HP71B IBOUND fooled

[Albert Chan]

(05-02-2022 01:42 AM)Albert Chan Wrote:  Unlike Kahan's transformation, getting u end-point involved solving cubics.
Luckily, with 0 ≤ x ≤ 1, only 1 u solution satisfied.

This solve cubics, x^3 = a*x + b, for x
Formula is general for (a,b), a ≠ 0, even complex numbers.
Note: constant k is to transform cubic to form, sin(3θ) = 3*sin(θ) - 4*sin(θ)^3

XCas> cubic(a,b) := {local k:=sqrt(4*a/3); k .* sin(([0,2*pi,-2*pi].-asin(4b/k^3))./3)}

3u² - 2u³ = x
3/u - 2 = x/u³
(1/u)³ = (3/x)(1/u) + (-2/x)

XCas> simplify(1 ./ cubic(3/x, -2/x))

\(\displaystyle \left[
\frac{\sqrt{x}}{2 \sin\left(\frac{\mathrm{asin}\left(\sqrt{x}\right)}{3}\right)},
\frac{\sqrt{x}}{2 \sin\left(\frac{\mathrm{asin}\left(\sqrt{x}\right)+2\cdot \pi}{3}\right)},
\frac{\sqrt{x}}{2 \sin\left(\frac{\mathrm{asin}\left(\sqrt{x}\right)-2\cdot \pi}{3}\right)}
\right]\)

Consider only denominators, with asin(√x)/3 = 0 .. pi/6
Denominator = 2*sin([(0 .. pi/6), (2/3*pi .. 5/6*pi), (-2/3*pi .. -pi/2)]) = [ (0 .. 1), (1..√3), (-2..-√3) ]

Only middle root gives correct u range:

3u² - 2u³ = x ≥ 0
u² = x/(3-2u) ≤ x/(3-2*max(u)) = x/1
0 ≤ u ≤ √x