常用幂级数展开

熟练掌握常见函数的幂级数展开以及推导过程是解决幂级数展开问题的关键，因此首先回顾一下常见函数的幂级数展开以及推导过程

指数与对数¶

• \(\displaystyle e^x = \sum\limits_{n = 0}^{\infty} \frac{x^n}{n!}\)

证明：用泰勒展开 \(f(x) = f(0) + f^{\prime}(0)x + f^{\prime\prime}(0)\frac{x^2}{2} + \cdots\) 即可，显然\(f^{(n)}(0) = 1\)。

• \(\displaystyle\ln(1+x) = \sum\limits_{n = 1}^{\infty} (-1)^{n-1} \frac{x^n}{n}\)

证明：同理也是直接使用泰勒展开，\(f(0) = 1, f^{\prime}(0) = \frac{1}{1+0} = 1, f^{\prime\prime}(0) = -\frac{1}{(1+0)^2} = -1 \cdots\)，同理可推得 \(\left[ \ln(1+x) \right]^{(n)}(0) = (-1)^{n-1}\)

三角函数¶

• \(\displaystyle\sin x = \sum\limits_{n = 0}^{\infty}(-1)^n \frac{x^{2n+1}}{(2n+1)!}\)

证明：由于 \((\sin x)^{(n)} = \sin \left( x + \frac{n\pi}{2} \right)\)，则

\[ a_n = \frac{f^{(n)}(0)}{n!} = \begin{cases} \frac{(-1)^{m}}{(2m+1)!}, & n = 2m+1\\ 0, & n = 2m \end{cases}\]

• \(\displaystyle\cos x = \sum\limits_{n = 0}^{\infty} (-1)^n \frac{x^{2n}}{(2n)!}\)

证明：同理根据 \((\cos x)^{(n)} = \cos \left( x + \frac{n\pi}{2} \right)\) 即可，或者用 \(\sin x = \sum\limits_{n = 0}^{\infty} (-1)^n \frac{x^{2n+1}}{(2n+1)!}\) 逐项求导。

反三角函数¶

• \(\displaystyle\arctan x = \sum\limits_{n = 0}^{\infty} \frac{(-1)^n x^{2n+1}}{2n+1}\)

证明：根据 常用导数公式可知 \((\arctan x)^{\prime} = \frac{1}{1+x^2}\)(具体原因见 反函数求导)，再使用 \(\frac{1}{1+x}\) 的幂级数展开可得到

\[ \frac{1}{1+x^2} = 1 - x^2 + x^4 + \cdots + (-1)^n x^{2n} + \cdots = \sum\limits_{n = 0}^{\infty} (-1)^n x^{2n} \]

使用逐项积分得到

\[ \arctan x = \int_0^x \frac{\mathrm{d} t}{1 + t^2} = \sum\limits_{n = 0}^{\infty} (-1)^n \frac{x^{2n+1}}{2n+1} \]

• \(\displaystyle\arcsin x = \sum\limits_{n = 0}^{\infty} \left( \frac{(2n)!}{2^{2n}(n!)^2} \right)\frac{x^{2n+1}}{2n+1} = \sum\limits_{n = 0}^{\infty} \frac{(2n-1)!!}{(2n)!!} \frac{x^{2n+1}}{2n+1}\)

证明：同理使用 \((\arcsin x)^{\prime} = \frac{1}{\sqrt{1+x^2}}\)，用 \((1+x)^{\alpha}\) 的幂级数展开即可。

1+x幂次¶

• \(\displaystyle\frac{1}{1+x} = \sum\limits_{n = 0}^{\infty} (-1)^nx^n\)
• \(\displaystyle\frac{1}{1-x} = \sum\limits_{n = 0}^{\infty}x^n\)
• \(\displaystyle(1+x)^{\alpha} = \sum\limits_{n = 0}^{\infty} {\alpha \choose n}x^n\)