# Question about iterations : $g(c,z) = a$ always has a solution where $z$ is strictly real?

Let $c$ be any positive real number.

Let $z$ be a complex number.

Let $g(c,z)$ be some locally analytic function that is the $c$ th iteration of the entire function $f(z)$ with $g(0,z)=f(z)$.

Let $a$ be a real number and $f(z)$ a function such that

$1)$ $a$ is not a fixpoint or cyclic point of $f(z)$.

$2)$ For the entire function $f(z)$ and the real $a$ we have that $g(c,z) = a$ always has a solution where $z$ is strictly real. (for any $c$)

What is the name for such an $a$ if any exist at all ?

How to find such $a,f(z)$ and $g(c,z)$ ?

Or when some are given ; find the others or prove that they cannot exist ?

Good free online references are welcome.

Im not sure how to put this into a clear short title ...

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What does "the $c$th iteration" mean when $c$ is not an integer? – Henning Makholm Nov 10 '12 at 22:45
Think of it in terms of locally analytic abel functions and their inverses ... (or like tetration and such). – mick Nov 10 '12 at 23:01
Does it hold that $g(c_1,(g(c_2,z))=g(c_1+c_2+1,z)$ for $c_1,c_2\not\in\mathbb{Z}$? – robjohn Nov 10 '12 at 23:17
Why the $+1$ ? – mick Nov 10 '12 at 23:23
I would divide the problem into three cases 1) case where f(z) is real valued, and $f^n(z)$ is real valued for all integers (or at least positive integers). 2) f(z) is not real valued, but for some value of n, $f^n(z)$ is real valued, and for all integers greater than n. 3) all other cases, it would seem there would have to be real values of $\alpha$ for which there is no solution of $f^c(z)=\alpha$, for any particular $g(c,z)$. – Sheldon L Nov 11 '12 at 0:15

Let as assume there is a Schroder function with a fixed point for $f(z)$, with $|\lambda|>1$ for the fixed point. It doesn't matter if the fixed point is real or complex, but if $f(z)$ is entire with this restriction, then the inverse of the Schroder function is also entire. There is also a Schroder equation for $|\lambda|<1$, but its inverse is not entire since $f^{-1}$ is not entire, and solutions with $|\lambda|=1$ can be very complicated. The Schroder function, with fixed point $z_0$, is a formal series, $\psi(z)=z+a_2 z^2+a_3 z^3..$ defined by $\psi(f(z)-z_0) = \lambda \psi(z-z_0)$, and the inverse of the Schroder function has $\psi^{-1}(\lambda z)+z_0= f(\psi^{-1}(z)+z_0)$. Then one possible starting point for an analytic solution for a particular value of z, using the op's notation for g is:
$g(x,z)= \psi^{-1}(\psi(z-z_0)\lambda^x+z_0)$
$g(0,z)=z$
$g(1,z)=f(z)$
$g(x,z)=f(g(x-1,z))$

However, the op also requires that for some real valued c, $g(c,z)=\alpha$. So, the algorithm is calculate $y=\log_\lambda \frac{\psi(\alpha)}{\psi(z)}$. Now, y could be a real number, and in that case we are done since we can choose c=y, since $g(y,z)=\alpha$. But if y is not real valued, then we can easily arrange some analytic 1-cyclic function, $\theta(z)$, such that $\theta(0)=0$, and for any completely arbitrarily chosen real number c that is not an integer, $c+\theta(c)=y$. This would seem to meet the op's requirements.

Then one solution to the op's problem is:
$g(x,z)= \psi^{-1}(\psi(z-z_0)\lambda^{x+\theta(x)}+z_0)$
$g(0,z)=z$
$g(c,z)=\alpha$

Note, this solution works for z and $\alpha$ being real numbers, but $g(x,z)$ may not be real valued for other values of x other than 0 and c. If the op is interested in a function that is real valued for real valued x, then this solution is not sufficient. If $g(x,z)$ is real valued for all integers>=0, and $f(z)$ is a real valued function, and $g(\lfloor(c))<\alpha$ and $g(\lceil(c))>\alpha$, then a very creative solution involves generating $\theta(z)$ via a Riemann mapping of the interval between floor(c) and ceil(c), so that the function $g(x,z)$ becomes real valued, and nicely behaved at the real axis and in the complex plane. Actually, the Riemann mapping itself is messy, and usually involves a singularity or branch point. But basically, calculate a Riemann mapping of $\exp(2\pi i g^{-1}(z))$ for the real interval $[z..f(z)]$, and use this Riemann mapping to generate a $\theta(x)$, which can be used in the above equation, with one particular value of c, such that $g(c,z)=\alpha$. I think this is equivalent to perturbed Fatou Coordinates. I can try to give more details if this real valued function with a complex fixed point is what the op is interested in.

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Thank you for your effort. Funny though I expected an answer from you. Your really into dynamical systems. As you might have expected the question in the OP is part of a bigger more complicated problem. Lets say for instance that $a$ is on a line in the complex plane rather than (condition) real. And f(z) should be exp(z). Then how does it work out ? in detail ? Once again this question was based on tetration actually ... – mick Nov 11 '12 at 23:24
I knew the riemann mapping was involved but did not want to confuse the average reader of the OP ... just saying :) – mick Nov 11 '12 at 23:25
Tetration has a unique solution, so $\alpha$ is either a real iterate of z, or it is not a real iterate of z. If $f(z)=\exp(z)$, and z is a real number, and $\alpha$ is a real number, then $c=g^{-1}(z)-g^{-1}(\alpha)$, is guaranteed to be a real number, but only after the Riemann mapping. If z and $\alpha$ are both complex, $\log_\lambda \frac{\psi(\alpha)}{\psi(z)}$ could be real or complex, but what we care about is $g^{-1}(z)-g^{-1}(\alpha)$, where g is a Riemann mapping at the real axis. – Sheldon L Nov 12 '12 at 14:24
For a complex $\alpha$, I don't know what easily definable line $\alpha$ would be on. The iterates, $[\alpha_0=\alpha,\alpha_1=\exp(\alpha),\alpha_2=\exp^2(\alpha),\alpha_3=\exp^3(‌​\alpha)]$ all have different imaginary values, so there is no "correct" path from $\alpha_n$ to $\alpha_{n+1}$. That's why it is easier to do the Riemann mapping at the real axis. Now, for iterating complex bases, or for iterating $x^2+c$ where c is complex, that is a problem whose solution I cannot formally define, although I can calculate it. It involves both fixed points, with separate Schroder/theta equations. – Sheldon L Nov 12 '12 at 15:51