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Differantiation proof question

  1. Feb 6, 2009 #1
    f(x) is differentiable twice at x_0
    prove that:

    [tex]
    f''(x_0 ) = \mathop {\lim }\limits_{x \to \infty } {{f(x_0 + h) - 2f(x_0 ) + f(x_0 - h)} \over {h^2 }}
    [/tex]
     
  2. jcsd
  3. Feb 6, 2009 #2

    CompuChip

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    If g(x) is differentiable at x_0, what is the definition of g'(x_0) ?
    Now apply this to g := f'.
     
  4. Feb 6, 2009 #3

    Mark44

    Staff: Mentor

    Re: differentiation proof question ..

    The limit should be as h approaches 0. You have it as x approaches [itex]\infty[/itex], which makes no sense, and besides, x doesn't even appear in the limit expression.
     
  5. Feb 8, 2009 #4
    this is the correct function:
    [tex]

    f''(x_0 ) = \mathop {\lim }\limits_{h \to \infty } {{f(x_0 + h) - 2f(x_0 ) + f(x_0 - h)} \over {h^2 }}

    [/tex]

    i dont know what means g'(x_0)
    if g(x) is differentiable at x_0
    ??
     
  6. Feb 8, 2009 #5

    CompuChip

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    No it's still not the correct function... the limit should be [itex]h \to 0[/itex] and not [itex]h \to \infty[/itex].

    Come on, you should know the definition of the derivative! Look it up in your textbook if you need to.

    Have you never seen the justification for the definition (with the first order approximation of the function and stuff)?
     
  7. Feb 8, 2009 #6
    i know tailor series approximations but its not telling
    what means g'(x_0)
    if g(x) is differentiable at x_0
    ??
     
  8. Feb 8, 2009 #7

    CompuChip

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    Surely you have seen

    [tex]g'(x_0) = \lim_{h \to 0} \frac{g(x_0 + h) - g(x_0)}{h}[/tex]
    or equivalently
    [tex]g'(x_0) = \lim_{x \to x_0} \frac{g(x) - g(x_0)}{|x - x_0|}[/tex]
    ?
     
  9. Feb 8, 2009 #8
    its a proof for a differentiability of a functions
    so if its differential twice at point x_0 then

    [tex]
    f''(x_0) = \lim_{x \to x_0} \frac{f'(x) - f'(x_0)}{|x - x_0|}
    [/tex]
    and i input f'(x_0) expression
     
  10. Feb 8, 2009 #9

    CompuChip

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    You better use the other one (with h) :P
     
  11. Feb 9, 2009 #10
    i tried to solve it but i cant get to the asked expression
    [tex]
    g'(x_0) = \lim_{h \to 0} \frac{g(x_0 + h) - g(x_0)}{h}\\
    [/tex]
    [tex]
    g''(x_0) = \lim_{h \to 0} \frac{g'(x_0 + h) - g'(x_0)}{h}\\
    [/tex]
    [tex]
    g''(x_0) = \lim_{h \to 0} \frac{\frac{g(x_0 + 2h) - g(x_0+h)}{h} - \frac{g(x_0 + h) - g(x_0)}{h}}{h}\\
    [/tex]
    [tex]
    g''(x_0) = \lim_{h \to 0} \frac{{g(x_0 + 2h) - g(x_0+h)} - {g(x_0 + h) - g(x_0)}}{h^2}\\
    [/tex]
    [tex]
    g''(x_0) = \lim_{h \to 0} \frac{{g(x_0 + 2h) - 2g(x_0+h)} { - g(x_0)}}{h^2}\\
    [/tex]
    ??
     
  12. Feb 9, 2009 #11

    HallsofIvy

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    It is also true that the derivative can be calculated as
    [tex]g'(x_0)= \lim_{h\rightarrow 0}\frac{g(x_0+ h/2)- g(x_0- h/2)}{h}[/tex]
    (this is the "centered" formula)_
    Try using that.
     
  13. Feb 10, 2009 #12
    It seems to me a more direct (and easy) method for tackling this problem involves applying L'Hôpital's Rule: the top approaches 0 as h approaches 0, as does the bottom. We know f is differentiable, so differentiate the top and bottom with respect to h. We can now consider the following limit (if it exists):

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

    As f is twice differentiable, and we have both numerator and denominator approaching 0, we can apply L'Hôpital's Rule again to end with:

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

    which exists and equals f''(x0).
     
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