期末考试数值计算方法15道程序题详解

//用辛普森公式计算p199页的例题-7#include<stdio.h>#include<math.h>doublef(double x){return pow(x,0.5);}void main(){ double x0,x1, m;x0=0.5;x1=1.0;printf("利用辛普森公式计算的结果为："); printf("\n");m=((x1-x0)*(f(x0)+f(x1)+4*f((x0+x1)/2)))/6; printf("%lf",m);printf("\n");}/用牛顿迭代法求x等于e的-x次方在.5附近的根。

#include"stdio.h"#include"math.h"#define E 2.71828182#define fn(x)x*pow(E,x)-1#define fn_(x)pow(E,x)+x*pow(E,x )#define N 10#define F 1e-8void main(){double x0=0;double x_=0;double x = 0; double tempx=0;int k=1;printf("键入x0的初值:\nx0=");scanf("%lf",&x0);while(1){x_=fn_(x0);//导数x =fn(x0);//函数值if(x_ == 0){ printf("newton iterationfail!");break;}else{tempx=x0- ( x/x_ );if( fabs(x0-tempx) < F){printf("newton iteration success!\nx=%.8lf\n\n",tempx); break;}else{if(k == N){printf("newton iteration fail!");break;}else{k++;x0=tempx;}}}}}//牛顿插入法#include<stdio.h>#define N 6#define M 4 doublex[N]={0.4,0.55,0.65,0.8,0.9,1.05};Double y[N]={0.41075,0.57815,0.69675,0.88811 ,1.02652,1.25382};double _w(int k,int i,int j){double sum=1;int ii=i;while(ii<=j){if(k!=ii)sum*=(x[k]-x[ii]);ii++;}return sum;}double D_value(int i,int j) {double sum=0;int k=i;while(k<=j){sum+=y[k]/_w(k,i,j);k++;}return sum;} double _Poly(double x0,int n){double sum=1;int i=0;while(i<=n){sum*=(x0-x[i]);i++;}return sum;}double Polynomial(double x0,int n) {if(n==0)return y[0];else returnPolynomial(x0,n-1)+D_value(0,n)*_Poly(x0,n-1);}void main(){double x0,y0;printf("x0="); scanf("%lf",&x0);getchar();y0=Polynomial(x0 ,M);printf("y=%lf\n",y0);}//拉格朗日差值法，三次差值#include<stdio.h>double L0(double x0,double x1,double x2,double x){return((x-x1)*(x-x2))/((x0-x1)*(x0-x2));}double L1(double x0,double x1,doublex2,double x){return((x-x0)*(x-x2))/((x1-x0)*(x1-x2));}double L2(double x0,double x1,doublex2,double x){return((x-x0)*(x-x1))/((x2-x0)*(x2-x1));}void main(){double x=0,x0=0,x1=0,x2=0;doubley0=0,y1=0,y2=0;printf("输入各项x的值:\n");scanf("%lf,%lf,%lf",&x0,&x1,&x2); printf("输入各项y的值:\n");scanf("%lf,%lf,%lf",&y0,&y1,&y2);getchar(); printf("输入x的值:\n");scanf("%lf",&x); getchar();printf("结果为:%lf\n",L0(x0,x1,x2,x)*y0+L1(x0,x1,x2,x)* y1+L2(x0,x1,x2,x)*y2);}//高斯消去法#include<stdio.h>#define size 3 double a[size][size+2];void guass(){int i,j;int n;for(n=1;n<size;n++)for(i=n;i<size;i++){//得到乘数a[i][size+1]=a[i][n-1]/a[n-1][n-1];//消去for(j=n-1;j<size+1;j++)a[i][j]=a[i][j]-a[i][size+1]*a[n-1][j];}}void main(){int i,j;printf("-----------高斯消去之前---------\n\n");for(i=0;i<size;i++) {for(j=0;j<size+1;j++){a[i][j]=rand()%10;if(a[i][j]<0)printf("%.2f\t",a[i][j]);else printf(" %.2f\t",a[i][j]);}printf("\n\n");} guass();printf("-----------高斯消去后----------\n\n");for(i=0;i<size;i++){for(j=0;j<size+1;j++){if(a[i][j]<0)printf( "%.2f\t",a[i][j]);else printf("%.2f\t",a[i][ j]);}printf("\n\n");}}//高斯赛德尔#include<stdio.h>#include<math.h>#define N 3#define E 1e-4double retx1(double x2,double x3){return 0.2*x2+0.1*x3+0.3;}double retx2(double x1,double x3){return 0.2*x1+0.1*x3+1.5;}double retx3(double x1,double x2){return 0.2*x1+0.4*x2+2;}void main() {double x1=0,x2=0,x3=0;while( fabs((x1-retx1(x2,x3)))>E ){x1=retx1(x2,x3);x2=retx2(x1,x3);x3=retx3(x1,x2);}printf("x1=%.5lf\nx2=%.5lf \nx3=%.5lf\n",x1,x2,x3);printf("\n高斯-塞德尔迭代。

第页例.27数值。

结果与答案相符\n");} //书本上例题2.20.用双点弦截法解x=e-x#include"stdio.h"#include"math.h"#define E 2.718281828459#define N 10#define XE 1e-6double fn(double x){return x*pow(E,x)-1;}void main(){double x0,x1,x2;int k;printf("input x0 and x1:\nx0=");scanf("%lf",&x0);getchar();printf("x1="); scanf("%lf",&x1);printf("\nx0=%lf,x1=%lf\n" ,x0,x1);k=1;while(1){x2=x1-(fn(x1)*(x1-x0)) /(fn(x1)-fn(x0));if(fabs(fn(x2))<XE) {printf("---success!---\nx=%lf\n",x2); break;}else{if(k==N){printf("--fail!--\n"); break;}else{k++;x0=x1;x1=x2;}}}}//LU分解#include<stdio.h>#define N 4intA[N][N]={{0,0,0,0},{0,2,2,3},{0,4,7,7},{0,-2,4,5}};int L[N][N]={0};int U[N][N]={0}; void dollitte_first()//r=1{int i=0;for(i=1;i<N;i++)L[i][i]=1;for(i=1;i<N;i++) U[1][i]=A[1][i];for(i=2;i<N;i++)L[i][1]=A[i][1]/U[1][1];}void dollitte_second(int r)//r>=2{if(r>=2 && r<= N){int i=0;for(i=r;i<N;i++) U[r][i]=A[r][i]-sumb_u(r,i);for(i=r+1;i<N;i ++)L[i][r]=(A[i][r]-sumb_l(r,i))/U[r][r];} Else printf("error! ---> r=%d\n",r);}int sumb_u(int r,int i){int k=1;int sum=0; for(k=1;k<=r-1;k++)sum+=L[r][k]*U[k][i]; return sum;}int sumb_l(int r,int i){int k=1;int sum=0;for(k=1;k<=r-1;k++) sum+=L[i][k]*U[k][r];return sum;}void main() {int i=1;int j=1;for(i=1;i<N;i++){if(i==1) dollitte_first();else dollitte_second(i);} printf("A:\n");for(i=1;i<N;i++){for(j=1;j<N ;j++){printf("\t%d",A[i][j]);}printf("\n\n" );}printf("L:\n");for(i=1;i<N;i++){for(j=1; j<N;j++){printf("\t%d",L[i][j]);}printf("\n \n");}printf("U:\n");for(i=1;i<N;i++){for(j =1;j<N;j++){printf("\t%d",U[i][j]);}printf( "\n\n");}printf("结果符合例题答案\n");}//雅克比迭代#include<stdio.h>#include<math.h>#define N 3#define E 1e-5;double retx1(double x2,double x3){return 0.2*x2+0.1*x3+0.3;}double retx2(double x1,double x3){return 0.2*x1+0.1*x3+1.5;}double retx3(double x1,double x2){return 0.2*x1+0.4*x2+2;}void main(){double x1=0,x2=0,x3=0;double xx1=0,xx2=0,xx3=0;while( fabs((x1-retx1(xx2,xx3)))>E ){x1=retx1(xx2,xx3);x2=retx2(xx1,xx3);x3=retx 3(xx1,xx2);xx1=x1;xx2=x2;x3=x3;}printf("x1=%.5lf\nx2=%.5lf\nx3=%.5lf\n",x1, x2,x3);printf("\n雅克比迭代。