10/24/2009, 02:15 AM
(This post was last modified: 10/24/2009, 03:55 AM by Base-Acid Tetration.)
I'm thinking that IF the abel function of f as a simple pole at L (fixed point), the f-iterational (superfunction of f) must decay to L as |z| -> infinity (no matter what the argument of z is), and that f(z) =/= L any z =/= complex infinity.
for example, 1/z, which has a simple pole at 0, is an abel function (also the iterational/superfunction) of z/(z+1), and z/(z+1) has a fixed point at zero. 1/z, being its own inverse, also decays to zero asymptotically as |z| -> infinity.
more complicated examples have the same pattern. z^-n's base function is \( \lbrace \operatorname{sgn}_{(1,2,...n)} \rbrace \)(z^n/(z^n-1))^1/n. the "sgn" thing is the symbol I invented for the nth roots of unity. (it's like the plus-minus sign.)
the inverse of z^-n is z^-1/n which has n branches, the k-th branch of which corresponds to the "k-th side" of z^-n's pole of order n. (at the k-th branch, where |z| is large is mapped to a "wedge" (which "points" to the pole at 0) whose angular measure is 2k*pi/n.)
for example, 1/z, which has a simple pole at 0, is an abel function (also the iterational/superfunction) of z/(z+1), and z/(z+1) has a fixed point at zero. 1/z, being its own inverse, also decays to zero asymptotically as |z| -> infinity.
more complicated examples have the same pattern. z^-n's base function is \( \lbrace \operatorname{sgn}_{(1,2,...n)} \rbrace \)(z^n/(z^n-1))^1/n. the "sgn" thing is the symbol I invented for the nth roots of unity. (it's like the plus-minus sign.)
the inverse of z^-n is z^-1/n which has n branches, the k-th branch of which corresponds to the "k-th side" of z^-n's pole of order n. (at the k-th branch, where |z| is large is mapped to a "wedge" (which "points" to the pole at 0) whose angular measure is 2k*pi/n.)
