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

Proof
Let the function $g$ be defined by:
 * $\displaystyle g \left({t}\right) = R_n \left({t}\right) - \frac{(t-a)^{n+1}}{(x-a)^{n+1}}R_n(x)$

Then:
 * $g^{\left({k}\right)} \left({a}\right)=0$ for $k=0, \ldots, n$, and $g \left({x}\right) = 0$.

Apply Rolle's Theorem successively to $g$, to $g'$, and so until $g^{\left({n}\right)}$.

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

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

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

Then:
 * $\displaystyle 0 = g^{\left({n+1}\right)} \left({\xi}\right) = f^{\left({n+1}\right)} \left({\xi}\right) - \frac{\left({n+1}\right)!}{(x-a)^{n+1}} R_n \left({x}\right)$

and the formula for $R_n \left({x}\right)$ follows.