Given $9$ points, $x$ and $y=f(x)$. Find $\int_{x1}^{x_9} f(x)dx$ using Romberg integration.
This is done as follows in the solution:
$T_{01}$
$T_{02}$ $T_{12}$
$T_{04}$ $T_{14}$ $T_{24}$
$T_{08}$ $T_{18}$ $T_{28}$ $T_{38}$
How does $T_{mn}$ relate to the number of points? Could e.g. $T_{4, 16}$ or $T_{2, 4}$ have been the final approximation as well?
I haven't seen this notation before but it looks like $$T_{0n}=\frac{b-a}{2n}f(a)+\frac{b-a}n\sum_{k=1}^{n-1}f\left(a+k\frac{b-a}n\right)+\frac{b-a}{2n}f(b)$$ And then $$T_{in}=\frac{2^{2i}T_{i-1,n}-T_{i-1,n-1}}{2^{2i}-1}$$ So you use all $9$ points to get $T_{08}$, only $5$, skipping every other one to get $T_{04}$ and so on down to $T_{01}$, which only uses the endpoints $a$ and $b$. The number of points used in $T_{in}$ is always $n+1$ and the number of Romberg steps taken is $i$ and the error is $O(h^{2i+2})$.
If you used $17$ data points you could evaluate $T_{4,16}$ with error of order $O(h^{10})$. $T_{24}$ has error $O(h^6)$ but only requires $5$ points. One cool aspect of Romberg is that you can see at each stage whether the results with large numbers of points and Romberg steps agree with each other so you have an idea of how well you might be converging to a reasonable estimate of the integral. Not to mention that it doesn't require any complicated constants to set up a quadrature subroutine, a characteristic shared by Clenshaw-Curtis quadrature.