函數(shù)近似計算的插值法hermite插值法-全文預(yù)覽
【正文】 101??????????xxxx)())(21( 201 xlxl ???即 將以上結(jié)果代入 3 0 0 1 1 0 0 1 1( ) ( ) ( ) ( ) ( )H x f x f x f x f x? ? ? ???? ? ? ?得兩個節(jié)點的三次 Hermite插值公式 3 0 0 1 1 0 0 1 1( ) ( ) ( ) ( ) ( )H x f x f x f x f x? ? ? ???? ? ? ?21 0 1( 1 2 ( ) ) ( )f l x l x? ? ?20 0 0( ) ( )f x x l x?? ? ? 21 1 1( ) ( )f x x l x?? ? ?20 1 0( 1 2 ( ) ) ( )f l x l x? ? ?110112 xxf xx??????????2010??????????xxxx? ?00f x x??? 2101??????????xxxx2010??????????xxxx? ?11f x x???001012 xxf xx??????????2101??????????xxxx二、兩點三次 Hermite插值的余項 兩點三次 Hermite插值的誤差為 )()()( 33 xHxfxR ??0)()()( 33 ??? iii xHxfxR0)()()( 33 ?????? iii xHxfxR1,0?i因此可設(shè)的二重零點均為 ,)(, 310 xRxx21203 )())(()( xxxxxKxR ???待定其中 )( xK21
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