How could I normalize the 2nd order central difference method?
The original function is F(A,B,C,D) and only one variable is varied at a time. ie

$F'' = \frac{F(A+\Delta A,B,C,D) - 2\times F(A,B,C,D) + F(A-\Delta A,B,C,D)}{\Delta A^2}$

I want to compare the sensitivity of each variable. But I would like to do it with normalizing the second derivatives.

  • $\begingroup$ Welcome to CS.SE! I find it hard to tell exactly what you are asking. What do you mean by "normalize"? What is wrong with the expression you already have for estimating the second derivative? What precisely do you mean by "sensitivity"? $\endgroup$ – D.W. Feb 2 '20 at 3:59
  • $\begingroup$ So F(A,B,C,D) is a lift to drag ratio for an airfoil. And A B C and D are shape coefficient that created the general shape of the airfoil. So i'm looking at how sensitive the lift to drag ratio is when an uncertainty is added to these shape coefficients (This uncertainty represents a manufacturing tolerance). $\endgroup$ – John Allen Feb 2 '20 at 19:11
  • $\begingroup$ And I'm comparing different airfoil designs. So different sets of shape coefficients that have different uncertainties. And id like to compare the sensitivities (or 2nd derivative) of each design. Where i would compare how adding uncertainty to B effected airfoil 1 and 2 respectively. $\endgroup$ – John Allen Feb 2 '20 at 19:15
  • $\begingroup$ But the problem was resolved! using the taylor series expansion, we are able to normalize the 2nd derivative to represent the % change in L/D when adding uncertainty to each coefficient independently. Also thank you for the welcome! 2nd Derivative *(0.5*uncertainty^2)/F(A,B,C,D) $\endgroup$ – John Allen Feb 2 '20 at 19:19
  • $\begingroup$ Please edit the question to clarify what you're asking, and make it self-contained and read well for someone who encounters it for the first time, rather than leaving clarifications in the comments. We don't want people to have to read the comments to understand what you're asking. If you figured out the answer on your own, you might also try leaving an answer. Normally I would expect that the sensitivity is measured using the first derivative, not the second derivative (unless for some reason you expect the first derivative to be zero). $\endgroup$ – D.W. Feb 2 '20 at 19:27

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