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A derivative problem

  1. Jun 10, 2004 #1
    Hi members,

    Could any one help me with the problem following?

    If x is a real number and P is a polynomial function, then


    lim {P(x+3h)+P(x-3h)-2P(x)}/h^2
    h->0


    A)0
    B)6P'(x)
    C)3P''(x)
    D)9P''(x)
    E) 00

    I guess D should be the answer, I need an explanation.

    Thank You
     
  2. jcsd
  3. Jun 10, 2004 #2

    arildno

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    Dearly Missed

    The easiest way is to use Taylor-series expansions of the terms P(x+3h), P(x-3h)
    to verify your guess.
    Then we have, for example:
    P(x+3h)=P(x)+P'(x)3h+1/2P''(x)(3h)^2+O(h^3)

    O(h^3) is a higher order term, i.e lim h->0 O(h^3)/h^2=0
     
  4. Jun 10, 2004 #3
    Thank you, arildno

    It couldn't be more wonderful solutions.
     
  5. Nov 12, 2010 #4
    It looks messy, but it's really no different from any other limit problem. What is typically the easiest way to find the limit of 0/0? L'Hopital's rule!

    Differentiate top and bottom with respect to h (not x!) twice, using chain rule for top terms. So the first differentiation gives [3P'(x+3h)-3P'(x-3h)]/2h (notice the third term has no h, so drops out). The second round, you get [9P''(x+3h)+9P''(x-3h)]/2. Then setting h=0 gives the desired answer.
     
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