# Tetration Forum

Full Version: The fermat superfunction
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Let f(x) = (x-1)^2 + 1.
If we want the half-iterate of f(x) for x>1 we can use
S(S^[-1](x)+1/2) where S is the superfunction of f(x).

For x>2 this superfunction is F(x) = 2^(2^x) + 1.
Hence I call that the fermat superfunction.

This fermat superfunction F(x) is entire so there are no other branches.

F^[-1] does have branches though but they are easy to understand.

Notice F(x) = 1 has no complex solution.
( 1 is the other fixpoint of f(x) )
But what do we do for the other nonparabolic fixpoint of f(x) ? What superfunction belongs there ?

Is this related to branches of F^[-1] ?

regards

tommy1729
f(x) = x has 2 solutions : 1 and 2.

The superfunction F(x) is entire and has the fixpoint (of f(x)) 1 @ complex infinity.

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 !

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

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 ?

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

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
(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 $F(z) = 2$. 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.

(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 $F(z)$ 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 $f(z)$.

(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 $1$. 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.

(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.
Hi mike3.

Its a bit helpful but I feel there are still issues.

The main one being F(x)=2 having multiple solutions.

Maybe its my lack of understanding but lets compare to sexp.

The sexp has all its fixpoints at +/- oo i.

If sexp had one of its fixpoints at say a + bi for 0 <a,b< oo then that would create a problem since that would require we have a straith line.
And hence sexp would then not be analytic near c + bi for any real c.

So what does that mean ?
Is not 2 a fixpoints of the half-iterate of f ?
Is there no line in this case ?

But whenever we get F(x)=2 , because of f(2)=2 we know that we get a local periodic behaviour. (period =< 1)

Guess similar question arise for most superfunctions of polynomials of degree 2 since they are all of double exponential nature.
(related : chebyshev polynomial , sinh(2^x) , mandelbrot , ... )

My confusion is probably because of the imho weird positions of solutions to F(x) = 2. ( because of ln branches ). In combination with the idea above ofcourse.

In other cases of superfunctions I can imagine the fixpoints (of the original function , not the superfunction) to be on other branches. Here that is problematic. Hence my special intrest.

regards

tommy1729