一元三次和四次方程的根式解法

无心人

http://www.qthome.org/bbs/pascal_view.asp?id=28 一元三次和四次方程的根式解法 基本信息： 提交人：tianyuan_0 提交日期：2006-1-17 23:44:40 点击数：2075 编辑 删除 程序简介： 前面看到了有人用二分法解一元三次方程.我正在编的程序也需要求解一元三次方程. 代数学早已解决了一元三次和一元四次方程的求解问题,因而再用二分法求解无疑是效率不高的. 下面的程序省略了主函数. 平台是Visual C++ 2005. 由于我没有在VC++2005的中找到操作符(这也是令人奇怪的一件事),我重写了complex类. 程序内容： namespace mathapp{ const double PAI=3.141592653589793626; public ref struct mathappEx:public System::Exception { public: int i; System::String^ Desp; mathappEx(int n,System::String^ s) { i=n; Desp=s; } }; ///complex ///////////////////////////////////////////////////////////////////////////// ///////////////////////////////////////////////////////////////////////////// ///////////////////////////////////////////////////////////////////////////// value class complex { public:double _im,_re; public: complex(double re,double im) { _im=im; _re=re; } // // //property // // property double mo { double get() { return System::Math::Sqrt(_im*_im+_re*_re); } } property double arg { double get() { if(_re==0) { if(_im>0) return PAI/2; else return -PAI/2; } else { if(_re>0) return System::Math::Atan(_im/_re); else return PAI+System::Math::Atan(_im/_re); } } } // //operator // static complex operator ^(complex c,double up) { double t=System::Math:ow(c.mo,up); return complex::complex(t*System::Math::Cos(c.arg*up),t*System::Math::Sin(c.arg*up)); } static complex operator +(complex left,complex right) { return complex::complex(left._re+right._re,left._im+right._im); } static complex operator +(complex left,double d) { return complex::complex(left._re+d,left._im); } static complex operator +(double d,complex left) { return complex::complex(d+left._re,left._im); } static complex operator -(complex left,complex right) { return complex::complex(left._re-right._re,left._im-right._im); } static complex operator-(double left,complex right) { return complex::complex(left-right._re,-right._im); } static complex operator-(complex right,double left) { return complex::complex(right._re-left,right._im); } static complex operator *(complex left,complex right) { return complex::complex(left._re*right._re-left._im*right._im,left._re*right._im+left._im*right._re); } static complex operator *(double left,complex right) { return complex::complex(left*right._re,left*right._im); } static complex operator *(complex left,double right) { return complex::complex(left._re*right,left._im*right); } static complex operator /(complex left,complex right) { double t=right._im*right._im+right._re*right._re; if (t==0) { mathappEx^ e=gcnew mathappEx(0,"除数为0"); throw e; } else { complex c=complex::complex(right._re,-right._im); return t*c*left; } } static complex operator /(complex left,double right) { if(right==0) { mathappEx^e=gcnew mathappEx(0,"除数为0"); throw e; } else { return complex::complex(left._re/right,left._im/right); } } static complex operator /(double left,complex right) { double t=right._im*right._im+right._re*right._re; if (t==0) { mathappEx^ e=gcnew mathappEx(0,"除数为0"); throw e; } else { complex c=complex::complex(right._re,-right._im); return c*left/t; } } static int operator==(complex __z1, complex __z2) { return __z1._re == __z2._re && __z1._im == __z2._im; } static int operator==(complex c,double d) { return (c._re==d)&&(c._im==0.0); } static int operator==(double d,complex c) { return (c._re==d)&&(c._im==0.0); } static int operator!=(complex __z1, complex __z2) { return __z1._re != __z2._re || __z1._im != __z2._im; } static int operator!=(complex c,double d) { return c._im!=0||c._re!=d; } static int operator!=(double d,complex c) { return c!=d; } complex operator +() { return *this; } complex operator -() { return complex::complex(-_re,-_im); } // //convert // virtual System::String^ ToString()override { if(_im>=0) return this->_re.ToString()+"+"+this->_im.ToString()+"i"; else return this->_re.ToString()+"-"+this->_im.ToString()+"i"; } }; const complex omiga(-0.5,0.86602540378443864676372317075294); const complex omiga2(-0.5,-0.86602540378443864676372317075294); static int Sign(complex c) { if(c._re>0) return 1; if(c._re=0) return 0; return -1; } //////////////////////////////////////////////////////////////////////////////////// //////////////////////////////////////////////////////////////////////////////////// //////////////////////////////////////////////////////////////////////////////////// ref class MathApp { public: MathApp(void); static int OneDegree(double x,double c,double &jie) { if(x==0) return 0; jie=-c/x; return 1; } static int QuadricOne(double xx,double x,double c,cli::array^ root) { if(xx==0) { double t; if(OneDegree(x,c,t)!=0) { root[0]._re=t; root[0]._im=0; return 1; } else return 0; } complex cc(x*x-4*xx*c,0); root[0]=(-x+(cc^0.5))/(2*xx); root[1]=(-x-(cc^0.5))/(2*xx); if(cc==0.0) return 1; return 2; } static int CubicOne(double xxx,double xx,double x,double c,cli::array^ root)//一元三次方程求解 { /*1.x^3+px+q=0的解： xk=w^k*(-q/2+((q/2)^2+(p/3)^3))^(1/2))^(1/3)+ w^(3-k)*(-q/2-((q/2)^2+(p/3)^3))^(1/2))^(1/3) 其中：w=cos(2*pi/3)+i*sin(2*pi/3),k=1,2,3 2.a*x^3+b*x^2+c*x+d=0 令：y=x-b/(3*a)化为1.*/ if(xxx==0) return QuadricOne(xx,x,c,root); xx=xx/xxx; x=x/xxx; c=c/xxx; double p=x*0.33333333333333333333333-xx*xx*0.1111111111111111111111111111111; double q=xx*xx*xx/27.0 -xx*x*0.166666666666666666666666666 +c*0.5; complex d(p*p*p+q*q,0.0); complex d1=d^0.5; d=(d1-q)^(0.33333333333333333333333333333); if(d.mo!=0) d1=-p/d; else d1=(-q-d1)^(0.333333333333333333333333333); root[0]=d+d1-xx*0.3333333333333333333333333333; root[1]=d*omiga+d1*omiga2-xx*0.3333333333333333333333333333; root[2]=d*omiga2+d1*omiga-xx*0.3333333333333333333333333333; return 3; } static int Quartic(double xxxx,double xxx,double xx,double x,double c,cli::array^ root)//一元四次方程求解 { if(xxxx==0) return CubicOne(xxx,xx,x,c,root); xxx=xxx/xxxx; xx=xx/xxxx; x=x/xxxx; c=c/xxxx; double p=-0.375000000*xxx*xxx+xx; double q=0.12500000*xxx*xxx*xxx-0.50*xxx*xx+x; double r=-0.01171875*xxx*xxx*xxx*xxx+0.0625*xxx*xxx*xx-0.25*xxx*x+c; cli::array^ carray=gcnew array(3); int j=CubicOne(1.0,-2*p,p*p-4*r,q*q,carray); for(j=0;j<3;j++) carray[j]=(-carray[j])^0.5; if(Sign(carray[0]*carray[1]*carray[2])*System::Math::Sign(q)<0) carray[0]=-carray[0]; xxx=0.250*xxx; root[0]=0.50*(carray[0]-carray[1]+carray[2])-xxx; root[1]=0.50*(carray[0]+carray[1]-carray[2])-xxx; root[2]=0.50*(-carray[0]+carray[1]+carray[2])-xxx; root[3]=0.50*(-carray[0]-carray[1]-carray[2])-xxx; return 1; } }; } //yaos注： 未验证正确性

mathe

其实用牛顿迭代法计算（3次和4次方程）的数值解效率已经很高了。

无心人

呵呵 实用的不优美 优美的不实用啊 对了mathe 知道5次以上代数方程的三角函数解法么？

wayne

5次以上代数方程的三角函数解法

google没找到，可否详之

mathe

不知道5次方程的三角函数解法。 不过我知道五次方程的解可以用椭圆函数来表示，这个可以google到的

plp626

5次以下的根式解法有现成的公式， 5次以上的用群论的知识根据所给的系数也是可以解出的