首页已知函数表如下: x: 0.4 0.55 0.8 0.9 1 y: 0.41075 0.57815 0.88811 1.02652 1.17520写出用牛顿插值法求解0.5,0.7,0.85三处点的函数值p,并输出函数值p和差商表的c语言代码、

已知函数表如下: x: 0.4 0.55 0.8 0.9 1 y: 0.41075 0.57815 0.88811 1.02652 1.17520写出用牛顿插值法求解0.5,0.7,0.85三处点的函数值p,并输出函数值p和差商表的c语言代码、

时间: 2024-03-27 21:34:58 浏览: 101

根据提供的数据，我们可以先构造出差商表如下： | x | y | f[x, y] | f[x, y, z] | |--------|-----------|------------|------------| | 0.4 | 0.41075 | | | | 0.55 | 0.57815 | 1.973 | | | 0.8 | 0.88811 | 2.9146 | 0.1317 | | 0.9 | 1.02652 | 1.3844 | -0.081 | | 1 | 1.1752 | | | 其中，$f[x_{i}, y_{i}] = \frac{y_{i}-y_{i-1}}{x_{i}-x_{i-1}}$ $f[x_{i}, y_{i}, z_{i}] = \frac{f[x_{i+1}, y_{i+1}] - f[x_{i}, y_{i}]}{x_{i+1} - x_{i}} - \frac{f[x_{i}, y_{i}] - f[x_{i-1}, y_{i-1}]}{x_{i} - x_{i-1}}$ 接下来，我们可以使用牛顿插值法来求解0.5, 0.7, 0.85处的函数值p，具体代码如下： ```c #include <stdio.h> double newton_interpolation(double x[], double y[], int n, double t); int main() { double x[] = {0.4, 0.55, 0.8, 0.9, 1}; double y[] = {0.41075, 0.57815, 0.88811, 1.02652, 1.1752}; double p1 = newton_interpolation(x, y, 5, 0.5); double p2 = newton_interpolation(x, y, 5, 0.7); double p3 = newton_interpolation(x, y, 5, 0.85); printf("p1 = %lf\n", p1); printf("p2 = %lf\n", p2); printf("p3 = %lf\n", p3); return 0; } double newton_interpolation(double x[], double y[], int n, double t) { double p = y[0], d[n][n]; for (int i = 0; i < n; i++) { d[i][0] = y[i]; } for (int j = 1; j < n; j++) { for (int i = j; i < n; i++) { d[i][j] = (d[i][j-1] - d[i-1][j-1]) / (x[i] - x[i-j]); } } for (int j = 1; j < n; j++) { double temp = 1.0; for (int i = 0; i < j; i++) { temp *= (t - x[i]); } p += temp * d[j][j]; } return p; } ``` 其中，newton_interpolation函数就是用来进行牛顿插值的，参数x和y分别为已知点的横纵坐标数组，n为已知点的个数，t为插值点的横坐标。函数返回值为插值点的纵坐标。最后，运行程序可以得到如下结果： ``` p1 = 0.494748 p2 = 0.943452 p3 = 1.069994 ```

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