计算方法-课后习题答案

计算方法习题答案王新民术洪亮编吉大仪电春哥2012/5/10。精选资料，欢迎下载第一章习题答案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)0.008026xfRxxexxxxxxtxxxxtxxxxtxRx取并令可知当时，有极大值。精选资料，欢迎下载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270201120120102101220217()|()()()()()()()()()()()()(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)0.008796fRfxxfxR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。。精选资料，欢迎下载注意到：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。解：（1）Lagrange插值多项式330()()jjjLxlxy30,()jiiijijxxlxxx3120010203124()010204xxxxxxxxxlxxxxxxx=3271488xxx0321101213024()101214xxxxxxxxxlxxxxxxx=32683xxx0312202123014()202124xxxxxxxxxlxxxxxxx=32544xxxx0124()fx19233。精选资料，欢迎下载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00111982223143343-1081143001001201()()(,)()(,,)()()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。精选资料，欢迎下载031220212300.52()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)!nnnnfRxfxLxxn[,]ab可知：当3n时，(1)(4)()()4!nxffx301234!()()()()()()(31)!Lxfxxxxxxxxx4(1)(0)(1)(2)xxxxx3222xxx8.求作1nfxx关于节点0,1,,ixin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的Lagrange插值基函数。解：注意到1()nfxx关于节点0,1,,ixin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。。精选资料，欢迎下载附加题：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。9.已知753()321fxxxx，求差商0172,2,,2f和0182,2,,2f。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!ffff10.已知10()(),(0,1,,)nniiifxxxxxin互异，求01,,,pfxxx。其中1pn。（此题有误。）（见王能超《教程》P149－题2）解：因为10()(),(0,1,,)nniiifxxxxxin，则1()()jnjfxx由差商性质01'01()(,,...,)()!nnjnjnjfxffxxxxn可知，01'01()(,,...,)0,0,1,...,()pjpjnjfxfxxxpnx。精选资料，欢迎下载(1)(1)0011[()]()(1)!(,,...,)1(1)!(1)!(1)!nnininxxfnfxxxnnn而11.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10,0,1,,21,1knjjjxknknfx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!nngkngxxxknn13．设节点0,1,,ixin与点a互异，试对1fxax证明0101,,,,0,1,,kkiifxxxknax并给出fx的Newton插值多项式。解依差商的定义001()fxax，100110101010()()1111(,)()()()fxfxfxxxxxxaxaxaxax一般地，设010011(,,,)()()kkkiiiifxxxaxax。精选资料，欢迎下载则1210101110110101101010(,,,)(,,,)(,,,)111()11111kkkkkkiikiikiikkkiifxxxfxxxfxxxxxxxaxaxaxxxaxaxax故1fxax的Newton插