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

Proof
Let $G$ be a real-valued function continuous on $\left[{a \,.\,.\, x}\right]$ and differentiable with non-vanishing derivative on $\left({a \,.\,.\, x}\right)$.

Let:
 * $F \left({t}\right) = f \left({t}\right) + \dfrac {f' \left({t}\right)} {1!} \left({x - t}\right) + \dotsb + \dfrac {f^{\left({n}\right)} \left({t}\right)} {n!} \left({x - t}\right)^n$

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

for some $\xi \in \left({a \,.\,.\, x}\right)$.

Note that the numerator:
 * $F \left({x}\right) - F \left({a}\right) = R_n$

is the remainder of the Taylor polynomial for $f \left({x}\right)$.

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

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

which was to be shown.

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