数值计算方法实验new

1、1.已知（100）=10,(121)=11,(144)=12,用拉格朗日插值法求（115）。C语言代码如下/* Note:Your choice is C IDE */#include stdio.h#define N 3void main() double xN=100,121,144; double yN=10,11,12; double s,t,y1=0,x1=115.0; int j,k; for(k=0;kN;k+) t=1.0; for(j=0;j1e-9); printf(x=%4.5f,x);即x=1.324723.用牛顿迭代法求(2x-4)x+3)x-6=0，x0=1.5C语

2、言代码如下/* Note:Your choice is C IDE */#include stdio.h#include math.hvoid main() float x1,x0,f,f1; x1=1.5; do x0=x1; f=(2*x0-4)*x0+3)*x0-6; f1=(6*x0-8)*x0+3; x1=x0-f/f1; while(fabs(x1-x0)=1e-5); printf (x=%8.7f,x1);输出结果即x的取值为24.用二分法求f(x)=x3-3x2+3x-2的根。可以算出f=3x2-6x+3=3(x-1)2即f(x)单调，即有一个根C语言代码如下/* Note:

3、Your choice is C IDE */#include stdio.h#include math.h#define f(x) (x*x*x-3*x*x+3*x-2)void main() double a=-10,b=10,c,eps=1e-5; int i=0;if(f(a)*f(b)0)printf(请重新输入a和b的值);while (b-a)eps) i+;c=(a+b)/2; if(f(c)=0) break; else if(f(a)*f(c)0) b=c; else a=c; printf(二分次数为%dn,i);printf(root=%fn,c);输出结果它的根准确为

4、2相对跟为1.9999985. 用高斯消去法解决方程组10x1-7x2+0x3=7的值。 5x1 -x2+5x3=6 -3x1+2x2+6x3=4C语言编程如下：/* Note:Your choice is C IDE */#include stdio.h#define N 3void main()float aNN;float bN;float s,akk,aik; int k,i,j; printf(请输入原始增广矩阵(按行输入):n); for(i=0;iN;i+) for(j=0;jN;j+) scanf(%f,&aij); scanf(%f,&bi); printf(你输入的原始增广

5、矩阵为：n); for(i=0;iN;i+)for(j=0;jN;j+) printf(%ft,aij); printf(%fn,bi); for(k=0;kN;k+) akk=akk; for(j=k;jN;j+) akj=akj/akk; bk=bk/akk; for(i=k+1;iN;i+) aik=aik; for(j=k;jN;j+) aij=aij-aik*akj; bi=bi-aik*bk; printf(第%d次消元:n,k); for(i=0;iN;i+) for(j=0;j=0;i-) s=0.0; for(j=2;ji;j-) s=s+aij*bj; bi=bi-s; f

6、or(i=0;iN;i+) printf(x%d=%ft,i,bi); 输出结果如下：6. 雅克比迭代求方程组10x1-x2-2x3=7.2 -x1+10x2-2x3=8.3 -x1-x2+5x3=4.2程序c语言如下：/* Note:Your choice is C IDE */#include stdio.h#define N 10#define n 3double func(double a,double b,double c,double x,double y)double fun;fun=a*x+b*y+c;return fun;void main()double func(doub

7、le,double,double,double,double); double x1N,x2N,x3N; double ann; double ann+1=10,-1,-2,7.2,-1,10,-2,8.3,-1,-1,5,4.2; int k,i,j; x10=0; x20=0.0; x30=0.0; printf(输入的线性方程组为：n); for(k=0;k3;k+) printf(%lf)x1+(%lf)x2+(%lf)x3=%lfn,ak0,ak1,ak2,ak3); for(k=0;kn;k+) akn=-akn; printf(迭代的线性方程组为：n); for(i=0;in;i

8、+) ann=aii; for(j=i;jn;j+) aij=aij+1; for(j=0;jn;j+) aij=-aij/ann; printf(x%d=(%lf)x+(%lf)x+(%lf)n,i,ai0,ai1,ai2); for(i=0;iN;i+) printf(迭代了%d次t,i+1); x1i+1=func(a00,a01,a02,x2i,x3i); x2i+1=func(a10,a11,a12,x1i,x3i); x3i+1=func(a20,a21,a22,x1i,x2i); printf(x1=%lftx2=%lftx3=%lfn,x1i+1,x2i+1,x3i+1); 结果输出：7. 追赶法求三角对角方程： 2x1 + x2 =1 x1 + 3x2 +x3 =2 x2 + x3 + x4 =2 2x3 + x4 =0C语言如下：/* Note:Your choice is C IDE */#include stdio.h#define N 4void main() int i; double cN-1=1,1,1; double bN=2,3,1,