(08/08/2021, 11:22 PM)JmsNxn Wrote: Also; I must've missed the challenge to construct an inverse abel function. I'll be hard pressed if the following method doesn't work. If it doesn't, I'll try another.

...

I've yet to find a function that didn't have exactly n roots about that fixed point though.

Hi James

Thanks for your patient and elaborate detailed explanation

But I'm afraid this case is blowing everyone's mind... because this case is so stubborn to be solved!

It's obvious that if such function has derivative at 0 , then it won't exist.

A quick check, first consider the case , and thus we can assume f has a Taylor series, we'll denote it as , simply plug it in we get , showing that if we want , we have to let , which is impossible. The case is with the same condition: let we have so in order to make it fit we have to solve which is self-contradicting.

I've done something resembles your dictation before, in which I denoted , we are hunting for an explicit calculation formula for

First of all, the function through your generalization, as , in practical calculation, we can transfer it by shifting to define .

As we use , the function has now an upper bound in the positive real axis which is .

We define , in practical calculation, using n=200 000 can promise a precision of 16 degrees except for is so close to . Then we find for which z, the function's value is exactly 0.5, we'll call it C. by mathematica 12.2, so we define the superfunction of f^2(z) as

, thus the value taken at integers is exactly .

Secondly, we can show that by law 3, or more simply calculation, that the super function of f should be asymptotically: . But to be noticed, function is always positive whenever z is positive, not T(z). And because T(z) is determined by function f, which has an oscillating multiplifier -1, T(z) should also behave oscillative. A quick check in mind is, Assume T(0)=0.5>0, so then T(1)=f(T(0))=-0.25<0, T(2)=0.3125>0, T(3)=-0.214844<0, and keeps this pattern all along the positive reals. By calculation, we find that denote briefly , W(0)=0.5, W(0.5)=0.358325>0 is not T(1)=-0.25, which also proves they are only asymptotic to each other.

And I got, finally, stuck here. I figured one way to generate T, as stated below:

This function T would fit the condition for all integer k, but may not fit for all z. This is as far as I reached.

ps. A generalization of any nonconstructable cases is hense obtainable following the same pattern.

I used both to get it converged to fit the equation

But it turned out that most P(z) would lead to the fact that T(z) does not converge to an analytic function except for real integers, instead, the process led T(z)=0 for all non-real-integers z.

This is a graph of T, where I set

Blue line- T(z) by P(z) definition

Red line- , divergent

Yellow line- , converge to 0

Green line- , converge to 0, and evidently it's way different from yellow line and others.

And here's the mathematica codes to generate T and other functions:

Code:

`Clear[F, IF, F2, IF2, A0, IAPrec16, IA2, IA]`

F[z_] := -z + z^2

IF[z_] := 1/2 (1 - Sqrt[1 + 4 z])

F2[z_] := z - 2 z^3 + z^4

IF2[z_] := 1/2 (1 - Sqrt[3 - 2 Sqrt[1 + 4 z]])

A0[z_] :=

1/2 Sqrt[1/

z] (1 + 1/(4 Sqrt[z]) + (5/64 - (11 Log[z])/64)/z^(3/2) - (

11 Log[z])/(32 z) + (

139/1024 - (231 Log[z])/1024 + (121 Log[z]^2)/1024)/z^(5/2) + (

77/512 - (121 Log[z])/512 + (363 Log[z]^2)/2048)/z^2 + (

2293/12288 - (6897 Log[z])/16384 + (1331 Log[z]^2)/4096 - (

6655 Log[z]^3)/65536)/z^3 + (

9959/49152 - (891 Log[z])/2048 + (10285 Log[z]^2)/32768 - (

1331 Log[z]^3)/16384)/z^(7/2) + (

783787/2359296 - (168355 Log[z])/196608 + (426767 Log[z]^2)/

524288 - (270193 Log[z]^3)/786432 + (14641 Log[z]^4)/262144)/z^(

9/2) + (1335835/4718592 - (580723 Log[z])/786432 + (

765325 Log[z]^2)/1048576 - (1039511 Log[z]^3)/3145728 + (

512435 Log[z]^4)/8388608)/z^4)

IAPrec16[z_] := N[Nest[IF2, A0[N[z, 50] + 200000], 200000], 16]

IA2[z_] := IAPrec16[z + N[1.666363609223794000000000`50, 50]]

IA[z_] := (Cos[Pi/2 z]^2) IA2[z/2] + (Sin[Pi/2 z]^2) F[

IA2[z/2 - 1/2]]

And ps, because of the oscillative behavior of T, the inverse of T, the Abel function of T, is multivalued and hard to define. So basically the computation of the second iterative root of f is very hard problem, if generated at the fixed point L=0.

Regards,

Leo