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   <subfield code="a">Second order iterative functional equations related to a competition equation</subfield>
   <subfield code="h">[Elektronische Daten]</subfield>
   <subfield code="c">[Peter Kahlig, Janusz Matkowski]</subfield>
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   <subfield code="a">The functional equation related to competition ([2]) $$f\left( \frac{x+y}{1-xy}\right) =\frac{f\left( x\right) +f\left(y\right)} {1+f\left( x\right) f\left( y\right)},\qquad x,y\in\mathbb{R}, xy\neq 1,$$ f x + y 1 - x y = f x + f y 1 + f x f y , x , y ∈ R , x y ≠ 1 , for y=cx with a fixed c&gt;0, leads to the equation $$f\left( \frac{\left( 1+c\right) x}{1-cx^{2}}\right) =\frac{f\left(x\right) +f\left( cx\right)} {1+f\left( x\right) f\left( cx\right)},\qquad x\in \mathbb{R}, \left\vert x \right\vert &lt;\frac{1}{\sqrt{c}}.$$ f 1 + c x 1 - c x 2 = f x + f c x 1 + f x f c x , x ∈ R , x &lt; 1 c . The case c=1 (a first order iterative functional equation) was treated in [3]. In this paper we consider the case c≠ 1 (when the equation is of the second order). We show that a function $${f:\mathbb{R} \rightarrow \mathbb{R},\,f\left( 0\right) =0}$$ f : R → R , f 0 = 0 , differentiable at the point 0 satisfies this functional equation iff there is a real p such that $${f=\tanh \circ \left( p\tan ^{-1} \right) }$$ f = tanh ∘ p tan - 1 which extends the main result of [3].</subfield>
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