函数展成幂级数

(2). f ( x) = ln(1 + x)
∞ 1 = ∑ (−1) n x n , (−1 < x < 1) 1 + x n =0
∞ ∞ xn x 1 n −1 dx = ∑ [ ∫ (−1) n x n dx] = ∑ (−1) ln(1 + x) = ∫ 0 1+ x 0 n n =1 n =0
∞
2.
1 展开为 x − 5的幂级数 2 x − 5x + 6 1 1 1 1 1 1 = = − = − 2 x − 5 x + 6 ( x − 2)( x − 3) x − 3 x − 2 ( x − 5) + 2 ( x − 5) + 3
∞ 1 1 1 1 1 1 = ⋅ − ⋅ = ∑ ( −1) n n +1 − n +1 ( x − 5) n 2 1 + x − 5 3 1 + x − 5 n =0 3 2 2 3 x−5 2 <1 ⇒ 3< x <7 x −5 <1 3 当x = 3时, 对应常数项级数发散；
x
(−1 < x ≤ 1)
(3). f ( x) = e
− x2
2 作变量替换 t = − x
e− x
2
t2 tn = e = 1+ t + + ⋅⋅⋅ + + ⋅⋅⋅ 2! n!
t 2
2n x4 x6 n x = 1 − x + − + ⋅ ⋅ ⋅ + (−1) + ⋅ ⋅ ⋅, (−∞ < x < ∞) 2! 3! n! x (4). f ( x) = 1+ x2 ∞ 1 x = x⋅ = x ⋅ ∑ ( −1) n x 2 n 1+ x2 1+ x2 n =0
f ′′(0) 2 f ( n ) (0) n (3) f (0) + f ′(0) x + x + ⋅⋅⋅ + x + ⋅⋅⋅ 2! n! 求出收敛半径R
(4) 在(-R,R)内,如果 lim Rn ( x) = 0 则 f(x) n →∞
例 将函数展开成 x 的幂级数
(1). f ( x) = e x f ( n ) ( x) = e x , (n = 1,2,...) f (0) = f ( n ) (0) = 1, (n = 1,2,...)
= ∑ (−1) n x 2 n +1 , (−1 < x < 1)
n =0
∞
1 例 将 分别展开成 x 的及 x－1 的幂级数 3− x
1 1 1 1 ∞ x n = ⋅ = ⋅ ( ) ① x 3 ∑ 3 3 − x 3 1− n =0 3 ∞ xn = ∑ n +1 , (−3 < x < 3) n =0 3
∞
∞
x n +1 xf ( x) = ∑ = e x −1 ⇒ n =1 ( n + 1)! 此时
n
( x − 1)e x + 1 f ' ( x) = ⇒ 2 x
1 1 3 (2) e = 1 + n + ( n ) + ( n ) 3 + ⋯ 2! 3! 1 2 24 n − n ∴ e > n ⇒ e < 2 n 24 由正项级数的比较判别法知原级数收敛.
第四节 函数展开成幂级数
第四节 函数展开成幂级数
前面研究的是幂级数的收敛域及和函数,现在反过 来,某个函数是否可以在某个区间内用幂级数表示 一. 泰勒级数 第三章研究过泰勒公式:
f ′′( x0 ) f ( x) = f ( x0 ) + f ′( x0 )( x − x0 ) + ( x − x0 ) 2 2! f ( n ) ( x0 ) 余项 + ⋅⋅⋅ + ( x − x0 ) n + Rn ( x) n! 其中f(x) 在 x0 的某邻域内具有n+1阶导数.
1 1 1 1 1 ∞ x −1 n ② 3 − x = 2 − ( x − 1) = 2 ⋅ x − 1 = 2 ⋅ ∑ ( 2 ) n=0 1− 2 ∞ ( x − 1) n = ∑ n +1 , (−1 < x < 3) 2 n =0
1 例 将 x 2 + 4 x + 3 展开成 x－1的幂级数
x2 xn 收敛半径 R = +∞ ∴1 + x + + ⋅ ⋅ ⋅ + + ⋅ ⋅ ⋅ 2! n! n+ n+1 eξ | x| | x | , (0 < ξ < x ) | Rn ( x) |= x n +1 < e ⋅ (n + 1)! (n + 1)! ∞ | x |n +1 有限 趋于零,因为 ∑ 收敛 n = 0 ( n + 1)! lim Rn ( x) = 0 所以 n→∞
牛顿二项式级数
2.间接展开法 利用已知的基本展开式和幂级数的性质 (1).逐项积分,逐项求导法 (2)变量替换法 (3)四则运算法
例 将函数展开成 x 的幂级数
(1). f ( x) = cos x
x3 x5 x 2 n +1 n −1 cos x = (sin x)′ = ( x − + − ⋅ ⋅ ⋅ + (−1) + ⋅ ⋅ ⋅)′ 3! 5! (2n + 1)! 2n x2 x4 n x = 1 − + − ⋅ ⋅ ⋅ + (−1) + ⋅ ⋅ ⋅, (−∞ < x < ∞) 2! 4! 2n!
当x = 7时，对应常数项级数发 散, ∴ 级数的收敛域为 (3,7).
x + ln(1 − x ) 3.∫ dx 2 0 x
1
1 n 1 2 1 3 1 n ln(1 − x) = −∑ x = − x − x − x − ⋯ − x − ⋯ , 2 3 n n =1 n 1 1 x + ln(1 − x ) 1 n−2 1 1 ∫0 x 2 dx = ∫0 (− 2 − 3 x − ⋯ − n x − ⋯)dx 1 1 x2 1 x n −1 = (− x − ⋅ − ⋯ − ⋅ − ⋯) |1 0 2 3 2 n n −1 = −1.
x2 xn e x = 1 + x + + ⋅ ⋅ ⋅ + + ⋅ ⋅ ⋅, (−∞ < x < +∞) n! 2!
(2). f ( x) = sin x = ∑ (−1)
n =0
∞
n −1
x 2 n +1 , (−∞ < x < ∞) (2n + 1)!
nπ ), (n = 1,2,...) 2 f ( n ) (0) = 0,1,0,−1,...(n = 0,1,2,...) (循环) x3 x5 x 2 n +1 n −1 ∴ x − + − ⋅ ⋅ ⋅ + (−1) + ⋅ ⋅ ⋅ 收敛半径 R = +∞ 3! 5! (2n + 1)! n +1 sin(ξ + π) | x |n +1 2 | Rn ( x) |= x n +1 ≤ (n + 1)! , (0 < ξ < x) (n + 1)! 0 所以 lim Rn ( x) = 0 f n ( x) = sin( x +
n →∞
(3).(1 + x ) = 1 + αx +
α
α (α − 1)
2! 3! α (α − 1)(α − 2) ⋅ ⋅ ⋅ (α − n + 1) n + ⋅⋅⋅ + x + ⋅ ⋅ ⋅, ( −1 < x < Biblioteka Baidu) n!
x +
2
α (α − 1)(α − 2)
x3
注: α>－1时,展式在 x =1成立; α>0时,展式在 x = －1成立.
( −1 < x < 3)
= ∑ (−1) (
n n =0
∞
1 2n+2
−
2
)( x − 1) n , 2 n+3
1
( −3 < x < 5)
(−1 < x < 3)
练习
1.将 ln(1 + x − 2 x 2 )展开成x的幂级数并求收敛域
ln(1 + x − 2 x 2 ) = ln(1 + 2 x)(1 − x) = ln(1 + 2 x) + ln(1 − x) ( −1) n −1 ⋅ 2 n n ∞ 1 n 1 =∑ x − ∑ x | x |< 2 n n =1 n =1 n 1 当x = − 时, 对应常数项级数收敛； 2 1 当x = 时，对应常数项级数发 散, 2 1 1 ∴ 级数的收敛域为 (− , ]. 2 2
2. 泰勒定理: 若f(x) 在x0 的某邻域内具有各阶导数, 则f(x)在 x0 的泰勒级数在该邻域内收敛于f(x)
⇔ lim Rn ( x) = 0
n→∞
(由泰勒公式很容易得出结论,证明略) 注: (1) 若f(x)在 x0 的泰勒级数收敛于f(x),即 ∞ f ( n ) ( x0 ) 泰勒展开式 f ( x) = ∑ ( x − x0 ) n n! n =0 则称 f(x)在 x0 可以展开成泰勒级数 (2) 如果函数可以展开成幂级数,则展开式唯一.
1 1 1 1 = = − x 2 + 4 x + 3 ( x + 1)( x + 3) 2(1 + x) 2(3 + x)
1 1 = − x −1 x −1 4(1 + ) 8(1 + ) 2 4
1 ∞ 1 ∞ x −1 n n x −1 n = ⋅ ∑ (−1) ( ) − ⋅ ∑ (−1) n ( ) 4 n=0 2 8 n=0 4
此时, f(x)可以用前n+1项近似表示,误差为 | Rn ( x ) | 由此引入泰勒级数: 1. 定义 若f(x)在 x0 的某邻域内具有各阶导数,则 f ′′( x0 ) f ( x0 ) + f ′( x0 )( x − x0 ) + ( x − x0 ) 2 2! f ( n ) ( x0 ) f(x)在 x0 的泰勒级数 + ⋅⋅⋅ + ( x − x0 ) n + ⋅ ⋅ ⋅ n! ∞ 泰勒系数 f ( n ) ( x0 ) n f ( x) ~ ∑ ( x − x0 ) n! n =0 ∞ f ( n ) ( 0) n x x0 = 0 f ( x) ~ ∑ 麦克劳林级数 n! n =0
∞
n 4. 证明 : (1) ∑ = 1. n =1 ( n + 1)!
1 (1) 记 f ( x) = ∑ xn ,则 n = 0 ( n + 1)!
∞ ∞
∞
(2) ∑ e − n 收敛.
n =1
∞
n f ' ( x) = ∑ x n −1 , n =1 ( n + 1)! ex −1 f ( x) = , x≠0 x n ∑ (n + 1)! = f ' (1) = 1. n =1
二. 函数展开成幂级数
x0 = 0麦克劳林级数
主要研究函数如何展开成 x 的幂级数. 1. 直接展开法
f ′( x), f ′′( x),⋅ ⋅ ⋅, f ( n ) ( x),⋅ ⋅ ⋅ (1) 求出
如果某阶导数不存 在,说明不能展开
(2) 求出
f (0), f ′(0), f ′′(0),⋅ ⋅ ⋅, f (n ) (0),⋅ ⋅ ⋅