HP50g simplifing a root

10092020, 02:31 PM
(This post was last modified: 10122020 06:30 AM by Albert Chan.)
Post: #20




RE: HP50g simplifing a root
(10092020 12:20 AM)Albert Chan Wrote: Unfortunately, our divisors based cube root simplify routines were flawed ! 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*AR,3) = a*a  (A/aa*a)/3 , a) XCas> find_all_a(81, 30^2 * 3) → [3, 3/2, 9/2] XCas> simplify( [3+2i*sqrt(3), 3/25i/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 branchcut In other words, simplified form should match numerical evaluated values. >>> (81 + 30j * 3**0.5) ** (1/3) (4.5+0.86602540378443837j) 

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