微积分笔记(13)——带 Peano 余项的 Taylor 公式

微积分笔记(13)——带 Peano 余项的 Taylor 公式

Taylor 公式——带 Peano 余项(续)

Taylor 公式证明

注意到 $P_n^{(k)}(0) = f^{(n)}(x_0),k = 0,1,2,\cdots,n$。

应用 L’Hospital 法则 $n-1$ 次。
$$
\lim_{\Delta x \to 0} \dfrac{f(x_0 + \Delta x) – P_n(\Delta x)}{\Delta x^n} \\
= \lim_{\Delta x \to 0} \dfrac{f^{\prime}(x_0 + \Delta x) – P_n^{\prime}(\Delta x)}{n\Delta x^{n-1}} \\
\vdots \\
= \lim_{\Delta x \to 0} \dfrac{f^{(n-1)}(x_0 + \Delta x) – f^{n-1}(x_0) – f^{n}(x_0) \Delta x}{n! \Delta x} \\
= \dfrac{1}{n!} \lim_{\Delta x \to 0} [\dfrac{f^{(n-1)}(x_0 + \Delta x) – f^{n-1}(x_0)}{\Delta x} -f^{(n)}(x_0)] \\
= 0 \ \square
$$

Maclaurin 展开($x_0 = 0$)

若 $f^{(n)}(0)$ 存在,则:
$$
f(x) = f(0) + \dfrac{f^{\prime}(0)}{1!} x + \dfrac{f^{\prime \prime}(0)}{2!} x^2 + \cdots + \dfrac{f^{(n)}(0)}{n!} x^n + o(x^n)
$$

应用

一些常见初等函数的 Maclaurin 展开。
$$
e^x = 1 + \dfrac{x}{1!} + \dfrac{x^2}{2!} + \cdots + \dfrac{x^n}{n!} + o(x^n) \\
\sin x = x – \dfrac{x^3}{3!} + \dfrac{x^5}{5!} – \cdots + (-1)^m \dfrac{x^{2m + 1}}{(2m + 1)!} + o(x^{2m + 2}) \\
\cos x = 1 – \dfrac{x^2}{2!} + \dfrac{x^4}{4!} – \cdots + (-1)^m \dfrac{x^{2m}}{(2m)!} + o(x^{2m + 1}) \\
\dfrac{1}{1 + x} = 1 – x + x^2 – x^3 + \cdots +(-1)^n x^n + o(x^n) \\
\ln (1 + x) = x – \dfrac{x^2}{2} + \dfrac{x^3}{3} – \cdots + (-1)^{n-1} \dfrac{x^n}{n} + o(x^n) \\
(1 + x)^a = 1 + ax + \dfrac{a(a – 1)}{2!} x^2 + \cdots + \dfrac{a(a – 1) \cdots (a – n + 1)}{n!} x^n + o(x^n)
$$

问题

已知 $f(x) = a_0 + a_1(x – x_0) + a_2(x – x_0)^2 + \cdots + a_n(x – x_0)^n + o((x – x_0)^n)$,是否 $a_k = \dfrac{f^{(k)}(x_0)}{k!},k=0,1,2,\cdots,n$?

唯一性:设 $f^{(n)}(x_0)$ 存在,则上述展开为 Taylor 展开,即 $a_k = \dfrac{f^{(k)}(x_0)}{k!},k=0,1,2,\cdots,n$。

证明:由 $f^{(n)}(x_0)$ 存在,取 $P_n(x – x_0)$ 为 $f$ 的 $n$ 阶 Taylor 多项式(在 $x_0$ 点),记 $Q_n(x – x_0) = a_0 + a_1(x – x_0) + a_2(x – x_0)^2 + \cdots + a_n(x – x_0)^n$,则:
$$
P_n(x – x_0) – Q_n(x – x_0) = o((x-x_0)^n) \\
\Rightarrow P_n(x – x_0) = Q_n(x – x_0) \ \square
$$

间接展开

$$
e^{x^2 + 2x} = e^{(x + 1)^2 – 1} \\
= e^{-1}(1 + \dfrac{(x + 1)^2}{1!} + \dfrac{(x + 1)^4}{2!} + \cdots + \dfrac{(x+1)^{2n}}{n!} + o((x+1)^{2n})) \\
\dfrac{1}{x^2 + 3x} = \dfrac{1}{3}(\dfrac{1}{x} – \dfrac{1}{x + 3}) \\
= \dfrac{1}{3}((\sum_{k = 0}^n (-1)^k (x-1)^k + o((x-1)^n))-(\dfrac{1}{4}\sum_{k = 0}^n (-1)^k (\dfrac{x-1}{4})^k + o((\dfrac{x-1}{4})^n))) \\
= \dfrac{1}{3} \sum_{k = 0}^n (-1)^k (1 – \dfrac{1}{4^{k+1}})(x – 1)^k + o((x – 1)^n) \\
\tan x = \dfrac{\sin x}{\cos x} \\
= \dfrac{x – \dfrac{x^3}{3!} + \dfrac{x^5}{5!} + o(x^6)}{1 – \dfrac{x^2}{2!} + \dfrac{x^4}{4!} + o(x^5)} \\
= x + \dfrac{x^3}{3} + \dfrac{2}{15} x^5 + o(x^5)
$$

 

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