Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question
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
up vote 2 down vote accepted

Let $f(z)=2^{g(z)}$ ,

Then $2^{g(z)}=2^{g(z/2)}2^{g(z-1)}$




Let $g(z)=\int_0^\infty2^{-zt}K(t)~dt$ ,

Then $\int_0^\infty2^{-zt}K(t)~dt-\int_0^\infty2^{-\frac{zt}{2}}K(t)~dt-\int_0^\infty2^{-(z-1)t}K(t)~dt=0$






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.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.