Java实现lagrange 插值法

￡神魔★判官ぃ 2022-03-10 04:58 205阅读 0赞

**一、题目：已知![1366638880_1329.png][]**

**1、请用两点 Lagrange插值 编程实现 ，用余项公式求误差；**

1.1）程序源代码

public class Chazhi {
    
    	public static void main(String[] args) {
    		lagrange();
    	}
    
    	public static void lagrange() {
    		double p = Math.atan(1) * 4;
    		double[] x = { p / 4, p / 3, (5 * p) / 18 };
    		double[] y = { Math.sqrt(2.0) / 2, Math.sqrt(3.0) / 2 };
    		double l = 0.0;
    		for (int j = 0; j < 2; j++) {
    			double s = 1.0;
    			for (int i = 0; i < 2; i++) {
    				if (i != j)
    					s = s * ((x[2] - x[i]) / (x[j] - x[i]));
    			}
    			l = l + s * y[j];
    		}
    		double m = (Math.abs(x[0]) / 2)
    				* Math.abs((x[2] - x[0]) * (x[2] - x[1]));
    		System.out.println("x=" + x[2]);
    		System.out.println("L=" + l);
    		System.out.println("误差为:" + m);
    	}
    }

1.2）程序结果:

![1366639031_2793.png][]

**2、请用三点 lagrange 插值 编程实现，用余项公式求误差；**

2.1）程序源代码：

public class Chazhi {
    
    	public static void main(String[] args) {
    		// lagrange();
    		lagrange_2();
    	}
    
    	public static void lagrange_2() {
    		double p = Math.atan(1) * 4;
    		double[] x = { p / 6, p / 4, p / 3, (5 * p) / 18 };
    		double[] y = { 0.5, Math.sqrt(2.0) / 2, Math.sqrt(3.0) / 2 };
    		double l = 0.0;
    		for (int j = 0; j < 3; j++) {
    			double s = 1.0;
    
    			for (int i = 0; i < 3; i++) {
    				if (i != j)
    					s = s * ((x[3] - x[i]) / (x[j] - x[i]));
    			}
    			l = l + s * y[j];
    		}
    		double M = (Math.abs(x[0]) / (2 * 3))
    				* Math.abs((x[3] - x[0]) * (x[3] - x[1]) * (x[3] - x[2]));
    		System.out.println("x=" + x[3]);
    		System.out.println("L=" + l);
    		System.out.println("误差为：" + M);
    	}
    }

![1366639149_2839.png][]

[1366638880_1329.png]: https://img-my.csdn.net/uploads/201304/22/1366638880_1329.png
[1366639031_2793.png]: https://img-my.csdn.net/uploads/201304/22/1366639031_2793.png
[1366639149_2839.png]: https://img-my.csdn.net/uploads/201304/22/1366639149_2839.png