數(shù)值分析課件第六章_數(shù)值插值方法

1、 5月月1日日 5月月31日日 6月月30日日日出日出 5：51 5：17 5：10日落日落 19：04 19：38 19：506614616135143131211322102210210.)(.)(.aaaaaaaaannnnnnnnnyxaxaayxaxaayxaxaa101111000010njinjinnnnnnnxxxxxxxxxxxxxxV121211020010)(111),()(jixxji)()(0010101xxxxyyyxL001110101yxxxxyxxxxxL)(01010110 xxxxxlxxxxxl)(,)(11001)()()(yxlyxlxLkjkjxl

2、jk01)(的線性組合得到，其系數(shù)分別為的線性組合得到，其系數(shù)分別為y0， y1。即。即顯然，顯然， l0 (x)及及l(fā)1 (x)也是線性插值多項(xiàng)式，在節(jié)也是線性插值多項(xiàng)式，在節(jié)點(diǎn)點(diǎn)x0,x1上滿足條件：上滿足條件：l0(x0)=1 , l0(x1)=0.l1(x0)=0 , l1(x1)=1.稱稱l0 (x)及及l(fā)1 (x)為線性插值基函數(shù)。為線性插值基函數(shù)。(j,k=0,1)即即kjkjxljk01)()()(210 xxxxAxl)(12010 xxxxA)()()(2010210 xxxxxxxxxl故故)()()(2101201xxxxxxxxxl)()()(1202102xxxxx

3、xxxxl2120210121012002010212yxxxxxxxxyxxxxxxxxyxxxxxxxxxL)()()()()()()(同理同理416, 39, 24734942499)(1xxxL6251347537952771.)()()(L4) 916)(416() 97)(47(3)169)(49()167)(47(2)164)(94()167)(97()7(72L6286. 2)6458. 27(取取x0=4, x1=9, x2=16（2）拋物插值：拋物插值：kjkjxljk01)(Lxlx ylx ylx ynnn( )( )( )( )0011nkiiikikxxxxxl0)

4、()()()()()()()()()(110110nkkkkkknkkkxxxxxxxxxxxxxxxxxl0( )( )nnk kkLxy lx)()()()()()()(110110nkkkkkknkkkxxxxxxxxxxxxxxxxxl)()!1()()()()(1)1(xnfxLxfxRnnnn,ba)()()(101nnxxxxxxx三、插值余項(xiàng)與誤差估計(jì)三、插值余項(xiàng)與誤差估計(jì)證明：證明： 因?yàn)橐驗(yàn)?()()(xLxfxRnn上顯然在插值節(jié)點(diǎn)為), 1 , 0(nixi)()()(iniinxLxfxRni, 1 , 0,0.1,)(個個零零點(diǎn)點(diǎn)上上至至少少有有在在因因此此 nba

5、xRn設(shè)設(shè)其中其中)()()(1xxKxRnn),()()(101nnxxxxxxx 為待定函數(shù))(xK)()()()()(1xxKxLxfxRnnn)()()()(1xxKxLxfnn0)()()()()(1txKtLtftnn若引入輔助函數(shù))(x則有0的區(qū)別的區(qū)別與與注意注意xt)(ix且)()()(1ininxxKxR0即個零點(diǎn)上至少有在區(qū)間若令因此,2,)(,nbatxxi,0)(xni, 1 , 0nixi, 2 , 1 , 0,0)(.)(,)(,)()(1也也可可微微則則可可微微因因此此若若為為多多項(xiàng)項(xiàng)式式和和由由于于txfxxLnn )()()()(1xxKxLxfnn)()(

6、)()(1ininixxKxLxf根據(jù)Rolle定理,;1),()(個零點(diǎn)個零點(diǎn)上有至少上有至少在區(qū)間在區(qū)間 nbat 再由Rolle定理,;),()(個零點(diǎn)個零點(diǎn)上有至少上有至少在區(qū)間在區(qū)間nbat 依此類推，.1)(,),(導(dǎo)數(shù)為零導(dǎo)數(shù)為零階階的的使得使得內(nèi)至少有一個點(diǎn)內(nèi)至少有一個點(diǎn)在區(qū)間在區(qū)間 ntba 0)()1(n)()1(tn)()()()()(1txKtLtftnn)()()()()1(1)1()1(txKtLtfnnnnn由于)()()()()()1(1)1()1()1(nnnnnnxKLf因此)!1()()()1(nxKfn0)!1()()()1(nfxKn)()()(1xx

7、KxRnn)()!1()(1)1(xnfnn所以1)1()(maxnnbxaMxf)()!1()(11xnMxRnnn)()(21)()(21)(1021xxxxfxfxR ),(10 xx)()()(61)(2102xxxxxxfxR ),(20 xx例例:225,169,144,)(三個節(jié)點(diǎn)為若xxf線性插值的余項(xiàng)為設(shè)LagrangexR)(1插值的余項(xiàng)為二次LagrangexR)(2解解:.)175(值值的的截截?cái)鄶嗾`誤差差近近似似線線性性和和二二次次插插值值做做試試估估計(jì)計(jì)用用fLagrangexxf21)(2341)( xxf2583)( xxf|)(|max2251692xfMx

8、|)169(| f 41014. 1|)(|max2251443xfMx |)144(| f 61051. 1|)(|22xN|)225175)(169175(|300|)(|33xN|)225175)(169175)(144175(|9300|)(|1xR22! 21NM3001014. 121421071. 1|)(|2xR33! 31NM93001051. 161631035. 2誤差更小。二次插值比線性插值的用時在求從以上分析可知Lagrange,175,例：例：已知函數(shù)已知函數(shù)f(x)的函數(shù)值列表如下：的函數(shù)值列表如下：列出一至三階的均差表。列出一至三階的均差表。解：解：4( )0.

9、41075 1.166(0.4)0.28(0.4)(0.55)0.19733(0.4)(0.55)(0.65)0.03134(0.4)(0.55)(0.65)(0.8)Nxxxxxxxxxxx4(0.596)(0.596)0.63192fN9554321041063. 3)596. 0(,)( xxxxxxfxR高次插值的病態(tài)性質(zhì)：高次插值的病態(tài)性質(zhì)：對于一個確定的區(qū)間，如果插值節(jié)點(diǎn)之間的距對于一個確定的區(qū)間，如果插值節(jié)點(diǎn)之間的距離較小，自然插值節(jié)點(diǎn)就增多，如果用一個多離較小，自然插值節(jié)點(diǎn)就增多，如果用一個多項(xiàng)式插值，自然次數(shù)就會升高，也就是說要用項(xiàng)式插值，自然次數(shù)就會升高，也就是說要用高次多

10、項(xiàng)式插值。高次多項(xiàng)式插值。但是否次數(shù)越高，插值多項(xiàng)式的逼近效果越好但是否次數(shù)越高，插值多項(xiàng)式的逼近效果越好呢？在某種程度上，插值多項(xiàng)式的次數(shù)越高，呢？在某種程度上，插值多項(xiàng)式的次數(shù)越高，截?cái)嗾`差會小一些，但并不是次數(shù)截?cái)嗾`差會小一些，但并不是次數(shù)n越大越好，越大越好，因?yàn)殡S著因?yàn)殡S著n增大，插值多項(xiàng)式并不一定收斂。增大，插值多項(xiàng)式并不一定收斂。211)(xxf5 10(0,1, )kkxknn 1201( )1( )1()()nnnjjjnjxL xxxxx1/211()2nnnxxx1/255nxn20世紀(jì)初，世紀(jì)初，Runge就給出了一個等距節(jié)點(diǎn)插值就給出了一個等距節(jié)點(diǎn)插值多項(xiàng)式不收斂的例

11、子。多項(xiàng)式不收斂的例子。下表列出了下表列出了n=2,4,20的的Ln(xn-1/2)和和R(xn-1/2)的值：的值：1/2()nf x1/2()nnLx1/2()nR x-5-4-3-2-1012345-0.500.511.52L10(t) f(t) f(x)()()(11)(1xlyxlyxLkkkkk11kkkkxxxxykkkkxxxxy11)(1xLnnnxxxxLxxxxLxxxxL1)1(121)1(110)0(1)()()(iiyxL)(1為插值點(diǎn)設(shè)*xx 1*kkxxx若*)(*1xLy 則*)()(1xLk11*kkkkxxxxykkkkxxxxy11*0*xx 若*)(*

12、1xLy 取*)()0(1xL1010*xxxxy0101*xxxxynxx *若*)(*1xLy 取*)()1(1xLnnnnnxxxxy11*11*nnnnxxxxy內(nèi)插外插外插-4-3-2-101234-1-0.8-0.6-0.4-0.200.20.40.60.81-4-3-2-101234-1-0.8-0.6-0.4-0.200.20.40.60.81-4-3-2-101234-1-0.8-0.6-0.4-0.200.20.40.60.81-4-3-2-101234-1-0.8-0.6-0.4-0.200.20.40.60.81-4-3-2-101234-1-0.8-0.6-0.4-0

13、.200.20.40.60.81的圖象分段線性插值)(1xLy 的一條折線實(shí)際上是連接點(diǎn)niyxkk, 1 , 0,),( )()(lim10 xfxLh上連續(xù)在若,)(baxf)()!1()(1)1(xnfnn)()()(xLxfxRnn)()()(11xLxfxR)()()(1xLxfk)(2)(1 kkxxxxf有關(guān)與且xxxxkk,1| )( |max| )(|max21)(11 kkkbxabxaxxxxxfxR224121hM 2281hM)()()()(1111)(2xlyxlyxlyxLkkkkkkk)()(11111kkkkkkkxxxxxxxxy1,2 , 1nk)()(2

14、xLk)()(1111kkkkkkkxxxxxxxxy)()(11111kkkkkkkxxxxxxxxy)()!1()(1)1(xnfnn)()()(xLxfxRnn)()()(22xLxfxR)()()(2xLxfk)()(6)(11 kkkxxxxxxf有關(guān)與且xxxxkk,11|)(|2xR| )()( |max| )(|max611111 kkkkxxxbxaxxxxxxxfkk3393261hM 33273hM由于由于那么分段二次插值那么分段二次插值L2(x)的余項(xiàng)為：的余項(xiàng)為：( )0.36,0.42,0.75,0.98,1.1.()f xx 求在處的近似值用分段線性、二次插值18

15、885. 187335. 069675. 057815. 041075. 030163. 005. 180. 065. 055. 040. 030. 0543210iiyxi在各節(jié)點(diǎn)處的數(shù)據(jù)為設(shè))(xf例例:)()(1xLk11kkkkxxxxykkkkxxxxy11解解:(1) 分段線性分段線性Lagrange插值的公式為插值的公式為1, 1 , 0nk)36. 0()0(1L4 . 03 . 04 . 036. 030163. 03 . 04 . 03 . 036. 041075. 036711. 0)42. 0()1(1L55. 04 . 055. 042. 041075. 04 . 0

16、55. 04 . 042. 057815. 043307. 0)75. 0()3(1L81448. 0)98. 0()4(1L10051. 1)1 . 1()4(1L05. 18 . 005. 11 . 187335. 08 . 005. 18 . 01 . 118885. 125195. 1)36. 0(f)42. 0(f)75. 0(f)98. 0(f)1 . 1(f同理)()(11111kkkkkkkxxxxxxxxy1,2 , 1nk)()(2xLk)()(1111kkkkkkkxxxxxxxxy)()(11111kkkkkkkxxxxxxxxy(2) 分段二次分段二次Lagrange

17、插值的公式為插值的公式為36686. 0)()36. 0)(36. 0(2010210 xxxxxxy)36. 0()1(2L)()36. 0)(36. 0(2101201xxxxxxy)()36. 0)(36. 0(1202102xxxxxxy)36. 0(f43281. 0)()42. 0)(42. 0(2010210 xxxxxxy)42. 0()1(2L)()42. 0)(42. 0(2101201xxxxxxy)()42. 0)(42. 0(1202102xxxxxxy81343. 0)()75. 0)(75. 0(5343543xxxxxxy)75. 0()4(2L)()75. 0

18、)(75. 0(5454534xxxxxxy)()75. 0)(75. 0(4535435xxxxxxy)42. 0(f)75. 0(f)98. 0(f)1 . 1(f09784. 1)98. 0()4(2L25513. 1)1 . 1()4(2L,)(,)(,)()(1212101nnnxxxxSxxxxSxxxxSxS三次樣條插值多項(xiàng)式的確定：三次樣條插值多項(xiàng)式的確定：jjjjjdxcxbxaxS23)()0()0( jjxSxS)0()0(jjxSxS)0()0(jjxSxS可得可得3n-3個方程，又由條件個方程，又由條件(3)jjyxS)(00()(),()()nnSxfxSxfx)(

19、)(),()(00nnxfxSxfxSj=0,1,n得得n+1個方程，共可得個方程，共可得4n-2個方程。個方程。要確定要確定4n個未知數(shù)，還差兩個方程。個未知數(shù)，還差兩個方程。通常在端點(diǎn)通常在端點(diǎn)x0 =a, xn =b處各附加一個條件，稱處各附加一個條件，稱邊界條件邊界條件，常見有兩種：，常見有兩種：(1)二階邊界條件：二階邊界條件：(2)一階邊界條件：一階邊界條件：共共4n個方程，可唯一地確定個方程，可唯一地確定4n個未知數(shù)。個未知數(shù)。 1 , 00 , 1)(222232112131xdxcxbxaxdxcxbxaxS11111dcba01d02d12222dcba21cc 21bb

20、02611ba02622ba023,2121212121ddccbbaa 1 , 023210 , 12321)(2323xxxxxxxSjjjjjjMhxxMhxxxS11)( 121122)(2)()(cMhxxMhxxxSjjjjjj213113)()(61)(cxcMxxMxxhxSjjjjjjjjjjjjjjjjjjjjhxxMhyhxxMhyMxxMxxhxS121213113)6()6()()(61)()(6)(1)()(21)(112112jjjjjjjjjjjMMhyyhMxxMxxhxSjjjjjjjjhyyMhMhxS1136)0(jjjjjjjjhyyMhMhxS11163)0(1111163)0(jjjjjjjjhyyMhMhxS)0()0(jjxSxS11111116336jjjjjjjjjjjjjjhyyMhMhhyyMhMh)(62111111111jjjjjjjjjjjjjjjjjhyyhyyhhMhhhMMhhh1jjjjhhhjjjjjhhh111),(6)(6111111jjjjjjjjjjjjxxxfhyyhyyhhdjjjjjjdMMM112nnnyxfxSyxfxS)()(,)()(000