数值计算方法》习题答案

课后题答案详解吉林大学《数值计算方法》《数值计算方法》第一章课后题答案1第一章习题答案1.已知(1)2,(1)1,(2)1fff−===，求()fx的Lagrange插值多项式。解：由题意知：()012012120010202110120122021201,1,2;2,1,1()()(1)(2)()()6()()(1)(2)()()2()()(1)(1)()()3(1)(2)(1)(2)()2162njjjxxxyyyxxxxxxlxxxxxxxxxxlxxxxxxxxxxlxxxxxxxxLxylx==−=====−−−−==−−−−+−==−−−−−+−==−−−−+−==×+×−∴∑()2(1)(1)131386xxxx+−+×=−+2.取节点01210,1,,2xxx===对xye−=建立Lagrange型二次插值函数，并估计差。解11201201210,1,;1,,2xxxyyeye−−======1)由题意知：则根据二次Lagrange插值公式得：02011201201021012202110.510.520.51()()()()()()()()()()()()()2(1)(0.5)2(0.5)4(1)(224)(43)1xxxxxxxxxxxxLxyyyxxxxxxxxxxxxxxxxexxeeexeex−−−−−−−−−−−−=++−−−−−−=−−+−−−=+−+−−+22)Lagrange根据余项定理，其误差为(3)2210122()1|()||()||(1)(0.5)|3!61max|(1)(0.5)|,(0,1)6()(1)(0.5),()330.50330.2113()61()0.2113(0.21131)(0.21130.5)0.008026xfRxxexxxxxxtxxxxtxxxxtxRxξξωξ−+≤≤==−−≤−−∈′=−−=−+=−==≤××−×−=∴取并令可知当时，有极大值3.已知函数yx=在4,6.25,9xxx===处的函数值，试通过一个二次插值函数求7的近似值，并估计其误差。解：0120124,6.25,9;2,2.5,3yxxxxyyy=======由题意知：（1）采用Lagrange插值多项式220()()jjjyxLxlxy==≈=∑《数值计算方法》第一章课后题答案2270201120120102101220217()|()()()()()()()()()()()()(76.25)(79)(74)(79)(74)(76.25)22.532.2552.252.752.7552.6484848xyLxxxxxxxxxxxxxyyyxxxxxxxxxxxx==≈−−−−−−=++−−−−−−−−−−−−=×+×+××−××=其误差为(3)25(3)25(3)2[4,9]2()(7)(74)(76.25)(79)3!3()83max|()|40.0117281|(7)|(4.5)(0.01172)0.008796fRfxxfxRξ−−=−−−==∴=又则（2）采用Newton插值多项式2()yxNx=≈根据题意作差商表：iix()ifx一阶差商二阶差商04216.252.5292932114495−224(7)2(74)()(74)(76.25)2.64848489495N=+×−+−×−×−≈4.设()()0,1,...,kfxxkn==，试列出()fx关于互异节点()0,1,...,ixin=的Lagrange插值多项式。注意到：若1n+个节点()0,1,...,ixin=互异，则对任意次数n≤的多项式()fx，它关于节点()0,1,...,ixin=满足条件(),0,1,...,iiPxyin==的插值多项式()Px就是它本身。可见，当kn≤时幂函数()(0,1,...,)kfxxkn==关于1n+个节点()0,1,...,ixin=的插值多项式就是它本身，故依Lagrange公式有()000(),0,1,...,nnnkkkijjjjjijiijxxxlxxxknxx===≠−=≡=−∑∑∏特别地，当0k=时，有()0001nnnijjjijiijxxlxxx===≠−=≡−∑∑∏而当1k=时有()000nnnijjjjjijiijxxxlxxxxx===≠⎛⎞−⎜⎟=≡⎜⎟−⎜⎟⎝⎠∑∑∏5.依据下列函数表分别建立次数不超过3的Lagrange插值多项式和Newton插值多项式，并验证插值多项式的唯一性。《数值计算方法》第一章课后题答案3解：（1）Lagrange插值多项式330()()jjjLxlxy==∑30,()jiiijijxxlxxx=≠−=−∏3120010203124()010204xxxxxxxxxlxxxxxxx−−−−−−=••=••−−−−−−=3271488xxx−+−−0321101213024()101214xxxxxxxxxlxxxxxxx−−−−−−=••=••−−−−−−=32683xxx−+0312202123014()202124xxxxxxxxxlxxxxxxx−−−−−−=••=••−−−−−−=32544xxx−+−0123303132012()404142xxxxxxxxxlxxxxxxx−−−−−−=••=••−−−−−−=323224xxx−+()()()()()()()()()()()()()()()()()()()()()()()()()()()()()()32222321240241901020410121401401223320212440414212313243685432848114511442xxxxxxLxxxxxxxxxxxxxxxxxxxxxx−−−−−−=×+×+−−−−−−−−−−−−×+×−−−−−−=−−+−+−+−−++−+=−+−+（2）Newton插值多项式kkx()kfx一阶差商二阶差商三阶差商00111982223143343-10-8-11/43001001201()()(,)()(,,)()()Nxfxfxxxxfxxxxxxx=+−+−−0123012(,,,)()()()fxxxxxxxxxx+−−−1118(0)3(0)(1)(0)(1)(2)4xxxxxx=+−+−−−−−−32114511442xxx=−+−+由求解结果可知：33()()LxNx=说明插值问题的解存在且唯一。6.已知由数据1(0,0),(0.5,),(1,3)(2,2)y和构造出的Lagrange插值多项式()3Lx的最高次项系数是6，试确定1y。解：31200102030.512()00.50102xxxxxxxxxlxxxxxxx−−−−−−=××=××−−−−−−=3277122xxx−+−+0321101213012()0.500.510.52xxxxxxxxxlxxxxxxx−−−−−−=××=××−−−−−−=328(32)3xxx−+x0124()fx19233《数值计算方法》第一章课后题答案4031220212300.52()1010.512xxxxxxxxxlxxxxxxx−−−−−−=××=××−−−−−−=32252xxx−+−012330313200.51()2020.521xxxxxxxxxlxxxxxxx−−−−−−=××=××−−−−−−=32111326xxx−+3()Lx中最高次项系数为：1810(1)(2)32633y×−++−×+×=⇒1174y=7.设()4fxx=，试利用Lagrange余项定理给出()fx以1,0,1,2−为节点的插值多项式()3Lx。解：由Lagrange余项定理(1)1()()()()()(1)!nnnnfRxfxLxxnξω++=−=+[,]abξ∈可知：当3n=时，(1)(4)()()4!nxffxξξ+===301234!()()()()()()(31)!Lxfxxxxxxxxx=−−−−−+4(1)(0)(1)(2)xxxxx=−+−−−3222xxx=+−8.设[]2(),fxCab∈且()()0fafb==，求证21max()()max()8axbaxbfxbafx≤≤≤≤′′≤−证明：以,ab为节点进行线性插值，得()()()xbxaLxfafbabba−−=+−−1由于()()0fafb==,故1()0Lx=。于是由''1()()()()(),2!ffxLxxaxbξ−=−−abξ有''()()()()2ffxxaxbξ=−−，令()()()[,]txxaxbxab=−−∈()2()0()2txxababxtx′=−+=+=∴时有极大值21max()=max()max()()21max()()()2221=()max()8axbaxbaxbaxbaxbfxfxxaxbababfxabbafx≤≤≤≤≤≤≤≤≤≤′′•−−++′′=•−−′′−∴9.求作()1nfxx+=关于节点()0,1,,ixin=L的Lagrange插值多项式，并利用插值余项定理证明()()10001nnnniiiiixlx+===−∑∏式中()ilx为关于节点()0,1,,ixin=L的Lagrange插值基函数。《数值计算方法》第一章课后题答案5解：注意到1()nfxx+=关于节点()0,1,,ixin=L的插值多项式为11000()()()nnnnninjjjjjijiijxxLxxxlxxx++===≠−==−∑∑∏其插值余项为()()()()()1111000()1!nnnnnnnjjiijiixxxlxxxxxn++++===−=−=−+∑∏∏据此令0x=即得()()10001nnnniiiiixlx+===−∑∏。附加题：设()ilx为关于节点()0,1,...,ixin=的Lagrange插值基函数，证明()01,000,1,2,...,nkiiikxlkn==⎧=⎨=⎩∑证明：据题4可知，()01niilx=≡∑令0x=，则有()001niil=≡∑。注意到()()00,1,2,...,nkiiixxlxkn=−≡=∑（证明见王能超数值简明教程145页题6）令0x=即有()000nniiixl==∑。10.已知753()321fxxxx=+++，求差商()0172,2,,2fL和()0182,2,,2fL。解：根据差商与微商的关系，有(7)017(8)018()7!(2,2,...,2)1,7!7!()0(2,2,...,2)08!8!ffffξξ======11.已知()10()(),(0,1,,)nniiifxxxxxinω+===−=∏L互异，求()01,,,pfxxxL。其中1pn≤+。（此题有误。）（见王能超《教程》P149－题2）解：因为()10()(),(0,1,,)nniiifxxxxxinω+===−=∏L，则1()()jnjfxxω+′′=由差商性质()()01'01()(,,...,)()!nnjnjnjfxffxxxxnξω=+==∑可知，01'01()(,,...,)0,0,1,...,()pjpjnjfxfxxxpnxω=+===∑(1)(1)0011[()]()(1)!(,,...,)1(1)!(1)!(1)!nnininxxfnfxxxnnnξξ++=+−+====+++∏而12.设首项系数为1的n次式()fx有n个互异的零点()1,2,,ixin=，证明《数值计算方法》第一章课后题答案6()10,0,1,,21,1knjjjxknknfx==−⎧=⎨=−′⎩∑L证明：按题设，()fx有表达式()()1niifxxx==−∏故原式左端()()111kknnjjnjjjjiiijxxfxxx===≠=′−∑∑∏注意到上式右端等于()kgxx=关于节点()1,2,...,ixin=的1n−阶差商()12,,...,ngxxx（见第10页2.1式）利用差商与导数的关系（见2.11式）得知()()()(){1120,0,1,...,2,,...,1,11!nngkngxxxknnξ−=−=