Group/Examples/inv x = 1 - x/Lemma 2
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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:
- $\map {g \circ f} x = x$
Proof
\(\ds \map {g \circ f} x\) | \(=\) | \(\ds \map g {\map f x}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 {1 + \exp \map f x}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 {1 + \map \exp {\ln \dfrac {1 - x} x} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 {1 + \dfrac {1 - x} x}\) | Exponential of Natural Logarithm | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac x {x + 1 - x}\) | multiplying both numerator and denominator by $x$ | |||||||||||
\(\ds \) | \(=\) | \(\ds x\) | simplifying |
$\Box$