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    Extrapolation formula for numerical derivative

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    Derive the extrapolation formula and show its error converges to zero more rapidly than does the error of (*).

    Please see the attachment for the formula.

    © BrainMass Inc. brainmass.com December 24, 2021, 10:52 pm ad1c9bdddf
    https://brainmass.com/math/derivatives/extrapolation-formula-numerical-derivative-514210

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    SOLUTION This solution is FREE courtesy of BrainMass!

    If f(x) is twice differentiable, we have:

    f(x+h) = f(x) + h f'(x) + h^2/2 f''(x) + o(h^2) (1)

    where the "little o" error term o(h^2) denotes a function that goes to zero for h to zero, faster than h^2: For every epsilon > 0 there exists a delta>0 such that:

    |o(h^2)| < epsilon h^2 when |h| < delta.

    If f(x) is 3 times differentiable then we know that o(h^2) will be proportional to h^3, but we don't need to make that asssumption here. From (1) it follows that:

    D_h f(x) = [f(x+h) - f(x)]/h = f'(x) + h/2 f''(x) + o(h) (2)

    where o(h) is, analogous to o(h^2), a term that goes to zero faster than any term proportional to h. Due to the presence of the term h/2 f''(x), the error in D_hf(x) as an estimate of the derivative is thus of order h (this is denoted in the big O notation as O(h), which means a term that goes to zero if h goes to zero as a constant times h). We can get rid of the contribution of h/2 f''(x) in this error as follows. If we make h twice as large, this term becomes twice as large, so if we take 2 times D_h minus D_{2h}, this term cancels:

    2 D_h f(x) - D_{2h}f(x) = f'(x) + o(h)

    We see that the error has been reduced to o(h), i.e. a term that goes to zero faster than anything that is proportional to h.

    This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

    © BrainMass Inc. brainmass.com December 24, 2021, 10:52 pm ad1c9bdddf>
    https://brainmass.com/math/derivatives/extrapolation-formula-numerical-derivative-514210

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