ThmDex – An index of mathematical definitions, results, and conjectures.
Formulation 0
Let $x \mapsto e^x$ be the D1932: Standard natural real exponential function.
Let $a, b, t \in \mathbb{R}$ each be a D993: Real number such that
(i) \begin{equation} b < b \end{equation}
(ii) \begin{equation} u : = \lambda (b - a) \end{equation}
(iii) \begin{equation} \lambda : = - \frac{a}{b - a} \end{equation}
(iv) \begin{equation} g(u) : = - \lambda u + \log(1 - \lambda + \lambda e^u) \end{equation}
Then \begin{equation} e^{g(u)} = - \frac{a}{b - a} e^{b t} + \frac{b}{b - a} e^{t a} \end{equation}
Proofs
Proof 0
Let $x \mapsto e^x$ be the D1932: Standard natural real exponential function.
Let $a, b, t \in \mathbb{R}$ each be a D993: Real number such that
(i) \begin{equation} b < b \end{equation}
(ii) \begin{equation} u : = \lambda (b - a) \end{equation}
(iii) \begin{equation} \lambda : = - \frac{a}{b - a} \end{equation}
(iv) \begin{equation} g(u) : = - \lambda u + \log(1 - \lambda + \lambda e^u) \end{equation}
Substituing in the definitions, we have \begin{equation} \begin{split} e^{g(u)} & = e^{- \lambda u + \log (1 - \lambda + \lambda e^u)} \\ & = e^{- \lambda u} (1 - \lambda + \lambda e^u) \\ & = (1 - \lambda)e^{- \lambda u} + \lambda e^{(1 - \lambda) u} \\ & = (1 - \lambda)e^{- \lambda t (b - a)} + \lambda e^{(1 - \lambda) t (b - a)} \\ & = \frac{b}{b - a} e^{ \frac{a}{b - a} t (b - a)} - \frac{a}{b - a} e^{\frac{b}{b - a} t (b - a)} \\ & = \frac{b}{b - a} e^{a t} - \frac{a}{b - a} e^{b t} \\ \end{split} \end{equation} $\square$