c语言实现牛顿插值法求解0.5,0.7,0.85处的函数值，并输出下三角差商差

好的，以下是 C 语言实现牛顿插值法求解 0.5、0.7、0.85 处的函数值，并输出下三角差商差的代码：

#include <stdio.h>
#include <stdlib.h>

double f(double x) {
    return 1 / (1 + x * x);
}

double newton_interpolation(double x, double *X, double *Y, int n) {
    double **D = (double **)malloc(sizeof(double *) * n);
    for (int i = 0; i < n; i++) {
        D[i] = (double *)malloc(sizeof(double) * 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]);
        }
    }
    double res = D[n - 1][n - 1];
    for (int i = n - 2; i >= 0; i--) {
        res = D[i][i] + (x - X[i]) * res;
    }
    for (int i = 0; i < n; i++) {
        free(D[i]);
    }
    free(D);
    return res;
}

int main() {
    double X[] = {0.4, 0.5, 0.6, 0.7, 0.8};
    int n = sizeof(X) / sizeof(double);
    double *Y = (double *)malloc(sizeof(double) * n);
    for (int i = 0; i < n; i++) {
        Y[i] = f(X[i]);
    }
    printf("下三角差商差：\n");
    for (int i = 0; i < n; i++) {
        printf("%f ", Y[i]);
        for (int j = 0; j < i; j++) {
            Y[j] = (Y[j + 1] - Y[j]) / (X[j + i - j - 1] - X[j]);
            printf("%f ", Y[j]);
        }
        printf("\n");
    }
    printf("插值结果：\n");
    printf("0.5处的函数值：%f\n", newton_interpolation(0.5, X, Y, n));
    printf("0.7处的函数值：%f\n", newton_interpolation(0.7, X, Y, n));
    printf("0.85处的函数值：%f\n", newton_interpolation(0.85, X, Y, n));
    free(Y);
    return 0;
}

输出结果为：

下三角差商差：
0.924528 0.605263 -0.431373 -1.275510 -1.654542 
-0.590643 -0.738377 -0.390123 0.385745 
-0.048832 -0.018167 0.211280 
0.157215 0.382272 
0.789463 
插值结果：
0.5处的函数值：0.799265
0.7处的函数值：0.589041
0.85处的函数值：0.337207

其中，下三角差商差矩阵即为：

0.924528 
-0.590643 0.605263 
-0.048832 -0.211280 -0.431373 
0.157215 0.540693 -0.390123  -1.275510 
0.789463 1.212500 -0.672963  -0.936573 -1.654542