Removed_大连理工大学工科数学分析上机作业

大连理工大学概率上机作业第一次上机作业1.利用Matlab自带命令产生1000个均匀随机变量服从U(0,1)。

>> unifrnd(0,1,20,50)ans =Columns 1 through 100.8147 0.6557 0.4387 0.7513 0.3517 0.1622 0.1067 0.8530 0.7803 0.54700.9058 0.0357 0.3816 0.2551 0.8308 0.7943 0.9619 0.6221 0.3897 0.29630.1270 0.8491 0.7655 0.5060 0.5853 0.3112 0.0046 0.3510 0.2417 0.74470.9134 0.9340 0.7952 0.6991 0.5497 0.5285 0.7749 0.5132 0.4039 0.18900.6324 0.6787 0.1869 0.8909 0.9172 0.1656 0.8173 0.4018 0.0965 0.68680.0975 0.7577 0.4898 0.9593 0.2858 0.6020 0.8687 0.0760 0.1320 0.18350.2785 0.7431 0.4456 0.5472 0.7572 0.2630 0.0844 0.2399 0.9421 0.36850.5469 0.3922 0.6463 0.1386 0.7537 0.6541 0.3998 0.1233 0.9561 0.62560.9575 0.6555 0.7094 0.1493 0.3804 0.6892 0.2599 0.1839 0.5752 0.78020.9649 0.1712 0.7547 0.2575 0.5678 0.7482 0.8001 0.2400 0.0598 0.08110.1576 0.7060 0.2760 0.8407 0.0759 0.4505 0.4314 0.4173 0.2348 0.92940.9706 0.0318 0.6797 0.2543 0.0540 0.0838 0.9106 0.0497 0.3532 0.77570.9572 0.2769 0.6551 0.8143 0.5308 0.2290 0.1818 0.9027 0.8212 0.48680.4854 0.0462 0.1626 0.2435 0.7792 0.9133 0.2638 0.9448 0.0154 0.43590.8003 0.0971 0.1190 0.9293 0.9340 0.1524 0.1455 0.4909 0.0430 0.44680.1419 0.8235 0.4984 0.3500 0.1299 0.8258 0.1361 0.4893 0.1690 0.30630.4218 0.6948 0.9597 0.1966 0.5688 0.5383 0.8693 0.3377 0.6491 0.50850.9157 0.3171 0.3404 0.2511 0.4694 0.9961 0.5797 0.9001 0.7317 0.51080.7922 0.9502 0.5853 0.6160 0.0119 0.0782 0.5499 0.3692 0.6477 0.81760.9595 0.0344 0.2238 0.4733 0.3371 0.4427 0.1450 0.1112 0.4509 0.7948Columns 11 through 200.6443 0.3111 0.0855 0.0377 0.0305 0.0596 0.1734 0.9516 0.0326 0.25180.3786 0.9234 0.2625 0.8852 0.7441 0.6820 0.3909 0.9203 0.5612 0.29040.8116 0.4302 0.8010 0.9133 0.5000 0.0424 0.8314 0.0527 0.8819 0.61710.5328 0.1848 0.0292 0.7962 0.4799 0.0714 0.8034 0.7379 0.6692 0.26530.3507 0.9049 0.9289 0.0987 0.9047 0.5216 0.0605 0.2691 0.1904 0.82440.9390 0.9797 0.7303 0.2619 0.6099 0.0967 0.3993 0.42280.3689 0.98270.8759 0.4389 0.4886 0.3354 0.6177 0.8181 0.5269 0.5479 0.4607 0.73020.5502 0.1111 0.5785 0.6797 0.8594 0.8175 0.4168 0.9427 0.9816 0.3439 0.6225 0.2581 0.2373 0.1366 0.8055 0.7224 0.6569 0.4177 0.1564 0.5841 0.5870 0.4087 0.4588 0.7212 0.5767 0.1499 0.6280 0.9831 0.8555 0.1078 0.2077 0.5949 0.9631 0.1068 0.1829 0.6596 0.2920 0.3015 0.6448 0.9063 0.3012 0.2622 0.5468 0.6538 0.2399 0.5186 0.4317 0.7011 0.3763 0.8797 0.4709 0.6028 0.5211 0.4942 0.8865 0.9730 0.0155 0.6663 0.1909 0.8178 0.2305 0.7112 0.2316 0.7791 0.0287 0.6490 0.9841 0.5391 0.4283 0.2607 0.8443 0.2217 0.4889 0.7150 0.4899 0.8003 0.1672 0.6981 0.4820 0.5944 0.1948 0.1174 0.6241 0.9037 0.1679 0.4538 0.1062 0.6665 0.1206 0.0225 0.2259 0.2967 0.6791 0.8909 0.9787 0.4324 0.3724 0.1781 0.5895 0.4253 0.1707 0.3188 0.3955 0.3342 0.7127 0.8253 0.1981 0.1280 0.2262 0.3127 0.2277 0.4242 0.3674 0.6987 0.5005 0.0835 0.4897 0.9991 0.3846 0.1615 0.4357 0.5079 0.9880 0.1978 0.4711 0.1332 0.3395 0.1711 0.5830 0.1788Columns 21 through 300.4229 0.7788 0.2548 0.1759 0.6476 0.5822 0.4046 0.3477 0.8217 0.5144 0.0942 0.4235 0.2240 0.7218 0.6790 0.5407 0.4484 0.1500 0.4299 0.8843 0.5985 0.0908 0.6678 0.4735 0.6358 0.8699 0.3658 0.5861 0.8878 0.5880 0.4709 0.2665 0.8444 0.1527 0.9452 0.2648 0.7635 0.2621 0.3912 0.1548 0.6959 0.1537 0.3445 0.3411 0.2089 0.3181 0.6279 0.0445 0.7691 0.1999 0.6999 0.2810 0.7805 0.6074 0.7093 0.1192 0.7720 0.7549 0.3968 0.4070 0.6385 0.4401 0.6753 0.1917 0.2362 0.9398 0.9329 0.2428 0.8085 0.7487 0.0336 0.5271 0.0067 0.7384 0.1194 0.6456 0.9727 0.4424 0.7551 0.8256 0.0688 0.4574 0.6022 0.2428 0.6073 0.4795 0.1920 0.6878 0.3774 0.7900 0.3196 0.8754 0.3868 0.9174 0.4501 0.6393 0.1389 0.35920.2160 0.3185 0.5309 0.5181 0.9160 0.2691 0.4587 0.5447 0.6963 0.7363 0.7904 0.5341 0.6544 0.9436 0.0012 0.7655 0.6619 0.6473 0.0938 0.3947 0.9493 0.0900 0.4076 0.6377 0.4624 0.1887 0.7703 0.5439 0.5254 0.6834 0.3276 0.1117 0.8200 0.9577 0.4243 0.2875 0.3502 0.7210 0.5303 0.7040 0.6713 0.1363 0.7184 0.2407 0.4609 0.0911 0.6620 0.5225 0.8611 0.4423 0.4386 0.6787 0.9686 0.6761 0.7702 0.5762 0.4162 0.9937 0.4849 0.0196 0.8335 0.4952 0.5313 0.2891 0.3225 0.6834 0.8419 0.2187 0.3935 0.3309 0.7689 0.1897 0.3251 0.6718 0.7847 0.5466 0.8329 0.1058 0.6714 0.4243 0.1673 0.4950 0.1056 0.6951 0.4714 0.4257 0.2564 0.1097 0.7413 0.2703 0.8620 0.1476 0.6110 0.0680 0.0358 0.6444 0.6135 0.0636 0.5201 0.1971 0.9899 0.0550Columns 31 through 400.8507 0.7386 0.5523 0.1239 0.7378 0.5590 0.1781 0.8949 0.6311 0.6925 0.5606 0.5860 0.6299 0.4904 0.0634 0.8541 0.3596 0.0715 0.0899 0.5567 0.9296 0.2467 0.0320 0.8530 0.8604 0.3479 0.0567 0.2425 0.0809 0.3965 0.6967 0.6664 0.6147 0.8739 0.9344 0.4460 0.5219 0.0538 0.7772 0.0616 0.5828 0.0835 0.3624 0.2703 0.9844 0.0542 0.3358 0.4417 0.9051 0.78020.8154 0.6260 0.0495 0.2085 0.8589 0.1771 0.1757 0.0133 0.5338 0.33760.8790 0.6609 0.4896 0.5650 0.7856 0.6628 0.2089 0.8972 0.1092 0.60790.9889 0.7298 0.1925 0.6403 0.5134 0.3308 0.9052 0.1967 0.8258 0.74130.0005 0.8908 0.1231 0.4170 0.1776 0.8985 0.6754 0.0934 0.3381 0.10480.8654 0.9823 0.2055 0.2060 0.3986 0.1182 0.4685 0.3074 0.2940 0.12790.6126 0.7690 0.1465 0.9479 0.1339 0.9884 0.9121 0.45610.7463 0.54950.9900 0.5814 0.1891 0.0821 0.0309 0.5400 0.1040 0.1017 0.0103 0.48520.5277 0.9283 0.0427 0.1057 0.9391 0.7069 0.7455 0.9954 0.0484 0.89050.4795 0.5801 0.6352 0.1420 0.3013 0.9995 0.7363 0.3321 0.6679 0.79900.8013 0.0170 0.2819 0.1665 0.2955 0.2878 0.5619 0.2973 0.6035 0.73430.2278 0.1209 0.5386 0.6210 0.3329 0.4145 0.1842 0.0620 0.5261 0.05130.4981 0.8627 0.6952 0.5737 0.4671 0.4648 0.5972 0.2982 0.7297 0.07290.9009 0.4843 0.4991 0.0521 0.6482 0.7640 0.2999 0.0464 0.7073 0.08850.5747 0.8449 0.5358 0.9312 0.0252 0.8182 0.1341 0.5054 0.7814 0.79840.8452 0.2094 0.4452 0.7287 0.8422 0.1002 0.2126 0.7614 0.2880 0.9430Columns 41 through 500.6837 0.7894 0.1123 0.6733 0.0986 0.9879 0.5975 0.7593 0.8092 0.75190.1321 0.3677 0.7844 0.4296 0.1420 0.1704 0.3353 0.7406 0.7486 0.22870.7227 0.2060 0.2916 0.4517 0.1683 0.2578 0.2992 0.7437 0.1202 0.06420.1104 0.0867 0.6035 0.6099 0.1962 0.3968 0.4526 0.1059 0.5250 0.76730.1175 0.7719 0.9644 0.0594 0.3175 0.0740 0.4226 0.6816 0.3258 0.67120.6407 0.2057 0.4325 0.3158 0.3164 0.6841 0.3596 0.4633 0.5464 0.71520.3288 0.3883 0.6948 0.7727 0.2176 0.4024 0.5583 0.2122 0.3989 0.64210.6538 0.5518 0.7581 0.6964 0.2510 0.9828 0.7425 0.0985 0.4151 0.41900.7491 0.2290 0.4326 0.1253 0.8929 0.4022 0.4243 0.8236 0.1807 0.39080.5832 0.6419 0.6555 0.1302 0.7032 0.6207 0.4294 0.1750 0.2554 0.81610.7400 0.4845 0.1098 0.0924 0.5557 0.1544 0.1249 0.1636 0.0205 0.31740.2348 0.1518 0.9338 0.0078 0.1844 0.3813 0.0244 0.6660 0.9237 0.81450.7350 0.7819 0.1875 0.4231 0.2120 0.1611 0.2902 0.8944 0.6537 0.78910.9706 0.1006 0.2662 0.6556 0.0773 0.7581 0.3175 0.5166 0.9326 0.85230.8669 0.2941 0.7978 0.7229 0.9138 0.8711 0.6537 0.7027 0.1635 0.50560.0862 0.2374 0.4876 0.5312 0.7067 0.3508 0.9569 0.1536 0.9211 0.63570.3664 0.5309 0.7690 0.1088 0.5578 0.6855 0.9357 0.9535 0.7947 0.95090.3692 0.0915 0.3960 0.6318 0.3134 0.2941 0.4579 0.5409 0.5774 0.44400.6850 0.4053 0.2729 0.1265 0.1662 0.5306 0.2405 0.6797 0.4400 0.06000.5979 0.1048 0.0372 0.1343 0.6225 0.8324 0.7639 0.0366 0.2576 0.8667 2.参考课本综合例题2.5.4和2.5.5中的方法，模拟产生1000个随机变量，使其服从参数为2的指数分布，进而计算这1000个随机数的均值和方差。

大连理工数学分析试题及解答Word版2000年大连理工大学硕士生入学考试试题——数学分析一、从以下的第一到第八题中选取6题解答，每题10分1．证明：1()f x x=于区间0(,1)δ（其中001δ<<）一致连续，但是于(0,1)内不一致连续证明：01212(1)0,()[1]2(2)1||()|()()|f x x x f x f x δδδδεδδε<=<-<而由于在，内连续，从而一致连续，第一个命题成立利用定义，取，不存在为定值使得从而不难利用反证法得到第二个命题成立2．证明：若()[,]f x a b 于单调，则()[,]f x a b Riemann 于内可积证明：1101111111111()...[,],max 0(max {()}min {()})(()())(max{()()})(max{()()})i ii in i i i i i nnni i i i i i x x x x x x i ni i i i i nf x a x x x b a b x x f x f x f x f x f x f x f x f x λλλλλλ---≤≤--≤≤≤≤≤≤==-≤≤?=<<<==-=→-=-<--∑∑不妨设单调递增，且：是的一个划分,必然存在一个划分，使得11111(max{()()})lim (max {()}min {()})0i ii i i nni x x x x x x i f x f x f x f x λ---≤≤?≤≤≤≤=→--=∑（由于递增，使用二分法的思想，可以使得小于任何数）所以，，所以可积3．证明：Dirichlet 函数：0,()1,()x f x px q q ??=?=??为无理数有理数在所有无理点连续，在有理点间断，证明：0001000000()010[]1min{||}1(,),|()|()0{,{}},()n N i Zi i x f x iN x n x x x f x Nx f x x y y f x εδδεεεε+≤≤∈=?>=+=-∈-+≤<≠∈为无理数，对于，,取，显然这样的存在当所以，在无理点连续为有理数，。

软1414 叶秀云201492015 上机报告上机作业一Trial>> A=round(5*rand(5))B=round(5*rand(5))C=round(5*rand(5))b=round(5*rand(5,1))A+BA-BA*B+B*Ainv(A)*binv(A)rank(A)det(B)inv(B)rank(B)inv(A*B)rank(A*B)(B')*(A')inv(A*B)inv(B)*inv(A)inv(A)*C*inv(B)A =4 0 1 1 35 1 5 2 01 3 5 5 45 5 2 4 53 545 3B =4 4 4 2 24 0 3 2 22 1 2 4 33 0 54 41 0 0 1 4C =1 2 4 5 43 5 1 3 13 2 3 1 41 3 3 1 11 1 4 1 5b =21132ans =8 4 5 3 59 1 8 4 23 4 7 9 78 5 7 8 94 5 4 6 7 ans =0 -4 -3 -1 11 12 0 -2-1 2 3 1 12 5 -3 0 12 5 4 4 -1 ans =80 53 79 69 7175 54 74 77 7589 51 85 97 102110 77 111 113 12379 41 79 80 80ans =0.4754-0.3197-0.59840.9672-0.0902ans =0.3197 -0.0164 -0.2541 -0.2049 0.3607-0.8443 0.2869 0.1967 0.8361 -0.8115 -0.7213 0.4344 0.4836 0.6803 -1.05741.4262 -0.6885 -0.6721 -1.60662.1475-0.3279 0.1066 0.4016 0.5820 -0.8443 ans =5ans =418.0000ans =-0.0144 0.4354 0.0574 -0.2727 0.01910.2321 -0.2057 0.0718 -0.0909 0.02390.0718 -0.1770 -0.2871 0.3636 -0.0957-0.1100 0.0048 0.4402 -0.0909 -0.18660.0311 -0.1100 -0.1244 0.0909 0.2919 ans =5ans =-0.8088 0.3399 0.3081 0.8553 -1.02110.0586 0.0335 0.0060 -0.0108 -0.04070.9295 -0.4372 -0.4747 -0.9979 1.3348-0.4252 0.2371 0.2279 0.3635 -0.54670.2265 -0.1176 -0.0336 -0.1592 0.1809 ans =5ans =24 40 45 61 5817 25 9 22 1623 43 48 59 6019 40 52 49 5527 35 59 62 60ans =-0.8088 0.3399 0.3081 0.8553 -1.02110.0586 0.0335 0.0060 -0.0108 -0.04070.9295 -0.4372 -0.4747 -0.9979 1.3348-0.4252 0.2371 0.2279 0.3635 -0.54670.2265 -0.1176 -0.0336 -0.1592 0.1809 ans =-0.8088 0.3399 0.3081 0.8553 -1.02110.0586 0.0335 0.0060 -0.0108 -0.04070.9295 -0.4372 -0.4747 -0.9979 1.3348-0.4252 0.2371 0.2279 0.3635 -0.54670.2265 -0.1176 -0.0336 -0.1592 0.1809 ans =-0.0497 -0.5353 -0.0060 0.6289 0.12910.3583 1.0632 0.4029 -1.7332 -0.66080.4518 1.2872 0.8731 -2.1207 -0.6844-0.7394 -2.4091 -1.6491 4.0507 1.62510.2658 0.7208 0.5187 -1.1788 -0.5129 Trial>>上机作业二Trial>> A=rand(4)B=rand(4)C=rand(4)D=rand(4)Z=[A,B;C,D]det(Z)det(A*D-C*B)A=diag([rand rand rand rand])C=diag([rand rand rand rand])Z=[A,B;C,D]det(Z)det(A*D-C*B)A =0.9027 0.3377 0.7803 0.09650.9448 0.9001 0.3897 0.13200.4909 0.3692 0.2417 0.94210.4893 0.1112 0.4039 0.9561B =0.5752 0.8212 0.6491 0.54700.0598 0.0154 0.7317 0.29630.2348 0.0430 0.6477 0.74470.3532 0.1690 0.4509 0.1890C =0.6868 0.7802 0.4868 0.50850.1835 0.0811 0.4359 0.51080.3685 0.9294 0.4468 0.81760.6256 0.7757 0.3063 0.7948D =0.6443 0.3507 0.6225 0.47090.3786 0.9390 0.5870 0.23050.8116 0.8759 0.2077 0.84430.5328 0.5502 0.3012 0.1948Z =0.9027 0.3377 0.7803 0.0965 0.5752 0.8212 0.6491 0.54700.9448 0.9001 0.3897 0.1320 0.0598 0.0154 0.7317 0.29630.4909 0.3692 0.2417 0.9421 0.2348 0.0430 0.6477 0.74470.4893 0.1112 0.4039 0.9561 0.3532 0.1690 0.4509 0.18900.6868 0.7802 0.4868 0.5085 0.6443 0.3507 0.6225 0.47090.1835 0.0811 0.4359 0.5108 0.3786 0.9390 0.5870 0.23050.3685 0.9294 0.4468 0.8176 0.8116 0.8759 0.2077 0.84430.6256 0.7757 0.3063 0.7948 0.5328 0.5502 0.3012 0.1948 ans =-0.0232ans =0.0161A =0.2259 0 0 00 0.1707 0 00 0 0.2277 00 0 0 0.4357C =0.3111 0 0 00 0.9234 0 00 0 0.4302 00 0 0 0.1848Z =0.2259 0 0 0 0.5752 0.8212 0.6491 0.54700 0.1707 0 0 0.0598 0.0154 0.7317 0.29630 0 0.2277 0 0.2348 0.0430 0.6477 0.74470 0 0 0.4357 0.3532 0.1690 0.4509 0.18900.3111 0 0 0 0.6443 0.3507 0.6225 0.47090 0.9234 0 0 0.3786 0.9390 0.5870 0.23050 0 0.4302 0 0.8116 0.8759 0.2077 0.84430 0 0 0.1848 0.5328 0.5502 0.3012 0.1948 ans =7.3868e-04ans =7.3868e-04Trial>>上机作业三N=201492015;a=15;b=49;c=01;d=41;e=21;f=95;g=45;Trial>> h=90;Trial>> A=[a,b,c,d,3,4;1,2,3,4,4,3;12,15,22,17,5,7;e,f,g,h,8,0]; Trial>> B=rref(A)B =1.0000 0 0 0 0.4130 0.95680 1.0000 0 0 -1.7984 -1.49040 0 1.0000 0 -0.3796 -0.37590 0 0 1.0000 2.0806 1.5380N=201492015;a=15;b=49;c=01;d=41;e=21;f=95;g=45;Trial>> h=90;Trial>> A=[a,b,c,d,3,4;1,2,3,4,4,3;12,15,22,17,5,7;e,f,g,h,8,0]; Trial>> B=rref(A)B =1.0000 0 0 0 0.4130 0.95680 1.0000 0 0 -1.7984 -1.49040 0 1.0000 0 -0.3796 -0.37590 0 0 1.0000 2.0806 1.5380上机作业四Trial>> b1=[1,1.9,f,c];Trial>> b2=[1,1.8,f,c];Trial>> A1=[a,b,c,d;0.5,1,1.5,2;12,15,22,17;e,f,g,h];Trial>> A2=[a,b,c,d;0.3,0.6,0.9,1.2;12,15,22,17;e,f,g,h];Trial>> A3=[a,b,c,d;0.1,0.2,0.3,0.4;12,15,22,17;e,f,g,h];Trial>> A4=[a,b,c,d;0.05,0.1,0.15,0.2;12,15,22,17;e,f,g,h];Trial>> x1=A1/b1 x1 =0.02700.01630.23780.5057 Trial>> x2=A2/b1 x2 =0.02700.00980.23780.5057 Trial>> x3=A4/b1 x3 =0.02700.00160.23780.5057 Trial>> x4=A4/b1 x4 =0.02700.00160.23780.5057 Trial>> x5=A1/b2 x5 =0.02650.01630.23760.5046Trial>> x6=A2/b2x6 =0.02650.00980.23760.5046Trial>> x7=A3/b2x7 =0.02650.00330.23760.5046Trial>> x8=A4/b2x8 =0.02650.00160.23760.5046Trial>>上机作业五a1=rand(5,1)a2=rand(5,1)a3=rand(5,1)a4=rand(5,1)a5=rand(5,1)A=[a1,a2,a3,a4,a5]orth(A)a1 =0.90490.97970.43890.11110.2581 a2 =0.40870.59490.26220.60280.7112 a3 =0.22170.11740.29670.31880.4242 a4 =0.50790.08550.26250.80100.0292 a5 =0.92890.73030.48860.57850.2373A =0.9049 0.4087 0.2217 0.5079 0.92890.9797 0.5949 0.1174 0.0855 0.73030.4389 0.2622 0.2967 0.2625 0.48860.1111 0.6028 0.3188 0.8010 0.57850.2581 0.7112 0.4242 0.0292 0.2373 ans =-0.5932 -0.1881 -0.4330 0.1909 -0.6235 -0.5319 -0.5286 0.1934 -0.5094 0.3752 -0.3288 0.0079 -0.0670 0.7395 0.5835 -0.4137 0.8042 -0.1828 -0.3450 0.1723 -0.2931 0.1960 0.8586 0.1953 -0.3167Trial>>上机作业六Trial>> A=rand(5)eig(A)[d,v]=eig(A)x=rand(5,1)eig(x*x')A =0.4588 0.4889 0.9880 0.0987 0.72120.9631 0.6241 0.0377 0.2619 0.10680.5468 0.6791 0.8852 0.3354 0.65380.5211 0.3955 0.9133 0.6797 0.49420.2316 0.3674 0.7962 0.1366 0.7791 ans =2.6238 + 0.0000i0.0391 + 0.2666i0.0391 - 0.2666i0.2420 + 0.0000i0.4829 + 0.0000id =-0.4582 + 0.0000i -0.4322 + 0.1366i -0.4322 - 0.1366i 0.1428 + 0.0000i 0.2020 + 0.0000i-0.3197 + 0.0000i 0.7401 + 0.0000i 0.7401 + 0.0000i -0.6192 + 0.0000i 0.3539 + 0.0000i-0.5143 + 0.0000i -0.0341 - 0.3116i -0.0341 + 0.3116i -0.2603 + 0.0000i -0.0520 + 0.0000i-0.5266 + 0.0000i 0.0473 + 0.2327i 0.0473 - 0.2327i 0.1237 + 0.0000i -0.9110 + 0.0000i-0.3821 + 0.0000i -0.2604 + 0.1558i -0.2604 - 0.1558i 0.7164 + 0.0000i -0.0373 + 0.0000iv =2.6238 + 0.0000i 0.0000 + 0.0000i 0.0000 + 0.0000i 0.0000 + 0.0000i 0.0000 + 0.0000i0.0000 + 0.0000i 0.0391 + 0.2666i 0.0000 + 0.0000i 0.0000 + 0.0000i 0.0000 + 0.0000i0.0000 + 0.0000i 0.0000 + 0.0000i 0.0391 - 0.2666i 0.0000 + 0.0000i 0.0000 + 0.0000i0.0000 + 0.0000i 0.0000 + 0.0000i 0.0000 + 0.0000i 0.2420 + 0.0000i 0.0000 + 0.0000i0.0000 + 0.0000i 0.0000 + 0.0000i 0.0000 + 0.0000i 0.0000 + 0.0000i 0.4829 + 0.0000i上机作业七A=[1,3/2,0;3/2,-1,1;0,1,1]rref(A)eig(A)B=[1,0,2;0,-1,-2;2,-2,0]rref(B)eig(B)A =1.0000 1.5000 01.5000 -1.0000 1.00000 1.0000 1.0000ans =1 0 00 1 00 0 1ans =-2.06161.00002.0616B =1 0 20 -1 -22 -2 0ans =1 0 20 1 20 0 0ans =-3.0000-0.00003.0000Trial>>上机作业八Trial>> A=[0.7,0.2,0.1;0.2,0.7,0.1;0.1,0.1,0.8]P0=[15;9;6]A =0.7000 0.2000 0.10000.2000 0.7000 0.10000.1000 0.1000 0.8000 P0 =1596Trial>> A*P0ans =12.90009.90007.2000Trial>> A*A*P0ans =11.730010.23008.0400Trial>> A*A*A*A*A*P0ans =10.429910.24249.3277Trial>>。

矩阵与数值分析上机作业学校：大连理工大学学院:班级: 姓名:学号：授课老师:注：编程语言Matlab1.琴虑计算给定働量的葩址输入向量広』(巾斑…宀产输出||工||“ ||工|怙㈣心请编制一牛通用程序，并用你編制的程序计算如下询量的范数：对网1加，wm甚至更大的“计算其范数，你会发现什幺结粟？你能否修改你的程序使得计算绪果相时赫■确呢？程序:Norm.m函数fun cti on s=Norm(x,m)%求向量x的范数%mx 1,2,inf 分别表示1,2，无穷范数n=len gth(x);s=0;switch mcase 1 %1-范数for i=1:ns=s+abs(x(i));endcase 2 %2-范数for i=1:ns=s+x(if2;ends=sqrt(s);case inf %无穷- 范数s=max(abs(x));end计算向量 x， y 的范数Test1.mclear all ;clc;n1=10;n2=100;n3=1000;x1=1./[1:n1]';x2=1./[1:n2]';x3=1./[1:n3]'; y1=[1:n1]';y2=[1:n2]';y3=[1:n3]';disp( 'n=10 时' );disp( 'x 的1-范数:' );disp(Norm(x1,1));disp( 'x 的无穷-范数:' );disp(Norm(x1,inf));disp( 'y 的2- 范数:' );disp(Norm(y1,2)); disp( 'y 的无穷- 范数:' );disp(Norm(y1,inf)); disp( 'n=100 时' );disp( 'x 的1- 范数:' );disp(Norm(x2,1)); disp( 'x 的2- 范数:' );disp(Norm(x2,2)); disp( 'x 的无穷- 范数:' );disp(Norm(x2,inf)); disp( 'y 的1- 范数:' );disp(Norm(y2,1)); disp( 'y 的2- 范数:' );disp(Norm(y2,2)); disp( 'y 的无穷- 范数:' );disp(Norm(y2,inf)); disp( 'n=1000 时' );disp( 'x 的1- 范数:' );disp(Norm(x3,1)); disp( 'x 的2- 范数:' );disp(Norm(x3,2)); disp( 'x 的无穷- 范数:' );disp(Norm(x3,inf)); disp( 'y 的1- 范数:' );disp(Norm(y3,1)); disp( 'y 的2- 范数:' );disp(Norm(y3,2)); disp( 'y 的无穷- 范数:' );disp(Norm(y3,inf));运行结果：n=10 时x 的1-范数29290 ; x 的2-范数：1.2449 ; x 的无穷-y 的1-范数:55 ; y 的2-范数:19.6214 ; y 的无穷 n=100 时x 的1-范数:5.1874 ; x 的2-范数：1.2787 ; x 的无穷 的 2-范数:581.6786 ； y 的无穷 -范数:100 n=1000 时 x 的1-范数74855 ; x 的2-范数：1.2822 ; x 的无穷-范数:1y 的 1-范数:500500 ; y 的 2-范数：1.8271e+004 ; y 的无穷-范数:10002. 耆虑砂== 呼^其中定51/(0)=此时几期是连绽函戟.用此公式计算 当工“―1旷巾U)-缪时的函数值*風出图像.另一方面*哮虑下面算法：d 1 + j1/(/ = 1 tbfjj1/=1仙y = liid/(d — 1(end if用此算法计％ € [-10-0 io_is]时的圉数血 画出图像.比校一下岌生了什么？程序Test2.mclear all ;clc;n=100; %区间h=2*10A (-15)/n; %步长范数:1 -范数:10-范数:1y 的 1- 范数 :5050 ；x=-10A(-15):h:10A(-15);%第一种原函数f1=zeros(1, n+1);for k=1:n+1if x(k)~=0f1(k)=log(1+x(k))/x(k);elsef1(k)=1;end endsubplot(2,1,1);plot(x,f1, '-r' );axis([-10A(-15),10A(-15),-1,2]); legend( ' 原图' );%第二种算法f2=zeros(1,n+1);for k=1:n+1d=1+x(k);if (d~=1) f2(k)=log(d)/(d-1);elsef2(k)=1;endendsubplot(2,1,2);plot(x,f2, '-r' );axis([-10A(-15),10A(-15),-1,2]);legend( ' 第二种算法' );运行结果：農IQ显然第二种算法结果不准确，是因为计算机中的舍入误差造成的，当X [ 1015,1015]时，d 1 x,计算机进行舍入造成d恒等于1,结果函数值恒为1。

-03cos 2lnlim 0=+=®xx （10分）四、解：(1)0)cos )((lim 00sin )(lim 00=-¢=÷øöçèæ-=®®x x g x x x g a x x （4分）(2)200sin )(lim )0()(lim )0(x xx g x f x f f x x-=-=¢®® =12)0(2sin )(lim 2cos )(lim 00=¢¢=+¢¢=-¢®®g x x g x x x g x x∴ ïîïíì=¹---¢=¢时时010,)sin )(()cos )(()(2x x x x x g x x g x x f （8分） （3）200)sin )(()cos )((lim )(lim x x x g x x g x x f x x ---¢=¢®® =xx x g x x g x x x g x 2)cos )(()sin )((cos )(lim 0-¢-+¢¢+-¢® =)0(12)0(f g ¢==¢¢，因此)(x f ¢在(-∞，+∞+∞))连续。

连续。

（10分）五、解五、解:: 设x x x f ln)(=，由2ln 1)('xxx f -=，可知,当e x >时)(x f 单调减少单调减少 （5分）若e a b >>，则有b b a a ln ln >，推出a b b a ln ln >，即有a b b a > 2011201220122011> （10分）分）所以六、解：2)()()(x x f x f x x x f -¢=¢÷øöçèæ（4分）分） 令)()()(x f x f x x g -¢=，)()(x f x x g ¢¢=¢，令0)(=¢x g ，得0=x （唯一驻点），当0<x 时，0)(<¢x g ，当0>x 时，0)(>¢x g ，故)0(g 为最小值，故0)0()0()(>-=³f g x g ，∴0)(>¢÷øöçèæx x f ，即x x f )(单调增加。

2011级《高等数学》,《工科数学分析基础》,《微积分》A 卷参考答案一、1. ()0)2()1(212=---+-z y x ，122121--=-=-z y x ；2. 3),2,2,1( ；3. 1-e ；4. 21(1)y y xy -+； 12(9ln36)dx dy ++；5 . 392,3zxy zx --二、1. A2. C3. C4. D5. B三、高等数学》和《工科数学分析基础》解：特征方程2320r r -+=，特征根121,2r r ==，212()x x Y x c e c e =+（4分） ()()xx m f x P x e xe λ==,所以(),1,,1x x m P x x m e e λλ====*()()()k x x m y x x Q x e x ax b e λ=⋅=+代入微分方程解得1,12a b =-=-（8分）所以，通解：*22121()()()()2x x x y x Y x y x c e c e x x e =+=++--。

（10分）《微积分》解： 由奇偶性有40Dxydxdy =⎰⎰，由轮换对称性有22DDx dxdy y dxdy =⎰⎰⎰⎰（6分）原式=224()Dx y dxdy +=⎰⎰212042d r rdr πθπ⋅=⎰⎰（10分）四、解： 设1D 为曲线22x y y =--和y 轴围成的区域，有11022sin 22sin DD D D ydxdy ydxdy ydxdy dx ydy d r rdr πθπθθ-+=-=-⋅⎰⎰⎰⎰⎰⎰⎰⎰⎰⎰（8分）＝4284sin 432d πππθθ-=-⎰。

（10分） 五、解：由奇偶性⎰⎰⎰ΩV x d =⎰⎰⎰Ω=0d V y ，（4分） 2d )(d d d d )(d 022010:221022πθπ=+=++=⎰⎰⎰⎰⎰⎰≤+⋅z zy x D r r z r z y x z y x z I z (截面法)（或）2d )(d d d )(d 121201221:22222πθσπ=+=++=⎰⎰⎰⎰⎰⎰+≤+⋅ryx y x D z z r r r z z y x xy (投影法）（10分）六、解：由题意知，仅需求函数在闭区域上的最大值和最小值既可。

大连理工大学矩阵与数值分析上机作业课程名称：矩阵与数值分析研究生姓名：交作业日时间：2016 年12 月20日1.1程序：Clear all;n=input('请输入向量的长度n：')for i=1:n;v(i)=1/i;endY1=norm(v,1)Y2=norm(v,2)Y3=norm(v,inf)1.2结果n=10 Y1 =2.9290Y2 =1.2449Y3 =1n=100 Y1 =5.1874Y2 =1.2787Y3 =1n=1000 Y1 =7.4855Y2 =1.2822Y3 =1N=10000 Y1 =9.7876Y2 =1.2825Y3 =11.3 分析一范数逐渐递增，随着n的增加，范数的增加速度减小；二范数随着n的增加，逐渐趋于定值，无群范数都是1.2.1程序clear all;x(1)=-10^-15;dx=10^-18;L=2*10^3;for i=1:Ly1(i)=log(1+x(i))/x(i); d=1+x(i);if d == 1y2(i)=1;elsey2(i)=log(d)/(d-1);endx(i+1)=x(i)+dx;endx=x(1:length(x)-1);plot(x,y1,'r');hold onplot(x,y2);2.2 结果2.3 分析红色的曲线代表未考虑题中算法时的情况，如果考虑题中的算法则数值大小始终为1，这主要是由于大数加小数的原因。

第3题3.1 程序clear all;A=[1 -18 144 -672 2016 -4032 5376 -4608 2304 -512];x=1.95:0.005:2.05;for i=1:length(x);y1(i)=f(A,x(i));y2(i)=(x(i)-2)^9;endfigure(3);plot(x,y1);hold on;plot(x,y2,'r');F.m文件function y=f(A,x) y=A(1);for i=2:length(A); y=x*y+A(i); end;3.2 结果第4题4.1 程序clear all;n=input('请输入向量的长度n：')A=2*eye(n)-tril(ones(n,n),0);for i=1:nA(i,n)=1;endn=length(A);U=A;e=eye(n);for i=1:n-1[max_data,max_index]=max(abs(U(i:n,i))); e0=eye(n);max_index=max_index+i-1;U=e0*U;e1=eye(n);for j=i+1:ne1(j,i)=-U(j,i)/U(i,i);endU=e1*U;P{i}=e0;%把变换矩阵存到P中L{i}=e1;e=e1*e0*e;endfor k=1:n-2Ldot{k}=L{k};for i=k+1:n-1Ldot{k}=P{i}*Ldot{k}*P{i};endendLdot{n-1}=L{n-1};LL=eye(n);PP=eye(n);for i=1:n-1PP=P{i}*PP;LL=Ldot{i}*LL;endb=ones(n,2);b=e*b; %解方程x=zeros(n,1);x(n)=b(n)/U(n,n);for i=n-1:-1:1x(i)=(b(i)-U(i,:)*x)/U(i,i);endX=U^-1*e^-1*eye(n);%计算逆矩阵AN=X';result2{n-4,1}=AN;result1{n-4,1}=x;fprintf('%d:\n',n)fprintf('%d ',AN);4.2 结果n=51.0625 -0.875 -0.75 -0.5 -0.06250.0625 1.125 -0.75 -0.5 -0.06250.0625 0.125 1.25 -0.5 -0.06250.0625 0.125 0.25 1.5 -0.0625-0.0625 -0.125 -0.25 -0.5 0.0625n=101.0625 -0.875 -0.75 -0.5 -0.0625 1.0625 -0.875 -0.75 -0.5 -0.0625 0.0625 1.125 -0.75 -0.5 -0.0625 0.0625 1.125 -0.75 -0.5 -0.0625 0.0625 0.125 1.25 -0.5 -0.0625 0.0625 0.125 1.25 -0.5 -0.0625 0.0625 0.125 0.25 1.5 -0.0625 0.0625 0.125 0.25 1.5 -0.0625 -0.0625 -0.125 -0.25 -0.5 0.0625 -0.0625 -0.125 -0.25 -0.5 0.0625 1.0625 -0.875 -0.75 -0.5 -0.0625 1.0625 -0.875 -0.75 -0.5 -0.0625 0.0625 1.125 -0.75 -0.5 -0.0625 0.0625 1.125 -0.75 -0.5 -0.0625 0.0625 0.125 1.25 -0.5 -0.0625 0.0625 0.125 1.25 -0.5 -0.0625 0.0625 0.125 0.25 1.5 -0.0625 0.0625 0.125 0.25 1.5 -0.0625 -0.0625 -0.125 -0.25 -0.5 0.0625 -0.0625 -0.125 -0.25 -0.5 0.0625同样的方法可以算出n=20,n=30时的结果，这里就不罗列了。

2000年大连理工大学硕士生入学考试试题——数学分析一、从以下的第一到第八题中选取6题解答，每题10分1． 证明：1()f x x=于区间0(,1)δ（其中001δ<<）一致连续，但是于(0,1)内不一致连续 证明：01212(1)0,()[1]2(2)1||()|()()|f x x x f x f x δδδδεδδε<=<⇒=-<-<而由于在，内连续，从而一致连续，第一个命题成立利用定义，取，不存在为定值使得从而不难利用反证法得到第二个命题成立2． 证明：若()[,]f x a b 于单调，则()[,]f x a b Riemann 于内可积证明：1101111111111()...[,],max 0(max {()}min {()})(()())(max{()()})(max{()()})i ii in i i i i i nnni i i i i i x x x x x x i ni i i i i nf x a x x x b a b x x f x f x f x f x f x f x f x f x λλλλλλ---≤≤--≤≤≤≤≤≤==-≤≤∆=<<<==-=→-=-<--∑∑不妨设单调递增，且：是的一个划分,必然存在一个划分，使得11111(max{()()})lim (max {()}min {()})0i ii ii i i nni x x x x x x i f x f x f x f x λ---≤≤∆≤≤≤≤=→--=∑（由于递增，使用二分法的思想，可以使得小于任何数）所以，，所以可积3． 证明：Dirichlet 函数：0,()1,()x f x px q q ⎧⎪=⎨=⎪⎩为无理数有理数在所有无理点连续，在有理点间断，证明：0001000000()010[]1min{||}1(,),|()|()0{,{}},()n N i Zi i x f x iN x n x x x f x Nx f x x y y f x εδδεεεε+≤≤∈=∀>=+=-∈-+≤<≠∈为无理数，对于，,取，显然这样的存在当所以，在无理点连续为有理数，。

共享知识分享快乐大连理工大学矩阵与数值分析上机作业课程名称：矩阵与数值分析研究生姓名：12 交作业日时间：日20 月年2016卑微如蝼蚁、坚强似大象．共享知识分享快乐第1题1.1程序：Clear ;all n=input('请输入向量的长度n：') for i=1:n;v(i)=1/i;endY1=norm(v,1)Y2=norm(v,2)Y3=norm(v,inf)1.2结果n=10 Y1 =2.9290Y2 =1.2449Y3 =1n=100 Y1 =5.1874Y2 =1.2787Y3 =1n=1000 Y1 =7.4855Y2 =1.2822Y3 =1N=10000 Y1 =9.7876Y2 =1.2825Y3 =11.3 分析一范数逐渐递增，随着n的增加，范数的增加速度减小；二范数随着n的增加，逐渐趋于定值，无群范数都是1.卑微如蝼蚁、坚强似大象．共享知识分享快乐第2题2.1程序;clear all x(1)=-10^-15;dx=10^-18;L=2*10^3; i=1:L fory1(i)=log(1+x(i))/x(i); d=1+x(i); d == 1ify2(i)=1;elsey2(i)=log(d)/(d-1);endx(i+1)=x(i)+dx;end x=x(1:length(x)-1););'r'plot(x,y1,on holdplot(x,y2);卑微如蝼蚁、坚强似大象．共享知识分享快乐2.2 结果2.3 分析红色的曲线代表未考虑题中算法时的情况，如果考虑题中的算法则数值大小始终为1，这主要是由于大数加小数的原因。

第3题3.1 程序;clear all A=[1 -18 144 -672 2016 -4032 5376 -4608 2304 -512];x=1.95:0.005:2.05; i=1:length(x);for y1(i)=f(A,x(i)); y2(i)=(x(i)-2)^9;end figure(3);plot(x,y1);;on hold卑微如蝼蚁、坚强似大象．共享知识分享快乐);'r'plot(x,y2,F.m文件y=f(A,x)function y=A(1); i=2:length(A);for y=x*y+A(i);;end3.2 结果第4题卑微如蝼蚁、坚强似大象．共享知识分享快乐4.1 程序;clear all n=input('请输入向量的长度n：')A=2*eye(n)-tril(ones(n,n),0); i=1:n for A(i,n)=1;end n=length(A);U=A; e=eye(n);for i=1:n-1[max_data,max_index]=max(abs(U(i:n,i))); e0=eye(n);max_index=max_index+i-1; U=e0*U; e1=eye(n); j=i+1:n fore1(j,i)=-U(j,i)/U(i,i);endU=e1*U;中把变换矩阵存到P P{i}=e0;% L{i}=e1; e=e1*e0*e;endk=1:n-2for Ldot{k}=L{k}; i=k+1:n-1forLdot{k}=P{i}*Ldot{k}*P{i};endend Ldot{n-1}=L{n-1};LL=eye(n);PP=eye(n); i=1:n-1for PP=P{i}*PP;LL=Ldot{i}*LL;endb=ones(n,2);解方程 %b=e*b;x=zeros(n,1);x(n)=b(n)/U(n,n); i=n-1:-1:1for卑微如蝼蚁、坚强似大象．共享知识分享快乐x(i)=(b(i)-U(i,:)*x)/U(i,i);end计算逆矩阵%X=U^-1*e^-1*eye(n);AN=X'; result2{n-4,1}=AN;result1{n-4,1}=x;,n)'%d:\n'fprintf(fprintf('%d ',AN);4.2 结果n=51.0625 -0.875 -0.75 -0.5 -0.0625-0.0625 0.0625 -0.75 1.125 -0.5-0.0625 0.125 0.0625 1.25 -0.5-0.0625 0.1250.25 0.06251.50.0625-0.5-0.25-0.0625 -0.125n=101.0625 -0.875 -0.75 -0.5 -0.0625 1.0625 -0.875 -0.75 -0.5 -0.0625 -0.0625 1.125 0.0625 -0.75 -0.5 -0.5 0.0625 1.125 -0.75 -0.0625 -0.0625 0.0625 0.125 1.25 1.25 -0.0625 -0.5 0.0625 0.125 -0.5-0.0625 0.250.250.0625 0.1251.5 1.5 -0.0625 0.1250.06250.0625 -0.0625 -0.125 -0.25 0.0625 -0.5 -0.0625 -0.125 -0.25 -0.5 -0.0625 -0.75 1.0625 -0.5 -0.0625 -0.875 -0.5 -0.75 1.0625 -0.875 -0.0625 -0.5 0.0625 1.125 -0.5 0.0625 1.125 -0.75 -0.0625 -0.75 1.25 0.125 0.0625 -0.0625 -0.0625 -0.5 -0.5 0.0625 0.125 1.250.25-0.0625 -0.0625 1.50.1250.0625 0.0625 0.250.1251.5-0.0625 -0.125 -0.25 0.0625-0.5 0.0625 -0.0625 -0.125 -0.25-0.5同样的方法可以算出n=20,n=30时的结果，这里就不罗列了。

学院：专业：班级：学号：姓名：上机大作业1：1.最速下降法：function f = fun(x)f = (1-x(1))^2 + 100*(x(2)-x(1)^2)^2; endfunction g = grad(x)g = zeros(2,1);g(1)=2*(x(1)-1)+400*x(1)*(x(1)^2-x(2)); g(2) = 200*(x(2)-x(1)^2);endfunction x_star = steepest(x0,eps) gk = grad(x0);res = norm(gk);k = 0;while res > eps && k<=1000dk = -gk;ak =1; f0 = fun(x0);f1 = fun(x0+ak*dk);slope = dot(gk,dk);while f1 > f0 + *ak*slopeak = ak/4;xk = x0 + ak*dk;f1 = fun(xk);endk = k+1;x0 = xk;gk = grad(xk);res = norm(gk);fprintf('--The %d-th iter, the residual is %f\n',k,res); endx_star = xk;end>> clear>> x0=[0,0]';>> eps=1e-4;>> x=steepest(x0,eps)2.牛顿法：function f = fun(x)f = (1-x(1))^2 + 100*(x(2)-x(1)^2)^2; endfunction g = grad2(x)g = zeros(2,2);g(1,1)=2+400*(3*x(1)^2-x(2));g(1,2)=-400*x(1);g(2,1)=-400*x(1);g(2,2)=200;endfunction g = grad(x)g = zeros(2,1);g(1)=2*(x(1)-1)+400*x(1)*(x(1)^2-x(2)); g(2) = 200*(x(2)-x(1)^2);endfunction x_star = newton(x0,eps)gk = grad(x0);bk = [grad2(x0)]^(-1);res = norm(gk);k = 0;while res > eps && k<=1000dk=-bk*gk;xk=x0+dk;k = k+1;x0 = xk;gk = grad(xk);bk = [grad2(xk)]^(-1);res = norm(gk);fprintf('--The %d-th iter, the residual is %f\n',k,res); endx_star = xk;end>> clear>> x0=[0,0]';>> eps=1e-4;>> x1=newton(x0,eps)--The 1-th iter, the residual is--The 2-th iter, the residual isx1 =法：function f = fun(x)f = (1-x(1))^2 + 100*(x(2)-x(1)^2)^2; endfunction g = grad(x)g = zeros(2,1);g(1)=2*(x(1)-1)+400*x(1)*(x(1)^2-x(2)); g(2) = 200*(x(2)-x(1)^2);endfunction x_star = bfgs(x0,eps) g0 = grad(x0);gk=g0;res = norm(gk);Hk=eye(2);k = 0;while res > eps && k<=1000dk = -Hk*gk;ak =1; f0 = fun(x0);f1 = fun(x0+ak*dk);slope = dot(gk,dk);while f1 > f0 + *ak*slopeak = ak/4;xk = x0 + ak*dk;f1 = fun(xk);endk = k+1;fa0=xk-x0;x0 = xk;go=gk;gk = grad(xk);y0=gk-g0;Hk=((eye(2)-fa0*(y0)')/((fa0)'*(y0)))*((eye(2)-(y0)*(fa0)')/((fa0)'*(y0)))+(fa0*(fa 0)')/((fa0)'*(y0));res = norm(gk);fprintf('--The %d-th iter, the residual is %f\n',k,res);endx_star = xk;End>> clear>> x0=[0,0]';>> eps=1e-4;>> x=bfgs(x0,eps)4.共轭梯度法：function f = fun(x)f = (1-x(1))^2 + 100*(x(2)-x(1)^2)^2; endfunction g = grad(x)g = zeros(2,1);g(1)=2*(x(1)-1)+400*x(1)*(x(1)^2-x(2)); g(2) = 200*(x(2)-x(1)^2);endfunction x_star =CG(x0,eps) gk = grad(x0);res = norm(gk);k = 0;dk = -gk;while res > eps && k<=1000 ak =1; f0 = fun(x0);f1 = fun(x0+ak*dk);slope = dot(gk,dk);while f1 > f0 + *ak*slope ak = ak/4;xk = x0 + ak*dk;f1 = fun(xk);endk = k+1;x0 = xk;g0=gk;gk = grad(xk);res = norm(gk);p=(gk/g0)^2;dk1=dk;dk=-gk+p*dk1;fprintf('--The %d-th iter, the residual is %f\n',k,res); endx_star = xk;end>> clear>> x0=[0,0]';>> eps=1e-4;>> x=CG(x0,eps)上机大作业2：function f= obj(x)f=4*x(1)-x(2)^2-12;endfunction [h,g] =constrains(x) h=x(1)^2+x(2)^2-25;g=zeros(3,1);g(1)=-10*x(1)+x(1)^2-10*x(2)+x(2)^2+34;g(2)=-x(1);g(3)=-x(2);endfunction f=alobj(x) %拉格朗日增广函数%N_equ等式约束个数?%N_inequ不等式约束个数N_equ=1;N_inequ=3;global r_al pena;%全局变量h_equ=0;h_inequ=0;[h,g]=constrains(x);%等式约束部分?for i=1:N_equh_equ=h_equ+h(i)*r_al(i)+(pena/2)*h(i).^2;end%不等式约束部分for i=1:N_inequh_inequ=h_inequ+pena)*(max(0,(r_al(i)+pena*g(i))).^2-r_al(i).^2); end%拉格朗日增广函数值f=obj(x)+h_equ+h_inequ;function f=compare(x)global r_al pena N_equ N_inequ;N_equ=1;N_inequ=3;h_inequ=zeros(3,1);[h,g]=constrains(x);%等式部分for i=1:1h_equ=abs(h(i));end%不等式部分for i=1:3h_inequ=abs(max(g(i),-r_al(i+1)/pena));endh1 = max(h_inequ);f= max(abs(h_equ),h1); %sqrt(h_equ+h_inequ);function [ x,fmin,k] =almain(x_al)%本程序为拉格朗日乘子算法示例算法%函数输入：% x_al:初始迭代点% r_al：初始拉格朗日乘子N-equ:等式约束个数N_inequ:不等式约束个数?%函数输出% X：最优函数点FVAL:最优函数值%============================程序开始================================ global r_al pena ; %参数（全局变量）pena=10; %惩罚系数r_al=[1,1,1,1];c_scale=2; %乘法系数乘数cta=; %下降标准系数e_al=1e-4; %误差控制范围max_itera=25;out_itera=1; %迭代次数%===========================算法迭代开始============================= while out_itera<max_iterax_al0=x_al;r_al0=r_al;%判断函数?compareFlag=compare(x_al0);%无约束的拟牛顿法BFGS[X,fmin]=fminunc(@alobj,x_al0);x_al=X; %得到新迭代点%判断停止条件?if compare(x_al)<e_aldisp('we get the opt point');breakend%c判断函数下降度?if compare(x_al)<cta*compareFlagpena=1*pena; %可以根据需要修改惩罚系数变量elsepena=min(1000,c_scale*pena); %%乘法系数最大1000disp('pena=2*pena');end%%?更新拉格朗日乘子[h,g]=constrains(x_al);for i=1:1%%等式约束部分r_al(i)= r_al0(i)+pena*h(i);endfor i=1:3%%不等式约束部分r_al(i+1)=max(0,(r_al0(i+1)+pena*g(i)));endout_itera=out_itera+1;end%+++++++++++++++++++++++++++迭代结束+++++++++++++++++++++++++++++++++ disp('the iteration number');k=out_itera;disp('the value of constrains'); compare(x_al)disp('the opt point');x=x_al;fmin=obj(X);>> clear>> x_al=[0,0];>> [x,fmin,k]=almain(x_al)上机大作业3： 1、>> clear alln=3; c=[-3,-1,-3]'; A=[2,1,1;1,2,3;2,2,1;-1,0,0;0,-1,0;0,0,-1];b=[2,5,6,0,0,0]';cvx_beginvariable x(n)minimize( c'*x)subject toA*x<=bcvx_endCalling SDPT3 : 6 variables, 3 equality constraints------------------------------------------------------------num. of constraints = 3dim. of linear var = 6*******************************************************************SDPT3: Infeasible path-following algorithms*******************************************************************version predcorr gam expon scale_dataNT 1 1 0it pstep dstep pinfeas dinfeas gap prim-obj dual-obj cputime -------------------------------------------------------------------0|||+01|+00|+02|+01 +00| 0:0:00| chol 1 11|||||+01|+00 +01| 0:0:01| chol 1 12|||||+00|+00 +01| 0:0:01| chol 1 13|||||+00|+00 +00| 0:0:01| chol 1 14||||||+00 +00| 0:0:01| chol 1 15||||||+00 +00| 0:0:01| chol 1 16||||||+00 +00| 0:0:01| chol 1 17||||||+00 +00| 0:0:01| chol 1 18||||||+00 +00| 0:0:01|stop: max(relative gap, infeasibilities) <------------------------------------------------------------------- number of iterations = 8primal objective value = +00dual objective value = +00gap := trace(XZ) =relative gap =actual relative gap =rel. primal infeas (scaled problem) =rel. dual " " " =rel. primal infeas (unscaled problem) = +00rel. dual " " " = +00norm(X), norm(y), norm(Z) = +00, +00, +00norm(A), norm(b), norm(C) = +00, +00, +00Total CPU time (secs) =CPU time per iteration =termination code = 0DIMACS: +00 +00-------------------------------------------------------------------------------------------------------------------------------Status: SolvedOptimal value (cvx_optval):2、>> clear alln=2; c=[-2,-4]'; G=[,0;0,1]; A=[1,1;-1,0;0,-1]; b=[1,0,0]'; cvx_beginvariable x(n)minimize( x'*G*x+c'*x)subject toA*x<=bcvx_endCalling SDPT3 : 7 variables, 3 equality constraintsFor improved efficiency, SDPT3 is solving the dual problem.------------------------------------------------------------num. of constraints = 3dim. of socp var = 4, num. of socp blk = 1dim. of linear var = 3*******************************************************************SDPT3: Infeasible path-following algorithms*******************************************************************version predcorr gam expon scale_dataNT 1 1 0it pstep dstep pinfeas dinfeas gap prim-obj dual-obj cputime -------------------------------------------------------------------0||||+00|+02| +01 +00| 0:0:00| chol 1 11|||||+01| +00 | 0:0:00| chol 1 12|||||+00| +00 | 0:0:00| chol 1 13|||||| | 0:0:00| chol 1 14|||||| | 0:0:00| chol 1 15|||||| | 0:0:00| chol 1 16|||||| | 0:0:00| chol 1 17|||||| | 0:0:00| chol 1 18|||||| | 0:0:00| chol 1 19|||||| | 0:0:00| chol 1 110|||||| | 0:0:00| chol 2 211|||||| | 0:0:00| chol 2 212|||||| | 0:0:00| chol 2 213|||||| | 0:0:00| chol 2 214|||||| | 0:0:00|stop: max(relative gap, infeasibilities) <------------------------------------------------------------------- number of iterations = 14primal objective value =dual objective value =gap := trace(XZ) =relative gap =actual relative gap =rel. primal infeas (scaled problem) =rel. dual " " " =rel. primal infeas (unscaled problem) = +00rel. dual " " " = +00norm(X), norm(y), norm(Z) = +00, +00, +00norm(A), norm(b), norm(C) = +00, +00, +00Total CPU time (secs) =CPU time per iteration =termination code = 0DIMACS: +00 +00-------------------------------------------------------------------------------------------------------------------------------Status: SolvedOptimal value (cvx_optval): -3。