HP50g simplifing a root

[CMarangon]

Hello!

Well, I conclude that HP50 does not have the formulas below, into memory.
Maybe, it can be included, in next ROM update.

Formula:
(A ± B⋅√k)^(1/3) = a ± b⋅√k

See also post #22, from Albert Chan, above.


Carlos (BR)

(10-09-2020 02:31 PM)Albert Chan Wrote:  

(10-09-2020 12:20 AM)Albert Chan Wrote:  Unfortunately, our divisors based cube root simplify routines were flawed !
Real rational root for a is a divisor of n, *but* possibly divided by 4 ...

Another problem is we had assumed there is only 1 real root for a.
Lets rearrange the cubic to match form x³ + 3px - 2q = 0

c³ = A² - R
4*a³ = 3*c*a + A
a³ + 3*(-c/4)*a - 2*(A/8) = 0

→ Cubic discriminant = p³ + q² = (-c/4)³ + (A/8)² = (-c³ + A²) / 64 = R/64
→ If R < 0, we got 3 real roots.

see https://proofwiki.org/wiki/Cardano%27s_Formula

Quote:XCas> find_cbrt(81,30,-3)       → \(\frac{9}{2} + \frac{i}{2}\cdot\sqrt{3}\)

XCas> find_all_a(A,R) := solve(surd(A*A-R,3) = a*a - (A/a-a*a)/3 , a)

XCas> find_all_a(81, 30^2 * -3)       → [-3, -3/2, 9/2]
XCas> simplify( [-3+2i*sqrt(3), -3/2-5i/2*sqrt(3), 9/2+i/2*sqrt(3)] .^ 3)

\(\qquad\quad [81 + 30i\sqrt3\;,\; 81 + 30i\sqrt3\;,\; 81 + 30i\sqrt3 \)

If we consider integer as simplest, then \(\sqrt[3]{81 + 30i\sqrt3} = -3+2i\sqrt3\)

Comment: I was wrong on above example.

x³ = y does not imply ³√y = x. We need to consider branch-cut
In other words, simplified form should match numerical evaluated values.

>>> (81 + 30j * 3**0.5) ** (1/3)
(4.5+0.86602540378443837j)