數(shù)值分析上機題matlab版東南大學教學培訓

1、 數(shù)值分析上機報告第一章一、題目精確值為。1) 編制按從大到小的順序，計算sn的通用程序。2) 編制按從小到大的順序，計算sn的通用程序。3) 按兩種順序分別計算,并指出有效位數(shù)。（編制程序時用單精度）4) 通過本次上機題，你明白了什么？二、通用程序clearn=input('please input an n (n>1):');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/

2、(n-a+2)2-1);endfprintf('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、求解結果ple

3、ase input an n (n>1):102the value of sn using different algorithms (n=100)_accurate calculation 0.740049caculate from large to small 0.740049caculate from small to large 0.740050_please input an n (n>1):104the value of sn using different algorithms (n=10000)_accurate calculation 0.749900cacula

4、te from large to small 0.749852caculate from small to large 0.749900_please input an n (n>1):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、結果分析 有效位數(shù) n 順序 100 10000 1000000從大到小633

5、從小到大566可以得出，算法對誤差的傳播又一定的影響，在計算時選一種好的算法可以使結果更為精確。從以上的結果可以看到從大到小的順序導致大數(shù)吃小數(shù)的現(xiàn)象，容易產(chǎn)生較大的誤差，求和運算從小數(shù)到大數(shù)算所得到的結果才比較準確。 第二章一、題目（1）給定初值及容許誤差，編制牛頓法解方程f(x)=0的通用程序。（2）給定方程,易知其有三個根a) 由牛頓方法的局部收斂性可知存在當時，newton迭代序列收斂于根x2*。試確定盡可能大的。b)試取若干初始值，觀察當時newton序列的收斂性以及收斂于哪一個根。（3）通過本上機題，你明白了什么？二、通用程序文件search.m%尋找最大的delta值%clear

6、%flag=1;k=1;x0=0;while flag=1 delta=k*10-6; x0=delta; k=k+1; m=0; flag1=1; while flag1=1 && m<=103 x1=x0-fx(x0)/dfx(x0); if abs(x1-x0)<10-6 flag1=0; end m=m+1; x0=x1; end if flag1=1|abs(x0)>=10-6 flag=0; endendfprintf('the maximun delta is %fn',delta); 文件fx.m% 定義函數(shù)f(x)functi

7、on fx=fx(x) fx=x3/3-x;文件dfx.m% 定義導函數(shù)df(x)function fx=dfx(x) fx=x2-1;文件newton.m% newton法求方程的根%clear%ef=10-6; %給定容許誤差10-6k=0;x0=input('please input initial value xo:');disp('k xk');fprintf('0 %fn',x0); flag=1;while flag=1 && k<=103 x1=x0-fx(x0)/dfx(x0); if abs(x1-x0)

8、<ef flag=0; end k=k+1; x0=x1;fprintf('%d %fn',k,x0); end 三、求解結果1.運行search.m文件結果為： the maximum delta is 0.774597即得最大的為0.774597，newton迭代序列收斂于根=0的最大區(qū)間為（-0.774597，0.774597）。2.運行newton.m文件在區(qū)間上各輸入若干個數(shù)，計算結果如下：區(qū)間上取-1000，-100，-50，-30，-10，-8，-7，-5，-3，-1.513 -1.732051please input initial value xo:-3

9、0k 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 -3.2128404 -2.3716535 -1.9229816 -1.7571757 -1.7325808 -1.7320519 -1.7320

10、51please 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 -51.39788014 -34.27822915 -22.87161816 -15.27694917 -10.22845918 -6.88478019

11、-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 -4.1073249 -2.91075510 -2.20018911 -1.84868712 -1.74223513 -1.73213914 -

12、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 value xo:-3k xk0 -3.0000001 -2.2500002 -1.8692313 -1.7458104 -1.7322125

13、-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 -1.7360237 -1.7320648 -1.7320519 -1.732051please input initial value xo:

14、-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 -1.7320518 -1.732051結果顯示，以上初值迭代序列均收斂于-1.732051，即根。在區(qū)間即區(qū)間（-1，-0.774597）上取-0.77459

15、8，-0.8，-0.85，-0.9，-0.99，計算結果如下：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 input initial value xo:-0.8k xk0 -0.8000001 0.9481482 -5.6253703 -3.872

16、6254 -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 2.0129153 1.7816624 1.7340495 1.7320546 1.73

17、20517 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計算結果顯示，迭代序列局部收斂于-1.732051，即根，局部收斂于1.730251，即根。在區(qū)間即區(qū)間（-0.774597，0.774597）上，由search.m的運行過程表明，在整個區(qū)間上均收

18、斂于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 value xo:0.8k xk0 0.8000001 -0.9481482 5.6253703 3.8726254 2.7661975 2.12136

19、76 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 -2.0129153 -1.7816624 -1.7340495 -1.7320546 -1.7320517 -1.732051please inp

20、ut 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在區(qū)間即區(qū)間（0.774597，1）上取0.774598，0.8，0.85，0.9，0.99，計算結果如下：計算結果顯示，迭代序列局部收斂于-1.732051，即根，局部收斂于1.730251，即根。please input

21、 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 value xo:1.5k xk0 1.5000001 1.8000002 1.7357143 1.7320624 1.7320515 1.7

22、32051區(qū)間上取100，60，20，10，7，6，4，3，1.5，計算結果如下: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 1.7320519 1.732051please input initial value xo:7k xk0 7.0000001 4.7638892 3.3

23、223183 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 100.0000001 66.6733342 44.4588913 29.6542634 19.7920165 13.2284476 8.86

24、96517 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 2.6851519 2.07836010 1.80296711 1.73602712 1.73206413 1.73205114 1.732051plea

25、se input initial value xo:20k xk0 20.0000001 13.3667502 8.9613233 6.0495474 4.146328結果顯示，以上初值迭代序列均收斂于1.732051，即根。綜上所述：(-,-1)區(qū)間收斂于-1.73205,(-1,)區(qū)間局部收斂于1.73205,局部收斂于-1.73205,(-,)區(qū)間收斂于0,(,1)區(qū)間類似于(-1,)區(qū)間，(1,)收斂于1.73205。通過本上機題，明白了對于多根方程，newton法求方程根時，迭代序列收斂于某一個根有一定的區(qū)間限制，在一個區(qū)間上，可能會局部收斂于不同的根。第三章一、題目列主元gauss

26、消去法對于某電路的分析，歸結為求解線性方程組。其中（1） 編制解n階線性方程組的列主元高斯消去法的通用程序；（2） 用所編程序線性方程組，并打印出解向量，保留5位有效數(shù)；二、通用程序% 列主元gauss消去法求解線性方程組%參數(shù)輸入n=input('please input the order of matrix a: n='); %輸入線性方程組階數(shù)nb=zeros(1,n);a=input('input matrix a (such as a 2 order matrix:1 2;3,4) :');b(1,:)=input('input the co

27、lumn vector b:'); %輸入行向量bb=b' c=a,b; %得到增廣矩陣%列主元消去得上三角矩陣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,1);x(n)=c(n,n+1)/c(n,n);for i=n-1:-1:

28、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陣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位有效數(shù)字輸出結果 please input the order of mat

29、rix a: n=4input 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:2-1

30、2-1以教材第123頁習題16驗證通用程序的正確性。執(zhí)行程序，輸入系數(shù)矩陣a和列向量b，結果如下：結果與精確解完全一致。三、求解結果執(zhí)行程序，輸入矩陣a（即題中的矩陣r）和列向量b(即題中的v)，得如下結果：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 -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 0.000