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$$ f(x^2-1)+2f\left(\frac{2x-1}{(x-1)^2}\right)=2-\frac{4}{x}+\frac{3}{x^2}, \ x>1. \ \ f(x) = ? $$

Don't know how to solve such equiations, help me please. Thank you.

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Are there any restrictions on $f$? In any event, you should try to write a little more out in words. Writing things like that in symbols alone is good enough when you're jotting things down for yourself, or saying out loud what the logical structure of what you write, but this is neither. – tomasz Nov 14 '12 at 22:49
up vote 2 down vote accepted

The solution is based on finding a change of variable $x\to\phi(x)$ such that $\phi(\phi(x))=x$ leading to a system of 2 equations in 2 variables. Let's rewrite $$\frac{2x-1}{\left(x-1\right)^{2}}=\frac{-x^{2}+2x-1+x^{2}}{\left(x-1\right)^{2}}=\frac{x^{2}-\left(x-1\right)^{2}}{\left(x-1\right)^{2}}=\frac{x^{2}}{\left(x-1\right)^{2}}-1$$ Which suggests that $$x\to\frac{x}{x-1}$$ might be a good candidate. Indeed $$\frac{2\left(\frac{x}{x-1}\right)-1}{\left(\frac{x}{x-1}-1\right)^{2}}=x^{2}-1$$ Making the change and not remembering to do this in the RHS as well we obtain: $$f\left(\frac{2x-1}{\left(x-1\right)^{2}}\right)+2f(x^2-1)=2-\frac{4}{x}\left(x-1\right)+\frac{3}{x^{2}}\left(x-1\right)^{2}$$

Now multiply this equation by 2 and subtract the original one

$$3f(x^{2}-1)=2-\frac{4}{x}\left(2\left(x-1\right)-1\right)+\frac{3}{x^{2}}\left(2\left(x-1\right)^{2}-1\right)=\\2-\frac{4}{x}\left(2x-3\right)+\frac{3}{x^{2}}\left(2x^{2}-4x+1\right)=\\2-8+\frac{12}{x}+6-\frac{12}{x}+\frac{3}{x^{2}}=\frac{3}{x^{2}}$$ $$f(x^{2}-1)=\frac{1}{(x^{2}-1)+1}$$ $$f(x)=\frac{1}{x+1}$$

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I used you method and obtained that $f(x)=\frac{1}{x+1}$. Thank you! – Gordon Nov 14 '12 at 23:21
sure, I got a sign wrong. Thanks – Valentin Nov 14 '12 at 23:35

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