This involves a proof to Taylor's Theorem. Assume $\displaystyle x >0$ and let $\displaystyle f$ be defined on $\displaystyle (a,b)$ where $\displaystyle a < 0 < b$ and suppose the n-th derivative $\displaystyle f^{(n)}$ exists on $\displaystyle (a,b)$. Then for each nonzero $\displaystyle x\in (a,b)$ there is some $\displaystyle y$ between $\displaystyle 0$ and $\displaystyle x$ such that $\displaystyle R_{n}(x)=\frac{f^{(n)}(y)}{n!}x^{n}$.

Let $\displaystyle M$ be the unique solution of $\displaystyle f(x)= \sum_{k=0}^{n-1}\frac{f^{(k)}(0)}{k!}x^{k} + \frac{Mx^{n}}{n!}$ and let $\displaystyle F(t) = f(t) + \sum_{k=0}^{n-1}\frac{(x-t)^{k}}{k!}f^{(k)}(t) + M\frac{(x-t)^{n}}{n!}$

Show that $\displaystyle F$ is differentiable on $\displaystyle [0,x]$ (I think this part is straightforward since $\displaystyle f$ is differentiable on $\displaystyle (a,b)$ and the other terms are simply polynomials). Show that $\displaystyle F'(t)= \frac{(x-t)^{n-1}}{(n-1)!}[f^{(n)}(t) - M]$ (not really sure how this works out, not great with combining series), show that $\displaystyle F(0)=F(x)$ (again, don't know), and finally show that there is $\displaystyle y\in (0,x)$ such that $\displaystyle f^{(n)}(y) = M$.

My professor showed an alternate proof in class which seems at least somewhat easier, but I am absolutely lost here. Thanks.