数值分析第五

1、(完好版)数值分析第五版答案(全)(完好版)数值分析第五版答案(全)50/50(完好版)数值分析第五版答案(全)第一章绪论1设x0,x的相对偏差为，求lnx的偏差。解：近似值x*的相对偏差为=er*e*x*xx*x*而lnx的偏差为elnx*lnx*lnx1e*x*从而有(lnx*)2设x的相对偏差为2%，求xn的相对偏差。解：设f(x)xn，则函数的条件数为Cp|xf(x)|f(x)又Qf(x)nxn1,Cp|xnxn1|nn又Qr(*)n)Cpr(*)xx且er(x*)为2r(x*)n)0.02n3以下各数都是经过四舍五入获取的近似数，即偏差限不超出最后一位的半个单位，试指出它们是几位有效

2、数字：x1*1.1021,x2*0.031,x3*385.6,x4*56.430,x5*71.0.解：x1*1.1021是五位有效数字；x2*0.031是二位有效数字；x3*385.6是四位有效数字；x4*56.430是五位有效数字；x5*71.0.是二位有效数字。4利用公式(2.3)求以下各近似值的偏差限：(1)x1*x2*x4*,(2)x1*x2*x3*,(3)x2*/x4*.此中x1*,x*2,x3*,x4*均为第3题所给的数。解：(x1*)11042(x2*)1102(x3*)1102(x4*)1102(x5*)11023131(1)(x1*x2*x4*)(x1*)(x2*)(x4*)

3、1104110311032103221.05(x1*x*2x3*)x1*x2*(x3*)x2*x3*(x1*)x1*x3*(x2*)1.10210.03111010.031385.611041.1021385.611032220.215(3)(x*2/x*4)x2*(x*4)x*4(x*2)x*420.031110356.4301103256.43056.43021055计算球体积要使相对偏差限为1，问胸襟半径R时同意的相对偏差限是多少？解：球体体积为V4R33则何种函数的条件数为RVR4R2Cpgg3V4R33r(V*)Cpgr(R*)3r(R*)又Qr(V*)1%1故胸襟半径R时同意的相对

4、偏差限为(?)=?1%=1?133006设Y028，按递推公式YnYn1783（n=1,2,）1100计算到Y100。若取78327.982（5位有效数字），试问计算Y100将有多大偏差？解：QYnYn117831100Y100Y99783100Y99Y981783100Y98Y971783100Y1Y017831001挨次代入后，有Y100Y0100783100即Y100Y0783，若取78327.982,Y100Y027.982(Y*)(Y)(27.982)110310002Y100的偏差限为1103。27求方程x256x10的两个根，使它最少拥有4位有效数字（78327.982）。解：x

5、256x10，故方程的根应为x1,228783故x1287832827.98255.982x1拥有5位有效数字x2287831117832827.9820.0178632855.982x2拥有5位有效数字N118当N充分大时，如何求2dx？11NxN12dxarctan(N1)arctanN解1xN设arctan(N1),arctanN。则tanN1,tanN.N11dxN1x2arctan(tan()tantanarctantang1tanarctanN1N(N1)N1arctanN21N19正方形的边长大体为了100cm，应如何丈量才能使其面积偏差不超出1cm2？解：正方形的面积函数为A(

6、x)x2(A*)2A*g(x*).当x*100时，若(A*)1,则(x*)11022故丈量中边长偏差限不超出0.005cm时，才能使其面积偏差不超出1cm210设S1gt2，假设g是正确的，而对t的丈量有0.1秒的偏差，证明当t增添时S的2绝对偏差增添，而相对偏差却减少。解：QS1gt2,t02(S*)gt2g(t*)当t*增添时，S*的绝对偏差增添(S*)r(S*)S*gt2g(t*)g(t*)2(t*)2t*当t*增添时，(t*)保持不变，则S*的相对偏差减少。11序列yn满足递推关系yn10yn11(n=1,2,),若y021.41（三位有效数字），计算到y10时偏差有多大？这个计算过程

7、稳固吗？解：Qy021.41(y0*)11022又Qyn10yn11y110y01(y1*)10(y0*)又Qy210y11(y2*)10(y1*)(y2*)102(y0*)(y10*)1010(y0*)101011022108计算到y10时偏差为1108，这个计算过程不稳固。212(21)6，取2，利用以下等式计算，哪一个获取的结果最好？计算f16,(322)3,12)3，99702。(21)(32解：设y(x1)6，若x2，x*1.4，则x*1101。2若经过16计算y值，则2(1)y*(x*1gx*1)7*67y*x*(x1)y*x*若经过(322)3计算y值，则y*(32x*)2gx*

8、36*y*gx*2xy*x*若经过1计算y值，则22)3(3y*(31*)4gx*2x1*(32x*)7yxy*x*经过(312)3计算后获取的结果最好。213f(x)ln(xx21),求f(30)的值。若开平方用6位函数表，问求对数时偏差有多大？若改用另一等价公式。ln(xx21)ln(xx21)计算，求对数时偏差有多大？解Qf(x)ln(xx21),f(30)ln(30899)设u899,yf(30)则u*u*142故y*u*u*gu*0.01673若改用等价公式ln(xx21)ln(xx21)则f(30)ln(30899)此时，y*u*u*u*59.98337第二章插值法1当x1,1,2

9、时，f(x)0,3,4,求f(x)的二次插值多项式。解：x01,x11,x22,f(x0)0,f(x1)3,f(x2)4;l0(x)(xx1)(xx2)1(x1)(x2)(x0 x1)(x0 x2)2l1(x)(xx0)(xx2)1(x1)(x2)(x1x0)(x1x2)6l2(x)(xx0)(xx1)1(x1)(x1)(x2x0)(x2x1)3则二次拉格朗日插值多项式为2L2(x)k0yklk(x)3l0(x)4l2(x)1(x1)(x2)42(x1)(x1)35x23x76232给出f(x)lnx的数值表X0.40.50.60.70.8lnx用线性插值及二次插值计算ln0.54的近似值。解

10、：由表格知，x00.4,x10.5,x20.6,x30.7,x40.8;f(x0)0.916291,f(x1)0.693147f(x2)0.510826,f(x3)0.356675f(x4)0.223144若采纳线性插值法计算ln0.54即f(0.54)，则0.50.540.6xx210(x0.6)l1(x)x2x1l2xx110(x0.5)(x)x1x2L1(x)f(x1)l1(x)f(x2)l2(x)6.93147(x0.6)5.10826(x0.5)L1(0.54)0.62021860.620219若采纳二次插值法计算ln0.54时，l0(xx1)(xx2)50(x0.5)(x0.6)(

11、x)x1)(x0 x2)(x0(xx0)(xx2)100(x0.4)(x0.6)l1(x)x0)(x1x2)(x1l2(xx0)(xx1)50(x0.4)(x0.5)(x)x0)(x2x1)(x2L2(x)f(x0)l0(x)f(x1)l1(x)f(x2)l2(x)500.916291(x0.5)(x0.6)69.3147(x0.4)(x0.6)0.51082650(x0.4)(x0.5)L2(0.54)0.615319840.6153203给全cosx,0ox90o的函数表，步长h1(1/60)o,若函数表拥有5位有效数字，研究用线性插值求cosx近似值时的总偏差界。解：求解cosx近似值时

12、，偏差可以分为两个部分，一方面，x是近似值，拥有5位有效数字，在此后的计算过程中产生必定的偏差流传；另一方面，利用插值法求函数cosx的近似值时，采纳的线性插值法插值余项不为0，也会有必定的偏差。所以，总偏差界的计算应综合以上双方面的要素。当0ox90o时，令f(x)cosx取x00,h(1)o1180108006060令xix0ih,i0,1,.,5400则x5400290o当xxk,xk1时，线性插值多项式为L1(x)f(xk)xxk1f(xk1)xxkxkxk1xk1xk插值余项为R(x)cosxL1(x)1f()(xxk)(xxk1)2又Q在建立函数表时，表中数据拥有5位有效数字，且c

13、osx0,1，故计算中有偏差流传过程。(f*(xk)11052R2(x)(f*(xk)xxk1(f*(xk1)xxk1xkxk1xk1xk(f*(xk)(xxk1xxk1)xkxk1xk1xk(f*(x)1(xk1xxx)khk(f*(xk)总偏差界为RR1(x)R2(x)1(cos)(xxk)(xxk1)(f*(xk)21(xxk)(xk1x)(f*(xk)21(1h)2(f*(xk)2211.0610810520.501061054设为互异节点，求证：n（1）xkjlj(x)xk(k0,1,L,n);j0nx)klj(x)（2）(xj0(k0,1,L,n);j0证明（1）令f(x)xknx

14、kjlj(x)。若插值节点为xj,j0,1,L,n，则函数f(x)的n次插值多项式为Ln(x)j0Rn(x)f(x)f(n1)()n1(x)插值余项为Ln(x)1)!(n又Qkn,f(n1)()0Rn(x)0nxkjlj(x)xkj0(k0,1,L,n);nx)klj(x)(2)(xjj0nn(Ckjxij(x)ki)lj(x)j0i0nnCki(x)ki(xijlj(x)i0j0又Q0in由上题结论可知nxkjlj(x)xij0n原式Cki(x)kixii0(xx)k0得证。5设f(x)C2a,b且f(a)f(b)0,求证：maxf(x)1(ba)2maxf(x).axb8axb解：令x0a

15、,x1b，以此为插值节点，则线性插值多项式为L1(x)f(x0)xx1f(x1)xx0 x0 x1xx0=f(a)xbf(b)xaabxa又Qf(a)f(b)0L1(x)0插值余项为R(x)f(x)L1(x)1f(x)(xx0)(xx1)2f(x)1f(x)(xx0)(xx1)2又Q(xx0)(xx1)12x)(xx0)(x121(x1x0)241(ba)24maxf(x)1(ba)2maxf(x).axb8axb6在4x4上给出f(x)ex的等距节点函数表，若用二次插值求ex的近似值，要使截断偏差不超出106，问使用函数表的步长h应取多少？解：若插值节点为xi1,xi和xi1，则分段二次插值

16、多项式的插值余项为R2(x)1f()(xxi1)(xxi)(xxi1)3!R2(x)1(xxi1)(xxi)(xxi1)maxf(x)64x4设步长为h，即xi1xih,xi1xihR2(x)1e42h33e4h3.63327若截断偏差不超出106，则R2(x)106e4h310627h0.0065.7若yn2n,求4yn及4yn.，解：依据向前差分算子和中心差分算子的定义进行求解。yn2n4yn(E1)4yn4j44j(1)Eynjj04j4(1)y4njjj04(1)j424jynj0j(21)4ynyn2n114yn(E2E2)4yn1(E2)4(E1)4ynE24ynyn22n28假如

17、f(x)是m次多项式，记f(x)f(xh)f(x)，证明f(x)的k阶差分kf(x)(0km)是mk次多项式，而且m1f(x)0（l为正整数）。解：函数f(x)的Taylor展式为f(xh)f(x)f(x)h1f(x)h2L1f(m)(x)hm1f(m1)()hm12m!(m1)!此中(x,xh)又Qf(x)是次数为m的多项式f(m1)()0f(x)f(xh)f(x)f(x)h1f(x)h2L1f(m)(x)hm2m!f(x)为m1阶多项式2f(x)(f(x)2f(x)为m2阶多项式依此过程递推，得kf(x)是mk次多项式f(x)是常数当l为正整数时，m1f(x)09证明(fkgk)fkgkg

18、k1fk证明(fkgk)得证fk1gk1fk1gk1gk1(fk1gk1fkfkgkfkgkfkgk1fkgk1fkgkfk)fk(gk1gk)fkgkgk1fkn1n110证明fkgkfngnf0g0gk1fkk0k0证明：由上题结论可知fkgk(fkgk)gk1fkn1fkgk0n1(fkgk)gk1fk)k0n1n1(fkgk)gk1fkk0k0Q(fkgk)fk1gk1fkgkn1(fkgk)k0(f1g1f0g0)(f2g2f1g1)L(fngnfn1gn1)fngnf0g0n1n1fkgkfngnf0g0gk1fkk0k0得证。n12yj11证明yny0j0n1n1证明2yj(yj

19、1yj)j0j0(y1y0)(y2y1)L(ynyn1)yny0得证。12若f(x)a0a1xLan1xn1anxn有n个不一样实根x1,x2,L,xn，nk0,0kn2;xj证明：j1f(xj)n01,kn1证明：Qf(x)有个不一样实根x1,x2,L,xn且f(x)a0a1xLan1xn1anxnf(x)an(xx1)(xx2)L(xxn)令n(x)(xx1)(xx2)L(xxn)nknk则xjxjj1f(xj)j1ann(xj)而n(x)(xx2)(xx3)L(xxn)(xx1)(xx3)L(xxn)L(xx1)(xx2)L(xxn1)n(xj)(xjx1)(xjx2)L(xjxj1)(

20、xjxj1)L(xjxn)令g(x)xk,nkgx1,x2,L,xnxjj1n(xj)nk则gx1,x2,L,xnxjj1n(xj)nk1又xjgx1,x2,L,xnf(xj)j1annk0,0kn2;xjj1f(xj)n01,kn1得证。13证明n阶均差有以下性质：（1）若F(x)cf(x)，则Fx0,x1,L,xncfx0,x1,L,xn;（2）若F(x)f(x)g(x)，则Fx0,x1,L,xnfx0,x1,L,xngx0,x1,L,xn.证明：nj)（1）Qfx1,x2,L,xnf(x(xjx0)L(xjxj1)(xjxj1)L(xjxn)j0nFx1,x2,L,xnj0nj0(x(x

21、F(xj)jx0)L(xjxj1)(xjxj1)L(xjxn)cf(xj)jx0)L(xjxj1)(xjxj1)L(xjxn)nf(xj)c()j0(xjx0)L(xjxj1)(xjxj1)L(xjxn)cfx0,x1,L,xn得证。(2)QF(x)f(x)nFx0,L,xnj0nj0g(x)(xjx0)L(xj(xjx0)L(xjF(xj)xj1)(xjxj1)L(xjxn)f(xj)g(xj)xj1)(xjxj1)L(xjxn)nf(xj)j0(xjx0)L(xjxj1)(xjxj1)L(xjxn)ng(xj)+)(xjx0)L(xjxj1)(xjxj1)L(xjj0 xn)fx0,L,x

22、ngx0,L,xn得证。14f(x)x7x43x1,求F20,21,L,27及F20,21,L,28。解：Qf(x)x7x43x1若ixi2,i0,1,8L则fx0,x1,L,xnf(n)()n!fx0,x1,L,x7f(7)()7!17!7!f(8)()0fx0,x1,L,x88!15证明两点三次埃尔米特插值余项是R3(x)f(4)()(xxk)2(xxk1)2/4!,(xk,xk1)解：若xxk,xk1，且插值多项式满足条件H3(xk)f(xk),H3(xk)f(xk)H3(xk1)f(xk1),H3(xk1)f(xk1)插值余项为R(x)f(x)H3(x)由插值条件可知R(xk)R(xk

23、1)0且R(xk)R(xk1)0R(x)可写成R(x)g(x)(xxk)2(xxk1)2此中g(x)是关于x的待定函数，现把x看作xk,xk1上的一个固定点，作函数(t)f(t)H3(t)g(x)(txk)2(txk1)2依据余项性质，有(xk)0,(xk1)0(x)f(x)H3(x)g(x)(xxk)2(xxk1)2f(x)H3(x)R(x)0(t)f(t)H3(t)g(x)2(txk)(txk1)22(txk1)(txk)2(xk)0(xk1)0由罗尔定理可知，存在(xk,x)和(x,xk1)，使(1)0,(2)0即(x)在xk,xk1上有四个互异零点。依据罗尔定理，(t)在(t)的两个零

24、点间最罕有一个零点，故(t)在(xk,xk1)内最罕有三个互异零点，依此类推，(4)(t)在(xk,xk1)内最罕有一个零点。记为(xk,xk1)使(4)()f(4)()H3(4)()4!g(x)0又QH3(4)(t)0g(x)f(4)(),(xk,xk1)4!此中依赖于xR(x)f(4)()(xxk)2(xxk1)24!分段三次埃尔米特插值时，若节点为xk(k0,1,L,n)，设步长为h，即xkx0kh,k0,1,L,n在小区间xk,xk1上R(x)f(4)()2(xxk1)24!(xxk)R(x)1f(4)()(xxk)2(xxk1)24!(x4!(x4!14!24xk)2(xk1x)2m

25、axf(4)(x)axbxkxk1x)22maxf(4)(x)2axbh4maxf(4)(x)axbh4maxf(4)(x)384axb16求一个次数不高于4次的多项式P（x），使它满足P(0)P(0)0,P(1)P(1)0,P(2)0解：利用埃米尔特插值可获取次数不高于4的多项式x00,x11y00,y11m00,m1111H3(x)yjj(x)mjj(x)j0j00(x)(12xx0)(xx1)2x0 x1x0 x1(12x)(x1)21(x)(12xx1)(xx0)2x1x0 x1x0(32x)x20(x)x(x1)21(x)(x1)x2H3(x)(32x)x2(x1)x2x32x2设P

26、(x)H3(x)A(xx0)2(xx1)2此中，A为待定常数QP(2)1P(x)x32x2Ax2(x1)2A14从而P(x)1x2(x3)2417设f(x)1/(1x2)，在5x5上取n10，按等距节点求分段线性插值函数Ih(x)，计算各节点间中点处的Ih(x)与f(x)值，并预计偏差。解：若x05,x105则步长h1,xix0ih,i0,1,L,101f(x)1x2在小区间xi,xi1上，分段线性插值函数为Ih(x)xxi1f(xi)xxif(xi1)xixi1xi1xi(xi1x)1(xxi)11xi21xi12各节点间中点处的Ih(x)与f(x)的值为当x4.5时，f(x)0.0471,

27、Ih(x)0.0486当x3.5时，f(x)0.0755,Ih(x)0.0794当x2.5时，f(x)0.1379,Ih(x)0.1500当x1.5时，f(x)0.3077,Ih(x)0.3500当x0.5时，f(x)0.8000,Ih(x)0.7500偏差maxf(x)Ih(x)h2maxf()8xixxi15x51又Qf(x)1x22xf(x)(1x2)2,6x22f(x)(1x2)324x24x3f(x)(1x2)4令f(x)0得f(x)的驻点为x1,21和x30f(x1,2)1,f(x3)22maxf(x)Ih(x)145x518求f(x)x2在a,b上分段线性插值函数Ih(x)，并预

28、计偏差。解：在区间a,b上，x0a,xnb,hixi1xi,i0,1,L,n1,hmaxhi0in1Qf(x)x2函数f(x)在小区间xi,xi1上分段线性插值函数为Ih(x)xxi1f(xi)xxif(xi1)xixi1xi1xi1xi2(xi1x)xi12(xxi)hi偏差为maxf(x)Ih(x)1maxf()ghi2xixxi18abQf(x)x2f(x)2x,f(x)2maxf(x)Ih(x)h24axb19求f(x)x4在a,b上分段埃尔米特插值，并预计偏差。解：在a,b区间上，x0a,xnb,hixi1xi,i0,1,L,n1,令hmaxhi0in1Qf(x)x4,f(x)4x3

29、函数f(x)在区间xi,xi1上的分段埃尔米特插值函数为Ih(x)(xxi1)2(12xxi)f(xi)xixi1xi1xi(xxi)2(12xxi1)f(xi1)xi1xixixi1(xxi1)2(xxi)f(xi)xixi1(xxixi1xixi4(xhi3xi431(xhi)2(xxi1)f(xi1)xi1)2(hi2x2xi)x)2(h2x2x1)iii34x2i(xxi1)2(xxi)hi4xi13(xxi)2(xxi1)h2i偏差为f(x)Ih(x)1f(4)()(xxi)2(xxi1)24!1maxf(4)()(hi)424axb2又Qf(x)x4f(4)(x)4!24maxf(

30、x)Ih(x)maxhi4h4axb0in1161620给定数据表以下：Xj0.250.300.390.450.53Yj0.50000.54770.62450.67080.7280试求三次样条插值，并满足条件：(1)S(0.25)1.0000,S(0.53)0.6868;(2)S(0.25)S(0.53)0.解：h0 x1x00.05h1x2x10.09h2x3x20.06h3x4x30.08Qhj1,hjjhj1hjjhjhj115,23,33,41145719,22,34,011457fx0,x1f(x1)f(x0)0.9540 x1x0fx1,x20.8533fx2,x30.7717fx

31、3,x40.7150(1)S(x0)1.0000,S(x4)0.6868d06f0)5.5200(fx1,x2h0d1fx1,x2fx0,x14.31576h0h1d2fx2,x3fx1,x23.26406h1h2d3fx3,x4fx2,x32.43006h2h3d46(f4fx3,x4)2.1150h3由此得矩阵形式的方程组为21M0529M11414322M25534M327712M45.52004.31573.26402.43002.1150求解此方程组得M02.0278,M11.4643M21.0313,M30.8070,M40.6539三次样条表达式为(xj1x)3Mj1(xxj)3

32、S(x)Mj6hj6hjMjhj2xj1xMj1hj2)xxj(j0,1,L,n1)(yj)(yj1hj6hj6将M0,M1,M2,M3,M4代入得6.7593(0.30 x)34.8810(x0.25)310.0169(0.30 x)10.9662(x0.25)x0.25,0.302.7117(0.39x)31.9098(x0.30)36.1075(0.39x)6.9544(x0.30)S(x)x0.30,0.392.8647(0.45x)32.2422(x0.39)310.4186(0.45x)10.9662(x0.39)x0.39,0.451.6817(0.53x)31.3623(x0.

33、45)38.3958(0.53x)9.1087(x0.45)x0.45,0.53(2)S(x0)0,S(x4)0d02f00,d14.3157,d23.2640d32.4300,d42f40040由此得矩阵动工的方程组为M0M4029014M14.3157322M23.2640553M32.4300027求解此方程组，得M00,M11.8809M20.8616,M31.0304,M40又Q三次样条表达式为S(x)Mj(xj1x)3(xxj)36hjMj16hj(yjMjhj2)xj1x(yj1Mj1hj2)xxj6hj6hj将M0,M1,M2,M3,M4代入得6.2697(x0.25)310(

34、0.3x)10.9697(x0.25)x0.25,0.303.4831(0.39x)31.5956(x0.3)36.1138(0.39x)6.9518(x0.30)x0.30,0.39S(x)2.3933(0.45x)32.8622(x0.39)310.4186(0.45x)11.1903(x0.39)x0.39,0.452.1467(0.53x)38.3987(0.53x)9.1(x0.45)x0.45,0.53221若f(x)Ca,b,S(x)是三次样条函数，证明：bf(x)abf(x)a2bS(x)2dxdxa2b2S(x)dx2aS(x)f(x)S(x)dx(2)若f(xi)S(xi)

35、(i0,1,L,n)，式中xi为插值节点，且ax0 x1Lxnb，则bS(x)f(x)S(x)dxaS(b)f(b)S(b)S(a)f(a)S(a)证明：bf(x)2dx(1)S(x)abf(x)2bS(x)2bf(x)S(x)dxadxadx2abf(x)2b2bS(x)f(x)S(x)dxadxS(x)dx2aa从而有b2b2dxf(x)dxS(x)aab2bf(x)S(x)dx2S(x)f(x)S(x)dxaab(2)S(x)f(x)S(x)dxabS(x)df(x)S(x)abbS(x)f(x)S(x)f(x)aaS(x)dS(x)S(b)f(b)S(b)S(a)f(a)S(b)f(b

36、)S(b)S(a)f(a)S(b)f(b)S(b)S(a)f(a)S(b)f(b)S(b)S(a)f(a)bS(a)S(x)an1S(a)S(xkk0n1xkS(a)S(k0S(a)f(x)S(x)dxxk1)gxk1f(x)S(x)dx2xkxk1)gf(x)xk12S(x)xk第三章函数迫近与曲线拟合1f(x)sin2x，给出0,1上的伯恩斯坦多项式B1(f,x)及B3(f,x)。解：Qf(x)sin,x0,12伯恩斯坦多项式为nkBn(f,x)f()Pk(x)k0n此中Pk(x)nxk(1x)nkk当n1时，P0(x)1(1x)0P1(x)xB1(f,x)f(0)P0(x)f(1)P1(

37、x)1(1x)sin(0)xsin022x当n3时，P0(x)1(1x)30P1(x)1x(1x)23x(1x)20P(x)3x2(1x)3x2(1x)21P3(x)3x3x33B(f,x)3f(k)P(x)3k0nk03x(1x)2gsin63x2(1x)gsinx3sin323x(1x)233x2(1x)x322533x3336x23x2221.5x0.402x20.098x32当f(x)x时，求证Bn(f,x)x证明：若f(x)x，则nf(k)Pk(x)Bn(f,x)k0nnknxk(1x)nk0nkknkn(n1)L(nk1)k(1x)nk0nk!xkn(nL(n1)(k1)1xk(1

38、x)nk1)k1(k1)!nn1x)nkkxk(1k11nn1xk1(1x)(n1)(k1)xk1k1xx(1x)n1x3证明函数1,x,L,xn线性没关证明：若a0a1xa2x2Lanxn0,xR分别取xk(k0,1,2,L,n)，对上式两端在0,1上作带权(x)1的内积，得1Ln11a00MOa10MMM11Lan0n12n1此方程组的系数矩阵为希尔伯特矩阵，对称正定非奇异，只有零解a=0。函数1,x,L,xn线性没关。4。计算以下函数f(x)关于C0,1的f,f1与f2：f(x)(x1)3,x0,1f(x)x1,2(3)f(x)xm(1x)n,m与n为正整数，(4)f(x)(x1)10e

39、x解：(1)若f(x)(x1)3,x0,1，则f(x)3(x1)20f(x)(x1)3在(0,1)内单调递加fmaxf(x)0 x1maxf(0),f(1)max0,11fmaxf(x)0 x1maxf(0),f(1)max0,1111f(x)6dx)22(101x)711(127077(2)若f(x)x1,x0,1，则2fmax1f(x)0 x121f10f(x)dx11)dx21(x2214f2(1(x03611f2(x)dx)201)2dx2(3)若f(x)xm(1x)n,m与n为正整数当x0,1时,f(x)0f(x)mxm1(1x)nxmn(1x)n1(1)xm1(1x)n1m(1nm

40、x)m当x(0,m)时,f(x)0nmmf(x)在(0,)内单调递减mnm当x(,1)时,f(x)0nmmf(x)在(,1)内单调递减。nmx(m,1)f(x)0nmfmaxf(x)0 x1maxf(0),f(m)nmmmgnn(mn)mnf1f(x)dx101xm(1x)ndx0(sin2t)m(1sin2t)ndsin2t02sin2mtcos2ngg0tcost2sintdtn!m!(nm1)!11fx)2ndx22x2m(10102sin4mtcos4ntd(sin2t)21022sin4m1tcos4n1tdt2(2n)!(2m)!2(nm)1!(4)若f(x)(x1)10ex当x0

41、,1时，f(x)0f(x)10(x1)9ex(x1)10(ex)(x1)9ex(9x)0f(x)在0,1内单调递减。fmaxf(x)0 x1maxf(0),f(1)210e1f10f(x)dx11)10exdx(x0(x10 x119xdx1)e010(x1)e010e11f2xdx22(x1)20e07(342)4e5。证明fgfg证明：f(fg)gfggfgfg6。对f(x),g(x)C1a,b，定义(1)(f,g)b(x)g(x)dxfa(2)(f,g)b(x)g(x)dxf(a)g(a)fa问它们能否构成内积。解：(1)令f(x)C（C为常数，且C0）则f(x)0而(,)b()()ff

42、fxfxdxa这与当且仅当f0时，(f,f)0矛盾不可以构成C1a,b上的内积。(2)若(f,g)bf(a)g(a)，则f(x)g(x)dxa(g,f)b(x)dxg(a)f(a)(f,g),Kg(x)fa(f,g)baf(a)g(a)af(x)g(x)dxbf(x)g(x)dxf(a)g(a)a(f,g)hC1a,b,则(fg,h)bg(x)h(x)dxf(a)g(a)h(a)f(x)abf(x)h(x)dxf(a)h(a)bg(a)h(a)af(x)h(x)dxa(f,h)(h,g)(f,f)b2dxf2(a)0f(x)a若(f,f)0，则b(x)2dx0,且f2(a)0faf(x)0,f

43、(a)0f(x)0即当且仅当f0时，(f,f)0.故可以构成C1a,b上的内积。7。令Tn*(x)Tn(2x1),x0,1，试证Tn*(x)是在0,1上带权(x)1的正交多xx2项式，并求T0*(x),T1*(x),T2*(x),T3*(x)。解：若Tn*(x)Tn(2x1),x0,1，则1Tn*(x)Tm*(x)P(x)dx011Tn(2x1)Tm(2x1)2dx0 xx令t(2x1)，则t1,1，且xt12，故1Tn*(x)Tm*(x)(x)dx01d(t1)Tn(t)Tm(t)11t1(t1)222211dtTn(t)Tm(t)112t又Q切比雪夫多项式Tk*(x)在区间0,1上带权(x

44、)1正交，且1x2x0,nm11Tn(x)Tm(x)dt2,nm012,nm0Tn*(x)是在0,1上带权(x)1的正交多项式。xx2又QT0(x)1,x1,1T0*(x)T0(2x1)1,x0,1QT1(x)x,x1,1T1*(x)T1(2x1)2x1,x0,1QT2(x)2x21,x1,1T*(x)T(2x1)222(2x1)218x28x1,x0,1QT3(x)4x33x,x1,1T3*(x)T3(2x1)4(2x1)33(2x1)32x348x218x1,x0,18。对权函数(x)1x2，区间1,1，试求首项系数为1的正交多项式n(x),n0,1,2,3.解：若(x)1x2，则区间1,

45、1上内积为(f,g)1f(x)g(x)(x)dx1定义0(x)1，则n1(x)(xn)n(x)nn1(x)此中n(xn(x),n(x)/(n(x),n(n(x),n(x)/(n1(x),(x,1)/(1,1)1x(1x2)dx11(1x2)dx101(x)x(x2,x)/(x,x)x3(1x2)dx1x2(1x2)dx10(x,x)/(1,1)1x2(1x2)dxn(x)1(x)11(1x2)dx11628532(x)x2252(x32x,x22)/(x22,x22)555512x)(x22)(1x2)dx(x3155122)(x22)(1x2)dx(x15502(x22,x22)/(x,x)

46、5512)(x22)(1x2)dx(x21551x2)dxx2(11136171670153(x)x32x217xx39x570149。试证明由教材式(2.14)给出的第二类切比雪夫多项式族n(x)是0,1上带权u(x)1x2的正交多项式。证明：sin(n1)arccosx若Un(x)1x2令xcos，可得1Um(x)Un(x)1x2dx11sin(m1)arccosxsin(n1)arccosx11x2dx0sin(m1)sin(n1)1cos2dsin(m1)sin(n1)d0当mn时，sin2(m1)d01cos2(m1)d022当mn时，sin(m1)sin(n1)d01cos(n1)

47、0sin(m1)dn11cos(n1)dsin(m1)n1m10n1cos(n1)cos(m1)dm1cos(m1)d1sin(n1)0n1n1m11)(n2sin(n1)dcos(m01)(m12sin(n1)sin(m1)dn)0101(m1)2sin(n1)sin(m1)d0n10又Qmn，故(m1)21n1sin(n1)sin(m1)d00得证。10。证明切比雪夫多项式Tn(x)满足微分方程(1x2)Tn(x)xTn(x)n2Tn(x)0证明：切比雪夫多项式为Tn(x)cos(narccosx),x1从而有TnTn(x)gg1sin(narccosx)n()1x2nsin(narcco

48、sx)1x2nsin(narccosx)n2cos(narccosx)(x)3x2(1x2)21(1x2)T(x)xT(x)n2T(x)nnnnxsin(narccosx)n2cos(narccosx)1x2nxsin(narccosx)n2cos(narccosx)1x20得证。11。假设f(x)在a,b上连续，求f(x)的零次最正确一致迫近多项式？解：Qf(x)在闭区间a,b上连续存在x1,x2a,b，使f(x1)minf(x),axbf(x2)maxf(x),axb取P1f(x1)f(x2)2则x1和x2是a,b上的2个轮流为“正”、“负”的偏差点。由切比雪夫定理知P为f(x)的零次最正

49、确一致迫近多项式。12。采纳常数a，使maxx3ax达到极小，又问这个解能否独一？0 x1解：令f(x)x3ax则f(x)在1,1上为奇函数maxx3ax0 x1maxx3ax1x1f又Qf(x)的最高次项系数为1，且为3次多项式。3(x)123T3(x)与0的偏差最小。(x)1T(x)x33x3434从而有a3413。求f(x)sinx在0,上的最正确一次迫近多项式，并预计偏差。2解：Qf(x)sinx,x0,2f(x)cosx,f(x)sinx0a1f(b)f(a)2,bacosx22,x2arccos20.88069f(x2)0.77118a0f(a)f(x2)f(b)f(a)gax22

50、ba20.10526于是得f(x)的最正确一次迫近多项式为P1(x)0.105262x即sinx0.105262x,0 x2偏差限为sinxP1(x)sin0P1(0)0.1052614。求f(x)ex0,1在0,1上的最正确一次迫近多项式。解：Qf(x)ex,x0,1f(x)ex,f(x)ex0a1f(b)f(a)e1baex2e1x2ln(e1)f(x2)ex2e1a0f(a)2f(x2)f(b)f(a)gax2ba21(e1)(eln(e1)21)2ln(e1)于是得f(x)的最正确一次迫近多项式为P1(x)e(e1)x1ln(e1)22(e1)x1e(e1)ln(e1)215。求f(x

51、)x43x31在区间0,1上的三次最正确一致迫近多项式。解：Qf(x)x43x31,x0,1令t2(x1)，则t1,12且x1t122f(t)(1t1)43(1t1)3122221(t410t324t222t9)16令g(t)16f(t)，则g(t)t410t324t222t9若g(t)为区间1,1上的最正确三次迫近多项式P3*(t)应满足maxg(t)P3*(t)min1t1当g(t)P*(t)1T(t)1(8t48t21)32348时，多项式g(t)P3*(t)与零偏差最小，故3*(t)g(t)13T4(t)210t325t222t738从而，f(x)的三次最正确一致迫近多项式为1*，则f

52、(x)的三次最正确一致迫近多项式为16P3(t)P3*(t)110(2x1)325(2x1)222(2x1)731685x35x21x1294412816。f(x)x，在1,1上求关于span1,x2,x4的最正确平方迫近多项式。解：Qf(x)x,x1,1若(f,g)1f(x)g(x)dx1且01,1x2,2x4，则22022,12,222,9(f,0)1,(f,1)1,(f,2)1,23(0,1)1,(0,2)21,2)2,(,57则法方程组为222135a02221a13572a222215793解得a00.1171875a11.640625a20.8203125故f(x)关于span1,

53、x2,x4的最正确平方迫近多项式为S*(x)a0a1x2a2x40.11718751.640625x20.8203125x417。求函数f(x)在指定区间上关于span1,x的最正确迫近多项式：(1)f(x)1,1,3;(2)f(x)ex,0,1;x(3)f(x)cosx,0,1;(4)f(x)lnx,1,2;解：(1)Qf(x)1,1,3;x若(f,g)3f(x)g(x)dx1且01,1x,，则有22,2260212,3(0,1)4,(f,0)ln3,(f,1)2,则法方程组为24a0ln3264a123从而解得a01.1410a10.2958故f(x)关于span1,x的最正确平方迫近多项

54、式为*S(x)a0a1x(2)Qf(x)ex,0,11若(f,g)f(x)g(x)dx0且01,1x,，则有21,121022,3(0,1)1,2(f,0)e1,(f,1)1,则法方程组为11a02e111a1123从而解得a00.1878a11.6244故f(x)关于span1,x的最正确平方迫近多项式为*S(x)a0a1x(3)Qf(x)cosx,x0,11若(f,g)f(x)g(x)dx0且01,1x,，则有21,21,02123(0,1)1,22(f,0)0,(f,1)2,则法方程组为11a002211a1223从而解得a01.2159a10.24317故f(x)关于span1,x的最

55、正确平方迫近多项式为S*(x)a0a1x1.21590.24317x(4)Qf(x)lnx,x1,2若(f,g)2f(x)g(x)dx1且01,1x,则有21,27,02123(0,1)3,2(f,0)2ln21,(f,1)2ln23,4则法方程组为13a02ln212337a12ln2423从而解得a00.6371a10.6822故f(x)关于span1,x最正确平方迫近多项式为S*(x)a0a1x0.63710.6822x18。f(x)sinx，在1,1上按勒让德多项式睁开求三次最正确平方迫近多项式。2解：Qf(x)sinx,x1,12按勒让德多项式P0(x),P1(x),P2(x),P3(x)睁开121(f(x),P0(x)sinxdxcosx012211xsinxdx8(f(x),P1