[VA] SRC #012d - Then and Now: Area

[Albert Chan]

(01-12-2023 07:31 AM)Werner Wrote:  Incidentally (and accidentally) I also found a tiny region -4.0851.. <= y <= -4.0492.. where the original inequality will hold (eg. y=-4.05 and x=0.004).

Most likely, the region was fitted with inequality expression, not the other way around.
I would not worry about this, especially since OP suggested locate region by eye.

If we plot this, say, y = -5 .. 5, above "dot" area does not even show.

(01-12-2023 09:34 AM)J-F Garnier Wrote:  What about the accuracy?
I have no reference to compare with, I can only trust the HP integral algorithm and the IBOUND estimations for each zone.

If integrand behave like a polynomial, IBOUND likely over-estimates true error.
IBOUND was based from relative error of current result vs previous, not true result vs current.

For polynomial-like integrand, we can get by with bigger eps (= less function evaluations)
That's the reason my code, with eps=10^-6, results in area of 10+ accuracy.

(01-11-2023 10:17 PM)Albert Chan Wrote:  Area = \(\displaystyle \int_0^{1/2} g(z)\;dz
= \int_0^1 g\!\left(\frac{x^3}{2}\right)·\left(\frac{3}{2}x^2\;dx \right)
\)

I expected at z ≈ 0, g(z) behaves like c1 + c2*cbrt(z)
To simplify, say g(z) = z^(1/3), and we substitute with z=x^3

∫(z^(1/3) dz) = ∫((x^3)^(1/3) * (3*x^2 dx)) = 3*∫(x^3 dx)

RHS turned to plain polynomial, easy to integrate with Romberg.