关于脑电波，心电波的波形分析

关于脑电波，心电波的波形分析

脑电波，心电波的波形由很多个高次方程的抛物线波形，指数函数图像和y=(x-a)/(x-b)方程图像的波形叠加而成。这是一个杂波信号，可以和宇宙空间中的无线电射频信号发生电磁感应。用这个现象可以增强脑电波心电波信号，利用这个原理可以通过把信号调制到这个杂波信号上，达到增强传递无线电信号的目的。

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脑电波，心电波的波形的波形函数如下：

 

   a     x

Y=x   *a     *(x-b)/(x-c)

它的图像如下：

脑电波的图形如下所示

心电波的图像如下

这几个图像度比较类似，都是高波峰尖锐图形。

把这个函数的波形调制到下面这个函数上面，作为电磁波函数发送出去，就可以达到降低电磁波传递衰减的效果。

 

函数如下

    -84iθ

ρ=e

 

    -84iθ      -7iθ

ρ=e       +11e

 

    -84iθ      -7iθ      4iθ

ρ=e       +11e     +9ee

 

    -84iθ      -7iθ      4iθ

ρ=e       +11e     +9ee    -7iiθ

 

这个函数的图像如下图所示。

 

另外一个函数如下：

ρ(t)=a(t)*│sin[n(t)θ+φ(t)]*sin[2n(t))θ+φ(t)]│

当前值:n=6.81,a=1.23,φ=2.27rad,

               a

同时高次方程y=x  +bx+c 的解可以通过下面的公式求出。

第一部分  一元三次方程的求解，卡丹公式。

推导过程可参见А.Г.УРОШ库洛什著高等代数教程1953年版，

41.三次与四次方程，

说明，计算一元三次立方根的卡尔丹公式如下：

  3    2                                    

y   +ay  +by+c=0                  (1)

设y=x+h,得

     3       2                               

(x+h)  +a(x+h)  +b(x+h)+c=0

 3        2     2           3                                         

x  +(3h+a)x  +(3h  +2ah+b)x+h  +bh+c=0

上面方程可转化为,

  3                                

x  +px+q=0                (3)

其中, y=x-a/3,                (2)

h=-a/3,

     2          2                            

p=3h  +b+2ah=b-a  /3,

 

    3        3                                     

q=h  +bh+c=-a  /27-ab/3+c,

只要求得方程(3)的根，那么我们根据(2)就可以得到方程(1)的根, 根据基本定理方程(3)有三个复数根，设x0是其中一个, 我们引入辅组未知量u来讨论多项式,

 

  第二部分 一元四次方程费拉里求根公式         

   4    3     2

 y   +ay   +by   +cy+d=0                   (13)

预先代以y=x-a/4化方程(13)为:

  4     2                       

x   +px   +qx+r=0

上式中h=-a/4, y=x-a/4,

 

     2          4   3           3      2               

p=6h   +3ah, r=h  +ah  +h+d, q=4h   +3ah  +c,

解得,

 

 

 

 

 

解得,

 

                                   p              q   

     2t  ±     2t   -4(       +t`    +         )        

         0          0      2       0    2   2t`     

                                                       0             

  u=y=x-a``/4=                                                    -

                            2  

其中,

         3                                3            

 

                           2        3                     2       3                                            

            -q         q       p         -q      q      p       

t  =          +             +        +         -           +         -p/3         

    0        2            4      27           2          4      27        

 

 

 

 

         3                                3            

 

                           2        3                     2       3                                            

            -q         q     p    2   -q       q    p       

t  =ε          +             +      +ε         -           +          -p/3         

 1          2            4      27           2          4      27        

 

         3                                3            

 

                           2        3                     2       3                                            

     2     -q          q    p          -q      q     p       

t  =ε          +             +      +ε         -           +         -p/3          

 2          2            4      27           2          4      27        

 

上式中,

          2                   

p````=-r+p   /4-p``/3,

 

       3             2      2                           

q````=-p  /27-p(-r+p  )/12-q``  /8

 

 

 

 

 

最后得到上面一元四次方程的解x=y+h=u+v+h=u+v-a/4,

 

                                   p              q   

     2t  ±     2t   -4(       +t    +          )        

         0          0      2       0    2   2t     

                                                       0             

 y=x-a/4=                                                     - -a/4

                            2  

其中,

         3                                3            

 

                           2        3                     2       3                                            

            -q           q       p           -q          q      p       

t  =          +             +        +         -           +         -p/3         

    0        2            4      27           2          4      27        

 

 

         3                                3            

 

                           2        3                     2       3                                            

            -q           q        p     2    -q          q      p       

t  =ε          +             +      +ε         -           +          -p/3         

 1          2            4      27           2          4      27        

 

 

         3                                3            

 

                           2        3                     2       3                                            

     2     -q            q        p          -q          q       p      

t  =ε          +             +      +ε         -           +         -p/3          

 2          2            4      27           2          4      27        

 

上式中,

        2          3          2      2                                             

p=-r+p   /4-p/3, q=-p  /27-p(-r+p   )/12-q  /8,

 

 

 

三.计算一元四次，五次方程的近似解法

1.计算一元四次方程的近似解

  4    3     2                  

x   +ax   +bx  +cx+d=0

 

       4                                           4           

假设x   的系数k ≠1, 可以给方程左边同时除以k，使x   的系数k变成1, 

设x=y+h，得

     4       3       2                                      

(y+h)  +a (y+h)  +b(y+h)  +c(y+h)+d=0                (1)

化简(1)得,

  4   3     2   2    3   4    3     2       2   3     2         2                         

y  +4h  y+6h  y  +4hy  +h  +ay  +3ah  y+3ahy  +ah  +by  +2bhy+bh  +cy+ch+d=0

 

  4       3     2         2    3     2         4    3    2           

y  +(4h+a)y  +(6h  +3ah+b)y  +(4h  +3ah  +2bh)y+h  +ah  +bh  +ch+d=0      (2)

 

设 a+4h=0,得

h=-a/4,

化简(2)得

  4    2        2     3     2         4    3    2                                      

y  +(6h  +3ah+b)y  +(4h  +3ah  +2bh)y+h  +ah  +bh  +ch+d=0

设y=u-v+w，得

    4     2            2     3     2           4    3      2                      

(u+v)  +(6h  +3ah+b)(u+v)  +(4h  +3ah  +2bh)(u+v)+h  +ah   +bh  +ch+d=0

 

  4    3     2  2    3   4    2         2      2              2        2     

u  +4uv  +6u  v  +4u  v+v  +(6h  +3ah+b)u  +2(6h  +3ah+b)uv+(6h  +3ah+b)v

 

    3     2          3      2        4     3   2

+(4h  +3ah  +2bh)u+(4h  +3ah  +2bh)v+h  +ah  +bh  +ch+d=0

 

   3    3        2      2             2            3     2         3                  

u[u  +4v  +6uv+4u  v+(6h  +3ah+b)u+2(6h  +3ah+b)v+(4h  +3ah  +2bh)]+v[v  +

 

   2            3     2         4   3     2                    

(6h  +3ah+b)v+(4h  +3ah  +2bh)]+h  +ah  +bh  +ch+d=0

 

所以，可以这样选取u,v使得

 

 

 

   3    3        2      2             2            3     2  

u[u  +4v  +6uv+4u  v+(6h  +3ah+b)u+2(6h  +3ah+b)v+(4h  +3ah  +2bh)]=0     (3)

{

   3    2            3     2         4   3     2  

v[v  +(6h  +3ah+b)v+(4h  +3ah  +2bh)]+h  +ah  +bh  +ch+d=0          (4)

由(4)得,

 

   3    2            3     2          4    3     2  

v[v  +(6h  +3ah+b)v+(4h  +3ah  +2bh)]=-(h  +ah  +bh  +ch+d=0)     

 

 

   3    2            3     2      

v[v  +(6h  +3ah+b)v+(4h  +3ah  +2bh)] 

                                     =1       

       4    3    2                                          

-(h  +ah  +bh  +ch+d)    

 

 

    3    2            3     2      

v[v  +(6h  +3ah+b)v+(4h  +3ah  +2bh)]        1       

                                     =     

         4    3    2                     100000                     

-100000(h  +ah  +bh  +ch+d)   

由于上面方程左右两边的值都小于0.0001，所以给方程左边除以一个数v，再给方程右边乘以一个数v，方程左右两边近似相等,

 

    3    2            3     2      

v[v  +(6h  +3ah+b)v+(4h  +3ah  +2bh)]           

                                     =0.00001v     

         4    3    2                                       

-100000(h  +ah  +bh  +ch+d)   

注意：

3    2            3     2              4     3    2                              

v  +(6h  +3ah+b)v+(4h  +3ah  +2bh)≈-0.01v(h  +ah  +bh  +ch+d)

 

       4    3     2                                                      

0.01v(h   +ah  +bh  +ch+d)中的0.0001的取值和方程系数a,b,c有关系有关系，

 

                               4    3    2                     

当a,b,c都小于100时，取0.01v(h   +ah  +bh  +ch+d),

                                                          4    3    2           

当a,b,c都大于100时，且a,b,c的值都小于1000时，取0.001v(h   +ah  +bh  +ch+d),

 

 

                                                            4    3    2           

当a,b,c都大于1100时，且a,b,c的值都小于10000时，取0.0001v(h   +ah  +bh  +ch+d),

其它情况依次类推, 所以,

  3     2              4     3   2            3     2                                   

v   +[(6h  +3ah+b)+0.01(h  +ah  +bh  +ch+d)]v+(4h  +3ah  +2bh)≈0

上面方程（5）可转化为,

   3                  

x   +px+q=0,

其中, x`=v,

      2              4    3    2                                               

p`=[(6h  +3ah+b)+0.01(h  +ah  +bh  +ch+d)]

 

      3     2                         

q`=4h  +3ah  +2bh

 

根据一元三次方程卡尔丹公式上面方程的根为:

        3                           3               

                                           

                     2       3                    2        3                           

          -q       q       p          q        q       p   

v   =       +        +         +   -      -        +  

    0      2        4      27            2        4       27      

 

        3                           3               

                                           

                     2       3                    2        3                           

          -q       q       p     2    q        q       p   

v  =ε       +        +      +ε   -      -        +  

    1      2        4      27           2        4      27      

 

        3                           3               

                                           

                     2       3                    2        3                           

       2   -q      q       p          q       q       p   

v  =ε       +        +      +ε   -      -        +  

    2      2        4      27           2        4      27      

 

其中,

 

  3

ε =1,

因为1开立方在复数平面内有3个根。分别是

ε  =1, ε  =-1/2+i√3/2, ε  =-1/2+i√3/2,

  0      1               2

推导过程可参见7.复数的方根,

由(3)得,

 

 3    3         2     2             2            3      2                                   

u  +4v  +6uv+4u  v+(6h  +3ah+b)u+2(6h  +3ah+b)v+(4h  +3ah  +2bh)=0       (9)

上面方程（9）可转化为,

   3      2                                      

y  +ay  +by+c=0                  (1)

 

其中, a``=4v,

 

      2                                        

b``=6h  +3ah+b+6v,

 

      3     2             3      2                         

c``=4v   +2(6h  +3ah+b)v+(4h  +3ah  +2bh)

上面方程可转化为,

   3                             

x   +px+q=0                   (3)

其中,

y=x-a``/3                 (2)

 

       2                    

p=-a   +b``,

 

q=-ab/3+c,

上面方程的根为:

        3                           3               

                                           

                     2       3                    2        3                           

          -q       q     p           q      q      p   

u   =       +        +         +   -      -        +         -a``/3     

    0      2        4      27            2       4      27      

 

        3                           3               

                                           

                     2       3                    2        3                           

          -q      q      p     2    q       q       p   

u  =ε       +        +      +ε   -      -        +         -a``/3 

1      2       4      27            2        4      27      

 

 

 

 

 

        3                           3               

                                           

                     2       3                    2        3                           

       2   -q     q       p          q      q       p   

u  =ε       +        +      +ε   -      -        +         -a``/3  

    2      2       4       27           2        4      27      

 

其中,

 

  3

ε =1,

因为1开立方在复数平面内有3个根。分别是

ε  =1, ε  =-1/2+i√3/2, ε  =-1/2+i√3/2,

  0      1               2

推导过程可参见7.复数的方根,

推导过程可参见А.Г.УРОШ库洛什著高等代数教程1953年版,

最后得到上面一元四次方程的解, x=y+h=u+v+h=u+v-a/4,

        3                           3               

                                           

                     2       3                    2        3                           

          -q       q     p            q       q      p   

x   =       +        +         +   -      -        +  

    0      2        4      27            2       4      27      

 

        3                           3               

                                           

                     2       3                    2        3                           

          -q      q      p           q      q       p  

+       +        +          +   -      -        +       -a``/3-a/4

      2       4      27            2        4      27      

 

 

 

 

 

 

 

 

 

 

 

 

 

 

        3                           3               

                                           

                     2       3                    2        3                           

           -q     q       p     2     q       q       p   

x  =ε       +        +      +ε   -      -        +  

    1      2       4       27           2        4      27      

 

 

 

        3                           3               

                                           

                     2       3                    2        3                           

          -q     q       p    2      q      q       p   

+ε       +        +       +ε    -      -        +       -a``/3-a/4

           2       4       27           2        4      27      

 

 

        3                           3               

                                           

                     2       3                    2        3                           

       2    -q     q       p          q       q       p   

x  =ε       +        +      +ε   -      -        +  

    2      2       4       27           2        4      27      

 

 

 

        3                           3               

                                           

                     2       3                    2        3                           

       2   -q     q       p          q      q       p   

+ε       +        +       +ε    -      -        +       -a``/3-a/4

           2       4       27           2        4      27      

 

2.计算一元五次方程的近似解

  5    4    3    2                                       

x   +ax  +bx  +cx  +dx+e=0           

 

      5                                           5      

假设x  的系数k ≠1, 可以给方程左边同时除以k，使x   的系数k变成1,

设x=y+h，得

     5        4       3       2                

(y+h)   +a (y+h)  +b(y+h)  +c(y+h)  +d (y+h)+e=0           (1)

化简(1)得,

  5      4     2   3     3  2   4     5   4      3      2  2     3                           

y   +5hy   +10h  y  +10h  y  +5h  y+h  +ay  +4ahy   +6ah  y  +4ah  y

 

     4    3        2     2     3    2          2                        

+ahn   +by   +3bhy   +3bh  y+bh  +cy  +2chy +ch  +dy+dh+e=0

 

  5       4         2     3     2     3         2     3    4               5         

y  +(a+5h)y  +(4a+10h  +b)y  +(6ah  +10h  +c+3bh)y  +(4ah  +5h  +2ch+3bh+d)y+h  

 

   4    3    2                            

+ah  +bh  +ch  +dh+e=0

 

设a+5h=0,得, h=-a/5, x=y-a/5,

化简(2)得,

  5        2     3     2     3         2     3     4              5     4            

y  +(4a+10h  +b)y  +(6ah  +10h  +c+3bh)y  +(4ah  +5h  +2ch+3bh+d)y+h  +ah

 

    3   2      

+bh  +ch   +dh+e=0

 

设 y=u+v，得

    5          2       3      2     3            2      3     4               

(u+v)  +(4a+10h  +b)(u+v)  +(6ah  +10h  +c+3bh)(u+v)  +(4ah   +5h  

 

                  5    4     3    2             

+2ch+3bh+d)(u+v)+h   +ah  +bh   +ch  +dh+e=0             (3)

 

因为,

     5    5      4     2   3      3   2    4    5          

(u+v)   =u   +5vu  +10v   u   +10v   u  +5v   u+v             (4)      

 

       2        3      3        2       2       3      2  3     2   2                

(4a+10h  +b)(u+v)   =4au   +12avu   +12av   u+4av   +10h  u  +30h  vu

 

     2  2       2   3   3       2     2     3                

+30h  v   u+10h   v  +bu  +3bvu  +3bv  u+bv               (5)     

 

    2     3            2     2       2       2  2     3  2     3                                 

(6ah  +10h  +c+3bh)(u+v)  =6ah  u+12ah  vu+6ah  v  +10h  u  +20h  vu

 

     3   2    2         2      2            2                         

+10h   v  +cu  +2cuv+cu   +3bhu  +6bhu+3bhu                  (6)  

 

    3    4                    3     4                   3     4                        

(4ah  +5h  +2ch+3bh+d)(u+v)=4ah  u+5h  u+2chu+3bhu+du+4ah  v+5h  v+2chv+3bhv+dv (7)

 

化简(3)得,

   4     3     2  2     3     4        2     2           2                   

u[u  +5vu  +10v  u  +10v  u+5v  +(4a+10h  +b)u  +3v(4a+10h  +b)u

 

   2        2        2      3              2      3                                                               

+3v  (4a+10h  +b) +(6ah  +10h  +c+3bh)u+2v(6ah  +10h  +c+3bh)

 

     3    4                4        2     2     2      3                  

+(4ah  +5h  +2ch+3bh+d)]+v[v  +(4a+10h  +b)v  +(6ah  +10h  +c+3bh)v

 

    3    4              5    4    3    2                     

+(4a  +5h  +2ch+3bh+d)]+h  +ah  +bh  +ch  +dh+e=0

 

所以，可以这样选取u,v，使得

   4     3     2  2     3     4        2     2           2       2        2                   

u[u  +5vu  +10v  u  +10v  u+5v  +(4a+10h  +b)u  +3v(4a+10h  +b)u+3v  (4a+10h  +b)

 

     2     3               2     3            3    4                             

+(6ah  +10h  +c+3bh)u+2v(6ah  +10h  +c+3bh)+(4ah  +5h  +2ch+3bh+d)]=0         (8)

 

{

   4          2    2      2     3            3                 5     4    3                  

v[v   +(4a+10h  +b)v  +(6ah  +10h  +c+3bh)v+(4a  +5h +2ch+3bh+d)]+h  +ah  +bh

 

   2                  

+ch  +dh+e=0                                                      (9)

 

由(9)得

   4        2     2      2    3            3    4                                            

v[v  +(4a+10h  +b)v  +(6ah  +10h  +c+3bh)v+(4a  +5h  +2ch+3bh+d)]

 

    5   4     3    2                     

=-(h  +ah  +bh  +ch  +dh+e)

 

 

   4        2     2      2    3            3    4                                            

v[v  +(4a+10h  +b)v  +(6ah  +10h  +c+3bh)v+(4a  +5h  +2ch+3bh+d)]

                                                              =1  

                  5    4    3    2                                               

               -(h  +ah  +bh  +ch  +dh+e)    

 

    4        2     2      2    3            3    4                                            

v[v  +(4a+10h  +b)v  +(6ah  +10h  +c+3bh)v+(4a  +5h  +2ch+3bh+d)]     1       

                                                              =  

                        5    4    3    2                                               

               -100000(h  +ah  +bh  +ch  +dh+e)                   100000             

 由于上面方程左右两边的值都小于0.0001，所以给方程左边除以一个数v，再给方程右边乘以一个数v，方程左右两边近似相等,

    4        2     2      2    3            3    4                                            

v[v  +(4a+10h  +b)v  +(6ah  +10h  +c+3bh)v+(4a  +5h  +2ch+3bh+d)]         

                                                              ≈0.00001v

                        5    4    3    2                                               

               -100000(h  +ah  +bh  +ch  +dh+e)         

 

  4        2    2      2     3            3    4                                       

v  +(4a+10h  +b)v  +(6ah  +10h  +c+3bh)v+(4a  +5h  +2ch+3bh+d)

 

         5    4    3     2     

≈-0.01v(h  +ah  +bh  +ch  +dh+e)  

 

注意：

       5    4    3    2                                      

0.001v(h  +ah  +bh  +ch  +dh+e)中的0.0001的取值和方程系数a,b,c有关系有关系，   

 

                                5   4    3    2                                      

 当a,b,c都小于100时，取0.01v(h  +ah  +bh  +ch  +dh+e)

 

 

当a,b,c都大于100时，且a,b,c的值都小于1000时，

 

          5    4   3    2                                   

取0.001v(h  +ah  +bh  +ch  +dh+e),

 

当a,b,c都大于1100时，且a,b,c的值都小于10000时，

 

           5    4    3    2                                                                                                

取0.0001v(h  +ah  +bh  +ch  +dh+e),

其它情况依次类推, 所以,

  4        2     2     2     3            3    4                                        

v  +(4a+10h  +b)v  +(6ah  +10h  +c+3bh)v+(4a  +5h  +2ch+3bh+d)

 

        5   4    3     2                

+0.01v(h  +ah  +bh  +ch  +dh+e)≈0

上面方程（10）可转化为,

  4     2                                     

x  +px  +qx+r=0

上式中,

v=x`,

         2              

p`=4a+10h  +b,

 

      2     3             5    4    3    2                  

q`=6ah  +10h  +c+3bh+0.01(h  +ah  +bh  +ch  +dh+e)

 

     4    3                                

r`=4a  +5h  +2ch+3bh+d

 

根据一元四次方程费拉里公式上面方程的根为:

 

                                   p              q   

     2t  ±     2t   -4(       +t`    +         )        

         0          0      2       0    2   2t`     

                                                       0             

v=x`=                                                      -

                            2  

其中,

         3                                3            

 

                           2        3                     2       3                                            

            -q         q       p         -q      q      p       

t  =          +             +        +         -           +         -p/3         

    0        2            4      27           2          4      27        

 

         3                                3            

 

                           2        3                     2       3                                            

            -q         q       p    2   -q       q     p       

t  =ε          +             +      +ε         -           +          -p/3         

 1          2            4      27           2          4      27        

 

         3                                3            

 

                           2        3                     2       3                                            

     2     -q          q      p          -q       q     p       

t  =ε          +             +      +ε         -           +         -p/3          

 2          2            4      27           2          4      27        

 

上式中,

         2                   

p```=-r+p   /4-p`/3,

 

       3          2      2                           

q```=-p  /27-p(-r+p  )/12-q`  /8

由(8)得

 

  4    3      2  2    3     4         2    2           2       2        2                       

u  +5vu  +10v  u  +10v  u+5v  +(4a+10h  +b)u  +3v(4a+10h  +b)u+3v  (4a+10h  +b)

 

     2     3               2     3            3     4                     

+(6ah  +10h  +c+3bh)u+2v(6ah  +10h  +c+3bh)+(4ah  +5h  +2ch+3bh+d)=0

 

 

  4    3     2         2    2      3           2       2     3                       

u  +5vu  +(10v  +4a+10h  +b)u  +[10v  +3v(4a+10h  +b)+6ah  +10h  +c+3bh]u

 

   4        2     3            3    4                           

+5v  +2v(6ah  +10h  +c+3bh)+(4ah  +5h  +2ch+3bh+d)=0              (11)

 

上面方程（11）可转化为,

         4     3       2                               

y  +ay  +by  +cy+d=0

上式中, a``=5v,

 

      2        2                        

b``=10v  +4a+10h  +b

 

       3           2        2     3                        

c``=10v   +3v(4a+10h  +b)+6ah  +10h  +c+3bh

 

      4       2     3            3    4                         

d``=5v  +2v(6ah  +10h  +c+3bh)+(4ah  +5h  +2ch+3bh+d)

预先代以y=x-a``/4化方程为:

 

         4     2                                  

x  +px  +qx+r=0

上式中,

h=-a/4,

       2             4       3                                               

p=6h  +3ah, r=h  +ah  +h+d, y=x-a``/4, 

 

       3       2         

q=4h  +3ah  +c``

解得,

 

                                   p              q   

     2t  ±     2t   -4(       +t`    +         )        

         0          0      2       0    2   2t`     

                                                       0             

  u=y=x-a``/4=                                                    -

                            2  

其中,

         3                                3            

 

                           2        3                     2       3                                            

            -q         q       p         -q      q      p       

t  =          +             +        +         -           +         -p/3         

    0        2            4      27           2          4      27        

 

         3                                3            

 

                           2        3                     2       3                                            

            -q         q     p    2   -q       q    p       

t  =ε          +             +      +ε         -           +          -p/3         

 1          2            4      27           2          4      27        

 

         3                                3            

 

                           2        3                     2       3                                            

     2     -q          q    p          -q      q     p       

t  =ε          +             +      +ε         -           +         -p/3          

 2          2            4      27           2          4      27        

 

上式中,

          2                   

p````=-r+p   /4-p``/3,

 

       3             2      2                           

q````=-p  /27-p(-r+p  )/12-q``  /8

最后得到上面一元四次方程的解x=y+h=u+v+h=u+v-a/4,

 

                                   p              q   

     2t  ±     2t   -4(       +t`    +         )        

         0          0      2       0    2   2t`     

                                                       0             

       x=                                                    -

                            2  

 

                                   p              q   

     2t  ±     2t -4(       +t``    +         )        

         0          0      2       0    2   2t``     

                                                       0             

       +                                                    - -a``/4-a/4

                            2  

3.由数学归纳法可知，计算一元n次方程近似解的公式如下

   n       3    2                          

x    +...+ax   +bx  +cx+d=0

 

       4                                            4       

假设x    的系数k ≠1, 可以给方程左边同时除以k，使x   的系数k变成1,

设x=y+h，得,

    n          3        2                     

(y+h)  +...+a (y+h)  +b(y+h)  +c(y+h)+d=0                 (1)

化简(1)得,

 n      n      3     2       2     3   2         2                              

y  +...+h  +...+ay  +3ah  y+3ahy  +ah  +by  +2bhy+bh  +cy+ch+d=0

 

  n       n-1           2          3          4     3    2                                         

y  +(nh+a)y   +(...+3ah+b)y   +(...+3ah  +2bh)y+...+h  +ah  +bh  +ch+d=0        (2)

 

设a+nh=0,得, h=-a/n

化简(2)得,

  n            2         3           4    3    2                   

y   +(...+3ah+b)y   +(...+3ah  +2bh)y+...+h  +ah  +bh  +ch+d=0

n次方程各项的系数可以通过二项式定理计算, 二项式展开公式如下:

    n    0  n  1  n-1       k  n-k  k      n  n                         

(a+b)  =C  a  +C  a   b+...+C   a   b  +...+C  b

         n     n           n             n

设y=u-v+w，得

    n               2         3              4    3    2         

(u+v)  +(...+3ah+b)(u+v)  +(...+3ah  +2bh)(u+v)+...+h  +ah  +bh  +ch+d=0

 

   n-1     3    3        2     2             2             3     2                              

u[u   +...+u  +4v  +6uv+4u  v+(6h  +3ah+b)u+2(6h  +3ah+b)v+(4h  +3ah  +2bh)]

 

    n-1     3    2            3     2         n     4    3     2                

+v[v   +...+v  +(6h  +3ah+b)v+(4h  +3ah  +2bh)]+h  +...+h  +ah  +bh  +ch+d=0

 

所以，可以这样选取u,v使得,

 

   n-1     3    3        2     2              2            3     2                                    

u[u   +...+u  +4v  +6uv+4u  v+(6h  +3ah+b)u+2(6h  +3ah+b)v+(4h  +3ah  +2bh)]=0 (3)

 

{

   n-1     3    2              3     2        n      4     3   2                     

v[v   +...+v  +(6h  +3ah+b)v  +(4h  +3ah  +2bh)]+h  +...+h  +ah  +bh  +ch+d=0   (4)

由(4)得,

   n-1     3    2              3     2         n      4     3   2                                

v[v   +...+v  +(6h  +3ah+b)v  +(4h  +3ah  +2bh)]=-(h  +...+h  +ah  +bh  +ch+d)

 

 

   n-1     3    2              3     2                                    

v[v   +...+v  +(6h  +3ah+b)v  +(4h  +3ah  +2bh)]

                                              =1

           n     4    3    2                                    

-(h  +...+h  +ah  +bh  +ch+d)      

 

   n-1     3    2              3     2                                    

v[v   +...+v  +(6h  +3ah+b)v  +(4h  +3ah  +2bh)]       1   

                                              =

           n     4    3    2                     100000              

-100000(h  +...+h  +ah  +bh  +ch+d)      

由于上面方程左右两边的值都小于0.0001，所以给方程左边除以一个数v，再给方程右边乘以一个数v，方程左右两边近似相等,

   n-1     3    2              3     2                                    

v[v   +...+v  +(6h  +3ah+b)v  +(4h  +3ah  +2bh)]          

                                              ≈0.00001v

           n     4    3    2                                

-100000(h  +...+h  +ah  +bh  +ch+d)      

注意：

 n-1     3     2             3      2              4    3    2                        

v   +...+v  +(6h  +3ah+b)v  +(4h  +3ah  +2bh)≈-0.01v(h  +ah  +bh  +ch+d)

 

       n       4   3    2                                                         

0.01v(h   +...+h  +ah  +bh  +ch+d)中的0.0001的取值和方程系数a,b,c有关系有关系，

 

                              n       4    3      2               

当a,b,c都小于100时，取0.01v(h   +...+h   +ah   +bh  +ch+d),

 

                                                      n       4    3    2        

当a,b,c都大于100时，且a,b,c的值都小于1000时，取0.001v(h   +...+h  +ah  +bh  +ch+d),

 

 

当a,b,c都大于1100时，且a,b,c的值都小于10000时，

            n      4   3      2                               

取0.0001v(h   +...+h  +ah  +bh   +ch+d)

其它情况依次类推, 所以,

 n-1     3     2              4    3     2           3     2               

v   +...+v  +[(6h  +3ah+b)+0.01(h  +ah  +bh  +ch+d)]v+(4h  +3ah  +2bh)≈0   (5)

上面方程（5）可转化为:

         n-1                                     

       x   +...+px+q=0

其中, x`=v,

 

       2             4     3    2                             

p`=[(6h  +3ah+b)+0.01(h   +ah  +bh  +ch+d)]

 

      3     2                     

q`=4h  +3ah  +2bh

根据一元n-1次方程求根公式:

         3                                3            

 

                           2        3                     2       3                                            

            -q           q       p         -q          q      p      

v  =          +             +        +         -           +              (6)    

    0        2            4      27           2          4      27        

 

         3                                3            

 

                           2        3                     2       3                                            

            -q           q     p       2   -q          q      p      

v  =ε          +             +      +ε         -           +              (7)  

 1          2            4      27           2          4      27        

 

         3                                3            

 

                           2        3                     2       3                                            

     2     -q            q      p          -q         q       p       

v  =ε          +             +      +ε         -           +              (8)

 2          2            4      27           2          4      27        

 

上式中,

  3                    

ε   =1,

因为1开立方在复数平面内有3个根。分别是

ε  =1, ε  =-1/2+i√3/2, ε  =-1/2+i√3/2,

   0      1             2

推导过程可参见7.复数的方根,

由(3)得,

 n-1     3    3        2      2             2            3     2                         

u   +...+u  +4v  +6uv+4u  v+(6h  +3ah+b)u+2(6h  +3ah+b)v+(4h  +3ah  +2bh)=0   (9)

上面方程（9）可转化为:

  n-1        2                       

y   +...+ay  +by+c=0         (1)

其中, a``=4v,

      2               3      2            3      2                              

b=6h  +3ah+b+6v, c=4v  +2(6h  +3ah+b)v+(4h  +3ah  +2bh)

上面方程可转化为:

           n-1                                         

x   +...+px+q=0             (3)

其中,

       y=x-a``/3       (2)

               2                            

        p=-a  +b , q=-ab/3+c``

上面方程的根为:

         3                                3            

 

                           2        3                     2       3                                            

            -q           q      p         -q        q      p      

u  =          +             +        +         -           +       -a``/3        

    0        2            4      27           2          4      27        

 

         3                                3            

 

                           2        3                     2       3                                            

            -q           q     p       2   -q          q      p      

u  =ε          +             +      +ε         -           +        -a``/3          

 1          2            4      27           2          4      27        

 

         3                                3            

 

                           2        3                     2       3                                            

     2     -q            q      p          -q         q       p       

u  =ε          +             +      +ε         -           +        -a``/3        

 2          2            4      27           2          4      27        

 

        3                     

其中, ε   =1,

因为1开立方在复数平面内有3个根。分别是:

ε  =1, ε  =-1/2+i√3/2, ε  =-1/2+i√3/2,

   0      1             2

 

 推导过程可参见7.复数的方根,

     推导过程可参见А.Г.УРОШ库洛什著高等代数教程1953年版,

最后得到上面一元n-1次方程的解, x=y+h=u+v+h=u+v-a/4,

         3                                3            

 

                           2        3                     2       3                                            

            -q           q      p          -q         q       p      

x  =          +             +        +         -           +              

    0        2            4      27           2          4      27        

 

         3                                3            

 

                           2       3                      2       3                                            

            -q          q     p           -q        q      p      

+          +            +      +           -           +        -a``/3-a/4      

            2            4      27           2          4      27        

 

         3                                3            

 

                           2        3                     2       3                                            

           -q            q      p     2    -q         q       p       

x  =ε          +             +      +ε         -           +             

 1          2            4      27           2          4      27        

 

 

         3                                3            

 

                           2        3                     2       3                                            

           -q           q      p     2    -q        q      p       

+ε          +            +      +ε         -           +        -a``/3-a/4     

            2            4      27           2          4      27        

 

 

                               

         3                                3            

 

                           2        3                     2       3                                            

      2    -q            q      p          -q         q       p       

x  =ε          +             +      +ε         -           +             

 2          2            4      27           2          4       27        

 

 

 

 

 

         3                                3            

 

                           2        3                     2       3                                            

       2    -q          q      p         -q        q      p       

+ε          +            +      +ε         -           +        -a``/3-a/4     

            2            4      27           2          4      27        

利用下面计算一元二次方程的模拟计算机电路，可以计算模拟出上面心电波，脑电波的类似高次方程的函数的波形。计算一元二次方程的电路如下：

 

 

一元二次方程ax²+bx+c=0（a≠0）,它的求根公式是x=[-b±√(b²-4ac)]/2a。

例如计算下面的一元二次方程x²+3x+2=0，根据求根公式，它的根是2,1。

利用上面的调压器电路产生一个不固定可变电压，电流保持1A的信号，用这个信号表示方程式的自变量X，再用上面的调压器电路产生一个DC3V,1A的信号，这个信号表示常数3，再用上面的调压器电路产生一个DC3V,1A的信号，这个信号表示常数2，最后利用乘方电路，乘法器电路，加法器电路将上面的几个电压连接到一起，最后的输出端接上电压表，电流表。调节电源电路里面的电位器，使表示X的电源输出得电压呈线性变化，电流保持不变为DC1A，最后电路输出端的电压为DC0V，1A时，这时表示X的电源输出的电压值为DC2V或者DC1V，表示方程的一个根是2，1。上面计算一元二次方程的电路如下。

 

 

计算带根号一元二次方程√(x²-5)- √(x²-8)-1=0,得根为2，-2.。利用上面的调压器电路产生一个不固定可变电压，电流保持1A的信号，用这个信号表示方程式的自变量X，再用上面的调压器电路产生一个DC5V,1A的信号，这个信号表示常数5，再用上面的调压器电路产生一个DC8V,1A的信号，这个信号表示常数8，最后利用乘方电路，开方电路，减法器电路将上面的几个电压连接到一起，最后的输出端接上电压表，电流表。调节电源电路里面的电位器，使表示X的电源输出得电压呈线性变化，电流保持不变为DC1A，最后电路输出端的电压为DC0V，1A时，这时表示X的电源输出的电压值为DC2V或者DC-2V，表示方程的一个根是2，-2。上面计算一元二次方程的电路如下。

 

电压开方电路

 

 

 

 

 

电压减法器

 

电压加法器