Group/Examples/inv x = 1 - x/Lemma 2

Lemma for Group Example: $x^{-1} = 1 - x$

Define $f: \openint 0 1 \to \R$ by:

$\map f x := \map \ln {\dfrac {1 - x} x}$

and $g: \R \to \openint 0 1$:

$\map g z := \dfrac 1 {1 + \exp z}$

Then:

$\map {g \circ f} x = x$

$\ds \map {g \circ f} x$

$=$

$\ds \map g {\map f x}$

$\ds $

$=$

$\ds \frac 1 {1 + \exp \map f x}$

$\ds $

$=$

$\ds \frac 1 {1 + \map \exp {\ln \dfrac {1 - x} x} }$

$\ds $

$=$

$\ds \frac 1 {1 + \dfrac {1 - x} x}$

$\ds $

$=$

$\ds \frac x {x + 1 - x}$

multiplying both numerator and denominator by $x$

$\ds $

$=$

$\ds x$

simplifying

$\Box$