# Differentiation proof

• May 9th 2009, 04:11 AM
Showcase_22
Differentiation proof
Quote:

Suppose $f: \mathbb{R} \rightarrow \mathbb{R}$ is twwice differentiable and $a \in \mathbb{R}$. Prove that $\forall \ h>0 \ \exists \ \xi \in(a-h,a+h)$ with

$f(a+h)-2f(a)+f(a-h)=h^2f''(\xi)$

Hint: consider the function $\phi$ given by

$\phi(t)=f(a+t)-2f(a)+f(a-t)-\left( \frac{t}{h} \right)^2(f(a+h)-2f(a)+f(a-h))$.
I decided to use the mean value theorem.

Let $\xi_1 \in(0,h]$ and $\xi_2 \in [-h,0)$.
Then $\xi=\xi_1 \cup \xi_2$.
(this is done so that $\xi \neq 0$).

MVT gives: $\phi'(t_0)=\frac{f(a+h)+f(a-h)-f(a+h)-f(a-h)-f(a)+2f(a)-f(a)}{h}=0$ using the interval $(0,h]$.

I have no idea why this helps me (if it does!).
• May 9th 2009, 05:36 PM
hkito
Hi

what does "Then $\xi=\xi_1 \cup \xi_2$" means?
What is $t_0$?
• May 9th 2009, 11:14 PM
redsoxfan325
Quote:

Originally Posted by Showcase_22
I decided to use the mean value theorem.

Let $\xi_1 \in(0,h]$ and $\xi_2 \in [-h,0)$.
Then $\xi=\xi_1 \cup \xi_2$.
(this is done so that $\xi \neq 0$).

MVT gives: $\phi'(t_0)=\frac{f(a+h)+f(a-h)-f(a+h)-f(a-h)-f(a)+2f(a)-f(a)}{h}=0$ using the interval $(0,h]$.

I have no idea why this helps me (if it does!).

I would probably do something like this:

We want to show that $\frac{f(a+h)+f(a-h)-2f(a)}{h^2}=f''(\xi)$, for some $\xi\in[a-h,a+h]$

Factor out $h$ and break up the remaining fraction:

$\frac{1}{h}\left(\frac{f(a+h)-f(a)}{h}-\frac{f(a)-f(a-h)}{h}\right)$

By the MVT, we now have $\frac{1}{h}[f'(t_2)-f'(t_1)]$, for some $t_1,t_2\in[a-h,a+h]$.

Distribute the $\frac{1}{h}$ through to get $\frac{f'(t_2)-f'(t_1)}{h}$.

Using the MVT again achieves the desired result.
• May 10th 2009, 01:11 AM
Showcase_22
but then can't $\xi=0$?

Either way, that's a pretty good proof!
• May 10th 2009, 09:23 AM
redsoxfan325
Why can't $\xi=0$? You're not dividing by it.

Also, another interesting fact that follows from above is that if you take

$\lim_{h\to 0}\frac{f(a+h)+f(a-h)-2f(a)}{h^2}$

you get out $f''(a)$. The reason is that we know $\xi\in[a-h,a+h]$, so as $h\to 0$, then $[a-h,a+h]\to[a,a]$, and if $\xi\in[a,a]$, then $\xi=a$.
• May 10th 2009, 09:28 AM
Showcase_22
I have a "comments on the test" sheet where one of my lecturers has written:

Quote:

"..and in the problem apply MVT to $\phi$ on the interval $[-h,h]$ which gives a $\xi$ which could turn out to be 0. The latter can lead to complications and it is easier to use the interval $[0,h]$".
I'm going to ask my supervisor on Tuesday why $\xi \neq 0$ and exactly what complications arise if you do. I expected that when I applied MVT to $\phi$ some reason would reveal itself. Unfortunately this hasn't happened!