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

• Euklidian-Space
In summary: i should note that the epsilon is something you should be able to make go to 0, but i am not sure how.
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}##

## 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

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

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 $$\lim_{h \to 0} \frac{\epsilon(h)}{h^2} = 0.$$

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

DEvens said:
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.

Last edited:
Euklidian-Space said:
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.

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

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

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

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

Euklidian-Space said:
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.

pasmith said:
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 !

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

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

Euklidian-Space said:
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}\ ?##

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

SammyS said:
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?

Euklidian-Space said:
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.

Last edited:
Euklidian-Space said:
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.

vela said:
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?

pasmith said:
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 $$\lim_{h \to 0} \frac{\epsilon(h)}{h^2} = 0.$$

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

DEvens said:
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?

Euklidian-Space said:
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).

DEvens said:
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 $$\lim_{h \to 0} \frac{\epsilon(h)}{h^2} = \lim_{h \to 0} \frac{f(a + h) - f(a) - hf'(a) - \frac12h^2f''(a)}{h^2} \\ = \lim_{h \to 0} \frac{f'(a + h) - f'(a) - hf''(a)}{2h} = \frac{f''(a) - f''(a)}{2} = 0$$ 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 $$\lim_{h \to 0} \frac{f(x + h) - f(x - h)}{2h} = f'(x).)$$

Euklidian-Space said:
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
$$\lim_{h \to 0} \frac{f(a+h)-2f(a)+f(a-h)}{h^2}.$$

vela said:
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}##

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

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.

Last edited:
Euklidian-Space said:
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 $f'(a - h)$ with respect to $h$ is $-f''(a - h)$, so you should have $$\lim_{h \to 0} \frac{f''(a + h) + f''(a - h)}{2}.$$ 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 $f''(a + h)$ does not exist for any $h \neq 0$. So l'Hopital's rule can only take you as far as $$\lim_{h \to 0} \frac{f'(a + h) - f'(a - h)}{2h}.$$

pasmith said:
The derivative of $f'(a - h)$ with respect to $h$ is $-f''(a - h)$, so you should have $$\lim_{h \to 0} \frac{f''(a + h) + f''(a - h)}{2}.$$ 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 $f''(a + h)$ does not exist for any $h \neq 0$. So l'Hopital's rule can only take you as far as $$\lim_{h \to 0} \frac{f'(a + h) - f'(a - h)}{2h}.$$

oh i think I see now

pasmith said:
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 $$\lim_{h \to 0} \frac{f(x + h) - f(x - h)}{2h} = f'(x).)$$
wait how do you get
$$\lim_{h \to 0} \frac{f'(a + h) - f'(a) - hf''(a)}{2h} = \frac{f''(a) - f''(a)}{2} = 0$$

Euklidian-Space said:
wait how do you get
$$\lim_{h \to 0} \frac{f'(a + h) - f'(a) - hf''(a)}{2h} = \frac{f''(a) - f''(a)}{2} = 0$$

$$\lim_{h \to 0} \frac{f'(a + h) - f'(a) - hf''(a)}{2h} = \frac12 \lim_{h \to 0} \frac{f'(a + h) - f'(a)}{h} - \frac{f''(a)}{2}$$ and by definition $$f''(a) = \lim_{h \to 0} \frac{f'(a + h) - f'(a)}{h}.$$

## 1. What does it mean for f''(a) to exist?

For f''(a) to exist means that the second derivative of the function f(x) at the point a exists, or in other words, the rate of change of the slope of the function at a point is well-defined.

## 2. Can f''(a) exist even if f'(a) does not?

Yes, it is possible for f''(a) to exist even if f'(a) does not exist. This is because the second derivative measures the rate of change of the slope, while the first derivative measures the rate of change of the function itself. So, while the function may not be continuous at a point, the slope of the function can still have a well-defined rate of change at that point.

## 3. How is f''(a) calculated?

To calculate f''(a), we use the limit definition of the second derivative, which is the limit as h approaches 0 of [f'(a+h) - f'(a)]/h. This gives us the instantaneous rate of change of the slope at the point a, which is the value of f''(a).

## 4. What does it mean if f''(a) is positive?

If f''(a) is positive, it means that the function is concave up at the point a, and the slope of the function is increasing. This means that the function is getting steeper as x increases, and the rate of change of the slope is positive.

## 5. Can f''(a) be negative?

Yes, f''(a) can be negative. This means that the function is concave down at the point a, and the slope of the function is decreasing. This means that the function is becoming less steep as x increases, and the rate of change of the slope is negative.

Replies
12
Views
588
Replies
20
Views
2K
Replies
7
Views
1K
Replies
6
Views
768
Replies
9
Views
4K
Replies
4
Views
1K
Replies
2
Views
1K
Replies
12
Views
2K
Replies
6
Views
1K
Replies
16
Views
1K