《分析05-插值法下》ppt課件

1、W Y5-1第五章 插值法W YW Y), 2 , 1 , 0( )()(niyxHyxHiiii1)( ,0)( ,0)( ,0)(0)( , 1)( ,0)(,0)(0)( ,0)( , 1)( ,0)(0)( ,0)( ,0)( , 1)()(),(),(),()()()()( )()()()()(1121110112221202112111011020100012101122110011221100 xHxHxHxHxhxhxhxhxhxhxhxhxhxhxhxhxHxhxhxhyxHyxhyxhyxhyxlyxlyxlyxlxH：都是三次多項(xiàng)式且滿(mǎn)足其中可設(shè)：按插值基函數(shù)的方法，并估

2、計(jì)誤差。且使的多項(xiàng)式求不超過(guò)三次已知為互異節(jié)點(diǎn)，設(shè)引例,)(),2 , 1 , 0()()(,)2 , 1 , 0()(,:11210yxHiyxHxHiyxfxxxiiiiW Y21202210220210221020210002210020110100)()()( )()()()()( )()(11)( )()()()(,)( 0)(, 0)( )(xxxxxxxxxhxxxxxxxxxhxxxxCxhxxxxCxhxhxxhxxhxhxh同理可求：于是求出而由可設(shè)的一階零點(diǎn)是而的二階零點(diǎn)是：首先求W Y求。滿(mǎn)足前面條件，即為所法求出的中，可以檢查按上述方代入將利用可設(shè)：為一階零點(diǎn)：對(duì)由可

3、設(shè)分別為其一階零點(diǎn)：對(duì))()()(),(),(),()()()(1)()(11)()()()(,)()()()()223)2()()()(2)()(1 ,)()(2)(0)()()(1)()(0)(1)()()()(,)(1210210210112101112101210122120120202110211201221201112021012212011200112112101210111111201201xHxHxHxhxhxhxxxxxxxxxxxHxxxxCxHxxxxxxCxHxxxxHxxxxxxxxxxxxxxxxxxxxhxxxxxxxxxxxxbxxxxxxxaxxbaxxxb

4、axxxxxaxxxxbaxxhxhxxxxbaxxhxxxhW Y)()()(3xHxfxR)()()( )()()(22103xxxxxxxxHxfxR)()()()()(22103xtxtxtxtRt0)(! 4)()()()4()4()4(xHfxxx)(!41)( )4(xfx因此可得)()(! 4)()()()(2210)4(3xxxxxxfxHxfxRx因而有：W YniiiiiyxHyxhxH0)()()(；), 2 , 1 , 0( , 1 0)(, 0)(), 2 , 1 , 0(0)( , 1 0)( )2(njjijixHxHnjxhjijixhjijijijiW Y.

5、)()()()(,)()()(112插值基函數(shù)為為待定系數(shù)其中Lagrangexxxxxlbaxlxxbaxhininiiii221212120)()()()()()(niiixxxxxxxxxxbxaxh), 2 , 1 , 0( )()()(21 ()()(21 0)(2)()(2)()()(1)()(0)(1)(222nixlxlxxxhxl abaxl abxlxlxxbaxblxhaxalxhxhxhiiiiiiiiiiiiiiiiiiiiiiiiiii可求出W Y),2 , 1 , 0( )()()(1)()(1)( )()()(222nixlxxxHCxClxHxHxlxxCxH

6、iiiiiiiiiiii利用niiiiiiiiiniiiiiyxlxxyxlxlxxyxHyxhxH0220)()()()()(21( )()()(2010112101002010101121010100)()()()(21)(21)(xxxxxxxHxxxxxxxHxxxxxxxxxhxxxxxxxxxh，W Y17)-(5 )(111001110000yhxxhyhxxhyhxxyhxxxH0) 1 () 1 ()0( , 1)0(0) 1 ()0() 1 ( , 1)0()1 ()( ,)1)(21 ()(:)()()()()(:11110000212011001100它們分別滿(mǎn)足：面兩

7、個(gè)函數(shù)實(shí)際計(jì)算中經(jīng)常用到下插值多項(xiàng)式為因此兩個(gè)節(jié)點(diǎn)的三次xxxxxxyxHyxHyxhyxhxHHermite011101101010100110101000)()()()()()(xxxxxxxHxxxxxxxHxxxxxhxxxxxh，W Y),( )()!22()()()()(21)22(baxnfxHxfxRnn18)-(5 )()(! 4)()(2120)4(xxxxfxRW YniixhxH0)()()()!22()()()(21)22(xnHxHxHnn1)( 20niixh：推論W Y931010iiiyyxxxxxxxxxhxxhxxhxxhxxxHxxxxxhxxxxxhx

8、xxxxxxhxxxxxhxxxxh31210 x )1 (9)1 (3)23( )9(3 10)()1 ()1 (1)(1 ()1 ()1 ()()23()1 (1)(1 (21 ()1 ()1)(21 ()( 1:232221101100022111210122010200001 代入：而解W Y1)( 10)0( )21(1)()1()1(2)(0, 1,0)0()0()()(,)()()( )()( )(,)(,)()(1,0, 1 :3222102110112102210102100100210 xxHHxxxHxxxxHxxxfxxfxxffHxfxHxxxxxxxNxxxxxxx

9、xxxxxxfxxxxfxfxHNewtonxxxxi因此代入其中確定即可由為待定參數(shù)插值?？稍O(shè)：還可以利用為等距節(jié)點(diǎn)，三點(diǎn)記為解W Y)0 , 2 , 1 , 0)(ninmmkxfykiikk),1 ,0( )(),0,1,(i )(mkyxHnyxHkkiiiimkinnmkxxxnmfxHxfxR01)2()()()!2()()()()(W Y22121212222221112111110000)3(4)()()()()2)(1()(,)1(4)( )2(1)2()1(1 ()(1, 10)1 (, 1)1 (1)2()1()( ,0)0()2()0( :)()(),( ,0)1 ,0

10、,( 1 0)()1 ,0 ,2, 1 ,0( 0)()2, 1 ,0 , 1 ,0( 0)()2, 1 ,0,( 1 0)()2(;4)1 ,0)( ),2, 1 ,0)( )1 (xxxHxhxhxHHermitexxxxHxxxhxxxxxxhbahhxxxbaxhhhhxhxHxhyyjiijijxHijxHijxhjiijijxhixHixhjijijijiii插值多項(xiàng)式為：因此所求類(lèi)似可求代入可設(shè)對(duì)不必求次多項(xiàng)式都是12 101010iiiyyxW Y)()(22cbxaxxxH4/9234112341481611) 1 (1)2(1) 1 (cbacbacbacbaHHH22)

11、3(4)(xxxH12 101010iiiyyxW YW Yyx0.5)(xfy )(10 xPy )(4xP1W YniiiinxlyyxL0)()()()()()()()()(xLxLxLxfxLxfnnnnniiinnxlyxLxL0)()()(W Y引起誤差增大。也隨之而增加。因此，多時(shí)并且當(dāng)節(jié)點(diǎn)增能很好地逼近就不能保證也增大如果時(shí)節(jié)點(diǎn)增多增大當(dāng)其中)()(,).()(,)(, )(max)()()!1()()()(101) 1(1101nnnnbxannnnnxxxxxxxfxLMnfMxxxxxxnMxLxfxRx0 x1x2x3x5x6x7xy)(76xln時(shí)的W YW YW Y

12、iniiiinnhhnixxhbxxxxa101110max) 1, 1 , 0( 記設(shè)給定節(jié)點(diǎn)W Y, )(111111iiiiiiiiiixxxxxxxyxxxxyxPx0 x1x2x3x4W Y) 12, 1 , 0( , )()( )()()()()(:,) 12, 1 , 0(,), 1 , 0( )( 222221222222122122212212222222212222122222nkxxxyxxxxxxxxyxxxxxxxxyxxxxxxxxxPnkxxnnixfykkkkkkkkkkkkkkkkkkkkkkkkkii得上分別作拋物插值在每個(gè)小區(qū)間為偶數(shù)且設(shè)已知W Y1111

13、1100)(iiiiiiiiiiiiiiyhhxxyhhxxyhxxyhxxxHx)2,W Y, )(max8)(! 2)()()(12111 iixxxiiixxxxfhxxxxfxPxfii, )(max8)()(21baxxfhxPxfbxa , )(max384)()(! 4)()()(14212)4(1 iixxxiiixxxxfhxxxxfxHxfii, )(max384)()(4baxxfhxHxfbxa W Y3362101212101101210210210218)(max8)()(11max)(max1)(ln)( : 過(guò)，即數(shù)表的步長(zhǎng)應(yīng)不超應(yīng)?。豪蒙厦婀烙?jì)式，欲使所以：

14、解hhxfhxPxfxxfxxfxxfxxxW YW YW Y點(diǎn)上，不僅函數(shù)本身是延續(xù)的，而且它的一階和點(diǎn)上，不僅函數(shù)本身是延續(xù)的，而且它的一階和二階導(dǎo)二階導(dǎo)數(shù)也是延續(xù)的。由此籠統(tǒng)出數(shù)學(xué)模型稱(chēng)為樣條函數(shù)也是延續(xù)的。由此籠統(tǒng)出數(shù)學(xué)模型稱(chēng)為樣條函數(shù)。數(shù)。S(x)mW Y1,2, 1 )0()0()0()0()0()0( nixSxSxSxSxSxSiiiiiiW YW YW Y 1010101011110000110011213110020300)0()0()0()0(10011) 1 (, 0)0(0)0(, 1) 1( 1 , 0 ,)(0 , 1 ,)()( 10 01 1 , 12bbcc

15、SSSScbaddcbaSSSSxdxcxbxaxSxdxcxbxaxSxSn導(dǎo)數(shù)的連續(xù)條件：再由內(nèi)節(jié)點(diǎn)處一、二階可得：由插值和函數(shù)連續(xù)條件并設(shè)：兩個(gè)子區(qū)間，分為，區(qū)間解：這里W Y 1 , 0 ,23210 , 1 ,2321)(0,23,21:,0260260) 1 (0) 1(23231010101010110010 xxxxxxxSCCbbaabbaababaSS從而得到問(wèn)題的解為：因此聯(lián)立可解得：而由自然邊界條件：W Y211331112211111111111116)(6)()(2)(2)()(,)()()()( ,)(,)() 1, 1 , 0()()(,), 1 , 0()(c

16、xcMhxxMhxxxScMhxxMhxxxSxxhMhxxMhxxMxxxxMxxxxxSxSxSxSMMxxMxSMxSnjxSxSxxniMxSjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjii 再積分一次得：將上式積分一次得：其中為線(xiàn)性為三次多項(xiàng)式的表達(dá)式作線(xiàn)性插值求出可以利用為三次多項(xiàng)式上在設(shè)W Y19)-(5 ) 1, 2 , 1 , 0(666)(6)()()()(6)(6666)(6)(12211133111112111121112212121113121311njhxxhMyhxxhMyMhxxMhxxxSxSxMxMhhxy

17、xycMMhhyycycxcMhycxcMhycxchMxxycxcMhxxxxxxjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjj中整理后得：代入可解出：W Y20)-(5 ) 1, 1 , 0()(62)(2)()(61612)(2)()(11122122111221njMMhhyyMhxxMhxxxShMyhhMyhMhxxMhxxxSjjjjjjjjjjjjjjjjjjjjjjjjjjjjW Y)()(jjjjxSxS)(62)()()(62)(1111111jjjjjjjjjjjjjjjjjjjjjjjjMMhhyyMh

18、xSxxxSMMhhyyMhxSxx)()(62)(1:)()()()(,)(1111111jjjjjjjjjjjjjjjjjjjjxSMMhhyyMhxSjjxSxSxSxSxS中令在在節(jié)點(diǎn)處一階導(dǎo)數(shù)連續(xù)W Y，即有：由)()(jjjjxSxS)(6211jjjjjjjjMMhhyyMh)(6211111jjjjjjjjMMhhyyMh1111636jjjjjjjMhMhhMh1, 2 , 1111njhyyhyyjjjjjj)(626:1111111111jjjjjjjjjjjjjjjjjjjhyyhyyhhMhhhMMhhhhh同乘以W Y22)-(5 ) 1, 2 , 1( 221)-

19、 5 ( )(6,1,111111111njcMMMhyyhyyhhchhhhhhjjjjjjjjjjjjjjjjjjjjjjjj令23)-(5 2 2 2 211121343323232212121101nnnnnncMMMcMMMcMMMcMMMW Y24)-(5 )(62)()(62)(11111010001000nnnnnnnnnMMhhyyMhybSMMhhyyMhyaS25)-(5 626211110001010nnnnnnnhyyyhMMyhyyhMMW YnnnnnnncccccMMMMM1210121011221102202022可用追趕法求解這是三對(duì)角方程組其中對(duì)照,6,6

20、, 1, 1),255(1110001000nnnnnnnhyyyhcyhyyhc。nnyMyM 0W Y )()()()(bSaSbsaS)(62)(62),()()255(),()(,)()(11111010001000 nnnnnnnnnMMhhyyMhMMhhyyMhbSaSbSaSMMbSaS即利用而由由110011011101100111101010110011110010106)(266666266662nnnnnnnnnnnnnnnnnnnnnnnnnhyyhyyMhhMhMhhyyhyyMhMhMhhMhhhyyhyyMhMhMhMhMhhW Y27)-(5 211211cM

21、MMnnnnnnnnncccccMMMMM13211321113322112222226)-(5 2 6261111001101101110010nnnnnnnnnnnnnnncMMMhyyhyyhhMMhhhMhhhhh記為同乘以W Y.6,1100110100101三對(duì)解方程組的追趕法其求解過(guò)程類(lèi)似于求解其中nnnnnnnnnnhyyhyyhhChhhhhhW Y2120111111100)1 ()(,)1)(21 ()()285() 1, 1 , 0( , ,)(xxxxxxnjmmhxxxmhhxxmhhxxhxxyhxxxSjjjjjjjjjjjjjjjjjjj 其中) 1, 1 ,

22、 0( ) 0() 0(),0() 0(njxSxSxSxSjjjjjjjjW Y46)(, 612)(:)(101111211020 xxxxhmhxxhmhxxhyhxxhyhxxxSjjjjjjjjjjjjjjjjj其中)2(2)(6 )0()1()0()1()0()2(2)(6 )1()0()1()0()0(11211102102111211102102jjjjjjjjjjjjjijjjjjjjjjjjjjjjijjmmhyyhhmhmhyhyxSmmhyyhhmhmhyhyxS W Y)(6)(6242)2(2)(6)2(2)(6:),0()0()2(2)(6)0(:1)0(1211

23、2111111121112111112111 jjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjyyhyyhmhmhhhhmhmmhyyhmmhyyhxSxSmmhyyhxSjjxS有由可得中令在) 1, 2 , 1( 2)()(3)()(32)( 2111111111111111njdmmmhhhyyhyyhhhhmhhhmmhhhhhhhjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjj同乘以W Y2830)-(5 2 2 2 2295 )()(1,1112134332323221212110111111nnnnnnjjjjjjjjjjjjjjj

24、jjjdmmmdmmmdmmmdmmmyyhyyhdhhhhhh）（令W YW Y3 , 2 )52573232946(31)(2 , 1 )351089124(31)( 1 , 0 )31411(3 ) 1(914937) 1(914919 ) 1(63280)0(63382 6) 1(3286)0(338)(),195(3170,3106,338,32878363621002/122/1002/122/1001223223123333032103210 xxxxxSxxxxxSxxxxxxxxxxxxxSMMMMMMMM得將此解代入式其解為W Y21)(21)(21329)(21)(213

25、2123122120112121yyyydyyyyd332,32 21221,2921221212121mmmmmm其解為32 )52573232946(3121 )351089124(3110 )31411(3)(23232xxxxxxxxxxxxxSW YW Y) 3 , 2 , 1 , 0(, )(max)()()4(4)()(lbaxxfhcxSxfbxallll.max/max,2,83,241,3845,max11132101稱(chēng)為分劃比其中jnjajnjajxjahhcccchhW YW YW Y 例1求證：存在三次多項(xiàng)式滿(mǎn)足下面函數(shù)表x210137y5616224376證： 由于

26、表中給出了六個(gè)點(diǎn)的函數(shù)值，根據(jù)Newton插值公式，普通情況下可以構(gòu)造出一個(gè)5次插值多項(xiàng)式，這個(gè)5次多項(xiàng)式必然滿(mǎn)足上面函數(shù)表。 構(gòu)造差商表如下：xy一階差一階差商商二階差二階差商商三階差三階差商商四階差四階差商商25611640021413120723431207376931520)(3xPW Y普通情況下，給定普通情況下，給定n+1個(gè)節(jié)點(diǎn)，可構(gòu)造一個(gè)個(gè)節(jié)點(diǎn)，可構(gòu)造一個(gè)n次插值次插值多項(xiàng)式，假設(shè)得到低于多項(xiàng)式，假設(shè)得到低于n次的插值多項(xiàng)式，稱(chēng)為次的插值多項(xiàng)式，稱(chēng)為“退化退化情況，利用情況，利用Newton插值，很容易檢查出能否為插值，很容易檢查出能否為“退化退化情況，由于利用差商差分表，當(dāng)某一

27、階差商差分情況，由于利用差商差分表，當(dāng)某一階差商差分為常數(shù)時(shí)，那么下一階差商差商必定為為常數(shù)時(shí)，那么下一階差商差商必定為0，此時(shí)必會(huì)，此時(shí)必會(huì)出現(xiàn)出現(xiàn)“退化情況。退化情況。)0)(1)(2(2) 1)(2(13)2(4056)(3xxxxxxxN 由上表知，由于四階差商以上均為0，所以這個(gè)5次多項(xiàng)式實(shí)踐上是3次多項(xiàng)式 ： 故存在三次多項(xiàng)式滿(mǎn)足是所給的函數(shù)表。故存在三次多項(xiàng)式滿(mǎn)足是所給的函數(shù)表。W Y 對(duì)區(qū)間 作等距劃分： ，分別取 以 為節(jié)點(diǎn)， 對(duì)函數(shù) ： 1 , 1), 1 , 0(1nkkhxknh2,20,10n22511)(xxf), 1 , 0(nkxk按下述方案進(jìn)展插值按下述方案進(jìn)

28、展插值計(jì)算，并比較其結(jié)果。計(jì)算，并比較其結(jié)果。方案方案I拉格朗日插值；方案拉格朗日插值；方案II分段線(xiàn)性插值；分段線(xiàn)性插值；方案方案III三次樣條插值。三次樣條插值。 解 這里只將求解后的部分?jǐn)?shù)值結(jié)果列于表1中。由此可以看到，拉格朗日插值的效果并沒(méi)有隨n增大而變化，與此相反，在區(qū)間端點(diǎn)附近，反而發(fā)生了猛烈的振蕩，即出現(xiàn)了龍格景象。而分段線(xiàn)性插值、三次樣條插值都能較好地逼近 ，且隨著n的增大，逼近效果更好，反映了分段線(xiàn)性插值和三次樣條插值的一致收斂性，防止了龍格景象的產(chǎn)生，從表中數(shù)據(jù)可以看到，三次樣條插值的精度比分段線(xiàn)性插值更高。)( xfW Y)(xPn)(1xP)(xSXf (x)拉格朗日插

29、值拉格朗日插值分段線(xiàn)性分段線(xiàn)性插值插值三次樣三次樣條插值條插值n =10n =20n =10n =20n =10n =200.150.640000.678990.636760.625000.650000.657470.643170.250.390240.342640.395090.425000.403850.380490.389420.350.246150.190580.238450.275000.253850.240550.246270.450.164940.234970.179760.175000.168970.167230.164860.550.116780.215590.080660.

30、125000.118970.117840.116790.650.086480.072600.202420.089710.087740.085890.086480.750.066390.231460.447050.069120.067150.066020.066390.850.052450.719463.454970.053730.052940.052600.052460.950.042440.9236239.952580.043550.042760.042480.04244W Y 給出函數(shù) 的函數(shù)表表2，試?yán)么藬?shù)表求使 的x值。 shxy 5y表表2X2.22.42.62.83.0Y4.45

31、75.4666.6958.19810.018 解 插值是利用函數(shù)y=f(x)的知數(shù)據(jù)求給定的自變量x所對(duì)應(yīng)的函數(shù)y 的近似值。而此題那么是求知函數(shù)值y 所對(duì)應(yīng)的自變量x之值。假設(shè)函數(shù)y=f(x)的反函數(shù) 存在，那么可把所給數(shù)據(jù)值y視為自變量取值，而把x的值視為函數(shù)值，對(duì)反函數(shù) 進(jìn)展插值，即可求得欲求的x，這樣的問(wèn)題稱(chēng)為反插值。 )(yx)(yxW Y)(yxiyi一階差商一階差商二階差商二階差商三階差商三階差商四階差商四階差商4.4572.25.4662.40.198226.6952.60.178730.015868.1982.80.160380.050.0013410.0183.00.143

32、860.011940.001180.00009由于由于為單增函數(shù)，所以其反函數(shù)存在，現(xiàn)用為單增函數(shù)，所以其反函數(shù)存在，現(xiàn)用牛頓插值法求解該問(wèn)題。首先構(gòu)造反函數(shù)的差商表牛頓插值法求解該問(wèn)題。首先構(gòu)造反函數(shù)的差商表3。shxy W Y根據(jù)差商表根據(jù)差商表可得可得的牛頓插值多項(xiàng)式：的牛頓插值多項(xiàng)式：)(yx)198. 8)(695. 6)(466. 5)(457. 4(00009. 0 )695. 6)(4665)(4574(001340 )466. 5)(457. 4(01586. 0)457. 4(19822. 02 . 2)(yyyyy.y.y.yyyyP3123. 2)5()5(Px從而可得

33、從而可得y=5所對(duì)應(yīng)的值所對(duì)應(yīng)的值x為為:反插值法還可用于方程反插值法還可用于方程f(x)=0的近似求根。對(duì)的近似求根。對(duì)函數(shù)函數(shù)y=f(x)進(jìn)展反插值，求進(jìn)展反插值，求y=0所對(duì)應(yīng)的所對(duì)應(yīng)的x值，即為值，即為方程方程f(x)=0的近似根。的近似根。W Y 知 ，試用線(xiàn)性插值求 的近似值，并估計(jì)插值誤差。12)144(,11)121(,10)100(fff)125(f解解要用線(xiàn)性插值要用線(xiàn)性插值f(x)求在點(diǎn)求在點(diǎn)x=125的值，可取的值，可取x0=121,x1=144為插值節(jié)點(diǎn)，記線(xiàn)性插值式為為插值節(jié)點(diǎn)，記線(xiàn)性插值式為p1(x)。經(jīng)計(jì)算易得。經(jīng)計(jì)算易得:17391.11)125()125(1 Pf 但是，由于不知道f(x)的解析式，故不能直接利用拉格朗日余項(xiàng)式做誤差估計(jì)。為此，下面用另外一種方法來(lái)估計(jì)誤差。設(shè)以x0=121, x2=100為節(jié)點(diǎn)的線(xiàn)性插值式為 ，那么有:)(1xP)(2)()()(1011xxxxfxPxf )(2)()()(2021xxxxfxPxf W Y其中其中均屬于由均屬于由x0,x1,x2和和x所決議的區(qū)間。假設(shè)所決議的區(qū)間。假設(shè)在該區(qū)間內(nèi)變化不大，那么將上面兩個(gè)式子相除