HP50g simplifing a root
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10-11-2020, 06:28 PM
(This post was last modified: 10-12-2020 05:19 AM by Albert Chan.)
Post: #24
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RE: HP50g simplifing a root
(10-09-2020 02:31 PM)Albert Chan Wrote: Lets rearrange the cubic to match form x³ + 3px - 2q = 0 Instead of cubic discriminant, we can show this with more familiar quadratic discriminant. Let f(x) = 4x³ - 3*c*x - A, so that f(a) = 0 f(x) = 4 * (x-a) * (x² + a*x + (a² - 3*c/4)) Quadratic discriminant = a² - 4*(a² - 3*c/4) = -3*(a² - c) = -3*r → if r < 0 (due to R < 0), f got 3 real roots. We can solve the quadratics, but unnecessary. If we have 1 solution, where (a ± √r)³ = A ± √R, we can get the other 2. (a ± √r)³ = ((a ± √r) × w)³ = ((a ± √r) / w)³, where w = (-1)^(2/3) Example, simplify ³√(36 + 20i√7) >>> from mpmath import * >>> A, B, k = 36, 20, -7 >>> z = A + B*sqrt(k) >>> q = cbrt(z); print q (3.79128784747792 + 1.27520055582102j) >>> c = cbrt(A*A-B*B*k); print c 16.0 q looks hopeless to simplify, but c is integer, thus possible. Perhaps the "nice" root is not the principle root ? >>> w = exp(2j/3*pi) >>> print q*w, q/w (-3.0 + 2.64575131106459j) (-0.791287847477919 - 3.92095186688561j) >>> a = -3 >>> f = lambda x: 4*x**3 - 3*c*x - A >>> print f(a) 0.0 >>> print a, (A/a - a*a)/3 -3 -7.0 We have ³√(36 + 20i√7) = (-3 + i√7) / (-1)^(2/3) However, converted to form a ± √r, expression look messier than original ³√(A + √R) |
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