# Proving Taylor's theorem up to 2nd derivative

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 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 $$f(1) = f(0) + f'(0) + \frac{f''(c_2)}{2}.$$
• You can even abstract out a "2nd order Rolles" (or any order) which gives Taylor's theorem after adding a correct polynomial, as in this blog post by Gowers gowers.wordpress.com/2014/02/11/… . The higher order Rolles gives some motivation for why you might want to set $F(1) = 0$. Apr 10, 2020 at 9:45
• I also don't feel totally comfortable with the introduction of $F$. But here's one comment: in the standard proof of the mean value theorem, we subtract a linear function from $f$ to obtain a function $F$ that satisfies $F(a) = 0$ and $F(b) = 0$. If we're already comfortable with that, then it is not a huge leap to try subtracting a quadratic function from $f$ to obtain a function $F$ that satisfies $F(a) = F'(a) = 0$ and $F(b) =0$. Apr 10, 2020 at 11:22