# Tetration Forum

Full Version: introducing TPID 16
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TPID 16

Let $f(z)$ be a nonpolynomial real entire function.
$f(z)$ has a conjugate primary fixpoint pair : $L + M i , L - M i.$
$f(z)$ has no other primary fixpoints then the conjugate primary fixpoint pair.
For $t$ between $0$ and $1$ and $z$ such that $Re(z) > 1 + L^2$ we have that
$f^{[t]}(z)$ is analytic in $z$.
$f^{[t]}(x)$ is analytic for all real $x > 0$ and all real $t \ge 0$ .
If $f^{[t]}(x)$ is analytic for $x = 0$ then :
$\frac{d^n}{dx^n} f^{[t]}(x) \ge 0$ for all real $x \ge 0$ , all real $t \ge 0$ and all integer $n > 0$.
Otherwise
$\frac{d^n}{dx^n} f^{[t]}(x) \ge 0$ for all real $x > 0$ , all real $t \ge 0$ and all integer $n > 0$.

Are there solutions for $f(z)$ ?
I conjecture yes.

regards

tommy1729
(06/07/2014, 11:03 PM)tommy1729 Wrote: [ -> ]TPID 16

Let $f(z)$ be a nonpolynomial real entire function.
$f(z)$ has a conjugate fixpoint pair : $L + M i , L - M i.$
$f(z)$ has no other fixpoints then the conjugate fixpoint pair.
...
such that $Re(z) > 1 + L^2$ we have that
...
If $f^{[t]}(x)$ is analytic for $x = 0$ then :
$\frac{d^n}{dx^n} f^{[t]}(x) \ge 0$ for all real $x \ge 0$ , all real $t \ge 0$ and all integer $n > 0$

I'm not sure how big a deal this is but if $f(x)-x$ has two solutions, than we expect an infinite number of fixed points for an entire function, the zeros of $f(x)-x$.

It sounds like the conjecture is that all of the derivatives are positive for values of x>0, under some conditions? Not sure I get the entire conjecture; as applied to tetration, it might be that L~=0.3+1.3i, so if sexp(z)>real(L)^2+1????? Jay observed that all of the odd derivatives of tetration are positive, and that all of the even derivatives start out negative, and eventually become positive, crossing the real axis exactly once. But this does not happen instantaneously, and I'm not sure how long it takes. For example, the 22nd derivative of tetration is negative at x=0, and goes positive at x=0.0078; the 42nd derivative is the first derivative that is negative at x=0.15; and doesn't go positive until x=0.15105. For the 100th derivative, the transition >0.31.

One other problem is that the "otherwise" statement here is identical to the then statement.
Quote:Otherwise
$\frac{d^n}{dx^n} f^{[t]}(x) \ge 0$ for all real $x > 0$ , all real $t \ge 0$ and all integer $n > 0$.
....
(06/08/2014, 04:05 PM)sheldonison Wrote: [ -> ]One other problem is that the "otherwise" statement here is identical to the then statement.

No. Larger or equal than 0 =/= larger than 0.
( for x)

You are mistaken.
Even if I shake ideas out of my sleeve I usually know what Im doing.

I was aware of Jay's observation.

regards

tommy1729
Oh sheldon sorry.
I meant primary fixpoints.

I only noticed today I did not write it.

regards

tommy1729
Is it required that there exists a superfunction of f(z) such that the superfunction is bounded within any (finite) radius centered at 0 ?

regards

tommy1729