functional equation $\dfrac {x + y + xy} 2 = f \big( f^{[-1]}(x) + f^{[-1]}(y) \big)$

Question

Consider

$$ \frac {x + y + xy} 2 = f \big( f^{[-1]}(x) + f^{[-1]}(y) \big) $$

Where $f^{[-1]}$ denotes the functional inverse of $f$.

How to find $f$?

How about the more General idea of finding $f$ for a given $g$?

$$ G(x,y) = f \big( f^{[-1]}(x) + f^{[-1]}(y) \big) $$

I assume partial derivatives could help. (But ?) Im looking for An easy way.

I assume using Taylor series?

How do I estimate upper and lower bounds for $f$? Or asymptotic ?

Answer 1

$(x + y + xy)/2 = f( f^{[-1]}(x) + f^{[-1]}(y) )\Rightarrow f^{[-1]}\left(\frac{x + y + xy}{2}\right) = f^{[-1]}(x) + f^{[-1]}(y) $

now assume $h(x)=f^{[-1]}(x)$

$h\left(\frac{x + y + xy}{2}\right) = h(x) + h(y)$ ----[1]

using partial differentiation with respect to $x$

$h'\left(\frac{x + y + xy}{2}\right).\left(\frac{1+y}{2}\right) = h'(x) $

put $x=0$

$h'\left(\frac{y }{2}\right).\left(\frac{1+y}{2}\right) = h'(0) $

$h'\left(\frac{y }{2}\right) = h'(0)\left(\frac{2}{1+y}\right) $

let $h'(0)=k$

$h'(x)=\frac{2k}{1+2x}$, now integrating with respect to $x$ both side will give us

$h(x)=k\ln(1+2x) +c$

by---[1] $h(0)=0\rightarrow c=0$

hence $h(x)=k\ln(1+2x)$ still we need one more given condition to find $k$

Answer 2

Substituting $ f ( x ) $ for $ x $ and $ f ( y ) $ for $ y $ in the original equation we get: $$ f ( x ) + f ( y ) + f ( x ) f ( y ) = 2 f ( x + y ) \tag 0 \label 0 $$ Letting $ x = 0 $ and $ y = 0 $ in \eqref{0} we have $ f ( 0 ) = 0 $. Now we put $ y = 0 $ in \eqref{0} and the result is $ f ( x ) = 0 $.

For the more general problem, the given equation is equivalent to $$ G \big( f ( x ) , f ( y ) \big) = f ( x + y ) \tag 1 \label 1 $$ Letting $ x = 0 $ and $ y = 0 $ in \eqref{1} you'll find an equation which may help you find $ f ( 0 ) $. Then you can let $ y = 0 $ in \eqref{1} and get an equation for $ f ( x ) $. Also, as the right-hand side of \eqref{1} is symmetric in $ x $ and $ y $, you can change the role of them and get $ G \big( f ( x ) , f ( y ) \big) = G \big( f ( y ) , f ( x ) \big) $, which may help if $ G $ is not symmetric in its arguments. If those were not enough, assuming differentiability for $ f $ and differentiability with respect to the second argument for $ G $, you can differentiate \eqref{1} with respect to $ y $ and get $ f ' ( y ) \cdot \partial_2 G \big( f ( x ) , f ( y ) \big) = f ' ( x + y ) $. Now letting $ y = 0 $ you'll find a differential equation which may help you find $ f ( x ) $. In the case that $ G $ is not symmetric in its arguments, you can do the same, changin the role of $ x $ and $ y $, and assuming the corresponding differentiability conditions. You can continue this way for higher order derivatives, and at some point you may find a solution for $ f $, depending on what $ G $ is.