The fermat superfunction - Printable Version +- Tetration Forum (https://math.eretrandre.org/tetrationforum) +-- Forum: Tetration and Related Topics (https://math.eretrandre.org/tetrationforum/forumdisplay.php?fid=1) +--- Forum: Mathematical and General Discussion (https://math.eretrandre.org/tetrationforum/forumdisplay.php?fid=3) +--- Thread: The fermat superfunction (/showthread.php?tid=809) The fermat superfunction - tommy1729 - 08/13/2013 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 RE: The fermat superfunction - tommy1729 - 03/23/2014 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 RE: The fermat superfunction - mike3 - 03/23/2014 (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. RE: The fermat superfunction - tommy1729 - 03/24/2014 Hi mike3. Thanks for your answer. 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