If $f$ is a twice differentiable function on $[0,1]$ then prove that $f(1) = f(0)+ f'(0)+\frac{1}{2} f''(\xi)$ for some $0<\xi<1$.
So essentially we have to show that $f(1) - f(0) - f'(0) = \frac{1}{2} f''(\xi)$ for some $\xi$. My attempt for above is as follows:
$f(1) - f(0) - f'(0) = f'(t_1) - f'(0) = f''(t_2)\cdot t_1$ for some $0<t_2<t_1<1$ by using the Mean Value Theorem twice. I am stuck there, can someone help me out from here.
Define $F:[0,1] \to \mathbb R$ by $$F(x) = f(x) - f(0) - f'(0)x - M x^2,$$ where $M$ is chosen so that $F(1) = 0$. Note that $F(0) = F'(0) = 0$. Our choice of $M$ has set us up to use Rolle's theorem, which implies that there exists $c_1 \in (0,1)$ such that $F'(c_1) = 0$. Again by Rolle's theorem, there exists $c_2 \in (0, c_1)$ such that $F''(c_2) =0$. But by differentiating $F$ we see that $F''(c_2) = f''(c_2) -2M$. It follows that $M = \frac{f''(c_2)}{2}$. Because $F(1) = 0$, we have $$ f(1) = f(0) + f'(0) + \frac{f''(c_2)}{2}. $$
I learned this elegant proof of Taylor's theorem with remainder from the book Mathematical Analysis by Browder. I just spent about an hour remembering / figuring out how the proof works. The proof is a simple and beautiful generalization of the standard proof of the mean value theorem.