數(shù)值分析課后復習(C語言學習知識編程實現(xiàn))

1、*L (10分)用Rombe喀方法il算積分=f/小的近似值.誤差不超過101#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、/ 2;for(j=1;j=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 積分結果 I lfn,tk0); break; elseh=h/2;k+;積分結果I a 17.36725E2,（10分）用簡單誥代法求解方程/華納尸伍 要求同一9加10%#include #include double f(double t) double ans;ans=pow(cos(t),1.0/3); return ans;void main() double x=

3、0,eslong=0.000001,x0;int N=20,i;printf(n 近似初值 x0 = %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(得到近似結果為x%lfnn,x,i); elseprintf(迭代失敗 n);近彳比初值 = 0-RS003B xl = 1.000000 x2 = 0.S14477 乂3 = 0.8S2051 jc4 = Q.8S978B x5 = B. 86,7394 - 0

4、.8&4822 x7 - 0,8&S&95 xS = U.S&S399 x?=98&54 xl0 = 。&65465 xll - 0,865477 X12 = 0.865473 x13 = 0.&B5474 X140,865474得引近似結果為 x仁0.865474，（10用四階RK方法求初值問題的數(shù)值解.取步長汁算結果取X位小數(shù).Jy = 1 十0 W 110） =0#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;

5、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;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 次輸出結果為 (x%d,y%d) = ( %.8f , %.8f )n,i+1,i+1,i+1,x,y); 刃值才Eir i i出結果為 讓荒果為 出結果為 Llji ffi 4-

6、,第e ,??；曲td結果為第砍前出結果為第瞅施出結果為我?融結C 時.QS0B00B0 . 6.0O0B00B0 ?C0.30000000,0.3090SS0G0.4SB00030,0.42133333?C3.5QB0QQ00,0.41666(7)0.80000003,0.97O66C67) 0.90000000 = 1.00000000 8 1.33333333 54. U0分）已知函散衣如卜，用三次插值求NL5）的近似值.XiL2294,65.8氏7此14.8433.71P 5S3679.2496 79#include void main()double datax4=1.229,4.6,

7、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;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 處的近似值為:y = %fn,x,y);RG在 X - l.5enO0 處的近似值為 m y - 17.7500005. U0分己知數(shù)據(jù)表如F，試求其最小二乘二次擬合多項式.1

8、.22344,55.66.77.XS.910.Q1L714.SJ72L4431.3944.7361.468L57105.11BL191#include 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;i9;i+)for(j=1;j2*m+1;j+)dataij=1;for(k=0;kj;k+) dataij=da

9、taij*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+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列增廣矩陣/for(i=0;im+1;i+)顯示方程組 /for(j=0;jm+2;j+)if(j!=m+1) printf(%f)a%d ,aij

10、,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+)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

11、; ) printf(方程組的解為：n);for(j=0;j稹1 * (33548= 31&2Q.1ES388方程組的解為:-12.976541卜1 = -2.B00515旦2 = 1,398787式合親項式為:P2(x) = + C-Z.000515x x2心(2。分)分別用列主元法和J&sbi迭代法求解方程組.F元 1 - 22 -= 3一2x1 + 10，- 3y5 = 15- F - 2xz + 5工3 = 10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;

12、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d行互換位置后如下n,i+1);for(j=0;j3;j+)printf(%f)x1 + (%f)x2 + (%f)x3 = %fn,aj0,aj1,aj2,aj3); if(a皿i!=0) for(j=i+1;j3;j+)a皿i=a皿/aii;for(k=i+1;k=3;k+) ajk=a皿k-aik*a皿i;a皿i

13、=0;printf(方程經(jīng)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);elsebreak;if(a22!=0&i=2)printf(方程化簡得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(方程組的解

14、為：n);for(j=0;jx2 + c-1.006008x3 = 3.000080 (-1.000009)x1 1 (-2.030000x2 *- 10.000060方程經(jīng)i次消元如F-2 . G0000Hxd + (13.900000x2 * C3.080000x3 =.GQD0QQka.00Q008xl + x2 + x3 - 10.5G0OQB (0.00600axl + x2 +- 2-560600方程第之行互換位置后如下：2 -000009x1 +-090009x2 *3.000003x3 = 15 .6000000,000000)x1 + x2 * x3 = 2.506066 (

15、0.000000xl + x2 + c-2.500000x3 - 10.500000 計程經(jīng)漱消元如FC-2 *1- m + C3 .000000?x3 = 15 .BOGOBGxZ + C6.500600x3 = 2.50O0OS (0.000000)x1 + (0.a00900x2 + 0.2857147x3 - 11-571425 質(zhì)程化簡得 C-2 . 060089)x1 1- C10-+ C-3= 15 .B06SB6xl + (-7.0O008Sx2 + C6.500800m3 = 2.50S6OS (0.0O00a0xl +*(0.285714M3 = I1-S71429方程組的

16、解為二 xi = iie.aaaaea x2 = 37.250000 x3 40.09003Jacobi 迭代法#include #include void main()double a37=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矩陣轉化為b矩陣ai6=aii;for(j=0;j3;j+)aij=-aij/ai6;ai3=ai3/ai6;aii=0;printf( 化為 b 矩陣如下n);for(i=0;i3;i+)printf

17、(%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+)aj5=ak4*ajk+aj5;aj5=aj5+aj3;norm=0;for(k=0;k3;k+)norm=norm+fabs(ak4-ak5);if(normerror)break;elsefor(k=0;k3;k+)ak4=ak5;if(normerror)printf(計算結果為n);for(i=0;i3;i+)printf( x%d = %.3fn,i+1,ai5);elseprintf(迭代失敗 n);K為b矩陣如下0.00Q8

18、39 2.j. BQ0000 3 . QOQB0Q 8.200000 0000006 6.3Q00O0 1.508QGO 0.200009 9.400000 3.090000 2.000000計算結果為xl = 118.000x2 = 37.250x3 = 49.500現(xiàn)場考試題目 1#include stdio.h#include math.hdouble 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(積分結果為I弋fn,T);題目 2#include st