计算方法C语言编程第二章2已知方程在区间[1，2]内有一根，试问用二分法求根，使其具有5位有效数字至少应二分多少次？【程序设计】#includemain(){int n=0; float x1=1.0,x2=2.0,x=1.0,x0;do{ x0=x; x=(x1+x2)/2; n++; if(x*x*x+x-4>0) x2=x; else x1=x;} while(fabs(x-x0)>0.00005); printf("N=%d\n",n);} 〖运行结果〗N=154用迭代法求的正根，要求准确到小数点后第5位。【程序设计】#includemain(){ float x0, x=1.5,y;do{ y=(log(x+0.2))/5; x0=x; x=exp(y);} while(fabs(x-x0)>0.000005);printf(" X=%f\n",x);}- 〖运行结果〗x=1.0447639用牛顿法求方程在x0=2附近的根，要求准确到小数点后第3位。【程序设计】#includemain(){ float x=2.0,x0; do{x0=x; x=x-(x*x*x-3*x-1)/(3*x*x-3); } while(fabs(x-x0)>0.0005);printf("X=%f\n",x);} 〖运行结果〗x=1.87938511.分别用单点和双点弦截法求方程-x-1=0在[1,1.5]内的根。要求|xn+1-xn|<=0.000005【程序设计】#includefloat f(float x){ float f; f=x*x*x-x-1; return f;}float g(float x1,float x2){ float g; g=(x1*f(x2)-x2*f(x1))/(f(x2)-f(x1)); return g;}main(){ float x1=1.0,x2=1.5,x,x0; x=x1; do{ x0=x; x=g(x1,x2); if(f(x)>0) x2=x; else x1=x;} while(fabs(x-x0)>0.000005);printf(" X=%f\n",x);} 〖运行结果〗x=1.324717第三章1.分别用列主元素消去法求解下列方程组.(计算取4位小数).
【程序设计】#define n 4main(){floata[n][n]={1.1161,0.1254,0.1397,0.1490,0.1582,1.1675,0.1768,0.1871,0.1968,0.2071,1.2168,0.2271,0.2368,0.2471,0.2568,1.2671},b[n]={1.5471,1.6471,1.7471,1.8471},k,x[n+1],y[n+1]; int i,j,t; y[n]=0.0;for(i=0;i=0;i--){ for(j=n-2;j>=0;j--) { if(a[i][j]!=0) b[i]=b[i]-y[j+1]*a[i][j+1]; else break;} y[i]=b[i]/a[i][j+1]; x[i+1]=y[i]; }for(i=1;i<=n;i++) printf("X%d=%f\n",i,x[i]);} 〖运行结果〗x1=1.040584x2=0.986957x3=0.935052x4=0.8812979设有方程组 取初始
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