广东工业大学实验报告.docx

1、广东工业大学实验报告广东工业大学实验报告 应用数学 学院 09信息计算 专业 （2） 班 成绩评定_学号 3109008538 姓名 刘艳明 教师签名_实验 题目 P150E9 第 9 周 星期 五 第12节编写：用牛顿、平分、成功失败法、黄金分割法和进退法求函数f(x)的极小点的计算程序，并求解1.牛顿法： #includeusing namespace std;class Newtondouble x;double f;double e;public:Newton(double x1,double e1)x=x1;e=e1;f=x*x*x*x+-16*x*x*x+30*x*x+-24*x+

2、8;void set(double x1,double e1)x=x1;e=e1; f=3*x*x*x*x-16*x*x*x+30*x*x-24*x+8;double getfirstf(double x) return (12*x*x*x-48*x*x+60*x-24);double getsecondf(double x)return (36*x*x-96*x+60);int main()double x0=3;int i=0;Newton a=Newton(x0,0.01);double f1=a.getfirstf(x0);double f2=a.getsecondf(x0);doub

3、le x1=x0-f1/f2;while(x1-x0)0.01|(x0-x1)0.01) i+;x0=x1;a.set(x0,0.01); f1=a.getfirstf(x0); f2=a.getsecondf(x0); coutthe x0 is:x0endl; coutthe x1 is:x1endl; x1=x0-f1/f2;couteeeeeeeeeeeeeeeeeeeeendl;coutthe last x0 is:x0endl;coutthe last x1 is:x1endl;cout迭代次数为：iendl;return 0;2.平分法：#includeusing namespa

4、ce std;double f2(double l,double o,double v,double e);double a,b,eps,sita;int maxstep=30;double x30;int main() couta; coutendl; coutb; coutendl; couteps; coutendl; coutsita; coutendl; f2(a,b,eps,sita); return 0;double f2(double l,double o,double v,double e) for(int i=0;imaxstep;i+) xi=(a+b)/2; cout

5、a ,b endl x*=: xiendl0) b=xi; else a=xi; while(abs(df1)eps|abs(b-a)sita) coutendl x*= xi; return 0; 3.成功失败法#includeusing namespace std;double funtion1(double b,double u,double g);double funtion2(double x);double chong;double h;double eps;double maxstep=30;int main() coutchong; coutendl; cout0): ; ci

6、nh; coutendl; cout0) : ; cineps; coutendl; funtion1(chong,h,eps); return 0;double funtion1(double b,double u,double g) double x0=chong; double f0=funtion2(x0); for(int i=0;imaxstep;i+) double f1=funtion2(x0+h); / cout h= hendl; couti+1step x0= x0 f0= f0 f1= f1endl; if(f1f0) x0=x0+h; f0=f1; h=2*h; el

7、se if(abs(h)eps) coutchongchong: the answer is that you want 0.0 x* = x0endlMin f(x)=funtion2(x0); return 0; else h=-0.25*h; double funtion2(double x) double f2=x*x*x*x*x+2*x*x*x*x-4*x*x*x+x*x+x+2; return f2;4.黄金分割#includeusing namespace std;double goldivide(double c,double h,double o);double f(doub

8、le x);double a30;double b30;double u30;double v30;double eps;int maxstep=30;int main() couta0; coutb0; couteps; goldivide(a0,b0,eps); return 0;double goldivide(double c,double h,double o) u0=a0+(1-0.618)*(b0-a0); v0=a0+0.618*(b0-a0); for(int k=0;kmaxstep;k+) coutak= ak bk= bkendl; if(bk-ak)eps) doub

9、le x0=0.5*(bk+ak); coutk step needendlx0=x0; cout f(x0)=f(vk) ak+1=uk; bk+1=bk; uk+1=vk; vk+1=ak+1+0.618*(bk+1-ak+1); f(vk+1); else ak+1=ak; bk+1=vk; vk+1=uk; uk+1=ak+1+(1-0.618)*(bk+1-ak+1); f(uk+1);double f(double x) double f1=x*x*x*x*x+2*x*x*x*x-4*x*x*x+x*x+x+2; return f1;5.不确定精度一维搜索#includedoubl

10、e f(double x1,double x2);double f1(double x1,double x2);double f2(double x1,double x2);int xk2=-1,1;int sk2=1,1;double xk12;double c1=0.1;double c2=0.5;double a=0;double b;double r=1;int main() using namespace std; double fk=f(xk0,xk1); double gk2=f1(xk0,xk1),f2(xk0,xk1); double gkTsk=gk0*sk0+gk1*sk

11、1; coutfk=fk; cout gkTsk=gkTskendl; for(;) xk10=xk0+r*sk0; xk11=xk1+r*sk1; double fk1=f(xk10,xk11); double gk12=f1(xk10,xk11),f2(xk10,xk11); double gk1Tsk=gk10*sk0+gk11*sk1; coutr=r; cout f(k+1)=fk1; cout g(k+1)Tsk=gk1Tsk=-c1*r*gkTsk) if(gk1Tsk=c2*gkTsk) double rk=r; coutrk=: rk; cout x(k+1)=: (xk10

12、,xk11); return 0; else a=r;r=min(2*r,0.5*(r+b); else b=r;r=0.5*(r+a); double f(double x1,double x2) double zf=100*(x2-x1*x1)*(x2-x1*x1)+(1-x1)*(1-x1); return zf;double f1(double x1,double x2) double zf1=-400*(x2-x1*x1)*x1-2*(1-x1); return zf1;double f2(double x1,double x2) double zf2=200*(x2-x1*x1);

13、 return zf2;6.进退法#includeusing namespace std;double dilemma(double z,double c);double f(double x);double t30;double h;double a;double b;int maxstep=30;int main() coutt0; coutendlh; dilemma(t0,h); return 0;double dilemma(double z,double c) coutf(t0)=f(t0)endl; t2=t0+h; coutf(t2)=f(t2)f(t0) h=-h; for(;) t1=t0+h; f(t1); coutf(t1)=f(t1)endl; if(f(t1)=f(t0) h=2*h; t2=t0; t0=t1; else coutf(t0) the endendl; a=min(t1,t2); b=max(t1,t2); break; coutits the answer you want ：oa,b; cout and t0=0.5*(a+b); return 0; double f(double x) double f1=x*x*x*x*x+2*x*x*x*x-4*x*x*x+x*x+x+2; return f1;