把1/z代入在0的级数

hbghlyj

H. A. Priestley, Complex Analysis, second edition(2003)
202页:
因为$$\sin z=\sum_{n=0}^{\infty}(-1)^n \frac{z^{2 n+1}}{(2 n+1) !}$$
把$1/z$代入得, 对于$0<|z|<\infty$,
$$
\sin \left(\frac{1}{z}\right)=\sum_{n=0}^{\infty}(-1)^n \frac{z^{-(2 n+1)}}{(2 n+1) !} .
$$
所以0是$\sin (1 / z)$的essential singularity, 上式既为在0的Laurent级数,又为在∞的Laurent级数.

1. Series[Sin[1/z], {z, Infinity, 3}]

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$\frac{1}{z}-\frac{1}{6 z^3}+O\left(\left(\frac{1}{z}\right)^4\right)$
像上面这样“把1/z代入在0的级数”需要函数在∞解析. sin是整函数所以在∞解析. 请看199页:
Beware bogus expansions! Replacing $z$ by $1 / z$ in the expansion for $\operatorname{cosec} z$ above appears to give
$$
\operatorname{cosec}(1 / z)=z\left(1+\frac{1}{3 ! z^2}+\mathbf{O}\left(z^{-4}\right)\right) .
$$
This is not valid. The substitution of $1 / z$ for $z$ overlooks the restriction to small $|z|$ imposed above to validate inverting the sine expansion. In fact there is no Laurent expansion about 0 , since there is no punctured disc $\mathrm{D}^{\prime}(0 ; \varepsilon)$ in which $\operatorname{cosec}(1 / z)$ is holomorphic. (In the terminology of 17.8 below, the singularity at 0 is not isolated.)
因为$\operatorname{cosec}(z)$在0的级数的收敛半径<∞, 所以$\operatorname{cosec}(z)$在∞不解析, 所以不能将$1/z$代入级数.
事实上$\operatorname{cosec}(z)$在0的级数的收敛半径是$\pi$, 也就是计算下面的极限, 见ProofWiki$\newcommand{\paren}[1]{\left( #1 \right)}$
$$\lim_{n \to \infty} {\frac {\frac {B_{2 n + 2} \paren {-1}^n x^{2 n + 1} 2 \paren {2^{2 n + 1} - 1} } {\paren {2 n + 2}!} } {\frac {B_{2 n} \paren {-1}^{n - 1} x^{2 n - 1} 2 \paren {2^{2 n - 1} - 1} } {\paren {2 n}!} } }=\frac 1 {\pi^2} x^2$$

hbghlyj

同样地, $\arcsin(z)$在0的级数\[\arcsin(z)=\sum_{k=0}^\infty\binom{-1/2}{k}(-1)^k\frac{z^{2k+1}}{2k+1}\]的收敛半径为1.(代入系数直接计算, 或者利用这帖计算的arcsin的分岐点)
$\arcsin(z)$在∞不解析, 所以不能代入$1/z$.

hbghlyj

事实上$\operatorname{cosec}(z)$在0的级数的收敛半径是$\pi$, 也就是计算下面的极限, 见ProofWiki

无需计算那个系数之比的极限. 因为cosec的奇点是$\{z:\sin(z)=0\}=\{kπ:k∈\Bbb Z\}$.
在$\{kπ:k∈\Bbb Z\}∖\{0\}$中, 距0最近是$±π$, 根据下面的定理(见MSE)在0的级数的收敛半径$≥π$, 另一方面$π$是最大的[如果半径大于$π$就把奇点包括进去了], 因此cosec在0的级数的收敛半径是$π$. 同样地, cot在0的级数的收敛半径是$π$.[同样在ProofWiki上是直接计算极限的]

If $f$ is analytic on an open disk $B(0, R) \subseteq \mathbb C$ where $R>0$ then the radius of convergence of its Taylor series is $\ge R$.

用这种计算收敛半径的方法可以估计$n$阶导数当$n$增大时的减小速度, 见$-\log(\log(1-x))$在$1-e$的各阶导数的渐近估计