I've seen multiple sources, claiming that the Richardson Extrapolation for the Romberg Integration (Trapezoidal method) is…
$$R_{i,j}=\frac{4^{j-1}R_{i+1,j-1}-R_{i,j-1}}{4^{j-1}-1}$$
if the intervall step size is $h = (b-a)/t$ with $t = 2^{\mathbb{N}_0}$.
My Professor told me the Richardson Extrapolation for the Romberg Integration (midpoint/rectangle method) is
$$R_{i,j}=\frac{9^{j}R_{i+1,j-1}-R_{i,j-1}}{9^{j}-1}$$
if the intervall step size is $h = (b-a)/t$ with $t = 3^{\mathbb{N}_0}$.
So my questions are:
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Is the formula for the Richardson Extrapolation of my prof true? I can't find it anywhere. Especially why it's $9^{j}$ and not $9^{j-1}$
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Is $t = 3^{\mathbb{N}_0}$ necessary for the midpoint method or could I do it with $t = 2^{\mathbb{N}_0}$?
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Is there any practical difference between using base 2 or base 3?
Best Answer
This is a question of the re-use of previously computed values. If you refine the subdivision by halving, then the old midpoints are no longer used in any of the following refinements. Still, you compute 2 new function values for each subdivided segment.
On the other hand, by reducing the segments to thirds, the old middle point is still the middle point of the middle sub-segment. So you still have to compute 2 new function values per segment, but as this is an overall finer subdivision, the error will thus be smaller accordingly.