泰勒展开式是数学中极其强大的函数近似工具。利用函数某个点的导数，来近似这个点附近的函数值。

意义：用一个 n n n次多项式来近似函数 f ( x ) f(x) f(x)在 a a a点附近的部分。

1. n n n阶泰勒公式： 若 f ( x ) f(x) f(x)有直到 n + 1 n+1 n+1的导数，则有 f ( x ) = f ( a ) + f ′ ( a ) ( x − a ) + f ′ ′ ( a ) ( x − a ) 2 2 ! + . . . + f ( n ) ( a ) ( x − a ) n n ! + R n ( x )       \bm {f(x)=f(a)+f'(a)(x-a)+f''(a)\frac{(x-a)^2}{2!}+...+f^{(n)}(a)\frac{(x-a)^n}{n!}+R_n(x) \ \ \ \ \ } f(x)=f(a)+f′(a)(x−a)+f′′(a)2!(x−a)2​+...+f(n)(a)n!(x−a)n​+Rn​(x)      其中 R n ( x ) = f ( n + 1 ) ( c ) ( n + 1 ) ! ( x − a ) n + 1 ,   c ∈ ( a , x ) \bm{R_n(x)=\frac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1},\ c\in(a,x)} Rn​(x)=(n+1)!f(n+1)(c)​(x−a)n+1, c∈(a,x)

R n ( x ) \bm{R_n(x)} Rn​(x)被称为拉格朗日余项。

2. n n n阶泰勒公式的推导

我们希望让n次多项式 (1) P n ( x ) = b 0 + b 1 ( x − a ) + b 2 ( x − a ) 2 + . . . + b n ( x − a ) n \bm{P_n(x)=b_0+b_1(x-a)+b_2(x-a)^2+...+b_n(x-a)^n \tag{1}} Pn​(x)=b0​+b1​(x−a)+b2​(x−a)2+...+bn​(x−a)n(1) 能够近似函数 f ( x ) f(x) f(x)在点 a a a附近的部分。 对 P n ( x ) P_n(x) Pn​(x)求各阶导数，有            P n ′ ( x ) = b 1 + 2 b 2 ( x − a ) + . . . + n b n ( x − a ) n − 1 \ \ \ \ \ \ \ \ \ \ P_n'(x)=b_1+2b_2(x-a)+...+nb_n(x-a)^{n-1}           Pn′​(x)=b1​+2b2​(x−a)+...+nbn​(x−a)n−1

           P n ′ ′ ( x ) = 2 ! b 2 + . . . + n ( n − 1 ) b n ( x − a ) n − 2 \ \ \ \ \ \ \ \ \ \ P_n''(x)=2!b_2+...+n(n-1)b_n(x-a)^{n-2}           Pn′′​(x)=2!b2​+...+n(n−1)bn​(x−a)n−2            . . . \ \ \ \ \ \ \ \ \ \ ...           ...

           P n ( n ) ( x ) = n ! b n \ \ \ \ \ \ \ \ \ \ P^{(n)}_n(x)=n!b_n           Pn(n)​(x)=n!bn​

让 P n ( x ) P_n(x) Pn​(x)满足以下条件：   (1) 当 x = a x=a x=a，该多项式的值和 f ( x ) f(x) f(x)的函数值相等，则            f ( a ) = P n ( a ) = b 0 \ \ \ \ \ \ \ \ \ \ f(a)=\textcolor{green}{P_n(a)=b_0}           f(a)=Pn​(a)=b0​   (2) 当 x = a x=a x=a，该多项式的各阶导数值和 f ( x ) f(x) f(x)的各阶导数值相等，则            f ′ ( a ) = P n ′ ( a ) = b 1 \ \ \ \ \ \ \ \ \ \ f'(a)=\textcolor{green}{P_n'(a)=b_1}           f′(a)=Pn′​(a)=b1​

           f ′ ′ ( a ) = P n ′ ′ ( a ) = 2 ! b 2 \ \ \ \ \ \ \ \ \ \ f''(a)=\textcolor{green}{P_n''(a)=2!b_2}           f′′(a)=Pn′′​(a)=2!b2​            . . . \ \ \ \ \ \ \ \ \ \ ...           ...            f ( n ) ( a ) = P n ( n ) ( a ) = n ! b n \ \ \ \ \ \ \ \ \ \ f^{(n)}(a)=\textcolor{green}{P_n^{(n)}(a)=n!b_n}           f(n)(a)=Pn(n)​(a)=n!bn​

即得 b 0 = f ( a ) , b 1 = f ′ ( a ) , b 2 = f ′ ′ ( a ) 2 ! , . . . b n = f ( n ) ( a ) n ! b_0=f(a), b_1=f'(a), b_2=\frac{f''(a)}{2!},...b_n=\frac{f^{(n)}(a)}{n!} b0​=f(a),b1​=f′(a),b2​=2!f′′(a)​,...bn​=n!f(n)(a)​ 代入到公式(1)，则有 f ( x ) = f ( a ) + f ′ ( a ) ( x − a ) + f ′ ′ ( a ) ( x − a ) 2 2 ! + . . . + f ( n ) ( a ) ( x − a ) n n ! + R n ( x )       \bm {f(x)=f(a)+f'(a)(x-a)+f''(a)\frac{(x-a)^2}{2!}+...+f^{(n)}(a)\frac{(x-a)^n}{n!}+R_n(x) \ \ \ \ \ } f(x)=f(a)+f′(a)(x−a)+f′′(a)2!(x−a)2​+...+f(n)(a)n!(x−a)n​+Rn​(x)     

3. 用微分中值定理证明拉格朗日余项： 构造一个函数 g ( t ) g(t) g(t)： g ( t ) = f ( x ) − f ( t ) − f ′ ( t ) ( x − t ) − f ′ ′ ( t ) ( x − t ) 2 2 ! − . . . − f ( n ) ( t ) ( x − t ) n n ! − R n ( x ) ( x − t ) n + 1 ( x − a ) n + 1 g(t)=f(x)-f(t)-f'(t)(x-t)-f''(t)\frac{(x-t)^2}{2!}-...-f^{(n)}(t)\frac{(x-t)^n}{n!}-R_n(x)\frac{(x-t)^{n+1}}{(x-a)^{n+1}} g(t)=f(x)−f(t)−f′(t)(x−t)−f′′(t)2!(x−t)2​−...−f(n)(t)n!(x−t)n​−Rn​(x)(x−a)n+1(x−t)n+1​ 则有            g ( x ) = 0 \ \ \ \ \ \ \ \ \ \ g(x)=0           g(x)=0            g ( a ) = R n − R n = 0 \ \ \ \ \ \ \ \ \ \ g(a)=R_n-R_n=0           g(a)=Rn​−Rn​=0 对 g ( t ) g(t) g(t)求导有

g ′ ( t ) = − f ′ ( t ) − [ f ′ ( t ) ( − 1 ) + f ′ ′ ( t ) ( x − t ) ] − [ f ′ ′ ( t ) ( x − t ) ( − 1 ) + f ′ ′ ′ ( t ) ( x − t ) 2 2 ! ] − . . . − [ f ( n ) ( t ) ( x − t ) n − 1 ( n − 1 ) ! ( − 1 ) + f ( n + 1 ) ( t ) ( x − t ) n n ! ] + R n ( x ) ( n + 1 ) ( x − t ) n ( x − a ) n + 1 g'(t)=\textcolor{red}{-f'(t)}-[\textcolor{red}{f'(t)(-1)}+\textcolor{darkorange}{f''(t)(x-t)}]-[\textcolor{darkorange}{f''(t)(x-t)(-1)}+\textcolor{blue}{f'''(t)\frac{(x-t)^2}{2!}}]-...\\ -[\textcolor{blue}{f^{(n)}(t)\frac{(x-t)^{n-1}}{(n-1)!}(-1)}+f^{(n+1)}(t)\frac{(x-t)^n}{n!}]+R_n(x)\frac{(n+1)(x-t)^n}{(x-a)^{n+1}} g′(t)=−f′(t)−[f′(t)(−1)+f′′(t)(x−t)]−[f′′(t)(x−t)(−1)+f′′′(t)2!(x−t)2​]−...−[f(n)(t)(n−1)!(x−t)n−1​(−1)+f(n+1)(t)n!(x−t)n​]+Rn​(x)(x−a)n+1(n+1)(x−t)n​ 相邻两项相互抵消，则 g ′ ( t ) = − f ( n + 1 ) ( t ) ( x − t ) n n ! + R n ( x ) ( n + 1 ) ( x − t ) n ( x − a ) n + 1 g'(t)=-f^{(n+1)}(t)\frac{(x-t)^n}{n!}+R_n(x)\frac{(n+1)(x-t)^n}{(x-a)^{n+1}} g′(t)=−f(n+1)(t)n!(x−t)n​+Rn​(x)(x−a)n+1(n+1)(x−t)n​ 根据罗尔定理，存在 c ∈ ( a , x ) c\in(a,x) c∈(a,x)，使得 g ′ ( c ) = 0 g'(c)=0 g′(c)=0，则 g ′ ( c ) = − f ( n + 1 ) ( c ) ( x − c ) n n ! + R n ( x ) ( n + 1 ) ( x − c ) n ( x − a ) n + 1 = 0 g'(c)=-f^{(n+1)}(c)\frac{(x-c)^n}{n!}+R_n(x)\frac{(n+1)(x-c)^n}{(x-a)^{n+1}}=0 g′(c)=−f(n+1)(c)n!(x−c)n​+Rn​(x)(x−a)n+1(n+1)(x−c)n​=0 两边同时除以 ( x − c ) n (x-c)^n (x−c)n，得 − ( f ( n + 1 ) ( c ) n ! + R n ( x ) ( n + 1 ) ( x − a ) n + 1 = 0 -\frac{(f^{(n+1)}(c)}{n!}+R_n(x)\frac{(n+1)}{(x-a)^{n+1}}=0 −n!(f(n+1)(c)​+Rn​(x)(x−a)n+1(n+1)​=0 所以 R n ( x ) = f ( n + 1 ) ( c ) ( n + 1 ) ! ( x − a ) n + 1 ,   c ∈ ( a , x ) R_n(x)=\frac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1},\ c\in(a,x) Rn​(x)=(n+1)!f(n+1)(c)​(x−a)n+1, c∈(a,x)

参考资料： 1.《高等数学》第六版（上册）.高等教育出版社, 2007, p. 275–282. 2. https://www.youtube.com/watch?v=NZBvVkGn8CU