Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

I was wondering if there is a continuous function such that $f(f(x)) = xf(x)$ for every positive number $x$.

share|cite|improve this question
$f(x) = 0$ for $\forall x$ :) – Patrick Li Sep 29 '12 at 5:14
You must have $f(1)=0$ or $1$. – Henry Sep 29 '12 at 5:57
Did you mean to require $f(x)$ to be positive as well? – Hurkyl Sep 30 '12 at 13:45
At the invertible places, the equation also reads $f(x)=x\cdot f^{-1}(x)\ $. – NikolajK Sep 30 '12 at 21:18

Sure. $$ f(x) = x^{\frac{1 + \sqrt 5}{2}} $$

share|cite|improve this answer
How to get that? – Patrick Li Sep 29 '12 at 5:16
try $x^\beta$ and solve for $\beta$ – Will Jagy Sep 29 '12 at 5:25
That's a clever idea. How about functions with other forms? – Patrick Li Sep 29 '12 at 5:38
Is there a proof that leads to f(x) = x^{\frac{1 + \sqrt 5}{2}} – Geokal Sep 29 '12 at 5:41
@Geokal: You can easily prove $f(x) = x^{\frac{1 + \sqrt 5}{2}}$. You cannot prove it is the only solution as there is at least one more which is similar but with a different coefficient. – Henry Sep 29 '12 at 5:51

This is not yet a full answer for the proof, but possibly it is a good step to one. I also think that the problem is not more than a standard exercise in some textbook, but since there is not yet a more qualified answer here, I'll do some naive try so far...

To save notation, let#s write the h'th iterate $\underset{h \text{ times }}{\underbrace {f(...f(f(x)))}}$ as $x_h$ and its p'th power as $x_h^p$ where we understand, that the superscript gets evalauted after the subscript.

Then we can state the sequence:
$$ x = x_{-2} \cdot x_{-1} \\ x = x_{-4} \cdot x_{-3}^2 \cdot x_{-2} \\ x = x_{-6} \cdot x_{-5}^3 \cdot x_{-4}^3 \cdot x_{-3} \\ x = x_{-8} \cdot x_{-7}^4 \cdot x_{-6}^6 \cdot x_{-5}^4\cdot x_{-4} \\ \cdots $$ We observe, that the exponents are the binomial coefficients if powers of 2 $(=(1+1))$ are expanded. Now the idea is, to hope, that we can introduce a limit and that we can assume, that in the limit the difference between the iterates become insignificant below some epsilon, such that we can write $$ x = \lim (x_{-2h})^{2^h} $$ If we assume, that $x_{-2h}<x_{-h}$ then we can even write $$ (x_{-2h})^{2^h}< x < (x_{-h})^{2^h} $$ or $$ (x_{-2h+1})^{2^h}< f(x) < (x_{-h+1})^{2^h} $$ and then $$ (x_{-\infty} + \epsilon_1)^{2^h}< f(x) < (x_{-\infty} + \epsilon_2)^{2^h} $$ and then from a vanishing difference $\epsilon_1 - \epsilon_2 $ deduce, that the h'th iterate of f is necessarily of the form of the h'th iterate of a power $ax^b$ with some fixed a and b . Here I'm stuck because I've not much experience with the formal handling of such limts, but perhaps this is an intuitive path where one can proceed further...

share|cite|improve this answer

In fact this belongs to a functional equation of the form

Let $\begin{cases}x=u(t)\\f=u(t+1)\end{cases}$ ,

Then $u(t+2)=u(t)u(t+1)$

Let $u(t)=e^{v(t)}$ ,

Then $e^{v(t+2)}=e^{v(t)}e^{v(t+1)}$


$v(t+2)=v(t)+v(t+1)+2n\pi i$ , $\forall n\in\mathbb{Z}$

$v(t+2)-v(t+1)-v(t)=2n\pi i$ , $\forall n\in\mathbb{Z}$

Let $v(t)=v_c(t)+A$ ,

Then $v_c(t+2)+A-(v_c(t+1)+A)-(v_c(t)+A)=2n\pi i$

$v_c(t+2)-v_c(t+1)-v_c(t)-A=2n\pi i$

$\therefore A=-2n\pi i$

For $v_c(t+2)-v_c(t+1)-v_c(t)=0$ ,

$v_c(t)=C_1(t)\left(\dfrac{1+\sqrt{5}}{2}\right)^t+C_2(t)\left(\dfrac{1-\sqrt{5}}{2}\right)^t$ , where $C_1(t)$ and $C_2(t)$ are arbitrary periodic functions with unit period

$\therefore v(t)=C_1(t)\left(\dfrac{1+\sqrt{5}}{2}\right)^t+C_2(t)\left(\dfrac{1-\sqrt{5}}{2}\right)^t-2n\pi i$ , $\forall n\in\mathbb{Z}$ , where $C_1(t)$ and $C_2(t)$ are arbitrary periodic functions with unit period

Hence $u(t)=e^{C_1(t)\left(\frac{1+\sqrt{5}}{2}\right)^t+C_2(t)\left(\frac{1-\sqrt{5}}{2}\right)^t-2n\pi i}$ , $\forall n\in\mathbb{Z}$ , where $C_1(t)$ and $C_2(t)$ are arbitrary periodic functions with unit period

$u(t)=e^{C_1(t)\left(\frac{1+\sqrt{5}}{2}\right)^t}e^{C_2(t)\left(\frac{1-\sqrt{5}}{2}\right)^t}$ , where $C_1(t)$ and $C_2(t)$ are arbitrary periodic functions with unit period

$\therefore\begin{cases}x=e^{C_1(t)\left(\frac{1+\sqrt{5}}{2}\right)^t}e^{C_2(t)\left(\frac{1-\sqrt{5}}{2}\right)^t}\\f=e^{C_1(t)\left(\frac{1+\sqrt{5}}{2}\right)^{t+1}}e^{C_2(t)\left(\frac{1-\sqrt{5}}{2}\right)^{t+1}}\end{cases}$ , where $C_1(t)$ and $C_2(t)$ are arbitrary periodic functions with unit period

share|cite|improve this answer
While a neat approach, there's still a fairly large gap between the solution to the problem you solved and the solution to the problem asked. This gap includes checking which of your solutions make $f$ a well-defined partial function of $x$, which solutions are real, and checking how (and if!) different solutions can be patched together to define an $f$ as a total function of $x$. (Also, the OP hasn't specified whether $f(x)$ is allowed to be non-positive which makes a difference, since he doesn't require the equation to hold for nonpositive $x$) – Hurkyl Sep 30 '12 at 21:09
@ Hurkyl:If we require f(x) to be positive ,as you said above, then are things easier ? – Geokal Oct 1 '12 at 18:50
@Geokal: It means the functional equation holds for every element of the image of $f$. If $f(x)$ is allowed to be negative, but also allow $f(f(x)) \neq x f(x)$ for negative $x$, things become trickier. – Hurkyl Oct 1 '12 at 19:16

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.