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

[Euklidian-Space]

Homework Statement

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

Homework Equations

The Attempt at a Solution

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

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

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

 

[DEvens]

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

 

[pasmith]

Use Taylor's theorem: If f is twice differentiable at a then <br /> f(a + h) = f(a) + hf&#039;(a) + \frac12 h^2f&#039;&#039;(a) + \epsilon(h) where <br /> \lim_{h \to 0} \frac{\epsilon(h)}{h^2} = 0.

 

[momoko]

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

 

[Euklidian-Space]

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

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.

 

[pasmith]

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.

<br /> f&#039;(a + h) = \lim_{k \to 0} \frac{f(a + h + k) - f(a + h)}{k}.<br />

 

[Euklidian-Space]

<br /> f&#039;(a + h) = \lim_{k \to 0} \frac{f(a + h + k) - f(a + h)}{k}.<br />

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


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

 

[pasmith]

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


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

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

 

[Noctisdark]

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

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 !

 

[pasmith]

it is clear that f'(a+h) = lim h->0 f(a+2h)-f(a+h) / h

False: \lim_{h \to 0} \frac{f(a + 2h) - f(a + h)}h = \lim_{h \to 0} \frac{f(a) + 2hf&#039;(a) - f(a) - hf&#039;(a)}h = f&#039;(a).

 

[SammyS]

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.

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 $\displaystyle \ f''(a) = \lim_{h\to 0}\frac{\,f'(a - h) - f'(a)\,}{-h}\ ?$

 

[Noctisdark]

False: \lim_{h \to 0} \frac{f(a + 2h) - f(a + h)}h = \lim_{h \to 0} \frac{f(a) + 2hf&#039;(a) - f(a) - hf&#039;(a)}h = f&#039;(a).

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

 

[Euklidian-Space]

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 $\displaystyle \ f''(a) = \lim_{h\to 0}\frac{\,f'(a - h) - f'(a)\,}{-h}\ ?$

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?

 

[Dick]

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?

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.

 

[vela]

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?

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.

 

[Euklidian-Space]

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.

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?

 

[DEvens]

Use Taylor's theorem: If f is twice differentiable at a then <br /> f(a + h) = f(a) + hf&#039;(a) + \frac12 h^2f&#039;&#039;(a) + \epsilon(h) where <br /> \lim_{h \to 0} \frac{\epsilon(h)}{h^2} = 0.

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

 

[Euklidian-Space]

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

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

 

[vela]

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?

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

 

[pasmith]

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

I only need that \lim_{h \to 0} \frac{\epsilon(h)}{h^2} = \lim_{h \to 0} \frac{f(a + h) - f(a) - hf&#039;(a) - \frac12h^2f&#039;&#039;(a)}{h^2} \\<br /> = \lim_{h \to 0} \frac{f&#039;(a + h) - f&#039;(a) - hf&#039;&#039;(a)}{2h} = \frac{f&#039;&#039;(a) - f&#039;&#039;(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 <br /> \lim_{h \to 0} \frac{f(x + h) - f(x - h)}{2h} = f&#039;(x).)

 

[vela]

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

If you prefer not to use Taylor's theorem, follow Dick's suggestion and apply L'Hopital's rule to
$$\lim_{h \to 0} \frac{f(a+h)-2f(a)+f(a-h)}{h^2}.$$

 

[Euklidian-Space]

If you prefer not to use Taylor's theorem, follow Dick's suggestion and apply L'Hopital's rule to
$$\lim_{h \to 0} \frac{f(a+h)-2f(a)+f(a-h)}{h^2}.$$

ok when i apply l'hopital's i get the following

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

we differentiate the numerator and denominator with respect to the limit variable right? doesn't seem like it is giving us $f''(a) = \lim_{h \to 0} \frac{f''(a + h) - f''(a)}{h}$

 

[vela]

What can you do to the numerator so that it contains combinations that look like the numerator in the definition of a derivative?

 

[Euklidian-Space]

Well I tried applying lhopitals again. But I get $\lim_{n\to0} \frac{f"(a+h) - f"(a-h)}{2}$. Which seems to be a little closer, but not quite.

 

[pasmith]

Well I tried applying lhopitals again. But I get $\lim_{h \to 0} \frac{f"(a+h) - f"(a-h)}{2}$. Which seems to be a little closer, but not quite.

The derivative of f&#039;(a - h) with respect to h is -f&#039;&#039;(a - h), so you should have <br /> \lim_{h \to 0} \frac{f&#039;&#039;(a + h) + f&#039;&#039;(a - h)}{2}. Now if you knew that f'' was continuous at a you would be done, but you don't. Also, we are only assuming that f is twice differentiable at a, so it is entirely possible that f&#039;&#039;(a + h) does not exist for any h \neq 0. So l'Hopital's rule can only take you as far as <br /> \lim_{h \to 0} \frac{f&#039;(a + h) - f&#039;(a - h)}{2h}.

 

[Euklidian-Space]

The derivative of f&#039;(a - h) with respect to h is -f&#039;&#039;(a - h), so you should have <br /> \lim_{h \to 0} \frac{f&#039;&#039;(a + h) + f&#039;&#039;(a - h)}{2}. Now if you knew that f'' was continuous at a you would be done, but you don't. Also, we are only assuming that f is twice differentiable at a, so it is entirely possible that f&#039;&#039;(a + h) does not exist for any h \neq 0. So l'Hopital's rule can only take you as far as <br /> \lim_{h \to 0} \frac{f&#039;(a + h) - f&#039;(a - h)}{2h}.

oh i think I see now

 

[Euklidian-Space]

I only need that \lim_{h \to 0} \frac{\epsilon(h)}{h^2} = \lim_{h \to 0} \frac{f(a + h) - f(a) - hf&#039;(a) - \frac12h^2f&#039;&#039;(a)}{h^2} \\<br /> = \lim_{h \to 0} \frac{f&#039;(a + h) - f&#039;(a) - hf&#039;&#039;(a)}{2h} = \frac{f&#039;&#039;(a) - f&#039;&#039;(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 <br /> \lim_{h \to 0} \frac{f(x + h) - f(x - h)}{2h} = f&#039;(x).)

wait how do you get
\lim_{h \to 0} \frac{f&#039;(a + h) - f&#039;(a) - hf&#039;&#039;(a)}{2h} = \frac{f&#039;&#039;(a) - f&#039;&#039;(a)}{2} = 0

 

[pasmith]

wait how do you get
\lim_{h \to 0} \frac{f&#039;(a + h) - f&#039;(a) - hf&#039;&#039;(a)}{2h} = \frac{f&#039;&#039;(a) - f&#039;&#039;(a)}{2} = 0

\lim_{h \to 0} \frac{f&#039;(a + h) - f&#039;(a) - hf&#039;&#039;(a)}{2h} = \frac12 \lim_{h \to 0} \frac{f&#039;(a + h) - f&#039;(a)}{h} - \frac{f&#039;&#039;(a)}{2} and by definition <br /> f&#039;&#039;(a) = \lim_{h \to 0} \frac{f&#039;(a + h) - f&#039;(a)}{h}.