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

  • Thread starter mathishard
  • Start date
  • Tags
    Derivative
In summary: and k=0 in the last equation to getf"(x)=\lim_{k\to 0}\frac{f(x+h+k)-f(x+h)}{k}=\lim_{k\to 0}\frac{f(x-h+k)-f(x-h)}{k}=\lim_{k\to 0}\frac{f(x+h)-f(x+h)}{k}=\lim_{k\to 0}\frac{f(x-h)+f(x-h)}{k}=0
  • #1
mathishard
1
0
2nd derivative proof..help please?

:confused:

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!
 
Physics news on Phys.org
  • #2
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).
 
  • #3
or you might want to think like this : since f'(xo) exists, it means that

[tex]f'(x_o) = \lim_{h\rightarrow\ 0}\frac{f(x_o+h)-f(x_o)}{h}[/tex] , now let

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

so

[tex] 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}}[/tex]
now let
[tex]h=x-x_o=>x_o=x-h[/tex] so we get

[tex]\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}}[/tex], this way we have found that the second derivative at any point x, and also at xo is:

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

hence for [tex] x=x_o[/tex] we have

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

P.S. Nice problem!
 
Last edited:
  • #4
I'm a bit confused...

[tex]F(x)=\frac{ f(x_o+h)-f(x_o)}{h}[/tex] 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.
 
  • #5
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?
 
  • #6
But [tex]x_o[/tex] is just some point isn't it? It's not a variable.

Also how is [tex]h=x-x_o=>x_o=x-h[/tex] determined?

To put latex into posts it's just tex and /tex in brackets
 
Last edited:
  • #7
Feldoh said:
I'm a bit confused...

[tex]F(x)=\frac{ f(x_o+h)-f(x_o)}{h}[/tex] 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
[tex]F(x_0)= \lim_{\stack h\rightarrow 0}\frac{ f(x_0+h)-f(x_0)}{h}[/tex]
or
[tex]F(x)= \lim_{\stack h\rightarrow 0}\frac{ f(x+h)-f(x)}{h}[/tex]
 
  • #8
Ah that makes a bit more sense, however I'm still not seeing the relationship between h, xo, and x that is used...
 
  • #9
Feldoh said:
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.
 
  • #10
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.
 
  • #11


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.
 
  • #12


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

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.
 
  • #13


Many thanks. Very helpful.
 

1. What is the purpose of a 2nd derivative proof?

A 2nd derivative proof is used to show that a function has a minimum, maximum, or inflection point at a specific value of x. It is also used to determine the concavity of a function.

2. How do you find the 2nd derivative of a function?

To find the 2nd derivative of a function, you must first find the 1st derivative and then take the derivative of that result. This will give you the 2nd derivative of the function.

3. Can a function have multiple 2nd derivatives?

Yes, a function can have multiple 2nd derivatives, just as it can have multiple 1st derivatives. This is because the 2nd derivative is the derivative of the 1st derivative, so each time you take the derivative, you will get a new derivative.

4. What is the difference between a 1st derivative and a 2nd derivative?

The 1st derivative of a function represents its rate of change, while the 2nd derivative represents the rate of change of the 1st derivative. In other words, the 1st derivative tells us how the function is changing, while the 2nd derivative tells us how the change of the function is changing.

5. How can a 2nd derivative be used to determine the concavity of a function?

If the 2nd derivative of a function is positive at a specific value of x, then the function is concave up at that point. If the 2nd derivative is negative, then the function is concave down. A 2nd derivative of 0 indicates a possible inflection point where the concavity changes.

Similar threads

Replies
2
Views
1K
Replies
1
Views
1K
Replies
2
Views
977
Replies
5
Views
2K
  • Calculus
Replies
9
Views
2K
Replies
1
Views
963
Replies
11
Views
973
Replies
1
Views
887
  • Programming and Computer Science
Replies
4
Views
4K
Back
Top