數(shù)值分析作業(yè)(C語言編程實現(xiàn))

1、#include <stdio.h>#include <math.h>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;j<k-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;fo

2、r(j=1;j<=k;j+)c=1;for(n=0;n<j;n+)c=c*4;tjk-j=(c*tj-1k-j+1-tj-1k-j)/(c-1);if(fabs(tk0-tk-10)<error)printf("n積分結(jié)果 I %lfn",tk0); break;elseh=h/2;k+;#include <stdio.h>#include <math.h>double f(double t)double ans;ans=pow(cos(t),1.0/3);return ans;void main()double x=0,eslon

3、g=0.000001,x0;int N=20,i;printf("n近似初值 x0 = %lfn",x);for(i=0;i<N;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("得到近似結(jié)果為 x %lfnn",x,i);elseprintf("迭代失敗n");#include <stdio.h>#include <math.h&g

4、t;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()N=(b-a)/h;x=a; printf("n 初值為 (x0,y0) = ( %.8f , %.8f )n",x,y);for(i=0;i<N;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

5、)/6;x=x+h; printf(" 第%d次輸出結(jié)果為 (x%d,y%d) = ( %.8f , %.8f )n",i+1,i+1,i+1,x,y);#include <stdio.h>void main()double datax4=1.2,2.9,4.6,5.8,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;j<i;j+)li=(x-dataxj)/(dataxi-dataxj)*li;for(j=i+1;j<=3

6、;j+)li=(x-dataxj)/(dataxi-dataxj)*li;y=y+datayi*li;printf("n f(x)在 x = %f 處的近似值為: y = %fn",x,y);#include <stdio.h>void main()double datay9=11.7,14.87,21.44,31.39,44.73,61.46,81.57,105.11,131.91;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;i<

7、;9;i+)for(j=1;j<2*m+1;j+)dataij=1;for(k=0;k<j;k+)dataij=dataij*dataxi;for(i=0;i<m+1;i+)for(j=0;j<m+1;j+)aij=0;for(k=0;k<9;k+) aij=aij+dataki+j;a00=9;a0m+1=0;for(i=0;i<9;i+)a0m+1=a0m+1+datayi;for(i=1;i<m+1;i+)aim+1=0;for(j=0;j<9;j+)p=datayj;for(k=0;k<i;k+)p=p*dataxj;aim+1=a

8、im+1+p; /生成m+1行，m+2列增廣矩陣/for(i=0;i<m+1;i+) /顯示方程組/for(j=0;j<m+2;j+)if(j!=m+1)printf("(%f)a%d ",aij,j);if(j!=m)printf("+ ");elseprintf("= %f n",aij);for(i=0;i<m;i+) /高斯消去法/ if(aii!=0)for(j=i+1;j<m+1;j+)aji=aji/aii;for(k=i+1;k<m+2;k+)ajk=ajk-aik*aji;elsebre

9、ak;if(amm!=0&&i=m)amm+1=amm+1/amm;for(i=2;i<=m+1;i+)for(j=1;j<i;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("方程組的解為:n");for(j=0;j<m+1;j+)printf("a%d = %fn",j,ajm+1);printf("擬合多項式為:n");printf("P%d(x) = (%f)

10、+ (%f)x + (%f)x2n",m,a0m+1,a1m+1,a2m+1);elseprintf("數(shù)據(jù)有誤！n");列主元素法#include <stdio.h>#include <math.h>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;i<2;i+) comp=fabs(aii);for(k=i;k<3;k+) /比較絕對值大小并進行主元列交換/if(fabs(aki)>=comp)n

11、row=k; comp=fabs(aki);for(j=0;j<=3;j+)mov=aij;aij=anrowj;anrowj=mov;printf("方程第%d行互換位置后如下n",i+1);for(j=0;j<3;j+)printf("(%f)x1 + (%f)x2 + (%f)x3 = %fn",aj0,aj1,aj2,aj3);if(aii!=0)for(j=i+1;j<3;j+)aji=aji/aii;for(k=i+1;k<=3;k+)ajk=ajk-aik*aji;aji=0;printf("方程經(jīng)%d次消

12、元如下n",i+1);for(j=0;j<3;j+)printf("(%f)x1 + (%f)x2 + (%f)x3 = %fn",aj0,aj1,aj2,aj3);elsebreak;if(a22!=0&&i=2)printf("方程化簡得n");for(i=0;i<3;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;j<i;j+)a3-i3

13、=a3-i3-a3-i3-j*a3-j3; a3-i3=a3-i3/a3-i3-i; printf("方程組的解為:n");for(j=0;j<3;j+)printf("x%d = %fn",j+1,aj3);elseprintf("數(shù)據(jù)有誤！n");Jacobi迭代法#include <stdio.h>#include <math.h>void main() double a37=1,-2,-1,3,-2,10,-3,15,-1,-2,5,10,error=0.000001,norm;int N=423

14、,i,j,k;a04=0,a14=0,a24=0;for(i=0;i<3;i+) /把a矩陣轉(zhuǎn)化為b矩陣/ai6=aii;for(j=0;j<3;j+)aij=-aij/ai6;ai3=ai3/ai6;aii=0;printf("化為b矩陣如下n");for(i=0;i<3;i+)printf("%f %f %f %fn",ai0,ai1,ai2,ai3);for(i=1;i<N;i+)for(j=0;j<3;j+)aj5=0;for(k=0;k<3;k+)aj5=ak4*ajk+aj5;aj5=aj5+aj3;nor

15、m=0;for(k=0;k<3;k+)norm=norm+fabs(ak4-ak5);if(norm<error)break;elsefor(k=0;k<3;k+)ak4=ak5; if(norm<error)printf("計算結(jié)果為n"); for(i=0;i<3;i+)printf(" x%d = %.3fn",i+1,ai5); elseprintf("迭代失敗n");現(xiàn)場考試題目1#include "stdio.h"#include "math.h"dou

16、ble f(double x)double ans;ans=exp(x);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("積分結(jié)果為I %fn",T);題目2#include "stdio.h"double f(double t,double s)double ans;ans=1+t