[原创] 数学多次方程计算( 漂亮有趣的曲线）

shuxuestar

此方程有三个实数根，从图像上可以判定， 有两复数根





水滴线放大一千倍验证数据y=0.22515的准确度很高，一般实际应用小数点后三位足矣...............






对于高次运算   那些生成图像的软件些微误差也显现出来了.................  

shuxuestar

解中有两复根........

shuxuestar

  我搞不懂求出的复数解对于实际应用有何数学意义？    算出五个解应该取哪一个解？这是以前从没遇到过的问题。

shuxuestar

这个方程还可以用通用方法：

         y^5 + (59/16)*y^4 + (35/8)*y^3 + (23/16)*y^2 - (5/16)*y - 1/16 = 0

由水滴线原构造图像极限微分分析知道其一个根一定为-1,

可列出式子：

(y + 1)*(y + a)*(y + b)*(y + c)*(y + d) = 0;
     
y^5 + (d + c + b + a + 1)*y^4 + ((c + b + a + 1)*d + (b + a + 1)*c + (a + 1)*b + a)*y^3 + (((b + a + 1)*c + (a + 1)*b + a)*d + ((a + 1)*b + a)*c + a*b)*y^2 + ((((a + 1)*b + a)*c + a*b)*d + a*b*c)*y + a*b*c*d=0

四个未知数，五个式子
d + c + b + a + 1 = 59/16;
(c + b + a + 1)*d + (b + a + 1)*c + (a + 1)*b + a = 35/8;
((b + a + 1)*c + (a + 1)*b + a)*d + ((a + 1)*b + a)*c + a*b = 23/16;
(((a + 1)*b + a)*c + a*b)*d + a*b*c = -5/16;
a*b*c*d = -1/16;

解得：

a = 0.1424525158280573,
b = 1.385099035289393-0.1736777642881022*%i,
c = (-3.227331054310445e-16*%i)-0.2251505873207669,
d = 0.1736777642880978*%i+1.385099035289394,

a = 0.1424525158280573,
b = 1.385099035289393-0.1736777642881022*%i,
c = 0.1736777642881025*%i+1.385099035289393,
d = 2.629370195425532e-17*%i-0.2251505873207669,

a = 0.1424525158280573,
b = 0.1736777642881022*%i+1.385099035289393,
c = (-1.720832135641837e-16*%i)-0.225150587320767,
d = 1.385099035289392-0.1736777642881044*%i,

a = 0.1424525158280573,
b = 0.1736777642881022*%i+1.385099035289393,
c = 1.385099035289393-0.173677764288102*%i,
d = 1.17615701758094e-17*%i-0.2251505873207669,

a = 0.1424525158280573,
b = -0.2251506024096386,
c = 0.1736777642881021*%i+1.385099035289393,
d = 1.385099035289393-0.1736777642881021*%i,

a = 0.1424525158280573,
b = -0.2251506024096386,
c = 1.385099035289393-0.1736777642881021*%i,
d = 0.1736777642881021*%i+1.385099035289393,

a = -0.2251506024096386,
b = 1.385099035289393-0.1736777642881047*%i,
c = 1.978830968109014e-16*%i+0.1424525167419793,
d = 0.1736777642881054*%i+1.38509903528939,

a = -0.2251506024096386,
b = 1.385099035289393-0.1736777642881047*%i,
c = 0.1736777642881045*%i+1.385099035289395,
d = 0.1424525167419804-8.093215926332813e-17*%i,

a = -0.2251506024096386,
b = 0.1736777642881047*%i+1.385099035289393,
c = 0.1424525167419793-1.952790784758447e-16*%i,
d = 1.38509903528939-0.1736777642881054*%i,

a = -0.2251506024096386,
b = 0.1736777642881047*%i+1.385099035289393,
c = 1.385099035289395-0.1736777642881045*%i,
d = 8.093215926332813e-17*%i+0.1424525167419804,

a = -0.2251506024096386,
b = 0.1424525158280573,
c = 0.1736777642881021*%i+1.385099035289393,
d = 1.385099035289393-0.173677764288102*%i,

a = -0.2251506024096386,
b = 0.1424525158280573,
c = 1.385099035289393-0.1736777642881021*%i,
d = 0.173677764288102*%i+1.385099035289393,

a = 0.1736777642881021*%i+1.385099035289393,
b = 1.385099035289393-0.1736777642881021*%i,
c = 1.011105876115419e-17*%i+0.1424525167419802,
d = (-7.173551862572483e-18*%i)-0.2251505873207669,

a = 0.1736777642881021*%i+1.385099035289393,
b = 1.385099035289393-0.1736777642881021*%i,
c = 1.764451685447473e-17*%i-0.2251505873207672,
d = 0.142452516741981-5.881913249886616e-17*%i,

a = 0.1736777642881021*%i+1.385099035289393,
b = (-1.746656052621252e-17*%i)-0.2251505873207671,
c = 2.356212063456478e-17*%i+0.1424525167419802,
d = 1.385099035289391-0.1736777642881019*%i,

a = 0.1736777642881021*%i+1.385099035289393,
b = (-1.746656052621252e-17*%i)-0.2251505873207671,
c = 1.385099035289394-0.1736777642881021*%i,
d = 5.209773353490402e-17*%i+0.1424525167419808,

a = 0.1736777642881021*%i+1.385099035289393,
b = 8.047487978295697e-18*%i+0.1424525167419803,
c = (-1.673203473017727e-16*%i)-0.2251505873207673,
d = 1.385099035289389-0.1736777642881038*%i,

a = 0.1736777642881021*%i+1.385099035289393,
b = 8.047487978295697e-18*%i+0.1424525167419803,
c = 1.385099035289393-0.1736777642881019*%i,
d = 1.086730445866192e-17*%i-0.225150587320767,

a = 1.385099035289393-0.1736777642881021*%i,
b = 0.1736777642881021*%i+1.385099035289393,
c = 0.1424525167419785-1.140046790000185e-17*%i,
d = 9.867485600533811e-18*%i-0.2251505873207663,

a = 1.385099035289393-0.1736777642881021*%i,
b = 0.1736777642881021*%i+1.385099035289393,
c = (-1.635510771562707e-17*%i)-0.2251505873207652,
d = 6.436808821582165e-17*%i+0.1424525167419751,

a = 1.385099035289393-0.1736777642881021*%i,
b = 1.789470031907838e-17*%i-0.2251505873207671,
c = 0.1424525167419802-1.188992712934366e-16*%i,
d = 0.1736777642881015*%i+1.385099035289391,

a = 1.385099035289393-0.1736777642881021*%i,
b = 1.789470031907838e-17*%i-0.2251505873207671,
c = 0.1736777642881021*%i+1.385099035289393,
d = 0.1424525167419805-3.527274165556456e-17*%i,

a = 1.385099035289393-0.1736777642881021*%i,
b = 0.1424525167419803-7.352047458820314e-18*%i,
c = (-3.069054477786506e-16*%i)-0.2251505873207674,
d = 0.1736777642880973*%i+1.385099035289388,

a = 1.385099035289393-0.1736777642881021*%i,
b = 0.1424525167419803-7.352047458820314e-18*%i,
c = 0.1736777642881024*%i+1.385099035289393,
d = 2.948363159852845e-17*%i-0.225150587320767


从中筛选出两实数解：

a = -0.2251506024096386,
b = 0.1424525158280573,

y1=-1；y2=0.2251506024096386; y3=-0.1424525158280573

shuxuestar

从计算结果看，abcd解组属于一个4个解的排列问题 24=4×3×2×1个  而方程的解组为组合选取问题

解的数目应选取其中任何一组各不相同的（因为a,b,c,d各个互换等价）：

a = -0.2251506024096386,
b = 0.1424525158280573,
c = 1.385099035289393-0.1736777642881021*%i,
d = 0.173677764288102*%i+1.385099035289393,

shuxuestar

我发现的一类数学流形参数方程为：

x=(acos(a)+b)(  1-l/( (a^2+b^2+2abcos(a))^(1/2) ) )
y=asin(a)(  1-l/( (a^2+b^2+2abcos(a))^(1/2) ) )

标准方程为：

y^6+(3*x^2-4*b*x-2*l^2+2*b^2-2*a^2)*y^4+(3*x^4-8*b*x^3+((-4*l^2)+8*b^2-4*a^2)*x^2+(4*b*l^2-4*b^3+4*a^2*b)*x+l^4+(2*b^2-2*a^2)*l^2+b^4-2*a^2*b^2+a^4)*y^2+x^6-4*b*x^5+((-2*l^2)+6*b^2-2*a^2)*x^4+(4*b*l^2-4*b^3+4*a^2*b)*x^3+(l^4+((-2*b^2)-2*a^2)*l^2+b^4-2*a^2*b^2+a^4)*x^2=0；



可见x从1-6 次不缺次.......... 此方程是关于x不缺次的二元六次方程...

http://www.mathchina.com/bbs/for ... p;extra=&page=2

shuxuestar

当y=定数C，方程变为x的标准一元六次方程:


x^6
-4*b*x^5
+((-2*l^2)+3*c^2+6*b^2-2*a^2)*x^4
+(4*b*l^2-8*b*c^2-4*b^3+4*a^2*b)*x^3
+(l^4+((-4*c^2)-2*b^2-2*a^2)*l^2+3*c^4+(8*b^2-4*a^2)*c^2+b^4-2*a^2*b^2+a^4)*x^2
+(4*b*c^2*l^2-4*b*c^4+(4*a^2*b-4*b^3)*c^2)*x
+c^2*l^4+((2*b^2-2*a^2)*c^2-2*c^4)*l^2+c^6+(2*b^2-2*a^2)*c^4+(b^4-2*a^2*b^2+a^4)*c^2
=0.


我每取一个"可解算y值"=c，就会生成一个abcl参数的x 一元六次方程，各位要知道一定解不出x，

  但是呢 通过参数方程这个中间人就可以解算出每一个方程的一两个实数解x，

难道不是破解了高次方程了吗？ 六次通过一定变换取值可以退化成一元五次，一元四次........

应该统统可解，大家认为呢？

shuxuestar

现在做一个高次方程出来：

为了方便和简单，y只能取特殊值如（0，1，-1，1/2）*a等简单值，四个参数y=c的取值并不妨碍六次方程的复杂性。

x^6
-4*b*x^5
+((-2*l^2)+3*c^2+6*b^2-2*a^2)*x^4
+(4*b*l^2-8*b*c^2-4*b^3+4*a^2*b)*x^3
+(l^4+((-4*c^2)-2*b^2-2*a^2)*l^2+3*c^4+(8*b^2-4*a^2)*c^2+b^4-2*a^2*b^2+a^4)*x^2
+(4*b*c^2*l^2-4*b*c^4+(4*a^2*b-4*b^3)*c^2)*x
+c^2*l^4+((2*b^2-2*a^2)*c^2-2*c^4)*l^2+c^6+(2*b^2-2*a^2)*c^4+(b^4-2*a^2*b^2+a^4)*c^2
=0.   

取y=c=0，

x^6-4*b*x^5-2*l^2*x^4+6*b^2*x^4-2*a^2*x^4+4*b*l^2*x^3-4*b^3*x^3+4*a^2*b*x^3+l^4*x^2-2*b^2*l^2*x^2-2*a^2*l^2*x^2+b^4*x^2-2*a^2*b^2*x^2+a^4*x^2=0

因式分解为：x^2*(x-l-b-a)*(x-l-b+a)*(x+l-b-a)*(x+l-b+a)=0

六个解：0；0；l+b+a；l+b-a；-l+b+a；-l+b-a； a,b,l实数，六个实数解的一元六次方程

若x不等于0（一般不等于0），式子两边同乘以1/x^2,等式不变：

(x-l-b-a)*(x-l-b+a)*(x+l-b-a)*(x+l-b+a)=0

x^4
-4*b*x^3
+((-2*l^2)+6*b^2-2*a^2)*x^2
+(4*b*l^2-4*b^3+4*a^2*b)*x
+l^4+((-2*b^2)-2*a^2)*l^2+b^4-2*a^2*b^2+a^4
=0

属于典型的一元四次不齐次不缺次方程

取a=1，b=2，l=3；试试 可任意取值的

x^4-8*x^3+4*x^2+48*x=0    （x不等于0情况，两边除去x）

x^3-8*x^2+4*x+48=0  一个典型三次方程

取a=1，b=2，l=4

x^4-8*x^3-10*x^2+104*x+105=0  一个典型四次方程

这些方程的解均在前面解的（l+b+a；l+b-a；-l+b+a；-l+b-a）中的数值中寻找

可以称为存在一组通解的降次方程组解法

不难知道abl任意取 此类方程组是无穷多的...................

shuxuestar

  我保守估计：a,b,l 三剑客 可以做到三生万物.............

相信a,b,l 三剑客和万能的原方程   慢慢找方法  一定可以解决 世纪难题 ！

shuxuestar

取：a=123,b=456,l=789

x^4-1824*x^3-27684*x^2+783798336*x+146981640960=0

因式分解为：(x-1368)*(x-1122)*(x+210)*(x+456)=0

四个实数解无论多复杂应该可生成出来 反过来知道方程求abl 知道解