Find the general solution to $f(z)=f(z/2)f(z-1)$ Find the general solution to $f(z)=f(z/2)f(z-1)$ where $z$ is a complex number.
 A: Let $f(z)=2^{g(z)}$ ,
Then $2^{g(z)}=2^{g(z/2)}2^{g(z-1)}$
$2^{g(z)}=2^{g(z/2)+g(z-1)}$
$g(z)=g(z/2)+g(z-1)$
$g(z)-g(z/2)-g(z-1)=0$
Let $g(z)=\int_0^\infty2^{-zt}K(t)~dt$ ,
Then $\int_0^\infty2^{-zt}K(t)~dt-\int_0^\infty2^{-\frac{zt}{2}}K(t)~dt-\int_0^\infty2^{-(z-1)t}K(t)~dt=0$
$\int_0^\infty2^{-zt}K(t)~dt-\int_0^\infty2^{-zt}K(2t)~d(2t)-\int_0^\infty2^{-zt}2^tK(t)~dt=0$
$\int_0^\infty2^{-zt}K(t)~dt-\int_0^\infty2\times2^{-zt}K(2t)~dt-\int_0^\infty2^{-zt}2^tK(t)~dt=0$
$\int_0^\infty2^{-zt}((1-2^t)K(t)-2K(2t))~dt=0$
$\therefore(1-2^t)K(t)-2K(2t)=0$
$K(2t)=\dfrac{(1-2^t)K(t)}{2}$
Let $\begin{cases}t_1=\log_2t\\K_1(t_1)=K(t)\end{cases}$ ,
Then $K_1(t_1+1)=\dfrac{(1-2^{2^{t_1}})K_1(t_1)}{2}$
$K_1(t_1)=\Theta(t_1)\prod\limits_{t_1}\dfrac{1-2^{2^{t_1}}}{2}$ , where $\Theta(t_1)$ is an arbitrary periodic function with unit period
$K(t)=\Theta(\log_2t)\left(\prod\limits_{t_1}\dfrac{1-2^{2^{t_1}}}{2}\right)(\log_2t)$ , where $\Theta(t)$ is an arbitrary periodic function with unit period
$\therefore f(z)=2^{\int_0^\infty\Theta(\log_2t)2^{-zt}\left(\prod\limits_{t_1}\frac{1-2^{2^{t_1}}}{2}\right)(\log_2t)~dt}$ , where $\Theta(t)$ is an arbitrary periodic function with unit period
But this may be only one of the group of the solution and may be not enough general. I have no idea about the exact number of groups of the solution in the general solution of the functional equation of this type, so I stop here.
