東南大學(xué)計(jì)算方法與實(shí)習(xí)實(shí)驗(yàn)報(bào)告

1、精選優(yōu)質(zhì)文檔-傾情為你奉上計(jì)算方法與實(shí)習(xí)實(shí)驗(yàn)報(bào)告學(xué)院:學(xué)號(hào):姓名： 完成日期:實(shí)習(xí)題一4、設(shè)，已知其精確值為12(32-1N-1N+1)。1）編制按從大到小的順序計(jì)算Sn的程序；2）編制按從小到大的順序計(jì)算Sn的程序；3）按兩種順序分別計(jì)算S1000，S10000，S30000，并指出有效位數(shù)。l 實(shí)驗(yàn)代碼C語(yǔ)言程序如下：#include <iostream>#include <cmath>using namespace std;int main()float Sn=0;int N;cin>>N;for(float j=2;j<=N;j+)Sn=1/(

2、j*j-1)+Sn;cout<<"從小到大計(jì)算的結(jié)果為"<<Sn<<endl;Sn=0;for(j=N;j>=2;j-)Sn=1/(j*j-1)+Sn;cout<<"從大到小計(jì)算的結(jié)果為"<<Sn<<endl;return 0;l 運(yùn)行窗口實(shí)習(xí)題二1、用牛頓法求下列方程的根：1）l 實(shí)驗(yàn)代碼C語(yǔ)言程序代碼如下：#include <iostream>#include <cmath>#define N 100#define eps 1e-6#define e

3、ta 1e-8using namespace std;float Newton(float f(float),float fl(float),float x0)float x1,d;int k=0;dox1=x0-f(x0)/fl(x0);if(k+>N|fabs(fl(x1)<eps)cout<<"發(fā)散"<<endl;break;d=fabs(x1)<1?x1-x0:(x1-x0)/x1;x0=x1;cout<<"x="<<x0<<endl;while(fabs(d)>

4、eps&&fabs(f(x1)>eta);return x1;float f(float x)return x+log10(x)-2;float fl(float x)return 1+1/x;void main()float x0,y0;cin>>x0;y0=Newton(f,fl,x0);cout<<"方程的根為"<<y0<<endl;l 運(yùn)行窗口實(shí)習(xí)題三1、用列主元消去法解方程組：1） l 實(shí)驗(yàn)代碼C語(yǔ)言程序代碼如下:#include<iostream>#include<cmath

5、>using namespace std;void ColPivot(float *c,int n,float x)int i,j,t,k;float p;for(i=0;i<=n-2;i+)k=i;for(j=i+1;j<=n-1;j+)if(fabs(*(c+j*(n+1)+i)>(fabs(*(c+k*(n+1)+i) k=j;if(k!=j)for(j=i;j<=n;j+)p=*(c+i*(n+1)+j);*(c+i*(n+1)+j)=*(c+k*(n+1)+j);*(c+k*(n+1)+j)=p;for(j=i+1;j<=n-1;j+)p=(*(c

6、+j*(n+1)+i)/(*(c+i*(n+1)+i);for(t=i;t<=n;t+)*(c+j*(n+1)+t)-=p*(*(c+i*(n+1)+t);for(i=n-1;i>=0;i-)for(j=n-1;j>=i+1;j-)(*(c+i*(n+1)+n)-=xj*(*(c+i*(n+1)+j);xi=*(c+i*(n+1)+n)/(*(c+i*(n+1)+i);int main()void ColPivot(float*,int,float);int i;float x4;float c45=1,1,0,3,4, 2,1,-1,1,1, 3,-1,-1,3,-3, -

7、1,2,3,-1,4;ColPivot(c0,4,x);for(i=0;i<=3;i+)printf("x%d=%fn",i,xi);return 0;l 運(yùn)行窗口4、編寫(xiě)用追趕法解三對(duì)角線性方程組的程序，并解下列方程組：2），其中A10x10= -4 1 1 -4 1 1 -4 1 . . . . . . 1 -4 1 1 -4b= -27 -15 -15l 實(shí)驗(yàn)代碼C語(yǔ)言程序如下：#include<iostream>#include<cmath>using namespace std;void ColPivot(float *c,int n

8、,float x)int i,j,t,k;float p;for(i=0;i<=n-2;i+)k=i;for(j=i+1;j<=n-1;j+)if(fabs(*(c+j*(n+1)+i)>(fabs(*(c+k*(n+1)+i) k=j;if(k!=j)for(j=i;j<=n;j+)p=*(c+i*(n+1)+j);*(c+i*(n+1)+j)=*(c+k*(n+1)+j);*(c+k*(n+1)+j)=p;for(j=i+1;j<=n-1;j+)p=(*(c+j*(n+1)+i)/(*(c+i*(n+1)+i);for(t=i;t<=n;t+)*(c+j

9、*(n+1)+t)-=p*(*(c+i*(n+1)+t);for(i=n-1;i>=0;i-)for(j=n-1;j>=i+1;j-)(*(c+i*(n+1)+n)-=xj*(*(c+i*(n+1)+j);xi=*(c+i*(n+1)+n)/(*(c+i*(n+1)+i);int main()void ColPivot(float*,int,float);int i;float x10;float c1011=-4,1,0,0,0,0,0,0,0,0,-27, 1,-4,1,0,0,0,0,0,0,0,-15, 0,1,-4,1,0,0,0,0,0,0,-15, 0,0,1,-4,

10、1,0,0,0,0,0,-15, 0,0,0,1,-4,1,0,0,0,0,-15, 0,0,0,0,1,-4,1,0,0,0,-15, 0,0,0,0,0,1,-4,1,0,0,-15, 0,0,0,0,0,0,1,-4,1,0,-15, 0,0,0,0,0,0,0,1,-4,1,-15, 0,0,0,0,0,0,0,0,1,-4,-15;ColPivot(c0,10,x);for(i=0;i<=9;i+)printf("x%d=%fn",i,xi);return 0;l 運(yùn)行窗口實(shí)習(xí)題四2、按下列數(shù)據(jù)Xi0.300.420.500.580.660.72Yi1.04

11、4031.084621.118031.156031.198171.23223作5次插值，并求X1=0.46，X2=0.55，X3=0.60時(shí)的函數(shù)近似值。l 實(shí)驗(yàn)代碼C語(yǔ)言程序如下：#include<iostream>#include <cstdio>using namespace std;#define N 5void Difference(float x,float y,int n)float *f=new floatn+1;int k,i;for(k=1;k<=n;k+)f0=yk;for(i=0;i<k;i+)fi+1=(fi-yi)/(xk-xi)

12、;yk=fk;delete f;return;int main()int i;float a,b,c,varx=0.46,vary=0.55,varz=0.60;float xN+1=0.30,0.42,0.50,0.58,0.66,0.72;float yN+1=1.04403,1.08462,1.11803,1.15603,1.19817,1.23223;Difference(x,y,N);a=yN;b=yN;c=yN;for(i=N-1;i>=0;i-)a=a*(varx-xi)+yi;for(i=N-1;i>=0;i-)b=b*(vary-xi)+yi;for(i=N-1;

13、i>=0;i-)c=c*(varz-xi)+yi;printf("Nn(%f)=%fn",varx,a);printf("Nn(%f)=%fn",vary,b);printf("Nn(%f)=%fn",varz,c);return 0;l 運(yùn)行窗口實(shí)習(xí)題六1、 用復(fù)化梯形公式和復(fù)化辛卜生公式計(jì)算積分I1（f）=dx。觀察n為多少時(shí)，所得近似值具有6位有效數(shù)字。l 實(shí)驗(yàn)代碼C語(yǔ)言程序如下：#include<iostream>#include<cmath>#include <iomanip>usi

14、ng namespace std;int main() double f1=0,f2=0,p=0,q=0,e=0.,a,b,h,x1,x2,x3; int i,j=1,n1=1,n2=1; cout<<"計(jì)算結(jié)果如下："<<endl; cout<<"以下為用復(fù)化梯形公式的計(jì)算結(jié)果"<<endl; cout<<setw(15)<<"f1"<<setw(15)<<"前后項(xiàng)差值"<<endl; for(i=0;i+

15、) a=0; b=3.14159/2; h=(b-a)/n1; for(j=1;j<=n1;j+) x1=a+(j-1)*h; x2=a+j*h; f1+=h/2*(sqrt(1+cos(x1)*cos(x1)+sqrt(1+cos(x2)*cos(x2); if(fabs(f1-p)<0.)break; cout<<setw(15)<<f1<<setw(15)<<fabs(f1-p)<<endl; if(fabs(f1-p)<0.)break; p=f1; n1+; f1=0; cout<<setpre

16、cision(7)<<" 用 復(fù) 化 梯 形 公 式 計(jì) 算 得 f1="<<f1<<", 迭 代 次 數(shù) 為 ： n="<<n1<<endl<<endl; cout<<"以下為用復(fù)化辛卜生公式的計(jì)算結(jié)果"<<endl; cout<<setw(15)<<"f1"<<setw(15)<<"前后項(xiàng)差值"<<endl; for(i=0;i+) a=

17、0; b=3.14159/2; h=(b-a)/n2; for(j=1;j<=n2;j+) x1=a+(j-1)*h; x2=a+j*h; x3=a+(j-0.5)*h; f2+=h/6*(sqrt(1+cos(x1)*cos(x1)+4*sqrt(1+cos(x3)*cos(x3)+sqrt(1+cos(x2)*cos(x2); if(fabs(f2-q)<0.)break; cout<<setw(15)<<f2<<setw(15)<<fabs(f2-q)<<endl; if(fabs(f2-q)<0.)break

18、; q=f2; n2+; f2=0; cout<<setprecision(7)<<" 用 復(fù) 化 辛 卜 生 公 式 計(jì) 算 得 f1="<<f2<<", 迭 代 次 數(shù) 為 ： n="<<n2<<endl; return 0; l 運(yùn)行窗口實(shí)習(xí)題七2、用四階龍格-庫(kù)塔公式（取h=0.1）解下列微分方程組初值問(wèn)題：（1） l 實(shí)驗(yàn)代碼C語(yǔ)言程序如下：#include<stdio.h>void Runge_Kutta(float(*f1)(float x,float y1,float y2),float(*f2)(float x,float y1,float y2),float a,float b,float y0,int N)float x=a,K1,K2,K3,K4,L1,L2,L3,L4,y1=y0,y2=y0;float h=(b-a)/N;int i;printf("x0=%fty10=%fty20=%fn",x,y1,y2);for(i=1;i<=N;i+)K1=(*f1)(x,y1,y2);L1=(