数学实验练习题参考答案

1、专业 姓名 学号 成绩第一次练习教学要求：熟练掌握Matlab软件的基本命令和操作，会作二维、三维几何图形，能够用Matlab软件解决微积分、线性代数与解析几何中的计算问题。补充命令vpa(x,n)显示x的n位有效数字，教材102页fplot(f(x),a,b)函数作图命令，画出f(x)在区间a,b上的图形在下面的题目中为你的学号的后3位（1-9班）或4位（10班以上）1.1 计算与程序：syms xlimit(1001*x-sin(1001*x)/x3,x,0)结果：1003003001/6程序：syms xlimit(1001*x-sin(1001*x)/x3,x,inf)结果：01.2

2、，求 程序：syms xdiff(exp(x)*cos(1001*x/1000),2)结果：-2001/1000000*exp(x)*cos(1001/1000*x)-1001/500*exp(x)*sin(1001/1000*x)1.3 计算程序：dblquad(x,y) exp(x.2+y.2),0,1,0,1)结果：2.139350195142281.4 计算程序：syms xint(x4/(10002+4*x2)结果：1/12*x3-1002001/16*x+1003003001/32*atan(2/1001*x)1.5 程序：syms xdiff(exp(x)*cos(1000*x)

3、,10)结果：-1009999759158992000960720160000*exp(x)*cos(1001*x)-10090239998990319040000160032*exp(x)*sin(1001*x)1.6 给出在的泰勒展式(最高次幂为4). 程序：syms xtaylor(sqrt(1001/1000+x),5)结果：1/100*10010(1/2)+5/1001*10010(1/2)*x-1250/1002001*10010(1/2)*x2+625000/1003003001*10010(1/2)*x3-390625000/1004006004001*10010(1/2)*x

4、41.7 Fibonacci数列的定义是，用循环语句编程给出该数列的前20项（要求将结果用向量的形式给出）。程序：x=1,1;for n=3:20 x(n)=x(n-1)+x(n-2);endx结果：Columns 1 through 10 1 1 2 3 5 8 13 21 34 55 Columns 11 through 20 89 144 233 377 610 987 1597 2584 4181 67651.8 对矩阵，求该矩阵的逆矩阵，特征值，特征向量，行列式，计算，并求矩阵（是对角矩阵），使得。程序与结果：a=-2,1,1;0,2,0;-4,1,1001/1000;inv(a)

5、0.50100100100100 -0.00025025025025 -0.50050050050050 0 0.50000000000000 0 2.00200200200200 -0.50050050050050 -1.00100100100100eig(a)-0.49950000000000 + 1.32230849275046i -0.49950000000000 - 1.32230849275046i 2.00000000000000p,d=eig(a)p = 0.3355 - 0.2957i 0.3355 + 0.2957i 0.2425 0 0 0.9701 0.8944 0.8

6、944 0.0000 注：p的列向量为特征向量d = -0.4995 + 1.3223i 0 0 0 -0.4995 - 1.3223i 0 0 0 2.0000 a6 11.9680 13.0080 -4.9910 0 64.0000 0 19.9640 -4.9910 -3.0100 1.9 作出如下函数的图形（注：先用M文件定义函数，再用fplot进行函数作图）：函数文件f.m: function y=f(x)if 0=x&x=1/2 y=2.0*x;else 1/2x&x f=inline(x+1000/x)/2);x0=3;for i=1:20;x0=f(x0);fprintf(%g

7、,%gn,i,x0);end运行结果：1,168.167 11,31.62282,87.0566 12,31.62283,49.2717 13,31.62284,34.7837 14,31.62285,31.7664 15,31.62286,31.6231 16,31.62287,31.6228 17,31.62288,31.6228 18,31.62289,31.6228 19,31.622810,31.6228 20,31.6228由运行结果可以看出，数列收敛，其值为31.6228。2.2 求出分式线性函数的不动点，再编程判断它们的迭代序列是否收敛。解：取m=1000.（1）程序如下：f=

8、inline(x-1)/(x+1000);x0=2;for i=1:20;x0=f(x0);fprintf(%g,%gn,i,x0);end运行结果：1,0.000998004 11,-0.0010012,-0.000999001 12,-0.0010013,-0.001001 13,-0.0010014,-0.001001 14,-0.0010015,-0.001001 15,-0.0010016,-0.001001 16,-0.0010017,-0.001001 17,-0.0010018,-0.001001 18,-0.0010019,-0.001001 19,-0.00100110,-

9、0.001001 20,-0.001001由运行结果可以看出，分式线性函数收敛，其值为-0.001001。易见函数的不动点为-0.001001（吸引点）。（2）程序如下：f=inline(x+1000000)/(x+1000);x0=2;for i=1:20;x0=f(x0);fprintf(%g,%gn,i,x0);end运行结果：1,998.006 11,618.3322,500.999 12,618.3023,666.557 13,618.3144,600.439 14,618.3095,625.204 15,618.3116,615.692 16,618.317,619.311 17,

10、618.3118,617.929 18,618.319,618.456 19,618.3110,618.255 20,618.31由运行结果可以看出，分式线性函数收敛，其值为618.31。易见函数的不动点为618.31（吸引点）。2.3 下面函数的迭代是否会产生混沌？（56页练习7（1）解：程序如下：f=inline(1-2*abs(x-1/2);x=;y=;x(1)=rand();y(1)=0;x(2)=x(1);y(2)=f(x(1);for i=1:100;x(1+2*i)=y(2*i);x(2+2*i)=x(1+2*i);y(2+2*i)=f(x(2+2*i);endplot(x,y,

11、r);hold on;syms x;ezplot(x,0,1/2);ezplot(f(x),0,1);axis(0,1/2,0,1); hold off运行结果：2.4 函数称为Logistic映射，试从“蜘蛛网”图观察它取初值为产生的迭代序列的收敛性，将观察记录填人下表，若出现循环，请指出它的周期3.33.53.563.5683.63.84序列收敛情况T=2T=4T=8T=9混沌混沌解：当=3.3时，程序代码如下：f=inline(3.3*x*(1-x);x=;y=;x(1)=0.5;y(1)=0;x(2)=x(1);y(2)=f(x(1);for i=1:1000;x(1+2*i)=y(2

12、*i);x(2+2*i)=x(1+2*i);y(1+2*i)=x(1+2*i);y(2+2*i)=f(x(2+2*i);endplot (x,y,r);hold on;syms x;ezplot(x,0,1);ezplot(f(x),0,1);axis(0,1,0,1);hold off运行结果：当=3.5时，上述程序稍加修改，得：当=3.56时，得：当=3.568时，得：当=3.6时，得：当=3.84时，得：2.5 对于Martin迭代，取参数为其它的值会得到什么图形？参考下表（取自63页练习13）mmm-m-mm-mm/1000-mm/1000m/10000.5m/1000m-mm/100

13、m/10-10-m/10174解：取m=10000；迭代次数N=20000；在M-文件里面输入代码：function Martin(a,b,c,N)f=(x,y)(y-sign(x)*sqrt(abs(b*x-c);g=(x)(a-x);m=0;0;for n=1:N m(:,n+1)=f(m(1,n),m(2,n),g(m(1,n); end plot(m(1,:),m(2,:),kx); axis equal在命令窗口中执行Martin（10000，10000，10000，20000），得：执行Martin（-10000，-10000，10000，20000），得：执行Martin（-10

14、000，10，-10000，20000），得：执行Martin（10，10，0.5，20000），得：执行Martin（10，10000，-10000，20000），得：执行Martin（100，1000，-10，20000），得：执行Martin（-1000，17，4，20000），得：2.6 能否找到分式函数（其中是整数）,使它产生的迭代序列（迭代的初始值也是整数）收敛到(对于为整数的学号，请改为求)。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.提示：教材54页练习4的一些分析。若分式线性函数的迭代收敛到指定的数，则为的不动点，因此化简得：。若为整数，易见。取满

15、足这种条件的不同的以及迭代初值进行编。解：取m=10000；根据上述提示，取：a=e=1,b=10000,c=1,d=0.程序如下（初值为1200）：f=inline(x+10000)/(x2+1);x0=1200;for i=1:100;x0=f(x0);fprintf(%g,%gn,i,x0);end运行结果如下：1,0.007777772,9999.43,0.0002000184,100005,0.00026,100007,0.00028,100009,0.000210,1000011,0.000212,1000013,0.000214,1000015,0.000216,1000017,

16、0.000218,1000019,0.000220,1000021,0.000222,1000023,0.000224,1000025,0.000226,1000027,0.000228,1000029,0.000230,1000031,0.000232,1000033,0.000234,1000035,0.000236,1000037,0.000238,1000039,0.000240,1000041,0.000242,1000043,0.000244,1000045,0.000246,1000047,0.000248,1000049,0.000250,1000051,0.000252,10

17、00053,0.000254,1000055,0.000256,1000057,0.000258,1000059,0.000260,1000061,0.000262,1000063,0.000264,1000065,0.000266,1000067,0.000268,1000069,0.000270,1000071,0.000272,1000073,0.000274,1000075,0.000276,1000077,0.000278,1000079,0.000280,1000081,0.000282,1000083,0.000284,1000085,0.000286,1000087,0.000

18、288,1000089,0.000290,1000091,0.000292,1000093,0.000294,1000095,0.000296,1000097,0.000298,1000099,0.0002100,10000若初值取为1000，运行结果：1,0.0112,9998.83,0.0002000364,100005,0.00026,100007,0.00028,100009,0.000210,1000011,0.000212,1000013,0.000214,1000015,0.000216,1000017,0.000218,1000019,0.000220,1000021,0.00

19、0222,1000023,0.000224,1000025,0.000226,1000027,0.000228,1000029,0.000230,1000031,0.000232,1000033,0.000234,1000035,0.000236,1000037,0.000238,1000039,0.000240,1000041,0.000242,1000043,0.000244,1000045,0.000246,1000047,0.000248,1000049,0.000250,1000051,0.000252,1000053,0.000254,1000055,0.000256,100005

20、7,0.000258,1000059,0.000260,1000061,0.000262,1000063,0.000264,1000065,0.000266,1000067,0.000268,1000069,0.000270,1000071,0.000272,1000073,0.000274,1000075,0.000276,1000077,0.000278,1000079,0.000280,1000081,0.000282,1000083,0.000284,1000085,0.000286,1000087,0.000288,1000089,0.000290,1000091,0.000292,

21、1000093,0.000294,1000095,0.000296,1000097,0.000298,1000099,0.0002100,10000若初值取为-1，运行结果：1,4999.52,0.00060013,100004,0.00025,100006,0.00027,100008,0.00029,1000010,0.000211,1000012,0.000213,1000014,0.000215,1000016,0.000217,1000018,0.000219,1000020,0.000221,1000022,0.000223,1000024,0.000225,1000026,0.0

22、00227,1000028,0.000229,1000030,0.000231,1000032,0.000233,1000034,0.000235,1000036,0.000237,1000038,0.000239,1000040,0.000241,1000042,0.000243,1000044,0.000245,1000046,0.000247,1000048,0.000249,1000050,0.000251,1000052,0.000253,1000054,0.000255,1000056,0.000257,1000058,0.000259,1000060,0.000261,10000

23、62,0.000263,1000064,0.000265,1000066,0.000267,1000068,0.000269,1000070,0.000271,1000072,0.000273,1000074,0.000275,1000076,0.000277,1000078,0.000279,1000080,0.000281,1000082,0.000283,1000084,0.000285,1000086,0.000287,1000088,0.000289,1000090,0.000291,1000092,0.000293,1000094,0.000295,1000096,0.000297

24、,1000098,0.000299,10000100,0.0002 第三次练习教学要求：理解线性映射的思想，会用线性映射和特征值的思想方法解决诸如天气等实际问题。3.1 对，求出的通项. 程序：A=sym(4,2;1,3);P,D=eig(A)Q=inv(P)syms n; xn=P*(D.n)*Q*1;2 结果：P = 2, -1 1, 1D = 5, 0 0, 2Q = 1/3, 1/3 -1/3, 2/3xn =2*5n-2n 5n+2n3.2 对于练习1中的，求出的通项. 程序：A=sym(2/5,1/5;1/10,3/10); %没有sym下面的矩阵就会显示为小数P,D=eig(A)

25、Q=inv(P)xn=P*(D.n)*Q*1;2 结果：P = 2, -1 1, 1D = 1/2, 0 0, 1/5Q = 1/3, 1/3 -1/3, 2/3xn = 2*(1/2)n-(1/5)n (1/2)n+(1/5)n3.3 对随机给出的，观察数列.该数列有极限吗? A=4,2;1,3;a=;x=2*rand(2,1)-1;for i=1:20 a(i,1:2)=x; x=A*x; end for i=1:20 if a(i,1)=0 else t=a(i,2)/a(i,1); fprintf(%g,%gn,i,t); endend 结论：在迭代18次后，发现数列存在极限为0.53

26、.4 对120页中的例子，继续计算.观察及的极限是否存在. （120页练习9） 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;x0=1;2;3;4;x=A*x0;for i=1:1:100a=max(x);b=min(x);m=a*(abs(a)abs(b)+b*(abs(a) 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;P,D=eig(A)P = -0.3779 -0.8848 -0.0832 -0.39

27、08 -0.5367 0.3575 -0.2786 0.4777 -0.6473 0.2988 0.1092 -0.7442 -0.3874 -0.0015 0.9505 0.2555D = 7.2300 0 0 0 0 1.1352 0 0 0 0 -11.2213 0 0 0 0 -5.8439结论：A的绝对值最大特征值等于上面的的极限相等，为什么呢？还有，P的第三列也就是-11.2213对应的特征向量和上题求解到的y也有系数关系，两者都是-11.2213的特征向量。3.6 设，对问题2求出若干天之后的天气状态，并找出其特点（取4位有效数字）. （122页练习12） A2=3/4,1/2,

28、1/4;1/8,1/4,1/2;1/8,1/4,1/4;P=0.5;0.25;0.25;for i=1:1:20 P(:,i+1)=A2*P(:,i);endPP = Columns 1 through 14 0.5000 0.5625 0.5938 0.6035 0.6069 0.6081 0.6085 0.6086 0.6087 0.6087 0.6087 0.6087 0.6087 0.6087 0.2500 0.2500 0.2266 0.2207 0.2185 0.2178 0.2175 0.2174 0.2174 0.2174 0.2174 0.2174 0.2174 0.2174

29、 0.2500 0.1875 0.1797 0.1758 0.1746 0.1741 0.1740 0.1739 0.1739 0.1739 0.1739 0.1739 0.1739 0.1739 Columns 15 through 21 0.6087 0.6087 0.6087 0.6087 0.6087 0.6087 0.6087 0.2174 0.2174 0.2174 0.2174 0.2174 0.2174 0.21740.1739 0.1739 0.1739 0.1739 0.1739 0.1739 0.1739结论：9天后，天气状态趋于稳定P*=（0.6087,0.2174,0

30、.1739）T3.7 对于问题2，求出矩阵的特征值与特征向量，并将特征向量与上一题中的结论作对比. （122页练习14） P,D=eig(A2)P = -0.9094 -0.8069 0.3437 -0.3248 0.5116 -0.8133 -0.2598 0.2953 0.4695D = 1.0000 0 0 0 0.3415 0 0 0 -0.0915分析：事实上，q=k(-0.9094, -0.3248, -0.2598)T均为特征向量，而上题中P*的3个分量之和为1，可令k(-0.9094, -0.3248, -0.2598)T=1，得k=-0.6696.有q=(0.6087, 0.

31、2174, 0.1739),与P*一致。3.8 对问题1，设为的两个线性无关的特征向量，若，具体求出上述的，将表示成的线性组合，求的具体表达式，并求时的极限，与已知结论作比较. （123页练习16） A=3/4,7/18;1/4,11/18;P,D=eig(A);syms k pk;a=solve(u*P(1,1)+v*P(1,2)-1/2,u*P(2,1)+v*P(2,2)-1/2,u,v);pk=a.u*D(1,1).k*P(:,1)+a.v*D(2,2).k*P(:,2) pk = -5/46*(13/36)k+14/23 5/46*(13/36)k+9/23或者：p0=1/2;1/2;

32、P,D=eig(sym(A);B=inv(sym(P)*p0 B = 5/46 9/23syms kpk=B(1,1)*D(1,1).k*P(:,1)+B(2,1)*D(2,2).k*P(:,2) pk = -5/46*(13/36)k+14/23 5/46*(13/36)k+9/23 vpa(limit(pk,k,100),10) ans = .6086956522 .3913043478结论：和用练习12中用迭代的方法求得的结果是一样的。第四次练习教学要求：会利用软件求勾股数，并且能够分析勾股数之间的关系。会解简单的近似计算问题。4.1 求满足，的所有勾股数，能否类似于（11.8），把它们

33、用一个公式表示出来？程序：for b=1:998 a=sqrt(b+2)2-b2); if(a=floor(a) fprintf(a=%i,b=%i,c=%in,a,b,b+2) endend运行结果：a=4,b=3,c=5a=6,b=8,c=10a=8,b=15,c=17a=10,b=24,c=26a=12,b=35,c=37a=14,b=48,c=50a=16,b=63,c=65a=18,b=80,c=82a=20,b=99,c=101a=22,b=120,c=122a=24,b=143,c=145a=26,b=168,c=170a=28,b=195,c=197a=30,b=224,c=2

34、26a=32,b=255,c=257a=34,b=288,c=290a=36,b=323,c=325a=38,b=360,c=362a=40,b=399,c=401a=42,b=440,c=442a=44,b=483,c=485a=46,b=528,c=530a=48,b=575,c=577a=50,b=624,c=626a=52,b=675,c=677a=54,b=728,c=730a=56,b=783,c=785a=58,b=840,c=842a=60,b=899,c=901a=62,b=960,c=962勾股数，的解是： 以下是推导过程：由，有显然，从而是2的倍数.设，代入上式得到：因为

35、，从而.4.2 将上一题中改为,分别找出所有的勾股数.将它们与时的结果进行比较，然后用公式表达其结果。（1）时通项：a=8,b=6,c=10a=12,b=16,c=20a=16,b=30,c=34a=20,b=48,c=52a=24,b=70,c=74a=28,b=96,c=100a=32,b=126,c=130a=36,b=160,c=164a=40,b=198,c=202a=44,b=240,c=244a=48,b=286,c=290a=52,b=336,c=340a=56,b=390,c=394a=60,b=448,c=452a=64,b=510,c=514a=68,b=576,c=58

36、0a=72,b=646,c=650a=76,b=720,c=724a=80,b=798,c=802a=84,b=880,c=884a=88,b=966,c=970（2）5时通项： a=15,b=20,c=25a=25,b=60,c=65a=35,b=120,c=125a=45,b=200,c=205a=55,b=300,c=305a=65,b=420,c=425a=75,b=560,c=565a=85,b=720,c=725a=95,b=900,c=905（3）6时通项a=12,b=9,c=15a=18,b=24,c=30a=24,b=45,c=51a=30,b=72,c=78a=36,b=1

37、05,c=111a=42,b=144,c=150a=48,b=189,c=195a=54,b=240,c=246a=60,b=297,c=303a=66,b=360,c=366a=72,b=429,c=435a=78,b=504,c=510a=84,b=585,c=591a=90,b=672,c=678a=96,b=765,c=771a=102,b=864,c=870a=108,b=969,c=975（4）7时通项a=21,b=28,c=35a=35,b=84,c=91a=49,b=168,c=175a=63,b=280,c=287a=77,b=420,c=427a=91,b=588,c=595a=105,b=784,c=791综上：当c-b=k为奇数时，通项当c-b=k为偶数时，通项4.3 对，（），对哪些存在本原勾股数？（140页练习12）程序：for k=1:200 for b=1:999 a=sqrt(b+k)2-b2); if(a=floor(a)&gcd(gcd(a,b),(b+k)=1) fprintf(%i,k); break; end endend运行结果：1,2,8,9,18,25,32,49,50,72,81,98,121,128,162,169,200,4.4 设方程(11.15)的解构成数列,观