Aequationes mathematicae

, Volume 89, Issue 1, pp 107–117

# Second order iterative functional equations related to a competition equation

• Peter Kahlig
• Janusz Matkowski
Open Access
Article

## Abstract

The functional equation related to competition ([2])
$$f\left( \frac{x+y}{1-xy}\right) =\frac{f\left( x\right) +f\left(y\right)} {1+f\left( x\right) f\left( y\right)},\qquad x,y\in\mathbb{R}, xy\neq 1,$$
for y = cx with a fixed c > 0, leads to the equation
$$f\left( \frac{\left( 1+c\right) x}{1-cx^{2}}\right) =\frac{f\left(x\right) +f\left( cx\right)} {1+f\left( x\right) f\left( cx\right)},\qquad x\in \mathbb{R}, \left\vert x \right\vert <\frac{1}{\sqrt{c}}.$$
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 $${f:\mathbb{R} \rightarrow \mathbb{R},\,f\left( 0\right) =0}$$, differentiable at the point 0 satisfies this functional equation iff there is a real p such that $${f=\tanh \circ \left( p\tan ^{-1} \right) }$$ which extends the main result of [3].

## Mathematics Subject Classification

Primary 39B12 39B22

## Keywords

Functional equation competition equation iterative functional equation of the second order differentiable solution solution depending on an arbitrary function

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