0.618法:1.用C或C++等编程语言编写0.618法的程序,并求解3min()21fxxx的近似最优解,初始搜索区间为0,3,区间精度为510.解:用0.618法的C语言程序为:#includestdio.h#includemath.hdoublefun(doublex){doublef;f=pow(x,3)-2*x+1;returnf;}doublegolds(doublea,doubleb){doublet,fp,fq,fa,fb,p,q,s,fs,ds,df,epsilon,alpha,delta,h;t=(sqrt(5)-1)/2;epsilon=1e-5;delta=1e-4;fa=fun(a);fb=fun(b);h=b-a;p=a+(1-t)*h;q=a+t*h;fp=fun(p);fq=fun(q);while((fabs(fb-fa)epsilon)&&(fabs(b-a)delta)){if(fp=fq){a=a;b=q;fb=fq;q=p;fq=fp;h=b-a;p=a+(1-t)*h;fp=fun(p);}else{a=p;b=b;fa=fp;p=q;fp=fq;h=b-a;q=a+t*h;fq=fun(q);}}ds=fabs(b-a);df=fabs(fb-fa);if(fp=fq){s=p;fs=fp;}else{s=q;fs=fq;}alpha=ds;printf(\n步长为:alpha=%lf\n\n,alpha);printf(极小值点:x=%lf\n\n,s);returnfs;}intmain(){doublefk,a,b;a=0.0;b=3.0;;fk=golds(a,b);printf(极小值为:fk=%lf\n\n,fk);}运行结果为: