数值计算——多项式插值

「爱情、让人受尽委屈。」 2022-06-16 02:54 340阅读 0赞

# 数值计算——多项式插值 #

**插值：**用某个函数去拟合给定的数据，该函数在给定的数据点的函数值与数据值相等。在离散数据的基础上补插连续函数，使得这条连续曲线通过全部给定的离散数据点。插值是离散函数逼近的重要方法，利用它可通过函数在有限个点处的取值状况，估算出函数在其他点处的近似值。可以用来估计未知位置上的数据点。

**多项式插值：**这是最常见的一种函数插值。在一般插值问题中，若选取Φ为n次多项式类，由插值条件可以唯一确定一个n次插值多项式满足上述条件。从几何上看可以理解为:已知平面上n+1个不同点，要寻找一条n次多项式曲线通过这些点。插值多项式一般有两种常见的表达形式，一个是拉格朗日插值多项式，另一个是牛顿插值多项式。

## 1、单项式基底插值 ##

多项式形式：P(t) = x1+x2t+x3t^2……+xnt^(n-1)

得到的矩阵形式为范德蒙矩阵。

这里是java代码，matlab代码在最后

package com.kexin.lab8;
    
    import java.text.DecimalFormat;
    import java.util.List;
    
    import com.kexin.lab2.LU;
    
    /**
     * 单项式插值
     * 
     * @author KeXin
     *
     */
    public class MonomiulBasis {
    
    	/**
    	 * 生成矩阵A
    	 * 
    	 * @return
    	 */
    	public static double[][] MakeA(int n, double[] t) {
    		double[][] a = new double[n][n];
    		for (int i = 0; i < n; i++) {
    			for (int j = 0; j < n; j++) {
    				a[i][j] = Math.pow(t[i], j);
    			}
    		}
    		return a;
    	}
    
    	@SuppressWarnings("rawtypes")
    	public static void main(String[] args) {
    		//书上例题测试
    //		double[] t = { -2, 0, 1 };
    //		double[] y = {-27,-1,0};
    		double[] t = {0,0.5,1.0,6.0,7.0,9.0};
    		double[] y = {0,1.6,2,2,1.5,0};
    		double[][] A = MakeA(t.length, t);
    		Print2DArray("A", A);
    		PrintArray("y", y);
    		List a = LU.myLU(A);
    		double[][] L = (double[][]) a.get(0);
    		double[][] U = (double[][]) a.get(1);
    		double[] result = LU.myLU(L, U, y);
    		PrintArray("x", result);
    	}
    	/**
    	 * 打印一维数组
    	 * 
    	 * @param result
    	 */
    	public static void PrintArray(String str, double[] result) {
    		DecimalFormat df = new DecimalFormat("#.000");
    		int n = result.length;
    		System.out.print(str + "\n[");
    		for (int i = 0; i < n; i++) {
    			System.out.print(df.format(result[i]) + "\t");
    		}
    		System.out.print(']');
    		System.out.println();
    	}
    
    	public static void Print2DArray(String str, double[][] result) {
    		DecimalFormat df = new DecimalFormat("#.000");
    		int n = result.length;
    		int m = result[0].length;
    		System.out.print(str + "\n");
    		for (int i = 0; i < n; i++) {
    			for (int j = 0; j < m; j++)
    				System.out.print(df.format(result[i][j]) + "\t");
    			System.out.println();
    		}
    	}
    
    }

## 2、拉格朗日插值 ##

当n增加时，单项式插值的病态性特别明显，而且计算量很大，所以在原基础上又出现了拉格朗日插值和牛顿插值法。另一种形式的多项式插值。表达式：

![20170503194510329][]

package com.kexin.lab8;
    
    /**
     * 拉格朗日插值法
     * 
     * @author KeXin
     *
     */
    public class Lagrange {
    
    	/* 拉格朗日插值法 */
    	public static double Lag_method(double X[], double Y[], double t) {
    		int m = X.length;
    		double result = 0.0;
    		for (int i2 = 0; i2 < m; i2++) {// 遍历Y
    			double u = 1;
    			for (int i3 = 0; i3 < m; i3++) {// 遍历X
    				if (i2 != i3) {
    					u = u * (t - X[i3]) / (X[i2] - X[i3]);
    				}
    			}
    			u = u * Y[i2];
    			result = result + u;
    		}
    
    		return result;
    	}
    	public static void main(String[] args) {
    		/* 插值点横纵坐标 */
    		double X[] = { -2, 0, 1 };
    		double Y[] = { -27, -1, 0 };
    		/*检验位置*/
    		double t = -2;
    		double result = Lag_method(X, Y, t);
    		System.out.println("拉格朗日插值法t = "+t+"求解得:"+result);
    	}
    }

## 3、牛顿插值 ##

![20170503195030978][]

package com.kexin.lab8;
    
    import java.text.DecimalFormat;
    import java.util.List;
    
    import com.kexin.lab2.LU;
    
    public class Newton {
    	public static double[][] MakeA(double[] t) {
    		int len = t.length;
    		double temp;
    		double[][] result = new double[len][len];
    		for (int i = 0; i < len; i++) {
    			for (int j = 0; j <= i; j++) {
    				temp = 1;
    				for (int k = 0; k < j; k++) {
    					temp *= t[i] - t[k];
    				}
    				result[i][j] = temp;
    			}
    		}
    		return result;
    	}
    
    	public static void main(String[] args) {
    		// 书上例题测试
    //		double[] t = { -2, 0, 1 };
    //		double[] y = { -27, -1, 0 };
    		//实验题目
    		double[] t = {0,0.5,1.0,6.0,7.0,9.0};
    		double[] y = {0,1.6,2,2,1.5,0};
    		double[][] A = MakeA(t);
    		Print2DArray("A", A);
    		PrintArray("y", y);
    		List a = LU.myLU(A);
    		double[][] L = (double[][]) a.get(0);
    		double[][] U = (double[][]) a.get(1);
    		double[] result = LU.myLU(L, U, y);
    		PrintArray("x", result);
    	}
    	/**
    	 * 打印一维数组
    	 * 
    	 * @param result
    	 */
    	public static void PrintArray(String str, double[] result) {
    		DecimalFormat df = new DecimalFormat("#.000");
    		int n = result.length;
    		System.out.print(str + "\n[");
    		for (int i = 0; i < n; i++) {
    			System.out.print(df.format(result[i]) + "\t");
    		}
    		System.out.print(']');
    		System.out.println();
    	}
    
    	public static void Print2DArray(String str, double[][] result) {
    		DecimalFormat df = new DecimalFormat("#.000");
    		int n = result.length;
    		int m = result[0].length;
    		System.out.print(str + "\n");
    		for (int i = 0; i < n; i++) {
    			for (int j = 0; j < m; j++)
    				System.out.print(df.format(result[i][j]) + "\t");
    			System.out.println();
    		}
    	}
    
    }

## 4、matlab代码 ##

拉格朗日插值：

function [L] = lagrange()
    x0 = [0,0.5,1,6,7,9]; 
    y0 =[0,1.6,2.0,2.0,1.5,0.0]; 
    syms x 
    for i=1:6
      a=1; 
      for j=1:6
        if j~=i 
         a=expand(a*(x-x0(j))); 
        end 
      end 
      b=1; 
      for k=1:6
        if k~=i 
         b=b*(x0(i)-x0(k)); 
        end 
      end 
      A(i)=expand(a/b); 
    end 
    L=0; 
    for p=1:6
      L=L+y0(p)*A(p); 
    end

牛顿插值：

function [] = Newton() 
    syms x 
    x0 = [0,0.5,1,6,7,9]; 
    y0 =[0,1.6,2.0,2.0,1.5,0.0];   
    for k=1:6
        b=0;
      for i=1:k 
        a=1; 
        for j=1:k 
          if j~=i 
            a=a*(x0(i)-x0(j)); 
          end 
        end 
        b=b+y0(i)/a; 
      end 
      A(k)=b; 
    end 
    B=[1,(x-x0(1)),(x-x0(1))*(x-x0(2)),(x-x0(1))*(x-x0(2))*(x-x0(3)),(x-x0(1))*(x-x0(2))*(x-x0(3))*(x-x0(4)),(x-x0(1))*(x-x0(2))*(x-x0(3))*(x-x0(4))*(x-x0(5))]; 
    L1=A.*B; 
    l=0;
    for m=1:6
      l=l+L1(m); 
    end 
    L=expand(l)   
    end

[20170503194510329]: /images/20220616/e6f0f4bd6eaa4f0e997be13e084fd78c.png
[20170503195030978]: /images/20220616/001b78a3ee9244978ce20a8dcf57b0b1.png