李慶揚(yáng)數(shù)值分析第五版習(xí)題答案清華大學(xué)出版社

緒論解：近似值x*的相對(duì)誤差為6e*=e*=x*一xrx*x*lnxelnxlnxlnx1e*x*解：設(shè)f(x)=xn，則函數(shù)的條件數(shù)為C=|xf'(x)|pf(x)pnrprexrr3．下列各數(shù)都是經(jīng)過四舍五入得到的近似數(shù)，即誤差限不超過最后一位的半個(gè)單位，試指12345123454．利用公式(2.3)求下列各近似值的誤差限：(1)x*+x*+x*,(2)x*x*x*,(3)x*/x*.12412324必必222222424111222123123231132111222242442x*243則何種函數(shù)的條件數(shù)為C===33p3rprrr故度量半徑R時(shí)允許的相對(duì)誤差限為e(R*)=110.33r36．設(shè)Y=28，按遞推公式Y(jié)=Y783(n=1,2,…)nn1100100010097100依次代入后，有Y=Y10017831000100100021002111x=28783==128+78328+27.98255.982x效數(shù)字2N充分大時(shí)，怎樣求jN+jN解：正方形的面積函數(shù)為A(x)=x22m110．設(shè)S=gt2，假定g是準(zhǔn)確的，而對(duì)t的測(cè)量有士0.1秒的誤差，證明當(dāng)t增加時(shí)S的2絕對(duì)誤差增加，而相對(duì)誤差卻減少。2rS*t*01000201021120012121y定。211121若通過16y*e(x*ex若通過(3-22)3計(jì)算y值，則6y*e(x*)=30y*e(x*)=1若通過計(jì)算y值，則(3+22)3111通過計(jì)算后得到的結(jié)果最好。解ln12故 若改用等價(jià)公式x.e(u*)第二章插值法012012l(x)=(x_x)(x_x)12=_1(x+1)(x_2)0(x_x)(x_x)21(x_x)(x_x)62(x_x)(x_x)32021則二次拉格朗日插值多項(xiàng)式為L(zhǎng)(x)=x2yl(x)kk023623Xlnx用線性插值及二次插值計(jì)算ln0.54的近似值。01234012341xx122xx21L(x)=f(x)l(x)+f(x)l(x)11221ln0.54時(shí)，l(x)=(xx)(xx)12=50(x0.5)(x0.6)0(xx)(xx)1(xx)(xx)2(xx)(xx)2021L(x)=f(x)l(x)+f(x)l(x)+f(x)l(x)20011222字，在此后的計(jì)算過程中產(chǎn)生一定的誤差傳播；另一方面，利用插值法求函數(shù)cosx的近似值時(shí)，采用的線性插值法插值余項(xiàng)不為0，也會(huì)有一定的誤差。因此，總誤差界的計(jì)算應(yīng)綜的因素。i0則x9054002當(dāng)xx,x時(shí)，線性插值多項(xiàng)式為kk1L(x)f(x)xxk1f(x)xxkkxxk1xxkk1k1k插值余項(xiàng)為R(x)cosxL(x)f()(xx)(xx)12kk1(f*(x))105k2kxxk1xxkkk1kRxfxxxkkxxk1xxkkk1k(f*(x))(xxk1xxk1)kxxxxkhk1k(f*(x))k2kk1k(cos)(xx)(xx)(f*(x2kk1kx2kk1k1(1h)2(f*(x))22k12(1)nxkl(x)xk(k0,1,n,)jj(2)n(xx)kl(x)0(k0,1,n,)jj證明(1)令f(x)xk若插值節(jié)點(diǎn)為x,j0,1,jklxnjj插值余項(xiàng)為R(x)f(x)L(x)f(n1)()(x)nn(n1)!n1又kn,nnxkl(x)xk(k0,1,n,)jjjjkjjkjj又0in由上題結(jié)論可知((jjj=0k01L(x)=f(x)x_x1+f(x)x_x010x_x1x_x10又又112011201201)()2又11(x_(l01Jl04104h解：若插值節(jié)點(diǎn)為x,x和x，則分段二次插值多項(xiàng)式的插值余項(xiàng)為i_1ii+1R(x)f()(xx)(xx)(xx)23!i1ii1RR(x)(xx)(xx)(xx)maxf(x)26i1ii14x4設(shè)步長(zhǎng)為h，即xxh,xxhi1ii1iR(x)1e42h33e4h3.2633272nnn解：根據(jù)向前差分算子和中心差分算子的定義進(jìn)行求解。y2nnnnj0jj0jj0jE4jyny4njy4jynnynnnny8．如果f(x)是m次多項(xiàng)式，記f(x)f(xh)f(x)，證明f(x)的k階差分函數(shù)f(x)的Taylor展式為m!(m+1)! 又f(x)是次數(shù)為m的多項(xiàng)式12m!=f,(x)h+f,(x)h2++f(m)(12m!kkkkk+1k明編(fg)=fg一fg=fg一fg+fg一fg=g(f一f)+f(g一g)kkkkkknn00k+1kkk=0證明：由上題結(jié)論可知fg(fg)gfkkkkk1kfgkkk0((fg)gf)kkk1kk0(fg)gfkkk1k(fg)fgfgkkk1k1kk(fg)kkk0(fgfg)(fgfg)(fgfg)11002211nnn1n1fgfgnn00fgfgfggfkknn00k1kjn0yjj1j(yy)(yy)(yy)1021nn1n001n1n12n證明：nxkj0,0kn2;j1f(xj)n1,kn112n且f(x)aaxaxn1axn01n1nf(x)a(xx)(xx)(xx)n2nxxxxx(xx)n12n則f()aj(x則f()aj(x)j1jj1nnjxxxxn23n13n2n-1：O,(x)=(x-x)(x-x)(x-x)(x-x)(x-x)njj1j2jj-1jj+1jn令g(x)=xk,g[x,x,,x]=xnxkj12nO,(x)j=1nj則g[x,x,,x]=xnxkj12nO,(x)j=1nj又：xnxkj=1g[x,x,,x]f,(x)a12nj=1jn(1)若F(x)=cf(x)，則F[x,x,,x]=cf[x,x,,x];01n01nFxfxgxFxxxfxxxg[x,x,,x].01n01n01n (1)f[x,x,,x]=xnf(xj)12n(x-x)(x-x)(x-x)(x-x)j=0j0jj-1jj+1jnF[x,x,,x]=xnF(xj)12n(x-x)(x-x)(x-x)(x-x)j=0j0jj-1jj+1jn=xncf(xj)(x-x)(x-x)(x-x)(x-x)j=0j0jj-1jj+1jn=c(xnf(xj))(x-x)(x-x)(x-x)(x-x)j=0j0jj-1jj+1jn01n77!7!0,x]=xnF(xj)n(x-x)(x-x)(x-x)(x-x)j=0j0jj-1jj+1jn=xnf(xj)+g(xj)(x-x)(x-x)(x-x)(x-x)j=0j0jj-1jj+1jn=xnf(xj))(x-x)(x-x)(x-x)(x-x)j=0j0jj-1jj+1jn(x-x)(x-x(x-x)(x-x)(x-x)(x-x)j=0j0jj-1jj+1jn0n0nFi01nn!010188!15．證明兩點(diǎn)三次埃爾米特插值余項(xiàng)是R(x)=f(4))(x2)-(xx2)!,x(x,)3kk+1k+k1若x=[x,x]，且插值多項(xiàng)式滿足條件3kk3kkH(x)=f(x),H，(x)=f，(x)插值余項(xiàng)為R(x)=f(x)-H(x)3由插值條件可知R(x)=R(x)=0kk+1Rxgxxxxxk其中g(shù)(x)是關(guān)于x的待定函數(shù)，現(xiàn)把x看成[x,x]上的一個(gè)固定點(diǎn)，作函數(shù)(t)=f(t)H(t)g(x)(tx)2(tx)2kk+1根據(jù)余項(xiàng)性質(zhì)，有3kk+1=f(x)H(x)R(x)3(t)=f(t)H(t)g(x)[2(tx)(tx)2+2(tx)(tx)2]k2即(x)在[x,x]上有四個(gè)互異零點(diǎn)。故(t)在(x,x)內(nèi)至少有三個(gè)互異零點(diǎn)，x3又H(4)(t)=03x4!kk+11其中依賴于xR(x)f(4)()(xx)2(xx)24!kk1分段三次埃爾米特插值時(shí)，若節(jié)點(diǎn)為x(k0,1,,n)，設(shè)步長(zhǎng)為h，即kk0kk1R(x)f(4)()(xx)2(xx)24!kk1R(R(x)f(4)()(xx)2(xx)24!kk1xx)2(xx)2maxf(4)(x)4!kk1axbaxb4!axb4axbh4maxf(4)(x)axb16．求一個(gè)次數(shù)不高于4次的多項(xiàng)式P(x)，使它滿足4axbh4maxf(4)(x)axb解：利用埃米爾特插值可得到次數(shù)不高于4的多項(xiàng)式0101013jjjj(x)(12xx0)(xx1)20xxxx0101(x)(12xx1)(xx0)21xxxx1010(x)x(x1)203301144h計(jì)算各節(jié)點(diǎn)間中點(diǎn)處的I(x)與f(x)值，并估計(jì)誤差。hi0在小區(qū)間[x,x]上，分段線性插值函數(shù)為I(x)=f(x)+xxif(x)hxxixxi+1ii+1i+1iii+1各節(jié)點(diǎn)間中點(diǎn)處的I(x)與f(x)的值為hhhhhhmaxf(x)I(x)h2maxf()ii+1又f(x)=xxxii+1又f(x)=1：f(x)=2x,(1+x2)2f(x)=6x22(1+x2)3ff(x)=,f(x)=21,223：maxf(x)I(x)5x5h418．求f(x)=x2在[a,b]上分段線性插值函數(shù)I(x)，并估計(jì)誤差。h0nii+1ii0in1if(x)=x2I(x)=f(x)+xxif(x)hxxixxi+1ii+1i+1ihii+1i+1ii差為maxf(x)I(x)maxf,()h2ii+1xxxh8abiii+1f(x)=x2：maxf(x)I(x)h2axbh40nii+1ii0in1ixI(x)=()2(1+2xxi)f(x)hxxxxiii+1i+1i+(xxi)2(1+2)f(x)xxxxi+1i+1iii+1+()2(xx)f,(x)xxii+()2(xx)f,(x)xxi+1i+1h3i+1iihih3iii+1hih2i+1ihi if(x)I(x)h4!ii+11maxf(4)()(hi)4axbaxb又f(x)=x4maxf(x)一I(x)maxhi4h4axbh0in一11616XjYj值，并滿足條件：232343hhjjj一1j53311425374924114253701223fxx]=0.71503404f0h120ff[xx]f[xx]d=6,,1201=4.31571h+hf[xf[xx]f[xx]d=6,,2312=3.2640h+hf[xf[xx]f[xx]d=6,,3423=2.4300hh23xx4h4343由此得矩陣形式的方程組為22512359 23725214472M0M1MMM4=求解此方程組得01234三次樣條表達(dá)式為Sx)=M+M(xx)3jj6hj+16hjjj6hj+16hjj ：將M,M,M,M,M代入得2340401234404由此得矩陣開工的方程組為040)|求解此方程組，得01234又三次樣條表達(dá)式為j6hj+16hjj+(y-Mh2jj)+(y-)x-xjj6hj+16hjj將M,M,M,M,M代入得234aaiii01naaaaa2132伯恩斯坦多項(xiàng)式為0(0)3(30(0)3(3)1(0)2(1)nfkPxnnk其中P(x)=(|n)|xk(1-x)n-kk(k)1101 (0)22P(x)=||(1-0(0)3nkk=0"""k=0"""6322222nnfkPxnnk|1||1|xnknxk–x)n–kk=0n(k)xnknnnkxkx)n–knk!xnn(n–1)–(k–1)+1]xk(1–x)n–k(k–1)!k=1(k–1)012n(|1|||(|1|||此方程組的系數(shù)矩陣為希爾伯特矩陣，對(duì)稱正定非奇異，4。計(jì)算下列函數(shù)f(x)關(guān)于C[0,1]的f,f與f：w12f=maxf(x)f=maxf(x)20xx07=72f=maxf(x)=f=j1f(x)dx21=4f=(j1f2(x)dx)2023=6mxf(x)在(0,m)x(m,1)f(x)x(m,1)f(x)0fmaxf(x)0x1nmnmmmnnf1f(x)dx10000n!m!2000fx遞減。f=maxf(x)==ef=j1f(x)dx000000e204e2faa而(f,f)=jbf,(x)f,(x)dxa2)若(f,g)=jbf,(x)g,(x)dx+f(a)g(a)，則aKahaaaaan多項(xiàng)式，并求T*(x),T*(x),T*(x),T*(x)。0123nn221jj1T*(x)T*(x)P(x)dxjnm02j1T*(x)T*(x)p(x)dx0nm=j1T(t)T(t)nm=j1T(t)T(t)nmTx[0,1]上帶權(quán)p(x)=1正交，且1T(x)T(x)d1T(x)T(x)dnnm0001122233n(f,g)=j1f(x)g(x)p(x)dx10(x)=(xa)(x)(x)n+1nnnn1中a=(x(x),(x))/((x),(x))=((x),(=((x),(x))/((x),(x))nnnn1nnnn1n10111111==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-155255j1(x2-2)(x2-2)(1+x2)dx-155-1==1670：Q(x)=x3-2x2-17x=x3-9x3570149。試證明由教材式(2.14)給出的第二類切比雪夫多項(xiàng)式族{u(x)}是[0,1]上帶權(quán)nn1-x2j1U(x)U(x)1-x2dxmn-1-11-x20002冗=20jsinm1)9d{1cos(n+1)9}0n+10n+10n+10n+1n+10(n+1)2d0n+1n+10010。證明切比雪夫多項(xiàng)式T(x)滿足微分方程nnnn切比雪夫多項(xiàng)式為n11nxxnnnf(x)在閉區(qū)間[a,b]上連續(xù)：存在x,x[a,b]，使121axbf(x)=maxf(x),2axb取P=[f(x)+f(x)]21212理知0x10x11x1=f323334343422f(b)一f(a)2a=,a2幾2幾2f(a)+f(x)f(b)一f(a)a+xa2一2aba2于是得f(x)的最佳一次逼近多項(xiàng)式為1幾即13f(b)一f(a)22f(a)+f(x)f(b)一f(a)a+xba22于是得f(x)的最佳一次逼近多項(xiàng)式為221令t=2(x一)，則t=[一1,1]2222222gtPt3332348||||||(a) ||||||(a) *(t)=g(t)-T(t)32348進(jìn)而，f(x)的三次最佳一致逼近多項(xiàng)式為P*(t)，則f(x)的三次最佳一致逼近多項(xiàng)式為3168若(f,g)=j1f(x)g(x)dx01202125229(f,v)=1,(f,v)=1,(f,v)=1,01223(v,v)=1,(v,v)=2,(v,v)=2,01025127則法方程組為 |||||()1|||| ( (a2)解得22S*(x)aax2ax412xosxfxlnxx若(f,g)3f(x)g(x)dx10122,226,0212301則法方程組為aln326043a12從而解得0101若(f,g)1f(x)g(x)dx00110212301201則法方程組為從而解得(a(a〈0a01若(f,g)=j1f(x)g(x)dx00102123vv,0121"2(f,v)=0,(f,v)1"2則法方程組為從而解得〈0la=–0.2431701若(f,g)=j2f(x)g(x)dx10102123vv3,012014則法方程組為從而解得(a=-(a〈0la=0.6822012""2x3212一1222則0x1幾22233幾4從而f(x)的三次最佳平方逼近多項(xiàng)式為300112233幾4幾4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，從而選擇線性方程則02121則法方程組為(614.7)(a)(280) (14.753.63)(b)(1078 (14.753.63)(b)(1078)從而解得a048b22.25376故物體運(yùn)動(dòng)方程為xiyjsabx2的經(jīng)驗(yàn)公式，并計(jì)算均方誤差。則021201則法方程組為55327a271.4從而解得a0.9726046b0.0500351jj時(shí)間時(shí)間t05101520253035404550553.874.154.374.514.584.624.64觀察所給數(shù)據(jù)的特點(diǎn)，采用方程兩邊同時(shí)取對(duì)數(shù)，則tltJtltJt02120101則法方程組為(11-0.603975)(a*)(-87.674095)從而解得kkkkkxkA1222j2322A44444444423，用輾轉(zhuǎn)相除法將R(x)=3x2+6x化為連分式。22x2+6x+6解R(x)=3x2+6x22x2+6x+6=3123943x+3x223!5!7! 得C=0,01211C==33!6411C==55!1206132231423324153342516即–1––1– (6|||||616160從而解得kjk–jkjj=0則00011202113031221360故1232!3!得得0122!21133!6213即1121613kjk一jkjj=0則00210113121126xk[f(x)S(x)]xk+1xk故R(x)=1+bx1aaaa1.確定下列求積公式中的特定參數(shù)，使其代數(shù)精度盡量高，并指明所構(gòu)造出的求積公式所具(1)jhf(x)dxAf(h)+Af(0)+Af(h);h101(2)j2hf(x)dxAf(h)+Af(0)+Af(h);2h101(3)j1f(x)dx[f(1)+2f(x)+3f(x)]/3;0求解求積公式的代數(shù)精度時(shí)，應(yīng)根據(jù)代數(shù)精度的定義，即求積公式對(duì)于次數(shù)不超過m的多m1次多項(xiàng)式就不準(zhǔn)確成立，進(jìn)行驗(yàn)證性求解。(1)若(1)jhf(x)dxAf(h)+Af(0)+Af(h)h101_101_1013_11從而解得 _h_h_101_101jhf(x)dx=jhx4dx=2h5_h_h5_1013_101h_101 (2)若j2hf(x)dx如Af(_h)+Af(0)+Af(h)_101_2h3-11從而解得(4-2h-2h-101-101-2hjhfxdxjhx4dx=64h5-2h-2h5-1013-101-2h-101-2h (3)若j1f(x)dx如[f(-1)+2f(x)+3f(x)]/32-12-12從而解得(x=-0.2899(x=0.689922lx=0.5266lx=22-122-1(4)若jhf(x)dx如h[f(0)+f(h)]/2+ah2[f,(0)-f,(h)]00xh0022003fhahffhhah2h3h32ah2321aahfxdxhxdx1h4004h[f(0)f(h)]/2h2[f(0)f(h)]h4h4h412244fxx4，則