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Nov 13, 2012 at 15:25 comment added Garabed Gulbenkian Thanks, Steve. That was exactly what I was looking for. So, to approximate the nth derivative we can use the n+1 points z(k),z(k+1) ...z(k+n) together with the values of f(z) at these points and let k approach infinity. To obtain the leading coefficient of the nth degree interpolating polynomial we have just to solve a system of n+1 linear equations whose matrix is a (finite) Vandermonde matrix -which always has a unique inverse. So it seems that these matrices are useful after all.
Nov 11, 2012 at 20:25 comment added Steve It is sort of sad that such basic observations are not usually given to our calculus students, or complex analysis students for that matter. It was many years after learning these subjects, and feeling somewhat dissatisfied with my state of knowledge, that I finally figured these things out for myself. This plus analyticity is, in my opinion, the real reason for the identity theorem.
Nov 11, 2012 at 20:23 comment added Steve Here is the recipe for f''(0) which you should be able to generalize. Take 3 points within a distance of $\delta$ from 0. Find the unique quadratic passing through all 3 points. Half the lead term of this quadratic is an approx of the second derivative. Form a sequence of such approximations as $/delta$ goes to 0. The limit of this sequence is the second derivative.
Nov 11, 2012 at 19:07 comment added Garabed Gulbenkian Steve, I believe you have the right approach to this problem. However one must still find a definition of the nth derivative of f(z) at z=0 as the limit of a formula involving finite differences which requires no information other than the data we have. Suppose z(i) is a real number for each positive integral i. Many of the approximations to the nth derivative of f(z) at z=0 require that we know f(z) at values of z between z(k) and z(k+1)-for some integers k-and we do not have this information.
Nov 11, 2012 at 18:36 vote accept Garabed Gulbenkian
Nov 9, 2012 at 20:37 history answered Steve CC BY-SA 3.0