Exhibit a function \(f: \mathbf{R} \to \mathbf{R}\) satisfying \[f(f(x)) = 2f(x) - x - 2\]
Of course, nothing linear works. Interestingly, every \(f^n(x)\) can be expressed in terms of \(f(x)\) and \(x\). Let's see what this expression is and use \(x \to f(x)\) \[f^n(x) = nf(x) - (n-1)x - n(n-1)\] Okay. \(f\) is injective. If \(f(x) - x = n\), then \(f^{n-1}(x) = x\). \[f^n(x) - xf(x) = (x + n)(f(x) - n + 1)\] Let \(g(x) = f(x) - x\). \[g(g(x) + x) - g(x) = -2\]
Interlude: coping with other people in a meeting and figuring out what things must look like and some more interesting ideas (the chains are real) than the ones presented above.