Solutions to an Involution Functional Equations

Question

I'm reading functional equations and how to solve them and I am very confused by one of the statements made

Let $$f(x) = f(\frac{x}{x-1})$$

then a solution would be

$$f(x) = cos log (x-1)$$

but evaluating the function

$$cos log(x-1) = cos log(\frac{x}{x-1} -1)$$ $$cos log(x-1) = cos log(\frac{1}{x-1})$$ which is not true? or am I overthinking something?

Follow up Question:

It is stated in the book that for

$$f(x) = f(a(x))$$ and $a(x)$ is an involution $f(f(f...(x)...) = x$

then $f(x)$ can be written as $$ T(x,a(x)) $$ where $T(a,b) = T(b,a)$

I dont know how to approach solving for the symmetric function $T(a,b)$.

Answer 1

The question about $\, f(x) = T(x, a(x))\,$ is very easy. The general result is that $$ f(x) = F(x \!+\! a(x), x\, a(x)) \tag1$$ satisfies the equation $\, f(x) = f(a(x))\,$ where $\,F(.,.)\,$ is any function provided that $\, a(a(x)) = x\,$ is for all $\,x\,$ is assumed.

A power series for which $\,f(f(x))=x,\,$ $\,f(0)=0,\,$ and $\,f'(0)\ne 0\,$ is given by $$ a(x) - -x + c_2 x^2 - c_2^2 x^3 + c_4 x^4 + (2 c_2^4 - 3c_2 c_4) x^5 + c_6 x^6 + O(x^7) \tag2$$ where $\,c_2,c_4,c_6,\dots\,$ are arbitrary. This implies that, except for $\,f(0)=0,\,$ the even part of $\,f(x)\,$ is arbitrary and the odd part is uniquely determined by the even part. Several examples of these are given in the OEIS. For example, the generating function of OEIS sequence A000245.

Answer 2

HINT.-For the solution $f(x)=\cos(\log(x-1))$ we can do as follows: Since $f(x) = f(\dfrac{x}{x-1})$ we can verify that $f(x) - f(\dfrac{x}{x-1})=0$.

In fact $$\cos(\log(x-1))-\cos(\log(\dfrac{1}{x-1})=-2\sin\frac{\log(x-1)+\log(\frac{1}{x-1})}{2}\sin\frac{\log(x-1)-\log(\frac{1}{x-1})}{2}$$ The first factor is equal to $0$ because its numerator is $\log 1=0$.