If f''(a) exists then f''(a) = ...

1. Jun 5, 2015

Euklidian-Space

1. The problem statement, all variables and given/known data
Let $I \subset \mathbb{R}$ be an open interval and let $f:I \rightarrow \mathbb{R}$ be differentiable on I. Show that if $f''(a)$ exists, then $f''(a) = \lim_{h \rightarrow 0}\frac{f(a + h) - 2f(a) + f(a - h)}{h^2}$

2. Relevant equations

3. The attempt at a solution

I tried tackling this by using limit definition of the derivative for $f''(a)$.

$$f''(a) = \lim_{h \rightarrow 0}\frac{f'(a + h) - f'(a)}{h}$$

and then maybe use limit definition again for the terms in the numerator above? maybe use L'hopitals's rule? Not sure

2. Jun 5, 2015

DEvens

Can you then write down the definition in terms of limits of $f'(a + h)$ and $f'(a)$ ?

3. Jun 5, 2015

pasmith

Use Taylor's theorem: If $f$ is twice differentiable at $a$ then $$f(a + h) = f(a) + hf'(a) + \frac12 h^2f''(a) + \epsilon(h)$$ where $$\lim_{h \to 0} \frac{\epsilon(h)}{h^2} = 0.$$

4. Jun 5, 2015

momoko

Using the limit definition of differentiation in the numerator and sone manipulation will give you the answer.

5. Jun 5, 2015

Euklidian-Space

Ok so the limit def of diff for the following

$$f'(a) = \lim_{h \rightarrow 0} \frac{f(a + h) - f(a)}{h}$$

$$f'(a + h) = \lim_{h \rightarrow 0} \frac{f(a + 2h) - f(a + h)}{h}$$

The def of f'(a + h) I am not too sure about.

Last edited: Jun 5, 2015
6. Jun 5, 2015

pasmith

$$f'(a + h) = \lim_{k \to 0} \frac{f(a + h + k) - f(a + h)}{k}.$$

7. Jun 5, 2015

Euklidian-Space

pasmith, How do you deal with that index of k when we are trying to simplify the numerator of

$$f''(a) = \lim_{h \rightarrow 0}\frac{f'(a + h) - f'(a)}{h}$$

8. Jun 5, 2015

pasmith

I don't; I would instead use the Taylor theorem approach I suggested in my earlier post.

9. Jun 5, 2015

Noctisdark

No need to use taylor, just "fuse" limits, using your notation, it is clear that f'(a+h) = lim h->0 f(a+2h)-f(a+h) / h and the same for f'(a), and write all the limits as one, show me you attempt so far !!

10. Jun 5, 2015

pasmith

False: $$\lim_{h \to 0} \frac{f(a + 2h) - f(a + h)}h = \lim_{h \to 0} \frac{f(a) + 2hf'(a) - f(a) - hf'(a)}h = f'(a).$$

11. Jun 5, 2015

SammyS

Staff Emeritus
What happens if you put these into your expression for $\ f''(a)\ ?$

You get something pretty close to what you're looking for.

Since the limit exists, it is the same as each of its one-sided limits. So, isn't it true that $\displaystyle \ f''(a) = \lim_{h\to 0}\frac{\,f'(a - h) - f'(a)\,}{-h}\ ?$

12. Jun 5, 2015

Noctisdark

Not false, of course lim h-> 0 f'(a+h) = f'(a), f is continous and free if corners, you haven't prooved or denied anything !

13. Jun 5, 2015

Euklidian-Space

When i put those into my expression for f''(a) i get this...

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

what on earth am I suppose to do with that monster?

14. Jun 5, 2015

Dick

You are given that $f$ is differentiable on an open interval. That gives you the right to apply l'Hopital once if you can show it's a 0/0 form. THEN apply definitions of differentiability.

Last edited: Jun 5, 2015
15. Jun 6, 2015

vela

Staff Emeritus
You really should be using two different variables for the limits.
$$\lim_{h_1\to 0} \frac{\lim_{h_2\to 0} \frac{[f(a + h_1 + h_2) - f(a + h_1)] - [f(a+h_2)-f(a)]}{h_2}}{h_1}.$$ The question you've run into now about how to combine the two limits into one is probably why pasmith suggested the other method using Taylor's theorem. I don't offhand see how to justify combining the limits.

16. Jun 7, 2015

Euklidian-Space

it seems even with the taylor method you get into the same problem.

$$f(a + h) = f(a) + hf'(a) + \frac{1}{2} h^2f''(a) + \epsilon (h)$$

solve for f''(a) and get

$$f''(a) = \frac{2f(a + h) - 2f(a) - 2hf'(a) - 2\epsilon (h)}{h^2}$$
so i guess at this point you would take limit of both sides, but what do you do about the 2hf'(a) in the numerator?

17. Jun 8, 2015

DEvens

How do you know Taylor's theorem is correct? Hint: You need the limit that the original post requires to be proved.

18. Jun 8, 2015

Euklidian-Space

DEvens, would you use the limit definition for differentiation to prove this? If so how would you resolve the different limit variables?

19. Jun 8, 2015

vela

Staff Emeritus
Hint: Consider both f(a+h) and f(a-h).

20. Jun 8, 2015

pasmith

I only need that $$\lim_{h \to 0} \frac{\epsilon(h)}{h^2} = \lim_{h \to 0} \frac{f(a + h) - f(a) - hf'(a) - \frac12h^2f''(a)}{h^2} \\ = \lim_{h \to 0} \frac{f'(a + h) - f'(a) - hf''(a)}{2h} = \frac{f''(a) - f''(a)}{2} = 0$$ using l'Hopital's Theorem and the definition of the second derivative. The OP's limit then follows.

(l'Hopital's Theorem would also work to establish the OP's limit directly, once one has proven that $$\lim_{h \to 0} \frac{f(x + h) - f(x - h)}{2h} = f'(x).)$$