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Inverse super-composition

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Inverse super-composition
JmsNxn Offline
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#2
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

\( \alpha_x(f^{\circ n}(x)) = n \)
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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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