I've been set the following question:
Suppose that the real-valued function $f$ is such that the $(n-1)^{th}$ derivative of $f$ exists and is continuous on $[0,h]$ and that the $n^{th}$ derivative exists on $(0,h)$. Consider the function $G$ defined by
$$G(t) = F(t) - \left( \frac{h-t}{h} \right)^p F(0) $$
where $F$ is given by
$$F(t) = f(h) - f(t) - (h-t)f'(t) - \dots - \frac{(h-t)^{n-1}}{(n-1)!}f^{(n-1)}(t) $$ and $p$ is a constant. By considering the derivative of $G$ and choosing $p$ appropriately, prove that there exist $\theta_1 ,\theta_2 $ such that
$$f(h) = f(0) + hf'(0) + \dots + \frac{h^{n-1}}{(n-1)!}f^{(n-1)}(0) + S_n $$
where
$$ S_n = \frac{h^n}{n!}f^{(n)}(\theta_1 h) = \frac{h^n}{(n-1)!}(1-\theta_2)^{n-1}f^{(n)} (\theta_2 h)$$
I've most recently been introduced to Taylor's theorem with the Lagrange form of the remainder, and I think I need to show that $F(0) = S_n$, but I'm really not sure how to do that.
I'd appreciate any help anyone could offer me.
You should be able to show that $S_n = \frac{h^n}{n!}f^{(n)}(\theta_1 h)$ by just applying Taylor's theorem directly about $0$ to get $$f(h) = f(0) + hf'(0) + \dots + \frac{h^{n-1}}{(n-1)!}f^{(n-1)}(0) + \frac{h^n}{n!}f^{(n)}(\xi)$$
for some $\xi$, and then set $\theta _1 = \frac{\xi}{h}$ since h is fixed.
I'm not sure how to approach showing that the other form given exists, so hopefully someone else can chime in to help with that.