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  • 作者:zhangzhan
  • 积分:1930
  • 等级:博士研究生
  • 2021/10/23 15:31:29
  • 中国面包师贴吧-楼主(阅:1597/回:0)三角函数模拟计算机2

    下面介绍泰勒公式的推导

    第一部分三角函数模拟计算机电路介绍

    第二部分使用六分仪测量经纬度的三角函数法

    第三部分,模拟三角函数计算机公式介绍

    函数为常数的条件      推导出反三角函数的计算公式

    用模拟计算机计算开方,看参考拉格郎奇公式中的近似公式的推导

    计算三角函数的公式1

    通过无穷小及无穷大的分级中的应用题3),我们得到。在角度不太大时,

    1-cos ψ=4(1-         1+cos ψ      )     (90)

                          2                    

                                                                2  

                      2                      1+      1-   (sin ψ)  

    1-     1-   (sin ψ)     =    4   (1-                                 )       (90)  

                                                      2

    由上面的式子组成模拟计算机的计算电路。

    计算方程式的解,可见计算方程式的近似解页                  比例法则,或称弦线法,依据波查诺-柯西第一定理

      牛顿法则,或称切线法则

    联合法

    下面的公式可以用于模拟计算机的计算电路

    计算三角函数的公式2

    通过127. 近似公式中的例题4),我们得到。设s是弧长,d是对应于它的弦,而δ是对应于半弧的弦(图53)。最后得到关于x,cos x,d,δ的四元一次方程组

                          2       3d*x   2

    (cos x)   =1-(   8δ-d  )         (202a)

                  2    

             (d*δ)

    cos x  =          -1        (202b)

               2

      2                

    d       4δ        d    2        2    2

               (       )+(        -        )    (1-cos)  =δ     (202c)

                    2        3x       6x

    d       4δ        d         1-cos x 2    4δ    d   2

                (       )+(      -        -               *6 )   =(     -     )  (202d)

                    2        3x       6x          2               3x    6x

    计算三角函数的公式3, 最后得到关于x,cos x,d,δ的四元一次方程组

                                 2    2

    2                 d   *x

    cos  x   =1-                                (203a)

                    16     2    2

                         f    +d

                    3

                          2  

              2   2     d  

    d  *(f   +       )

                        4

    cos x=                        -1             (203b)

                 2

                                           2

    d        2           d        

                (       ) +  f    =                       (203c)

                    2               2(cos x+1)

                                                   (203d)

                                              2             2            2        2

                      2                 4    f    +    1   d         4   f  +  1  d    

    d          2        3              4            3         4            

                (       ) +  cos   x  *                           =            

                    2                              2                      2  

                                                x                        x          

    计算三角函数的公式4,详细推导过程可参见戴劳公式125例题

                         3      5                 2m-1

                        x     x         m-1      x        2m

                           +      -…+(-1)              +o(x     )

    3!    5!              (2m-1)!

    tan x=sin x/cos x=        

    2      4                 2m

                         x     x         m      x          2m+1

                      1-     +      -…+(-1)              +o(x     )

    2!    4!              (2m)!

                                           m-1

                         2*2!   2*4!    2*6!        (-1)    (2m)!         (2m)!  

                      =      -       +         -…+(             -        

                                                 m

    x      3!x      5!x          (2m)!x         (-1)    (2m-1)!x    

    2      4                 2m

                         x     x         m      x          2m+1

                cos x= 1-     +      -…+(-1)              +o(x     )

    2!    4!              (2m)!

                         3      5                      2m-1

                        x     x               m-1     x         2m

                sin x=x-           +      -…+(-1)              +o(x     )

    3!    5!                   (2m-1)!

    详细推导见初等函数的展开

                         3      5                  2m-1

                        x     x          m-1     x         2m

               sh x=x+      +      + …+(-1)              +o(x     )

    3!    5!               (2m-1)!

    2      4                 2m

                         x     x         m      x          2m+1

                ch x= 1+     +      +…+(-1)              +o(x     )

    2!    4!              (2m)!

    计算三角函数的拉格朗奇插值法,详细推导过程可见计算三角函数的插值法, 例如

                   ω(x)                    π  

    sin ( 31°)≈                        sin (    )  

                ω`(x     )(x-x     )         6

                      m      m

                            m!                           π

    =                                             sin(      )

         [m![(m-1)!((m-2)!(…(1!+0))+(m-2))+(m-1)]+m!]m         6

                           (m-1)!                    π

    =                                         sin(      )

         [m![(m-1)!((m-2)!(…(1!+0))+(m-2))+(m-1)]m         6

                        1*2*3…30                  π

    =                                       sin(      )

         31!(30!(29!(28!(…1!+1)+28!)+29!)+30!)+31!       6

                   ω(x)                    π  

    cos ( 31°)≈                       cos (    )  

                ω`(x     )(x-x     )         6

                      m      m

                        1*2*3…30                  π

    =                                       cos(      )

         31!(30!(29!(28!(…1!+1)+28!)+29!)+30!)+31!       6

        21           ω(x)             20

    e    ≈                        e

                ω`(x     )(x-x     )      

                      m      m

                     1*2*3…21                  20

    =                                       e

         21!(20!(19!(18!(…1!+1)+18!)+19!)+20!)+21!      

    计算三角函数的带余项的拉格朗奇插值法,详细推导过程可见计算三角函数的插值法, 例如.

                                                  (m+1)  π  

                                                sin     (     )    

                   ω(x)                π                 6    

    sin ( 31°)≈                   * sin(      )+                    w(x)              

    ω`(x   )(x-x   )          6           (m+1)!    

                     m     m

                                                                         π  

                                                                     cos (     )    

                                 m!                            π         6

             =                                           * sin(     )+            m!

              [m![(m-1)! ((m-2)! (...(1!+0)) +(m-2)) +(m-1)]+m!]m!        6       (m+1)!        

                                                                       π  

                                                                  cos (     )    

                                (m-1)!                      π           6

             =                                        * sin(     )+            m!

               [(m-1)! ((m-2)! (...(1!+0)) +(m-2)) +(m-1)]+m!         6       (m+1)!      

                                                                         π                

                                                                    cos(     )  

                             1*2*3...*29                      π           6

             =                                        * sin(      ) +             30!

                 30!(29!(28!(...1!+1)+28!)+29!)+30!          6              31!        

                                                  (m+1)   π  

                                                cos     (     )    

                   ω(x)                π                 6    

    cos ( 31°)≈                   * cos(     )+                    w(x)              

    ω`(x   )(x-x   )          6           (m+1)!    

                     m     m

                                                                   π                

                                                                sin(     )  

                             1*2*3...*29                π           6

             =                                   * sin(      ) +             30!

                 30!(29!(28!(...1!+1)+28!)+29!)+30!          6           31!        

                                              20  (m+1)

      21            ω(x)           20      (e    )

    e      ≈                   * e    +               w(x)      

    ω`(x   )(x-x   )              (m+1)!

                     m     m      

                                                                     20   (m+1)                                              

                                 m!                        20     (e     )

             =                                           *e    +    

                m![(m-1)! ((m-2)! (...(1!+0)) +(m-2)) +(m-1)]+m!              (m+1)!

                                                                  20

                             1*2*3...*21                  20      e          

             =                                        *e    +         m!

                21!(20!(19!(18!(...1!+1)+18!)+19!)+20!)+21!            (m+1)!  

    计算三角函数的埃尔密特公式插值法,详细推导过程可见计算三角函数的插值法, 例如。



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