例 7.30求下列函数的幂级数展开式 $$(a) \frac{1}{\sqrt{(1-x^2)}} \quad (b) \frac{1}{(1-x)(1+3x)} \quad (c) \frac{\ln(1+x)}{1+x}$$

解 (a) 使用二项式级数 (7.16)，取 $n = -\frac{1}{2}$ 得 $$\begin{aligned} \frac{1}{\sqrt{(1+x)}} &= (1+x)^{-1/2} \ &= 1 + \frac{(-\frac{1}{2})}{1!}x + \frac{(-\frac{1}{2})(-\frac{3}{2})}{2!}x^2 + \frac{(-\frac{1}{2})(-\frac{3}{2})(-\frac{5}{2})}{3!}x^3 + \dots \quad (-1 < x < 1) \end{aligned}$$

现在将 $x$ 替换为 $-x^2$ 即得所需结果 $$\begin{aligned} \frac{1}{\sqrt{(1-x^2)}} &= 1 + \frac{(\frac{1}{2})}{1!}x^2 + \frac{(\frac{1}{2})(\frac{3}{2})}{2!}x^4 + \frac{(\frac{1}{2})(\frac{3}{2})(\frac{5}{2})}{3!}x^6 + \dots \quad (-1 < x < 1) \ &= 1 + \frac{1}{2}x^2 + \frac{1 \cdot 3}{2 \cdot 4}x^4 + \frac{1 \cdot 3 \cdot 5}{2 \cdot 4 \cdot 6}x^6 + \dots \quad (-1 < x < 1) \end{aligned}$$

(b) 表示为部分分式 $$\frac{1}{(1-x)(1+3x)} = \frac{\frac{1}{4}}{1-x} + \frac{\frac{3}{4}}{1+3x}$$

由图 7.13 的表 $$\frac{1}{1-x} = 1 + x + x^2 + x^3 + \dots + x^n + \dots \quad (-1 < x < 1)$$

并将 $x$ 替换为 $3x$ 代入 (7.15) 得 $$\frac{1}{1+3x} = 1 - (3x) + (3x)^2 - (3x)^3 + \dots + (-1)^n (3x)^n + \dots \quad (-\frac{1}{3} < x < \frac{1}{3})$$

因此 $$\begin{aligned} \frac{1}{(1-x)(1+3x)} &= \frac{1}{(1-x)(1+3x)} \ &= \frac{1}{4}[1+x+x^2+x^3+\dots] + \frac{3}{4}[1-3x+9x^2-27x^3+\dots] \quad (-\frac{1}{3} < x < \frac{1}{3}) \ &= 1 - 2x + 7x^2 - 20x^3 + \dots + \frac{1}{4}(1 + (-1)^n 3^{n+1})x^n + \dots \quad (-\frac{1}{3} < x < \frac{1}{3}) \end{aligned}$$

(c) 利用级数 $\ln(1+x)$ 以及 $(1+x)^{-1}$ 由 (7.18) 和 (7.15)， $$\begin{aligned} \frac{\ln(1+x)}{1+x} &= (x - \frac{1}{2}x^2 + \frac{1}{3}x^3 - \frac{1}{4}x^4 + \dots)(1-x+x^2-x^3+\dots) \ &= x - (1 + \frac{1}{2})x^2 + (1 + \frac{1}{2} + \frac{1}{3})x^3 - (1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4})x^4 + \dots \ &\quad (-1 < x < 1) \end{aligned}$$