文章目录
- 泰勒展开
- 定义
- 证明
- 举例
- e^x在x=0处展开
- sinx 在x=0处展开
- cosx 在x=0处展开
- (1+x)^a在x=0处展开
- ln(1+x)在x=0处展开
- 求极限
- 例一
泰勒展开 定义 设 n n n是一个正整数，如果定义在一个包含 a 的区间上的函数 f 在 a 点处 n+1 次可导，那么对于这个区间上的任意 x，都有：
f ( x ) = f ( a ) + f ′ ( a ) 1 ! ( x ? a ) + f ′ ′ ( a ) 2 ! ( x ? a ) 2 + f ′ ′ ′ ( a ) 3 ! ( x ? a ) 3 + ? + f n ( x ) n ! ( x ? a ) n + R n ( x ) f(x)=f(a)+{f' (a) \over 1! }(x-a)+{f' ' (a) \over 2!}(x-a)^2+{f' ' ' (a) \over 3!}(x-a)^3+\cdots+{f^n(x) \over n!}(x-a)^n + R_n(x) f(x)=f(a)+1!f′(a)?(x?a)+2!f′′(a)?(x?a)2+3!f′′′(a)?(x?a)3+?+n!fn(x)?(x?a)n+Rn?(x)
其中的多项式称为函数在a 处的泰勒展开式，剩余的 R n ( x ) R_{n}(x) Rn?(x)是泰勒公式的余项，是 ( x ? a ) n (x-a)^n (x?a)n的高阶无穷小。
证明 f ( x ) = f ( x 0 ) + f ′ ( x 0 ) 1 ! ( x ? x 0 ) + R n ( x ) f(x) = f(x0)+{f' (x0) \over 1!}(x-x0) +R_{n}(x) f(x)=f(x0)+1!f′(x0)?(x?x0)+Rn?(x)
求证：
l i m x → x 0 f ( x ) ? f ( x 0 ) ? f ′ ( x 0 ) ( x ? x 0 ) x ? x 0 = 0 lim_{x \to x0}{f(x)-f(x0)-f' (x0)(x-x0) \over x-x0} = 0 limx→x0?x?x0f(x)?f(x0)?f′(x0)(x?x0)?=0
根据洛必达法则
l i m x → x 0 f ( x ) ? f ( x 0 ) ? f ′ ( x 0 ) ( x ? x 0 ) x ? x 0 = l i m x → x 0 ( f ( x ) ? f ( x 0 ) ? f ′ ( x 0 ) ( x ? x 0 ) ) ′ ( x ? x 0 ) ′ = l i m x → x 0 f ′ ( x ) ? f ( x 0 ) ′ 1 = 0 lim_{x \to x0}{f(x)-f(x0)-f' (x0)(x-x0) \over x-x0} = lim_{x \to x0}{(f(x)-f(x0)-f' (x0)(x-x0))' \over (x-x0)' }=lim_{x \to x0} {f' (x) -f(x0)' \over 1} = 0 limx→x0?x?x0f(x)?f(x0)?f′(x0)(x?x0)?=limx→x0?(x?x0)′(f(x)?f(x0)?f′(x0)(x?x0))′?=limx→x0?1f′(x)?f(x0)′?=0
归纳演绎，泰勒公式得证
举例 e^x在x=0处展开 e x = e x 0 + e x 0 ( x ? x 0 ) + e x 0 ( x ? x 0 ) 2 2 ! + e x 0 ( x ? x 0 ) 3 3 ! + ? e^x = e^{x_0}+e^{x_0}{(x-x_0)}+{e^{x_0}(x-x_0)^2 \over 2!}+{e^{x_0}(x-x_0)^3 \over 3!}+ \cdots ex=ex0?+ex0?(x?x0?)+2!ex0?(x?x0?)2?+3!ex0?(x?x0?)3?+?
在0点处的展开
e x = 1 + x + x 2 2 ! + x 3 3 ! + x 4 4 ! + x 5 5 ! + ? e^x = 1+x+{x^2 \over 2!}+{x^3 \over 3!}+{x^4 \over 4!}+{x^5 \over 5!} + \cdots ex=1+x+2!x2?+3!x3?+4!x4?+5!x5?+?
sinx 在x=0处展开 sin ? x = sin ? 0 + sin ? ′ 0 ( x ? 0 ) 1 ! + sin ? ′ ′ 0 ( x ? 0 ) 2 2 ! + sin ? ′ ′ ′ 0 ( x ? 0 ) 3 3 ! + sin ? ′ ′ ′ ′ 0 ( x ? 0 ) 4 4 ! + sin ? ′ ′ ′ ′ ′ 0 ( x ? 0 ) 5 5 ! + ? \sin x = {\sin 0}+{\sin ' 0(x-0) \over 1!}+{\sin ' ' 0(x-0)^2 \over 2!}+{\sin ' ' ' 0(x-0)^3 \over 3!}+{\sin ' ' ' ' 0(x-0)^4 \over 4!}+{\sin ' ' ' ' ' 0(x-0)^5 \over 5!}+ \cdots sinx=sin0+1!sin′0(x?0)?+2!sin′′0(x?0)2?+3!sin′′′0(x?0)3?+4!sin′′′′0(x?0)4?+5!sin′′′′′0(x?0)5?+?
= 0 + cos ? 0 ( x ) 1 ! + ? sin ? 0 ( x ? 0 ) 2 2 ! + ? cos ? 0 ( x ? 0 ) 3 3 ! + sin ? 0 ( x ? 0 ) 4 4 ! + cos ? 0 ( x ? 0 ) 5 5 ! + ? = {0}+{\cos 0(x) \over 1!}+{-\sin 0(x-0)^2 \over 2!}+{-\cos 0(x-0)^3 \over 3!}+{\sin 0(x-0)^4 \over 4!}+{\cos 0(x-0)^5 \over 5!}+ \cdots =0+1!cos0(x)?+2!?sin0(x?0)2?+3!?cos0(x?0)3?+4!sin0(x?0)4?+5!cos0(x?0)5?+?
= 0 + 1 ( x ) 1 ! + ? 1 ( x ) 3 3 ! + 1 ( x ) 5 5 ! + ? = {0}+{1(x) \over 1!}+{-1(x)^3 \over 3!}+{1(x)^5 \over 5!}+ \cdots =0+1!1(x)?+3!?1(x)3?+5!1(x)5?+?
cosx 在x=0处展开 cos ? x = cos ? 0 + cos ? ′ 0 ( x ? 0 ) 1 ! + cos ? ′ ′ 0 ( x ? 0 ) 2 2 ! + cos ? ′ ′ ′ 0 ( x ? 0 ) 3 3 ! + cos ? ′ ′ ′ ′ 0 ( x ? 0 ) 4 4 ! + cos ? ′ ′ ′ ′ ′ 0 ( x ? 0 ) 5 5 ! + ? + cos ? ′ ′ ′ ′ ′ ′ 0 ( x ? 0 ) 6 6 ! + ? \cos x = {\cos 0}+{\cos ' 0(x-0) \over 1!}+{\cos ' ' 0(x-0)^2 \over 2!}+{\cos ' ' ' 0(x-0)^3 \over 3!}+{\cos ' ' ' ' 0(x-0)^4 \over 4!}+{\cos ' ' ' ' ' 0(x-0)^5 \over 5!}+ \cdots+{\cos ' ' ' ' ' ' 0(x-0)^6 \over 6!}+ \cdots cosx=cos0+1!cos′0(x?0)?+2!cos′′0(x?0)2?+3!cos′′′0(x?0)3?+4!cos′′′′0(x?0)4?+5!cos′′′′′0(x?0)5?+?+6!cos′′′′′′0(x?0)6?+?
= 1 + ? sin ? 0 ( x ) 1 ! + ? cos ? 0 ( x ? 0 ) 2 2 ! + sin ? 0 ( x ? 0 ) 3 3 ! + cos ? 0 ( x ? 0 ) 4 4 ! + ? sin ? 0 ( x ? 0 ) 5 5 ! + ? + ? cos ? 0 ( x ? 0 ) 6 6 ! + ? = {1}+{-\sin 0(x) \over 1!}+{-\cos 0(x-0)^2 \over 2!}+{\sin 0(x-0)^3 \over 3!}+{\cos 0(x-0)^4 \over 4!}+{-\sin 0(x-0)^5 \over 5!}+ \cdots+{-\cos 0(x-0)^6 \over 6!}+ \cdots =1+1!?sin0(x)?+2!?cos0(x?0)2?+3!sin0(x?0)3?+4!cos0(x?0)4?+5!?sin0(x?0)5?+?+6!?cos0(x?0)6?+?
= 1 + ? 1 ( x ) 2 2 ! + ( x ) 4 4 ! + ? 1 ( x ) 6 6 ! + ? = {1}+{-1(x)^2 \over 2!}+{(x)^4 \over 4!}+{-1(x)^6 \over 6!}+ \cdots =1+2!?1(x)2?+4!(x)4?+6!?1(x)6?+?
(1+x)^a在x=0处展开 【泰勒展开范例】 ( 1 + x ) α , α ∈ ? (1+x)^\alpha, \alpha \in \Re (1+x)α,α∈?
( 1 + x ) α = ( 1 + 0 ) α + ( 1 + 0 ) ′ α ( x ? 0 ) 1 1 ! + ( 1 + 0 ) ′ ′ α ( x ? 0 ) 2 2 ! + ( 1 + 0 ) ′ ′ ′ α ( x ? 0 ) 3 3 ! + ( 1 + 0 ) ′ ′ ′ ′ α ( x ? 0 ) 4 4 ! (1+x)^\alpha = {(1+0)^\alpha+{(1+0)^{' \alpha}(x-0)^1 \over 1!}}+{(1+0)^{' ' \alpha}(x-0)^2 \over 2!}+{(1+0)^{' ' ' \alpha}(x-0)^3 \over 3!}+{(1+0)^{' ' ' ' \alpha}(x-0)^4 \over 4!} (1+x)α=(1+0)α+1!(1+0)′α(x?0)1?+2!(1+0)′′α(x?0)2?+3!(1+0)′′′α(x?0)3?+4!(1+0)′′′′α(x?0)4?
= ( 1 ) + α ( x ) 1 1 ! + α ( α ? 1 ) ( x ) 2 2 ! + α ( α ? 1 ) ( α ? 2 ) ( x ) 3 3 ! + α ( α ? 1 ) ( α ? 2 ) ( α ? 3 ) ( x ) 4 4 ! = {(1)+{\alpha(x)^1 \over 1!}}+{{\alpha (\alpha -1)}(x)^2 \over 2!}+{{\alpha (\alpha -1)(\alpha -2)}(x)^3 \over 3!}+{{\alpha (\alpha -1)(\alpha -2)(\alpha -3)}(x)^4 \over 4!} =(1)+1!α(x)1?+2!α(α?1)(x)2?+3!α(α?1)(α?2)(x)3?+4!α(α?1)(α?2)(α?3)(x)4?
ln(1+x)在x=0处展开 f ( x ) = f ( 0 ) + f ′ ( 0 ) 1 ! + f ′ ′ ( 0 ) 2 ! + f ′ ′ ′ ( 0 ) 3 ! + ? f(x) = f(0)+{f' (0) \over 1!}+{f' ' (0) \over 2!}+{f' ' ' (0) \over 3!}+\cdots f(x)=f(0)+1!f′(0)?+2!f′′(0)?+3!f′′′(0)?+?
其中
ln ? ′ ( 1 + x ) = 1 1 + x \ln ' (1+x) = {1 \over 1+x} ln′(1+x)=1+x1?
ln ? ′ ′ ( 1 + x ) = ( 1 1 + x ) ′ = ( ? 1 ) 1 ( 1 + x ) 2 \ln ' ' (1+x) = ({1 \over 1+x})' = (-1){1 \over (1+x)^2} ln′′(1+x)=(1+x1?)′=(?1)(1+x)21?
ln ? ′ ′ ′ ( 1 + x ) = ( ( ? 1 ) 1 ( 1 + x ) 2 ) ′ = ( ? 1 ) ( ? 2 ) 1 ( 1 + x ) 3 \ln ' ' ' (1+x) = ((-1){1 \over (1+x)^2})' = (-1)(-2){1 \over (1+x)^3} ln′′′(1+x)=((?1)(1+x)21?)′=(?1)(?2)(1+x)31?
所以
ln ? ( 1 + x ) = x ? x 2 2 + x 3 3 ? x 4 4 + x 5 5 + ? \ln (1+x) = x - {x^2 \over 2}+{x^3 \over 3}-{x^4 \over 4}+{x^5 \over 5}+\cdots ln(1+x)=x?2x2?+3x3??4x4?+5x5?+?
求极限 例一 lim ? x → 0 e x ? 1 ? x ? x 2 sin ? x sin ? x ? x ? cos ? x \lim_{x \to 0}{{e^x-1-x-{x \over 2}\sin x} \over {\sin x -x \cdot \cos x}} limx→0?sinx?x?cosxex?1?x?2x?sinx?
原式等于：
1 + x + x 2 2 ! + x 3 3 ! + x 4 4 ! + x 5 5 ! + ? ? 1 ? x ? x 2 ( 1 ( x ) 1 ! + ? 1 ( x ) 3 3 ! + 1 ( x ) 5 5 ! + ? ) 1 ( x ) 1 ! + ? 1 ( x ) 3 3 ! + 1 ( x ) 5 5 ! + ? ? x ( 1 + ? 1 ( x ) 2 2 ! + ( x ) 4 4 ! + ? 1 ( x ) 6 6 ! + ? ) {{1+x+{x^2 \over 2!}+{x^3 \over 3!}+{x^4 \over 4!}+{x^5 \over 5!} + \cdots} -1 -x -{x \over 2}({{1(x) \over 1!}+{-1(x)^3 \over 3!}+{1(x)^5 \over 5!}+ \cdots})} \over{{{1(x) \over 1!}+{-1(x)^3 \over 3!}+{1(x)^5 \over 5!}+ \cdots} -x({{1}+{-1(x)^2 \over 2!}+{(x)^4 \over 4!}+{-1(x)^6 \over 6!}+ \cdots})} 1!1(x)?+3!?1(x)3?+5!1(x)5?+??x(1+2!?1(x)2?+4!(x)4?+6!?1(x)6?+?)1+x+2!x2?+3!x3?+4!x4?+5!x5?+??1?x?2x?(1!1(x)?+3!?1(x)3?+5!1(x)5?+?)?
= x 3 3 ! + x 4 4 ! + x 5 5 ! + ? ? x 2 ( ? 1 ( x ) 3 3 ! + 1 ( x ) 5 5 ! + ? ) ? 1 ( x ) 3 3 ! + 1 ( x ) 5 5 ! + ? ? x ( ? 1 ( x ) 2 2 ! + ( x ) 4 4 ! + ? 1 ( x ) 6 6 ! + ? ) ={{{x^3 \over 3!}+{x^4 \over 4!}+{x^5 \over 5!} + \cdots} -{x \over 2}({{-1(x)^3 \over 3!}+{1(x)^5 \over 5!}+ \cdots})} \over{{{-1(x)^3 \over 3!}+{1(x)^5 \over 5!}+ \cdots} -x({{-1(x)^2 \over 2!}+{(x)^4 \over 4!}+{-1(x)^6 \over 6!}+ \cdots})} 3!?1(x)3?+5!1(x)5?+??x(2!?1(x)2?+4!(x)4?+6!?1(x)6?+?)=3!x3?+4!x4?+5!x5?+??2x?(3!?1(x)3?+5!1(x)5?+?)?
= x 3 3 ! + x 4 4 ! + x 5 5 ! + ? ? x 2 ( ? 1 ( x ) 3 3 ! + 1 ( x ) 5 5 ! + ? ) ? 1 ( x ) 3 3 ! + 1 ( x ) 5 5 ! + ? ? x ( ? 1 ( x ) 2 2 ! + ( x ) 4 4 ! + ? 1 ( x ) 6 6 ! + ? ) ={{{x^3 \over 3!}+{x^4 \over 4!}+{x^5 \over 5!} + \cdots} -{x \over 2}({{-1(x)^3 \over 3!}+{1(x)^5 \over 5!}+ \cdots})} \over{{{-1(x)^3 \over 3!}+{1(x)^5 \over 5!}+ \cdots} -x({{-1(x)^2 \over 2!}+{(x)^4 \over 4!}+{-1(x)^6 \over 6!}+ \cdots})} 3!?1(x)3?+5!1(x)5?+??x(2!?1(x)2?+4!(x)4?+6!?1(x)6?+?)=3!x3?+4!x4?+5!x5?+??2x?(3!?1(x)3?+5!1(x)5?+?)?
= x 3 3 + R n ( x ) x 3 3 + R n ( x ) ={{x^3 \over 3} +R_n(x)} \over{{x^3 \over 3} +R_n(x)} 3x3?+Rn?(x)=3x3?+Rn?(x)?
= 1 =1 =1
