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:
 * $\displaystyle F \left({t}\right) = f \left({t}\right) + \frac{f' \left({t}\right)}{1!}(x-t) + \cdots + \frac{f^{(n)} \left({t}\right)}{n!}(x-t)^n$

By the Cauchy Mean Value Theorem:
 * $\displaystyle \frac{F'(\xi)}{G'(\xi)} = \frac{F(x) - F(a)}{G(x) - G(a)} \qquad (1)$

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

Note that the numerator $F(x)-F(a)=R_n$ is the remainder of the Taylor polynomial for $f(x)$.

On the other hand, computing $F^{\prime} (\xi)$:
 * $\displaystyle F'(\xi) = f'(\xi) - f'(\xi) + \frac{f(\xi)}{1!}(x-\xi) - \frac{f(\xi)}{1!}(x-\xi) + \cdots + \frac{f^{(n+1)}(t)}{n!}(x-\xi)^n = \frac{f^{(n+1)}(\xi)}{n!}(x-\xi)^n$

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

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 $G \left({t}\right) = t - a$.