Taylor's Theorem/One Variable/Proof by Cauchy Mean Value Theorem

Proof
Let $G$ be a real-valued function continuous on $\closedint a x$ and differentiable with non-vanishing derivative on $\openint a x$.

Let:
 * $\map F t = \map f t + \dfrac {\map {f'} t} {1!} \paren {x - t} + \dotsb + \dfrac {\map {f^{\paren n} } t} {n!} \paren {x - t}^n$

By the Cauchy Mean Value Theorem:
 * $(1): \quad \dfrac {\map {F'} x} {\map {G'} \xi} = \dfrac {\map F x - \map F a} {\map G x - \map G a}$

for some $\xi \in \openint a x$.

Note that the numerator:
 * $\map F x - \map F a = R_n$

is the remainder of the Taylor polynomial for $\map f x$.

On the other hand, computing $\map {F'} \xi$:
 * $\map {F'} \xi = \map {f'} \xi - \map {f'} \xi + \dfrac {\map {f} \xi} {1!} \paren {x - \xi} - \dfrac {\map {f} \xi} {1!} \paren {x - \xi} + \dotsb + \dfrac {\map {f^{\paren {n + 1} } } t} {n!} \paren {x - \xi}^n = \dfrac {\map {f^{\paren {n + 1} } } \xi} {n!} \paren {x - \xi}^n$

Putting these two facts together and rearranging the terms of $(1)$ yields:
 * $R_n = \dfrac {\map {f^{\paren {n + 1} } } \xi} {n!} \paren {x - \xi}^n \dfrac {\map G x - \map G a} {\map {G'} \xi}$

which was to be shown.

Note that the Lagrange Form of the Remainder comes from taking $\map G t = \paren {x - t}^{n + 1}$ and the given Cauchy Form of the Remainder comes from taking $\map G t = t - a$.