I have a homework question that I don't know how to answer. It goes like this:
Give complete details on the derivation of the five-point centered approximation to the second derivative of a function $f(x)$ for the example below. Also, give complete details on using Taylor expansions to determine the leading error term, as given in the example below.
With $n = 4$, $m = 2$, and $x = x_2$, and reference interval $x_0 = -2h$, $x_1 = -h$, $x_2 = 0$, $x_3 = h$, $x_4 = 2h$, we have $$f''(x_2) \cong \sum_{i=0}^4 f_i l''_i(x_2)$$ Here $$l_0(x) = \frac{(x - x_1)(x - x_2)(x - x_3)(x - x_4)}{(x_0 - x_1)(x_0 - x_2)(x_0 - x_3)(x_0 - x_4)}$$ $$=\frac{(x - x_1)(x - x_2)(x - x_3)(x - x_4)}{24h^4}$$
Differentiating, and setting $x$ equal to $x_2$, we find $$l''_0(x_2)=\frac{-1}{12h^2}$$ Similarly $$l''_1(x_2)=\frac{16}{12h^2}, l''_2(x_2)=\frac{-30}{12h^2}, l''_3(x_2)=\frac{16}{12h^2}, l''_4(x_2)=\frac{-1}{12h^2}$$
Hence we have the five point finite difference approximation $$f''(x_2)\cong\frac{-f_0+16f_1-30f_2+16f_3-f_4}{12h^2}$$
By Taylor expansion, one can show that the leading error term is $$\frac{h^4f^{(6)}(x_2)}{90}$$
I have no clue how to answer this question. Some help will be much appreciated. Thank you!
Edit: To be clear, I just don't understand what they mean by "give complete details".