数值分析上机题(matlab版).doc

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1、数值剖析上机讲演姓名：许泽玮学号：180422第一章一、标题精确值为。1) 体例按从年夜到小的次序，盘算SN的MATLAB次序。2) 体例按从小到年夜的次序，盘算SN的MATLAB次序。3) 按两种次序分不盘算,并指出无效位数。体例次序时用单精度4) 经过本次上机题，你清楚了什么？二、MATLAB次序clearN=input(PleaseInputanN(N1):);AccurateValue=single(0-1/(N+1)-1/N+3/2)/2);SN1=single(0);fora=2:N;SN1=SN1+1/(a2-1);endSN2=single(0);fora=2:N;SN2=SN

2、2+1/(N-a+2)2-1);endfprintf(N=%d)n,N);disp()fprintf(Theexactvalueoftheinputis%fn,AccurateValue);fprintf(fromlargetosmall%fn,SN1);fprintf(fromsmalltolarge%fn,SN2);disp()三、求解后果PleaseInputanN(N1):102ThevalueofSnusingdifferentalgorithms(N=100)AccurateCalculation0.740049fromlargetosmall0.740049fromsmallto

3、large0.740050PleaseInputanN(N1):104ThevalueofSnusingdifferentalgorithms(N=10000)AccurateCalculation0.749900fromlargetosmall0.749852fromsmalltolarge0.749900PleaseInputanN(N1):106ThevalueofSnusingdifferentalgorithms(N=1000000)AccurateCalculation0.749999fromlargetosmall0.749852fromsmalltolarge0.749999四

4、、后果剖析无效位数n次序102104106从年夜到小633从小到年夜566从上述后果可知，差别的算法对偏差的传达是有妨碍的，在盘算时选一种好的算法能够使后果更为精确。从以上的后果能够看到从年夜到小的次序招致年夜数吃小数的景象，轻易发生较年夜的偏差，求跟运算从小数到年夜数算所失掉的后果才比拟精确。第二章一、标题1给定初值及允许偏差，体例牛顿法解方程f(x)=0的MATLAB次序。2给定方程,易知其有三个根a) 由牛顿办法的部分收敛性可知存在事先，Newton迭代序列收敛于根x2*。试断定尽能够年夜的。b)试取假定干初始值，不雅看事先Newton序列的收敛性以及收敛于哪一个根。3经过本上机题，你清

5、楚了什么？二、MATLAB次序文件search.m%寻寻最年夜的delta值%clear%flag=1;k=1;x0=0;whileflag=1delta=k*10-6;x0=delta;k=k+1;m=0;flag1=1;whileflag1=1&m=103x1=x0-fx(x0)/dfx(x0);ifabs(x1-x0)=10-6flag=0;endendfprintf(Themaximundeltais%fn,delta);文件fx.m%界说函数f(x)functionFx=fx(x)Fx=x3/3-x;文件dfx.m%界说导函数df(x)functionFx=dfx(x)Fx=x2-1

6、;文件Newton.m%Newton法求方程的根%clear%ef=10-6;%给定允许偏差10-6k=0;x0=input(PleaseinputinitialvalueXo:);disp(kXk);fprintf(0%fn,x0);flag=1;whileflag=1&k=103x1=x0-fx(x0)/dfx(x0);ifabs(x1-x0)efflag=0;endk=k+1;x0=x1;fprintf(%d%fn,k,x0);end三、求解后果后果为：Themaximumdeltais0.774597即得最年夜的为0.774597，Newton迭代序列收敛于根=0的最年夜区间为-0.7

7、74597，0.774597。在区间上各输入假定干个数，盘算后果如下：区间上取-1000，-100，-50，-30，-10，-8，-7，-5，-3，-1.5PleaseinputinitialvalueXo:-3kXk0-3.0000001-2.2500002-1.8692313-1.7458104-1.7322125-1.7320516-1.732051PleaseinputinitialvalueXo:-1.5kXk0-1.5000001-1.8000002-1.7357143-1.7320624-1.7320515-1.732051PleaseinputinitialvalueXo:-8

8、kXk0-8.0000001-5.4179892-3.7393793-2.6849344-2.0782465-1.8029286-1.7360237-1.7320648-1.7320519-1.732051PleaseinputinitialvalueXo:-7kXk0-7.0000001-4.7638892-3.3223183-2.4355334-1.9529155-1.7646306-1.7329317-1.7320518-1.732051PleaseinputinitialvalueXo:-5kXk0-5.0000001-3.4722222-2.5241803-1.9960684-1.7

9、766185-1.7336746-1.7320537-1.7320518-1.732051后果表现，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，盘算后果如下：PleaseinputinitialvalueXo:-0.774598kXk0-0.77459810.7746052-0.77464530.7748844-0.77632450.7850496-0.84064171.35018781.99383091.775963101.733628111.732053121.732051131.

10、732051PleaseinputinitialvalueXo:-0.8kXk0-0.80000010.9481482-5.6253703-3.8726254-2.7661975-2.1213676-1.8182927-1.7378228-1.7320799-1.73205110-1.732051PleaseinputinitialvalueXo:0.85kXk00.8500001-1.4753752-1.8194443-1.7379694-1.7320815-1.7320516-1.732051PleaseinputinitialvalueXo:-0.9kXk0-0.90000012.557

11、89522.01291531.78166241.73404951.73205461.73205171.732051PleaseinputinitialvalueXo:-0.99kXk0-0.990000132.505829221.691081314.49152149.70723856.54090664.46496673.13384082.32607591.902303101.752478111.732403121.732051131.732051盘算后果表现，迭代序列部分收敛于-1.732051，即根，部分收敛于1.730251，即根。在区间即区间-0.774597，0.774597上，由se

12、arch.m的运转进程阐明，在全部区间上均收敛于0，即根。PleaseinputinitialvalueXo:0.774598kXk00.7745981-0.77460520.7746453-0.77488440.7763245-0.78504960.8406417-1.3501878-1.9938309-1.77596310-1.73362811-1.73205312-1.73205113-1.732051PleaseinputinitialvalueXo:0.8kXk00.8000001-0.94814825.62537033.87262542.76619752.12136761.8182

13、9271.73782281.73207991.732051101.732051PleaseinputinitialvalueXo:0.85kXk00.8500001-1.4753752-1.8194443-1.7379694-1.7320815-1.7320516-1.732051PleaseinputinitialvalueXo:0.9kXk00.9000001-2.5578952-2.0129153-1.7816624-1.7340495-1.7320546-1.7320517-1.732051PleaseinputinitialvalueXo:0.99kXk00.9900001-32.5

14、058292-21.6910813-14.4915214-9.7072385-6.5409066-4.4649667-3.1338408-2.3260759-1.90230310-1.75247811-1.73240312-1.73205113-1.732051在区间即区间0.774597，1上取0.774598，0.8，0.85，0.9，0.99，盘算后果如下：盘算后果表现，迭代序列部分收敛于-1.732051，即根，部分收敛于1.730251，即根。PleaseinputinitialvalueXo:4kXk04.00000012.84444422.16372431.83428141.74

15、000751.73210561.73205171.732051PleaseinputinitialvalueXo:3kXk03.00000012.25000021.86923131.74581041.73221251.73205161.732051PleaseinputinitialvalueXo:1.5kXk01.50000011.80000021.73571431.73206241.73205151.732051区间上取100，60，20，10，7，6，4，3，1.5，盘算后果如下:后果表现，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,

16、(-1,)区间部分收敛于1.73205,部分收敛于-1.73205,(-,)区间收敛于0,(,1)区间相似于(-1,)区间，(1,)收敛于1.73205。经过本上机题，清楚了关于多根方程，Newton法求方程根时，迭代序列收敛于某一个根有必定的区间限度，在一个区间上，能够会部分收敛于差别的根。第三章一、标题列主元Gauss消去法关于某电路的剖析，归纳为求解线性方程组。此中（1） 体例解n阶线性方程组的列主元高斯消去法的MATLAB次序；（2） 用所编次序线性方程组，并打印出解向量，保存5位无效数；二、MATLAB次序%列主元Gauss消去法求解线性方程组%参数输入n=input(Pleasei

17、nputtheorderofmatrixA:n=);%输入线性方程组阶数nb=zeros(1,n);A=input(InputmatrixA(suchasa2ordermatrix:12;3,4):);b(1,:)=input(Inputthecolumnvectorb:);%输入行向量bb=b;C=A,b;%失掉增广矩阵%列主元消去得上三角矩阵fori=1:n-1maximum,index=max(abs(C(i:n,i);index=index+i-1;T=C(index,:);C(index,:)=C(i,:);C(i,:)=T;fork=i+1:n%列主元消去ifC(k,i)=0C(k

18、,:)=C(k,:)-C(k,i)/C(i,i)*C(i,:);endendend%回代求解%x=zeros(n,1);x(n)=C(n,n+1)/C(n,n);fori=n-1:-1:1x(i)=(C(i,n+1)-C(i,i+1:n)*x(i+1:n,1)/C(i,i);endA=C(1:n,1:n);%消元后失掉的上三角矩阵disp(Theupperteianguularmatrixis:)fork=1:nfprintf(%f,A(k,:);fprintf(n);enddisp(Solutionoftheequations:);fprintf(%.5gn,x);%以5位无效数字输入后果P

19、leaseinputtheorderofmatrixA:n=4InputmatrixA(suchasa2ordermatrix:12;3,4)121-2253-2-2-2351323Inputthecolumnvectorb:47-102.0000005.0000003.000000-2.0000000.0000003.0000006.0000003.0000000.0000000.0000000.500000-0.5000000.0000000.0000000.0000003.000000Solutionoftheequations:2-12-1以课本第123页习题16验证MATLAB次序的

20、准确性。履行次序，输入系数矩阵A跟列向量b，后果如下：后果与精确解完整分歧。三、求解后果履行次序，输入矩阵A即题中的矩阵R跟列向量b(即题中的V)，得如下后果：PleaseinputtheorderofmatrixA:n=9InputmatrixA(suchasa2ordermatrix:12;3,4):31-13000-10000-1335-90-1100000-931-100000000-1079-30000-9000-3057-70-500000-747-300000000-3041000000-50027-2000-9000-229Inputthecolumnvectorb:-1527

21、-230-2012-771031.000000-13.0000000.0000000.0000000.000000-10.0000000.0000000.0000000.0000000.00000029.548387-9.0000000.000000-11.000000-4.1935480.0000000.0000000.0000000.0000000.00000028.258734-10.000000-3.350437-1.2772930.0000000.0000000.0000000.0000000.0000000.00000075.461271-31.185629-0.4519990.0

22、000000.000000-9.0000000.0000000.0000000.0000000.00000044.602000-7.1796950.000000-5.000000-3.5779940.0000000.0000000.0000000.000000-0.00000045.873193-30.000000-0.784718-0.5615430.0000000.0000000.0000000.000000-0.000000-0.00000021.380698-0.513187-0.3672360.0000000.0000000.0000000.000000-0.000000-0.000

23、0000.00000026.413085-2.4199960.0000000.0000000.0000000.000000-0.000000-0.0000000.0000000.00000027.389504Solutionoftheequations:-0.289230.34544-0.71281-0.22061-0.43040.15431-0.0578230.50.29023由上述后果得：第四章一、标题二、MATLAB次序三、求解后果1、数据输入Inputn:n=10Inputx:012345678910Inputy:2.513.304.044.705.225.545.785.405.57

24、5.705.80Inputthederivativeofy(0):0.5Inputthederivativeofy(n):0.22、盘算后果第一章一、标题常微分方程初值咨询题数值解1体例RK4办法的通用次序；2体例AB4办法的通用次序由RK4供给初值；3体例AB4-AM4猜测校订办法的通用次序由RK4供给初值；4体例带改良的AB4-AM4猜测校订办法的通用次序由RK4供给初值;5关于初值咨询题取步长,使用14中的四种办法进展盘算，并将盘算后果跟精确解作比拟；6经过本上机题，你能失掉哪些论断？二、C+次序解：源次序如下：#include#include#include#includeofstre

25、amoutfile(data.txt);/此处界说函数f(x,y)的表白式/用户能够本人设定所需求求得函数表白式doublef1(doublex,doubley)doublef1;f1=(-1)*x*x*y*y;returnf1;/此处界说求函数精确解的函数表白式doublef2(doublex)doublef2;f2=3/(1+x*x*x);returnf2;/此处为精确求函数解的通用次序voidaccurate(doublea,doubleb,doubleh)doublex100,accurate100;x0=a;inti=0;outfile输入函数精确值的次序后果:n;doxi=x0+i

26、*h;accuratei=f2(xi);outfileaccuratei=accuratein;i+;while(i(b-a)/h+1);/此处为经典Runge-Kutta公式的通用次序voidRK4(doublea,doubleb,doubleh,doublec)inti=0;doublek1,k2,k3,k4;doublex100,y100;y0=c;x0=a;outfile输入经典Runge-Kutta公式的次序后果:n;doxi=x0+i*h;k1=f1(xi,yi);k2=f1(xi+h/2),(yi+h*k1/2);k3=f1(xi+h/2),(yi+h*k2/2);k4=f1(x

27、i+h),(yi+h*k3);yi+1=yi+h*(k1+2*k2+2*k3+k4)/6;outfileyi=yin;i+;while(i(b-a)/h+1);/此处为4阶Adams显式办法的通用次序voidAB4(doublea,doubleb,doubleh,doublec)doublex100,y100,y1100;doublek1,k2,k3,k4;y0=c;x0=a;outfile输入4阶Adams显式办法的次序后果:n;for(inti=0;i=2;i+)xi=x0+i*h;k1=f1(xi,yi);k2=f1(xi+h/2),(yi+h*k1/2);k3=f1(xi+h/2),(

28、yi+h*k2/2);k4=f1(xi+h),(yi+h*k3);yi+1=yi+h*(k1+2*k2+2*k3+k4)/6;intj=3;y10=y0;y11=y1;y12=y2;y13=y3;doxj=x0+j*h;y1j+1=y1j+(55*f1(xj,y1j)-59*f1(xj-1,y1j-1)+37*f1(xj-2,y1j-2)-9*f1(xj-3,y1j-3)*h/24;outfiley1j=y1jn;j+;while(j(b-a)/h+1);/主函数voidmain(void)doublea,b,h,c;coutabhc;accurate(a,b,h);RK4(a,b,h,c);

29、AB4(a,b,h,c);三、求解后果ixiyRKyABy(xi)|y(xi)-y(RK)|00333010.12.9972.9972.9971.87138e-00720.22.976192.976192.976193.91665e-00730.32.921132.921132.921137.58342e-00740.42.819552.818392.819551.61101e-00650.52.666662.664672.666673.17735e-00660.62.46712.46522.467115.00551e-00670.72.23382.233082.23385.77233e-00

30、680.81.984121.984951.984134.12954e-00690.91.735111.737041.735111.15554e-007101.01.500011.502191.55.80668e-006111.11.287011.288761.2871.13075e-005121.21.099721.100721.099711.54242e-005131.30.9383970.938710.938381.77272e-005141.40.80130.8011350.8012821.83754e-005151.50.6857320.6853350.6857141.78e-005心得：本上机题让我更熟习了种种求微分方程的数值办法，经典Runge-Kutta公式、AB4办法以及AB4-AM4猜测校订办法求解公式的精度是差别的。此中经典Runge-Kutta公式、四阶Adams显式办法AB4存在4阶精度。