Functional Equation Fun

Relevant wiki: Functional Equations - Problem Solving

We are going to use the method of contradiction to prove that the given equation does not have solutions on the domain described in the question. Let us assume that such a function $f$ exists and $\phi (x)= \frac{6x-5}{4x-2}.$ By using the notation $\phi^n$ for the composition of $\phi$ with itself $n$ times, we obtain that $\phi ^2(x)=\frac{4x-5}{4x-4},$ $\phi ^3(x)=\frac{2x-5}{4x-6},$ and $\phi ^4(x)=x,$ for all values of $x$ in the domain of $f$ . We can also see that all the values of the functions $\phi (x),$ $\phi^2 (x),$ $\phi^3 (x),$ and $\phi^4(x)=x$ are in the domain of $f.$ Substituting into the functional equation $x$ by $\phi (x),$ $\phi^2 (x),$ and $\phi^3 (x),$ respectively, we get three new equations that together with the original one form the following system: $\begin{aligned} f(x)+f(\phi(x)) & =x \\ f(\phi(x))+f(\phi^2(x))& =\phi(x)\\ f(\phi^2(x))+f(\phi^3(x))& =\phi^2(x)\\ f(\phi^3(x))+f(x)& =\phi^3(x) \end{aligned}$ Now,you can notice that for any $x$ in the domain of $f$ the following is true: $f(\phi^3(x))+f(x)=f(x)+f(\phi(x)) -(f(\phi(x))+f(\phi^2(x)))+f(\phi^2(x))+f(\phi^3(x))=x-\phi (x)+\phi^2 (x)\neq\phi^3 (x).$ This is the contradiction, so our proof is complete.

Compañero Presa has written a careful systematic solution (+1), as usual. Being in a hurry, I did the problem by considering an example:

$f(2)+f(7/6)=2,$ $f(7/6)+f(3/4)=7/6,f(3/4)+f(-1/2)=3/4, f(-1/2)+f(2)=-1/2$ . If we add the first and third equation and subtract the second and fourth, we find $0=\frac{25}{12}$ , a contradiction. Thus no such function exists.

We can suppose that a function like that exists. In that case, consider $x=0, \frac{5}{2}, \frac{5}{4},\frac{5}{6}$ in the domain of f. We obtain the following system: $f(0)+f(\frac{5}{2})=0$ ; $f(\frac{5}{2})+f(\frac{5}{4})=\frac{5}{2}$ ; $f(\frac{5}{4})+f(\frac{5}{6})=\frac{5}{4}$ ; $f(\frac{5}{6})+f(0)=\frac{5}{6}$ . The system has no solution, therefore, the function is not well defined for 4 values in the domain of f, a contradiction.

I found the same contradiction as Angela's ( the way I stumbled on it is prettier however !!).

$f(0)=-f(\frac{5}{2})$

$=-\frac{5}{2} + f(\frac{5}{4})$

$=-\frac{5}{2} + \frac{5}{4} - f(\frac{5}{6})$

$=-\frac{5}{2} + \frac{5}{4} - \frac{5}{6} + f(0)$

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