【正文】 ???kkkkkkkk yxxxxyxxxxxL ( 2 . 3 ) ),()()( 2 . 2 )( ,)( 1111111xlyxlyxLxxxxxlxxxxxlkkkkkkkkkkkk??????????????則所求線性插值多項(xiàng)式）（令 ( 2 . 3 ) ),()()( 2 . 2 )( ,)( 1111111xlyxlyxLxxxxxlxxxxxlkkkkkkkkkkkk??????????????則所求線性插值多項(xiàng)式）（令.1,)(0)( 0)(1)( )()(11111線性插值基函數(shù)稱為，并滿足也是線性插值多項(xiàng)式，和其中?????????kkkkkkkkkkxlxlxlxlxlxl.幾何意義.)( ,)( ,)( )(, ,21122112211??????????kkkkkkkkkyxLyxLyxLxLxxxn，滿足二次插值多項(xiàng)式，要求假定給定插值節(jié)點(diǎn)時(shí)再考察( 2 . 4 ) 1.)(0)(0)( 0,)(1)(0)( 0)(0)(1)( )()(),(11111111111111????????????????????????????kkkkkkkkkkkkkkkkkkkkkxlxlxlxlxlxlxlxlxlxlxlxl，，并滿足是二次函數(shù)，和采用基函數(shù)法，基函數(shù)??111 1 1( ) ( ) ( ) .( ) ( )kkkk k k kx x x xlxx x x x??? ? ?????? 同理 .))(())(()( ,))(())(()( 111111111kkkkkkkkkkkkkkxxxxxxxxxlxxxxxxxxxl???????????????????.幾何圖示 ( 2 . 5 ) ),()()()( 11112 xlyxlyxlyxL kkkkkk ???? ???項(xiàng)式于是，所求二次插值多. ))(())(( ))(())(( ))(())(()( ,111111111111112kkkkkkkkkkkkkkkkkkkkkxxxxxxxxyxxxxxxxxyxxxxxxxxyxL?????????????????????????????也就是( 2 .7 ) ),1,0,( ,1 ,0)( ),(nkikikixlxlnikk????????滿足次個(gè)仍采用基函數(shù)法，求一 插值基函數(shù) ),())(()()( 110 nkkkk xxxxxxxxAxl ????? ?? ??可知二、拉格朗日插值多項(xiàng)式 ( 2 .6 ) ).,1,0( ,)( )(1 10niyxLxLnxxxniinnn????????，滿足次插值多項(xiàng)式要求，個(gè)插值節(jié)點(diǎn)對(duì)于給定的一般情況， ),())(()()( 110 nkkkk xxxxxxxxAxl ????? ?? ??可知( 2 . 8 ) ),1,0( )())(()()())(()()( ,1)(110110nkxxxxxxxxxxxxxxxxxlAxlnkkkkkknkkkkkk????????????????????于是得到由 ).(,),(),(1,11010xlxlxlnnxxxnnn?? 朗日基函數(shù)拉格次個(gè)上的個(gè)節(jié)點(diǎn)從而得到在 ??.)(( 2 . 9 ) )( )( 0拉格朗日插值多項(xiàng)式次稱為次插值多項(xiàng)式于是，所求nxLxlyxLnnnkkkn ???( 2 . 8 ) ),1,0( )())(()()())(()()( ,0110110nkxxxxxxxxxxxxxxxxxxxxxlnkjj jkjnkkkkkknkkk?????????????????????????也就是( 2 . 1 0 ) )())(()( 101 nn xxxxxxx ????? ??引入記號(hào))())(()()( 1101 nkkkkkkkn xxxxxxxxx ?????? ??? ???則得( 2 . 1 1 ) )()()( )( 0 11?? ?????nk knknkn xxxxyxL??于是.,0 , 1 , ,)( ,10nmxxlx mnkkmk ?????得由定理1)(0???nkk xl 由插值多項(xiàng)式的存在唯一性 取 m=0得： 練習(xí) 給定數(shù)據(jù)表 xi 0 1 2 3 yi 0 1 5 14 求三次拉格朗日插值多項(xiàng)式 L3(x). 123)2)(1(14)1(12)3)(1(5)2()1(1)3)(2(10??????????????????? xxxxxxxxx)(14)(5)(1)(0)(32103 xlxlxlxlxLn????????? 解 并代入數(shù)據(jù)表值得）中，?。涸冢?.12)(1(616 )132( 2?????? xxxxxx0 1 100 1 1( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )njk k nkjk k k k k k n k jjkxxx x x x x x x xlx x x x x x x x x x???????? ? ? ???? ? ? ? ??思考 : 進(jìn)一步求函數(shù) x= ? 思考 : 誤差估計(jì) ? ))(()()( 3 ?????? Lf. ),(( 2 . 1 4 ) ,)()!1()()()()( ],[ ,)( 2 . 61)()( ,),()( ,],[)( 0)1(10)1()(xbaxxnfxLxfxRbaxbxxxanxfxLbaxfbaxfnjjnnnnnnn且依賴于其中插值余項(xiàng)則對(duì)于任何的插值多項(xiàng)式足條件上的滿個(gè)節(jié)點(diǎn)在是內(nèi)存在在設(shè)上連續(xù)在設(shè)????????????????????定理2三、插值余項(xiàng)與誤差估計(jì) ( 2 . 1 6 ) |,)(|)!1(|)(| ,|)(|ma x1n11)1(xnMxRMxfnnnnbxa??????????則若 證明： P26 ( 2 . 1 7 ) ],[ ),)(()(21)()(21)( ,1101021xxxxxxfxfxRn??????????????線性插值余項(xiàng)時(shí)當(dāng)( 2 . 1 8 ) ],[ ),)()(()(61)( ,2202102 xxxxxxxxfxRn???????????拋物插值余項(xiàng)時(shí)當(dāng)例 1 已知 =, =, = ,用線性插值計(jì)算和 拋物插值計(jì)算 , 并估計(jì)誤差 . 0 . 3 3 0 3 6 5 . )()( i n 0010101?????????xxxyyyL解： |,))((|2|)(| 1021 xxxxMxR ???. . 3 3 3 521|)(|, i n|)(|m a x511220??????????????RxxfMxxx例 1 已知 =, =, = ,用拋物插值計(jì)算 , 并估計(jì)誤差 . 02122 0 10 1 0 2 1 0 1 20122 0 2 12( ) ( )( ) ( )( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) sin 367 ( 367 ) 303 65.x x x xx x x xL x y yx x x x x x x xx x x xyx x x xL??????? ? ? ??????? ? ?解 ：. . 8 2 861|)(|,)c o s ( |,))()((|6|)(|620321032??????????????RxMxxxxxxMxR167。 方法三、采用分段光滑插值，即樣條插值。 ② 計(jì)算 Mj (追趕法等 ) 。