数值分析报告作业C语言编程实现.docx

1、数值分析报告作业C语言编程实现#include #include double f(double x) double ans; ans=exp(x); return ans;void main() double a=1,b=3,error=0.0001,t2020,h,c; int i,j,k,m,n; h=b-a; t00=h*(f(a)+f(b)/2; k=1; while(1) t0k=0; m=1; for(j=0;jk-1;j+) m=m*2; for(i=1;i=m;i+) t0k=t0k+h*f(a+(i-0.5)*h); t0k=(t0k+t0k-1)/2; for(j=1;j

2、=k;j+) c=1; for(n=0;nj;n+) c=c*4; tjk-j=(c*tj-1k-j+1-tj-1k-j)/(c-1); if(fabs(tk0-tk-10)error) printf(n积分结果 I %lfn,tk0); break; else h=h/2; k+; #include #include double f(double t) double ans; ans=pow(cos(t),1.0/3); return ans;void main() double x=0,eslong=0.000001,x0; int N=20,i; printf(n近似初值 x0 = %

3、lfn,x); for(i=0;iN;i+) x0=x; x=f(x); printf( x%d = %lfn,i+1,x); if(fabs(x-x0)eslong) break; if(fabs(x-x0)eslong) printf(得到近似结果为 x %lfnn,x,i); else printf(迭代失败n);#include #include double a=0,b=1,x,y=0,h=0.1,k1,k2,k3,k4;int i,N;double f(double t,double s) double ans; ans=1+t*t; return ans;void main()

4、N=(b-a)/h; x=a; printf(n 初值为 (x0,y0) = ( %.8f , %.8f )n,x,y); for(i=0;iN;i+) k1=f(x,y); k2=f(x+h/2,y+h*k1/2); k3=f(x+h/2,y+h*k2/2); k4=f(x+h,y+h*k3); y=y+h*(k1+2*(k2+k3)+k4)/6; x=x+h; printf( 第%d次输出结果为 (x%d,y%d) = ( %.8f , %.8f )n,i+1,i+1,i+1,x,y); #include void main() double datax4=1.2,2.9,4.6,5.8,

5、datay10=14.84,33.71,58.36,79.24,l3,x=1.5,y; int i,j; y=0; for(i=0;i=3;i+) li=1; for(j=0;ji;j+) li=(x-dataxj)/(dataxi-dataxj)*li; for(j=i+1;j=3;j+) li=(x-dataxj)/(dataxi-dataxj)*li; y=y+datayi*li; printf(n f(x)在 x = %f 处的近似值为: y = %fn,x,y);#include void main() double datay9=11.7,14.87,21.44,31.39,44.

6、73,61.46,81.57,105.11,131.91; int m=2,i,j,k; double p,data94,a34,datax9=1.2,2.3,3.4,4.5,5.6,6.7,7.8,8.9,10.0; for(i=0;i9;i+) for(j=1;j2*m+1;j+) dataij=1; for(k=0;kj;k+) dataij=dataij*dataxi; for(i=0;im+1;i+) for(j=0;jm+1;j+) aij=0; for(k=0;k9;k+) aij=aij+dataki+j; a00=9; a0m+1=0; for(i=0;i9;i+) a0m+

7、1=a0m+1+datayi; for(i=1;im+1;i+) aim+1=0; for(j=0;j9;j+) p=datayj; for(k=0;ki;k+) p=p*dataxj; aim+1=aim+1+p; /生成m+1行，m+2列增广矩阵/ for(i=0;im+1;i+) /显示方程组/ for(j=0;jm+2;j+) if(j!=m+1) printf(%f)a%d ,aij,j); if(j!=m) printf(+ ); else printf(= %f n,aij); for(i=0;im;i+) /高斯消去法/ if(aii!=0) for(j=i+1;jm+1;j+

8、) aji=aji/aii; for(k=i+1;km+2;k+) ajk=ajk-aik*aji; else break; if(amm!=0&i=m) amm+1=amm+1/amm; for(i=2;i=m+1;i+) for(j=1;ji;j+) am+1-im+1=am+1-im+1-am+1-im+1-j*am+1-jm+1; am+1-im+1=am+1-im+1/am+1-im+1-i; printf(方程组的解为:n); for(j=0;jm+1;j+) printf(a%d = %fn,j,ajm+1); printf(拟合多项式为:n); printf(P%d(x) =

9、(%f) + (%f)x + (%f)x2n,m,a0m+1,a1m+1,a2m+1); else printf(数据有误！n);列主元素法#include #include void main() double a34=1,-2,-1,3,-2,10,-3,15,-1,-2,5,10,mov,comp; int i,j,k,nrow; for(i=0;i2;i+) comp=fabs(aii); for(k=i;k=comp) nrow=k; comp=fabs(aki); for(j=0;j=3;j+) mov=aij; aij=anrowj; anrowj=mov; printf(方程第

10、%d行互换位置后如下n,i+1); for(j=0;j3;j+) printf(%f)x1 + (%f)x2 + (%f)x3 = %fn,aj0,aj1,aj2,aj3); if(aii!=0) for(j=i+1;j3;j+) aji=aji/aii; for(k=i+1;k=3;k+) ajk=ajk-aik*aji; aji=0; printf(方程经%d次消元如下n,i+1); for(j=0;j3;j+) printf(%f)x1 + (%f)x2 + (%f)x3 = %fn,aj0,aj1,aj2,aj3); else break; if(a22!=0&i=2) printf(

11、方程化简得n); for(i=0;i3;i+) printf(%f)x1 + (%f)x2 + (%f)x3 = %fn,ai0,ai1,ai2,ai3); a23=a23/a22; for(i=2;i=3;i+) for(j=1;ji;j+) a3-i3=a3-i3-a3-i3-j*a3-j3; a3-i3=a3-i3/a3-i3-i; printf(方程组的解为:n); for(j=0;j3;j+) printf(x%d = %fn,j+1,aj3); else printf(数据有误！n);Jacobi迭代法#include #include void main() double a37

12、=1,-2,-1,3,-2,10,-3,15,-1,-2,5,10,error=0.000001,norm; int N=423,i,j,k; a04=0,a14=0,a24=0; for(i=0;i3;i+) /把a矩阵转化为b矩阵/ ai6=aii; for(j=0;j3;j+) aij=-aij/ai6; ai3=ai3/ai6; aii=0; printf(化为b矩阵如下n); for(i=0;i3;i+) printf(%f %f %f %fn,ai0,ai1,ai2,ai3); for(i=1;iN;i+) for(j=0;j3;j+) aj5=0; for(k=0;k3;k+)

13、aj5=ak4*ajk+aj5; aj5=aj5+aj3; norm=0; for(k=0;k3;k+) norm=norm+fabs(ak4-ak5); if(normerror) break; else for(k=0;k3;k+) ak4=ak5; if(normerror) printf(计算结果为n); for(i=0;i3;i+) printf( x%d = %.3fn,i+1,ai5); else printf(迭代失败n);现场考试题目1#include stdio.h#include math.hdouble f(double x) double ans; ans=exp(x

14、); return(ans);void main() double a=-1,b=1,error=0.0001,m=1,h,T0,T,F; int k; h=(b-a)/2; T0=h*(f(a)+f(b); while(1) F=0; for(k=1;k=pow(2.0,m-1);k+) F=F+f(a+(2*k-1)*h); T=T0/2+h*F; if(fabs(T-T0)error) break; m+; h=h/2; T0=T; printf(积分结果为I %fn,T);题目2#include stdio.hdouble f(double t,double s) double ans; ans=1+t*t; return(ans);void main() double a=0,b=1,h=0.2,x0=0,y0=0,x,k1,k2,k3,y; int N,n; N=(b-a)/h; for(n=1;n=N;n+) x=x0+h; k1=f(x0,y0); k2=f(x0+h/2,y0+h/2*k1); k3=f(x0+h,y0-h*k1+2*h*k2); y=y0+h/6*(k1+4*k2+k3); printf(第%d次输出结果为(%.8f,%.8f)n,n,x,y); x0=x; y0=y;