HP71B IBOUND fooled

[Albert Chan]

(05-02-2022 02:57 PM)Albert Chan Wrote:  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)}

For cubic(a,b), (sin,asin) → (cos,acos), we *still* get solutions of cubic, albeit in different order.

XCAS> cubic2(a,b) := {local k:=sqrt(4*a/3); k .* cos(([0,2*pi,-2*pi].-acos(4b/k^3))./3)}

If (a,b) are real, complex roots, if any, always comes in conjugate pairs.
With above symmetry, if asin(), acos() are real, this implied at least 2 real roots --> all cubic roots real.

(cos, acos) can easily mapped to (cosh, acosh)

cosh(z) = cos(±i*z)            // both cosh and cos are even function.
acosh(z) = acos(z) * ±i      // pick sign to produce non-negative real part

Example, for value of Super Golden Ratio

x^3 = x^2 + 1
(x-1/3)^3 = (x-1/3) * (1/3) + (29/27)      // depressed cubic

(x-1/3) = 2/3 * cos((2*n*pi - acos(29/2))/3)

29/2 > 1, cubic had only 1 real root, for n=0:

x = 2/3*cosh(acosh(29/2)/3) + 1/3 ≈ 1.465571232