Let $f:\mathbb{R}\backslash\left\{0,1\right\}\to \mathbb{R}$ be a function such that $$f\left(x\right)+f\left(\frac{x-1}{x}\right)=1+x$$ Find $f\left(x\right)$.
I tried plugging for $x$ as $\frac{x-1}{x}$, but that didn't lead me anywhere. Could someone suggest a more helpful substitution?
Actually, you need to substitute it twice. It is due to the fact that the transformation $\phi(x) := \frac{x-1}{x} = 1-\frac{1}{x}$ is an element of order 3 with respect to composition, meaning $\phi \circ \phi \circ \phi = \mathrm{id}$. In consequence, you are facing three unknowns, namely : $$ \left\{\begin{array}{ll} f_1 &:= f(\phi(x)) &= f(1-\frac{1}{x}) \\ f_2 &:= f(\phi(\phi(x))) &= f(\frac{1}{1-x}) \\ f_3 &:= f(\phi(\phi(\phi(x)))) &= f(x) \end{array} \right. $$ Then, the functional equation you were provided permits to generate a system of three equations for these unknowns : $$ \left\{\begin{array}{lll} f(x) + f(\phi(x)) = 1-x \\ f(\phi(x)) + f(\phi(\phi(x))) = 1-\phi(x) = \frac{1}{x} \\ f(\phi(\phi(x)) + f(\phi(\phi(\phi(x)))) = 1-\phi(\phi(x)) = -\frac{x}{1-x} \end{array} \right. $$ It can be recasted matricially in the following way : $$ \begin{pmatrix} 0 & 1 & 1 \\ 1 & 1 & 0 \\ 1 & 0 & 1 \end{pmatrix} \begin{pmatrix} f_1 \\ f_2 \\ f_3 \end{pmatrix} = \begin{pmatrix} 1-x \\ \frac{1}{x} \\ -\frac{x}{1-x} \end{pmatrix} $$ I let you invert this matrix and solve it for $f_3 = f(x)$.
Of course, you could have done the same substitutions "manually", i.e. without introducing all the new notations; however, here you have a procedure to tackle similar problems when the transformation $\phi$ belongs to a finite group.
