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I was playing around with a problem and arrived at the following functional equation,

$$xf(x)-1 = f\left(3-\frac{1}{x-1}\right),$$

where $x$ is a real number. I know such a function must satisfy $f(2) = 1$, and based on the problem this arose from, I believe $f(3) = \frac{\pi}{2}-1$, but I know nothing beyond that at the moment.

Is there a good resource that goes into examples and into the theory of functional equations, like the one above?

Thanks in advance for your help!

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The equation by itself leaves too much freedom. Without additional constraints, it does not pinpoint the function you had in mind, whatever that could be.

The general solution goes like this: define $f(x)$ arbitrarily for $x\in(3,\infty)$, then by finding $f\left(3-{1\over x-1}\right)$ from $f(x)$ extend the domain to $(2{1\over2},3)$, then go on to $(2{1\over3},2{1\over2})$, etc. Then apply the equation in reverse and find $f(x)$ from $f\left(3-{1\over x-1}\right)$, and in this manner reconstruct the function on $(-\infty,1)$, then on $(1,{3\over2})$ and so on. Thus the function will be defined on all $\mathbb R$; with relatively little effort, it can also be made continuous almost everywhere.

On the other hand, if you want $f(x)$ to behave nicely around $x=2$, that brings us up to an entirely new level of difficulty. The obvious first step would be to express it in a form of power series: $f(x) = 1 + c_1(x-2) + c_2(x-2)^2 + c_3(x-2)^3+\dots$, right? Well, after finding the first few $c_i$, we notice they look suspiciously like the terms of A014307 (except for alternating signs), which is a pity, since it promises a power series with zero convergence radius. This still leaves the chances that the function can be treated in a more sophisticated way I'm not aware of.

So it goes.

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$\because$ The general solution of $T(x+1)=3-\dfrac{1}{T(x)-1}$ is $T(x)=\dfrac{2x+3+2\Theta(x)}{x+1+\Theta(x)}$ , where $\Theta(x)$ is an arbitrary periodic function with unit period (with reference to https://www.wolframalpha.com/input/?i=T(x%2B1)%3D3-1%2F(T(x)-1))

$\therefore\left(2+\dfrac{1}{x+1}\right)f\left(2+\dfrac{1}{x+1}\right)-1=f\left(3-\dfrac{1}{2+\dfrac{1}{x+1}-1}\right)$

$f\left(3-\dfrac{x+1}{x+2}\right)-\left(2+\dfrac{1}{x+1}\right)f\left(2+\dfrac{1}{x+1}\right)=-1$

$f\left(2+\dfrac{1}{x+2}\right)-\left(2+\dfrac{1}{x+1}\right)f\left(2+\dfrac{1}{x+1}\right)=-1$

$f\left(2+\dfrac{1}{x+1}\right)-\left(2+\dfrac{1}{x}\right)f\left(2+\dfrac{1}{x}\right)=-1$

$f\left(2+\dfrac{1}{x}\right)=\dfrac{2^x(\Theta(x)-\sqrt\pi)}{\Gamma(x)}+\dfrac{x}{2}~\Gamma\left(x+\dfrac{1}{2}\right)~_2\tilde{F}_1\left(1,x+1;x+\dfrac{3}{2};\dfrac{1}{2}\right)$ , where $\Theta(x)$ is an arbitrary periodic function with unit period (with reference to https://www.wolframalpha.com/input/?i=f(x%2B1)-(2%2B1%2Fx)f(x)%3D-1)

$f(x)=\dfrac{2^\frac{1}{x-2}\left(\Theta\left(\dfrac{1}{x-2}\right)-\sqrt\pi\right)}{\Gamma\left(\dfrac{1}{x-2}\right)}+\dfrac{1}{2(x-2)}~\Gamma\left(\dfrac{1}{x-2}+\dfrac{1}{2}\right)~_2\tilde{F}_1\left(1,\dfrac{1}{x-2}+1;\dfrac{1}{x-2}+\dfrac{3}{2};\dfrac{1}{2}\right)$ , where $\Theta(x)$ is an arbitrary periodic function with unit period

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