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作业：已知二元函数试以（0，0）为初始点，分别使用Newton法和最速下降法求函数f(x)的极小点，要求编写计算程序求解（编程语言不限），并给出计算结果。注：（1）程序及计算结果需要用A4纸打印。 （2）作业在11月22日上交。（3）自己独立完成，严禁互相抄袭。若发现抄袭，平时成绩按照零分计算。用Newton法和最速下降法求函数f(x)的极小点 java实现编程2010-11-14 11:28同学让帮忙做数值分析，要求用编码实现Newton法和最速下降法，觉得对不少朋友应该有用，所以代码写完之后贴到这里，供大家参考。题目如下：已知二元函数试以（0，0）为初始点，分别使用Newton法和最速下降法求函数f(x)的极小点，要求编写计算程序求解（编程语言不限），并给出计算结果。算法描述如下：用java实现的算法代码如下：1.牛顿法：package Test_Math;public class NewTon_way /* param args* 这是牛顿法求极值* author 竹下听箫* date 2010.11.13*/* Old_fun()是原函数*/public double Old_fun(int coeff, double x) return (double)coeff0*Math.pow(x0,2)+coeff1*Math.pow(x1,2)+coeff2*x0*x1+coeff3*x0+coeff4*x1;/*grads()是求得的导函数*/public void grads(int coeff, double x,double grads_x) grads_x0 = (double)2*coeff0*x0+coeff2*x1+coeff3; grads_x1 = (double)2*coeff1*x1+coeff2*x0+coeff4;/*求矩阵*/public void Matrix(int coeff, double temp_matrix) temp_matrix0 = (double)2.0*coeff0; temp_matrix1 = (double)coeff2; temp_matrix2 = (double)coeff2; temp_matrix3 = (double)2.0*coeff1;/*对应的逆矩阵*/public void invertH(double temp_matrix, double invert) double deter; deter = temp_matrix0*temp_matrix3-temp_matrix1*temp_matrix2; invert0 = temp_matrix3/deter; invert1 = -temp_matrix1/deter; invert2 = -temp_matrix2/deter; invert3 = temp_matrix0/deter; public static void main(String args) / TODO Auto-generated method stub double eps = 0.00001 ; /精度值 int coeff = new int5;/这是初始函数的系数值数组，可以自己设定 coeff0 =2; coeff1 =1; coeff2 =2; coeff3 =1; coeff4 =-1; double x = new double2;/初始的值 x0 = 0; x1 = 0; double grads_x = new double2; double matrix = new double4; double invert = new double4; double last_val = 0; NewTon_way newTon = new NewTon_way(); newTon.grads(coeff, x, grads_x);/求导函数 while(Math.sqrt(Math.pow(grads_x0,2)+Math.pow(grads_x1,2)eps) newTon.Matrix(coeff, matrix);/求矩阵 newTon.invertH(matrix, invert); x0 = x0-(invert0*grads_x0+invert1*grads_x1); x1 = x1-(invert2*grads_x0+invert3*grads_x1); newTon.grads(coeff, x, grads_x); last_val = newTon.Old_fun(coeff, x); System.out.println(最优解为：); for(int i=0;ieps); return init_Step;public static void main(String args) / TODO Auto-generated method stub double eps = 0.00001 ; /精度值 int coeff = new int5;/这是初始函数的系数值数组，可以自己设定 coeff0 =2; coeff1 =1; coeff2 =2; coeff3 =1; coeff4 =-1; double x = new double2;/初始的值 x0 = 0; x1 = 0; double grads_x = new double2; double init_Step; double last_val = 0; Fast_way fast_way = new Fast_way(); fast_way.grads(coeff, x, grads_x);/求导函数 while(Math.sqrt(Math.pow(grads_x0,2)+Math.pow(grads_x1,2)eps) init_Step = fast_way.by_Step(coeff, x, grads_x, eps); x0 = x0 - init_Step*grads_x0; x1 = x1 - init_Step*grads_x1; fast_way.grads(coeff, x, grads_x); last_val = fast_way.