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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.

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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.

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    $\begingroup$ 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 $F(1) = 0$. $\endgroup$ Apr 10, 2020 at 9:45
  • $\begingroup$ I really like the proof :). But is there anyway to proceed from where I left without construction of such auxiliary function. $\endgroup$
    – manifold
    Apr 10, 2020 at 10:52
  • $\begingroup$ Or how do people come up with such functions? $\endgroup$
    – manifold
    Apr 10, 2020 at 10:59
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    $\begingroup$ 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 $F(b) =0$. $\endgroup$
    – littleO
    Apr 10, 2020 at 11:22
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    $\begingroup$ Also, if a proof this simple and elegant is out there waiting to be discovered, and hundreds of mathematicians scour this realm of ideas obsessively, they will inevitably stumble over the proof eventually. $\endgroup$
    – littleO
    Apr 10, 2020 at 11:23

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