用逐阶微分法推导泰勒展开式若函数f (x ) 在x 0的某个邻域内有n 阶导数,则f (x 0) 存在且在x 0处连续。如此,根据一阶导数的定义,有
f (x ) −f (x 0) =f (x 0) lim x →x 0x −x 0
将其改写成含无穷小量o 1的表达式,即
f (x ) =f (x 0) +f (x 0)(x −x 0) +o 1(x −x 0)
如此,则有
f (x ) −f (x 0) −f (x 0)(x −x 0) o 1=x −x 0
考察以下极限,并运用洛必达法则,有
o 1f (x ) −f (x 0) −f (x 0)(x −x 0) =lim lim x →x 0x −x 0x →x 0(x −x 0) 2
f (x ) −f (x 0) 1f (x ) −f (x 0) =lim =lim x →x 02(x −x 0) 2x →x 0x −x 0
1=f (x 0) 2
再将此结果改写成含无穷小量o 2的表达式,即
1o 1=f (x 0)(x −x 0) +o 2(x −x 0) 2
代入得
1f (x ) =f (x 0) +f (x 0)(x −x 0) +f (x 0)(x −x 0) 2+o 2(x −x 0) 2
2
1
2
又由此式,得
1 f (x ) −f (x 0) −f (x 0)(x −x 0) −2f (x 0)(x −x 0) 2
o 2=(x −x 0) 2
进一步,有
f (x ) −f (x 0) −f (x 0)(x −x 0) −1f (x 0)(x −x 0) 2o 2lim =lim x →x 0x −x 0x →x 0(x −x 0) 3
f (x ) −f (x 0) −f (x 0)(x −x 0) =lim x →x 03(x −x 0) 2
f (x ) −f (x 0) 11 f (x ) −f (x 0) =lim =f (x 0) =lim x →x 03·2(x −x 0) 3·2x →x 0x −x 03·2
所以,将其改写成含无穷小量o 3的表达式,并代入得
11 f (x ) =f (x 0)+f (x 0)(x −x 0)+f (x 0)(x −x 0) 2+f (x 0)(x −x 0) 3+o 3(x −x 0) 3
23·2
重复此法,直至将f (x ) 展开至n 次多项式,即得泰勒展开式。
用逐阶微分法推导泰勒展开式若函数f (x ) 在x 0的某个邻域内有n 阶导数,则f (x 0) 存在且在x 0处连续。如此,根据一阶导数的定义,有
f (x ) −f (x 0) =f (x 0) lim x →x 0x −x 0
将其改写成含无穷小量o 1的表达式,即
f (x ) =f (x 0) +f (x 0)(x −x 0) +o 1(x −x 0)
如此,则有
f (x ) −f (x 0) −f (x 0)(x −x 0) o 1=x −x 0
考察以下极限,并运用洛必达法则,有
o 1f (x ) −f (x 0) −f (x 0)(x −x 0) =lim lim x →x 0x −x 0x →x 0(x −x 0) 2
f (x ) −f (x 0) 1f (x ) −f (x 0) =lim =lim x →x 02(x −x 0) 2x →x 0x −x 0
1=f (x 0) 2
再将此结果改写成含无穷小量o 2的表达式,即
1o 1=f (x 0)(x −x 0) +o 2(x −x 0) 2
代入得
1f (x ) =f (x 0) +f (x 0)(x −x 0) +f (x 0)(x −x 0) 2+o 2(x −x 0) 2
2
1
2
又由此式,得
1 f (x ) −f (x 0) −f (x 0)(x −x 0) −2f (x 0)(x −x 0) 2
o 2=(x −x 0) 2
进一步,有
f (x ) −f (x 0) −f (x 0)(x −x 0) −1f (x 0)(x −x 0) 2o 2lim =lim x →x 0x −x 0x →x 0(x −x 0) 3
f (x ) −f (x 0) −f (x 0)(x −x 0) =lim x →x 03(x −x 0) 2
f (x ) −f (x 0) 11 f (x ) −f (x 0) =lim =f (x 0) =lim x →x 03·2(x −x 0) 3·2x →x 0x −x 03·2
所以,将其改写成含无穷小量o 3的表达式,并代入得
11 f (x ) =f (x 0)+f (x 0)(x −x 0)+f (x 0)(x −x 0) 2+f (x 0)(x −x 0) 3+o 3(x −x 0) 3
23·2
重复此法,直至将f (x ) 展开至n 次多项式,即得泰勒展开式。