldonison - 04/26/2015, 05:08 AM RE: How it looks (i.θ)ₐ - by marraco - 04/20/2015, 03:46 AM RE: How it looks (i.θ)ₐ - by marraco - 04/23/2015, 02:21 AM RE: How it looks (i.θ)ₐ - by marraco - 04/25/2015, 07:52 PM

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