L a g r a n g e \qquad Lagrange Lagrange和 N e w t o n Newton Newton插值法都要求与被插值函数在插值节点上的函数值相等，现在如果再将要求收敛一下，还要求在节点上它们的一阶导数甚至高阶导数也相等呢？这时我们该怎么处理这类问题嘞？所以此时我们引入了今天的主题 H e r m i t e Hermite Hermite插值法。

\qquad 给定 n + 1 n+1 n+1个相异的插值节点，不妨设 a ≤ x 0 < x 1 < . . . < x n ≤ b a\leq x_0< x_1αj​(xk​)=δjk​βj​(xk​)=0​αj′​(xk​)=0βj′​(xk​)=δjk​​

\qquad 利用 α j , β j \alpha_j,\beta_j αj​,βj​ ( j = 0 , 1 , 2 , . . . , n ) (j=0,1,2,...,n) (j=0,1,2,...,n)来构造多项式 H 2 n + 1 ( x ) = ∑ j = 0 n [ y j α j ( x ) + m j β j ( x ) ] H_{2n+1}(x)=\sum\limits_{j=0}^{n}[y_j\alpha_j(x)+m_j\beta_j(x)] H2n+1​(x)=j=0∑n​[yj​αj​(x)+mj​βj​(x)]由于 α j , β j ∈ ℘ 2 n + 1 \alpha_j,\beta_j\in \wp_{2n+1} αj​,βj​∈℘2n+1​,所以 H 2 n + 1 ∈ ℘ 2 n + 1 H_{2n+1}\in\wp_{2n+1} H2n+1​∈℘2n+1​，由基函数条件可以得到 H 2 n + 1 H_{2n+1} H2n+1​满足插值条件，后结合n次 L a g r a n g e Lagrange Lagrange插值多项式的基函数：

l i ( x ) = ∏ i = 0 , i ≠ j n ( x − x j x j − x i ) , \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad l_i(x)=\prod\limits_{i=0,i\neq j}^n\Large (\frac{x-x_j}{x_j-x_i}), li​(x)=i=0,i​=j∏n​(xj​−xi​x−xj​​), j = 0 , 1 , 2... n j=0,1,2...n j=0,1,2...n 在加上简易推导（嘿嘿兴趣的可以自己去搞一下哦）即可得到对应插值多项式 其中应用最为广泛的就是三次 H e r m i t e Hermite Hermite插值多项式： H 3 ( x ) = f ( x k ) α k ( x ) + f ( x k + 1 ) α k + 1 ( x ) + f ′ ( x k ) β k ( x ) + f ′ ( x k + 1 ) β k + 1 ( x ) H_3(x)=f(x_k)\alpha_k(x)+f(x_{k+1})\alpha_{k+1}(x)+f'(x_k)\beta_k(x)+f'(x_{k+1})\beta_{k+1}(x) H3​(x)=f(xk​)αk​(x)+f(xk+1​)αk+1​(x)+f′(xk​)βk​(x)+f′(xk+1​)βk+1​(x)其中的基函数 α j ( x ) , β j ( x ) \alpha_j(x),\beta_j(x) αj​(x),βj​(x)分别是个啥呢？

{ α k ( x ) = ( 1 + 2 x − x k x k + 1 − x k ) ( x − x k + 1 x k − x k + 1 ) 2 α k + 1 ( x ) = ( 1 + 2 x − x k + 1 x k − x k + 1 ) ( x − x k x k + 1 − x k ) 2 \begin{cases} \alpha_k(x)=(1+2\Large \frac{x-x_k}{x_{k+1}-x_k})(\frac{x-x_{k+1}}{x_k-x_{k+1}})^2\\ \\ \alpha_{k+1}(x)=(1+2\Large \frac{x-x_{k+1}}{x_k-x_{k+1}})(\frac{x-x_k}{x_{k+1}-x_k})^2 \end{cases} ⎩⎪⎪⎪⎨⎪⎪⎪⎧​αk​(x)=(1+2xk+1​−xk​x−xk​​)(xk​−xk+1​x−xk+1​​)2αk+1​(x)=(1+2xk​−xk+1​x−xk+1​​)(xk+1​−xk​x−xk​​)2​

{ β k ( x ) = ( x − x k ) ( x − x k + 1 x k − x k + 1 ) 2 β k ( x ) = ( x − x k + 1 ) ( x − x k x k + 1 − x k ) 2 \begin{cases} \beta_k(x)=({x-x_k})\Large(\frac{x-x_{k+1}}{x_k-x_{k+1}})^2\\ \\ \beta_k(x)=({x-x_{k+1}})\Large(\frac{x-x_k}{x_{k+1}-x_k})^2 \end{cases} ⎩⎪⎪⎨⎪⎪⎧​βk​(x)=(x−xk​)(xk​−xk+1​x−xk+1​​)2βk​(x)=(x−xk+1​)(xk+1​−xk​x−xk​​)2​

\qquad 设 f ∈ C 2 n + 2 [ a , b ] , x 0 , x 1 , x 2 , . . . , x n ∈ [ a , b ] f\in C^{2n+2}[a,b],x_0,x_1,x_2,...,x_n\in[a,b] f∈C2n+2[a,b],x0​,x1​,x2​,...,xn​∈[a,b]为相异节点， H 2 n + 1 H_{2n+1} H2n+1​为满足插值条件的2n+1次 H e r m i t e Hermite Hermite插值多项式，那么对于任意的 x ∈ [ a , b ] x\in[a,b] x∈[a,b]存在 ξ = ξ ( x ) ∈ ( a , b ) \xi=\xi(x)\in(a,b) ξ=ξ(x)∈(a,b)使得：

\qquad \qquad \qquad \qquad R 2 n + 1 ( x ) = f ( x ) − H 2 n + 1 ( x ) = f ( 2 n + 2 ) ( ξ ) ( 2 n + 2 ) ! ω n + 1 2 ( x ) R_{2n+1}(x)=f(x)-H_{2n+1}(x)= \Large \frac{f^{(2n+2)}(\xi)}{(2n+2)!}\omega^2_{n+1}(x) R2n+1​(x)=f(x)−H2n+1​(x)=(2n+2)!f(2n+2)(ξ)​ωn+12​(x)

其中 ω n + 1 ( x ) = ∏ i = 0 n ( x − x i ) \omega_{n+1}(x)=\prod\limits_{i=0}^n(x-x_i) ωn+1​(x)=i=0∏n​(x−xi​)

老规矩，看不懂公式没关系，我们之间看看咱会不会用公式就行了。先会用然后才有机会弄懂这玩意到底在干嘛。 咱自己也上来手写一道试试看：