It seems you are looked at divided differences.
Let f be a continuous function and its third derivative exists. I need to show that [ -f(x+2h) + 4f(x+h) - 3f(x)] / 2h = f’(x) + O[h^2].
I know how to show that -f(x+2h) + 4f(x+h) - 3f(x) = 2h f’(x) by using the Taylor expansions:
f(x+2h) = f(x) + 2h f’(x) + [(2h)^2] f’’(x) / 2! + O[(2h)^3],
f(x+h) = f(x) + h f’(x) + [h^2] f’’(x) / 2! + O[h^3].
Note that O[h^(n+1)] is the remainder of the Taylor expansion, and O[h^(n+1)] = f^(n+1)(z_n) h^(n+1) / (n+1)!, where z_n is some number between x and c (since h = x-c).
How can I prove that we get O[h^2] in the first equation above? I cannot prove it by directly using the definition above. Please help me answer this question. If you know any links that may help me, please give me.
It seems you are looked at divided differences.
In general if is defined in the neighborhood of then -th divided difference is . So for this expression is (in numerator) . This means if is (infinitly) differenciable in the neighborhood of then where as by Taylor's theorem. Furthermore, where as by Taylor's theorem. This means this difference is simply . Thus .