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  1. #include <stdio.h>
  2. #include <math.h>
  3.  
  4. int main() {
  5.     double x[8] = {0.15708, 0.23982, 0.37400, 0.57120, 0.82674, 1.04720, 1.23200, 1.43452};
  6.     double y[8] = {0.98769, 0.97138, 0.93087, 0.84125, 0.67728, 0.50000, 0.33236, 0.13586};
  7.  
  8.     double Sx2 = 0, Sx4 = 0, Sy = 0, Sx2y = 0;
  9.  
  10.     for (int i = 0; i < 8; i++) {
  11.         double x2 = x[i] * x[i];
  12.         Sx2 += x2;
  13.         Sx4 += x2 * x2;
  14.         Sy += y[i];
  15.         Sx2y += x2 * y[i];
  16.     }
  17.  
  18.     double A11 = 8,     A12 = Sx2;
  19.     double A21 = Sx2,   A22 = Sx4;
  20.     double B1 = Sy,     B2 = Sx2y;
  21.     double det = A11 * A22 - A12 * A21;
  22.     double a1 = (B1 * A22 - A12 * B2) / det;
  23.     double a2 = (A11 * B2 - B1 * A21) / det;
  24.  
  25.     printf("(1): y = a1 + a2*x^2 計算結果\n");
  26.     printf("求められた係数:\n");
  27.     printf("  a1 = %.6f\n", a1);
  28.     printf("  a2 = %.6f\n", a2);
  29.  
  30.     double E = 0;
  31.     for (int i = 0; i < 8; i++) {
  32.         double yhat = a1 + a2 * x[i] * x[i];
  33.         double diff = y[i] - yhat;
  34.         E += diff * diff;
  35.     }
  36.  
  37.     printf("\n近似式による偏差の2乗和 (誤差 E):\n");
  38.     printf("  E = %.10f\n", E);
  39.  
  40.     printf("\n近似式による y の予測値 :\n");
  41.     for (int i = 0; i < 8; i++) {
  42.         double yhat = a1 + a2 * x[i] * x[i];
  43.         printf("%f\n", yhat);
  44.     }
  45.  
  46.     return 0;
  47. }
Success #stdin #stdout 0.01s 5288KB
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(1): y = a1 + a2*x^2 計算結果
求められた係数:
  a1 = 0.985005
  a2 = -0.423999

近似式による偏差の2乗和 (誤差 E):
  E = 0.0016966201

近似式による y の予測値 :
0.974543
0.960619
0.925697
0.846667
0.695202
0.520035
0.341448
0.112479
https://ideone.com/yguABU
language:
C (gcc 8.3)
created:
16 hours ago
visibility:
public

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