数值分析上机题matlab版东南大学教学培训

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1、 数值分析上机报告第一章一、题目精确值为。1) 编制按从大到小的顺序，计算SN的通用程序。2) 编制按从小到大的顺序，计算SN的通用程序。3) 按两种顺序分别计算,并指出有效位数。（编制程序时用单精度）4) 通过本次上机题，你明白了什么？二、通用程序clearN=input(Please Input an N (N1):);AccurateValue=single(0-1/(N+1)-1/N+3/2)/2);Sn1=single(0);for a=2:N; Sn1=Sn1+1/(a2-1);endSn2=single(0);for a=2:N; Sn2=Sn2+1/(N-a+2)2-1);en

2、dfprintf(The value of Sn using different algorithms (N=%d)n,N);disp(_)fprintf(Accurate Calculation %fn,AccurateValue);fprintf(Caculate from large to small %fn,Sn1);fprintf(Caculate from small to large %fn,Sn2);disp(_)三、求解结果Please Input an N (N1):102The value of Sn using different algorithms (N=100)_

3、Accurate Calculation 0.740049Caculate from large to small 0.740049Caculate from small to large 0.740050_Please Input an N (N1):104The value of Sn using different algorithms (N=10000)_Accurate Calculation 0.749900Caculate from large to small 0.749852Caculate from small to large 0.749900_Please Input

4、an N (N1):106The value of Sn using different algorithms (N=1000000)_Accurate Calculation 0.749999Caculate from large to small 0.749852Caculate from small to large 0.749999_ 四、结果分析 有效位数 n 顺序 100 10000 1000000从大到小633从小到大566可以得出，算法对误差的传播又一定的影响，在计算时选一种好的算法可以使结果更为精确。从以上的结果可以看到从大到小的顺序导致大数吃小数的现象，容易产生较大的误差，

5、求和运算从小数到大数算所得到的结果才比较准确。 第二章一、题目（1）给定初值及容许误差，编制牛顿法解方程f(x)=0的通用程序。（2）给定方程,易知其有三个根a) 由牛顿方法的局部收敛性可知存在当时，Newton迭代序列收敛于根x2*。试确定尽可能大的。b)试取若干初始值，观察当时Newton序列的收敛性以及收敛于哪一个根。（3）通过本上机题，你明白了什么？二、通用程序文件search.m%寻找最大的delta值%clear%flag=1;k=1;x0=0;while flag=1 delta=k*10-6; x0=delta; k=k+1; m=0; flag1=1; while flag1

6、=1 & m=103 x1=x0-fx(x0)/dfx(x0); if abs(x1-x0)=10-6 flag=0; endendfprintf(The maximun delta is %fn,delta); 文件fx.m% 定义函数f(x)function Fx=fx(x) Fx=x3/3-x;文件dfx.m% 定义导函数df(x)function Fx=dfx(x) Fx=x2-1;文件Newton.m% Newton法求方程的根%clear%ef=10-6; %给定容许误差10-6k=0;x0=input(Please input initial value Xo:);disp(k

7、Xk);fprintf(0 %fn,x0); flag=1;while flag=1 & k=103 x1=x0-fx(x0)/dfx(x0); if abs(x1-x0)ef flag=0; end k=k+1; x0=x1;fprintf(%d %fn,k,x0); end 三、求解结果1.运行search.m文件结果为： The maximum delta is 0.774597即得最大的为0.774597，Newton迭代序列收敛于根=0的最大区间为（-0.774597，0.774597）。2.运行Newton.m文件在区间上各输入若干个数，计算结果如下：区间上取-1000，-100，

8、-50，-30，-10，-8，-7，-5，-3，-1.513 -1.732051Please input initial value Xo:-30k Xk0 -30.0000001 -20.0222472 -13.3815443 -8.9711294 -6.0560005 -4.1505036 -2.9375247 -2.2150468 -1.8547149 -1.74323610 -1.73215811 -1.73205112 -1.732051Please input initial value Xo:-10k Xk0 -10.0000001 -6.7340072 -4.5905703 -

9、3.2128404 -2.3716535 -1.9229816 -1.7571757 -1.7325808 -1.7320519 -1.732051Please input initial value Xo:-10000k Xk0 -10000.0000001 -6666.6667332 -4444.4445893 -2962.9632094 -1975.3090315 -1316.8730256 -877.9158567 -585.2779978 -390.1864709 -260.12602210 -173.41991111 -115.61711812 -77.08384513 -51.3

10、9788014 -34.27822915 -22.87161816 -15.27694917 -10.22845918 -6.88478019 -4.68877220 -3.27480721 -2.40771422 -1.93975023 -1.76125924 -1.73276225 -1.73205126 -1.732051Please input initial value Xo:-100k Xk0 -100.0000001 -66.6733342 -44.4588913 -29.6542634 -19.7920165 -13.2284476 -8.8696517 -5.9892318

11、-4.1073249 -2.91075510 -2.20018911 -1.84868712 -1.74223513 -1.73213914 -1.73205115 -1.732051Please input initial value Xo:-50k Xk0 -50.0000001 -33.3466722 -22.2511253 -14.8641054 -9.9544585 -6.7039606 -4.5710137 -3.2005208 -2.3645159 -1.91970310 -1.75640511 -1.73254812 -1.732051Please input initial

12、value Xo:-3k Xk0 -3.0000001 -2.2500002 -1.8692313 -1.7458104 -1.7322125 -1.7320516 -1.732051Please input initial value Xo:-1.5k Xk0 -1.5000001 -1.8000002 -1.7357143 -1.7320624 -1.7320515 -1.732051Please input initial value Xo:-8k Xk0 -8.0000001 -5.4179892 -3.7393793 -2.6849344 -2.0782465 -1.8029286

13、-1.7360237 -1.7320648 -1.7320519 -1.732051Please input initial value Xo:-7k Xk0 -7.0000001 -4.7638892 -3.3223183 -2.4355334 -1.9529155 -1.7646306 -1.7329317 -1.7320518 -1.732051Please input initial value Xo:-5k Xk0 -5.0000001 -3.4722222 -2.5241803 -1.9960684 -1.7766185 -1.7336746 -1.7320537 -1.73205

14、18 -1.732051结果显示，以上初值迭代序列均收敛于-1.732051，即根。在区间即区间（-1，-0.774597）上取-0.774598，-0.8，-0.85，-0.9，-0.99，计算结果如下：Please input initial value Xo:-0.774598k Xk0 -0.7745981 0.7746052 -0.7746453 0.7748844 -0.7763245 0.7850496 -0.8406417 1.3501878 1.9938309 1.77596310 1.73362811 1.73205312 1.73205113 1.732051Please

15、 input initial value Xo:-0.8k Xk0 -0.8000001 0.9481482 -5.6253703 -3.8726254 -2.7661975 -2.1213676 -1.8182927 -1.7378228 -1.7320799 -1.73205110 -1.732051Please input initial value Xo:0.85k Xk0 0.8500001 -1.4753752 -1.8194443 -1.7379694 -1.7320815 -1.7320516 -1.732051Please input initial value Xo:-0.

16、9k Xk0 -0.9000001 2.5578952 2.0129153 1.7816624 1.7340495 1.7320546 1.7320517 1.732051Please input initial value Xo:-0.99k Xk0 -0.9900001 32.5058292 21.6910813 14.4915214 9.7072385 6.5409066 4.4649667 3.1338408 2.3260759 1.90230310 1.75247811 1.73240312 1.73205113 1.732051计算结果显示，迭代序列局部收敛于-1.732051，即

17、根，局部收敛于1.730251，即根。在区间即区间（-0.774597，0.774597）上，由search.m的运行过程表明，在整个区间上均收敛于0，即根。Please input initial value Xo:0.774598k Xk0 0.7745981 -0.7746052 0.7746453 -0.7748844 0.7763245 -0.7850496 0.8406417 -1.3501878 -1.9938309 -1.77596310 -1.73362811 -1.73205312 -1.73205113 -1.732051Please input initial valu

18、e Xo:0.8k Xk0 0.8000001 -0.9481482 5.6253703 3.8726254 2.7661975 2.1213676 1.8182927 1.7378228 1.7320799 1.73205110 1.732051Please input initial value Xo:0.85k Xk0 0.8500001 -1.4753752 -1.8194443 -1.7379694 -1.7320815 -1.7320516 -1.732051Please input initial value Xo:0.9k Xk0 0.9000001 -2.5578952 -2

19、.0129153 -1.7816624 -1.7340495 -1.7320546 -1.7320517 -1.732051Please input initial value Xo:0.99k Xk0 0.9900001 -32.5058292 -21.6910813 -14.4915214 -9.7072385 -6.5409066 -4.4649667 -3.1338408 -2.3260759 -1.90230310 -1.75247811 -1.73240312 -1.73205113 -1.732051在区间即区间（0.774597，1）上取0.774598，0.8，0.85，0.

20、9，0.99，计算结果如下：计算结果显示，迭代序列局部收敛于-1.732051，即根，局部收敛于1.730251，即根。Please input initial value Xo:4k Xk0 4.0000001 2.8444442 2.1637243 1.8342814 1.7400075 1.7321056 1.7320517 1.732051Please input initial value Xo:3k Xk0 3.0000001 2.2500002 1.8692313 1.7458104 1.7322125 1.7320516 1.732051Please input initial

21、 value Xo:1.5k Xk0 1.5000001 1.8000002 1.7357143 1.7320624 1.7320515 1.732051区间上取100，60，20，10，7，6，4，3，1.5，计算结果如下:6 2.2136057 1.8541268 1.7431369 1.73215610 1.73205111 1.732051Please input initial value Xo:10k Xk0 10.0000001 6.7340072 4.5905703 3.2128404 2.3716535 1.9229816 1.7571757 1.7325808 1.7320

22、519 1.732051Please input initial value Xo:7k Xk0 7.0000001 4.7638892 3.3223183 2.4355334 1.9529155 1.7646306 1.7329317 1.7320518 1.732051Please input initial value Xo:6k Xk0 6.0000001 4.1142862 2.9150683 2.2025784 1.8496505 1.7423926 1.7321427 1.7320518 1.732051Please input initial value Xo:100k Xk0

23、 100.0000001 66.6733342 44.4588913 29.6542634 19.7920165 13.2284476 8.8696517 5.9892318 4.1073249 2.91075510 2.20018911 1.84868712 1.74223513 1.73213914 1.73205115 1.732051Please input initial value Xo:60k Xk0 60.0000001 40.0111142 26.6907493 17.8188454 11.9167625 8.0008486 5.4185467 3.7397368 2.685

24、1519 2.07836010 1.80296711 1.73602712 1.73206413 1.73205114 1.732051Please input initial value Xo:20k Xk0 20.0000001 13.3667502 8.9613233 6.0495474 4.146328结果显示，以上初值迭代序列均收敛于1.732051，即根。综上所述：(-,-1)区间收敛于-1.73205,(-1,)区间局部收敛于1.73205,局部收敛于-1.73205,(-,)区间收敛于0,(,1)区间类似于(-1,)区间，(1,)收敛于1.73205。通过本上机题，明白了对于多

25、根方程，Newton法求方程根时，迭代序列收敛于某一个根有一定的区间限制，在一个区间上，可能会局部收敛于不同的根。第三章一、题目列主元Gauss消去法对于某电路的分析，归结为求解线性方程组。其中（1） 编制解n阶线性方程组的列主元高斯消去法的通用程序；（2） 用所编程序线性方程组，并打印出解向量，保留5位有效数；二、通用程序% 列主元Gauss消去法求解线性方程组%参数输入n=input(Please input the order of matrix A: n=); %输入线性方程组阶数nb=zeros(1,n);A=input(Input matrix A (such as a 2 ord

26、er matrix:1 2;3,4) :);b(1,:)=input(Input the column vector b:); %输入行向量bb=b; C=A,b; %得到增广矩阵%列主元消去得上三角矩阵for i=1:n-1 maximum,index=max(abs(C(i:n,i); index=index+i-1; T=C(index,:); C(index,:)=C(i,:); C(i,:)=T; for k=i+1:n %列主元消去 if C(k,i)=0 C(k,:)=C(k,:)-C(k,i)/C(i,i)*C(i,:); end endend% 回代求解 %x=zeros(n

27、,1);x(n)=C(n,n+1)/C(n,n);for i=n-1:-1:1 x(i)=(C(i,n+1)-C(i,i+1:n)*x(i+1:n,1)/C(i,i);endA=C(1:n,1:n); %消元后得到的上三角矩阵disp(The upper teianguular matrix is:)for k=1:n fprintf(%f ,A(k,:); fprintf(n);enddisp(Solution of the equations:);fprintf(%.5gn,x); %以5位有效数字输出结果 Please input the order of matrix A: n=4In

28、put matrix A (such as a 2 order matrix:1 2;3,4)1 2 1 -22 5 3 -2-2 -2 3 51 3 2 3Input the column vector b:4 7 -1 02.000000 5.000000 3.000000 -2.000000 0.000000 3.000000 6.000000 3.000000 0.000000 0.000000 0.500000 -0.500000 0.000000 0.000000 0.000000 3.000000 Solution of the equations:2-12-1以教材第123页习

29、题16验证通用程序的正确性。执行程序，输入系数矩阵A和列向量b，结果如下：结果与精确解完全一致。三、求解结果执行程序，输入矩阵A（即题中的矩阵R）和列向量b(即题中的V)，得如下结果：Please input the order of matrix A: n=9Input matrix A (such as a 2 order matrix:1 2;3,4):31 -13 0 0 0 -10 0 0 0-13 35 -9 0 -11 0 0 0 00 -9 31 -10 0 0 0 0 00 0 -10 79 -30 0 0 0 -90 0 0 -30 57 -7 0 -5 00 0 0 0

30、-7 47 -30 0 00 0 0 0 0 -30 41 0 00 0 0 0 -5 0 0 27 -20 0 0 -9 0 0 0 -2 29Input the column vector b:-15 27 -23 0 -20 12 -7 7 1031.000000 -13.000000 0.000000 0.000000 0.000000 -10.000000 0.000000 0.000000 0.000000 0.000000 29.548387 -9.000000 0.000000 -11.000000 -4.193548 0.000000 0.000000 0.000000 0.

31、000000 0.000000 28.258734 -10.000000 -3.350437 -1.277293 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 75.461271 -31.185629 -0.451999 0.000000 0.000000 -9.000000 0.000000 0.000000 0.000000 0.000000 44.602000 -7.179695 0.000000 -5.000000 -3.577994 0.000000 0.000000 0.000000 0.000000 -0.000000

32、 45.873193 -30.000000 -0.784718 -0.561543 0.000000 0.000000 0.000000 0.000000 -0.000000 -0.000000 21.380698 -0.513187 -0.367236 0.000000 0.000000 0.000000 0.000000 -0.000000 -0.000000 0.000000 26.413085 -2.419996 0.000000 0.000000 0.000000 0.000000 -0.000000 -0.000000 0.000000 0.000000 27.389504 Sol

33、ution of the equations:-0.289230.34544-0.71281-0.22061-0.43040.15431-0.0578230.201050.29023 由上述结果得： 第四章一、题目二、通用程序三、求解结果1、数据输入Input n: n=10Input x:0 1 2 3 4 5 6 7 8 9 10Input y:2.51 3.30 4.04 4.70 5.22 5.54 5.78 5.40 5.57 5.70 5.80Input the derivative of y(0):0.5Input the derivative of y(n):0.22、计算结果第五章一、题目二、通用程序三、运行结果第六章一、题目二、通用程序1、RK4方法的通用程序2、AB4方法的通用程序3、AB4- AB4预测校正方法的通用程序4、带改进的AB4- AB4预测校正方法的通用程序三、结果比较四、结论24教-资