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Let $f : \mathbb{R}\setminus\{0\} \to \mathbb{R}$ be a non-constant function such that $f(y) + f\left(\frac{1}{y}\right) = 0$. I found that $f(y) = h(\log|y|)$ will be a solution , where $h$ is an odd function. Does any other solution exist ?

The need of this functional equation arose in order to get non-polynomial solutions of $g(x)g\left(\frac{1}{x}\right) = g(x) + g\left(\frac{1}{x}\right)$, assuming $g(x)$ to be nonzero I saw that $\frac{1}{g(x)} = h(\log|y|) + \frac{1}{1\pm y^n}$ would be a non-polynomial solution and I think any solution for the function $f$, as conditioned, would yield a non-polynomial solution for $g(x)$.

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    $\begingroup$ Let $f(\pm 1)=0$, let $f(x)$ be anything at all for $|x|>1$, and define $f(x)=-f(1/x)$ for $0<|x|<1$. $\endgroup$ – mjqxxxx Oct 10 '12 at 5:27
  • $\begingroup$ Your family of functions has the property $f(y)=f(-y)$. But our conditions make it possible to define the function differently for positive than for negative. For example, you could use $\log y$ for $y$ positive, $(\log(-y))^3$ for $y$ negative. $\endgroup$ – André Nicolas Oct 10 '12 at 5:54
  • $\begingroup$ In fact, mjqxxxx's family of solutions is the complete solution set to your functional equation. $\endgroup$ – Hurkyl Oct 10 '12 at 7:13
  • $\begingroup$ Probably $f(y)=\log y$ or $\ln y$. This is because: $f(y)+f\left(\frac1{y}\right)=\log y + \log \frac1{y}=\log y - \log y=0$ $\endgroup$ – Shahar May 1 '14 at 21:58
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EqWorld is our right partner.

In fact this functional equation belongs to the form of http://eqworld.ipmnet.ru/en/solutions/fe/fe1121.pdf.

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

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  • $\begingroup$ Many , many thanks for the links. $\endgroup$ – Souvik Dey Oct 10 '12 at 12:22
  • $\begingroup$ A misprint in the "general solution" given? $\endgroup$ – GEdgar Oct 10 '12 at 13:32

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