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  1. #include <stdio.h>
  2. #include <math.h>
  3.  
  4. #define EPS 0.00000001
  5. #define N   1000
  6.  
  7. typedef struct {
  8.   double ans;
  9.   int iter;
  10. } Ans;
  11.  
  12. double f(double x)
  13. {
  14.   return 0.5 - x + 0.2 * sin(x);
  15. }
  16.  
  17. /* f'(x) */
  18. double f2(double x)
  19. {
  20.   return -1.0 + 0.2 * cos(x);
  21. }
  22.  
  23. double g(double x)
  24. {
  25.   return 0.5 + 0.2 * sin(x);
  26. }
  27.  
  28. Ans linear_iteration_method(double x, double (*f)(double))
  29. {
  30.   double x0;
  31.   int i = 0;
  32.   Ans ans;
  33.  
  34.   do {
  35.     x0 = x;
  36.     x = f(x);
  37.     i++;
  38.   } while (fabs(x - x0) > EPS);
  39.  
  40.   ans.ans = x;
  41.   ans.iter = i;
  42.  
  43.   return ans;
  44. }
  45.  
  46. Ans aitken_deltasquare_process(double a[], double b[])
  47. {
  48.   int i;
  49.   double dx1, dx2;
  50.   Ans ans;
  51.  
  52.   for (i = 0; i < N - 2; i++) {
  53.     dx1 = a[i + 1] - a[i];
  54.     dx2 = a[i + 2] - a[i + 1];
  55.     b[i] = a[i + 2] - dx2 * dx2 / (dx2 - dx1);
  56.     if (fabs(b[i] - a[i + 2]) <= EPS)
  57.       break;
  58.   }
  59.  
  60.   ans.ans = b[i];
  61.   ans.iter = i + 1;
  62.  
  63.   return ans;
  64. }
  65.  
  66. Ans newton_method(double x, double (*f)(double), double (*f2)(double))
  67. {
  68.   int i = 0;
  69.   double x0;
  70.   Ans ans;
  71.  
  72.   do {
  73.     x0 = x;
  74.     x = x - f(x) / f2(x);
  75.     i++;
  76.   } while (fabs(x0 - x) > EPS);
  77.  
  78.   ans.ans = x;
  79.   ans.iter = i;
  80.  
  81.   return ans;
  82. }
  83.  
  84. int main(void)
  85. {
  86.   int i;
  87.   double x0 = 0.2;
  88.   Ans ans;
  89.   static double a[N], b[N];
  90.  
  91.   puts("線形反復法");
  92.   ans = linear_iteration_method(x0, g);
  93.   printf("solution = %.8f, iteration = %d\n", ans.ans, ans.iter);
  94.  
  95.   puts("エイトケンのデルタ2乗法");
  96.   /* 元数列の生成 */
  97.   a[0] = x0;
  98.   for (i = 1; i < N; i++)
  99.     a[i] = g(a[i - 1]);
  100.   ans = aitken_deltasquare_process(a, b);
  101.   printf("solution = %.8f, iteration = %d\n", ans.ans, ans.iter);
  102.  
  103.   puts("ニュートン法");
  104.   ans = newton_method(x0, f, f2);
  105.   printf("solution = %.8f, iteration = %d\n", ans.ans, ans.iter);
  106.  
  107.   return 0;
  108. }
  109.  
Success #stdin #stdout 0.01s 1732KB
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Standard input is empty
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線形反復法
solution = 0.61546817, iteration = 11
エイトケンのデルタ2乗法
solution = 0.61546817, iteration = 9
ニュートン法
solution = 0.61546817, iteration = 4
https://ideone.com/g5vNd
language:
C (gcc 8.3)
created:
12 years ago
visibility:
public

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