泰勒展开范例
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→x0x−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→x0x−x0f(x)−f(x0)−f′(x0)(x−x0)=limx→x0(x−x0)′(f(x)−f(x0)−f′(x0)(x−x0))′=limx→x01f′(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
( 1 + x ) α , α ∈ ℜ (1+x)^\alpha, \alpha \in \Re (1+x)α,α∈ℜ
( 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→0sinx−x⋅cosxex−1−x−2xsinx
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