Re: Characterizing the second derivative


"LordBeotian" <pokipsy76@xxxxxxxx> ha scritto

Suppose you have a linear function f : R->R.
If you take any two ponts x1, x2 then computing
(f(x2)-f(x1))/(x2-x1)
you obtain a value which does not depend on x1, x2, and is equal to f'(x) (which is indeed constant).

Now take a quadratic function f : R->R.
Is there a formula involving a small set of points and their images such that the result is equal to f''(x) (which is constant for f) regardless of the choice of the points.

It was a question, I forgot the question mark.

.

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