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Chris Hare

Derivative Via Extrapolation

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Consider the following example of my problem... For an unknown function f(x), if given these points f(0.3) = ... f(0.5) = ... f(0.7) = ... f(0.9) = ... f(1.1) = ... f(1.3) = ... f(1.5) = ... is it possible to approximate f''(0.9) to within 0(h^6), or is 0(n^4) the limit??? I know that f''(x) can be approximated with h = 0.2 and h = 0.4 with the central-difference formula (err = 0(h^2)), and then extrapolated using std extrapolation to within 0(n^4) of the real value. If I had f(0.1) and f(1.7) I could approximate the differential with h = 0.8, extrapolate with h = 0.4 and h = 0.8, and finally use a second-order extrapolation to get with O(n^6), but I don''t have f(0.1) & f(1.7). I guess what I''m asking is if there is a way around the 2 : 1 ration of h''s to extrapolate, is there an additional approximation that can be used to extrapolate or something. ** Note ** I''m looking for a general case solution to implement in a program, but just the theory behind the solution would be fine. Wizza Wuzza?

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Guest Anonymous Poster
If it is a completely unknown function, there is no way to figure out what the values might be. The function could just be discontinous random numbers.

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Guest Anonymous Poster
2nd order accuracy:
f''(x) = (f(x+h)-f(x-h))/(2*h)

4th order accuracy:
f''(x) = (8*(f(x+h)-f(x-h)) - (f(x+2*h)-f(x-2*h)))/(12*h)

6th order accuracy:
f''(x) = (45*(f(x+h)-f(x-h)) - 9*(f(x+2*h)-f(x-2*h)) + (f(x+3*h)-f(x-3*h)))/(60*h)

etc...

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