I was wondering if there is a continuous function such that $f(f(x)) = xf(x)$ for every positive number $x$.
How to find the function $f$ given $f(f(x)) = xf(x)$?
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0$f(x) = 0$ for $\forall x$ :) – 2012-09-29
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1You must have $f(1)=0$ or $1$. – 2012-09-29
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1Did you mean to require $f(x)$ to be positive as well? – 2012-09-30
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0At the invertible places, the equation also reads $f(x)=x\cdot f^{-1}(x)\ $. – 2012-09-30
3 Answers
Sure. $$ f(x) = x^{\frac{1 + \sqrt 5}{2}} $$
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0How to get that? – 2012-09-29
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2try $x^\beta$ and solve for $\beta$ – 2012-09-29
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0That's a clever idea. How about functions with other forms? – 2012-09-29
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0Is there a proof that leads to f(x) = x^{\frac{1 + \sqrt 5}{2}} – 2012-09-29
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0@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. – 2012-09-29
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0@Henry : I wanted a proof that leads to the f above with no doubts that it is the only one . Question : What coefficient ? – 2012-09-29
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0@Geokal: If you followed Will Jagy's comment then you would find it. As a clue, he solved $\beta^2=1+\beta$. But Patrick Li has also given an answer. – 2012-09-29
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0In fact your solution is my solution that take $C_2(t)=0$ and thus should be not enough general. – 2012-10-01
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}
In fact this belongs to a functional equation of the form http://eqworld.ipmnet.ru/en/solutions/fe/fe2315.pdf.
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 $e^{v(t+2)}=e^{v(t)}e^{v(t+1)}$
$e^{v(t+2)}=e^{v(t)+v(t+1)}$
$v(t+2)=v(t)+v(t+1)+2n\pi i$ , $\forall n\in\mathbb{Z}$
$v(t+2)-v(t+1)-v(t)=2n\pi i$ , $\forall n\in\mathbb{Z}$
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
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0While 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 $x$) – 2012-09-30
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0@ Hurkyl:If we require f(x) to be positive ,as you said above, then are things easier ? – 2012-10-01
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0@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. – 2012-10-01