文章目录
一、幂级数是什么？二、常见的幂级数三、幂级数重要结论：在收敛域内可逐项求导和积分四、用幂级数求数项级数的和五、用幂级数求数列通项公式六、用幂级数解微分方程
一、幂级数是什么？
由常数项乘以幂函数组成的无穷级数：
∑
n
=
0
∞
c
n
x
n
=
c
0
+
c
1
x
+
c
2
x
2
+
c
3
x
3
+
…
\sum_{n=0}^{\infty} c_{n} x^{n}=c_{0}+c_{1} x+c_{2} x^{2}+c_{3} x^{3}+\ldots
n=0∑∞cnxn=c0+c1x+c2x2+c3x3+… 它在
(
−
R
,
R
)
(-R,R)
(−R,R)内一定收敛，通常是否取到端点要具体讨论。这里
R
=
1
/
lim
n
→
∞
∣
c
n
+
1
c
n
∣
R=1\Big/\lim _{n \rightarrow \infty}\left|\frac{c_{n+1} }{c_{n}}\right|
R=1/limn→∞∣∣∣cncn+1∣∣∣，且
R
R
R 可以为0或无穷大。
二、常见的幂级数
1
1
−
x
=
1
+
x
+
x
2
+
x
3
+
x
4
+
…
=
∑
n
=
0
∞
x
n
,
x
∈
(
−
1
,
1
)
\frac{1}{1-x} = 1+x+x^{2}+x^{3}+x^{4}+\ldots =\sum_{n=0}^{\infty} x^{n}, \quad x \in(-1,1)
1−x1=1+x+x2+x3+x4+…=n=0∑∞xn,x∈(−1,1)
e
x
=
1
+
x
+
x
2
2
!
+
x
3
3
!
+
x
4
4
!
+
…
=
∑
n
=
0
∞
x
n
n
!
,
x
∈
R
e^{x} =1+x+\frac{x^{2}}{2 !}+\frac{x^{3}}{3 !}+\frac{x^{4}}{4 !}+\ldots =\sum_{n=0}^{\infty} \frac{x^{n}}{n !}, \quad x \in R
ex=1+x+2!x2+3!x3+4!x4+…=n=0∑∞n!xn,x∈R
cos
x
=
1
−
x
2
2
!
+
x
4
4
!
−
x
6
6
!
+
x
8
8
!
−
⋯
=
∑
n
=
0
∞
(
−
1
)
n
x
2
n
(
2
n
)
!
,
x
∈
R
\cos x =1-\frac{x^{2}}{2 !}+\frac{x^{4}}{4 !}-\frac{x^{6}}{6 !}+\frac{x^{8}}{8 !}-\cdots =\sum_{n=0}^{\infty}(-1)^{n} \frac{x^{2 n}}{(2 n) !}, \quad x \in R
cosx=1−2!x2+4!x4−6!x6+8!x8−⋯=n=0∑∞(−1)n(2n)!x2n,x∈R
sin
x
=
x
−
x
3
3
!
+
x
5
5
!
−
x
7
7
!
+
x
9
9
!
−
⋯
=
∑
n
=
1
∞
(
−
1
)
(
n
−
1
)
x
2
n
−
1
(
2
n
−
1
)
!
,
x
∈
R
\sin x=x-\frac{x^{3}}{3 !}+\frac{x^{5}}{5 !}-\frac{x^{7}}{7 !}+\frac{x^{9}}{9 !}-\cdots =\sum_{n=1}^{\infty}(-1)^{(n-1)} \frac{x^{2 n-1}}{(2 n-1) !}, \quad x \in R
sinx=x−3!x3+5!x5−7!x7+9!x9−⋯=n=1∑∞(−1)(n−1)(2n−1)!x2n−1,x∈R
ln
(
1
+
x
)
=
x
−
x
2
2
+
x
3
3
−
x
4
4
+
x
5
5
−
⋯
=
∑
n
=
1
∞
(
−
1
)
(
n
−
1
)
x
n
n
=
or
∑
n
=
1
∞
(
−
1
)
n
+
1
x
n
n
,
(
x
∈
(
−
1
,
1
]
)
\ln (1+x) =x-\frac{x^{2}}{2}+\frac{x^{3}}{3}-\frac{x^{4}}{4}+\frac{x^{5}}{5}-\cdots =\sum_{n=1}^{\infty}(-1)^{(n-1)} \frac{x^{n}}{n} \stackrel{\text { or }}{=} \sum_{n=1}^{\infty}(-1)^{n+1} \frac{x^{n}}{n},\quad (x \in(-1,1] )
ln(1+x)=x−2x2+3x3−4x4+5x5−⋯=n=1∑∞(−1)(n−1)nxn= or n=1∑∞(−1)n+1nxn,(x∈(−1,1])
tan
−
1
x
=
x
−
x
3
3
+
x
5
5
−
x
7
7
+
x
9
9
−
⋯
=
∑
n
=
1
∞
(
−
1
)
(
n
−
1
)
x
2
n
−
1
2
n
−
1
,
(
x
∈
(
−
1
,
1
]
)
\tan ^{-1} x \quad =x-\frac{x^{3}}{3}+\frac{x^{5}}{5}-\frac{x^{7}}{7}+\frac{x^{9}}{9}-\cdots =\sum_{n=1}^{\infty}(-1)^{(n-1)} \frac{x^{2 n-1}}{2 n-1} ,\quad (x \in(-1,1] )
tan−1x=x−3x3+5x5−7x7+9x9−⋯=n=1∑∞(−1)(n−1)2n−1x2n−1,(x∈(−1,1])
1
+
x
=
1
+
x
2
−
x
2
8
+
x
3
16
+
…
,
x
≥
−
1
\sqrt{1+x}=1+\frac{x}{2}-\frac{x^{2}}{8}+\frac{x^{3}}{16}+\ldots,\quad x \geq -1
1+x
=1+2x−8x2+16x3+…,x≥−1
(
1
+
x
)
k
=
1
+
k
x
+
k
(
k
−
1
)
2
!
x
2
+
k
(
k
−
1
)
(
k
−
2
)
3
!
x
3
+
…
=
∑
n
=
0
∞
(
k
n
)
x
n
,
∣
x
∣
<
1
(1+x)^{k}=1+k x+\frac{k(k-1)}{2 !} x^{2}+\frac{k(k-1)(k-2)}{3 !} x^{3}+\ldots=\sum_{n=0}^{\infty}\left(\begin{array}{l} k \\ n \end{array}\right) x^{n}, \quad |x|< 1
(1+x)k=1+kx+2!k(k−1)x2+3!k(k−1)(k−2)x3+…=n=0∑∞(kn)xn,∣x∣<1
暂时写上这些，有空再来补充。
三、幂级数重要结论：在收敛域内可逐项求导和积分
例1：(
∣
x
∣
<
1
|x|<1
∣x∣<1)
1
(
1
−
x
)
2
=
d
d
x
(
1
1
−
x
)
=
d
d
x
(
∑
n
=
0
∞
x
n
)
=
∑
n
=
1
∞
n
x
n
−
1
=
1
+
2
x
+
3
x
2
+
4
x
3
+
⋯
\begin{aligned} \frac{1}{(1-x)^{2}} &=\frac{d}{d x}\left(\frac{1}{1-x}\right) \\ &=\frac{d}{d x}\left(\sum_{n=0}^{\infty} x^{n}\right) \\ &=\sum_{n=1}^{\infty} n x^{n-1} \\ &=1+2 x+3 x^{2}+4 x^{3}+\cdots \end{aligned}
(1−x)21=dxd(1−x1)=dxd(n=0∑∞xn)=n=1∑∞nxn−1=1+2x+3x2+4x3+⋯ 例2：(
∣
x
∣
<
1
|x|<1
∣x∣<1)
ln
(
1
+
x
)
=
∫
1
1
+
x
d
x
=
∫
(
∑
n
=
0
∞
(
−
1
)
n
x
n
)
d
x
=
(
∑
n
=
0
∞
(
−
1
)
n
x
n
+
1
n
+
1
)
+
C
=
∑
n
=
0
∞
(
−
1
)
n
x
n
+
1
n
+
1
=
x
−
x
2
2
+
x
3
3
−
x
4
4
+
⋯
\begin{aligned} \ln (1+x) &=\int \frac{1}{1+x} d x \\ &=\int\left(\sum_{n=0}^{\infty}(-1)^{n} x^{n}\right) d x \\ &=\left(\sum_{n=0}^{\infty}(-1)^{n} \frac{x^{n+1}}{n+1}\right)+C \\ &=\sum_{n=0}^{\infty}(-1)^{n} \frac{x^{n+1}}{n+1} \\ &=x-\frac{x^{2}}{2}+\frac{x^{3}}{3}-\frac{x^{4}}{4}+\cdots \end{aligned}
ln(1+x)=∫1+x1dx=∫(n=0∑∞(−1)nxn)dx=(n=0∑∞(−1)nn+1xn+1)+C=n=0∑∞(−1)nn+1xn+1=x−2x2+3x3−4x4+⋯ 例3： 已知
d
d
x
tan
−
1
(
x
)
=
1
1
+
x
2
\frac{d}{d x} \tan ^{-1}(x)=\frac{1}{1+x^{2}}
dxdtan−1(x)=1+x21，
1
1
+
x
2
=
∑
n
=
0
∞
(
−
x
2
)
n
=
∑
n
=
0
∞
(
−
1
)
n
x
2
n
\frac{1}{1+x^{2}}=\sum_{n=0}^{\infty}\left(-x^{2}\right)^{n}=\sum_{n=0}^{\infty}(-1)^{n} x^{2 n}
1+x21=∑n=0∞(−x2)n=∑n=0∞(−1)nx2n，那么 KaTeX parse error: No such environment: eqnarray at position 8: \begin{̲e̲q̲n̲a̲r̲r̲a̲y̲}̲ \tan^{-… 由于
tan
−
1
(
0
)
=
0
\tan^{-1}(0)=0
tan−1(0)=0，因此
C
=
0
C=0
C=0，所以得到上面的公式：
tan
−
1
(
x
)
=
∑
n
=
0
∞
(
−
1
)
n
x
2
n
+
1
2
n
+
1
=
x
−
x
3
3
+
x
5
5
−
x
7
7
+
⋯
\tan^{-1}(x)= \sum_{n=0}^\infty(-1)^n\frac{x^{2n+1}}{2n+1}=x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + \cdots
tan−1(x)=n=0∑∞(−1)n2n+1x2n+1=x−3x3+5x5−7x7+⋯ 例4： 从
arctanh
x
\operatorname{arctanh} x
arctanhx 出发我们能推出些啥结论? (注意不是
arctan
\arctan
arctan, 而是
tanh
(
x
)
=
sinh
(
x
)
cosh
(
x
)
=
e
x
−
e
−
x
e
x
+
e
−
x
\tanh (x)=\frac{\sinh (x)}{\cosh (x)}=\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}
tanh(x)=cosh(x)sinh(x)=ex+e−xex−e−x的反函数)
f
(
x
)
=
arctanh
x
f
′
(
x
)
=
1
1
−
x
2
f
′
′
(
x
)
=
2
x
(
1
−
x
2
)
2
f
′
′
′
(
x
)
=
2
(
1
−
x
2
)
2
−
8
x
2
(
1
−
x
2
)
3
\begin{aligned} f(x) &=\operatorname{arctanh} x \\ f^{\prime}(x) &=\frac{1}{1-x^{2}} \\ f^{\prime \prime}(x) &=\frac{2 x}{\left(1-x^{2}\right)^{2}} \\ f^{\prime \prime \prime}(x) &=\frac{2}{\left(1-x^{2}\right)^{2}}-\frac{8 x^{2}}{\left(1-x^{2}\right)^{3}} \end{aligned}
f(x)f′(x)f′′(x)f′′′(x)=arctanhx=1−x21=(1−x2)22x=(1−x2)22−(1−x2)38x2 注意到
arctanh
x
=
1
1
−
x
2
=
∑
n
=
0
∞
x
2
n
,
(
∣
x
∣
<
1
)
\operatorname{arctanh} x = \frac{1}{1-x^2} = \sum_{n=0}^{\infty} x^{2n},(|x|<1)
arctanhx=1−x21=∑n=0∞x2n,(∣x∣<1)， 可以得出：
arctanh
x
=
∫
0
x
(
1
+
t
2
+
t
4
+
…
)
d
t
=
x
+
x
3
3
+
x
5
5
+
…
\operatorname{arctanh} x=\int_{0}^{x}\left(1+t^{2}+t^{4}+\ldots\right) d t=x+\frac{x^{3}}{3}+\frac{x^{5}}{5}+\ldots
arctanhx=∫0x(1+t2+t4+…)dt=x+3x3+5x5+… 当然还没结束：
f
(
x
)
=
log
1
+
x
1
−
x
=
1
2
log
(
1
+
x
)
−
1
2
log
(
1
−
x
)
=
1
2
∫
0
x
(
1
1
+
t
+
1
1
−
t
)
d
t
=
∫
0
x
d
t
1
−
t
2
=
∫
0
x
(
1
+
t
2
+
t
4
+
…
)
d
t
=
x
+
x
3
3
+
x
5
5
+
…
=
arctanh
x
\begin{aligned} f(x) &=\log \sqrt{\frac{1+x}{1-x}}\\ &=\frac{1}{2} \log (1+x)-\frac{1}{2} \log (1-x) \\ &=\frac{1}{2} \int_{0}^{x}\left(\frac{1}{1+t}+\frac{1}{1-t}\right) d t \\ &=\int_{0}^{x} \frac{d t}{1-t^{2}} \\ &=\int_{0}^{x}\left(1+t^{2}+t^{4}+\ldots\right) d t \\ &=x+\frac{x^{3}}{3}+\frac{x^{5}}{5}+\ldots \\ &=\operatorname{arctanh} x \end{aligned}
f(x)=log1−x1+x
=21log(1+x)−21log(1−x)=21∫0x(1+t1+1−t1)dt=∫0x1−t2dt=∫0x(1+t2+t4+…)dt=x+3x3+5x5+…=arctanhx
四、用幂级数求数项级数的和
例5： 求
∑
n
=
1
∞
n
3
n
\sum_{n=1}^{\infty} \frac{n}{3^{n}}
∑n=1∞3nn
f
(
x
)
=
∑
n
=
0
∞
n
x
n
f
(
x
)
x
=
∑
n
=
0
∞
n
x
n
−
1
∫
f
(
x
)
x
d
x
=
C
+
∑
n
=
0
∞
x
n
∫
f
(
x
)
x
d
x
=
C
+
1
1
−
x
d
d
x
∫
f
(
x
)
x
d
x
=
−
1
(
1
−
x
)
2
f
(
x
)
x
=
1
(
1
−
x
)
2
f
(
x
)
=
x
(
1
−
x
)
2
\begin{aligned} \quad f(x)&= \sum_{n=0}^{\infty} nx^n \\ \quad \frac{f(x)}{x}&= \sum_{n=0}^{\infty} nx^{n-1} \\ \quad \int \frac{f(x)}{x} \: dx&= C + \sum_{n=0}^{\infty} x^n \\ \quad \int \frac{f(x)}{x} \: dx&= C + \frac{1}{1 - x} \\ \quad \frac{d}{dx} \int \frac{f(x)}{x} \: dx&=-\frac{1}{(1 - x)^2} \\ \quad \frac{f(x)}{x}&= \frac{1}{(1 - x)^2} \\ \quad f(x)&= \frac{x}{(1 - x)^2} \end{aligned}
f(x)xf(x)∫xf(x)dx∫xf(x)dxdxd∫xf(x)dxxf(x)f(x)=n=0∑∞nxn=n=0∑∞nxn−1=C+n=0∑∞xn=C+1−x1=−(1−x)21=(1−x)21=(1−x)2x 由于它在
∣
x
∣
<
1
\mid x \mid < 1
∣x∣<1内收敛，因此可令
x
=
1
3
x=\frac{1}{3}
x=31，有
f
(
1
3
)
=
1
3
(
2
3
)
2
=
∑
n
=
0
∞
n
3
n
=
∑
n
=
1
∞
n
3
n
=
3
4
f\left ( \frac{1}{3} \right ) = \frac{\frac{1}{3}}{\left ( \frac{2}{3} \right )^2} = \sum_{n=0}^{\infty} \frac{n}{3^n} = \sum_{n=1}^{\infty} \frac{n}{3^n} = \frac{3}{4}
f(31)=(32)231=n=0∑∞3nn=n=1∑∞3nn=43
例6：
∑
n
=
0
∞
(
n
+
1
)
2
π
n
\sum_{n=0}^{\infty} \frac{(n+1)^2}{\pi^n}
∑n=0∞πn(n+1)2
f
(
x
)
=
∑
n
=
0
∞
(
n
+
1
)
2
x
n
∫
f
(
x
)
d
x
=
C
+
∑
n
=
0
∞
(
n
+
1
)
x
n
+
1
∫
f
(
x
)
d
x
x
=
C
x
+
∑
n
=
0
∞
(
n
+
1
)
x
n
∫
∫
f
(
x
)
d
x
x
d
x
=
D
+
∫
C
x
d
x
+
∑
n
=
0
∞
x
n
+
1
∫
∫
f
(
x
)
d
x
x
d
x
=
D
+
∫
C
x
d
x
+
x
∑
n
=
0
∞
x
n
∫
∫
f
(
x
)
d
x
x
d
x
=
D
+
∫
C
x
d
x
+
x
1
−
x
∫
f
(
x
)
d
x
x
=
C
x
+
(
1
−
x
)
(
1
)
−
x
(
−
1
)
(
1
−
x
)
2
∫
f
(
x
)
d
x
x
=
C
x
+
1
(
1
−
x
)
2
∫
f
(
x
)
d
x
=
C
+
x
(
1
−
x
)
2
f
(
x
)
=
(
1
−
x
)
2
(
1
)
−
x
(
2
(
1
−
x
)
(
−
1
)
)
(
1
−
x
)
4
f
(
x
)
=
(
1
−
x
)
2
+
2
x
(
1
−
x
)
(
1
−
x
)
4
f
(
x
)
=
(
1
−
x
)
+
2
x
(
1
−
x
)
3
\begin{aligned} \quad f(x)&= \sum_{n=0}^{\infty} (n + 1)^2 x^n \\ \quad \int f(x) \: dx&= C + \sum_{n=0}^{\infty} (n+1)x^{n+1} \\ \quad \frac{\int f(x) \: dx}{x}&= \frac{C}{x} + \sum_{n=0}^{\infty} (n+1) x^n \\ \quad \int \frac{\int f(x) \: dx}{x} \: dx&= D + \int \frac{C}{x} \: dx + \sum_{n=0}^{\infty} x^{n+1} \\ \quad \int \frac{\int f(x) \: dx}{x} \: dx&= D + \int \frac{C}{x} \: dx + x \sum_{n=0}^{\infty} x^n \\ \quad \int \frac{\int f(x) \: dx}{x} \: dx&= D + \int \frac{C}{x} \: dx + \frac{x}{1 - x} \\ \quad \frac{\int f(x) \: dx}{x}&= \frac{C}{x} + \frac{(1 - x)(1) - x(-1)}{(1 - x)^2} \\ \quad \frac{\int f(x) \: dx}{x}&= \frac{C}{x} + \frac{1}{(1 - x)^2} \\ \quad \int f(x) \: dx&= C + \frac{x}{(1 - x)^2} \\ \quad f(x)&= \frac{(1 - x)^2(1) - x(2(1 - x)(-1))}{(1 - x)^4} \\ \quad f(x)&= \frac{(1 - x)^2 + 2x(1 - x)}{(1 - x)^4} \\ \quad f(x)&= \frac{(1 - x) + 2x}{(1 - x)^3} \end{aligned}
f(x)∫f(x)dxx∫f(x)dx∫x∫f(x)dxdx∫x∫f(x)dxdx∫x∫f(x)dxdxx∫f(x)dxx∫f(x)dx∫f(x)dxf(x)f(x)f(x)=n=0∑∞(n+1)2xn=C+n=0∑∞(n+1)xn+1=xC+n=0∑∞(n+1)xn=D+∫xCdx+n=0∑∞xn+1=D+∫xCdx+xn=0∑∞xn=D+∫xCdx+1−xx=xC+(1−x)2(1−x)(1)−x(−1)=xC+(1−x)21=C+(1−x)2x=(1−x)4(1−x)2(1)−x(2(1−x)(−1))=(1−x)4(1−x)2+2x(1−x)=(1−x)3(1−x)+2x 由于级数
f
(
x
)
=
∑
n
=
0
∞
(
n
+
1
)
2
x
n
f(x) = \sum_{n=0}^{\infty} (n + 1)^2 x^n
f(x)=∑n=0∞(n+1)2xn的收敛域也是
∣
x
∣
<
1
\mid x \mid < 1
∣x∣<1，因此可令
x
=
1
π
x=\frac{1}{\pi}
x=π1，于是有
f
(
1
π
)
=
∑
n
=
0
∞
(
n
+
1
)
2
π
n
=
(
1
−
1
π
)
+
2
π
(
1
−
1
π
)
3
\quad f \left ( \frac{1}{\pi} \right ) = \sum_{n=0}^{\infty} \frac{(n + 1)^2}{\pi^n} = \frac{\left (1 - \frac{1}{\pi} \right ) + \frac{2}{\pi}}{\left (1 - \frac{1}{\pi} \right )^3}
f(π1)=n=0∑∞πn(n+1)2=(1−π1)3(1−π1)+π2
五、用幂级数求数列通项公式
幂级数还可以用来求数列的通项公式，基本的思路为：
这种方法又称母函数法，通常见于组合数学。
例7. 求Fibonacci数列的通项公式：
先设：
s
(
x
)
=
∑
k
=
0
∞
F
k
x
k
s(x)=\sum_{k=0}^{\infty} F_{k} x^{k}
s(x)=k=0∑∞Fkxk
考虑Fibonacci的定义
F
n
=
F
n
−
1
+
F
n
−
2
F_n=F_{n-1}+F_{n-2}
Fn=Fn−1+Fn−2，有
s
(
x
)
=
∑
k
=
0
∞
F
k
x
k
=
F
0
+
F
1
x
+
∑
k
=
2
∞
(
F
k
−
1
+
F
k
−
2
)
x
k
=
x
+
∑
k
=
2
∞
F
k
−
1
x
k
−
∑
k
=
2
∞
F
k
−
2
x
k
=
x
+
x
∑
k
=
0
∞
F
k
x
k
+
x
2
∑
k
=
0
∞
F
k
x
k
=
x
+
x
s
(
x
)
+
x
2
s
(
x
)
\begin{aligned} s(x) &=\sum_{k=0}^{\infty} F_{k} x^{k} \\ &=F_{0}+F_{1} x+\sum_{k=2}^{\infty}\left(F_{k-1}+F_{k-2}\right) x^{k} \\ &=x+\sum_{k=2}^{\infty} F_{k-1} x^{k}-\sum_{k=2}^{\infty} F_{k-2} x^{k} \\ &=x+x \sum_{k=0}^{\infty} F_{k} x^{k}+x^{2} \sum_{k=0}^{\infty} F_{k} x^{k} \\ &=x+x s(x)+x^{2} s(x) \end{aligned}
s(x)=k=0∑∞Fkxk=F0+F1x+k=2∑∞(Fk−1+Fk−2)xk=x+k=2∑∞Fk−1xk−k=2∑∞Fk−2xk=x+xk=0∑∞Fkxk+x2k=0∑∞Fkxk=x+xs(x)+x2s(x) 解出
s
(
x
)
s(x)
s(x)
s
(
x
)
=
x
1
−
x
−
x
2
=
−
x
x
2
+
x
−
1
s(x)=\frac{x}{1-x-x^{2}}=\frac{-x}{x^{2}+x-1}
s(x)=1−x−x2x=x2+x−1−x 又：
x
1
−
x
−
x
2
=
x
(
1
−
φ
x
)
(
1
−
ψ
x
)
=
1
5
1
−
φ
x
+
−
1
5
1
−
ψ
x
=
1
5
(
1
1
−
φ
x
−
1
1
−
ψ
x
)
\frac{x}{1-x-x^{2}}=\frac{x}{(1-\varphi x)(1-\psi x)} =\frac{\frac{1}{\sqrt{5}}}{1-\varphi x}+\frac{-\frac{1}{\sqrt{5}}}{1-\psi x}=\frac{1}{\sqrt{5}}\left(\frac{1}{1-\varphi x}-\frac{1}{1-\psi x}\right)
1−x−x2x=(1−φx)(1−ψx)x=1−φx5
1+1−ψx−5
1=5
1(1−φx1−1−ψx1) 注意到
φ
,
ψ
\varphi,\psi
φ,ψ很好求，它们是分母的根：
φ
=
1
+
5
2
,
ψ
=
1
−
5
2
\varphi=\frac{1+\sqrt{5}}{2}, \quad \psi=\frac{1-\sqrt{5}}{2}
φ=21+5
,ψ=21−5
再代回上式得到：
s
(
x
)
=
1
5
(
1
1
−
φ
x
−
1
1
−
ψ
x
)
=
1
5
(
∑
n
=
0
∞
φ
n
x
n
−
∑
n
=
0
∞
ψ
n
x
n
)
=
1
5
(
∑
n
=
0
∞
(
φ
n
−
ψ
n
)
x
n
)
=
∑
k
=
0
∞
F
k
x
k
\begin{aligned} s(x) &=\frac{1}{\sqrt{5}}\left(\frac{1}{1-\varphi x}-\frac{1}{1-\psi x}\right) \\ &=\frac{1}{\sqrt{5}}\left(\sum_{n=0}^{\infty} \varphi^{n} x^{n}-\sum_{n=0}^{\infty} \psi^{n} x^{n}\right) \\ &=\frac{1}{\sqrt{5}}\left(\sum_{n=0}^{\infty}\left(\varphi^{n}-\psi^{n}\right) x^{n}\right)=\sum_{k=0}^{\infty} F_{k} x^{k} \end{aligned}
s(x)=5
1(1−φx1−1−ψx1)=5
1(n=0∑∞φnxn−n=0∑∞ψnxn)=5
1(n=0∑∞(φn−ψn)xn)=k=0∑∞Fkxk 因此：
F
n
=
φ
n
−
ψ
n
5
=
φ
n
−
(
−
φ
)
−
n
5
=
1
5
[
(
1
+
5
2
)
n
−
(
1
−
5
2
)
n
]
F_{n}=\frac{\varphi^{n}-\psi^{n}}{\sqrt{5}}=\frac{\varphi^{n}-(-\varphi)^{-n}}{\sqrt{5}}=\frac{1}{\sqrt{5}}\left[\left(\frac{1+\sqrt{5}}{2}\right)^n-\left(\frac{1-\sqrt{5}}{2}\right)^n\right]
Fn=5
φn−ψn=5
φn−(−φ)−n=5
1[(21+5
)n−(21−5
)n]
六、用幂级数解微分方程
例8. 考虑一个简单的方程：
y
′
′
+
y
=
0
y'' + y = 0
y′′+y=0 设它的解为一个幂级数：
y
(
x
)
=
∑
n
=
0
∞
a
n
x
n
y\left( x \right) = \sum\limits_{n = 0}^\infty {{a_n}{x^n}}
y(x)=n=0∑∞anxn 考虑：
y
′
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x
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=
∑
n
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∞
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a
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y
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a
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y'\left( x \right) = \sum\limits_{n = 1}^\infty {n{a_n}{x^{n - 1}}} \hspace{0.25in}y''\left( x \right) = \sum\limits_{n = 2}^\infty {n\left( {n - 1} \right){a_n}{x^{n - 2}}}
y′(x)=n=1∑∞nanxn−1y′′(x)=n=2∑∞n(n−1)anxn−2 带回原方程得到：
∑
n
=
2
∞
n
(
n
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1
)
a
n
x
n
−
2
+
∑
n
=
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∞
a
n
x
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=
0
\sum\limits_{n = 2}^\infty {n\left( {n - 1} \right){a_n}{x^{n - 2}}} + \sum\limits_{n = 0}^\infty {{a_n}{x^n}} = 0
n=2∑∞n(n−1)anxn−2+n=0∑∞anxn=0 对第一项的系数进行一下调整：
∑
n
=
0
∞
(
n
+
2
)
(
n
+
1
)
a
n
+
2
x
n
+
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=
0
∞
a
n
x
n
=
0
\sum\limits_{n = 0}^\infty {\left( {n + 2} \right)\left( {n + 1} \right){a_{n + 2}}{x^n}} + \sum\limits_{n = 0}^\infty {{a_n}{x^n}} = 0
n=0∑∞(n+2)(n+1)an+2xn+n=0∑∞anxn=0
实际上就是将
∑
n
=
2
∞
n
(
n
−
1
)
a
n
x
n
−
2
\sum\limits_{n = 2}^\infty {n\left( {n - 1} \right){a_n}{x^{n - 2}}}
n=2∑∞n(n−1)anxn−2 中的
n
n
n 换成
n
−
2
n-2
n−2.
整理得到：
∑
n
=
0
∞
[
(
n
+
2
)
(
n
+
1
)
a
n
+
2
+
a
n
]
x
n
=
0
\sum\limits_{n = 0}^\infty {\left[ {\left( {n + 2} \right)\left( {n + 1} \right){a_{n + 2}} + {a_n}} \right]{x^n}} = 0
n=0∑∞[(n+2)(n+1)an+2+an]xn=0 接下来就有意思了，由于上面的幂级数必须为零，它又是无穷级数，因此每项系数必为0：
(
n
+
2
)
(
n
+
1
)
a
n
+
2
+
a
n
=
0
,
n
=
0
,
1
,
2
,
…
\left( {n + 2} \right)\left( {n + 1} \right){a_{n + 2}} + {a_n} = 0,\hspace{0.25in}n = 0,1,2, \ldots
(n+2)(n+1)an+2+an=0,n=0,1,2,… 由上式可以总结出以下规律：
a
2
k
=
(
−
1
)
k
a
0
(
2
k
)
!
,
k
=
1
,
2
,
…
a
2
k
+
1
=
(
−
1
)
k
a
1
(
2
k
+
1
)
!
,
k
=
1
,
2
,
…
a_{2 k}=\frac{(-1)^{k} a_{0}}{(2 k) !}, \quad k=1,2, \ldots \quad a_{2 k+1}=\frac{(-1)^{k} a_{1}}{(2 k+1) !}, \quad k=1,2, \ldots
a2k=(2k)!(−1)ka0,k=1,2,…a2k+1=(2k+1)!(−1)ka1,k=1,2,… 接下来的事情虽然看似麻烦，但仍然清楚：
y
(
x
)
=
∑
n
=
0
∞
a
n
x
n
=
a
0
+
a
1
x
+
a
2
x
2
+
a
3
x
3
+
⋯
+
a
2
k
x
2
k
+
a
2
k
+
1
x
2
k
+
1
+
⋯
=
a
0
+
a
1
x
−
a
0
2
!
x
2
−
a
1
3
!
x
3
+
⋯
+
(
−
1
)
k
a
0
(
2
k
)
!
x
2
k
+
(
−
1
)
k
a
1
(
2
k
+
1
)
!
x
2
k
+
1
+
⋯
\begin{aligned}y\left( x \right) & = \sum\limits_{n = 0}^\infty {{a_n}{x^n}} \\ & = {a_0} + {a_1}x + {a_2}{x^2} + {a_3}{x^3} + \cdots + {a_{2k}}{x^{2k}} + {a_{2k + 1}}{x^{2k + 1}} + \cdots \\ & = {a_0} + {a_1}x - \frac{{{a_0}}}{{2!}}{x^2} - \frac{{{a_1}}}{{3!}}{x^3} + \cdots + \frac{{{{\left( { - 1} \right)}^k}{a_0}}}{{\left( {2k} \right)!}}{x^{2k}} + \frac{{{{\left( { - 1} \right)}^k}{a_1}}}{{\left( {2k + 1} \right)!}}{x^{2k + 1}} + \cdots \end{aligned}
y(x)=n=0∑∞anxn=a0+a1x+a2x2+a3x3+⋯+a2kx2k+a2k+1x2k+1+⋯=a0+a1x−2!a0x2−3!a1x3+⋯+(2k)!(−1)ka0x2k+(2k+1)!(−1)ka1x2k+1+⋯ 再重新整理：
y
(
x
)
=
a
0
{
1
−
x
2
2
!
⋯
+
(
−
1
)
k
x
2
k
(
2
k
)
!
+
⋯
}
+
a
1
{
x
−
x
3
3
!
+
⋯
+
(
−
1
)
k
(
2
k
+
1
)
!
x
2
k
+
1
+
⋯
}
=
a
0
∑
k
=
0
∞
(
−
1
)
k
x
2
k
(
2
k
)
!
+
a
1
∑
k
=
0
∞
(
−
1
)
k
x
2
k
+
1
(
2
k
+
1
)
!
\begin{aligned}y\left( x \right) & = {a_0}\left\{ {1 - \frac{{{x^2}}}{{2!}} \cdots + \frac{{{{\left( { - 1} \right)}^k}{x^{2k}}}}{{\left( {2k} \right)!}} + \cdots } \right\} + {a_1}\left\{ {x - \frac{{{x^3}}}{{3!}} + \cdots + \frac{{{{\left( { - 1} \right)}^k}}}{{\left( {2k + 1} \right)!}}{x^{2k + 1}} + \cdots } \right\}\\ & = {a_0}\sum\limits_{k = 0}^\infty {\frac{{{{\left( { - 1} \right)}^k}{x^{2k}}}}{{\left( {2k} \right)!}}} + {a_1}\sum\limits_{k = 0}^\infty {\frac{{{{\left( { - 1} \right)}^k}{x^{2k + 1}}}}{{\left( {2k + 1} \right)!}}} \end{aligned}
y(x)=a0{1−2!x2⋯+(2k)!(−1)kx2k+⋯}+a1{x−3!x3+⋯+(2k+1)!(−1)kx2k+1+⋯}=a0k=0∑∞(2k)!(−1)kx2k+a1k=0∑∞(2k+1)!(−1)kx2k+1 注意到：
cos
(
x
)
=
∑
n
=
0
∞
(
−
1
)
n
x
2
n
(
2
n
)
!
sin
(
x
)
=
∑
n
=
0
∞
(
−
1
)
n
x
2
n
+
1
(
2
n
+
1
)
!
\cos \left( x \right) = \sum\limits_{n = 0}^\infty {\frac{{{{\left( { - 1} \right)}^n}{x^{2n}}}}{{\left( {2n} \right)!}}} \hspace{0.25in}\sin \left( x \right) = \sum\limits_{n = 0}^\infty {\frac{{{{\left( { - 1} \right)}^n}{x^{2n + 1}}}}{{\left( {2n + 1} \right)!}}}
cos(x)=n=0∑∞(2n)!(−1)nx2nsin(x)=n=0∑∞(2n+1)!(−1)nx2n+1 因此：
y
(
x
)
=
c
1
cos
(
x
)
+
c
2
sin
(
x
)
y\left( x \right) = {c_1}\cos \left( x \right) + {c_2}\sin \left( x \right)
y(x)=c1cos(x)+c2sin(x)
这个例子只是一个比较简单的问题。当然完全可以直接用简单的方法求解。但这种方法在求解其它更为复杂的非线性方程的时候十分有用，因为对于许多方程它并不一定有初等表达式。但如果其幂级数解存在的话，那么就可以对它进行近似计算。这种数值计算方法通常又比普通的差分法要精确得多，目前也是计算数学界较为主流的一种方法。
参考资料：
http://people.math.sc.edu/girardi/m142/handouts/10sTaylorPolySeries.pdf
https://math.berkeley.edu/~neu/undergrad_chap1.pdf
https://www.math.cuhk.edu.hk/course_builder/1516/math1010c/Power_series.pdf
https://web.ma.utexas.edu/users/m408s/CurrentWeb/LM14-3-10.php
[http://math.caltech.edu/syye/teaching/courses/Ma8_2015/Lecture%20Notes/ma8_wk10.pdf](http://math.caltech.edu/syye/teaching/courses/Ma8_2015/Lecture Notes/ma8_wk10.pdf)
https://tutorial.math.lamar.edu/classes/de/seriessolutions.aspx