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$f(x)+f(\frac{x-1}{x}) = x+1$ is given.
find $f(x)$.

I did tried to change it into another form buy substituting $x$ with $\frac{x}{x-1}$
The result were $f(\frac{1}{x})+f(\frac{x}{x-1})=\frac{2x-1}{x-1}$

I don't know what to do next.

Cettt
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Chutchatut Sutichavengkul
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  • $f(x)=\dfrac12\left[x+\dfrac1x+\dfrac1{1-x}\right]$. –  Dec 20 '16 at 06:43
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    Define a linear map $A$ on the space of (real-valued, I guess?) functions via $(Af)(x) = f(\frac{x-1}{x})$. Then you can check that $A^3 = 1$. Now use the fact that $2f = (1+A^3)(f) = (1-A+A^2)(1+A)(f) = (1-A+A^2)(x+1)$ to solve for $f$. – K.K. Dec 20 '16 at 06:45

1 Answers1

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All you have to do is make the substitution again. $$f(x)+f\bigg(\frac{x-1}{x}\bigg)=x+1$$ $$\implies f\bigg(\frac{x-1}{x}\bigg)+f\bigg(\frac{1}{1-x}\bigg)=\frac{x-1}{x}+1$$ $$\implies f\bigg(\frac{1}{1-x}\bigg)+f(x)=\frac{1}{1-x}+1$$ Now subtract the first and second equations to get $$f(x)-f\bigg(\frac{1}{1-x}\bigg)=x-\frac{x-1}{x}$$ Then add this to the third equation to get $$f(x)+f(x)=x-\frac{x-1}{x}+\frac{1}{1-x}+1$$ $$2f(x)=x-\frac{x-1}{x}+\frac{1}{1-x}+1$$ $$2f(x)=x-1+\frac{1}{x}+\frac{1}{1-x}+1$$ $$2f(x)=x+\frac{1}{x}+\frac{1}{1-x}$$ $$2f(x)=x+\frac{1}{x(1-x)}$$ $$\color{red}{f(x)=\frac{x^2(1-x)+1}{2x(1-x)}}$$

Franklin Pezzuti Dyer
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