Displaying 1 result from an estimated 1 matches for "cardangeneralisation".
2024 Oct 10
0
Discriminant of a cubic polynomial
...ediary quantities
p = (d + sqrt(discr))^(1/n);
q = (d - sqrt(discr))^(1/n);
# Roots:
p + q # Base-Root
m = cos(2*pi/n) + 1i * sin(2*pi/n); # root of unity
p * m^c(0, seq(n-1)) + q * m^c(0, rev(seq(n-1))); # All Roots
See on my GitHub page:
https://github.com/discoleo/R/blob/master/Math/Polynomials.CardanGeneralisation.R
Sincerely,
Leonard
[[alternative HTML version deleted]]