[JAVA程式語言]範例1-4 通過下面10 點的Lagrange內插法求P(1.5)之值?

[JAVA程式語言]範例1-4 通過下面10 點的Lagrange內插法求P(1.5)之值?

n=9    xa=1.5

 x           f(x)
=============
1.0       0.000
1.2       0.182
1.7       0.531
2.0       0.693
2.2       0.788
2.7       0.993
3.0       1.099
3.2       1.163
3.7       1.308

4.0       1.386
=============

/* ex1-4.java: Lagrange Interpolation Algorithm
 * Read in data file of ex1-4.dat which has n point values
 * and the value of interpolating point xa. Based on Lagrange
 * Interpolation algorithm to compute p(xa) and output its value.
 * (x[i],f[i]):given points and n+1 are number of points
 * Ln,k(x)=l=summation of (x-x[i])/(x[k]-x[i]).
 * p(x)=ff=L(x)*f(x[k])
 */

// 範例1-4 通過下面10點的Lagrange內插法求P(1.5)之值?
import java.util.Scanner;

public class Main {
    public static void main(String []args) {
        Scanner scanner = new Scanner(System.in);
        float[]  x, f ;
        float xa , l ,ff;
        int i,k,n;
        x = new float[10]; // 利用new指令產生物件
        f = new float[10]; // 利用new指令產生物件

        n=scanner.nextInt();
        xa=scanner.nextFloat();
        System.out.printf("n=%2d\txa=%2.3f",n,xa);
        System.out.println();
        for(k=0;k<=n;k++)
        {
            //Scanner scanner = new Scanner(System.in);
            x[k]=scanner.nextFloat();
            f[k]=scanner.nextFloat();
            //System.out.printf("x[k]=%2.3f, f[k]=%2.3f",x[k],f[k]);
            //System.out.println();
        }
       
        ff=0;
        for(k=0;k<=n;k++){
           l=1;
           i=0;
           do{
        if(i !=k) {
                l=l*(xa-x[i])/(x[k]-x[i]);
        }
        i=i+1;
        //System.out.printf("l=%f n=%2d i=%2d",l, n, i);
                //System.out.println();
           } while (i<=n);
        ff=ff+l*f[k];
        System.out.printf("l=%f k=%2d The value of p(%2.5f)=%2.5f",l, k,xa ,ff);
        System.out.println();
        }
        System.out.printf("The value of p(%f)=%2.5f",xa ,ff);
    }
}

STDIN
9    1.5
1.0  0.000
1.2  0.182
1.7  0.531
2.0  0.693
2.2  0.788
2.7  0.993
3.0  1.099
3.2  1.163
3.7  1.308

4.0  1.386

輸出畫面

$javac Main.java
$java -Xmx128M -Xms16M Main
n= 9 xa=1.500
l=-0.034722 k= 0 The value of p(1.50000)=0.00000
l=0.194792 k= 1 The value of p(1.50000)=0.03545
l=1.876255 k= 2 The value of p(1.50000)=1.03174
l=-2.578125 k= 3 The value of p(1.50000)=-0.75490
l=1.947917 k= 4 The value of p(1.50000)=0.78006
l=-0.846154 k= 5 The value of p(1.50000)=-0.06017
l=0.749199 k= 6 The value of p(1.50000)=0.76320
l=-0.328125 k= 7 The value of p(1.50000)=0.38159
l=0.022222 k= 8 The value of p(1.50000)=0.41066
l=-0.003257 k= 9 The value of p(1.50000)=0.40614
The value of p(1.500000)=0.40614