Let $a,b,c,$ and $d$ be real numbers such that at least one of $c$ and $d$ is non-zero. Let $ f:\mathbb{R}\to\mathbb{R}$ be a function defined as $f(x)=\frac{ax+b}{cx+d}$. Suppose that for all $x\in\mathbb{R}$, we have $f(x) \neq x$. Prove that if there exists some real number $a$ for which $f^{1387}(a)=a$, then for all $x$ in the domain of $f^{1387}$, we have $f^{1387}(x)=x$. Notice that in this problem, \[f^{1387}(x)=\underbrace{f(f(\cdots(f(x)))\cdots)}_{\text{1387 times}}.\] Hint. Prove that for every function $g(x)=\frac{sx+t}{ux+v}$, if the equation $g(x)=x$ has more than $2$ roots, then $g(x)=x$ for all $x\in\mathbb{R}-\left\{\frac{-v}{u}\right\}$.
Problem
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Tags: function, algebra proposed, algebra
mahanmath
22.09.2010 22:25
Barzil 1990 - P5
Mahdi_Mashayekhi
07.04.2022 17:17
Let $g(x) = f^{1387}(a)$. Let $a$ be such that $f^{1387}(a)=a$. Let $f(a) = c \neq a$. we have $f(f^{1387}(a)) = f^{1387}(f(a)) = f(a)$. Let $f(f(a)) = k \neq c$. Assume $k = a$ then $f^{1389}(a) = f(f(a)) = a = f^{2}(f^{1387}(a)) = f^{1387}(a) = ... = f(a)$ which gives contradiction. Now for $3$ distinct values ${a,c,k}$ we have $g(x) = x$. Now is the time to use the hint. we need to prove $g(x)$ is in from of $\frac{sx+t}{ux+v}$. $f^{2}(x) = \frac{a(\frac{ax+b}{cx+d}) + b}{c(\frac{ax+b}{cx+d}) + d} = \frac{a^2x + ab + bcx + bd}{cax + cb + dcx + d^2} = \frac{(a^2 + bc)x + (ab + bd)}{(ca + dc)x + (cb + d^2)}$ so $f^{2}(x)$ has the form $\frac{ax+b}{cx+d}$ as well, with this approach $f^{1387}(a)$ has this form too so $g(x)$ is in form of $\frac{sx+t}{ux+v}$. Note that quadratic equation $ux^2 + (v-s)x - t = 0$ has $2$ roots but we had $3$ values so $u = t = 0$ and $v = s$ so $g(x)=\frac{sx+t}{ux+v} = \frac{sx}{v} = x$.