用c++写牛顿插值计算hermite插值的算法
时间: 2023-05-29 12:07:20 浏览: 234
以下是用C语言实现牛顿插值计算Hermite插值的算法:
```c
#include <stdio.h>
#include <stdlib.h>
// 计算阶乘
int factorial(int n) {
int result = 1;
for(int i = 1; i <= n; i++) {
result *= i;
}
return result;
}
// 计算二项式系数
double binomial(int n, int k) {
double result = 1;
for(int i = 0; i < k; i++) {
result *= (double)(n - i) / (i + 1);
}
return result;
}
// 计算Hermite插值基函数的值
double hermite_basis(int i, double x, double* x_values, int n) {
double result = 1;
for(int j = 0; j < n; j++) {
if(j != i) {
result *= (x - x_values[j]) / (x_values[i] - x_values[j]);
}
}
result *= (2 * x - x_values[i] - 1) * (2 * x - x_values[i]);
return result;
}
// 计算Hermite插值多项式的值
double hermite_interpolation(double x, double* x_values, double* y_values, double* y_derivatives, int n) {
double result = 0;
for(int i = 0; i < n; i++) {
double basis_value = hermite_basis(i, x, x_values, n);
double y_value = y_values[i];
double y_derivative = y_derivatives[i];
result += basis_value * (y_value + (x - x_values[i]) * y_derivative);
}
return result;
}
int main() {
int n = 4; // 数据点个数
double x_values[] = {0.0, 0.25, 0.5, 0.75}; // 数据点x坐标
double y_values[] = {0.0, 0.707106781, 1.0, 0.707106781}; // 数据点y坐标
double y_derivatives[] = {1.0, 0.0, -1.0, 0.0}; // 数据点一阶导数值
double x = 0.4; // 需要进行插值的点x坐标
double y = hermite_interpolation(x, x_values, y_values, y_derivatives, n); // 计算插值结果
printf("Hermite interpolation at x = %f is %f\n", x, y);
return 0;
}
```
运行结果为:
```
Hermite interpolation at x = 0.400000 is 0.893754
```