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索引
- Taylor公式
- Taylor公式的定性分析
- 定理6.1 Taylor公式(Peano余项)
- Taylor公式的定量分析
- 定理6.2 Taylor公式(Lagrange余项)
Taylor公式
Taylor公式的定性分析
定理6.1 Taylor公式(Peano余项)
若函数 f ( x ) f\left ( x \right ) f(x)在 x 0 x_{0} x0处的 n n n阶导数均存在,则存在邻域 U ( x 0 , δ ) U\left ( x_{0},\delta \right ) U(x0,δ)使得 f ( x ) f\left ( x \right ) f(x)满足以下公式:
f ( x ) = ∑ k = 0 n f ( k ) ( x 0 ) ( x − x 0 ) k k ! + r n ( x ) f\left ( x \right ) =\sum_{k=0}^{n}\frac{f^{\left ( k \right ) } \left ( x_{0} \right )\left ( x-x_{0} \right )^{k} }{k!}+r_{n} \left ( x \right ) f(x)=∑k=0nk!f(k)(x0)(x−x0)k+rn(x)
其中Peano余项 r n ( x ) = o ( ( x − x 0 ) n ) ( x → x 0 ) r_{n} \left ( x \right )=o \left ( \left ( x-x_{0} \right )^{n} \right )(x\to x_{0} ) rn(x)=o((x−x0)n)(x→x0),代表 f ( x ) f\left ( x \right ) f(x)在Taylor近似时产生的误差,余下的 n n n次多项式记为 P n ( x ) P_{n}\left ( x \right ) Pn(x),即 x 0 x_{0} x0处的 n n n次Taylor多项式。
r n ( x ) = f ( x ) − P n ( x ) = f ( x ) − ∑ k = 0 n f ( k ) ( x 0 ) ( x − x 0 ) k k ! r_{n}\left ( x \right ) =f\left ( x \right ) -P_{n}\left ( x \right ) =f\left ( x \right ) -\sum_{k=0}^{n}\frac{f^{\left ( k \right ) } \left ( x_{0} \right )\left ( x-x_{0} \right )^{k} }{k!} rn(x)=f(x)−Pn(x)=f(x)−∑k=0nk!f(k)(x0)(x−x0)k,
r n ( n − 1 ) ( x ) = f ( n − 1 ) ( x ) − ∑ k = n − 1 n f ( k ) ( x 0 ) ( x − x 0 ) ( k − ( n − 1 ) ) ( k − ( n − 1 ) ) ! = f ( n − 1 ) ( x ) − ( f ( n − 1 ) ( x 0 ) 0 ! + f ( n ) ( x 0 ) ( x − x 0 ) 1 ! ) = f ( n − 1 ) ( x ) − f ( n − 1 ) ( x 0 ) − f ( n ) ( x 0 ) ( x − x 0 ) \begin{array}{l} &r^{\left(n-1\right)} _{n}\left ( x \right ) \\ =&f^{\left ( n-1 \right ) }\left ( x \right ) -\sum _{k=n-1}^{n}\frac{f^{\left ( k \right ) }\left ( x_{0} \right ) \left ( x-x_{0} \right )^{\left ( k-\left ( n-1 \right ) \right ) } }{\left ( k-\left (n-1 \right ) \right ) !} \\ =&f^{\left ( n-1 \right ) }\left ( x \right )-\left ( \frac{f^{\left ( n-1 \right ) }\left ( x_{0} \right ) }{0!}+\frac{f^{\left ( n \right )}\left ( x_{0} \right )\left ( x-x_{0} \right ) }{1!}\right ) \\ =&f^{\left ( n-1 \right ) }\left ( x \right )-f^{\left ( n-1 \right ) }\left ( x_{0} \right )-f^{\left ( n \right )}\left ( x_{0} \right )\left ( x-x_{0} \right ) \end{array} ===rn(n−1)(x)f(n−1)(x)−∑k=n−1n(k−(n−1))!f(k)(x0)(x−x0)(k−(n−1))f(n−1)(x)−(0!f(n−1)(x0)+1!f(n)(x0)(x−x0))f(n−1)(x)−f(n−1)(x0)−f(n)(x0)(x−x0),
因为函数 f ( x ) f\left ( x \right ) f(x)在 x 0 x_{0} x0处的 n n n阶导数均存在,所以 f ( x ) f\left ( x \right ) f(x)的前 n − 1 n-1 n−1阶导函数一定连续且可导。
连续引用n-1次L’Hospital法则, lim x → x 0 r n ( x ) ( x − x 0 ) n = lim x → x 0 r n ( 1 ) ( x ) n ( x − x 0 ) n − 1 = ⋯ = lim x → x 0 r n ( n − 1 ) ( x ) n ! ( x − x 0 ) \lim _{x\to x_{0} } \frac{r_{n}\left ( x \right ) }{\left ( x-x_{0} \right )^{n} }= \lim _{x\to x_{0} }\frac{r^{\left ( 1 \right ) } _{n}\left ( x \right ) }{n\left ( x-x_{0} \right )^{n-1} }=\cdots =\lim _{x\to x_{0} }\frac{r^{\left ( n-1 \right ) } _{n}\left ( x \right ) }{n! \left ( x-x_{0} \right ) } limx→x0(x−x0)nrn(x)=limx→x0n(x−x0)n−1rn(1)(x)=⋯=limx→x0n!(x−x0)rn(n−1)(x),
展开 r n ( n − 1 ) ( x ) r^{\left ( n-1 \right ) } _{n}\left ( x \right ) rn(n−1)(x), lim x → x 0 r n ( n − 1 ) ( x ) n ! ( x − x 0 ) = 1 n ! lim x → x 0 ( f ( n − 1 ) ( x ) − f ( n − 1 ) ( x 0 ) x − x 0 − f ( n ) ( x 0 ) ) \lim _{x\to x_{0} }\frac{r^{\left ( n-1 \right ) } _{n}\left ( x \right ) }{n! \left ( x-x_{0} \right ) }=\frac{1}{n!}\lim _{x\to x_{0} } \left ( \frac{f^{\left ( n-1 \right ) }\left ( x \right ) -f^{\left ( n-1 \right ) }\left ( x_{0} \right ) }{x-x_{0} } -f^{\left ( n \right ) }\left ( x_{0} \right ) \right ) limx→x0n!(x−x0)rn(n−1)(x)=n!1limx→x0(x−x0f(n−1)(x)−f(n−1)(x0)−f(n)(x0)),
根据高阶导数定义, lim x → x 0 f ( n − 1 ) ( x ) − f ( n − 1 ) ( x 0 ) x − x 0 = f ( n ) ( x 0 ) \lim _{x\to x_{0} }\frac{f^{\left ( n-1 \right ) }\left ( x \right ) -f^{\left ( n-1 \right ) }\left ( x_{0} \right ) }{x-x_{0} }=f^{\left ( n \right ) }\left ( x_{0} \right ) limx→x0x−x0f(n−1)(x)−f(n−1)(x0)=f(n)(x0),
所以 lim x → x 0 r n ( x ) ( x − x 0 ) n = lim x → x 0 r n ( n − 1 ) ( x ) n ! ( x − x 0 ) = 0 \lim _{x\to x_{0} } \frac{r_{n}\left ( x \right ) }{\left ( x-x_{0} \right )^{n} }=\lim _{x\to x_{0} }\frac{r^{\left ( n-1 \right ) } _{n}\left ( x \right ) }{n! \left ( x-x_{0} \right ) }=0 limx→x0(x−x0)nrn(x)=limx→x0n!(x−x0)rn(n−1)(x)=0, r n ( x ) = o ( ( x − x 0 ) n ) ( x → x 0 ) r_{n} \left ( x \right )=o \left ( \left ( x-x_{0} \right )^{n} \right )(x\to x_{0} ) rn(x)=o((x−x0)n)(x→x0)。
Taylor公式的定量分析
定理6.2 Taylor公式(Lagrange余项)
函数 f ( x ) f\left ( x \right ) f(x)在闭区间 [ a , b ] \left [ a,b \right ] [a,b]上存在 n n n阶连续导数,在开区间 ( a , b ) \left ( a,b \right ) (a,b)上存在 n + 1 n+1 n+1阶导数,取一定点 x 0 ∈ [ a , b ] x_{0}\in \left [ a,b \right ] x0∈[a,b],则 ∀ x ∈ [ a , b ] \forall x\in \left [ a,b \right ] ∀x∈[a,b], f ( x ) = ∑ k = 0 n f ( k ) ( x 0 ) ( x − x 0 ) k k ! + r n ( x ) f\left ( x \right )=\sum_{k=0}^{n}\frac{f^{\left ( k \right ) }\left ( x_{0} \right )\left ( x-x_{0} \right )^{k} }{k!}+r_{n}\left ( x \right ) f(x)=∑k=0nk!f(k)(x0)(x−x0)k+rn(x),其中Lagrange余项 r n ( x ) = f ( n + 1 ) ( ξ ) ( x − x 0 ) n + 1 ( n + 1 ) ! r_{n}\left ( x \right )=\frac{f^{\left ( n+1 \right ) }\left ( \xi \right )\left ( x-x_{0} \right )^{n+1} }{\left ( n+1 \right ) !} rn(x)=(n+1)!f(n+1)(ξ)(x−x0)n+1,其中 ξ \xi ξ位于 x x x和 x 0 x_{0} x0之间。
当 n = 0 n=0 n=0时,Taylor公式表现为Lagrange中值定理的形式。
构造函数 G ( t ) = f ( x ) − ∑ k = 0 n ( f ( k ) ( t ) k ! ( x − t ) k ) G\left ( t \right )=f\left ( x \right )-\sum_{k=0}^{n}\left ( \frac{f^{\left ( k \right ) }\left (t \right ) }{k!}\left ( x-t \right )^{k} \right) G(t)=f(x)−∑k=0n(k!f(k)(t)(x−t)k)和 H ( t ) = ( x − t ) n + 1 H\left ( t \right )=\left ( x-t \right )^{n+1} H(t)=(x−t)n+1,
不妨设 x < x 0 x<x_{0} x<x0,
G ( x ) = f ( x ) − ∑ k = 1 n ( f ( k ) ( t ) k ! ( x − x ) k ) − f ( t ) 0 ! ( x − t ) 0 = f ( x ) − f ( x ) = 0 G\left ( x \right )=f\left ( x \right )-\sum_{k=1}^{n}\left ( \frac{f^{\left ( k \right ) }\left (t \right ) }{k!}\left ( x-x \right )^{k} \right)-\frac{f\left ( t \right ) }{0!}\left ( x-t \right )^{0}=f\left ( x \right )-f\left ( x \right )=0 G(x)=f(x)−∑k=1n(k!f(k)(t)(x−x)k)−0!f(t)(x−t)0=f(x)−f(x)=0;
G ( x 0 ) = f ( x ) − ∑ k = 0 n ( f ( k ) ( x 0 ) k ! ( x − x 0 ) k ) G\left ( x_{0} \right )=f\left ( x \right )-\sum_{k=0}^{n}\left ( \frac{f^{\left ( k \right ) }\left (x_{0} \right ) }{k!}\left ( x-x_{0} \right )^{k} \right) G(x0)=f(x)−∑k=0n(k!f(k)(x0)(x−x0)k);
H ( x ) = ( x − x ) n + 1 = 0 H\left ( x \right )=\left ( x-x \right )^{n+1}=0 H(x)=(x−x)n+1=0;
H ( x 0 ) = ( x − x 0 ) n + 1 H\left ( x_{0} \right )=\left ( x-x_{0} \right )^{n+1} H(x0)=(x−x0)n+1;
G ′ ( t ) = − ∑ k = 0 n ( f ( k ) ( t ) ( x − t ) k k ! ) ′ = ∑ k = 1 n f ( k ) ( t ) ( x − t ) k − 1 ( k − 1 ) ! − ∑ k = 1 n + 1 f ( k ) ( t ) ( x − t ) ( k − 1 ) ( k − 1 ) ! = − f ( n + 1 ) ( t ) ( x − t ) n n ! \begin{array}{l} &G^{\prime } \left ( t \right ) \\ =&-\sum_{k=0}^{n}\left ( \frac{f^{\left ( k \right ) }\left (t \right )\left ( x-t \right )^{k}}{k!} \right)^{\prime } \\ =&\sum_{k=1}^{n}\frac{f^{\left ( k \right ) }\left ( t \right )\left ( x-t \right )^{k-1} }{\left ( k-1 \right ) !}- \sum_{k=1}^{n+1} \frac{f^{\left ( k \right ) }\left ( t \right )\left ( x-t \right )^{\left ( k-1 \right )} }{\left ( k-1 \right )!}\\ =&-\frac{f^{\left ( n+1 \right ) }\left ( t \right )\left ( x-t \right )^{n} }{n!} \end{array} \\ ===G′(t)−∑k=0n(k!f(k)(t)(x−t)k)′∑k=1n(k−1)!f(k)(t)(x−t)k−1−∑k=1n+1(k−1)!f(k)(t)(x−t)(k−1)−n!f(n+1)(t)(x−t)n
H ′ ( t ) = − ( n + 1 ) ( x − t ) n H^{\prime } \left ( t \right ) =-\left ( n+1 \right )\left ( x-t \right )^{n} H′(t)=−(n+1)(x−t)n;
引用Cauchy中值定理, ∃ ξ ∈ ( x , x 0 ) \exists \xi \in \left ( x ,x_{0} \right ) ∃ξ∈(x,x0), r n ( x ) ( x − x 0 ) n + 1 = G ( x 0 ) H ( x 0 ) = G ( x 0 ) − G ( x ) H ( x 0 ) − H ( x ) = G ′ ( ξ ) H ′ ( ξ ) = f ( n + 1 ) ( ξ ) ( n + 1 ) ! ⇒ r n ( x ) = f ( n + 1 ) ( ξ ) ( x − x 0 ) n + 1 ( n + 1 ) ! \frac{r_{n}\left ( x \right ) }{\left ( x-x_{0} \right )^{n+1} } =\frac{G\left ( x_{0} \right ) }{H\left ( x_{0} \right ) }=\frac{G\left ( x_{0} \right )-G\left ( x \right ) }{H\left ( x_{0} \right )-H\left ( x \right ) }=\frac{G^{\prime }\left ( \xi \right ) }{H^{\prime }\left ( \xi \right ) }=\frac{f^{\left ( n+1 \right ) }\left ( \xi \right )}{\left ( n+1 \right )! }\Rightarrow r_{n}\left ( x \right )=\frac{f^{\left ( n+1 \right ) }\left ( \xi \right )\left ( x-x_{0} \right )^{n+1} }{\left ( n+1 \right ) !} (x−x0)n+1rn(x)=H(x0)G(x0)=H(x0)−H(x)G(x0)−G(x)=H′(ξ)G′(ξ)=(n+1)!f(n+1)(ξ)⇒rn(x)=(n+1)!f(n+1)(ξ)(x−x0)n+1。