數(shù)值分析考試

一、插值法(P50):1、己知函數(shù)在下列各點的值為0.20.40.60.81.0fM0.980.920.810.640.38用4次牛頓插值法對數(shù)據(jù)進行插值。functionf=lagfun(x)a=[0.2,0.4,0.6,0.8,1.0];b=[0.98,0.92,0.81,0.64,0.38];fori=l:5L⑴=1;forj=l:5ifJ=1L⑴=L(i)*(x-a(j))/(a(i)-a(j));endendendf=0；fori=l:5f=f+L(i)*b(i);end執(zhí)行文件x0=[0.2,0.4,0.6,0.8,1.0];y0=[0.98,0.92,0.81,0.64,0.38];plot(xO,yO,'o')holdongridonfplot(Jlagfun，,[0,1]);holdonx=0:0.1:1;plot(x,newton(xO,yO,x),'r');

1.牛頓插值以及三次樣條插值(第一個實驗題)此題要求利用給定點，及給定點的函數(shù)值進行牛頓插值以及三次樣條插值。a.牛頓插值要實現(xiàn)牛頓插值，要用到以下代碼%調用格式：yi=Lagran_(x,y,xi)%x,y數(shù)組形式的數(shù)據(jù)豪functionfi=Lagran_(x,f,xi)fi=zeros(size(xi));npl=length(f);fori=l:nplz二ones(size(xi));forj=l:nplifi~=j,z=z.*(xi-x(j))/(x(i)-x(j));endendfi=fi+z*f(i);end二、曲線擬合(P95):16、觀測物體的直線運動，得出以卜.數(shù)據(jù):時間t/s00.91.93.03.95.0距離s.zm010305080110求運動方程。解：被觀測物體的運動距離與運動時間大體為線性函數(shù)關系，從而選擇線性方程s=a-^bt令C)=spn〃{l,f}則"；=6，|虹=53.63,(%,0)=14.7,(%,s)=280,(°")=1078,則法方程組為(614.7><280J4.753.63；J078,從而解得。=-7.855048'b=22.25376故物體運動方程為S=22.25376J7.855048MATLAB程序：x0=[00.91.933.95];y0=[010305080110];a=polyfit(x0.y0J)x=0:0.1:5;y=polyval(a,x);%計算擬合多項式在x的值plot(x0,y0,*,x,y)運行結果：n=22.2538-7.855012017、己知實驗數(shù)據(jù)如下:192531384419.032349.073.397.8用最小二乘法求形如)，=。+bx2的經驗公式，并計算均方誤差。解：若s=a+bx2,貝ije=spo〃{l,r}則帆卜5,|礎=7277699,(%,0)=5327,(/,%)=271.4,。*)=369321.5,則法方程組為(55327、/\a_<271.4、,53277277699,0_(369321.5,從而解得a=0.9726046=0.0500351故)，=0.9726046+0.0500351F均方誤差為5=[￡()，(易)一刀=0.1226j=O18、在某化學反應中，由實驗得分解物質濃度與時間關系如卜.:時間t/s0510152025303540455055濃度y/(xl°T)0L272.162.863.443.874.154.374.514.584.624.64用最小二乘法求y=解：觀察所給數(shù)據(jù)的特點，采用方程-by=ae，,(a,b>0)兩邊同時取對數(shù)，則]1bIny=lna--取中=sps"l,-L=bi>\x=--則S=a+i)x腐卜11，1就=°?°62321,(%,0)=-0.603975,(%,/)=-87.674095,(^,/)=5.032489,則法方程組為,11-0.603975Y/)』一87.674095、、—0.6039750.062321人歹J一(5.032489/從而解得=-7.5587812//=7.4961692因此=5.2151048/?=//=7.49616927.4961692.??y=5.215104阮

X/(x)011/80.99739781/40.98961583/80.97672671/20.95885105/80.93615563/40.90885167/80.877192510.841409例三、對于函數(shù)/(a-)=—,給出n=8時X的函數(shù)表，使用復合梯形公式及復合辛普森公式計算枳分并估計誤差以及比較其精度。三、數(shù)值積分與數(shù)值微分(例三、對于函數(shù)/(a-)=—,給出n=8時X的函數(shù)表，使用復合梯形公式及復合辛普森公式計算枳分并估計誤差以及比較其精度。1復合梯形fhtfunctionT_n=fht(a,b,n)h=(b-a)/n;fork=0:nx(k+l)=a+k*h;ifx(k+l)==0x(k+l)=10A(-10);endendT_l=h/2*(fx(x(l))+fx(x(n+l)));fori=2:nF(i)=h*fx(x(i));endT_2=sum(F)T_n=T_l+T_2文件2fx.mfunctiony=fx(x)y=sin(x)./x;endmatlab中輸入的?fht(0,1,8)0.83060.9457ans=0.9457>>formatlong?fht(0,1,8)T_2=0.830598927032208T_n=0.945690863582701ans=0.945690863582701>>formatshort?fht(0,1,8)復合桑普森復合S_P_S.m文件functionS_n=S_P_S(a,bzn)h=(b-a)/n;fork=0:nx(k+l)=a+k*h;x_k(k+l)=x(k+l)+l/2*h;if(x(k+l)==0)|(x_k(k+l)==0)x(k+l)=10A(-10);x_k(k+l)=10A(-10);endendS_l=(fx(x(l))+fx(x(n+l)))*(h/6);fori=2:nF1(i)=h/3*fx(x(i));endforj=l:nF_2(j)=2*h/3*fx(x_k(j));endS_2=sum(F_l)+sum(F_2);S_n=S_l+S_2;文件名fx.xfunctiony=fx(x)y=sin(x)./x;end在matlab輸入?S_l=s_p_s(0,1.8)s_l=0.9461四、用牛頓法解方程(223):用牛頓法求解方程x3-x2-l=0在1.5附近的根(要求誤差《10-6)。利用Newton法求方程的根。function[x_star,index,it]=Newton(fun,x,ep,it_max)%求解非線性方程的Newton法,其中%fun(x)—需要求根的函數(shù)，%第一個分量是函數(shù)值，第二個分量是導數(shù)值%x…初始點。%ep…精度，當l(x(k)-x(k-l)l<ep時，終止計算。省缺為le?5%it_max…最大迭代次數(shù)，省缺為100%x.star-當?shù)晒r，輸出方程的根，%當?shù)r，輸出最后的迭代值。%index―當index=l時，表明迭代成功，%當index=0時，表明迭代失敗(迭代次數(shù)>=it.max)。%it-迭代次數(shù)。ifnargin<4it_max=100;endifnargin<3ep=le-5;endindex=0;k=l;whilek<=it_maxxl=x;f=feval(fun,x);ifabs(f(2))<epbreak;endx=x-f(l)/f(2);ifabs(x-xl)<epindex=l;break;enclk=k+l;endx_star=x;it=k;在Matlab窗口輸入：fun=in!ine(,[xA3-x-13*xA2-l]t);[x_star,indexJt]=Newton(funJ.5)五、雅克比迭代法與高斯一賽德爾迭代法(187):給出線性方程組，寫出計算格式判斷兩種方法的收斂性(譜半徑)解方程(精度)需要多少次達到精度5x.4-2匚+x,=-12A—J設方程組+4叢+2人\=20,2x.-3xy+10x,=3考察用雅可比迭代法，高斯-賽德爾迭代法解此方程組的收斂性；用雅可比迭代法及高斯-賽德爾迭代法解此方程組，要求當<1。7時迭代終止?！?21、[解](a)由系數(shù)矩陣-142為嚴格對角占優(yōu)矩陣可知，使用雅可比、高2-310X/斯-賽德爾迭代法求解此方程組均收斂。[精確解為X】=_4,叢=3叫=2](b)使用雅可比迭代法：x{k+l)=D~\L+U)x(k)+D~[b‘02!5(02P5555￡-102X⑴+￡2001X⑴+544422-30■331\71\713八、瓦3ioJ<10>使用高斯-賽德爾迭代法:"町=（。_匕）一以工⑶+（。_匕）-%\7O10O43O->72O-23/，‘\.1OJO-11-43一40?！?02一:I+\-//12000ooO111OO-11-43一40120-設方程組設方程組（a）x.+04〕+0.4x.=1&■J<0.4i]+x2+O.&q=2;0.4i]+0.8x2+心=3X+2x2一2x3=1(b)x.+x.+x=1；

▲■J2i]+2x2+X3=10.40.40.40.4、10.8可知,0.81'1[解](a)由系數(shù)矩陣A=0.40.4(°-0.4—0.4)<0-0.4一0.4、B0=D~\L+U)=-0.40-0.8—-0.40一0.8〔J[-0.4-0.801/廠0.4一0.8°;由0.42\AI-B0\=0.40.40.420.80.8A=(2-0.8)(22+0.82-0.32)可知,=(2-0.8)(2+0.4-7048)(2+0.4+V048)=0p（5o）=O.4+VO48>b從而雅可比迭代法不收斂。<100'-1fo—0.4-0.4)G=(D-L)-p—0.41000-0.8.0.40.8〔°00\z/r100Y0-0.4-0.4、fo-0.4-0.4、—-0.41000—0.8=00.16-0.64-0.4-0.8100〔°0.160.8,\7\/20.40.4,由0.642-0.16-0.162-0.8="2—0"+0.1152)=可知,\AI-G\=00=2(Z-0.48+VO.1152)(2-0.48-Jo.1152)=0P(G)=0.48+VoJBT<1,從而高斯-塞德爾迭代法收斂。4(b)由系數(shù)矩陣A=122-2、1可知,17-22、22、B°=DSU)=i-10-1—-10-1廠2-20z廠2-20>\由22|2Z-B0|=1222-21=23=0可知，p(Bo)=02從而雅可比迭代法收斂。100、■?0-22)G=(D-LylU—11000-1,221/000/‘100)—22、-22)—-11000—-1—02-3￥-2ij1。0074-2/由2\AI-G\=002-2人一23=4(矛+8)=人以一27^)(/1+27^)=0可知,-42+2p(G)=2皿>1,從而高斯-塞德爾迭代法不收斂。程序代碼function[x,i,G]=kb(Azbfmax,eps)D=diag(diag(A));ID=inv(D);J=ID*(-A+D);f=ID*b;xO=zeros(rank(A)f1);x=xO;G(l,:)=xO1;fori=l:maxk=x;x=J*x+f;G(i+1,:)=x!;n=norm((x-k),inf);ifn<=epsbreak;endendG.1在niatlab中輸入A=[1023;2101;3110];?b=[141120]r;?inax=100;?eps=1.0e-006;?[x,i.G]=gs(A,b,max,eps)C=001.40000.82001.49800.78660.79291.68470.73600.78431.70080.73290.78331.70180.73280.78331.70180.73280.78331.70180.73280.78331.70180.73280.78331.7018-0.2000-0.30000.73280.78331.70180.73280.78331.7018-0.20000.04000.0560-03000-0.04000.0940六、常微分方程初值問題數(shù)值解法（318）:用經典四階K-R方法解初值問題），，=-50.v+50x-+2x,0MxVl,

步長分別取h=0.1,0.025,0.01，計算各步長R數(shù)值解在x=0.4,0.6、0.8處的誤差，指出誤差值隨步長的變化規(guī)律(己知真解y=^e~5Qx+x2)o這時經典R-K公式為解：取步長h二0.1,程序如下：%經典的四階R-K方法clear;F='-50*y+50*x"2+2*x';a=0;b=l;h=0.1;n=(b~a)/0.1;X=a:O.l:b;Y二zeros(1,n+1);Y(l)=l/3;fori=l:nx=X(i);y=Y(i);Kl=h*eval(F);x=x+h/2;y=y+Kl/2;K2=h*eval(F);y=Y(i)+K2/2;K3=h*eval(F);x=X(i)+h;y=Y(i)+K3;K4=h*eval(F);Y(i+1)=Y(i)+(K1+2*K2+2*K3+K4)/6;end%準確值temp=L];f=dsolve(，Dyh50*y+50*x"2+2*x','y(0)=l/3*,'x');df=zeros(1,n+1);fori=l:n+ltemp=subs(f,'x',X(i));df(i)=double(vpa(temp));enddispC步長四階經典R-K法準確值')；disp([X',Y',df']);運行結果:步長四階經典R-K法準確值

00.000000000010000.000000000020000.000000000030000.000000000010000.000000000050000.000000000060000.000000000070000.000000000080000.000000000090000.00000000010000%畫圖觀察結果figure;0.000000000033330.000000000460550.000000006306250.000000086404940.000001184363000.000016235451100.000222560671340.003050935427780.041823239217100.運行結果:步長四階經典R-K法準確值00.000000000010000.000000000020000.000000000030000.000000000010000.000000000050000.000000000060000.000000000070000.000000000080000.000000000090000.00000000010000%畫圖觀察結果figure;0.000000000033330.000000000460550.000000006306250.000000086404940.000001184363000.000016235451100.000222560671340.003050935427780.041823239217100.573326903478097.859356300837710.000000000033330.000000000001220.000000000001000.000000000009000.000000000016000.000000000025000.000000000036000.000000000049000.000000000061000.000000000081000.000000000