泰勒中值定理
条件:f(x)在x=x0领域内(n+1)f(x)在x=x_0领域内(n+1)f(x)在x=x0领域内(n+1)阶可导
结论:f(x)=Pn(x)+Rn(x)⟶Pn(x)为主项,Rn(x)为次项f(x)=P_n(x)+R_n(x) \longrightarrow P_n(x)为主项,R_n(x)为次项f(x)=Pn(x)+Rn(x)⟶Pn(x)为主项,Rn(x)为次项
Pn(x)=f(x0)+f′(x0)(x−x0)+f′′(x0)2!(x−x0)2+...+f(n)(x0)n!(x−x0)n\LARGE P_n(x)=f(x_0)+f'(x_0)(x-x_0)+\frac{f''(x_0)}{2!}(x-x_0)^2+...+\frac{f^{(n)}(x_0)}{n!}(x-x_0)^nPn(x)=f(x0)+f′(x0)(x−x0)+2!f′′(x0)(x−x0)2+...+n!f(n)(x0)(x−x0)n
Rn(x)={f(n+1)(ξ)(n+1)!⟹拉格朗日型预项o((x−x0)n)⟹皮亚诺型余项 \huge R_n(x)= \begin{cases} \frac{f^{(n+1)(\xi)}}{(n+1)!} \Longrightarrow 拉格朗日型预项 \\ o((x-x_0)^n) \Longrightarrow 皮亚诺型余项 \end{cases} Rn(x)=⎩⎪⎪⎪⎪⎨⎪⎪⎪⎪⎧(n+1)!f(n+1)(ξ)⟹拉格朗日型预项o((x−x0)n)⟹皮亚诺型余项
麦克劳林公式
ex=1+x+x22!+x33!+...++xnn!+o(xn)e^x = 1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+...++\frac{x^n}{n!}+o(x^n)ex=1+x+2!x2+3!x3+...++n!xn+o(xn)
sinx=x−x33!+x55!+...+o(x7)\sin x = x -\frac{x^3}{3!}+\frac{x^5}{5!}+...+o(x^7)sinx=x−3!x3+5!x5+...+o(x7)
cosx=1−x22!+x44!+...+o(x6)\cos x= 1-\frac{x^2}{2!}+\frac{x^4}{4!}+...+o(x^6)cosx=1−2!x2+4!x4+...+o(x6)
11−x=1+x+x2+x3+...+xn+o(xn)\frac{1}{1-x} = 1 +x +x^2 +x^3+...+x^n+o(x^n)1−x1=1+x+x2+x3+...+xn+o(xn)
11+x=1−x+x2−x3+...+(−1)nxn+o(xn)\frac{1}{1+x} = 1-x+x^2-x^3+...+(-1)^nx^n+o(x^n)1+x1=1−x+x2−x3+...+(−1)nxn+o(xn)
ln(1+x)=x−x22+x33−x44+o(x4)\ln(1+x) = x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+o(x^4)ln(1+x)=x−2x2+3x3−4x4+o(x4)
(1+x)a=1+ax+a(a−1)2!x2+a(a−1)(a−2)3!x3+...+o(x3)(1+x)^a = 1+ax+\frac{a(a-1)}{2!}x^2+\frac{a(a-1)(a-2)}{3!}x^3+...+o(x^3)(1+x)a=1+ax+2!a(a−1)x2+3!a(a−1)(a−2)x3+...+o(x3)
使用泰勒展开公式求极限
例题1:limx→0x−sinxx3例题1:\underset{x\to 0}{\lim} \frac{x-\sin x}{x^3}例题1:x→0limx3x−sinx
解:
1.把sinx展开到阶,跟分母同阶\sin x 展开到阶,跟分母同阶sinx展开到阶,跟分母同阶
原式=limx→0x−(x−x33!+o(x3))x3=x33!+o(x3)x3原式=\underset{x \to 0}{\lim} \frac{x-(x-\frac{x^3}{3!}+o(x^3))}{x^3}=\frac{\frac{x^3}{3!}+o(x^3)}{x^3}原式=x→0limx3x−(x−3!x3+o(x3))=x33!x3+o(x3)
=limx→016x3+o(x3)x3=16=\underset{x \to 0}{\lim}\frac{\frac{1}{6}x^3+o(x^3)}{x^3}=\frac{1}{6}=x→0limx361x3+o(x3)=61