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带Peano余项的Taylor公式

带Peano余项的Taylor公式:


若函数$f$在点$x_0$存在直至$n$阶导数,则有
$$
f(x)=f(x_0)+f’(x_0)(x-x_0)+\frac{f’’(x_0)}{2!}(x-x_0)^2+\cdots+\frac{f^{(n)}(x_0)}{n!}(x-x_0)^n+o((x-x_0)^n).
$$


下面我们采用积分的方法来证明该公式.为此,我们只证明$n=3$的特殊情形,一般情形完全可以类推.


若函数$f$在点$x_0$存在直至$3$阶导数,则有
\begin{equation}\label{eq:1}
f(x)=f(x_0)+f’(x_0)(x-x_0)+\frac{f’’(x_0)}{2!}(x-x_0)^2+\frac{f^{(3)}(x_0)}{3!}(x-x_0)^3+o((x-x_0)^3).
\end{equation}


证明:由于$f$在$x_0$处$3$阶可导,因此
\begin{equation}\label{eq:2}
f^{(2)}(x)=f^{(2)}(x_0)+f^{(3)}(x_0)(x-x_0)+o_{1}(x-x_0),
\end{equation}
将\eqref{eq:2}两边从$x_0$到$x$进行积分,可得
\begin{equation}
\label{eq:3}
\int_{x_0}^xf^{(2)}(t)dt=f^{(2)}(x_{0})(x-x_0)+\frac{1}{2}f^{(3)}(x_{0})(x-x_0)^{2}+\int_{x_0}^xo_1(t-x_0)dt.
\end{equation}
由微积分第一基本定理,\eqref{eq:3}可以化为
$$
f’(x)-f’(x_0)=f^{(2)}(x_{0})(x-x_0)+\frac{1}{2}f^{(3)}(x_{0})(x-x_0)^{2}+\int_{x_0}^xo_1(t-x_0)dt,
$$
也即,
\begin{equation}
\label{eq:4}
f’(x)=f’(x_{0})+f^{(2)}(x_{0})(x-x_0)+\frac{1}{2}f^{(3)}(x_{0})(x-x_0)^{2}+\int_{x_0}^xo_1(t-x_0)dt,
\end{equation}
其中$\int_{x_0}^xo_1(t-x_0)dt$至少是关于$x-x_0$的二阶无穷小量,这是因为由L’Hospital法则,
$$
\lim_{x\to x_0}\frac{\int_{x_0}^x
o_1(t-x_0)dt}{(x-x_0)^{2}}=\lim_{x\to x_0}\frac{o_1(x-x_0)}{2(x-x_0)}=0.
$$
再将式\eqref{eq:4}两边积分,可得
\begin{equation}
\label{eq:5}
\int_{x_0}^xf’(t)dt=f’(x_0)(x-x_0)+\frac{1}{2}(x-x_0)^2f^{(2)}(x_0)+\frac{1}{6}(x-x_0)^3f^{(3)}(x_0)+\int_{x_{0}}^{x}\int_{x_0}^po_1(t-x_0)dtdp,
\end{equation}
由微积分第一基本定理,式\eqref{eq:5}可以写为
$$
f(x)-f(x_0)=f’(x_0)(x-x_0)+\frac{1}{2}(x-x_0)^2f^{(2)}(x_0)+\frac{1}{6}(x-x_0)^3f^{(3)}(x_0)+\int_{x_{0}}^{x}\int_{x_0}^po_1(t-x_0)dtdp,
$$

$$
f(x)=f(x_0)+f’(x_0)(x-x_0)+\frac{1}{2}(x-x_0)^2f^{(2)}(x_0)+\frac{1}{6}(x-x_0)^3f^{(3)}(x_0)+\int_{x_{0}}^{x}\int_{x_0}^po_1(t-x_0)dtdp.
$$
其中$\int_{x_{0}}^{x}\int_{x_0}^po_1(t-x_0)dtdp$至少是关于$x-x_0$的三阶无穷小量,这是因为由L’Hospital法则,
$$
\lim_{x\to
x_0}\frac{\int_{x_{0}}^{x}\int_{x_0}^po_1(t-x_0)dtdp}{(x-x_0)^3}=\lim_{x\to
x_0}\frac{\int_{x_0}^xo_1(t-x_0)dt}{3(x-x_0)^2}=\lim_{x\to x_0}\frac{o_1(x-x_0)}{6(x-x_0)}=0.
$$