插值法與最小二乘法-在線瀏覽
2025-06-16 07:50本頁(yè)面
【正文】 kk k k kx x x xyx x x x??? ? ??????122 2 1( ) ( )( ) ( )kkkk k k kx x x xyx x x x??? ? ??????上式稱為分段二次 Lagrange插值 ],[*,* 1?? kk xxxx 且為插值點(diǎn)若顯然 ,插值區(qū)間 ],[],[211 ??? kkkk xxxx 和 *x都包含( 1 )2* ( * )ky L x??那 么 ()2* ( * )ky L x?還是 8 一般 則更接近且若 ,*,* 1 kkk xxxxx ???( 1 )2* ( * )ky L x??則更接近且若 ,*,* 11 ???? kkk xxxxx()2* ( * )ky L x?1 , 2 , , 1kn??則含若 ),*(* 01 xxxx ??1 , , 2kn??( 0 )2* ( * )y L x?則含若 ),*(* 1 nn xxxx ?? ? ( 2 )2* ( * )ny L x??時(shí)使用的方法是外插和 nxxxx ?? ** 09 *x *x *x *x *x *x0x 1x 2?kx 1?kx kx 1?kx 1?nx nx*x(0 )2 ( *)Lx ( 2)2kL ?( 1)2kL ? ()2kL( 2 )2 ( * )nLx?? ?外插 內(nèi) 插 外插 10 )()!1( )( 1)1(xnf nn???? ??)()()( xPxfxR nn ??的余項(xiàng)為那么分段二次插值 )(2 xL)()()( 22 xLxfxR ?? )()( )(2 xLf k??12() ( ) ( ) ( )6 k k kf x x x x x x??????? ? ? ?2, [ , ] .kkx x x? ??|)(| 2 xR2121 m a x | ( ) | m a x | ( ) ( ) ( ) |6 kk k k ka x b x x xf x x x x x x x? ??? ? ? ????? ? ? ? ? ?33 93261 hM ??? 33273 hM?2. 分段二次插值的誤差估計(jì) 由于 11 ，次插值用分段線性、二處的近似值在求)(,)( ?xxf1 8 8 8 7 3 3 9 6 7 7 8 1 1 0 7 0 1 6 543210iiyxi在各節(jié)點(diǎn)處的數(shù)據(jù)為設(shè) )( xf例 : )()(1 xL k11?????kkkk xxxxykkkk xxxxy?????11解 : (1). 分段線性 Lagrange插值的公式為 1,1,0 ?? nk ?12 )()0(1L 0 16 ??? 1 0 7 ????)()1(1L 1 0 7 ??? 7 8 1 ???43 30 ?)()3(1L ?)()4(1L ?)()4(1L 7 3 3 ??? 8 8 8 ?????)(f?)(f?)(f?)(f?)(f同理 13 (2). 分段二次 Lagrange插值的公式為 ?))(())((2022210 xxxxxxy?????( 0 )2 (0 .3 6 )L))(())((2101201 xxxxxxy?????))(())((1202102 xxxxxxy??????)(f1212( ) ( )( ) ( )kkkk k k kx x x xyx x x x?????????0 , 2 , , 2kn??)()(2 xL k 211 1 2( ) ( )( ) ( )kkkk k k kx x x xyx x x x??? ? ??????122 2 1( ) ( )( ) ( )kkkk k k kx x x xyx x x x??? ? ??????14 ?))(())((2022210 xxxxxxy?????( 0 )2 (0 .4 2 )L))(())((2101201 xxxxxxy?????))(())((1202102 xxxxxxy??????))(())((5343543 xxxxxxy?????( 3 )2 (0 .7 5)L))(())((5454534 xxxxxxy?????))(())((4535435 xxxxxxy??????)(f?)(f?)(f ?)(?(
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