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2nd derivative proofhelp ?

  1. Mar 8, 2008 #1
    2nd derivative proof..help plz?


    hi, this will be my first post on the forum, although i in the past have looked over it!
    um, this is NOT a hw 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 havent been able to get the expression they ask for??? am i missing something?? (a trick)? thanks!
  2. jcsd
  3. Mar 8, 2008 #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).
  4. Mar 8, 2008 #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], lets try to find F'(x)


    [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: Mar 8, 2008
  5. May 13, 2008 #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.
  6. May 14, 2008 #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?
  7. May 14, 2008 #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: May 14, 2008
  8. May 15, 2008 #7


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    Staff Emeritus
    Science Advisor

    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]
    [tex]F(x)= \lim_{\stack h\rightarrow 0}\frac{ f(x+h)-f(x)}{h}[/tex]
  9. May 15, 2008 #8
    Ah that makes a bit more sense, however I'm still not seeing the relationship between h, xo, and x that is used...
  10. May 16, 2008 #9

    well h is the distance from x_0 to any point x.
  11. May 17, 2008 #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.
  12. Oct 22, 2009 #11
    Re: 2nd derivative proof..help plz?

    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.
  13. Oct 22, 2009 #12


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    Staff Emeritus
    Science Advisor

    Re: 2nd derivative proof..help plz?

    I would do this.
    [tex]f"(x)= \lim_{h\to 0}\frac{f'(x+h)- f'(x)}{h}[/tex]
    [tex]f'(x+h)= \lim_{k\to 0}\frac{f(x+h+k)- f(x+h)}{k}[/tex]
    [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.
  14. Oct 24, 2009 #13
    Re: 2nd derivative proof..help plz?

    Many thanks. Very helpful.
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