二次曲面とトーラス(torus) (2回目)
2022/12/5(月) 二次曲面とトーラス (2回目) トーラス(torus)は四次方程式なので 4次方程式の解(フェラーリの公式) の求め方の解説です Quartic equation ax 4 + bx 3 + cx 2 + dx 2 + e = 0を解く 変形 B = b/a, C = c/a, D = d/a, E = e/aと置くと x 4 + Bx 3 + Cx 2 + Dx + E = 0 ここで x = X - B/4と置くと(X 3 の項を消すため) x 4 + Bx 3 + Cx 2 + Dx + E = (X-B/4) 4 +B(X-B/4) 3 +C(X-B/4) 2 +D(X-B/4)+E = X 4 - 4X 3 (B/4) + 6X 2 (B/4) 2 - 4X(B/4) 3 + (B/4) 4 + BX 3 - 3BX 2 (B/4) + 3BX(B/4) 2 - B(B/4) 3 + CX 2 - 2CX(B/4) + C(B/4) 2 + DX - DB/4 + E = X 4 + (-B + B)X 3 + {6(B/4) 2 - 3B(B/4) + C}X 2 + {-4(B/4) 3 + 3B(B/4) 2 - 2C(B/4) + D}X + {(B/4) 4 - B(B/4) 3 - D(B/4) + E} = X 4 + {6(B/4) 2 - 12(B/4) 2 + C}X 2 + {-4(B/4) 3 + 12(B/4) 3 - 2C(B/4) + D}X + {(B/4) 4 - 4(B/4) 4 - D(B/4) + E} = X 4 + {-6(B/4) 2 + C}X 2 + {8(B/4) 3 - 2C(B/4) + D}X + {-3(B/4) 4 - D(B/4) + E} s = B/4 p = -6s 2 + C q = 8s 3 - 2Cs + D r = -3s 4 - Ds + E と置くと X 4 + pX 2 + qX + r = 0 q ＝ 0の時、(X 4 + pX 2 + r = 0) X 2 = {-p±√(p 2 -4r)}/2 X = ±√[{
