南邮 数学实验

1、第一次练习题1.求解下列各题：1)>> limit(1834*x-sin(1834*x)/(x3)ans =3084380852/3 2)>> diff(exp(x)*cos(1834*x/1000.0),10)ans = 3）int(x4/(18342+4*x2),x)ans =1/12*x3-840889/4*x+771095213/4*atan(1/917*x)4）将在展开(最高次幂为8). >> taylor(sqrt(1834/1000.0+x),9,x)ans =2.求矩阵 的逆矩阵 及特征值和特征向量。逆矩阵：>> A=-2,1,1

2、;0,2,0;-4,1,1834;inv(A)ans = -0.5005 0.2501 0.0003 0 0.5000 0 -0.0011 0.0003 0.0005特征值：>> A=-2,1,1;0,2,0;-4,1,1834;eig(A)ans = 1.0e+003 * -0.0020 1.8340 0.0020特征向量：>> A=-2,1,1;0,2,0;-4,1,1834;P,D=eig(A)P = -1.0000 -0.0005 0.2425 0 0 0.9701 -0.0022 -1.0000 0.0000D = 1.0e+003 * -0.0020 0 0

3、 0 1.8340 0 0 0 0.00203.已知分别在下列条件下画出的图形：,分别为(在同一坐标系上作图);>> x=-2:1/50:2;y1=1/(sqrt(2*pi)*1834/600)*exp(-x.2/(2*(1834/600)2);y2=1/sqrt(2*pi)*1834/600*exp(-(x+1).2/(2*(1834/600)2);y3=1/sqrt(2*pi)*1834/600*exp(-(x-1).2/(2*(1834/600)2);plot(x,y1,x,y2,x,y3)4.画 （1） >>t=0:pi/1000:20;u=0:pi/10000

4、:2;x=u.*sin(t);y=u.*cos(t);z=100.*t/1834;plot3(x,y,z) (2) >>ezmesh('sin(1834*x*y)',0,3,0,3)（3） >>t,u=meshgrid(0:.01*pi:2*pi,0:.01*pi:2*pi);x=sin(t).*(1834./100+cos(u);y=cos(t).*(1834./100+cos(u);z=sin(u);surf(x,y,z)5对于方程，先画出左边的函数在合适的区间上的图形，借助于软件中的方程求根的命令求出所有的实根,找出函数的单调区间,结合高等数学的知

5、识说明函数为什么在这些区间上是单调的,以及该方程确实只有你求出的这些实根。最后写出你做此题的体会.>> subplot(2,1,1);ezplot('x5-1834/200*x-.1');subplot(2,1,2);ezplot('x5-1834/200*x-.1');axis(-4 4 -.1 .1);solve('x5-1834/200*x-.1=0') ans = 第二次练习题1. 统计1到以内可以写为两个素数之和的偶数与奇数的个数.>>function vector=judge(x); vector=; i=2;

6、 while isempty(vector)&(i<ceil(x/2) if isprime(i)&isprime(x-i) vector=i,x-i; endm=1834oddnumber=0; for i=1:2:m; if isempty(judge(i)=0 oddnumber=oddnumber+1; end end evennumber=0; for i=2:2:m; if isempty(judge(i)=0 evennumber=evennumber+1; end end oddnumber evennumberans =oddnumber = 281 e

7、vennumber = 9152.设 是否收敛？若收敛，其值为多少？精确到17位有效数字。其中.(注意为精确的有理数)>> format long;x=1;stopc=1;eps=1e-17;n=1;p=7+1834/1000;while stopc>eps x=x+1/np;n=n+1;stopc=1/np;end nxn = 85x = 2.002257920470133.设，数列是否收敛？若收敛，其值为多少？精确到6位有效数字。>> format long;x=3;eps=1e-6;n=1;stopc=1;while x-stopc>eps stopc

8、=x; x=(x+1834/x)/2; n=n+1;endnxn = 3x = 1.565686833062037e+0024.能否找到分式函数以及分式函数,使它产生的迭代序列收敛到(对于为整数的学号，请改为求。收敛时要求精确到17位有效数字。有一个要求：必须全部是整数)?并研究如果迭代收敛,那么迭代的初值与收敛的速度有什么关系.写出你做此题的体会.>>function y=fex2_4_1(x,a,b,c,d,e) y=(a*x2+b*x+c)/(d*x+e);x=-10000; answer=; vpa(x,17) for a=-10000:10000 for b=-10000

9、:10000 for c=-10000:10000 for d=-10000:10000 for e=-10000:100000 if vpa(fex2_4_1(x,a,b,c,d,e),17)=x answer=x,a,b,c,d,e end end end end end end第三次练习题练习4>>A=4,2,;1,3; a=rand; b=rand; x=a;b t=; for i=1:40 x=A*x; t(i,1:2)=i,x(1,1)/x(2,1); end x =t = 1.00000000000000 2.59365859734352 4.000000000000

10、00 2.03205658249965 6.00000000000000 2.00508342517318 7.00000000000000 2.00203130487201 8.00000000000000 2.00081219198689 9.00000000000000 2.00032482403086 11.00000000000000 2.00005196711849 12.00000000000000 2.00002078663135 13.00000000000000 2.00000831461797 14.00000000000000 2.00000332584166 15.0

11、0000000000000 2.00000133033578 16.00000000000000 2.00000053213417 17.00000000000000 2.00000021285365 18.00000000000000 2.00000008514145 19.00000000000000 2.00000003405658 20.00000000000000 2.00000001362263 21.00000000000000 2.00000000544905 22.00000000000000 2.00000000217962 23.00000000000000 2.0000

12、0000087185 24.00000000000000 2.00000000034874 25.00000000000000 2.00000000013950 26.00000000000000 2.00000000005580 27.00000000000000 2.00000000002232 28.00000000000000 2.00000000000893 29.00000000000000 2.00000000000357 30.00000000000000 2.00000000000143 31.00000000000000 2.00000000000057 32.000000

13、00000000 2.00000000000023 33.00000000000000 2.00000000000009 34.00000000000000 2.00000000000004 35.00000000000000 2.00000000000001 36.00000000000000 2.00000000000001 37.00000000000000 2.00000000000000 38.00000000000000 2.00000000000000 39.00000000000000 2.00000000000000 40.00000000000000 2.000000000

14、00000练习7>>T=; syms n; n=input('n='); B=1/10*62,581,511,-1012;-100,-957,-844,1668;560,5394,4761,-9388;230,2218,1958,-3859; x=1;3;-2;2; P,D=eig(B) disp('xn=x*T') T=P*Dn*P'P =D = 0.69999999999187 0 0 0 0 -0.30000000004505 0 0 0 0 0.20000000005212 0xn=x*TT =练习8>>format lo

15、ng; A=2.1,3.4,-1.2,2.3;0.8,-0.3,4.1,2.8;2.3,7.9,-1.5,1.4;3.5,7.2,1.7,-9.0 x=1;2;3;4 for i=1:300 fprintf('i=%5i',i) x=A*x endi= 300x = NaN NaN NaN NaNans=i= 295 x = NaN Inf NaN -Inf练习9>>format long A=2.1,3.4,-1.2,2.3;0.8,-0.3,4.1,2.8;2.3,7.9,-1.5,1.4;3.5,7.2,1.7,-9.0 x=1;2;3;4 y=; for i

16、=1:10000000 fprintf('i=%5i',i) x=A*x y=x./max(x) x=y; end练习12>>A=3/4,1/2,1/4;1/8,1/4,1/2;1/8,1/4,1/4; P=.5;.25;.25; for i=1:10 fprintf('Day %2i',i+1) P=A*P; vpa(P,4) endDay 2 ans = .5625 .2500 .1875 Day 3 ans = .5938 .2266 .1797 Day 4 ans = .6035 .2207 .1758 Day 5 ans = .6069 .

17、2185 .1746 Day 6 ans = .6081 .2178 .1741 Day 7 ans = .6085 .2175 .1740 Day 8 ans = .6086 .2174 .1739 Day 9 ans = .6087 .2174 .1739练习13>>t=; A=3/4,7/18;1/4,11/18; p,d=eig(A); t=p(:,1)./0.6087;.3913; vpa(t,4)ans = 1.382 1.382练习14>>t=; A=3/4,1/2,1/4;1/8,1/4,1/2;1/8,1/4,1/4; P=.5;.25;.25; fo

18、r i=1:10 P=A*P; end p,d=eig(A); t=p(:,1)./P; vpa(t,4)ans = -1.494 -1.494 -1.494第四次练习题1.设一个355毫升的易拉罐是圆柱体,上底面与下底面的厚度分别为侧面厚度的倍与倍.问在圆柱体的高度与上下底面的半径为多少时,该易拉罐所用的材料最省.(求解时取)>>a=input('a=') b=input('b=') syms s r ; s=volume(a,b,0.00001) r=0.00001; for i=0.00001:.001:10 if s>volume(a,b,i); r=i; s=volume(a,b,i); end end fprintf('r=%10.5f smin=%10.5f',r,s)ans=r=2.79801 smin= 121.168622.分别取,运行程序,你能否验证所得到的解是最优解?>> a=1834/300; b=1834/400; syms s r ; s=volume(a,b,0.00001) r=0.00001; f