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§ 3 § 3 §3 泰勒公式 多项式函数是各类函数中最简单的一种, 用多项式逼近函数是近似计算和理论分析的一个重要内容. 一、带有佩亚诺型余项的泰勒公式 我们在学习导数和微分概念时已经知道, 如果函数 f f f 在点 x 0 x_{0} x0 可导, 则有 f ( x ) = f ( x 0 ) + f ′ ( x 0 ) ( x − x 0 ) + o ( x − x 0 ) . f(x)=f\left(x_{0}\right)+f^{\prime}\left(x_{0}\right)\left(x-x_{0}\right)+o\left(x-x_{0}\right) . f(x)=f(x0)+f′(x0)(x−x0)+o(x−x0). 即在点 x 0 x_{0} x0 附近, 用一次多项式 f ( x 0 ) + f ′ ( x 0 ) ( x − x 0 ) f\left(x_{0}\right)+f^{\prime}\left(x_{0}\right)\left(x-x_{0}\right) f(x0)+f′(x0)(x−x0) 逼近函数 f ( x ) f(x) f(x) 时, 其误差为 ( x − (x- (x− x 0 x_{0} x0 ) 的高阶无穷小是. 然而在很多场合, 取一次多项式逼近是不㿟的, 往往需要用二次或高干二次的多项式去逼近, 并要求误差为 o ( ( x − x 0 ) n ) o\left(\left(x-x_{0}\right)^{n}\right) o((x−x0)n), 其中 n n n 为多项式的次数. 为此,我们考察任一 n n n 次多项式 p n ( x ) = a 0 + a 1 ( x − x 0 ) + a 2 ( x − x 0 ) 2 + ⋯ + a n ( x − x 0 ) n . p_{n}(x)=a_{0}+a_{1}\left(x-x_{0}\right)+a_{2}\left(x-x_{0}\right)^{2}+\cdots+a_{n}\left(x-x_{0}\right)^{n} . pn(x)=a0+a1(x−x0)+a2(x−x0)2+⋯+an(x−x0)n. 逐次求它在点 x 0 x_{0} x0 的各阶导数,得到 p n ( x 0 ) = a 0 , p n ′ ( x 0 ) = a 1 , p n ′ ′ ( x 0 ) = 2 ! a 2 , ⋯ , p n ( n ) ( x 0 ) = n ! a n , p_{n}\left(x_{0}\right)=a_{0}, p_{n}^{\prime}\left(x_{0}\right)=a_{1}, p_{n}^{\prime \prime}\left(x_{0}\right)=2 ! a_{2}, \cdots, p_{n}^{(n)}\left(x_{0}\right)=n ! a_{n}, pn(x0)=a0,pn′(x0)=a1,pn′′(x0)=2!a2,⋯,pn(n)(x0)=n!an, 即 a 0 = p n ( x 0 ) , a 1 = p n ′ ( x 0 ) 1 ! , a 2 = p n ′ ′ ( x 0 ) 2 ! , ⋯ , a n = p n ( n ) ( x 0 ) n ! . a_{0}=p_{n}\left(x_{0}\right), a_{1}=\frac{p_{n}^{\prime}\left(x_{0}\right)}{1 !}, a_{2}=\frac{p_{n}^{\prime \prime}\left(x_{0}\right)}{2 !}, \cdots, a_{n}=\frac{p_{n}^{(n)}\left(x_{0}\right)}{n !} . a0=pn(x0),a1=1!pn′(x0),a2=2!pn′′(x0),⋯,an=n!pn(n)(x0). 由此可见, 多项式 p n ( x ) p_{n}(x) pn(x) 的各项系数由其在点 x 0 x_{0} x0 的各阶导数值所唯一确定. 对于一般函数 f f f, 设它在点 x 0 x_{0} x0 存在直到 n n n 阶的导数. 由这些导数构造一个 n n n 次多项式 T n ( x ) = f ( x 0 ) + f ′ ( x 0 ) 1 ! ( x − x 0 ) + f ′ ′ ( x 0 ) 2 ! ( x − x 0 ) 2 + ⋯ + f ( n ) ( x 0 ) n ! ( x − x 0 ) n , \begin{aligned} T_{n}(x)= & f\left(x_{0}\right)+\frac{f^{\prime}\left(x_{0}\right)}{1 !}\left(x-x_{0}\right)+\frac{f^{\prime \prime}\left(x_{0}\right)}{2 !}\left(x-x_{0}\right)^{2}+\cdots+ \\ & \frac{f^{(n)}\left(x_{0}\right)}{n !}\left(x-x_{0}\right)^{n}, \end{aligned} Tn(x)=f(x0)+1!f′(x0)(x−x0)+2!f′′(x0)(x−x0)2+⋯+n!f(n)(x0)(x−x0)n, 称为函数 f f f 在点 x 0 x_{0} x0 的泰勒 (Taylor) 多项式, T n ( x ) T_{n}(x) Tn(x) 的各项系数 f ( k ) ( x 0 ) k ! ( k = 1 , 2 , ⋯ , n ) \frac{f^{(k)}\left(x_{0}\right)}{k !}(k=1,2, \cdots, n) k!f(k)(x0)(k=1,2,⋯,n) 称 为泰勒系数. 由上面对多项式系数的讨论, 易知 f ( x ) f(x) f(x) 与其泰勒多项式 T n ( x ) T_{n}(x) Tn(x) 在点 x 0 x_{0} x0 有相同的函数值和相同的直至 n n n 阶导数值, 即 f ( k ) ( x 0 ) = T n ( k ) ( x 0 ) , k = 0 , 1 , 2 , ⋯ , n . f^{(k)}\left(x_{0}\right)=T_{n}^{(k)}\left(x_{0}\right), k=0,1,2, \cdots, n . f(k)(x0)=Tn(k)(x0),k=0,1,2,⋯,n. 下面将要证明 f ( x ) − T n ( x ) = o ( ( x − x 0 ) n ) f(x)-T_{n}(x)=o\left(\left(x-x_{0}\right)^{n}\right) f(x)−Tn(x)=o((x−x0)n), 即以 (2) 式所示的泰勒多项式逼近 f ( x ) f(x) f(x) 时,其误差为关于 ( x − x 0 ) n \left(x-x_{0}\right)^{n} (x−x0)n 的高阶无穷小量. 定理 6.9 若函数 f f f 在点 x 0 x_{0} x0 存在直至 n n n 阶导数, 则有 f ( x ) = T n ( x ) + o ( ( x − x 0 ) n ) f(x)=T_{n}(x)+o\left(\left(x-x_{0}\right)^{n}\right) f(x)=Tn(x)+o((x−x0)n), 即 f ( x ) = f ( x 0 ) + f ′ ( x 0 ) ( x − x 0 ) + f ′ ′ ( x 0 ) 2 ! ( x − x 0 ) 2 + ⋯ + f ( n ) ( x 0 ) n ! ( x − x 0 ) n + o ( ( x − x 0 ) n ) . \begin{aligned} f(x)= & f\left(x_{0}\right)+f^{\prime}\left(x_{0}\right)\left(x-x_{0}\right)+\frac{f^{\prime \prime}\left(x_{0}\right)}{2 !}\left(x-x_{0}\right)^{2}+\cdots+ \\ & \frac{f^{(n)}\left(x_{0}\right)}{n !}\left(x-x_{0}\right)^{n}+o\left(\left(x-x_{0}\right)^{n}\right) . \end{aligned} f(x)=f(x0)+f′(x0)(x−x0)+2!f′′(x0)(x−x0)2+⋯+n!f(n)(x0)(x−x0)n+o((x−x0)n). 证 设 R n ( x ) = f ( x ) − T n ( x ) , Q n ( x ) = ( x − x 0 ) n , R_{n}(x)=f(x)-T_{n}(x), Q_{n}(x)=\left(x-x_{0}\right)^{n}, Rn(x)=f(x)−Tn(x),Qn(x)=(x−x0)n, 现在只要证 lim x → x 0 R n ( x ) Q n ( x ) = 0. \lim \limits_{x \rightarrow x_{0}} \frac{R_{n}(x)}{Q_{n}(x)}=0 . x→x0limQn(x)Rn(x)=0. 由关系式 (3) 可知, R n ( x 0 ) = R n ′ ( x 0 ) = ⋯ = R n ( n ) ( x 0 ) = 0 , R_{n}\left(x_{0}\right)=R_{n}^{\prime}\left(x_{0}\right)=\cdots=R_{n}^{(n)}\left(x_{0}\right)=0, Rn(x0)=Rn′(x0)=⋯=Rn(n)(x0)=0, 并易知 Q n ( x 0 ) = Q n ′ ( x 0 ) = ⋯ = Q n ( n − 1 ) ( x 0 ) = 0 , Q n ( n ) ( x 0 ) = n ! . Q_{n}\left(x_{0}\right)=Q_{n}^{\prime}\left(x_{0}\right)=\cdots=Q_{n}^{(n-1)}\left(x_{0}\right)=0, Q_{n}^{(n)}\left(x_{0}\right)=n ! . Qn(x0)=Qn′(x0)=⋯=Qn(n−1)(x0)=
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