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I'm implementing the Finite Different 2nd and 3rd derivatives in my research and naturally I'm looking for the most efficient approach.

From https://en.wikipedia.org/wiki/Finite_difference#cite_ref-WilmottHowison1995_1-0 I found a more efficient formula that takes 2 function evaluations out of the picture. But I'm not sure what simplification or substitution was implemented. Can anyone help me out? See the Image below.

enter image description here

Thank you!

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  • $\begingroup$ Try to move this question in the mathematics community. $\endgroup$ – james watt Sep 24 '20 at 12:05
  • $\begingroup$ @JamesWatt: I think the question is completely appropriate here. $\endgroup$ – davidhigh Sep 24 '20 at 15:34
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The two quoted approximations to the mixed (second) derivative are simply different formulas. There is no direct way to arrive from one at the other.

The first formula evaluates the derivative on the points $\Big((x-h,y-k), (x-h,y+k), (x+h,y-k),(x+h,y+k)\Big)$, the second formula uses another stencil that involves points that are also used by the first derivative.

The basic idea for both is the same, however. Take the given stencil, and approximate the derivative to highest degree possible for this setup. By doing this, you end up with the quoted formulas.

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