I'd like to solve functional equation: $f(x+y)=\frac{f(x)+f(y)}{1-f(x)f(y)}.$
I've managed to get: $f(0)=0,f(n)=0$ for all $n\in N$; $f(\frac{1}{2})=0$; $f(-x)=-f(x)$. I'll be grateful for any help.
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3Hint: think about $\tan(x+y)$. – Anurag A Apr 24 '16 at 17:37
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1As per the above comment, if $k\ne 0$, and if the domain of $f $ is $R$ minus the set of odd multiples of $\pi /2 k$, then $\tan k x$ is a solution, so I don't see how you can get $f(1)=0$. Is the domain $R$ ? And how did you get $f(1)=0$? Are you assuming $f$ is continuous? – DanielWainfleet Apr 24 '16 at 18:00
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Without loss of generality, we may look for $f$ in the form of $\tan(g(x))$. Then $g(x+y)=g(x)+g(y)+\pi n$, which looks less complicated. – Ivan Neretin May 01 '16 at 09:04
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You have to careful in these kind of problems. For example, if we assume $\;f(x):=k\;$ for some constant $k$, then $\;k=0\;$ or $\;k^2=-1\;.$ If we assume $\;f(x):=\sum_{n=0}^\infty c_nx^n\;$ then solving for the coefficients gives $\;f(x)=\tan(cx)\;$ for some constant $c.\;$ If $f$ is not assumed to be continuous then there are many "bad" solutions similar to Cauchy's functional equation. See also MSE question 423492 for equations related to Cauchy's.
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