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Find the general solution to $f(z)=f(z/2)f(z-1)$ where $z$ is a complex number.

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As a sidenote , i wonder if $f(z)=f(z/2)f(3z+1)$ relates to collatz ? –  mick Sep 21 '12 at 19:07
    
what is the class of admissible functions? –  Valentin Sep 21 '12 at 19:11
    
@Valentin: No particular restrictions. Preferably analytic in every closed contour when bounded in that same closed contour. Also preferred that $f(z)$ maps reals to reals. I know that 0 is a solution of course but everything else is welcome. –  mick Sep 21 '12 at 19:17
    
I wonder if Laplace could kick things off, probably not: $$L\left[e^{t}\phi\left(t\right)\right]=f\left(z-1\right)$$ $$L\left[2\phi\left(2t\right)\right]=f\left(\frac{z}{2}\right)$$ $$f\left(z\right)=2\int_{0}^{\infty}dte^{-zt}\int_{0}^{t}d\tau e^{\tau-t}\phi\left(\tau-t\right)\phi\left(2\tau\right)$$ –  Valentin Sep 21 '12 at 19:43

1 Answer 1

up vote 2 down vote accepted

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.

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It's not necessarily true that $g(z) = g(z/2) + g(z-1)$: rather $g(z) = g(z/2) + g(z-1) + 2 \pi i n$ for some integer $n$ (which, as long as $g$ is continuous, is constant)$. –  Robert Israel Sep 21 '12 at 23:08
    
@Robert, that is equivalent to $g(z)-2\pi i n$ being a solution of the log functional equation with no $n$. Changes of $n$ do not change $f$, so this is just some minor gauge freedom. –  zyx Sep 22 '12 at 10:28

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