【数学】泰勒公式推导(佩亚诺余项)

每次忘了都要查半天,自己推导一遍记录下来


Step 1

f ( x ) f(x) f(x)存在n阶导数,根据一阶导数定义:

lim ⁡ x → x 0 f ( x ) − f ( x 0 ) x − x 0 = f ′ ( x ) \lim\limits_{x\rightarrow x_0}\cfrac{f(x)-f(x_0)}{x-x_0}=f'(x) xx0limxx0f(x)f(x0)=f(x)

等式脱帽
f ( x ) − f ( x 0 ) x − x 0 = f ′ ( x 0 ) + o \cfrac{f(x)-f(x_0)}{x-x_0}=f'(x_0)+o xx0f(x)f(x0)=f(x0)+o
f ( x ) − f ( x 0 ) = ( x − x 0 ) f ′ ( x 0 ) + o ( x − x 0 ) f(x)-f(x_0)=(x-x_0)f'(x_0)+o(x-x_0) f(x)f(x0)=(xx0)f(x0)+o(xx0)

化简得
{ f ( x ) = f ( x 0 ) + ( x − x 0 ) f ′ ( x 0 ) + o ( x − x 0 ) ( 1 ) o ( x − x 0 ) = f ( x ) − f ( x 0 ) − ( x − x 0 ) f ′ ( x 0 ) ( 2 ) \begin{cases} f(x)=f(x_0)+(x-x_0)f'(x_0)+o(x-x_0)(1)\\ o(x-x_0)= f(x)-f(x_0)-(x-x_0)f'(x_0)(2) \end{cases} {f(x)=f(x0)+(xx0)f(x0)+o(xx0)1o(xx0)=f(x)f(x0)(xx0)f(x0)2


Step 2

根据式 ( 2 ) (2) (2)
lim ⁡ x → x 0 o ( x − x 0 ) ( x − x 0 ) 2 = lim ⁡ x → x 0 f ( x ) − f ( x 0 ) − ( x − x 0 ) f ′ ( x 0 ) ( x − x 0 ) 2 \lim\limits_{x\rightarrow x_0}\cfrac{o(x-x_0)}{(x-x_0)^2}=\lim\limits_{x\rightarrow x_0}\cfrac{f(x)-f(x_0)-(x-x_0)f'(x_0)}{(x-x_0)^2} xx0lim(xx0)2o(xx0)=xx0lim(xx0)2f(x)f(x0)(xx0)f(x0) ( 0 0 型 ) \big(\frac{0}{0} 型\big) (00)

使用洛必达法则
lim ⁡ x → x 0 o ( x − x 0 ) ( x − x 0 ) 2 = lim ⁡ x → x 0 f ′ ( x ) − f ′ ( x 0 ) 2 ( x − x 0 ) = lim ⁡ x → x 0 f ′ ′ ( x ) 2 = f ′ ′ ( x 0 ) 2 ( 常 数 ) \lim\limits_{x\rightarrow x_0}\cfrac{o(x-x_0)}{(x-x_0)^2}=\lim\limits_{x\rightarrow x_0}\cfrac{f'(x)-f'(x_0)}{2(x-x_0)}=\lim\limits_{x\rightarrow x_0}\cfrac{f''(x)}{2}=\cfrac{f''(x_0)}{2}(常数) xx0lim(xx0)2o(xx0)=xx0lim2(xx0)f(x)f(x0)=xx0lim2f(x)=2f(x0)()

所以得出 o ( x − x 0 ) o(x-x_0) o(xx0) ( x − x 0 ) 2 (x-x_0)^2 (xx0)2是等价无穷小
o ( x − x 0 ) ( x − x 0 ) 2 = 1 2 f ′ ′ ( x 0 ) + o \cfrac{o(x-x_0)}{(x-x_0)^2}=\cfrac{1}{2}f''(x_0)+o (xx0)2o(xx0)=21f(x0)+o
o ( x − x 0 ) = 1 2 ( x − x 0 ) 2 f ′ ′ ( x 0 ) + o ( x − x 0 ) 2 o(x-x_0)=\cfrac{1}{2}(x-x_0)^2f''(x_0)+o(x-x_0)^2 o(xx0)=21(xx0)2f(x0)+o(xx0)2

带入式 ( 1 ) (1) (1)
{ f ( x ) = f ( x 0 ) + ( x − x 0 ) f ′ ( x 0 ) + 1 2 ( x − x 0 ) 2 f ′ ′ ( x 0 ) + o ( x − x 0 ) 2 ( 3 ) o ( x − x 0 ) 2 = f ( x ) − f ( x 0 ) − ( x − x 0 ) f ′ ( x 0 ) − 1 2 ( x − x 0 ) 2 f ′ ′ ( x 0 ) ( 4 ) \begin{cases} f(x)=f(x_0)+(x-x_0)f'(x_0)+\cfrac{1}{2}(x-x_0)^2f''(x_0)+o(x-x_0)^2(3)\\ o(x-x_0)^2= f(x)-f(x_0)-(x-x_0)f'(x_0)-\cfrac{1}{2}(x-x_0)^2f''(x_0)(4) \end{cases} f(x)=f(x0)+(xx0)f(x0)+21(xx0)2f(x0)+o(xx0)23o(xx0)2=f(x)f(x0)(xx0)f(x0)21(xx0)2f(x0)4


Step 3

根据式 ( 4 ) (4) (4)
lim ⁡ x → x 0 o ( x − x 0 ) 2 ( x − x 0 ) 3 = lim ⁡ x → x 0 f ( x ) − f ( x 0 ) − ( x − x 0 ) f ′ ( x 0 ) − 1 2 ( x − x 0 ) 2 f ′ ′ ( x 0 ) ( x − x 0 ) 3 \lim\limits_{x\rightarrow x_0}\cfrac{o(x-x_0)^2}{(x-x_0)^3}=\lim\limits_{x\rightarrow x_0}\cfrac{f(x)-f(x_0)-(x-x_0)f'(x_0)-\cfrac{1}{2}(x-x_0)^2f''(x_0)}{(x-x_0)^3} xx0lim(xx0)3o(xx0)2=xx0lim(xx0)3f(x)f(x0)(xx0)f(x0)21(xx0)2f(x0) ( 0 0 型 ) \big(\frac{0}{0} 型\big) (00)

使用洛必达法则
lim ⁡ x → x 0 o ( x − x 0 ) 2 ( x − x 0 ) 3 = 2 次 lim ⁡ x → x 0 f ′ ′ ( x ) − f ′ ′ ( x 0 ) 3 ⋅ 2 ( x − x 0 ) = lim ⁡ x → x 0 f ′ ′ ′ ( x ) 6 = f ′ ′ ′ ( x 0 ) 6 ( 常 数 ) \lim\limits_{x\rightarrow x_0}\cfrac{o(x-x_0)^2}{(x-x_0)^3}\xlongequal{2次}\lim\limits_{x\rightarrow x_0}\cfrac{f''(x)-f''(x_0)}{3\cdot2(x-x_0)}=\lim\limits_{x\rightarrow x_0}\cfrac{f'''(x)}{6}=\cfrac{f'''(x_0)}{6}(常数) xx0lim(xx0)3o(xx0)22 xx0lim32(xx0)f(x)f(x0)=xx0lim6f(x)=6f(x0)()

所以得出 o ( x − x 0 ) 2 o(x-x_0)^2 o(xx0)2 ( x − x 0 ) 3 (x-x_0)^3 (xx0)3是等价无穷小
o ( x − x 0 ) 2 ( x − x 0 ) 3 = 1 6 f ′ ′ ′ ( x 0 ) + o \cfrac{o(x-x_0)^2}{(x-x_0)^3}=\cfrac{1}{6}f'''(x_0)+o (xx0)3o(xx0)2=61f(x0)+o
o ( x − x 0 ) 2 = 1 6 ( x − x 0 ) 3 f ′ ′ ′ ( x 0 ) + o ( x − x 0 ) 3 o(x-x_0)^2=\cfrac{1}{6}(x-x_0)^3f'''(x_0)+o(x-x_0)^3 o(xx0)2=61(xx0)3f(x0)+o(xx0)3

带入式 ( 1 ) (1) (1)
{ f ( x ) = f ( x 0 ) + ( x − x 0 ) f ′ ( x 0 ) + 1 2 ( x − x 0 ) 2 f ′ ′ ( x 0 ) + 1 6 ( x − x 0 ) 3 f ′ ′ ′ ( x 0 ) + o ( x − x 0 ) 3 ( 5 ) o ( x − x 0 ) 3 = f ( x ) − f ( x 0 ) − ( x − x 0 ) f ′ ( x 0 ) − 1 2 ( x − x 0 ) 2 f ′ ′ ( x 0 ) − 1 6 ( x − x 0 ) 3 f ′ ′ ′ ( x 0 ) ( 6 ) \begin{cases} f(x)=f(x_0)+(x-x_0)f'(x_0)+\cfrac{1}{2}(x-x_0)^2f''(x_0)+\cfrac{1}{6}(x-x_0)^3f'''(x_0)+o(x-x_0)^3(5)\\ o(x-x_0)^3= f(x)-f(x_0)-(x-x_0)f'(x_0)-\cfrac{1}{2}(x-x_0)^2f''(x_0)-\cfrac{1}{6}(x-x_0)^3f'''(x_0)(6) \end{cases} f(x)=f(x0)+(xx0)f(x0)+21(xx0)2f(x0)+61(xx0)3f(x0)+o(xx0)35o(xx0)3=f(x)f(x0)(xx0)f(x0)21(xx0)2f(x0)61(xx0)3f(x0)6


结论

以此类推
o ( x − x 0 ) n − 1 = 1 n ! ( x − x 0 ) n f ( n ) ( x 0 ) + o ( x − x 0 ) n o(x-x_0)^{n-1}=\cfrac{1}{n!}(x-x_0)^nf^{(n)}(x_0)+o(x-x_0)^{n} o(xx0)n1=n!1(xx0)nf(n)(x0)+o(xx0)n
f ( x ) = f ( x 0 ) + ( x − x 0 ) f ′ ( x 0 ) + 1 2 ( x − x 0 ) 2 f ′ ′ ( x 0 ) + 1 6 ( x − x 0 ) 3 f ′ ′ ′ ( x 0 ) + ⋯ + 1 n ! ( x − x 0 ) n f ( n ) ( x 0 ) + o ( x − x 0 ) n f(x)=f(x_0)+(x-x_0)f'(x_0)+\cfrac{1}{2}(x-x_0)^2f''(x_0)+\cfrac{1}{6}(x-x_0)^3f'''(x_0)+\cdots+\cfrac{1}{n!}(x-x_0)^nf^{(n)}(x_0)+o(x-x_0)^n f(x)=f(x0)+(xx0)f(x0)+21(xx0)2f(x0)+61(xx0)3f(x0)++n!1(xx0)nf(n)(x0)+o(xx0)n


常用麦克劳林展开式

( 1 ) e x = 1 + x + x 2 2 ! + ⋯ + x n n ! + o ( x n ) = ∑ n = 0 ∞ x n n ! (1)e^x=1+x+\cfrac{x^2}{2!}+\cdots+\cfrac{x^n}{n!}+o(x^n)=\displaystyle\sum_{n=0}^{\infin}\cfrac{x^n}{n!} 1ex=1+x+2!x2++n!xn+o(xn)=n=0n!xn
( 2 ) sin ⁡ x = x − x 3 3 ! + ⋯ + ( − 1 ) n x 2 n + 1 ( 2 n + 1 ) ! + o ( x 2 n + 1 ) = ∑ n = 0 ∞ ( − 1 ) n x 2 n + 1 ( 2 n + 1 ) ! (2)\sin x=x-\cfrac{x^3}{3!}+\cdots+(-1)^n\cfrac{x^{2n+1}}{(2n+1)!}+o(x^{2n+1})=\displaystyle\sum_{n=0}^{\infin}(-1)^n\cfrac{x^{2n+1}}{(2n+1)!} 2sinx=x3!x3++(1)n(2n+1)!x2n+1+o(x2n+1)=n=0(1)n(2n+1)!x2n+1
( 3 ) cos ⁡ x = 1 − x 2 2 ! + ⋯ + ( − 1 ) n x 2 n ( 2 n ) ! + o ( x 2 n ) = ∑ n = 0 ∞ ( − 1 ) n x 2 n ( 2 n ) ! (3)\cos x=1-\cfrac{x^2}{2!}+\cdots+(-1)^n\cfrac{x^{2n}}{(2n)!}+o(x^{2n})=\displaystyle\sum_{n=0}^{\infin}(-1)^n\cfrac{x^{2n}}{(2n)!} 3cosx=12!x2++(1)n(2n)!x2n+o(x2n)=n=0(1)n(2n)!x2n
( 4 ) ln ⁡ ( 1 + x ) = x − x 2 2 + ⋯ + ( − 1 ) n − 1 x n ( n ) + o ( x n ) = ∑ n = 1 ∞ ( − 1 ) n − 1 x n n . ( x > − 1 ) (4)\ln (1+x)=x-\cfrac{x^2}{2}+\cdots+(-1)^{n-1}\cfrac{x^{n}}{(n)}+o(x^n)=\displaystyle\sum_{n=1}^{\infin}(-1)^{n-1}\cfrac{x^{n}}{n }.(x>-1) 4ln(1+x)=x2x2++(1)n1(n)xn+o(xn)=n=1(1)n1nxn.(x>1) [注]: 从n=1开始
( 5 ) 1 1 − x = 1 + x + x 2 + ⋯ + x n + o ( x n ) = ∑ n = 0 ∞ x n (5)\cfrac{1}{1-x}=1+x+x^2+\cdots+x^n+o(x^n)=\displaystyle\sum_{n=0}^\infin x^n 51x1=1+x+x2++xn+o(xn)=n=0xn
( 6 ) 1 1 + x = 1 − x + x 2 − ⋯ + ( − 1 ) n x n + o ( x n ) = ∑ n = 0 ∞ ( − 1 ) n x n (6)\cfrac{1}{1+x}=1-x+x^2-\cdots+(-1)^nx^n+o(x^n)=\displaystyle\sum_{n=0}^\infin (-1)^nx^n 61+x1=1x+x2+(1)nxn+o(xn)=n=0(1)nxn
( 7 ) ( 1 + x ) α = 1 + α x + α ( α − 1 ) 2 x 2 + o ( x 2 ) ( x → 0 ) (7)(1+x)^\alpha=1+\alpha x+\cfrac{\alpha(\alpha-1)}{2}x^2+o(x^2)(x\rightarrow0) 7(1+x)α=1+αx+2α(α1)x2+o(x2)(x0)
( 8 ) tan ⁡ x = x + 1 3 x 3 + o ( x 3 ) ( x → 0 ) (8)\tan x = x+\cfrac{1}{3}x^3+o(x^3)(x\rightarrow0) 8tanx=x+31x3+o(x3)(x0)
( 9 ) arcsin ⁡ x = x + 1 6 x 3 + o ( x 3 ) ( x → 0 ) (9)\arcsin x=x+\cfrac{1}{6}x^3+o(x^3)(x\rightarrow0) 9arcsinx=x+61x3+o(x3)(x0)
( 10 ) arctan ⁡ x = x − 1 3 x 3 + o ( x 3 ) ( x → 0 ) (10)\arctan x=x-\cfrac{1}{3}x^3+o(x^3)(x\rightarrow0) 10arctanx=x31x3+o(x3)(x0)

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