【正文】 ? Spline：分段低次 , 自身光滑 , f 的導(dǎo)數(shù)只在邊界給出。 ? Newton ? Ln(x)， 只是形式不同；節(jié)點(diǎn)等距或漸增節(jié)點(diǎn)時(shí)方便處理。 由該區(qū)間上的 S[j](x) 算出 f(x) 的近似值。 ③ 寫(xiě)出樣條函數(shù) S(x)。 f(x) H(x) S(x) 三次樣條插值的算法步驟： ① 計(jì)算 ?j , ? j , gj 。 “ matlab中的 spline(x,y)” 二、三次樣條插值函數(shù)的建立 .用三彎矩法和三轉(zhuǎn)角法求三次樣條插值函數(shù)常( 7 .6 ) ],)()([)( 0????njjjjj xmxfxs ??.,數(shù)，得到三次樣條插值函對(duì)角方程組，求出的三，可得關(guān)于連續(xù)性條件和邊界條件由插值條件jjmm埃爾米特插值多項(xiàng)式，根據(jù)分段三次三轉(zhuǎn)角法：假定 ,),0()( njmxs jj ????. ,0,)( 1 jjjjj xxhnjMxs ?????? ??三彎矩法：令）（則 ].,[ ,)( 111 ??? ??????? jjjjjjjj xxxMhxxMhxxxsP42 ,2)(2)()(11221 cMhxxMhxxxsjjjjjj ????????? ,6)(6)()(211331 cxcMhxxMhxxxsjjjjjj ????????,61)( ,61)( 1211121212 ???? ???????? jjjjjjjjjj ycxcMhxsycxcMhxs.)(61),(61 11112111 ?????? ???????? jjjjjjjjjjjjjjjj MxMxhhxyxycMMhhyyc)( ,)6()6( 6)(6)()(211121331jjjjjjjjjjjjjjjjhxxhMyhxxhMyMhxxMhxxxs???????????????jjjjjjjjjjjj hMMhyyMhxxMhxxxs62)(2)()( 111221 ?????????? ????)0()0(: , 0 ????? jjn xsxsMM 要用導(dǎo)數(shù)連續(xù)條件為了求 ?,63)0( 11jjjjjjjjhyyMhMhxs ??????? ??,36)0( 111jjjjjjjjhyyMhMhxs ?????? ???,1,1 ,6361111111 ???????????????? njhyyhyyMhMhhMhjjjjjjjjjjjjj ?.36)0( 11111????? ??????jjjjjjjjhyyMhMhxs].,[6, ,1,1 ,2 1111111?????????????????jjjjjjjjjjjjjjjjjjxxxfdhhhhhhnjdMMM????其中?00001010000110000:)(6263)0(dfhyyhMMfhyyMhMhxs???????????????nnnnnnnn dhyyfhMM :)(62 1111 ??????????在第一邊界條件下：nnnnnnnnn fhyyMhMhxs ?????????????1111136)0()( ,2122121101101111?????????????????????????????????????????????????????nnnnnnddddMMMM?????????.,00 nn fMfM ??????在第二邊界條件下：)( ,222211220111221122221????????????????????????????????????????????????????????????????nnnnnnnnnfdddfdMMMM?????????????].,[6,2 )()(110100111100xxxfdhhhhhhdMMMMMxSxSnnnnnnnnnnnnnnn???????????????????其中，并且，就有意在第三邊界條件下：注)( ,2222121121112211?????????????????????????????????????????????????????nnnnnnnnddddMMMM?????????????系數(shù)矩陣為嚴(yán)格對(duì)角占優(yōu)陣，方程組有唯一解。 7 樣條插值 問(wèn)題背景 ? .,)(, 2)。 方法四、采用有理逼近。 211)(xxf ??二、分段線性插值 所謂分段線性插值就是用通過(guò)插值點(diǎn)的折線段逼近 f(x). .)(,],[)( ( 3 ) ,1,0,)( ( 2 ) ],[)( ( 1 ) )(,m a x , ,11010分段線性函數(shù)xIxxxInkfxIbaCxIxIhhxxhffbxxxahkkhkkhhhkkkkknn則稱(chēng)上是線性函數(shù)在每個(gè)小區(qū)間滿足求折線函數(shù)記上的函數(shù)值已知節(jié)點(diǎn)????????????????解決 : 方法一、采用分段低次插值 方法二、在各節(jié)點(diǎn)處不僅給出其函數(shù)值，還給出其各階導(dǎo) 數(shù)值，即分段 Hermite插值法。0)( ,)( nkjxxxxjkkjkjkjjkkj ????????????????? ( 5 .3 ) . )]()([)(012 ??? ???njjjjjn xfxfxH ?? ),()()( 2 xlbaxx jj ???令( 5 .4 ) ).(1)(21)( 20xlxxxxx jnjkk kjjj??????????????? ???? ),()()( 2 xlxxAx jjj ???令( 5 . 5 ) ).()()( 2 xlxxx jjj ??? ? 1)(0 kjnjkkjj xxxl ??? ???. ),(( 5 . 6 ) ),()!22()()()()( ),(22),()(2)22(12)22(1xbaxnfxHxfxRxfnbaxfnnnn且依賴(lài)于其中插值余項(xiàng)為則階導(dǎo)數(shù)內(nèi)有在插值區(qū)間若?????????????. )()()(1)(21 )]()([)(0 0220012? ???? ?????????????????????????njnjjjjjjnjkk kjjnjjjjjnfxlxxfxlxxxxxfxfxH ??爾米特插值多項(xiàng)式：時(shí)，應(yīng)用廣泛的三次埃當(dāng) 1?n .)()( 2121)(12010102101012010101021010103fxxxxxxfxxxxxxfxxxxxxxxfxxxxxxxxxH?????????????????????????????????????????????????????????? ).,( ,)()(!4)( )()()( 102120)4(33xxxxxxfxHxfxR????????余項(xiàng)為P38 .)()(),()(),()(),()( ,],[)( 11221100的插值多項(xiàng)式及其余項(xiàng)試求滿足條件上有四階連續(xù)導(dǎo)數(shù)在若xfxHxfxHxfxHxfxHbaxf??????例4另一類(lèi)三次 Hermite插值多項(xiàng)式見(jiàn) P36. .)()(),()(),()( ,],[)( 001100的插值多項(xiàng)式及其余項(xiàng)試求滿足條件上有三階連續(xù)導(dǎo)數(shù)在若xfxHxfxHxfxHbaxf?????練習(xí)).)((],[)()()( ))(()()( ),()(),()( 1010001011100xxxxAxxfxxxfxHxxxxAxNxHxfx