The fermat superfunction

[mike3]

f(x) = x has 2 solutions : 1 and 2.

The superfunction F(x) is entire and has the fixpoint (of f(x)) 1 @ +/- oo i.

Also the other fixpoint 2 occurs for F(x) at - oo.

so everything seems to work out nicely.

But it might be deception !

F(x) = 2
<=> 2^(2^x) +1 = 2
<=> 2^(2^x) = 1

This has MANY solutions.
And that fact seems to make life hard.
We cannot blame singularities now since F(x) is ENTIRE !

Yes, there exist other points where \( F(z) = 2 \). But this doesn't mean there are additional regular superfunctions beyond the 2, as \( f(z) \) has only 2 fixed points, and the number of regular superfunctions depends on the number of fixed points of \( f \), not the number of times \( F \) attains the value of a fixed point.

Does this imply that F(x) is pseudoperiodic or something ? Or does the functional equation fail ? Both seem to weird to be true.

No, neither -- in fact \( F(z) \) is periodic with imaginary period \( \frac{2\pi i}{\log(2)} \) since that is the period of \( 2^z \). You can check the functional equation holds by simply plugging \( F(z) \) into \( f(z) \).

Now f(x) has 2 fixpoints. So maybe we need 2 superfunctions ?
One seems entire , but to what fixpoint does it belong ?
How does the other superfunction behave ?

Yes, there will be another superfunction at the fixed point \( 1 \). The Fermat superfunction is for the fixed point \( 2 \), as can be seen by evaluating the limit at \( -\infty \).

However, I'm not sure this other superfunction is all that exciting. The fixed point \( 1 \) has \( f'(1) = 0 \) and so is a superattracting fixed point -- in fact the rate of convergence there is quadratic. This means that the superfunction at that point will approach it double-exponentially toward \( +\infty \).

I haven't really studied the case of a superattracting fixed point, but if I were to wager a guess, I'd suggest this superfunction is \( 2^{-2^x} + 1 \)... which is nothing more than an imaginary translation of the first superfunction by a shift value of \( \frac{\pi i}{\log(2)} \). There doesn't seem to be much room for what else it could be, under the constraint of being a double-exponential function with double-exponental approach to the fixed point.

What about those methods where we use 2 fixpoints such as the analytic sickel between two fixpoints based on fatou ?

There is no sickel between two fixed points on the real line for a real-valued-on-the-real-line function, or more precisely it is "degenerate". But if you mean the "merged superfunction" method... If I am right, then the two superfunctions are nothing more than translations of each other, and I'd believe -- though I'm not sure -- that the merge method would not give anything new -- essentially, they are already "merged"!

This seems to be as puzzling as tetration itself , hence like I said this is imho " deception ". It is more complicated then it looks.

Seems having the entire property does not solve all issues !

Keep in mind that an answer like " oh thats just because of the log branches " is not a " real " answer.

I was aware of this for a long time but I was waiting for a response to my first post. Since it did not come I felt the need to explain more.
Maybe you agree on the opinion that this is a serious important topic now.

regards

tommy1729

Hopefully, this will be an illuminating response.