Taylor's Theorem/One Variable/Proof by Rolle's Theorem

Proof
Let the function $g$ be defined as:
 * $\map g t = \map {R_n} t - \dfrac {\paren {t - a}^{n + 1} } {\paren {x - a}^{n + 1} } \map {R_n} x$

Then:
 * $\map {g^{\paren k} } a = 0$

for $k = 0, \dotsc, n$, and $\map g x = 0$.

Apply Rolle's Theorem successively to $g, g', \dotsc, g^{\paren n}$.

Then there exist:
 * $\xi_1, \ldots, \xi_{n + 1}$

between $a$ and $x$ such that:
 * $\map {g'} {\xi_1} = 0, \map {g''} {\xi_2} = 0, \ldots, \map {g^{\paren {n + 1} } } {\xi_{n + 1} } = 0$

Let $\xi = \xi_{n + 1}$.

Then:
 * $0 = \map {g^{\paren {n + 1} } } \xi = \map {f^{\paren {n + 1} } } \xi - \dfrac {\paren {n + 1}!} {\paren {x - a}^{n + 1} } \map {R_n} x$

and the formula for $\map {R_n} x$ follows.