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 Inverse super-composition JmsNxn Long Time Fellow Posts: 291 Threads: 67 Joined: Dec 2010 11/25/2016, 08:55 PM This is just the abel function which satisfies $\alpha(f(x)) = \alpha(x) + 1$ and then if $\alpha_x(z) = \alpha(z) - \alpha(x)$ we have $\alpha_x(f^{\circ n}(x)) = n$ « Next Oldest | Next Newest »

 Messages In This Thread Inverse super-composition - by Xorter - 11/24/2016, 12:53 PM RE: Inverse super-composition - by JmsNxn - 11/25/2016, 08:55 PM RE: Inverse super-composition - by Xorter - 12/23/2016, 01:33 PM RE: Inverse super-composition - by JmsNxn - 12/23/2016, 08:12 PM RE: Inverse super-composition - by Xorter - 12/24/2016, 09:53 PM RE: Inverse super-composition - by sheldonison - 12/25/2016, 04:16 AM RE: Inverse super-composition - by Xorter - 12/25/2016, 04:38 PM RE: Inverse super-composition - by sheldonison - 12/25/2016, 08:35 PM RE: Inverse super-composition - by Xorter - 12/25/2016, 10:23 PM RE: Inverse super-composition - by sheldonison - 12/26/2016, 07:10 AM RE: Inverse super-composition - by Xorter - 01/12/2017, 04:19 PM RE: Inverse super-composition - by Xorter - 05/26/2018, 12:00 AM

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