1. Oct 26, 2008

### TaylorWatts

Suppose f is defined in a neighborhood of x, and suppose f '' (x) exists. Show that:

lim [f(x+h)+f(x-h)-2f(x)] / h^2 = f''(x).
h->0

Show by an example that that the limit may exist even if f '' (x) may not. (hint: use lHopital's Theorem).

Proof:

f '' (x) exists implies f ' (x) exists and is differentiable which implies f(x) exists and is differentiable.

Thus,

f '' (x) =

lim [f ' (x) - f ' (t)] / (x-t) (def of derivative)
x->t

Doing a substitution h = x - t

lim [f ' (x+h) - f ' (x)] / h (def of derivative)
h->0

= lim [f(x+h) - f(x) - (f(x) - f(x-h)] / h / h
h->0

= lim [f(x+h) + f(x-h) - 2f(x)] / h^2 (*)
h->0

QED

What I'm having trouble with is the example of where the limit exists but f '' (x) does not. Either using the original limit/derivative definition or the end of the proof (*) I get 0/0 or another indeterminate form.

2. Oct 26, 2008

### HallsofIvy

Staff Emeritus
The very simplest function I know that does not have a derivative at one point is |x|, which is not differentiable at x=0. The anti-derivative of that, i.e. f(x) such that f'(x)= |x|, does not have a second derivative at x= 0. Will that work?

3. Oct 26, 2008

### TaylorWatts

I tried both absolute value of x and square root of x and still get 0/0 or if I apply l'Hopital's Rule I get infinity. Which also means the limit does not exist.

4. Oct 26, 2008

### HallsofIvy

Staff Emeritus
So you have not tried what I suggested? Why not?

5. Oct 26, 2008

### TaylorWatts

I did.

The problem is if f ' (x) = |x| then f '' (0) does not exist, but neither does:

lim [f (0 + h) + f (0 - h) - 2 f (0)] / h^2, unless I am missing something here???
h->0

6. Oct 26, 2008

### HallsofIvy

Staff Emeritus
That limit does indeed exist. What is f(x) in this case?

7. Oct 26, 2008

### TaylorWatts

I see my mistake.

Thanks!