Proving the Second Derivative Using Limit Definition: A Step-by-Step Guide

[mathishard]

2nd derivative proof..help please?



hi, this will be my first post on the forum, although i in the past have looked over it!
um, this is NOT a homework problem, but is a problem in my textbook that i attempted to do.

it asks to show that if , for a function f, a second derivative exists at x0
to prove that

f''(x0) = lim h->0 [f(x0+h)-f(x0-h)-2f(x0)] / h^2
...At first i thought this would be easy, just using
f ' (x0) = limh->0 ( f(x0+h)-f(x0)) / h

and f''(x0) = lim (f'(x0+h)-f'(x0))/h

but somehow i haven't been able to get the expression they ask for? am i missing something?? (a trick)? thanks!

 

[WHOAguitarninja]

Try this...pull a 1/h to the outside, and consider how you might be able to rewrite the fraction as two more useful fractions added together (or subtracted).

 

[sutupidmath]

or you might want to think like this : since f'(xo) exists, it means that

$$f'(x_o) = \lim_{h\rightarrow\ 0}\frac{f(x_o+h)-f(x_o)}{h}$$ , now let

$$F(x)=\frac{ f(x_o+h)-f(x_o)}{h}$$, let's try to find F'(x)

so

$$f''(x)=F'(x)=\lim_{h\rightarrow\ 0} \frac{F(x_o+h)-F(x_o)}{h}=\lim_{h\rightarrow\ 0} \frac{\frac{f(x_o+2h)-f(x_o+h)}{h}-\frac{f(x+h)-f(x)}{h}}{h}=\lim_{h\rightarrow\ 0}\frac{f(x_o+2h)-2f(x_o+h)-f(x_o)}{h^{2}}$$
now let
$$h=x-x_o=>x_o=x-h$$ so we get

$$\lim_{h\rightarrow\ 0} \frac{f(x-h+2h)-2f(x-h+h)-f(x-h)}{h^{2}}=\lim_{h\rightarrow\ 0}\frac{f(x+h)-2f(x)-f(x-h)}{h^{2}}$$, this way we have found that the second derivative at any point x, and also at xo is:

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

hence for $$x=x_o$$ we have

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

P.S. Nice problem!

 

[Feldoh]

I'm a bit confused...

$$F(x)=\frac{ f(x_o+h)-f(x_o)}{h}$$ but doesn't that make F(x) a constant for all x?

Sorry to bring up a dead thread but I was actually wondering this as well.

 

[swapnilster]

No that equation is a standard one to get the derivative of any function using the first principle.
Btw how do you guys get those latex or whatever images into your answers?

 

[Feldoh]

But $$x_o$$ is just some point isn't it? It's not a variable.

Also how is $$h=x-x_o=>x_o=x-h$$ determined?

To put latex into posts it's just tex and /tex in brackets

 

[HallsofIvy]

I'm a bit confused...

$$F(x)=\frac{ f(x_o+h)-f(x_o)}{h}$$ but doesn't that make F(x) a constant for all x?

Sorry to bring up a dead thread but I was actually wondering this as well.

It would have been better to write
$$F(x_0)= \lim_{\stack h\rightarrow 0}\frac{ f(x_0+h)-f(x_0)}{h}$$
or
$$F(x)= \lim_{\stack h\rightarrow 0}\frac{ f(x+h)-f(x)}{h}$$

 

[Feldoh]

Ah that makes a bit more sense, however I'm still not seeing the relationship between h, xo, and x that is used...

 

[sutupidmath]

Ah that makes a bit more sense, however I'm still not seeing the relationship between h, xo, and x that is used...

well h is the distance from x_0 to any point x.

 

[Feldoh]

Oh ok, that was sort of what I was thinking. In my class we've always used points x and x+h to define the derivative, that's why I was a bit confused.

 

[mbsl5]

I know this is an old thread, but hoping someone can clarify something. I can follow the proof from sutupidmath. But in the proof, I'm thinking there should be two distinct limits working in the equations, one because the definition for F(x) should contain it (as HallsofIvy defined it later), and then again to define F'(x). How would the proof need to change to address the two limits?

I have seen other notes that suggest using the mean value theorem for this proof. Although I understand the MVT, I have not been able to see how to use it.

Thanks for any help.

 

[HallsofIvy]

I would do this.
$$f"(x)= \lim_{h\to 0}\frac{f'(x+h)- f'(x)}{h}$$
But
$$f'(x+h)= \lim_{k\to 0}\frac{f(x+h+k)- f(x+h)}{k}$$
and
$$f'(x)= \lim_{k\to 0}\frac{f(x+k)- f(x)}{k}$$

Put those into the the first equation and simplify. Since those must be true for h and k approaching 0 in any way, take h= k.

 

[mbsl5]

Many thanks. Very helpful.