(03/23/2014, 12:15 AM)tommy1729 Wrote: 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 . But this doesn't mean there are additional regular superfunctions beyond the 2, as has only 2 fixed points, and the number of regular superfunctions depends on the number of fixed points of , not the number of times attains the value of a fixed point.

(03/23/2014, 12:15 AM)tommy1729 Wrote: 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 is periodic with imaginary period since that is the period of . You can check the functional equation holds by simply plugging into .

(03/23/2014, 12:15 AM)tommy1729 Wrote: 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 . The Fermat superfunction is for the fixed point , as can be seen by evaluating the limit at .

However, I'm not sure this other superfunction is all that exciting. The fixed point has 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 .

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 ... which is nothing more than an imaginary translation of the first superfunction by a shift value of . 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.

(03/23/2014, 12:15 AM)tommy1729 Wrote: 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"!

(03/23/2014, 12:15 AM)tommy1729 Wrote: 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.