Comments and discussion on Valentin's 4th "Then and Now" - Area

[Albert Chan]

(01-21-2023 10:15 AM)J-F Garnier Wrote:  I solved the issue by splitting the main integral into 4 pieces in my solution.
In that way, it is possible to concentrate the samples in the regions that need them, and avoid wasting time with useless samples in the other regions.

This is what HP Prime integrate does, by zeroed-in to the region that need them.
However, this adaptive routine does not work well with fuzzy numbers. (see here)

(01-11-2023 10:17 PM)Albert Chan Wrote:  Let a = 0.831971149978, a+b = a+B/2 = 2.82740261413
Infinite slopes at y = 0 and a, moved to z = 0

Area = \(\displaystyle \int_0^{a+b} f(y)\;dy
= \int_0^{1/2} \Big[\big(f(a\,z) + f(a- a\,z)\big)·a \;+\; f(a+B\,z)·B\Big] \,dz
\)

Let g(z) = RHS integrand, and substitute z = x³/2, to make z=0 infinite slope, down to 0.
INTEGRAL built-in u-transform should turn curve to bell-shaped, easy to integrate.      (*)

Area = \(\displaystyle \int_0^{1/2} g(z)\;dz
= \int_0^1 g\!\left(\frac{x^3}{2}\right)·\left(\frac{3}{2}x^2\;dx \right)
\)

My approach was by folding back to 1 integrand, with really bad region to the left.
It has the advantage of flattening the curve, and possibly cancelled out singularities a bit.

Except the ends, folded curve look like a straight line, with dowward slope: ◜◝◟ = ◜ + ◜ + ◟ ≈ ╲
We can even estimate size of main area by 1 g(z) evaluation

>FNG(0.25) * 0.5
2.07508490846

BTW, a+b = a+B/2 = 2.82740261413 meant b = 2.82740261413-a, B = 2*b
Better: (b,B) such that the relation is true ... sometimes it is hard to isolate variable.

Thanks to JFG analysis, we have:    f(ε) ≈ 2 * ⁶√(ε)
Based from plots, I have the other: f(a ± ε) ≈ f(a) - 7/8 * ³√(±ε)

→ g(ε) ≈ f(a)*(a+B) + (2a) ⁶√(ε) - (B-a)*(7/8) ³√(ε) ≈ 5.715 + 1.664 ⁶√(ε) - 2.764 ³√(ε)

Note that sign of ε terms are opposite, cancelling each other somewhat.
g(ε) started curving downward when ε > 0.0000207