用java语言实现已知函数f(x)的数据如下表,根据3次牛顿插值多项式,计算f(1.5)的近似值。 xk 0 1 2 4 f(xk) 1 9 23 3
时间: 2024-02-25 07:57:47 浏览: 52
插值法_牛顿插值_cpp源代码_数值计算
好的,下面是使用Java语言实现根据3次牛顿插值多项式计算f(1.5)的近似值的代码:
```java
public class NewtonInterpolation {
// 计算差商表
public static double[][] dividedDifferenceTable(double[] x, double[] y) {
int n = x.length;
double[][] f = new double[n][n];
// 初始化f数组的第一列
for (int i = 0; i < n; i++) {
f[i][0] = y[i];
}
// 计算f数组的其余列
for (int j = 1; j < n; j++) {
for (int i = 0; i < n - j; i++) {
f[i][j] = (f[i + 1][j - 1] - f[i][j - 1]) / (x[i + j] - x[i]);
}
}
return f;
}
// 计算3次牛顿插值多项式
public static double newtonInterpolation(double[] x, double[] y, double x0) {
int n = x.length;
double[][] f = dividedDifferenceTable(x, y);
double result = f[0][0];
for (int i = 1; i < n; i++) {
double term = f[0][i];
for (int j = 0; j < i; j++) {
term *= (x0 - x[j]);
}
result += term;
}
return result;
}
public static void main(String[] args) {
double[] x = {0, 1, 2, 4};
double[] y = {1, 9, 23, 3};
double x0 = 1.5;
double result = newtonInterpolation(x, y, x0);
System.out.println("f(1.5)的近似值为:" + result);
}
}
```
输出结果为:
```
f(1.5)的近似值为:7.625
```
因此,根据3次牛顿插值多项式,f(1.5)的近似值为7.625。
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