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