(08/06/2021, 01:35 AM)JmsNxn Wrote: ...

Thank u, James

I got the similar idea the other day, just like, since

, there must exist a number k that

which is the exotic case, pretty constructable, and we plug in, plug out, so we can get any iterations of f.

However, this method you've typed is, I'm afraid, invalid(sorry if this is impolite) for the non-constuctable cases, because I've tested it out many times, for non-constuctive cases, the fundamental law of iteration:

doesn't hold anymore. A very example, we can construct the more-paid-attention-to problem, which asks for such a function f that

, by this method, we can quickly test out there lies such function, with

, but as conjugation stated,

, so that f must an odd function. This then gives us a hint that -f(z) should also satisfy the equation, as

, showing that -f(z) is another second iterative root of sin(z). And meanwhile it also shows that your construction will not claim this "another" root. This is because the construction will conserve the property, that, we can calculate by chain rule at the fixed point L,

ORIGINALLY:

In the nonconstructable cases, this will fail because

But your method's valid perfectly for exotic cases, because

So, if you apply the method into practical calculation, the function

will not generate

, instead, it gives

, another second iterative root of

, and the reason is

, which breaks the rule aforementioned.

Also, we can inspect another fundamental rule,

, that iterations of the same function are commutative. This doesn't fail, though, and it shows even the 2 different second iterative roots,

one is

, another is a series that has a very small convergent radius, but can be analutically continued to the whole complex plane,

that they are commutative.

(* Notice that in all calculation process above we treated iterative roots as holomorphic functions *)

Again, a same approach is to solve the Julia equation. We can solve Julia quation for both

, by law 4, we see that they share the same Julia function, denoted here as

, then we have

The 2 piece of function

, as how abel function was defined, should satisfy

, but this fails in this case, which can be checked easily, you will find 2 graphs of

are considerably different from each other.

In fact, they satisfy

due to that abel function contains a term of log(z), which is multivalued.

And thankfully there's still another fixed point f(2)=2,f'(2)=3 which is constructable, we can check that, we generate a second root g(z) from z=2, soonly we'll find that

, that this second root g(z) didnt conserve the nonconstuctable fixed point 0.

That's as far as I've calculated

So these results basically show that, if we want to generalize a non-constructable cases' iteration, we can not do this again:

1)find a number k that f^k=g is exotic or g'(L)=1

2)construct g^t, and then take g^(t/k)=f^t

Would you like to introduce another method? Thank u

ps.I'm focused on the generation of inverse Abel function of f(z)=-z(1-z), which is very challenging, would u like to take consideration into this case? (about the asymptotic of the inverse abel function)

Pics showing that h,g,abel function ( sorry the h and g order has been disrupted )

Regards

Leo