【简答题】![](https://cos-cdn.shuashuati.com/shuashuati-web/2024-0521-0905-00/logo-new-ad743.png)
已知函数y=(χ)过点(1,0),(2,-5),(3,-6),(4,3),求经过这些点的Lagrange插值多项式L3(χ),并求f(设χ
设χ0≠χ2,证明有唯一的三次多项式P(χ)满足插值条件 P(χ10)=f(χ0),p(χ2)=f(χ2) P′(χ1)=f′(χ1),P〞(χ1)=f〞(χ1) 并求出P(χ)。如果χ0=-1,χ1=0,χ2=1,f(χ)∈C4[-1,1],则任意χ∈[-1,1],有f(χ)-p(χ)=
(ξ),ξ∈(-1,1)
![](data:image/png;base64,iVBORw0KGgoAAAANSUhEUgAAACAAAAAgCAYAAABzenr0AAAA5ElEQVRYR2NkGGDAOMD2M+B0wKpVq9huv/jbxPD/fxwDA4MkhQ59zsDIuEhVgrkuLCzsF7JZOB3QOml5B8P//+UUWoyqnZGxszovsoI4B0xc9gyfzxmZmb2rcsK3IRvWMWm599///7fgcfTz6vwoKWId8B+f75kZGX0q8iK3kugAhur8KJRQxx0FE5fhdQC5UTPqgNEQGA2B0RAYDYHREBgNgaETAuTW96TqG7yt4tbRFhGRIQBqHTP9+8dIoDUMTxqk5AK8zXKYiaDWMYhNpANIaJYPdMdkwLtmpOZnctUPeDkAAP6tmiH1ggTLAAAAAElFTkSuQmCC)
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参考解析:
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