Differentiation proof

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

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

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

Hint: consider the function $\displaystyle \phi$ given by

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

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

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

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

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

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

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

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

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

I would probably do something like this:

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

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

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

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

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

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

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

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

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

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