注意：一阶三点公式 P 2 ′ ( x 0 t h ) P_{2}' (x_{0} th) P2′(x0 th)(p119 4.61)是由 P 2 ( x 0 t h ) P_{2} (x_{0} th) P2(x0 th)对 x x x求导得到的.(原书写的对x求导)

d d x P 2 ( x 0 + t h ) = d d t P 2 ( x 0 + t h ) × d d t d t d x ( x 0 + t h ) \frac{d}{dx} P_{2} (x_{0}+th)= \frac{d}{dt}P_{2} (x_{0}+th)\times\frac{d}{dt}\frac{dt}{dx}(x_{0}+th) dxd​P2​(x0​+th)=dtd​P2​(x0​+th)×dtd​dxdt​(x0​+th) d d t d t d x ( x 0 + t h ) = 1 h \frac{d}{dt}\frac{dt}{dx}(x_{0}+th)=\frac{1}{h} dtd​dxdt​(x0​+th)=h1​ 关于 t t t求导的部分比较简单，不再展开

二阶三点公式是通过对一阶三点公式 P 2 ′ ( x 0 + t h ) P_{2}' (x_{0}+th) P2′​(x0​+th)(p119 4.61)关于 t 求导得到的。同理有： P 2 ′ ′ ( x 0 + t h ) = d d x P 2 ′ ( x 0 + t h ) = d d t P 2 ′ ( x 0 + t h ) × d d t d t d x ( x 0 + t h ) P_{2}'' (x_{0}+th)=\frac{d}{dx} P_{2}' (x_{0}+th)= \frac{d}{dt}P_{2}' (x_{0}+th)\times\frac{d}{dt}\frac{dt}{dx}(x_{0}+th) P2′′​(x0​+th)=dxd​P2′​(x0​+th)=dtd​P2′​(x0​+th)×dtd​dxdt​(x0​+th) d d t d t d x ( x 0 + t h ) = 1 h \frac{d}{dt}\frac{dt}{dx}(x_{0}+th)=\frac{1}{h} dtd​dxdt​(x0​+th)=h1​ 因此二阶插值多项式有系数 1 h 2 \frac{1}{h^2} h21​，关于 t t t求导的部分比较简单，不再展开

二阶三点公式的余项来自对二阶三点公式的泰勒展开。

二阶三点公式： P 2 ′ ′ ( x 0 + t h ) = 1 h 2 [ f ( x 0 ) − 2 f ( x 1 ) − f ( x 2 ) ] P_{2}'' (x_{0}+th)=\frac{1}{h^2}[f(x_0)-2f(x_1)-f(x_2)] P2′′​(x0​+th)=h21​[f(x0​)−2f(x1​)−f(x2​)]

将 f ( x 0 ) f(x_0) f(x0​)在 x = − h x=-h x=−h处做泰勒展开： f ( x 0 ) = f ( x 1 − h ) = f ( x 1 ) − h f ′ ( x 1 ) + h 2 2 ! f ′ ′ ( x 1 ) − h 3 3 ! f ′ ′ ′ ( x 1 ) + h 4 4 ! f ′ ′ ′ ′ ( x 1 ) . . . . f(x_0)=f(x_1-h)=f(x_1)-hf'(x_1)+\frac{h^2}{2!}f''(x_1)-\frac{h^3}{3!}f'''(x_1)+\frac{h^4}{4!}f''''(x_1).... f(x0​)=f(x1​−h)=f(x1​)−hf′(x1​)+2!h2​f′′(x1​)−3!h3​f′′′(x1​)+4!h4​f′′′′(x1​)....

将 f ( x 2 ) f(x_2) f(x2​)在 x = h x=h x=h处做泰勒展开： f ( x 2 ) = f ( x 1 + h ) = f ( x 1 ) + h f ′ ( x 1 ) + h 2 2 ! f ′ ′ ( x 1 ) + h 3 3 ! f ′ ′ ′ ( x 1 ) + h 4 4 ! f ′ ′ ′ ′ ( x 1 ) . . . . f(x_2)=f(x_1+h)=f(x_1)+hf'(x_1)+\frac{h^2}{2!}f''(x_1)+\frac{h^3}{3!}f'''(x_1)+\frac{h^4}{4!}f''''(x_1).... f(x2​)=f(x1​+h)=f(x1​)+hf′(x1​)+2!h2​f′′(x1​)+3!h3​f′′′(x1​)+4!h4​f′′′′(x1​)....

f ( x ) f(x) f(x)在a点处(或以a为中心)进行泰勒展开的公式： f ( x + a ) = a n n ! f n ( x ) , n = 0 , 1 , 2 , 3... f(x+a)=\frac{a^n}{n!}f^n(x),n=0,1,2,3... f(x+a)=n!an​fn(x),n=0,1,2,3... f ( x − a ) = ( − 1 ) n a n n ! f n ( x ) , n = 0 , 1 , 2 , 3... f(x-a)=(-1)^n\frac{a^n}{n!}f^n(x),n=0,1,2,3... f(x−a)=(−1)nn!an​fn(x),n=0,1,2,3...

则 1 h 2 [ f ( x 0 ) − 2 f ( x 1 ) − f ( x 2 ) ] = 1 h 2 [ − 2 f ( x 1 ) + f ( x 1 ) − h f ′ ( x 1 ) + h 2 2 ! f ′ ′ ( x 1 ) − h 3 3 ! f ′ ′ ′ ( x 1 ) + h 4 4 ! f ′ ′ ′ ′ ( x 1 ) − [ f ( x 1 ) + h f ′ ( x 1 ) + h 2 2 ! f ′ ′ ( x 1 ) + h 3 3 ! f ′ ′ ′ ( x 1 ) + h 4 4 ! f ′ ′ ′ ′ ( x 1 ) . . . ] ] = f ′ ′ ( x 1 ) + h 4 12 f ′ ′ ′ ′ ( x 1 ) . . . \frac{1}{h^2}[f(x_0)-2f(x_1)-f(x_2)]\\ =\frac{1}{h^2}[ -2f(x_1)+f(x_1)-hf'(x_1)+\frac{h^2}{2!}f''(x_1)-\frac{h^3}{3!}f'''(x_1)+\frac{h^4}{4!}f''''(x_1)-\\ [f(x_1)+hf'(x_1)+\frac{h^2}{2!}f''(x_1)+\frac{h^3}{3!}f'''(x_1)+\frac{h^4}{4!}f''''(x_1)...] ]\\ =f''(x_1)+\frac{h^4}{12}f''''(x_1)... h21​[f(x0​)−2f(x1​)−f(x2​)]=h21​[−2f(x1​)+f(x1​)−hf′(x1​)+2!h2​f′′(x1​)−3!h3​f′′′(x1​)+4!h4​f′′′′(x1​)−[f(x1​)+hf′(x1​)+2!h2​f′′(x1​)+3!h3​f′′′(x1​)+4!h4​f′′′′(x1​)...]]=f′′(x1​)+12h4​f′′′′(x1​)... 在 h 4 4 ! f ′ ′ ′ ′ ( x 1 ) \frac{h^4}{4!}f''''(x_1) 4!h4​f′′′′(x1​)处截断， f ′ ′ ( x 1 ) = 1 h 2 [ f ( x 0 ) − 2 f ( x 1 ) − f ( x 2 ) ] + h 4 12 f ′ ′ ′ ′ ( ε ) , ε ∈ ( x , x 1 ) f''(x_1)=\frac{1}{h^2}[f(x_0)-2f(x_1)-f(x_2)]+\frac{h^4}{12}f''''( \varepsilon ) ,\varepsilon \in (x,x_1) f′′(x1​)=h21​[f(x0​)−2f(x1​)−f(x2​)]+12h4​f′′′′(ε),ε∈(x,x1​)