《数值分析》第四章答案

1、- - -习题41.给定f(x)=託在x=100,121,1443点处的值，试以这3点建立f(x)的2次(抛物)插值公式，利用插值公式亡H5求的近似值并估计误差。再给69=13建立3次插值公式，给出相应的结果。“1-11_33-5解：f(x)=、；xf(x)=x2，f(x)=-x2,f(x)=x2,248-15x-2，f(115)=10.72380529x=100，x1=121，x2=144，x3=169y0=10，y1=11，y2=12，y3=13*L(x)=y丄2+y_20(x0-x1)(x0-x2)1(x1-x0)(x1-x2)2(x2-x0)(x2-x1)(x-x1)(x-x2)(x-

2、x0)(x-x2)(x-x0)(x-x1)12_+y02+y011(x-x)(x-x)21012L(115)=10乂(115-121)(115-144)乂(115-100)(115-144)2()=(100-121)(100-144)(121-100)(121-144)丄12乂(115-100)(115-121)x(144-100)(144-121)=10 x(-6)(-29)+11x=0.1631x10-2=0.001631实际误差f(115)-L2(115)=0.1045x10-2x(-29)+12x(-21)(-44)21x(-23)44x23=1.88312+9.90683-1.067

3、19=10.72276心)f(x)-L2(x)=(x-x)(x-x/(x-x2)，100144|f(115)-L2(115)|6max|f(x)|(115-100)x(115-121)x(115-44)|6100 x144(x-x1)(x-x2)(x-x3)(x-x0)(x-x2)(x-x3)30(x0-x1)(x0-x2)(x0-x3)1(x1-x0)(x1-x2)(x1-x3)(x-x)(x-x)(x-x)(x-x)(x-x)(x-x)+y013+y0122(x2-x0)(x2-x1)(x2-x3)3(x3-x0)(x3-x1)(x3-x2)L(115)_10X(115-121)x(115

4、-144)x(115-169)，)_(100-121)x(100-144)x(100-169)+11x(115-100)x(115-144)x(115-169)X(121-100)x(121-144)x(121-169)+12x(115-100)x(115-121)x(115-169)X(144-100)x(144-121)x(144-169)+13x(115-100)X(115一121)X(115一144)(169-100)x(169-121)x(169-144)=10 x(-6)X(-29)X(-54)+11x15X(-29)X(-54)(-21)x(-44)x(-69)21x(-23)x

5、(-48)+12X15x(-6)x(-54)+13x15x(-6)x(-29)44x23x(-25)69x48x25=1.473744+11.145186-2.305138+0.409783=10.723571f(x)-L3(x)=f节)(x-x0)(x-x1)(x-x2)(x-x3),100V1694!f(115)-L3(115)24XX=丄x兰x10-7X241615x(-6)x(-29)x(-54)|10-7X|(115-10)(115-121)(115-144)(115-169)_0.5505x10-3_0.0005505实际误差f(115)-L2(115)_0.23429x10-32

6、.设x.为互异节点(j_0,1,n)求证：jn(1)工xkl(x)_xk(k_0,1,n);jjj_0n(2)Z(x.-x)l.(x)=0(k=1,n)。j=o解：(1)考虑函数g(x)=xk,(0kn)以x,x，,x为插值节点的h01nn次插值多项式，由插值余项公式有nxk-Zxkl.(xk)(n+1)X=兀(xx)=0(n+1)!i=0iZxkl(x)=xk,0kn(2)法1当1kn时Z(xx)kl(x)=ZZCl(x)l(x)kll(x).k.=0.=0l=0k=ZC；(x)k1-x1=(x+(x)k=0k=0法2设g(x)=(xt)k，1kn考虑它的n次插值多项式有Z(xt)k1(x)

7、=(xt)k，1kn.=0令t=x得Z(xx)k1(x)=0，1kn.=04.设f(x)=C2a,b，且f(a)=f(b)=0，求证:max|f(x)l(ba)2-max|frr(x)|解：考虑f(x)以x=a,x=b为节点的一次插值多项式L1(x)，则xbxaL1(x)=f(a)+f(b)=01abbaf(x)=f(x)L(x)=:L21(xa)(xb),当xea,b时gg(a,b)于是max|f(x)|axbmax|(xa)(xb)=(ba)2axb8max|f(x)|axbxga,bmax|f(x)|(b)max|f(x)|axb8axb法2设f(x)在cga,b处达到最大值，如果c=a

8、或c=b则结论显然成立，现设cg(a,b)则有广(c)=01f(a)=f(c)+2(ac)2f(g)=0%g(a,c)1f(b)=f(c)+2(bc)2f(g2)=0e(c,b)当cg(a,a+b)时,21f(c)|=才ac)2f(g1)(ba)28max|f(x)|axb,b)时,f(c)|=2(ba)2f(g2)证明:nxk00kn-2Ej=ii=1-i丰j丿f(x.)=a0n(x.-xi)i=1i丰j因而工丄=丄工j=if(xj)a0j=10(x-x)j一i=1i丰jxkj(*)记g(x)=xk，则k0 x.=0g(kj)j=10(xj-xi)j=10(xj-xi)i=1i=1i丰ji丰

9、j=gk,x2x=gn-1確)(n-1)!JO0kn-21k=n-1将上式代入(*)得r，nx：ij=丿1ao法2jf9考虑g(x)以X,x2，,x为插值节点的n-1次插值多项式,k12n则有ixkn(x-xi)/n(Xj-xi)二xk，j=1Ji=1IJ1l丰j0kn-1比较两边xn-1的系数，得/_=j=1(x.x.)jii=1i丰j1k=n-100kn-2x134679y9764316.设有函数值表试求各阶差商，并写出Newton插值多项式。解：134679976431-1-1-1-1-10000N5(x)=9+(-1)(x-1)7.设f(x)=x7+x4+3x+1，求f20,2i,27

10、及f20,2i,28。解：f20,21,27=f(7)G)7=1，f202,27,28=011.设x,x,x互不相同，01n作2n+1次多项式a(x)满足ia.(x.),a.(x.)二0(0jn)ijijij作2n+1多项式0.(x)满足i卩.(x.)二0,卩(x.)=5.(0jn)ijijij解：由条件a.(x.)=0,a(x.)二00jn,j丰ii.i.可设a.(x)=A.+B(x-x.)12(x)再由a(x)=1得Al2(x)=A=1对a.(x)求导得a(x)=Bl2(x)+A.+B(x-x.)a1.(x)1(x)由a(x)=B+2Al(x)=B+21(x)=0得B=-21(x).于是a

11、.(x)=1-21.(x.)1.2(x)2)由0.(x.)=0，0.n.0.(x.)=0,0jn,j主i可设0.(x)=C.(x-x.)1.2(x)求导得0(x)=C.12(x)+(x-x.)-21.(x)1(x)iiiiii求0(x.)=1得.C=1.于是0.(x)=(x-x.)1.2(x)13.给定f(x)=ex。设x=0是4重插值节点，x=1是单重插值节点,解：f(0)1，f(1)1，f(1)1，f(1)1，f(1)e011101丄21011168125301e2e2e21e试求相应的Hermite插值公式，并估计误差(xg0,1)。H4118(x)1+x+x2+x3+(e)x4263R

12、(x)f(i2(x0)4(x1)空x4(x1)TOC o 1-5 h z5!5!eR(x)$x4(x1)e412718maxR(x)(_)4x_x0.081920.001860 x15!5512014.在a,b上求插值多项式H3(x)，使得H3(a)f(a),H；(a)f(a),H；(a)fa),H；，(b)f(b)解：作H2(x)满足H2(a)f(a)H2(a)f(a)，H2(a)f(a)贝H2(x)f(a)+广(a)(xa)+1f(a)(xa)222令g(x)H3(x)H2(x)，(*)贝g(a)0，g(a)0，g(a)0又g(x)为3次多项式，故g(x)A(xa)3代入（*）得H3（x）

13、=H2（x）+g（x）1=f（a）+f（a）（X-a）+2门a）（X-a）2+A（X-a）3（*）求2阶导数得H3(x)=f(a)+bA(x-a)由HJb)=f(b)得f(a)+bA(b-a)=f(b)解得A=1竺一6b-a因而1H3(x)=f(a)+f(a)(x-a)+f(a)(x-a)2321f(b)-f(a)6b-a(x-a)316.设f（x）=，在-1x1上取n=20，按等距节点求分段1+25x2线性插值函数九（x），计算各相邻节点间中点处的/（x）与f（x）的值,hh并计算误差。2解：h=20=0.1，xi=-1+ih=-1+-1i，0i201x=(x.+x)，0i19.1oii+1

14、i+f(x)-f(x)I(x)=f(x)+-il1(x-x)，xxx，i=0,1,2,19TOC o 1-5 h zhiiii+1/Vi+1i1Ih(x1)=-f(x.)+f(x.+1)，0i19i+-2112各相邻节点间中点处的In(x)的值f(x)的值及误差列于下表 -i012345678910111213141516171819h(x1)+2f(x1)i+2f(X1)-In(X1)-0.950.04276020.04244030.0003199-0.850.05294120.05245900.0004822-0.750.06714760.06639000.0007576-0.650.08

15、773580.08648650.0012493-0.550.11896550.11678830.0021772-0.450.16896550.16494850.0040170-0.350.25381620.24615380.0076924-0.250.40384620.39024390.0136023-0.150.650.640.01-0.050.050.150.250.350.450.550.650.750.850.950.90.90.650.40384620.25384620.16896550.11896550.08773580.06714760.05294120.04276020.941

16、17650.94117650.640.39024390.24615380.164948540.1167883J0.0864865J0.06639000.05245900.04244030.04117650.04117650.010.01360230.00769240.0040170.00217720.00124930.00075760.00048220.0003199+2+217.欲使线性插值具有4位有效数字。在区间0,2上列出函数esinx的具有五位有效数字的等距节点的函数值表，问步长最多可取多大？解：x=ih,0inif(x)=esinx,f(x)=esinx-cosx,f(x)=esin

17、xcos2x-esinxsinx=esinx1-sinx-sin2x=eu1-u-u2，u=sinx=g(u)当xg0,2时，ug0,1L1(x)=f(xi)+x-xi+1x-x.ii+1x-xi厶(x)=f(x)xi+1-x+f(x1)x-xi八丿八hx.-x+1i+1hf()-L1()=f()-L1()+L1()-L1()1-=2fGi)(x-xi)(x-xi+1)+f(xi)-f(x,)亠厂+f(xi+1)-f(xi+1)x-ximaxxixxig(u)f(x)-L1(x)maxf(x)-L1(x)xixxi+11h2e+1x10-48211xxxxTOC o 1-5 h z-h2max

18、|f(x)|+-x10-4x_i1_+i8x.xx.J2hhii+1eu(1-u-u2)+eu(-1-2u)=eu(-3u-u2)=-u(3+u)eu0g(0)=1,g(1)=-e,max|f(x)=max|g(u)=e0 x20u11h2e+1x10-41x10-31h2e10-3-10-4=9x10-4822476X10-2h即只要h6X10_218.求f(x)=x4在0,5上的分段3次Hermite插值，并估计误差(h=1)。解：x=i,i=0,1,2,5i当xg(x.,x)时iz+1f(4)()f(x)H3i)=(xx.)2(xx.1)2，i=0,123,4.TOC o 1-5 h z

19、34!ii+1=(xx)2(xx)2，xgx,xii+1ii+1H(i)=x4(xi)2(xi1)2，xgx,xii+119.给定下列函数值表i012x.346yi602y11求3次样条插值函数。解：x0=3,x1=4,x2=6y0=6y1=0,y2=2y0=1y2=-1h0=x1x0=1,h1=x2x1=2u1fx0,x1,x1=7,7fx0,x，x2=3,fx1,x2,x2=1解得M033466-7731-6186可=-28667,M146=15.333,M322=-10.6673将M0,M和M2代入插值函数表达式中得4322xe3,4xe4,66+(x-3)-(x-3)2+(x-3)3,

20、-17(x-4)+23(x-4)2-(x-4)3,30.观测物体的直线运动，得出以下数据：ts00.91.93.03.95.0s/m010305180111求运动方程。解：设S(t)=C0+Ct+Cj2,p0(t)=1,91(t)=t,p2(t)=t2厂0、厂0、厂0、10.90.8110-1-1.9-3.61-30p=,p=,p=,S=0113.0295113.915.218015.0丿25丿屮何)=53.63(备徨)=951.032(P；,5)=4567.2(叽，叽)=6,(叽,)=14.7,(0,2)=218.907(叽,02)=53.63(申0,s)=282(p,s)=1086正规方程

21、为53.63218.907951.03228210864567.2614.714.753.6353.63218.907Co=-0-6184，C1=11-1607，C2=2.2683s(t)=-0.6184+11.160t+2.2683t2均方误差s-卜J(s(ti)si)2=19.024531.用最小二乘法，求一个形如y=a+bx2的经验公式，使它与下列数据拟合，并计算均方误差：X1925313844y19.032.349.073.397.8解：i12345Xi1925313844yi19.032.349.073.397.80(X)=1,w1(x)=X20,申0)=爲0(X.)2=5,(申0,

22、竹)=爲0(xm0(x2)=5327i=1i=1仰,昭)=7277699,仰,y)=271.41,y)=369321.5正规方程组_55327_a271.4-_53277277699_b_369321.5_列主元Gauss消去法得到a=0.050035,b=0.972748经验公式y=0.050035+0.972748x2,_5:工(y(x.)-y.)2=0.1225711i=133.设已知一组实验数据x2.22.63.44.01.0y6561545090试用最小二乘法确定拟合模型y=axb中的参数a,b。解：yi2.2652.6613.4544.0501.090y=axblny=lna+bl

23、nx令Y=lny，t=lnx，则有Y=c+ct，其中c=lna，c=b0101实验数据转化为i12345t=lnxii0.3420.4150.5310.6020Y=lnyii1.8131.7851.7321.6991.954正规方程组5c+1.89C=8.9831.89C+0.933554q=3.303311c=0.421,c=1.95610lna=1.956a=e1.956=7.071b=c=0.4211经验公式为y=7.071e0.42134.试用最小二乘法，求解下列超定方程组广X+2x2=42x+x2=52x+2x2=6X+2x2=23X1X2=4解：将该方程组两边同时左乘以at，得112121223145624193_X1_36_-X2-25-31314解得x2=1.42802，x1=1.6692621最小二乘解为166926、J.42802丿35.求a,b，使佇sinx-(a+bx)2dx为最小，并与26题