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3.2 泰勒公式
2025-12-03
高等数学
数学

§3.2\S 3.2 泰勒公式#

记 Tn(x)=a0+a1(x−x0)+a2(x−x0)2+⋯+an(x−x0)n(≈f(x))T_n(x)=a_0+a_1(x-x_0)+a_2(x-x_0)^2+\cdots+a_n(x-x_0)^n(\approx f(x)) 为泰勒公式，aka_k 为系数，且 ak=1k!f(k)(x0)a_k=\dfrac{1}{k!}f^{(k)}(x_0)

一、带有佩亚诺型余项的泰勒公式#

  • 定理：若 f(x)f(x) 在 U(x0)U(x_0) 内 nn 阶可导，则 f(x)=f(x0)+11!f′(x0)(x−x0)+12!f′′(x0)(x−x0)+⋯+1n!f(n)(x0)(x−x0)+o[(x−x0)n]高阶无穷小Rn(x)f(x)=f(x_0)+\dfrac{1}{1!}f'(x_0)(x-x_0)+\dfrac{1}{2!}f''(x_0)(x-x_0)+\cdots+\dfrac{1}{n!}f^{(n)}(x_0)(x-x_0)+\underset{高阶无穷小R_n(x)}{o[(x-x_0)^n]}

证：只需说明 lim⁡x→0Rn(x)(x−x0)n=0\displaystyle \lim_{x\to 0}\dfrac{R_n(x)}{(x-x_0)^n}=0

Rn(x0)=f(x0)−Tn(x0)=0R_n(x_0)=f(x_0)-T_n(x_0)=0

Rn(x)−0(x−x0)n−0=Rn(x)−Rn(x0)(x−x0)n−(x0−x0)n=R′n(ξ1)S′n(ξ1)\dfrac{R_n(x)-0}{(x-x_0)^n-0}=\dfrac{R_n(x)-R_n(x_0)}{(x-x_0)^n-(x_0-x_0)^n}=\dfrac{R'n(\xi_1)}{S'n(\xi_1)}

应用 n−1n-1 次柯西中值定理得 Rn′(ξ)Sn′(ξ)=⋯=Rn(n−1)(ξn−1)Sn(n−1)(ξn−1)=f(n−1)(x0)−T(n−1)(x0)n!(ξn−1−x0)=1n!(f(n−1)ξn−1−f(n−1)(x0)ξn−1−x0−f(n)x0)→0\dfrac{R'_n(\xi)}{S'_n(\xi)}=\cdots=\dfrac{R_n^{(n-1)}(\xi_{n-1})}{S_n^{(n-1)}(\xi_{n-1})}=\dfrac{f^{(n-1)}(x_0)-T^{(n-1)}(x_0)}{n!(\xi_{n-1}-x_0)}=\dfrac{1}{n!}(\dfrac{f^{(n-1)}\xi_{n-1}-f^{(n-1)}(x_0)}{\xi_{n-1}-x_0}-f^{(n)}x_0)\to 0

Sn′=(x−x0)nS'_n=(x-x_0)^{n}

f(x)=f(x0)+11!f′(x0)(x−x0)+12!f′′(x0)(x−x0)+⋯+1(n−1)!f(n−1)(x0)(x−x0)+o[(x−x0)n]f(x)=f(x_0)+\dfrac{1}{1!}f'(x_0)(x-x_0)+\dfrac{1}{2!}f''(x_0)(x-x_0)+\cdots+\dfrac{1}{(n-1)!}f^{(n-1)}(x_0)(x-x_0)+o[(x-x_0)^n]

ak=f(k)(x0)k!⇒f(k)(x0)=k!aka_k=\dfrac{f^{(k)}(x_0)}{k!}\Rightarrow f^{(k)}(x_0)=k!a_k

  • 当 x0=0x_0=0 时，f(x)=f(0)+11!f′(0)(x−0)+12!f′′(0)(x−0)+⋯+1(n−1)!f(n−1)(0)(x−0)+o(xn)f(x)=f(0)+\dfrac{1}{1!}f'(0)(x-0)+\dfrac{1}{2!}f''(0)(x-0)+\cdots+\dfrac{1}{(n-1)!}f^{(n-1)}(0)(x-0)+o(x^n) 成为Maclaurin[麦克劳林]公式

常见函数的佩亚诺型麦克劳林展开式#

  1. y=exy=e^x

解：f(k)=e0=1f^{(k)}=e^0=1

ex=1+11!x+12!x2+⋯+1n!xn+o(xn)e^x=1+\dfrac{1}{1!}x+\dfrac{1}{2!}x^2+\cdots+\dfrac{1}{n!}x^n+o(x^n)

  1. sin⁡x=x−x33!+x55!−x77!+⋯+(−1)k−11(2k−1)!x2k−1+o(x2k)\sin x=x-\dfrac{x^3}{3!}+\dfrac{x^5}{5!}-\dfrac{x^7}{7!}+\cdots+(-1)^{k-1}\dfrac{1}{(2k-1)!}x^{2k-1}+o(x^{2k})

  2. cos⁡x=1−x22!+x44!−⋯+(−1)k1(2k)!x2k+o(x2k)\cos x=1-\dfrac{x^2}{2!}+\dfrac{x^4}{4!}-\cdots+(-1)^k\dfrac{1}{(2k)!}x^{2k}+o(x^{2k})

  3. 11−x=1+x+x2+x3+⋯+xn+o(xn)\dfrac{1}{1-x}=1+x+x^2+x^3+\cdots+x^n+o(x^n)

  4. 11+x=1−x+x2−x3+⋯+(−1)nxn+o(xn)\dfrac{1}{1+x}=1-x+x^2-x^3+\cdots+(-1)^nx^n+o(x^n)

  5. ln⁡(1+x)=x−x22+x33−⋯+(−1)n−1xnn+o(xn)\ln (1+x)=x-\dfrac{x^2}{2}+\dfrac{x^3}{3}-\cdots+(-1)^{n-1}\dfrac{x^n}{n}+o(x^n)

  6. ln⁡(1−x)=−x−x22−x33−⋯−xnn+o(xn)\ln (1-x)=-x-\dfrac{x^2}{2}-\dfrac{x^3}{3}-\cdots-\dfrac{x^n}{n}+o(x^n)

  7. (1+x)α=1+αx+α(α−1)2!x2+⋯+α(α−1)⋯(α−n+1)n!xn+o(xn)(1+x)^\alpha=1+\alpha x+\dfrac{\alpha (\alpha -1)}{2!}x^2+\cdots+\dfrac{\alpha(\alpha-1)\cdots(\alpha-n+1)}{n!}x^n+o(x^n)

例题#

  1. f(x)=ln⁡xf(x)=\ln x，求其在 x=2x=2 处的泰勒公式

解： f(x)=a0+a1(x−2)+a2(x−2)2+⋯+an(x−2)n+o[(x−2)n]f(x)=a_0+a_1(x-2)+a_2(x-2)^2+\cdots+a_n(x-2)^n+o[(x-2)^n]

ak=f(k)(2)k!a_k=\dfrac{f^{(k)}(2)}{k!}

f(x)=ln⁡x=ln⁡(2+x−2)=ln⁡[2(1+x−22)]=ln⁡2+ln⁡(1+x−22)=ln⁡2+x−22−12⋅(x−22)2+13⋅(x−22)3−⋯+(−1)n−11n⋅(x−22)n+o[(x−2)n]\begin{align*} f(x)=\ln x=\ln(2+x-2)&=\ln[2(1+\dfrac{x-2}{2})]\\ &=\ln 2+\ln(1+\dfrac{x-2}{2})\\ &=\ln 2+\dfrac{x-2}{2}-\dfrac{1}{2}\cdot (\dfrac{x-2}{2})^2+\dfrac{1}{3}\cdot (\dfrac{x-2}{2})^3-\cdots+(-1)^{n-1}\dfrac{1}{n}\cdot (\dfrac{x-2}{2})^n+o[(x-2)^n] \end{align*}
  1. f(x)=1x2+5x+6f(x)=\dfrac{1}{x^2+5x+6}，求其在 x=1x=1 处的泰勒公式

解： f(x)=a0+a1(x−1)+a2(x−1)2+⋯+an(x−1)n+o[(x−1)n]f(x)=a_0+a_1(x-1)+a_2(x-1)^2+\cdots+a_n(x-1)^n+o[(x-1)^n]

f(x)=1(x+2)(x+3)=1x+2−1x+3=13+(x−1)−14+(x−1)=13(11+x−13)−14(11+x−14)后续过程略\begin{align*} f(x)=\dfrac{1}{(x+2)(x+3)}&=\dfrac{1}{x+2}-\dfrac{1}{x+3}\\ &=\dfrac{1}{3+(x-1)}-\dfrac{1}{4+(x-1)}\\ &=\dfrac{1}{3}\left(\dfrac{1}{1+\frac{x-1}{3}}\right)-\dfrac{1}{4}\left(\dfrac{1}{1+\frac{x-1}{4}}\right)\\ 后续过程略 \end{align*}
lim⁡x→0x−sin⁡xx3=lim⁡x→0x−[x−x33!+o(x3)]x3=16\begin{align*} \lim_{x\to 0}\dfrac{x-\sin x}{x^3}&=\lim_{x\to 0}\dfrac{x-[x-\frac{x^3}{3!}+o(x^3)]}{x^3}\\ &=\dfrac{1}{6} \end{align*}
lim⁡x→0exsin⁡x−x(1+x)x3=lim⁡x→0(1+x1!+x22!+x33!+o(x3))(x−x33!+o(x3))−x(1+x)x3=lim⁡x→0x+x2+(12−16)x3+o(x3)+x−x2x3=13\begin{align*} \lim_{x\to 0}\dfrac{e^x\sin x-x(1+x)}{x^3}&=\lim_{x\to 0}\dfrac{(1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+o(x^3))(x-\frac{x^3}{3!}+o(x^3))-x(1+x)}{x^3}\\ &=\lim_{x\to 0}\dfrac{x+x^2+(\frac{1}{2}-\frac{1}{6})x^3+o(x^3)+x-x^2}{x^3}\\ &=\dfrac{1}{3} \end{align*}

二、带有拉格朗日型余项的泰勒公式#

  • 定理[泰勒中值定理]：若 f(x)f(x) 在 [a,b][a,b] 上 nn 阶可导且连续，(a,b)(a,b) 上存在 n+1n+1 阶导数，则对于任意 x,x0∈[a,b]x,x_0 \in [a,b]，存在 ξ∈(a,b)\xi \in (a,b) 使得
f(x)=f(x0)+f′(x0)1!(x−x0)+⋯+f(n)(x0)n!(x−x0)n+f(n+1)(ξ)(n+1)!(x−x0)n+1f(x)= f(x_0)+\dfrac{f'(x_0)}{1!}(x-x_0)+\cdots+\dfrac{f^{(n)}(x_0)}{n!}(x-x_0)^n+\dfrac{f^{(n+1)}(\xi)}{(n+1)!}(x-x_0)^{n+1}
  • x0=0x_0=0 时的麦克劳林公式 (ξ∈(0,x))(\xi \in (0,x))
f(x)=f(0)+f′(0)1!x+⋯+f(n)(0)n!xn+f(n+1)(ξ)(n+1)!xn+1或f(n+1)(θx)(n+1)!xn+1θ∈(0,1)f(x)=f(0)+\dfrac{f'(0)}{1!}x+\cdots+\dfrac{f^{(n)}(0)}{n!}x^n+\underset{或\dfrac{f^{(n+1)}(\theta x)}{(n+1)!}x^{n+1}\quad \theta \in(0,1)}{\dfrac{f^{(n+1)}(\xi)}{(n+1)!}x^{n+1}}

常见函数的拉格朗日型麦克劳林展开式#

  1. ex=1+11!x+12!x2+⋯+1n!xn+eθx(n+1)!xn+1(0<θ<1,x∈R)e^x=1+\dfrac{1}{1!}x+\dfrac{1}{2!}x^2+\cdots+\dfrac{1}{n!}x^n+\dfrac{e^{\theta x}}{(n+1)!}x^{n+1}(0\lt \theta \lt 1,x\in R)

  2. sin⁡x=x−x33!+x55!−x77!+⋯+(−1)m−11(2m−1)!x2m−1+(−1)mcos⁡θx(2m+1)!x2m+1(0<θ<1,x∈R)\sin x=x-\dfrac{x^3}{3!}+\dfrac{x^5}{5!}-\dfrac{x^7}{7!}+\cdots+(-1)^{m-1}\dfrac{1}{(2m-1)!}x^{2m-1}+(-1)^m\dfrac{\cos \theta x}{(2m+1)!}x^{2m+1}(0\lt \theta \lt 1,x\in R)

  3. cos⁡x=1−x22!+x44!−⋯+(−1)mk1(2m)!x2m+(−1)m+1cos⁡θx(2m+2)!x2m+2(0<θ<1,x∈R)\cos x=1-\dfrac{x^2}{2!}+\dfrac{x^4}{4!}-\cdots+(-1)^mk\dfrac{1}{(2m)!}x^{2m}+(-1)^{m+1}\dfrac{\cos \theta x}{(2m+2)!}x^{2m+2}(0\lt \theta \lt 1,x\in R)

  4. ln⁡(1+x)=x−x22+x33−⋯+(−1)n−1xnn+(−1)nxn+1(n+1)(1+θx)n+1(0<θ<1,x>−1)\ln (1+x)=x-\dfrac{x^2}{2}+\dfrac{x^3}{3}-\cdots+(-1)^{n-1}\dfrac{x^n}{n}+(-1)^n\dfrac{x^{n+1}}{(n+1)(1+\theta x)^{n+1}}(0\lt \theta \lt 1,x\gt -1)

  5. (1+x)α=1+αx+α(α−1)2!x2+⋯+α(α−1)⋯(α−n+1)n!xn+α(α−1)⋯(α−n)(n+1)!(1+θx)−n−1xn+1(0<θ<1,x>−1)(1+x)^\alpha=1+\alpha x+\dfrac{\alpha (\alpha -1)}{2!}x^2+\cdots+\dfrac{\alpha(\alpha-1)\cdots(\alpha-n+1)}{n!}x^n+\dfrac{\alpha(\alpha-1)\cdots(\alpha-n)}{(n+1)!}(1+\theta x)^{-n-1}x^{n+1}(0\lt \theta \lt 1,x\gt -1)

  6. 11−x=1+x+x2+x3+⋯+xn+xn+1(1−θx)n+2(0<θ<1,∣x∣<1)\dfrac{1}{1-x}=1+x+x^2+x^3+\cdots+x^n+\dfrac{x^{n+1}}{(1-\theta x)^{n+2}}(0\lt \theta \lt 1,|x|\lt 1)

3.2 泰勒公式
https://gitee.com/jason_ren/advanced-math-note
作者
Jason Ren
发布于
2025-12-03
许可协议
CC BY-SA 4.0

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