Numerical integration methods

[Albert Chan]

(08-01-2022 03:43 AM)Wes Loewer Wrote:  For instance, why *(err1/err2)^2 ? Why not *(err1/err2) or *(err1/err2)^3 ? Was ^2 derived mathematically, or experimentally? I'm guessing it was experimentally determined to give a reasonable error approximation.

And why does the 3rd calculation use 6 nodes? Why not 5 or 7?

This is my guess ...

err1 = k1 * h^15
err2 = k2 * h^7

integral error estimate = err1*(err1/err2)^2 = k1*(k1/k2)^2 * h^(15+8*2) = k * h^31

If (k, k1, k2) are similar in size, constant term matched too.

Exponents cannot be picked in random; it had to produce O(h^31) on the right.
If we use 5 or 7 nodes, instead of square, it would need some non-integer exponents.