Proof of f''(a): Continuous Differentiation at a

[gaborfk]

Homework Statement

Prove that if f''(x) exists and is continuous in some neighborhood of a, than we can write


[tex]
f''(a)= \lim_{\substack{h\rightarrow 0}}\frac{f(a+h)- 2f(a)+f(a-h)}{h^2}
[/tex]

The Attempt at a Solution

I just proved in the first part of the question, not posted, that

[tex]
f'(x)= \lim_{\substack{h\rightarrow 0}}\frac{f(x+h)- f(x-h)}{2h}
[/tex]

but I am not sure how to use it for the 2nd derivative, since there is no "formula" in my book to get started with for the second derivative limit, but only for the first derivative.

Thank you in advance

 

[lzkelley]

f''(x) is the same thing as g'(x) st g(x) = f'(x).
Try something like that.

 

[gaborfk]

Can you be a little bit more specific please?
Thank you

 

[lzkelley]

f '(x)= g(x)

then, to find f ''(x), you just need to find g(g(x)) right?

 

[Nick89]

In the first part of the question you have shown that:
[tex]f'(x)= \lim_{\substack{h\rightarrow 0}}\frac{f(x+h)- f(x-h)}{2h}[/tex]

Now, you know the second derivative is just the derivative of the first derivative, right?

So why not try using the exact same equation, except using [itex]f''(x)[/itex] instead of [itex]f'(x)[/itex], and using [itex]f'(x)[/itex] instead of [itex]f(x)[/itex]

This yields:
[tex]f''(x)= \lim_{\substack{h\rightarrow 0}}\frac{f'(x+h)- f'(x-h)}{2h}[/tex]

 

[HallsofIvy]

f '(x)= g(x)

then, to find f ''(x), you just need to find g(g(x)) right?

No, not right.

 

[BryanP]

Definitely not right.

This is just a plug and chug question. You already have the equation for the derivative of a function f(x).

 

[lzkelley]

Doing what nick said - which is correct - is the same as taking g(g(x))... you take the derivative of the derivative to get the second derivative... weird huh?

 

[BryanP]

I was confused when I first read what you wrote.

But anyways, if you have some function f(x), then the derivative is given by some F(x) for some x.

We have this:
[tex]
f'(x)= \lim_{\substack{h\rightarrow 0}}\frac{f(x+h)- f(x-h)}{2h}
[/tex]

Then if you want the derivative of f'(x), let x = f'(x) and feed it into F(x). This gives you F(f'(x)).

I'm guessing that's what you meant initially.

 

[lurflurf]

lim_{x->a}(f(x)-A-Bx)/h^2=0
iff A=f'(a) B=f''(a)
find
lim_{x->a}(f(x)-f'(a)-xf''(a))/h^2=0
with f'(a) f''(a) replaced by your expressions
or use L'Hopitals rule or taylor expansion is you know of them

 

[lurflurf]

lim_{x->a}(f(x)-A-Bx)/h^2=0
iff A=f'(a) B=f''(a)
find
lim_{x->a}(f(x)-f'(a)-xf''(a))/h^2=0
with f'(a) f''(a) replaced by your expressions
or use L'Hopitals rule or taylor expansion is you know of them

should be

lim_{x->a}(f(x)-A-Bx-Cx^2)/h^2=0
iff A=f(a) B=f'(a) C=f''(a)
find
lim_{x->a}(f(x)-f(a)-f'(a)-xf''(a))/h^2=0
with f'(a) f''(a) replaced by your expressions
or use L'Hopitals rule or taylor expansion is you know of them

 

[HallsofIvy]

Doing what nick said - which is correct - is the same as taking g(g(x))... you take the derivative of the derivative to get the second derivative... weird huh?

NO, it is NOT right. Yes, the second derivative is the derivative of the derivative, so what Nick89 said is correct but that is NOT what g(g(x)) means! For example if f(x)= x², then g(x)= f'(x)= 2x. g(g(x))= g(2x)= 2(2x)= 4x which is NOT f"(x).

 

[HallsofIvy]

L'Hopital's rule or Taylor expansion is overkill.

Gaborfk, you say you have already proved that:
[tex]f'(a)= \lim_{\stack{h\rightarrow 0}} \frac{f(a+h)- f(a-h)}{2h}[/tex]
Great! That's the hard part! Now use Nick89's original suggestion: the second derivative is the "derivative of the derivative" so
[tex]f"(a)= \lim_{\stack{h\rightarrow 0}} \frac{f'(a+h)- f'(a-h)}{2h}[/tex]
and
[tex]f'(a+h)= \lim_{\stack{h\rightarrow 0}} \frac{f((a+h)+h)- f((a-h)+h)}{2h}[/tex]
[tex]= \lim_{\stack{h\rightarrow 0}} \frac{f(a+2h)- f(a)}{2h}[/tex]

Do the same for f'(a-h) and put them into that formula for f".