(完整版)数值分析第五版答案(全)

第一章绪论解：近似值x*的相对误差为6=e*=e*=x*一xrx*x*而lnx的误差为e(lnx*)=lnx*一lnx必1e*x*2．设x的相对误差为2%，求xn的相对误差。解：设f(x)=xn，则函数的条件数为C=|xf'(x)|pf(x)pnrpr且e(x*)为2rr3．下列各数都是经过四舍五入得到的近似数，即误差限不超过最后一位的半个单位，试指12345123454．利用公式(2.3)求下列各近似值的误差限：(1)x*+x*+x*,(2)x*x*x*,(3)x*/x*.12412324必必222222424111222123123231132111222242442x*243则何种函数的条件数为C===33p3rprrrx具有5位有效数字2R许的相对误差限为ᵰᵅ(ᵄ∗)=∗16．设Y=28，按递推公式Y=Y783(n=1,2,…)0nn11001001解：Y=Y783nn11001YY831989711101依次代入后，有Y=Y10078300021211111则tanN1,tanN.cm使其面积误差不超过1cm2？解：正方形的面积函数为A(x)x2(A*)2A*(x*).2cm，才能使其面积误差不超过1cm210．设S1gt2，假定g是准确的，而对t的测量有0.1秒的误差，证明当t增加时S的2绝对误差增加，而相对误差却减少。2(S*)gt2(t*)11rS*=2t*nnn一101000201021120012121y人108，这个计算过程不稳定。2112116y*c(x*2.53y*c(x*)=6y*c(x*)=30y*c(x*)=1若通过计算y值，则(3+22)3111通过计算后得到的结果最好。解2故 若改用等价公式.e(u*)第二章插值法0120120(x一x)(x一x)21(x一x)(x一x)62(x一x)(x一x)32021则二次拉格朗日插值多项式为Lx2yl(x)kk023537623-0.916291-0.510826-0.693147-0.356675-0.223144Xxln.54的近似值。01234012341221L(x)=f(x)l(x)+f(x)l(x)112210(x一x)(x一x)1(x一x)(x一x)2(x一x)(x一x)2021L(x)=f(x)l(x)+f(x)l(x)+f(x)l(x)20011222字，在此后的计算过程中产生一定的误差传播；另一方面，利用插值法求函数cosx的近似值时，采用的线性插值法插值余项不为0，也会有一定的误差。因此，总误差界的计算应综的因素。i0几54002(1)令f(x)=xk插值余项为2kk+1k2kkhk+1kkkk+1k2kk+1k22k12jjj=0jjj=0若插值节点为x,j=0,1,jnfxnLxx若插值节点为x,j=0,1,jnjjj=0又kn,njjj=0jjj=0kjjj=0i=0kjji=0j=0jjj=0k01Lxfxxxfxxx010n((11201011l201Jl201J4104解：若插值节点为x,x和x，则分段二次插值多项式的插值余项为23!i一1ii+1i一1ii+1i2633272nnn解：根据向前差分算子和中心差分算子的定义进行求解。4y(E1)4ynnj0jnj0j4njj0jnnynnnny8．如果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，并且m1f(x)0(l为正整数)。函数f(x)的Taylor展式为f(xh)f(f(xh)f(x)f(x)h1f(x)h21f(m)(x)hm1f(m1)()hm1又f(x)是次数为m的多项式ff(x)hf(x)h2f(m)(x)hm2f(x)(f(x))得kf(x)是mk次多项式9．证明(fg)=fg+gfkkkkk+1k明(fg)=fgfg=fgfg+fgfg=g(ff)+f(gg)gf+fgfg+gfk10．证明fg=fgfggfkknn00k+1kkk=0证明：由上题结论可知fg=(fg)gf：fgkk=((fg)gf)=(fg)gf(fg)=fgfgkkk+1k+1kk：(fg)kk=(fgfg)+(fgfg)++(fgfg)11002211nnn1n1=fgfgnn00：fg=fgfggfkknn00k+1kkkjn0j=0一x一xjj+1jj=0j=01nn一1n0nn12xxx12n12nn2nn12nxkjaO,(x)j=1jj=1nnj一一n23n13nn1一Ox=(x一x)(x一x)(x一x)(x一x)(xx)一njj1j2jj一1jj+1jn令g(x)=xk,g[x,x,,x]=xnxkj12nO,(x)j=1njj=1njf,(x)a12n又：xnxkj=1g[f,(x)a12nj=1jni=(j))()()()]n(,,=(j))()()()]n(,,,n2[,,]n(j)0()()()()n0],00j(1)若F(x)=cf(x)，则F[x,x,,x]cf[x,x,,x];01n01n(2)若F(x)=f(x)+g(x)，则F[x,x,,x]=f[x,x,,x]g[x,x,,x].01n01n01n()(1)[]fxj(x()(1)[]fxj(x-x)(x-x)fxx(x-x)jn1(x-x)jn12jj-jj-jj+011Fxxx=xFxx-xx-xx-xx-x=jj=jjj-jj+jnj00cf(xj(x-x)jn(x-x(x-x)jn(x-x)j=0j0jj-1jj+1=cx(n)=cx(n)()(j)(()(j)(jx-xx-xx-xx-x=-+j1=-+j1jj1n001n：Fxx=xFxx-xx-xx-xx-xj=jj=jjj-jj+jn=xnfx=xnfx+gxjx-xx-xx-xx-x0()0()()()0j=jjj-jj+jn0ffxj))(x-x)j)(x-x)jn()0(x-xjx-xx-xjj+01g(xj)01g(xj)n)n)+x)0(xj0jx)()x-xx)0(xj0jx)()x-x]x-x-[-nj1jj-nnk0101nn!010177!7!0188!15．证明两点三次埃尔米特插值余项是kkkk若x=[x,x]，且插值多项式满足条件H(x)=f(x),H,(x)=f,(x)3kk3kkH(x)=f(x),H,(x)=f,(x)3现把x看成[x,x]上的一个固定点，作函数kk1根据余项性质，有3kk+13kk+12即,(x)在[x,x]上有四个互异零点。故,(t)在(x,x)内至少有三个互异零点，x3又H(4)(t)=03：g(x)=f(4)(),(x,x)4!kk+1其中依赖于xkk+1kk0kk+1R(x)=f(4)()(xx)2(xx)24!kk+1RRxf(4)()(xx)2(xx)2kk+1(xx)2(xx)2maxf(4)(x)4!kk1axbaxb4!axb4axbh4maxf(4)(x)axb16．求一个次数不高于4次4axbh4maxf(4)(x)axb解：利用埃米尔特插值可得到次数不高于4的多项式010101x3jjjj(x)(12xx0)(xx1)20xxxx0101(x)(12xx1)(xx0)21xxxx1010(x)x(x1)201H(x)(32x)x2(x1)x2x32x23设P(x)H(x)A(xx)2(xx)201P(x)x32x2Ax2(x1)21A44h计算各节点间中点处的I(x)与f(x)值，并估计误差。hi0在小区间[x,x]上，分段线性插值函数为I(x)=f(x)+xxif(x)hxxixxi+1ii+1i+1iii+1各节点间中点处的I(x)与f(x)的值为hhhhhhmaxf(x)I(x)h2maxf,()ii+1xxxh85x5ii+1又又f(x)=f,(x)=24x24x3(1+x2)4ff(x)=,f(x)=21,223：maxf(x)I(x)5x5h418．求f(x)=x2在[a,b]上分段线性插值函数I(x)，并估计误差。h0nii+1ii0in1i=f(x)x2=I(x)=xxi+1f(x)+xxif(x)hxxixxi+1ii+1i+1ihii+1i+1ii差为maxf(x)I(x)maxf()h2ii+1xxxh8abiii+1f(x)=x2：maxf(x)I(x)h2axbh419．求f(x)=x4在[a,b]上分段埃尔米特插值，并估计误差。0nii+1ii0in1iI(x)=(xxi+1)2(1+2xxi)f(x)hxxxxiii+1i+1i+(xxi)2(1+2xxi+1)f(x)xxxxi+1i+1iii+1xxiiii+1+(xxi)2(xx)f,(x)xxi+1i+1h3i+1iihih3iii+1hih2i+1ihih2ii+1hif(x)I(x)h4!ii+11maxf(4)()(hi)4axb又axb又f(x)x4=：maxf(x)I(x)maxhi4h4axbh0in1161667085000547762457280XjYj值，并满足条件：M0M1M2M3MM0M1M2M3M4=1232343hh=j一1,入=jj一1jj一1j533=,=,=,=111425374924114253701223fxx]=0.715034040h120ff[xx]一f[xx]d71h+hf[xf[xx]一f[xx]h+hf[xf[xx]一f[xx]hh234h4343由此得矩阵形式的方程组为99232711225472求解此方程组得01234三次样条表达式为S(x)=M+M(x-x)3jj6hj+16hjjjjnj6hj+16hjj：将M,M,M,M,M代入得123404001234404由此得矩阵开工的方程组为0490)|求解此方程组，得01234又三次样条表达式为j6hj+16hjj+(y-Mh2jj)+(y-)x-xjj6hj+16hjj将M,M,M,M,M代入得1234aaiii01njbSxfxS,(x)]dxaaaaaxkxka=S,(x)[f,(x)S,(x)]bjb[f,(x)S,(x)]d[S,(x)]aaa2第三章函数逼近与曲线拟合2132伯恩斯坦多项式为nnk其中P(x)nxk(1x)nkkk1101022xPxPx(1x)3P(x)P(x)x3x33nkkk6322222| ()xxn (k1x-(1n| ()xxn (k1x-(1n(x)----(n1)(k1)-nn (a)()nk若nBfxxnfkP(x)nnk()||kk()||kkn(1)xx-n-knk0k=0k--+=((knnnkxnxk-n-k!--+=((knnnkxnxk-n-k!k0(1)(k-1)!nk-)nnk-)k1=|k120122(|1|)|()(|1|)|()a00|1|1||||||此方程组的系数矩阵为希尔伯特矩阵，对称正定非奇异，4。计算下列函数f(x)关于C[0,1]的f,f与f：w2w(2)f(2)f(x)=x-,2f=maxf(x)f=maxf(x)2007=72f=maxf(x)=f=j1f(x)dx021=4200236mf(x)在(0,m)内单调递减f(x)在(0,m)内单调递减f(x)在(m,1)内单调递减。f(x)在(m,1)内单调递减。x(m,1)f(x)0fmaxf(x)0x1nmnmmmnnf1f(x)dx10000n!m!2000==fx[0,1]内单调递减。f=maxf(x)==ef=j1f(x)dx0000e204e2ffxgxdxa而(f,f)=jbf,(x)f,(x)dxaaaaVheCab则aaa2211_t2a11nnn项式，并求T*(x),T*(x),T*(x),T*(x)。0123nnj1T*(x)T*(x)P(x)dx0nmnmx_x22j1T*(x)T*(x)p(x)dx0nm=j1T(t)T(t)_1nm=j1T(t)T(t)_1nm又切比雪夫多项式{T*(x)}在区间[0,1]上带权p(x)=1正交，且k1_x2j1T(x)T(x)j1T(x)T(x)d1_t2|2_1n_1n011001122233n(f,g)=j1f(x)g(x)p(x)dx0nnnnn一1中=(x(x),(x))/((x),(x))nnnnn=((x),(x))/((x),(x))011111jj1==85325a=(x3-2x,x2-2)/(x2-2,x2-2)25555j1(x3-2x)(x2-2)(1+x2)dx-155=j1(x2-2)(x2-2)(1+x2)dx-155x255j1(x2-2)(x2-2)(1+x2)dx-155-1==1670：Q(x)=x3-2x2-17x=x3-9x3570149。试证明由教材式(2.14)给出的第二类切比雪夫多项式族{u(x)}是[0,1]上带权nn1-x2j1U(x)U(x)1-x2dxmn-1-11-x20002冗=20jsinm1)9d{1cos(n+1)9}0n+10n+10n+10n+1n+1m0(n+1)20n+1n+10010。证明切比雪夫多项式T(x)满足微分方程nnnn切比雪夫多项式为n)212cos(arccos))212cos(arccos)nn32222T,x=nxT,x=nxnnxn12=nx=nx212T,x=nxnxn(1x)x2222nnn2222f(x)在闭区间[a,b]上连续：存在x,x[a,b]，使121axbf(x)=maxf(x),2axb取P=[f(x)+f(x)]21212理知1为f(x)的零次最佳一致逼近多项式。12。选取常数a，使maxx3ax达到极小，又问这个解是否唯一？0x10x11x1=f3233343434x22f(b)一f(a)2a=,a2几2几22f(a)+f(x)f(b)一f(a)a+x2a=2一a于是得f(x)的最佳一次逼近多项式为1几即13f(b)一f(a)222f(a)+f(x)f(b)一f(a)a+x2a=2一2于是得f(x)的最佳一次逼近多项式为2222222223332348||||||(a)||||||(a) 32348进而，f(x)的三次最佳一致逼近多项式为P*(t)，则f(x)的三次最佳一致逼近多项式为163若(f,g)=j1f(x)g(x)dx-101202125229(f,v)=1,(f,v)=1,(f,v)=1,01223(v,v)=1,(v,v)=2,(v,v)=2,01025127则法方程组为 |||||()1|||| ( (a2)解得22S*(x)aax2ax412xfxcosxfxlnx;x若(f,g)3f(x)g(x)dx10122,226,0212301(f,)ln3,(f,)2,01则法方程组为24aln3260从而解得011若(f,g)1f(x)g(x)dx00110212301201则法方程组为从而解得(a(a〈0la1.62441若(f,g)=j1f(x)g(x)dx001021230121"2(f,v)=0,(f,v)1"2则法方程组为从而解得〈0la=–0.2431701若(f,g)=j2f(x)g(x)dx10102123vv3,012014则法方程组为从而解得(a=-(a〈0la=0.6822118。f(x)=sin"x，在[-1,1]上按勒让德多项式展开求三次最佳平方逼近多项式。2""20123212一1222则01几22233几4从而f(x)的三次最佳平方逼近多项式为300112233几4几400被观测物体的运动距离与运动时间大体为线性函数关系，从而选择线性方程则021201则法方程组为(614.7)(a)(280) (14.753.63)(b)(1078) (14.753.63)(b)(1078)从而解得a855048b22.25376故物体运动方程为544544xi49.049.0yj用最小二乘法求形如sabx2的经验公式，并计算均方误差。则021201则法方程组为55327a271.4从而解得a0.9726046b0.0500351jj得分解物浓度与时间关系如下：时间时间t050505浓度01.274.642.162.86474.154.374.514.584.62观察所给数据的特点，采用方程两边同时取对数，则tltJtltJt021201则法方程组为(11-0.603975)(a*)(-87.674095)从而解得f(4,3,2,1,0,1,2,3),用FFT算法求{c}的离散谱。kkk4xxkkxkA18220222C2042204+22j23A4444444442220000733365823，用辗转相除法将R(x)=3x2+6x化为连分式。22x2+6x+6解R(x)=3x2+6x22x2+6x+6323x+220.753!5!7!得C=0,012133!611==55!1206132231423324153342516即–1––1– (6|||||–6011–60|||||从而解得kjk–jkj=0则00011202113031221360故又又则x1232!3!得01C==22!233!6213即1121613kjk一jkj=000210113121126故Rbx12113_1_1第四章数值积分与数值微分1.确定下列求积公式中的特定参数，使其代数精度尽量高，并指明所构造出的求积公式所具_101_10120求解求积公式的代数精度时，应根据代数精度的定义，即求积公式对于次数不超过m的多项式均能准确地成立，但对于m+1次多项式就不准确成立，进行验证性求解。 (1)若(1)jhf(x)dx则Af(_h)+Af(0)+Af(h)_101h_11从而解得(4_h_h101_101jhf(x)dx=jhx4dx=2h5_h_h5_1013_101hjhfxdxAfhAfAf(h)_101 (2)若j2hf(x)dx~Af(_h)+Af(0)+Af(h)_101_2h_113_11从而解得4 _2h_2h101_101_2h0j2hf(x)dx=j2hx4dx=64h53 2122从而解得122lx=0.5266lx22220002hffhah2[f(0)f(h)]=1h22003h[f(0)+f(h)]/2+ah2[f(0)f(h)]=1h32ah22321jhfxdxjhxdx=1h4004hffh2+1h2[f(0)f(h)]=1h41h4=1h412244jhfxdx=jhx4dx=1h5005h[f(0)+f(h)]/2+1h2[f(0)f(h)]=1h51h5=1h512236jhf(x)dxh[f(0)+f(h)]/2+1h2[f(0)f(h)],12jhfxdx必h[f(0)+f(h)]/2+1h2[f(0)f(h)]122.分别用梯形公式和辛普森公式计算下列积分：复化梯形公式为0x10(1)n=8,a=0,b=1,h=1,f(x)=x84+x2复化梯形公式为Thfafxfb0.1114082k复化辛普森公式为810x复化梯形公式为102k复化辛普森公式为kk10kk复化梯形公式为42k复化辛普森公式为k=02k=146k=02k=1636T=h[f(a)+25f(x)+f(b)]=1.0356262k复化辛普森公式为Shfafxfx)+f(b)]=1.03577k66k3。直接验证柯特斯教材公式(2。4)具有5交代数精度。柯特斯公式为jbf(x)dx=ba[7f(x)+32f(x)+12f(x)+32f(x)+7f(x)]a9001234a90ba[7f(x)+32f(x)+12f(x)+32f(x)+7f(x)]=ba9001234aa2ba[7f(x)+32f(x)+12f(x)+32f(x)+7f(x)]=1(b2a2)900123