无论是Newton插值还是Lagrange插值，都只能在数值本身上满足插值函数与数据节点的重合，Hermite插值则要求其导数值相等。

设在节点 a ⩽ x 0 ⩽ x 1 ⩽ … ⩽ x n ⩽ b a\leqslant x_0\leqslant x_1 \leqslant\ldots\leqslant x_n\leqslant b a⩽x0​⩽x1​⩽…⩽xn​⩽b上， y j = f ( x j ) , m j = f ′ ( x j ) , j = 0 , 1 , … , n y_j=f(x_j),m_j=f'(x_j),j=0,1,\ldots,n yj​=f(xj​),mj​=f′(xj​),j=0,1,…,n，则插值多项式的条件为

H ( x j ) = y j , H ′ ( x j ) = m j , j = 0 , 1 , … , n H(x_j)=y_j,H'(x_j)=m_j,j=0,1,\ldots,n H(xj​)=yj​,H′(xj​)=mj​,j=0,1,…,n

总共有 2 n + 2 2n+2 2n+2个等式，可唯一确定一个次数不大于 2 n + 1 2n+1 2n+1的多项式 H 2 n + 1 ( x ) H_{2n+1}(x) H2n+1​(x)，设其表达式可通过基函数 α j , β j \alpha_j,\beta_j αj​,βj​写成

H 2 n + 1 ( x ) = ∑ j = 0 n [ y j α j + m j β j ] H_{2n+1}(x)=\displaystyle\sum_{j=0}^n[y_j\alpha_j+m_j\beta_j] H2n+1​(x)=j=0∑n​[yj​αj​+mj​βj​]

则 α j , β j \alpha_j,\beta_j αj​,βj​需要满足

{ α j ( x k ) = δ j k α j ′ ( x k ) = 0 β j ( x k ) = 0 β j ′ ( x k ) = δ j k \left\{\begin{aligned} \alpha_j(x_k) &= \delta_{jk}\\ \alpha_j'(x_k)&= 0\\ \beta_j(x_k) &= 0\\ \beta_j'(x_k) & = \delta_{jk} \end{aligned}\right. ⎩ ⎨ ⎧​αj​(xk​)αj′​(xk​)βj​(xk​)βj′​(xk​)​=δjk​=0=0=δjk​​

两个基函数的条件与Lagrange插值中的基函数相似，由于目标函数为 2 n + 1 2n+1 2n+1次，所以假设

{ α j = ( a x + b ) l j 2 ( x ) β j = ( c x + d ) l j 2 ( x ) \left\{\begin{aligned} \alpha_j = (ax+b)l_j^2(x)\\ \beta_j = (cx+d)l_j^2(x) \end{aligned}\right. {αj​=(ax+b)lj2​(x)βj​=(cx+d)lj2​(x)​

结合二者所满足的基函数条件，可得

{ α j ( x j ) = ( a x j + b ) l j 2 ( x j ) = 1 α j ′ ( x j ) = l j ( x j ) [ a l j ( x j ) + 2 ( a x j + b ) l j ′ ( x j ) ] = 0 β j ( x j ) = ( c x j + d ) l j 2 ( x j ) = 0 β j ′ ( x j ) = l j ( x j ) [ c l j ( x j ) + 2 ( c x j + d ) l j ′ ( x j ) ] = 1 \left\{\begin{aligned} \alpha_j(x_j)&=(ax_j+b)l_j^2(x_j)=1\\ \alpha_j'(x_j)&=l_j(x_j)[al_j(x_j)+2(ax_j+b)l_j'(x_j)]=0\\ \beta_j(x_j)&=(cx_j+d)l_j^2(x_j)=0\\ \beta_j'(x_j)&=l_j(x_j)[cl_j(x_j)+2(cx_j+d)l_j'(x_j)]=1 \end{aligned}\right. ⎩ ⎨ ⎧​αj​(xj​)αj′​(xj​)βj​(xj​)βj′​(xj​)​=(axj​+b)lj2​(xj​)=1=lj​(xj​)[alj​(xj​)+2(axj​+b)lj′​(xj​)]=0=(cxj​+d)lj2​(xj​)=0=lj​(xj​)[clj​(xj​)+2(cxj​+d)lj′​(xj​)]=1​

解得

{ α j = [ 1 − 2 （ x − x j ) ∑ k = 0 , k ≠ j n 1 x j − x k ] l j 2 ( x ) β j = ( x − x j ) l j 2 ( x ) \left\{\begin{aligned} \alpha_j &= [1-2（x-x_j)\displaystyle\sum_{k=0,k\not=j}^n\frac{1}{x_j-x_k}]l_j^2(x)\\ \beta_j &= (x-x_j)l_j^2(x) \end{aligned}\right. ⎩ ⎨ ⎧​αj​βj​​=[1−2（x−xj​)k=0,k=j∑n​xj​−xk​1​]lj2​(x)=(x−xj​)lj2​(x)​

令 s j = ∑ k = 0 , k ≠ j 1 x j − x k s_j=\displaystyle\sum_{k=0,k\not=j}\frac{1}{x_j-x_k} sj​=k=0,k=j∑​xj​−xk​1​,则Hermite插值公式可写为

H 2 n + 1 ( x ) = ∑ j = 0 n { [ y j + 2 s j y j x j − x j m j ] ⋅ l j 2 ( x ) + [ − 2 s j y j + m j ] ⋅ x l j 2 ( x ) } H_{2n+1}(x)=\displaystyle\sum_{j=0}^n\{ [y_j+2s_jy_jx_j-x_jm_j]\cdot l_j^2(x) +[-2s_jy_j+m_j]\cdot xl_j^2(x) \} H2n+1​(x)=j=0∑n​{[yj​+2sj​yj​xj​−xj​mj​]⋅lj2​(x)+[−2sj​yj​+mj​]⋅xlj2​(x)}

其代码为

验证

由于引入了一组随即变量代表导数，所以拟合得到的多项式函数自然与输入的多项式不同