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Which function verified that: $f(x)+f(\frac1x)=\frac1a; a$-constant value?

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Does f have to be continious? – Amr Nov 9 '12 at 9:48

Let $g \colon (1, \infty) \to \mathbb R$ be any function. Define $f \colon (0, \infty) \to \mathbb R$ by \[ f(x) = \begin{cases} g(x) & x > 1\\[2mm] \frac 1{2a} & x = 1\\[2mm] \frac 1a -g\left(\frac 1x\right) & x < 1 \end{cases} \] Then $f$ is as wished.

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$f(x)=\frac{1}{2a}+\log{x^n}$ is a solution.

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EqWorld is our right partner.

In fact this functional equation belongs to the form of

The general solution is $f(x)=C\left(x~,\dfrac{1}{x}\right)+\dfrac{1}{2a}$ , where $C(u,v)$ is any antisymmetric function.

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