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Jul 17, 2019 at 14:14 comment added Iosif Pinelis More generally, if $f$ is not an involution, this suggests that the answer will depend on the structure of the orbits $\{x,f(x),f(f(x)),\dots\}$. If the orbits are finite, then I think we can reason similarly.
Jul 17, 2019 at 14:13 comment added v0511 One way I can think of extending your approach is by assuming that $f^{-1}$ is contractive (which also means it has a fixed point). Then one can make use of the expressions for $\phi(x), \phi(f^{-1}(x)), \phi(f^{-1}\circ f^{-1}(x)) \dots$ to at least approximately evaluate $p(x)$
Jul 17, 2019 at 13:52 comment added v0511 Thanks. This is interesting. Can something be said when f is not an involution?
Jul 17, 2019 at 3:21 history edited Iosif Pinelis CC BY-SA 4.0
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Jul 17, 2019 at 0:46 history edited Iosif Pinelis CC BY-SA 4.0
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Jul 17, 2019 at 0:24 history answered Iosif Pinelis CC BY-SA 4.0