计算方法课后习题答案

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习题一3.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。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()fx192330123303132012()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。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.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!ffxLxxaxbab有''()()()()2ffxxaxb,令()()()[,]txxaxbxab()2()0()2txxababxtx时有极大值21max()=max()max()()21max()()()2221=()max()8axbaxbaxbaxbaxbfxfxxaxbababfxabbafx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则1210101110110101101010(,,,)(,,,)(,,,)111()11111kkkkkkiikiikiikkkiifxxxfxxxfxxxxxxxaxaxaxxxaxaxax故1fxax的Newton插值多项式为001001011001100101100()()(,)()(,,,)()()()()()()1()()()()()1nnnnnknikikiNxfxfxxxxfxxxxxxxxxxxxxxxxxaxaxaxaxaxaxxxaxax16.求作满足条件1(0)1,(0),(1)2,(1)2.2HHHH的插值多项式Px。解法1:根据三次Hermite插值多项式:22001130101011010220100110110()(12)()(12)()()()()()xxxxxxxxHxyyxxxxxxxxxxxxxxyxxyxxxx并依条件1(0)1,(0),(1)2,(1)2.2HHHH,得2222331()(12)(1)2(32)(1)2(1)211122Hxxxxxxxxxxx解法2:由于010,1xx,故可直接由书中(3.9)式,得''3001100112222311211232112211122HxAxyAxyBxyBxyxxxxxxxxxx18.求作满足条件333301,12,29,13HHHH的插值多项式3Hx,并估计其误差。解法1:由已知条件x012y129y3用基函数方法构造3xH。令300112211xAxyAxyAxyBxyH其中,0121,,,AxAxAxBx均为三次多项式,且满足条件0000AAAAAAAAABBBBAAA依条件可设2012xCxxA,由00=1,A可得:2011C=-,1222xxxA同理,212112,1,122xxxxxxxxxxAAB231121221232xxxxxxxxH21192xx31x误差为:4233124!fxfxHxxxxR解法2:用承袭性构造3xH由条件33301,12,29HHH先构造一个二次多项式2()Nx作差商表:iix()iPx一阶差商二阶差商001112122973于是有:22()11(0)3(0)(1)321Nxxxxxx令所求插值多项式32012()()()()xNxcxxxxxxH利用剩下的一个插值条件313H,得21101231()()()Nxcxxxxfx由此解出31211012()341()()1012fxNxcxxxx故有32()()(1)(2)1PxNxxxxx19.求作满足条件33000,1,1,2kkiiHxfxiHxfxk的插值多项式Px。并给出插值余项。解:令320200003202fxHxfxfxxxxxHxHxcxx利用插值条件311Hxfx定出:1230fxHxcxx注意到这里0x是三重零点,1x是单零点,故插值余项为433014!ffxHxxxxx20.求作次数4的多项式Px,使满足条件01,10,02,110,140PPPPP并列出插值余项。解法1:由于在0x处有直到一阶导数值的插值条件,所以它是“二重节点”;而在1x处有直到二阶导数值的插值条件所以1x是“三重节点”。因此利用重节点的差商公式:011011,,...,1,,...,,,...,,!kkkxxxxkfxfxxxfxxxxk+=lim可以作出差商表ixifx一阶二阶三阶四阶00111-1-1000-21101039206115根据Newton插值多项式,有2000000102220011010011101,,,,,,(),,,,()Pxfxfxxxxfxxxxxfxxxxxxxxfxxxxxxxxx22221236(1)5(1),Pxxxxxxx且插值余项为352115!fxPxfxx第二章答案1.计算下列函数fx关于0,1C的12,,fff:注:max,axbffx1baffxdx,1222baffxdx3101112231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