论文部分内容阅读
利用卡当(Carden)公式可以解一般的三次方程,这里介绍一种用双曲正弦、双曲余弦和余弦来解三次方程的方法。任给三次方程x~3+bx~2+ex+d=0,利用代换y=x+1/3b,得出缺二次项的三次方程: y~3+py=q,其中p=1/3(3e-b~2),q=(9be-27d-2b~3)/27。再令y=hz,代入上式,得h~3z~3+phz=q,用k乘方程两边,得kh~3z~3+kphz=kq。令kh~3=4,pkh==3,这样从h=4|p_|/3~(1/2),k=3/h|p|,使上述方程取形为下面两种形式: 4z~3+3z=c,或4z~3-3z=c,其中c=kq。对于第一个方程,利用sh3θ=4sh~3θ++3shθ令z=shθ,解得z=sh(1/3Arshc)。对于第二个方程,当c≥1时,可利用
Using the Carden formula to solve ordinary cubic equations, a method for solving cubic equations using hyperbolic sine, hyperbolic cosine and cosine is introduced here. Given the cubic equation x~3+bx~2+ex+d=0, use the substitution y=x+1/3b to find the cubic equation missing the quadratic term: y~3+py=q, where p=1 /3(3e-b~2), q=(9be-27d-2b~3)/27. Let y = hz, substituting into the above equation, get h~3z~3+phz=q, multiply k by two sides of equation, get kh~3z~3+kphz=kq. Let kh~3=4, pkh==3, so from h=4|p_|/3~(1/2), k=3/h|p|, make the above equation take the following two forms: 4z ~3+3z=c, or 4z~3-3z=c, where c=kq. For the first equation, let z=shθ using sh3θ=4sh~3θ++3shθ and solve for z=sh(1/3Arshc). For the second equation, when c ≥ 1, available