Apply Richardson extrapolation to
$ S_{2h} = \frac{2h}{3}[f(x_0) + 4f(x_2) + f(x_4)]$
$S_h =\frac{h}{3}[f(x_0) + 4f(x_1) + 2f(x_2) + 4f(x_3) + f(x_4)] $ to derive a Newton–Cotes formula based on the $5$ equally spaced points $x_0, x_1 = x_0 + h, x_2 = x_0 + 2h, x_3 = x_0 + 3h, x_4 = x_0 + 4h $ for the approximation of$\int_{x_0}^{x_4}f(x) dx$. Verify that this new formula integrates exactly any quintic polynomial.
This was one of the optional questions on my numerical analysis sheet and I am really confused about how to approach it (We didn't have time to look at it). I am OK, with the basics if the Richardson Extrapolation method, where you increase the least degree in the series of the approximation making it a closer one. Here, however, we don't have series to work with. I was thinking about using the Euler–Maclaurin formula but it seems like an overkill. Every answer/explanation/hint is appreciated.
Note I am not looking just for the solution since the deadline has passed away long time ago. I just want to be able to use this approach freely. Thanks in advance
Richardson extrapolation posits that you have a symmetric analytic function $t \mapsto F(t)$. You wish to know $F(0)$. Analyticity means:
$F(h) = F(0) + h^2 F^{(2)}(0) + h^4 F^{(4)}(0) + \ldots$
and
$F(2 h) = F(0) + 2^2 h^2 F^{(2)}(0) + 2^4 h^4 F^{(4)}(0) + \ldots$.
In our case you have the values for $F(h)$ and $F(2 h)$, which are $S_h$ and $S_{2 h}$, respectively. But how you obtain $F(h)$ and $F(2 h)$ is actually irrelevant for extrapolation.
You have two equations and as many unknowns as you wish. Drop the higher-order terms, here all those with $h^4$ and more.
Solve the resulting system for $F(0)$ in terms of $F(h)$ and $F(2 h)$.