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Finite difference approximation for third order partials?

  1. Nov 12, 2011 #1
    I'm attempting to perform interpolation in 3 dimensions and have a question that hopefully someone can answer.

    The derivative approximation is simple in a single direction:

    df/dx(i,j,k)= [f(i+1,j,k) - f(i-1,j,k)] / 2

    And I know that in the second order:

    d2f/dxdy(i,j,k)= [f(i+1,j+1,k) - f(i+1,j,k) - f(i,j+1,k) + f(i-1,j-1,k)] / 4

    The final item I need is the third order approximation, and I'm not sure how to scale the first two into a third variable.

    d3f/dxdydz(i,j,k)= ?

    Can anyone shed some light on this?

    Thanks in advance!
  2. jcsd
  3. Nov 13, 2011 #2


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    can't you just apply Taylor series? You want
    \frac{\partial^{3}f}{\partial x\partial y\partial z}
  4. Nov 13, 2011 #3
    Yes but I don't have the function itself; I only have its value at various points i-2, i-1, i, i+1, i+2, etc.
  5. Nov 13, 2011 #4


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    Just apply Taylors theorem in 1D to each variable in turn.
  6. Nov 14, 2011 #5
    You already have
    d2f/dxdy(i,j,k) = [f(i+1,j+1,k) - f(i+1,j,k) - f(i,j+1,k) + f(i-1,j-1,k)] / 4
    so you have to do the first order derivation for z, which means for indexes in short-hand

    [(k->k+1) - (k->k-1)]/2

    By the way, congratulations for having chosen the symetrical representation of the first derivative, it is much more accurate than [(k->k+1) - ()].

    So lets do it:
    d3f(x,y,z)/dxdydz = {[f(i+1,j+1,k+1) - f(i+1,j,k+1) - f(i,j+1,k+1) + f(i-1,j-1,k+1)]
    - [f(i+1,j+1,k-1) - f(i+1,j,k-1) - f(i,j+1,k-1) + f(i-1,j-1,k-1)]}/8 =
    [f(i+1,j+1,k+1) - f(i+1,j,k+1) - f(i,j+1,k+1) + f(i-1,j-1,k+1)
    - f(i+1,j+1,k-1) + f(i+1,j,k-1) + f(i,j+1,k-1) - f(i-1,j-1,k-1)]/8
    That's it.
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