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Let $f$ be continuous on $\mathbb{R}$. Then how to find all continuous functions satisfying $f(f(x))=f(x)+x$

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The polynomial f(x)=((1+√5)/2)x is one solution. –  yjj Sep 7 '10 at 0:28
    
Did you create this problem yourself? –  ShreevatsaR Sep 7 '10 at 1:31
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An [unsourced] tag would be useful. –  T.. Sep 7 '10 at 2:01
    
@ShreevatsaR: Yes, i posted a similar problem where one is asked to find all function such that $f(x^k)=f^{k}(x)$, that was the motivation for this problem –  anonymous Sep 7 '10 at 4:29
    
possible duplicate of Solving (quadratic) equations of iterated functions, such as f(f(x))=f(x)+x –  anon Oct 22 '10 at 19:57
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2 Answers 2

up vote 7 down vote accepted

This one is a problem from a journal or from competitions at the level of the Putnam contest (see reference below).

Hint: $g(x) = x + Af(x)$ satisfies $g(f^n(x))=A^ng(x)$ when $A^2 = A + 1$; consider the cases $n \to \pm \infty$.

Source for a similar problem, with solution: http://books.google.com/books?id=-CNbGp2ZFXUC&pg=PA21

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I like the book you mentioned. –  Jack Sep 2 '11 at 23:00
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In fact this belongs to a functional equation of the form http://eqworld.ipmnet.ru/en/solutions/fe/fe1220.pdf.

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

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

$u(t+2)-u(t+1)-u(t)=0$

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

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

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So, (assuming you have the details right,) any solution to the original problem yields a solution to the problem you solved. There's more work to be done to solve the original problem, though! –  Hurkyl Sep 30 '12 at 21:14
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