F(t)?f(t)?H(t)?wn(t)wr(t)[f(x)?H(x)]
wn(x)wr(x)由条件(1)知
F(x)?F(x0)?F(x1)???F(xn)?0 F?(x0)?F?(x1)???F?(xr)?0
即F?t??0有n?r?1个单根x,xr?1,xr?2,…,xn和r+1个二重根x,x1,…,xr
由Rolle定理知,
\'F(t)至少有n+r+1个零点 F(t)在(a,b)内至少有n+r+2个零点,
依此类推可知:F(n?r?2)(t)在(a,b)内至少有一个零点 因此
f(n?r?2)(?)?(n?r?2)![f(x)?H(x)]?0
wn(x)wr(x)f(n?r?2)(?)wn(x)wr(x) 即得 f(x)?H(x)?(n?r?2)!
若r?n,则响应的Hermite差值多项式为
H(x)??hk(x)f(xk)??hk(x)f?(xk) -----(11)
k?0k?0nn其中
2?(xk)]lknhk(x)?[1?2(x?xk)lkn(x),k?0,1,?,nhk(x)?(x?xk)l(x),k?0,1,?,n2kn
f(2n?2)(?)2wn(x),??(a,b) 余项公式为:f(x)?H(x)?(2n?2)!特别当r?n?1时,插值条件为:H(xi)?f(xi),H?(xi)?f?(xi),i?0,1
由此得三次Hermite插值多项式:
H(x)?h0(x)f(x0)?h1(x)f(x1)?h0(x)f?(x0)?h1(x)f?(x1)-----(12)
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h0(x)?(1?2 h0(x)?(x?x0)(
多项式(12)常用作分段低次插值,称为分段三次Hermite插值
x?x02x?x12) h1(x)?(x?x1)() x0?x1x1?x0x?x0x?x12x?x1x?x02)() h1(x)?(1?2)() x1?x0x0?x1x0?x1x1?x02 两点三次Hermite插值及其余项
2.1 两点三次Hermite插值 先考虑只有两个节点的插值问题:
?,y1?,两个设f(x)在节点x0,x1处的函数值为y0,y1,在节点x0,x1处的一阶导数值为y0节点最高可以用3次Hermite多项式H3(x),作为插值函数H3(x)应满足插值条件:
H3(x0)?y0 H3(x1)?y1 ?(x0)?y0? H3?(x1)?y1? H3 H3(x)应用四个插值基函数表示,设H3(x)的插值基函数为hi(x),i?0,1,2,3
H3(x)?a0h0(x)?a1h1(x)?a2h2(x)?a3h3(x)
希望插值系数与Lagrange插值一样简单重新假设
??0(x)?y1??1(x) H3(x)?y0?0(x)?y1?1(x)?y0?(x)?y0?0?(x)?y1?1?(x)?y0??0?(x)?y1??1?(x) H3其中
?(x1)?0 ?(x0)?0 ?0?0(x0)?1 ?0(x1)?0 ?0?1(x0)?0 ?1(x1)?1 ?1?(x0)?0 ?1?(x1)?0
?(x1)?0 ?(x0)?1 ?0?0(x0)?0 ?0(x1)?0 ?0?1(x0)?0 ?1(x1)?0 ?1?(x0)?0 ?1?(x1)?1
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可知 x1是?0(x)的二重零点,即可假设?0(x)?(x?x1)2(ax?b)
?(x0)?0 由 ?0(x0)?1 ?0
可得 a??
?2x0?2x1?0(x)?(x?x1)(ax?b)?(x?x1)???? 323?(x?x)(x?x)(x?x)010101??222x021 b??(x0?x1)2(x0?x1)3(x0?x1)32x0(x?x1)2?2x??1??? 2?(x0?x1)?x0?x1x0?x1??x?x0??x?x1?2??1?2????(1?2l1(x))?l0(x)………. Lagrange插值基函数
x1?x0??x0?x1???x?x0??x?x1?2即 ?0(x)?(1?2l1(x))?l0(x)??1?2???
x?xx?x10??01??22?x?x1?类似可得 ?1(x)?(1?2l0(x))?l12(x)??1?2?
x?x01???x?x1? ?0(x)?(x?x0)?l02(x)??x?x0???
x?x?01??x?x0? ?1(x)?(x?x1)?l12(x)??x?x1???
?x1?x0?22将以上结果代入
??0(x)?y1??1(x) H3(x)?y0?0(x)?y1?1(x)?y0
得两个节点的三次Hermite插值公式
??0(x)?y1??1(x) H3(x)?y0?0(x)?y1?1(x)?y022?(x?x0)?l0?(x?x1)?l12(x)?y0(1?2l1(x))?l0(x)?y1(1?2l0(x))?l12(x)?y0(x)?y1
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???x?x1??x?x0?x?x0??x?x1?x?x1????y0?1?2????y1?1?2??y0?x?x0????y1?x?x1???x?xx?xx?xx?x10??01?01????01??x1?x0?222
2.2两点三次Hermite插值的余项
两点三次Hermite插值的误差为
R3(x)?f(x)?H3(x) R3(xi)?f(xi)?H3(xi)?0
?(xi)?f?(xi)?H3?(xi)?0 i?0,1 R3x0,x1均为R3(x)的二重零点,因此可设:
R3(x)?K(x)(x?x0)2(x?x1)2
其中K(x)待定
构造辅助函数 ?(t)?f(t)?H3(t)?K(x)(t?x0)2(t?x1)2
?(xi)?f(xi)?H3(xi)?K(x)(xi?x0)2(xi?x1)2?0 ?(xi)?f(xi)?H3(xi)?K(x)(xi?x0)2(xi?x1)2?0 i?0,1
均是二重根
?(x)?f(x)?H3(x)?K(x)(x?x0)2(x?x1)2?0
因此?(t)至少有5个零点
连续使用4次Rolle定理,可得,至少存在一点??[x0,x1],使得 ?(4)(?)?0
即?(4)(?)?f(4)(?)?4!K(x)?0
f(4)(?)K(x)?
4!所以,两点三次Hermite插值的余项为
f(4)(?)R3(x)?(x?x0)2(x?x1)2 x0???x1
4! 9
以上分析都能成立吗?
当f(4)(x)在[x0,x1]上存在时,上述余项公式成立
三次埃尔米特插值多项式
设y?f?x?是区间[a, b]上的实函数, x0,x1是[a, b]上相异两点, 且 x0<x1,
y?f?x?在xi上的函数值和一阶导数值分别为yi?f?xi? ?i?0,1?和mi? f?xi??i?0,1?,
求三次多项式H3?x?, 使其满足:
??H3(xi)?yi(i?0,1) ?'H(x)?m?i?3iH3(x)称为三次埃尔米特插值多项式。
误差估计
定理4 设f(x)在包含x0、x1的区间[a,b]内存在四阶导数,则当x∈[a,b]时有余项
R3(x)?f(x)?H3(x)?x0?x?x11(4)22 (??(a,b)且与x有关) f(?)(x?x0)(x?x1)4!设M4?maxf(4)(x) 则当x??x0,x1?时,
余项有如下估计式(误差限)R3(x)?M44h 384
二重Hermite插值多项式
常用的Hermite插值为mi=2 的情况,即给定的插值节点?xi?f(x)?C2??a,b??,及插值节点?xi?ni?0ni?0 均为二重节点,更具体些
,若有H2n?1(x)?P2n?1 满足H2n?1(xi)?f(xi)
的二重Hermite插值?x?为f?x?关于节点?xi?i?0 n''H2n?1(xi)?f(xi),i?0,1,…,n,就称H2n ?多项式。
[x0,x1]上存在且连续时,上述余项公式成立
Hermite插值优点:
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Hermite插值的若干问题研究
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