第三章-插值法-Hermite插值

第三章插值法第五节Hermite插值埃尔米特插值埃尔米特Hermite插值问题其中),,1,0(nixi互异，im为正整数，记，10mmnii)(xfy给定)1()1(1)1(0)1()1()1(1)1(0101010)()()(nimnmmmnnnfffxffffxffffxfxxxx函数值表及各阶导数值表如下：),()()()(ikikxfxP)1,,1,0;,,1,0(imkni寻求m次多项式P(x)使满足插值条件：Hermite插值问题共有m+1个条件)()(),()(iiiixfxPxfxP我们只讨论的情形。).(15一讨论Hermite插值问题],[)(1baCxfy函数表及导数表nnnyyyxfyyyxfxxxx101010)()(已知),,1,0()()(1212niyxHyxHiiniin其中互异,寻求),,1,0(nixi12n次多项式使满足)(12xHn插值条件:).(25],[)(1baCxf且已知)(xf函数表及导数表，如果12n次多项式满足插值条件(5.2).则存在唯一次数不超过)(12xHn证明：先证唯一性。下证存在性。（用构造法，同构造L-插值多项式的方法）),,1,0(,0)(,0,1)(nkxjkjkxkjkj时当时当的12n次多项式。),,.1,0(),(njxj第一，求Hermite插值基函数1.求满足插值条件：问题定理其中C为待定常数,,于是可令由为)(xj的二重零点且njjxxxxx,,,,,,11101)(jjx).(35)(xljnjiiijixxxx0(5.3)式求导，得)()(]1)([2)()(2xlxlxxcxclxjjjjj，得由0)(jjx)()(2)()(02jjjjjjjjxlxlxclx)(2)()(2jjjjjjxlxlxlc所以),,1,0()(]1)(21[)(20njxlxxxxxjnjiiijjj))(1(jxxc）（xlj2njiiijxx012221212120)()()()()(njjxxxxxxxxxx221212120221212120)()()()()()()()()()())(1()(njjjjjjjnjjjjxxxxxxxxxxxxxxxxxxxxxxcx221212120221212120)()()()()()()()()()(njjjjjjjnjjxxxxxxxxxxxxxxxxxxxx(2)已知，001000)(0000)(10xfxfxxxxxnj由于njjxxxxx,,,,,,1110为)(xj的二重零点且,0)(jjx])()()()[(12212120njjjjjjxxxxxxxxA又由1)(jjx，则有221212120)()()()()()(1njjjjjjjjjxxxxxxxxxxAx).(45212120)()())(()(jjjxxxxxxxxAx221)()(njxxxx则可令),,1,0(),()()(2njxlxxxjjj于是求12n次多项式),,2,1,0()(njxj,使满足插值条件：),,1,0(,0)(nkxkj时当时当jkjkxkj,0,1)(第二，求多项式)(12xHnnjjjjjnyxyxxH012)')()(()().(55injjijjijinyyxyxxH')')(')('()('012),,1,0(ni（满足插值条件(5.2)的多项式）)(12xHnnjijijiijinyyxyxxH012)')()(()(事实上,有即(5.5)式是满足插值条件(5.2)的插值多项式.所以存在2n+1次多项式满足插值条件(5.2).#；)()()(2xlxxxjjj),,1,0(),(),(njxxjj为Hermite插值基函数,即其中ijinjiijxxxxxl0)(？；)()1)(21()(20xlxxxxxjnjiiijjjHermite插值余项)(12xHbn）（为Hermite插值多项式，证明与拉格朗日余项公式证明类似.举例时再证.存在，于设),()(],,[)()()22(12baxfbaCxfann],,[(baxi),,,1,0互异ixni有关。且与xba),(),()!22()(21)22(xnfnn22120)22()()()()!22()(nnxxxxxxnf)()()(1212xHxfxRnn则定理二带导数的两点插值（特例:）1n],[)(1baCxf函数表及导数表111)(')(kkkkkkmmxfyyxfxxx已知).(75求3次多项式)(3xH使满足插值条件:11331133)(',)(')(,)(kkkkkkkkmxHmxHyxHyxH).(85).(95)(3xH存在且唯一，表达式为)()()()()(11113xmxmxyxyxHkkkkkkkk21111))(21()(kkkkkkkxxxxxxxxx211))(()(kkkkkxxxxxxx,))(()(2111kkkkkxxxxxxx],[1kkxxx2111))(21()(kkkkkkkxxxxxxxxx其中).(105；)()()(2xlxxxjjj；)()1)(21()(20xlxxxxxjniijjjjiijinjiijxxxxxl0)(问题结论余项公式为:2124334)()(!)()()()()(kkxxxxfxHxfxR例已知)(xfy函数表及导数表1210210)()(fxfyyyxfxxxx1133')(')2,1,0(,)(fxpiyxPii)(3xP使满足插值条件：求次数不超过3的多项式已知),(),,(),,(221100yxyxyx三点，由牛顿插值多项式，可确定2次多项式，在此基础上，增加了节点，则增加三次项即可，并使前三个插值条件不受影响。分析解：))((,,)(,)()(1021001002xxxxxxxfxxxxfxfxP221100,,,,,yxyxyx的2次牛顿插值多项式为过3点21210,,xxxxxf2110,,,xxxxf1021001003)(,,,)()(xxxxxxxfxxxxfxfxP))(()(,,,)(210101210101xxxxxxxxxfxxfxfA12101012101013))(()(,,,)(fxxxxAxxxxxfxxfxP由---带重节点的牛顿插值多项式1021001003)(,,,)()(xxxxxxxfxxxxfxfxP设所求多项式为113)(fxP确定A.再由条件],[11xxf重节点定义)(1xf011011,,xxxxfxxf110,,xxxf21210110,,,,xxxxxfxxxf重节点定义))()((210xxxxxxA))()(](,,,,[2102110xxxxxxxxxxf)(01xx)(01xx插值余项（误差估计）:存在。，）（)(],[)(43xfbaCxfxxx且依赖于20,。)())((!4)()()()(2210)4(3xxxxxxfxPxfxR;00)(,)2,1,0(1右端时）当（iixRixx,)2,1,0(2时）当（ixxi)())()(()()()(22103xtxtxtxktPtft构造函数（作辅助函数):至少有四个互异根)(t至少有一个根，）（)(4t0,420）（使即至少存在一点）（xx!4)(4xkf）（）（!4)(4）（）（fxk)())((!4)()(2210)4(xxxxxxfxR设)())()(()(2210xxxxxxxkxR,其中)(xk为待定函数。的根是的根，且为)()(,,,1210txtxxxx则条件结论证明P.8815作业:(1)理解H-插值多项式的构造方法（基函数法与例的方法）；(2)能根据具体条件求出插值多项式及插值余项。Hermite插值（以mi=2,i=0,1,…,n为例）；)()()(2xlxxxjjj),,1,0(),(),(njxxjj为Hermite插值基函数,即其中)())(()(xlxxxxxjnjiiijjj20121ijinjiijxxxxxl0)(njjjjjnyxyxxH012)')()(()(Hermite插值余项：有关。且与xba),(),()!()()(xnfnn21222222120)22()()()()!22()(nnxxxxxxnf)()()(1212xHxfxRnn结论