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Math Help - Proof of Approximation Formula for a Second Derivative

  1. #1
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    Proof of Approximation Formula for a Second Derivative

    Using the formula [f(a+h)-f(a-h)]/2h = f prime of a, prove that
    [f(a+h) - 2f(a) + f(a-h)]/h^2 = f double prime of a.

    I get [f(a+h) - 2f(a) + f(a-h)]/2h^2

    Could someone show me the proof of this? It's driving me crazy.

    My work
    [[2f(a+h) - 2f(a)] - [f(a+h) - f(a-h)]]/2h/h = the above (I assume incorrect) answer
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  2. #2
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    I think that
    f'(a+h/2)=[f(a+h)-f(a)]/h
    f'(a-h/2)=[f(a)-f(a-h)]/h
    f''(a)=[f'(a+h/2)-f'(a-h/2)]/h
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  3. #3
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    Quote Originally Posted by zzzoak View Post
    I think that
    f'(a+h/2)=[f(a+h)-f(a)]/h
    f'(a-h/2)=[f(a)-f(a-h)]/h
    f''(a)=[f'(a+h/2)-f'(a-h/2)]/h
    Genius!
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