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Let : $$f(x)+f\left(\frac{1-x}{x} \right)=x$$ then: $$f(x)=?$$

My Try :

$x\to \dfrac{1-x}{x}$

\begin{align}f\left(\dfrac{1-x}{x}\right)+f\left(\frac{1-\dfrac{1-x}{x}}{\dfrac{1-x}{x}} \right)&=\dfrac{1-x}{x}\\ f\left(\dfrac{1-x}{x}\right)+f\left(\dfrac{2x-1}{x^2} \right)&=\dfrac{1-x}{x}\end{align}

And $u=\dfrac{1-x}{x}$

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

Now what ?

lioness99a
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Almot1960
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    Does this question help? https://math.stackexchange.com/questions/557925/functional-equation-f-left-fracx-1x-right-f-left-frac11-x-right?rq=1, in particular the top answer – lioness99a May 24 '17 at 12:22
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    The term $\frac{2x-1}{x^2}$ is wrong, but even with the correct calculation, continuing to plug in $\frac{1-x}{x}$ for $x$ does not seem to help, as the matrix $\begin{bmatrix}-1&1\1&0\end{bmatrix}$ has eigenvalues $\frac{-1\pm\sqrt{5}}{2}$, which are not primitive roots of unity. Unless there is a typo in the question, and it should be $\frac{x-1}{x}$ rather than $\frac{1-x}{x}$. – Batominovski May 24 '17 at 12:36
  • @Batominovski. Please write complete.. thank you . – Almot1960 May 24 '17 at 12:46
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    What do you mean? I told you that your method should not work. Are you asking me to write a solution using a method I don't think it works? – Batominovski May 24 '17 at 12:48
  • @Batominovski I think he's trying to say use a different method. – Arbuja May 24 '17 at 13:13

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