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- 1970-1-1
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发表于 2019-1-7 21:35:54
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function x = Solve3Polynomial(a, b, c, d)
% 范盛金. 一元三次方程的新求根公式与新判别法[J]. 海南师范学院学报,1989,2(2):91-98.
A = b*b - 3*a*c; if abs(A) < 1e-14; A = 0; end
B = b*c - 9*a*d; if abs(B) < 1e-14; B = 0; end
C = c*c - 3*b*d; if abs(C) < 1e-14; C = 0; end
DET = B*B - 4*A*C; if abs(DET) < 1e-14; DET = 0; end
if (A == 0) && (B == 0)
x1 = -c/b; x2 = x1 ; x3 = x1;
end
if DET > 0
Y1 = A*b + 1.5*a*(-B + sqrt(DET));
Y2 = A*b + 1.5*a*(-B - sqrt(DET));
y1 = nthroot(Y1,3); y2 = nthroot(Y2,3);
x1 = (-b-y1-y2)/(3*a);
vec1 = (-b + 0.5*(y1 + y2))/(3*a);
vec2 = 0.5*sqrt(3)*(y1 - y2)/(3*a);
x2 = complex(vec1, vec2);
x3 = complex(vec1, -vec2);
clear Y1 Y2 y1 y2 vec1 vec2;
end
if DET == 0 && (A ~= 0) && (B ~= 0)
K = (b*c-9*a*d)/(b*b - 3*a*c); K = round(K,14);
x1 = -b/a + K; x2 = -0.5*K; x3 = x2;
end
if DET < 0
sqA = sqrt(A);
T = (A*b - 1.5*a*B)/(A*sqA);
theta = acos(T);
csth = cos(theta/3);
sn3th = sqrt(3)*sin(theta/3);
x1 = (-b - 2*sqA*csth)/(3*a);
x2 = (-b + sqA*(csth + sn3th))/(3*a);
x3 = (-b + sqA*(csth - sn3th))/(3*a);
clear sqA T theta csth sn3th;
end
x = [x1; x2; x3];
end |
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