[理学]数值方法第2版答案.doc

C语言编程习题 第二章 习题2-2 5. 用二分法编程求 6x4 -40x2+9=0 的所有实根。 #include stdio.h #include math.h #define N 10000 double A,B,C; double f(double x) { return (A*x*x*x*x+B*x*x+C); } void BM(double a,double b,double eps1,double eps2) { int k; double x,xe; double valuea = f(a); double valueb = f(b); if (valuea 0 valueb 0 || valuea 0 valueb 0) return; printf(Finding root in the range: [%.3lf, %.3lf]\n, a, b); for(k=1;k=N;k++) { x=(a+b)/2; xe=(b-a)/2; if(fabs(xe)eps2 || fabs(f(x))eps1) { printf(The x value is:%g\n,x); printf(f(x)=%g\n\n,f(x)); return; } if(f(a)*f(x)0) b=x; else a=x; } printf(No convergence!\n); } int main() { double a,b,eps1,eps2,step,start; printf(Please input A,B,C:\n); scanf(%lf %lf %lf,A,B,C); printf(Please input a,b, step, eps1,eps2:\n); scanf(%lf %lf %lf %lf %lf,a,b,step,eps1,eps2); for (start=a; (start+step) = b; start += step) { double left = start; double right = start + step; BM(left, right, eps1, eps2); } return 0; } 运行： Please input A,B,C: 6 -40 9 Please input a,b, step, eps1,eps2: -10 10 1 1e-5 1e-5 Finding root in the range: [-3.000, -2.000] The x value is:-2.53643 f(x)=-0Finding root in the range: [-1.000, 0.000] The x value is:-0.482857 f(x)=0Finding root in the range: [0.000, 1.000] The x value is:0.482857 f(x)=0Finding root in the range: [2.000, 3.000] The x value is:2.53643 f(x)=-0有时若把判别语句 if(fabs(xe)eps2 || fabs(f(x))eps1) 改为 if(fabs(xe)eps2 fabs(f(x))eps1) 会提高精度，对同一题运行结果： Finding root in the range: [-3.000, -2.000] The x value is:-2.53644 f(x)=-4.26496e-007 Finding root in the range: [-1.000, 0.000] The x value is:-0.482861 f(x)=-7.3797e-006 Finding root in the range: [0.000, 1.000] The x value is:0.482861 f(x)=-7.3797e-006 Finding root in the range: [2.000, 3.000] The x value is:2.53644 f(x)=-4.26496e-007 习题2-3 5. 请用埃特金方法编程求出x=tgx在4.5（弧度）附近的根。 #include