§2.6-Hermite插值

1第六节Hermite插值2©2009,HenanPolytechnicUniversity2§6Hermite插值第二章插值法许多实际问题不但要求插值函数p(x)在插值节点处与被插函数f(x)有相同的函数值p(xi)=f(xi)(i=0,1,2,…,n),而且要求在有些节点或全部节点上与f(x)的导数值也相等，甚至要求高阶导数值也相等，能满足这种要求的插值问题就称为埃尔米特插值(Hermite)。法1822-19013©2009,HenanPolytechnicUniversity3§6Hermite插值第二章插值法2.6.1三次埃尔米特插值多项式设y=f(x)是区间[a,b]上的实函数,x0,x1是[a,b]上相异两点,且x0＜x1,y=f(x)在xi上的函数值和一阶导数值分别为yi=f(xi)(i=0,1)和mi=f(xi)(i=0,1),求三次多项式H3(x),使其满足：33()(0,1)()iiiiHxyiHxmH3(x)称为三次埃尔米特插值多项式。4©2009,HenanPolytechnicUniversity4§6Hermite插值第二章插值法构造三次埃尔米特插值多项式如下:300110011()()()()()Hxyxyxmxmx条件函数函数值导数值x0x1x0x10(x)10001(x)01000(x)00101(x)0001构造三次埃尔米特插值多项式如下:300110011()()()()()Hxyxyxmxmx模仿拉格朗日插值多项式的构造思想5©2009,HenanPolytechnicUniversity5§6Hermite插值第二章插值法由0)()(1010xx可将它写成2100))](([)(xxxxbax21000)(11)(xxax，得由，所以）（，得再由3100020)(xxbx21010100)](21[)(xxxxxxxxx20101011)](21[)(xxxxxxxxx）（将同理10xx6©2009,HenanPolytechnicUniversity6§6Hermite插值第二章插值法，可令同样由0)()()(101000xxx2100))(()(xxxxcx，再由1)(00x210)(1xxc得,))(()(210100xxxxxxx201011))(()(xxxxxxx7©2009,HenanPolytechnicUniversity7§6Hermite插值第二章插值法210100))(()(xxxxxxx201011))(()(xxxxxxx2201000022101111()[12()]()()()()()[12()]()()()()xlxlxxxxlxxlxlxxxxlx,)](21[)(21010100xxxxxxxxx,)](21[)(20101011xxxxxxxxx即)(),(10xlxl插值点的Lagrange),(),,(1100yxyx为以一次基函数.8©2009,HenanPolytechnicUniversity8§6Hermite插值第二章插值法可得满足条件的三次埃尔米特插值多项式为300110011()()()()()Hxyxyxmxmx2200110110010110220100110110[12]()[12]()()()()()xxxxxxxxyyxxxxxxxxxxxxmxxmxxxxxx9©2009,HenanPolytechnicUniversity9§6Hermite插值第二章插值法定义已知n+1个互异点上的函数值和导数值,若存在一个次数不超过2n+1的多项式H(x)，满足则称H(x)为f(x)的2n+1次埃尔米特(Hermite)插值。bxxxan10)(ixf)(ixf),,1,0()()(),()(nixfxHxfxHiiii2.6.2一般埃尔米特插值多项式10©2009,HenanPolytechnicUniversity10§6Hermite插值第二章插值法H2n+1(x)=a0+a1x+a2x2+…+a2n+1x2n+1由2n+2个条件来确定2n+2个系数a0,a1,a2,…a2n+1显然非常复杂,所以要用求Lagrange插值多项式的基函数的方法,求插值基函数i(x)及i(x)(i=0,1,2,…,n)共有2n+2个,设每一个基函数为次数不超过2n+1次的多项式，且满足条件上式给出了2n+2个条件，可惟一确定一个次数不超过2n+1的多项式:11©2009,HenanPolytechnicUniversity11§6Hermite插值第二章插值法0)(10)(0)(10)(jiijjijiijjixjijixxjijix(i,j=0,1,2,…,n)Hermite插值多项式可写成插值基函数表示的形式niiiiinxfxxfxxH012)()()()()(12©2009,HenanPolytechnicUniversity12§6Hermite插值第二章插值法)(1)(21)(20xlxxxxxjnjkkkjjj)()()(2xlxxxjjj根据插值条件可求出和)(xj),1,0)((njxj13©2009,HenanPolytechnicUniversity13§6Hermite插值第二章插值法)()()()()(1)(21)(0220012jnjjjjjnjnjkkkjjnxfxlxxxfxlxxxxxHH2n+1(x)为满足条件的2n+1次Hermite插值多项式。),,1,0()()(),()(nixfxHxfxHiiii14©2009,HenanPolytechnicUniversity14§6Hermite插值第二章插值法定理满足插值条件的Hermite插值多项式是惟一的。),,1,0()()(),()(nixfxHxfxHiiii)(12xHn)(12xHn)()()(1212xHxHxnn证:设和都满足上述插值条件,令则每个节点均为的二重根,),,1,0(nkxk)(x15©2009,HenanPolytechnicUniversity15§6Hermite插值第二章插值法)(x0)(x)()(1212xHxHnn)(x即有2n+2个根，但是不高于2n+1次的多项式，所以惟一性得证。16©2009,HenanPolytechnicUniversity16§6Hermite插值第二章插值法定理若f(x)在a,b上存在2n+2阶导数，则2n+1次Hermite插值多项式的余项为)()!22()()()()(2)22(1212xnfxHxfxRnnn)())(()(10nxxxxxxx),(ba其中定理的证明可仿照Lagrange插值余项的证明方法17©2009,HenanPolytechnicUniversity17§6Hermite插值第二章插值法例2已知f(x)=x1/2及其一阶导数的数据见下表,用埃尔米特插值公式计算1251/2的近似值,并估计其截断误差.x121144f(x)1112f'(x)1/221/24解23121144()1112144121121144xxHx21441211212121144144121xx2112114422144121121144xx2114412124121144144121xx18©2009,HenanPolytechnicUniversity18§6Hermite插值第二章插值法得3125(125)11.18035H由2/7)4(1615)(xxf可求得223323151(125)4193841615190.00001238412111R2233322221112()2219144265212123231112114414412122232423Hxxxxxxxxx19©2009,HenanPolytechnicUniversity19§6Hermite插值第二章插值法要求在1个节点x0处直到m0阶导数都重合的插值多项式即为Taylor多项式00)(!)(...))(()()(000)(000mmxxmxfxxxfxfx其余项为)1(00)1(00)()!1()()()()(mmxxmfxxfxR注：20©2009,HenanPolytechnicUniversity20§6Hermite插值第二章插值法§3HermiteInterpolation例：设x0x1x2,已知f(x0)、f(x1)、f(x2)和f’(x1),求多项式P(x)满足P(xi)=f(xi)，i=0,1,2，且P’(x1)=f’(x1),并估计误差。模仿Lagrange多项式的思想，设解：首先，P的阶数=3210iii3())(()()()xhx1f’xhxfxPh0(x)有根x1,x2，且h0’(x1)=0x1是重根。)()()(22100xxxxCxh又:h0(x0)=1C0)()()()()(202102210xxxxxxxxxhh2(x)h1(x)有根x0,x2))()(()(201xxxxBAxxh由余下条件h1(x1)=1和h1’(x1)=0可解。与h0(x)完全类似。(x)h1有根x0,x1,x2h1))()(()(2101xxxxxxCxh1又:’(x1)=1C1可解。其中hi(xj)=ij,hi’(x1)=0,(xi)=0,’(x1)=1h1h1),())()(()()()(221033xxxxxxxKxPxfxR!4)()()4(xfxK与Lagrange分析完全类似