Return to Answer

Actually, there are many other cases : if $f$ is a surjective map from $C$ to $C$ such that $f(az)=f(z)^1+c$, we get the simple formula $z_n=f(a^nZ0)$, where $f(Z_0)=z_0$. So the question is : for which values of $a$ and $c$ does such an $f$ exists, with perhaps some more constraints on $f$ such as smoothness. It turns out that for any $c$ large enough (at least of module greater than 1) there exists exactly one such entire function $f$, with $f'(0)=1$ and $a=2f(0)=1+sqrt(1-4c)$ (this is shown by an elementary, but tedious calculation of the coefficients of the series by recurrence (done in French here, page 8)), followed by a simple argument of domination of those coefficients, proving convergence for $R>0$ ; then, analytic continuation shows that the radius of convergence is infinite) ; $f$ is (almost) surjective by Picard theorem. Of course, those $f$ are never "usual" or elementary functions, except for $c=0$ or $c=-2$, but so what?

Actually, there are many other cases : if $f$ is a surjective map from $C$ to $C$ such that $f(az)=f(z)^2+c$, we get the simple formula $z_n=f(a^nZ0)$, where $f(Z_0)=z_0$. So the question is : for which values of $a$ and $c$ does such an $f$ exists, with perhaps some more constraints on $f$ such as smoothness. It turns out that for any $c$ large enough (at least of module greater than 1) there exists exactly one such entire function $f$, with $f'(0)=1$ and $a=2f(0)=1+\sqrt{1-4c}$ (this is shown by an elementary, but tedious calculation of the coefficients of the series by recurrence (done in French here, page 8)), followed by a simple argument of domination of those coefficients, proving convergence for $R>0$ ; then, analytic continuation shows that the radius of convergence is infinite) ; $f$ is (almost) surjective by Picard theorem. Of course, those $f$ are never "usual" or elementary functions, except for $c=0$ or $c=-2$, but so what?

Actually, there are many other cases : if $f$ is a surjective map from $C$ to $C$ such that $f(az)=f(z)^1+c$, we get the simple formula $z_n=f(a^nZ0)$, where $f(Z_0)=z_0$. So the question is : for which values of $a$ and $c$ does such an $f$ exists, with perhaps some more constraints on $f$ such as smoothness. It turns out that for any $c$ large enough (at least of module greater than 1) there exists exactly one such entire function $f$, with $f'(0)=1$ and $a=2f(0)=1+sqrt(1-4c)$ (this is shown by an elementary, but tedious calculation of the coefficients of the series by recurrence (done in French here, page 8)), followed by a simple argument of domination of those coefficients by an exponential, iirc) ; $f$ is (almost) surjective by Picard theorem. Of course, those $f$ are never "usual" or elementary functions, except for $c=0$ or $c=-2$, but so what?

Actually, there are many other cases : if $f$ is a surjective map from $C$ to $C$ such that $f(az)=f(z)^1+c$, we get the simple formula $z_n=f(a^nZ0)$, where $f(Z_0)=z_0$. So the question is : for which values of $a$ and $c$ does such an $f$ exists, with perhaps some more constraints on $f$ such as smoothness. It turns out that for any $c$ large enough (at least of module greater than 1) there exists exactly one such entire function $f$, with $f'(0)=1$ and $a=2f(0)=1+sqrt(1-4c)$ (this is shown by an elementary, but tedious calculation of the coefficients of the series by recurrence (done in French here, page 8)), followed by a simple argument of domination of those coefficients, proving convergence for $R>0$ ; then, analytic continuation shows that the radius of convergence is infinite) ; $f$ is (almost) surjective by Picard theorem. Of course, those $f$ are never "usual" or elementary functions, except for $c=0$ or $c=-2$, but so what?

Actually, there are many other cases : if $f$ is a surjective map from $C$ to $C$ such that $f(az)=f(z)^1+c$, we get the simple formula $z_n=f(a^nZ0)$, where $f(Z_0)=z_0$. So the question is : for which values of $a$ and $c$ does such an $f$ exists, with perhaps some more constraints on $f$ such as smoothness. It turns out that for any $c$ large enough (at least of module greater than 1) there exists exactly one such entire function $f$, with $f'(0)=1$ and $a=2f(0)=1+sqrt(1-4c)$ (elementary, but tedious calculation of the coefficients of the series by recurrence, followed by a simple argument of domination of those coefficients by an exponential, iirc) . Of course, those $f$ are never "usual" functions, except for $c=0$ or $c=-2$, but so what?

Actually, there are many other cases : if $f$ is a surjective map from $C$ to $C$ such that $f(az)=f(z)^1+c$, we get the simple formula $z_n=f(a^nZ0)$, where $f(Z_0)=z_0$. So the question is : for which values of $a$ and $c$ does such an $f$ exists, with perhaps some more constraints on $f$ such as smoothness. It turns out that for any $c$ large enough (at least of module greater than 1) there exists exactly one such entire function $f$, with $f'(0)=1$ and $a=2f(0)=1+sqrt(1-4c)$ (this is shown by an elementary, but tedious calculation of the coefficients of the series by recurrence (done in French here, page 8)), followed by a simple argument of domination of those coefficients by an exponential, iirc) ; $f$ is (almost) surjective by Picard theorem. Of course, those $f$ are never "usual" or elementary functions, except for $c=0$ or $c=-2$, but so what?