Differentiation proof

**Showcase_22** · May 9th 2009, 04:11 AM

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!).

**hkito** · May 9th 2009, 05:36 PM

Hi

what does "Then $\xi=\xi_1 \cup \xi_2$ " means?
What is $t_0$ ?

**redsoxfan325** · May 9th 2009, 11:14 PM

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.

**Showcase_22** · May 10th 2009, 01:11 AM

but then can't $\xi=0$ ?

Either way, that's a pretty good proof!

**redsoxfan325** · May 10th 2009, 09:23 AM

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 $\xi=a$ .

**Showcase_22** · May 10th 2009, 09:28 AM

I have a "comments on the test" sheet where one of my lecturers has written:

"..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!

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