用割线法求方程根的程序-yiteyi-C++库

割线法用于求方程f（x）=0的根。它从根的两个不同估计值x1和x2开始。这是一个迭代过程，涉及到根的线性插值。如果两个中间值之间的差值小于收敛因子，则迭代停止。

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例如：

Input : equation = x³ + x - 1         x1 = 0, x2 = 1, E = 0.0001Output : Root of the given equation = 0.682326         No. of iteration=5

算法

Initialize: x1, x2, E, n         // E = convergence indicatorcalculate f(x1),f(x2)if(f(x1) * f(x2) = E); //repeat the loop until the convergence    print 'x0' //value of the root    print 'n' //number of iteration}else    print "can not found a root in the given interval"

推荐：请尝试你的方法 {IDE} 首先，在进入解决方案之前。

C++

                         // C++ Program to find root of an                       
                         // equations using secant method                       
                         #include <bits/stdc++.h>                       
                         using                                     namespace                                     std;                       
                         // function takes value of x and returns f(x)                       
                         float                                     f(                                     float                                     x)                       
                         {                       
                                                 // we are taking equation as x^3+x-1                       
                                                 float                                     f =                                     pow                                     (x, 3) + x - 1;                       
                                                 return                                     f;                       
                         }                       
           
                         void                                     secant(                                     float                                     x1,                                     float                                     x2,                                     float                                     E)                       
                         {                       
                                                 float                                     n = 0, xm, x0, c;                       
                                                 if                                     (f(x1) * f(x2) < 0) {                       
                                                 do                                     {                       
                                                 // calculate the intermediate value                       
                                                 x0 = (x1 * f(x2) - x2 * f(x1)) / (f(x2) - f(x1));                       
           
                                                 // check if x0 is root of equation or not                       
                                                 c = f(x1) * f(x0);                       
           
                                                 // update the value of interval                       
                                                 x1 = x2;                       
                                                 x2 = x0;                       
           
                                                 // update number of iteration                       
                                                 n++;                       
           
                                                 // if x0 is the root of equation then break the loop                       
                                                 if                                     (c == 0)                       
                                                 break                                     ;                       
                                                 xm = (x1 * f(x2) - x2 * f(x1)) / (f(x2) - f(x1));                       
                                                 }                                     while                                     (                                     fabs                                     (xm - x0) >= E);                                     // repeat the loop                       
                                                 // until the convergence                       
           
                                                 cout <<                                     "Root of the given equation="                                     << x0 << endl;                       
                                                 cout <<                                     "No. of iterations = "                                     << n << endl;                       
                                                 }                                     else                       
                                                 cout <<                                     "Can not find a root in the given interval"                                     ;                       
                         }                       
           
                         // Driver code                       
                         int                                     main()                       
                         {                       
                                                 // initializing the values                       
                                                 float                                     x1 = 0, x2 = 1, E = 0.0001;                       
                                                 secant(x1, x2, E);                       
                                                 return                                     0;                       
                         }                       

JAVA

                         // Java Program to find root of an                       
                         // equations using secant method                       
                         class                                     GFG {                       
                                   
                                                 // function takes value of x and                       
                                                 // returns f(x)                       
                                                 static                                     float                                     f(                                     float                                     x) {                       
                                   
                                                 // we are taking equation                       
                                                 // as x^3+x-1                       
                                                 float                                     f = (                                     float                                     )Math.pow(x,                                     3                                     )                       
                                                 + x -                                     1                                     ;                       
                                   
                                                 return                                     f;                       
                                                 }                       
                                   
                                                 static                                     void                                     secant(                                     float                                     x1,                                     float                                     x2,                       
                                                 float                                     E) {                       
                                   
                                                 float                                     n =                                     0                                     , xm, x0, c;                       
                                                 if                                     (f(x1) * f(x2) <                                     0                                     )                       
                                                 {                       
                                                 do                                     {                       
                                   
                                                 // calculate the intermediate                       
                                                 // value                       
                                                 x0 = (x1 * f(x2) - x2 * f(x1))                       
                                                 / (f(x2) - f(x1));                       
                                   
                                                 // check if x0 is root of                       
                                                 // equation or not                       
                                                 c = f(x1) * f(x0);                       
                                   
                                                 // update the value of interval                       
                                                 x1 = x2;                       
                                                 x2 = x0;                       
                                   
                                                 // update number of iteration                       
                                                 n++;                       
                                   
                                                 // if x0 is the root of equation                       
                                                 // then break the loop                       
                                                 if                                     (c ==                                     0                                     )                       
                                                 break                                     ;                       
                                                 xm = (x1 * f(x2) - x2 * f(x1))                       
                                                 / (f(x2) - f(x1));                       
                                   
                                                 // repeat the loop until the                       
                                                 // convergence                       
                                                 }                                     while                                     (Math.abs(xm - x0) >= E);                       
                                   
                                                 System.out.println(                                     "Root of the"                                     +                       
                                                 " given equation="                                     + x0);                       
                                   
                                                 System.out.println(                                     "No. of "                       
                                                 +                                     "iterations = "                                     + n);                       
                                                 }                       
                                   
                                                 else                       
                                                 System.out.print(                                     "Can not find a"                       
                                                 +                                     " root in the given interval"                                     );                       
                                                 }                       
                                   
                                                 // Driver code                       
                                                 public                                     static                                     void                                     main(String[] args) {                       
                                   
                                                 // initializing the values                       
                                                 float                                     x1 =                                     0                                     , x2 =                                     1                                     , E =                                     0                                     .0001f;                       
                                                 secant(x1, x2, E);                       
                                                 }                       
                         }                       
           
                         // This code is contributed by Anant Agarwal.                       

Python3

                         # Python3 Program to find root of an                       
                         # equations using secant method                       
           
                         # function takes value of x                       
                         # and returns f(x)                       
                         def                                     f(x):                       
                                   
                                                 # we are taking equation                       
                                                 # as x^3+x-1                       
                                                 f                                     =                                     pow                                     (x,                                     3                                     )                                     +                                     x                                     -                                     1                                     ;                       
                                                 return                                     f;                       
           
                         def                                     secant(x1, x2, E):                       
                                                 n                                     =                                     0                                     ; xm                                     =                                     0                                     ; x0                                     =                                     0                                     ; c                                     =                                     0                                     ;                       
                                                 if                                     (f(x1)                                     *                                     f(x2) <                                     0                                     ):                       
                                                 while                                     True                                     :                       
                                   
                                                 # calculate the intermediate value                       
                                                 x0                                     =                                     ((x1                                     *                                     f(x2)                                     -                                     x2                                     *                                     f(x1))                                     /                       
                                                 (f(x2)                                     -                                     f(x1)));                       
           
                                                 # check if x0 is root of                       
                                                 # equation or not                       
                                                 c                                     =                                     f(x1)                                     *                                     f(x0);                       
           
                                                 # update the value of interval                       
                                                 x1                                     =                                     x2;                       
                                                 x2                                     =                                     x0;                       
           
                                                 # update number of iteration                       
                                                 n                                     +                                     =                                     1                                     ;                       
           
                                                 # if x0 is the root of equation                       
                                                 # then break the loop                       
                                                 if                                     (c                                     =                                     =                                     0                                     ):                       
                                                 break                                     ;                       
                                                 xm                                     =                                     ((x1                                     *                                     f(x2)                                     -                                     x2                                     *                                     f(x1))                                     /                       
                                                 (f(x2)                                     -                                     f(x1)));                       
                                   
                                                 if                                     (                                     abs                                     (xm                                     -                                     x0) < E):                       
                                                 break                                     ;                       
                                   
                                                 print                                     (                                     "Root of the given equation ="                                     ,                       
                                                 round                                     (x0,                                     6                                     ));                       
                                                 print                                     (                                     "No. of iterations = "                                     , n);                       
                                   
                                                 else                                     :                       
                                                 print                                     (                                     "Can not find a root in "                                     ,                       
                                                 "the given interval"                                     );                       
           
                         # Driver code                       
           
                         # initializing the values                       
                         x1                                     =                                     0                                     ;                       
                         x2                                     =                                     1                                     ;                       
                         E                                     =                                     0.0001                                     ;                       
                         secant(x1, x2, E);                       
           
                         # This code is contributed by mits                       

C#

                         // C# Program to find root of an                       
                         // equations using secant method                       
                         using                                     System;                       
           
                         class                                     GFG {                       
                                   
                                                 // function takes value of                       
                                                 // x and returns f(x)                       
                                                 static                                     float                                     f(                                     float                                     x)                       
                                                 {                       
                                   
                                                 // we are taking equation                       
                                                 // as x^3+x-1                       
                                                 float                                     f = (                                     float                                     )Math.Pow(x, 3)                       
                                                 + x - 1;                       
                                                 return                                     f;                       
                                                 }                       
                                   
                                                 static                                     void                                     secant(                                     float                                     x1,                                     float                                     x2,                       
                                                 float                                     E)                       
                                   
                                                 {                       
                                   
                                                 float                                     n = 0, xm, x0, c;                       
                                                 if                                     (f(x1) * f(x2) < 0)                       
                                                 {                       
                                                 do                                     {                       
                                   
                                                 // calculate the intermediate                       
                                                 // value                       
                                                 x0 = (x1 * f(x2) - x2 * f(x1))                       
                                                 / (f(x2) - f(x1));                       
                                   
                                                 // check if x0 is root of                       
                                                 // equation or not                       
                                                 c = f(x1) * f(x0);                       
                                   
                                                 // update the value of interval                       
                                                 x1 = x2;                       
                                                 x2 = x0;                       
                                   
                                                 // update number of iteration                       
                                                 n++;                       
                                   
                                                 // if x0 is the root of equation                       
                                                 // then break the loop                       
                                                 if                                     (c == 0)                       
                                                 break                                     ;                       
                                                 xm = (x1 * f(x2) - x2 * f(x1))                       
                                                 / (f(x2) - f(x1));                       
                                   
                                                 // repeat the loop until                       
                                                 // the convergence                       
                                                 }                                     while                                     (Math.Abs(xm - x0) >= E);                       
                                   
                                                 Console.WriteLine(                                     "Root of the"                                     +                       
                                                 " given equation="                                     + x0);                       
                                   
                                                 Console.WriteLine(                                     "No. of "                                     +                       
                                                 "iterations = "                                     + n);                       
                                                 }                       
                                   
                                                 else                       
                                                 Console.WriteLine(                                     "Can not find a"                                     +                       
                                                 " root in the given interval"                                     );                       
                                                 }                       
                                   
                                                 // Driver code                       
                                                 public                                     static                                     void                                     Main(String []args)                       
                                                 {                       
                                   
                                                 // initializing the values                       
                                                 float                                     x1 = 0, x2 = 1, E = 0.0001f;                       
                                                 secant(x1, x2, E);                       
                                                 }                       
                         }                       
           
                         // This code is contributed by vt_m.                       

PHP

                         <?php                       
                         // PHP Program to find root of an                       
                         // equations using secant method                       
           
                         // function takes value of x                       
                         // and returns f(x)                       
                         function                                     f(                                     $x                                     )                       
                         {                       
                                   
                                                 // we are taking equation                       
                                                 // as x^3+x-1                       
                                                 $f                                     = pow(                                     $x                                     , 3) +                                     $x                                     - 1;                       
                                                 return                                     $f                                     ;                       
                         }                       
           
                         function                                     secant(                                     $x1                                     ,                                     $x2                                     ,                                     $E                                     )                       
                         {                       
                                                 $n                                     = 0;                                     $xm                                     ;                       
                                                 $x0                                     ;                                     $c                                     ;                       
                                                 if                                     (f(                                     $x1                                     ) * f(                                     $x2                                     ) < 0)                       
                                                 {                       
                                                 do                                     {                       
                                   
                                                 // calculate the intermediate value                       
                                                 $x0                                     = (                                     $x1                                     * f(                                     $x2                                     ) -                                     $x2                                     *                       
                                                 f(                                     $x1                                     )) / (f(                                     $x2                                     ) - f(                                     $x1                                     ));                       
           
                                                 // check if x0 is root                       
                                                 // of equation or not                       
                                                 $c                                     = f(                                     $x1                                     ) * f(                                     $x0                                     );                       
           
                                                 // update the value of interval                       
                                                 $x1                                     =                                     $x2                                     ;                       
                                                 $x2                                     =                                     $x0                                     ;                       
           
                                                 // update number of iteration                       
                                                 $n                                     ++;                       
           
                                                 // if x0 is the root of equation                       
                                                 // then break the loop                       
                                                 if                                     (                                     $c                                     == 0)                       
                                                 break                                     ;                       
                                                 $xm                                     = (                                     $x1                                     * f(                                     $x2                                     ) -                                     $x2                                     * f(                                     $x1                                     )) /                       
                                                 (f(                                     $x2                                     ) - f(                                     $x1                                     ));                       
                                   
                                                 // repeat the loop                       
                                                 // until the convergence                       
                                                 }                                     while                                     (                                     abs                                     (                                     $xm                                     -                                     $x0                                     ) >=                                     $E                                     );                       
                                   
                                                 echo                                     "Root of the given equation="                                     .                                     $x0                                     .                                     ""                                     ;                       
                                                 echo                                     "No. of iterations = "                                     .                                     $n                                     ;                       
                                   
                                                 }                                     else                       
                                                 echo                                     "Can not find a root in the given interval"                                     ;                       
                         }                       
           
                         // Driver code                       
                         {                       
                                   
                                                 // initializing the values                       
                                                 $x1                                     = 0;                                     $x2                                     = 1;                       
                                                 $E                                     = 0.0001;                       
                                                 secant(                                     $x1                                     ,                                     $x2                                     ,                                     $E                                     );                       
                                                 return                                     0;                       
                         }                       
           
                         // This code is contributed by nitin mittal.                       
                         ?>                       

Javascript

                         <script>                       
                         // JavaScript Program to find root of an                       
                         // equations using secant method                       
           
                         // function takes value of x and returns f(x)                       
                         function                                     f(x)                       
                         {                       
                                                 // we are taking equation as x^3+x-1                       
                                                 let f = Math.pow(x, 3) + x - 1;                       
                                                 return                                     f;                       
                         }                       
           
                         function                                     secant(x1, x2, E)                       
                         {                       
                                                 let n = 0, xm, x0, c;                       
                                                 if                                     (f(x1) * f(x2) < 0) {                       
                                                 do                                     {                       
                                                 // calculate the intermediate value                       
                                                 x0 = (x1 * f(x2) - x2 * f(x1)) / (f(x2) - f(x1));                       
           
                                                 // check if x0 is root of equation or not                       
                                                 c = f(x1) * f(x0);                       
           
                                                 // update the value of interval                       
                                                 x1 = x2;                       
                                                 x2 = x0;                       
           
                                                 // update number of iteration                       
                                                 n++;                       
           
                                                 // if x0 is the root of equation then break the loop                       
                                                 if                                     (c == 0)                       
                                                 break                                     ;                       
                                                 xm = (x1 * f(x2) - x2 * f(x1)) / (f(x2) - f(x1));                       
                                                 }                                     while                                     (Math.abs(xm - x0) >= E);                                     // repeat the loop                       
                                                 // until the convergence                       
           
                                                 document.write(                                     "Root of the given equation="                                     + x0.toFixed(6) +                                     "<br>"                                     );                       
                                                 document.write(                                     "No. of iterations = "                                     + n +                                     "<br>"                                     );                       
                                                 }                                     else                       
                                                 document.write(                                     "Can not find a root in the given interval"                                     );                       
                         }                       
           
                         // Driver code                       
                                                 // initializing the values                       
                                                 let x1 = 0, x2 = 1, E = 0.0001;                       
                                                 secant(x1, x2, E);                       
           
           
                         // This code is contributed by Surbhi Tyagi.                       
                         </script>                       

输出：

Root of the given equation = 0.682326No. of iterations = 5

时间复杂度=O（1）

参考 https://en.wikipedia.org/wiki/Secant_method 本文由 尼蒂什·库马尔 .如果你喜欢GeekSforgek，并想贡献自己的力量，你也可以使用写极客。组织或者把你的文章寄去评论-team@geeksforgeeks.org.看到你的文章出现在Geeksforgeks主页上，并帮助其他极客。如果您发现任何不正确的地方，或者您想分享有关上述主题的更多信息，请写下评论。

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