Now I compared Daniels method with Andrews method for the base (hyperbolic case).

Daniel's approach is to consider the fixed point of , and to determine the unique hyperbolic iterate there and then set . The lower fixed point of for is 2 (and the upper fixed point is not reachable from 1 by iterations, so it is not of importance for our method ).

Because I have no actual formula to compute the series expansion for the hyperbolic iterate, I used an iterative formula (which can be found in [1] and has quite some similarity with Jay's approach):

where f is assumed to have its fixed point at 0 and is the derivative at the fixed point, which is in our case

. Of course there are some demands on the function f for the formula to be valid, but they are satisfied by our f, particularely .

In the usual way we can move the fixed point to 0 by conjugating and after iteration move it back to its original place by inverse conjugating. Resulting in this case in the formula

and

i.e.

And now guess how Andrew's and Daniel's slog compare! (At least on the picture, I didnt start exacter numerical computations)

[1] M. C. Zdun, Regular fractional iterations, Aequationes Mathematicae 28 (1985), 73-79

Daniel's approach is to consider the fixed point of , and to determine the unique hyperbolic iterate there and then set . The lower fixed point of for is 2 (and the upper fixed point is not reachable from 1 by iterations, so it is not of importance for our method ).

Because I have no actual formula to compute the series expansion for the hyperbolic iterate, I used an iterative formula (which can be found in [1] and has quite some similarity with Jay's approach):

where f is assumed to have its fixed point at 0 and is the derivative at the fixed point, which is in our case

. Of course there are some demands on the function f for the formula to be valid, but they are satisfied by our f, particularely .

In the usual way we can move the fixed point to 0 by conjugating and after iteration move it back to its original place by inverse conjugating. Resulting in this case in the formula

and

i.e.

And now guess how Andrew's and Daniel's slog compare! (At least on the picture, I didnt start exacter numerical computations)

[1] M. C. Zdun, Regular fractional iterations, Aequationes Mathematicae 28 (1985), 73-79