數(shù)值分析考試

一、插值法(P50):1、己知函數(shù)在下列各點(diǎn)的值為0.20.40.60.81.0fM0.980.920.810.640.38用4次牛頓插值法對(duì)數(shù)據(jù)進(jìn)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(gè)實(shí)驗(yàn)題)此題要求利用給定點(diǎn)，及給定點(diǎn)的函數(shù)值進(jìn)行牛頓插值以及三次樣條插值。a.牛頓插值要實(shí)現(xiàn)牛頓插值，要用到以下代碼%調(diào)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(cè)物體的直線運(yùn)動(dòng)，得出以卜.數(shù)據(jù):時(shí)間t/s00.91.93.03.95.0距離s.zm010305080110求運(yùn)動(dòng)方程。解：被觀測(cè)物體的運(yùn)動(dòng)距離與運(yùn)動(dòng)時(shí)間大體為線性函數(shù)關(guān)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(yùn)動(dòng)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(jì)算擬合多項(xiàng)式在x的值plot(x0,y0,*,x,y)運(yùn)行結(jié)果：n=22.2538-7.855012017、己知實(shí)驗(yàn)數(shù)據(jù)如下:192531384419.032349.073.397.8用最小二乘法求形如)，=。+bx2的經(jīng)驗(yàn)公式，并計(jì)算均方誤差。解：若s=a+bx2,貝ije=spo〃{l,r}則帆卜5,|礎(chǔ)=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(xué)反應(yīng)中，由實(shí)驗(yàn)得分解物質(zhì)濃度與時(shí)間關(guān)系如卜.:時(shí)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(diǎn)，采用方程-by=ae，,(a,b>0)兩邊同時(shí)取對(duì)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(duì)于函數(shù)/(a-)=—,給出n=8時(shí)X的函數(shù)表，使用復(fù)合梯形公式及復(fù)合辛普森公式計(jì)算枳分并估計(jì)誤差以及比較其精度。三、數(shù)值積分與數(shù)值微分(例三、對(duì)于函數(shù)/(a-)=—,給出n=8時(shí)X的函數(shù)表，使用復(fù)合梯形公式及復(fù)合辛普森公式計(jì)算枳分并估計(jì)誤差以及比較其精度。1復(fù)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(fù)合桑普森復(fù)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(gè)分量是函數(shù)值，第二個(gè)分量是導(dǎo)數(shù)值%x…初始點(diǎn)。%ep…精度，當(dāng)l(x(k)-x(k-l)l<ep時(shí)，終止計(jì)算。省缺為le?5%it_max…最大迭代次數(shù)，省缺為100%x.star-當(dāng)?shù)晒r(shí)，輸出方程的根，%當(dāng)?shù)r(shí)，輸出最后的迭代值。%index―當(dāng)index=l時(shí)，表明迭代成功，%當(dāng)index=0時(shí)，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?tīng)柕?187):給出線性方程組，寫(xiě)出計(jì)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(dá)到精度5x.4-2匚+x,=-12A—J設(shè)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?tīng)柕ń獯朔匠探M的收斂性；用雅可比迭代法及高斯-賽德?tīng)柕ń獯朔匠探M，要求當(dāng)<1。7時(shí)迭代終止?！?21、[解](a)由系數(shù)矩陣-142為嚴(yán)格對(duì)角占優(yōu)矩陣可知，使用雅可比、高2-310X/斯-賽德?tīng)柕ㄇ蠼獯朔匠探M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?tīng)柕?"町=（。_匕）一以工⑶+（。_匕）-%\7O10O43O->72O-23/，‘\.1OJO-11-43一40?！?02一:I+\-//12000ooO111OO-11-43一40120-設(shè)方程組設(shè)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?tīng)柕ㄊ諗俊?(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?tīng)柕ú皇諗?。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(wèn)題數(shù)值解法（318）:用經(jīng)典四階K-R方法解初值問(wèn)題），，=-50.v+50x-+2x,0MxVl,

步長(zhǎng)分別取h=0.1,0.025,0.01，計(jì)算各步長(zhǎng)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(zhǎng)的變化規(guī)律(己知真解y=^e~5Qx+x2)o這時(shí)經(jīng)典R-K公式為解：取步長(zhǎng)h二0.1,程序如下：%經(jīng)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(zhǔn)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(zhǎng)四階經(jīng)典R-K法準(zhǔn)確值')；disp([X',Y',df']);運(yùn)行結(jié)果:步長(zhǎng)四階經(jīng)典R-K法準(zhǔn)確值

00.000000000010000.000000000020000.000000000030000.000000000010000.000000000050000.000000000060000.000000000070000.000000000080000.000000000090000.00000000010000%畫(huà)圖觀察結(jié)果figure;0.000000000033330.000000000460550.000000006306250.000000086404940.000001184363000.000016235451100.000222560671340.003050935427780.041823239217100.運(yùn)行結(jié)果:步長(zhǎng)四階經(jīng)典R-K法準(zhǔn)確值00.000000000010000.000000000020000.000000000030000.000000000010000.000000000050000.000000000060000.000000000070000.000000000080000.000000000090000.00000000010000%畫(huà)圖觀察結(jié)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