Sharp EL-W506T vs. Sharp EL-W516T

[Albert Chan]

(02-01-2020 12:20 PM)Mjim Wrote:  The shifted equation though starts at 0 though instead of 15.667 (natural log of earths radius). Technically you are spreading the divisions used for the Simpson algorithm over nearly twice the integration length, but it appears to me likely that if you were to squash that area into the same integration length as the first green area equation, the peak would still only probably reach about half the height of the non-shifted equation, and the area would still be better distributed.

"Integration length" does not matter. Only the shape of curve matters.
In other words, with the same shape, scaling does nothing to area convergence.

For fair comparison, we should normalize the plots with same base, say 0 to 1.

\(\int _a ^b f(x) dx = \int _0 ^1 (b-a) f(a + (b-a)x) dx \)

XCas> t(f, low, high) := (high-low) * f(low + x*(high-low))
XCas> t1 := t(x -> c/e^x, log(a), log(b))
XCas> t2 := t(x -> c*e^x / (e^x+a-1)^2, 0, log(b-a+1))
XCas> [a, b, c] := [6.371e6, 9.4607304725808e15, 3.98589196e17]
XCas> plot([t1,t2], x=0..1, color=[green, magenta])

If done correctly, both curve plots should have the same area.

My guess t2 converge faster is because it is flatter, curve get better quadratic fit.