# 魏尔施特拉斯分解定理

${\displaystyle f(z)=}$${\displaystyle z^{m}e^{g(z)}}$${\displaystyle \prod _{n=1}^{\infty }}$${\displaystyle (1-{\tfrac {z}{a_{n}}})}$${\displaystyle e^{{\tfrac {z}{a_{n}}}+{\tfrac {1}{2}}({\tfrac {z}{a_{n}}})^{2}+{\tfrac {1}{3}}({\tfrac {z}{a_{n}}})^{3}+\cdots +{\tfrac {1}{h}}({\tfrac {z}{a_{n}}})^{h}}}$${\displaystyle =z^{m}e^{g(z)}\prod _{n=1}^{\infty }E_{p_{n}}\left({\frac {z}{a_{n}}}\right)}$

## 基本因子

{\displaystyle {\begin{aligned}&E_{n}(z)=(1-z)\exp \left(h_{n}(z)\right)\\&h_{n}(z)={\begin{cases}0&{\text{ if }}n=0,\\{\frac {z^{1}}{1}}+{\frac {z^{2}}{2}}+\cdots +{\frac {z^{n}}{n}}&{\text{ otherwise. }}\end{cases}}\end{aligned}}}

1. ${\displaystyle |z|<1}$ 的情况下，${\displaystyle h_{n}(z)}$ 可被展开${\displaystyle {\frac {1}{1-z}}=1+z^{1}+z^{2}+z^{3}+\cdots }$ 。接着两边同时积分，可得{\displaystyle {\begin{aligned}\int {\frac {1}{1-z}}dz=-\log(1-z)={\frac {z^{1}}{1}}+{\frac {z^{2}}{2}}+{\frac {z^{3}}{3}}+\cdots \end{aligned}}} 。所以${\displaystyle h_{n}(z)}$ 的极限可以表示为${\displaystyle h_{\infty }(z)=\lim _{n\to \infty }h_{n}(z)=-\log(1-z)}$
2. 因为${\displaystyle (1-z)=\exp(\log(1-z))=\exp \left(-h_{\infty }(z)\right)}$ ，所以${\displaystyle {\frac {1}{1-z}}=\exp \left(h_{\infty }(z)\right)}$
3. 如果将${\displaystyle h_{\infty }(z)}$ ${\displaystyle h_{n}(z)}$ 之间的差额定义为新的级数${\displaystyle r_{n}(z)=h_{\infty }(z)-h_{n}(z)}$
4. 利用2.与3.改写${\displaystyle E_{n}(z)}$ 的定义式：{\displaystyle {\begin{aligned}&E_{n}(z)=(1-z)\exp \left(h_{n}(z)\right)=(1-z)\exp \left(h_{\infty }(z)-r_{n}(z)\right)\\&\quad =(1-z){\frac {1}{1-z}}\exp \left(-r_{n}(z)\right)=\exp \left(-r_{n}(z)\right)\end{aligned}}} 。改写后的基本因子定义式${\displaystyle E_{n}(z)=\exp \left(-r_{n}(z)\right)}$ 将会在后续引理的证明中用到。
5. 将3.的关系写成级数形式：${\displaystyle r_{n}(z)={\frac {z^{n+1}}{n+1}}+{\frac {z^{n+2}}{n+2}}+{\frac {z^{n+3}}{n+3}}+\cdots =\sum _{k=n+1}^{\infty }{\frac {z^{n+1}}{n+1}}={\frac {z^{n+1}}{n+1}}\sum _{k=0}^{\infty }{\frac {n+1}{n+1+k}}z^{k}}$

{\displaystyle {\begin{aligned}\left|1-E_{n}(z)\right|\leq |z|^{n+1}\end{aligned}}} 成立。

${\displaystyle n=0}$ 时，${\displaystyle |1-z|\leq |z|}$ 显而易见。所以只讨论${\displaystyle n\geq 1}$ 的情况。

i) 将引理左边的部分（不带绝对值）定义为一个新函数${\displaystyle u_{n}(z)=1-E_{n}(z)=1-(1-z)\exp h_{n}(z)}$ 。后续称此式为式${\displaystyle (1)}$

{\displaystyle {\begin{aligned}u_{n}(z)=1-E_{n}(z)=1-\exp \left(-r_{n}(z)\right)=1-\exp \left(-{\frac {z^{n+1}}{n+1}}\sum _{k=0}^{\infty }{\frac {n+1}{n+1+k}}z^{k}\right)=1-\exp(-{\frac {z^{n+1}}{n+1}})\exp {(-{\frac {z^{n+2}}{n+1}}{\frac {n+1}{n+2}})}\exp {(-{\frac {z^{n+3}}{n+1}}{\frac {n+1}{n+3}})}...\end{aligned}}}

${\displaystyle u'_{n}(z)=a_{0}(n+1)z^{n}+a_{1}(n+2)z^{n+1}+a_{2}(n+3)z^{n+2}+...}$

ii) 将式${\displaystyle (1)}$ 直接微分，可得

{\displaystyle {\begin{aligned}&u_{n}^{\prime }(z)=-E_{n}^{\prime }(z)=\exp h_{n}(z)-(1-z)h_{n}^{\prime }(z)\exp h_{n}(z)\\&\quad =\exp h_{n}(z)-(1-z){\frac {1-zn}{1-z}}\exp h_{n}(z)=z^{n}\exp h_{n}(z)\end{aligned}}}

${\displaystyle u_{n}^{\prime }(z)=z^{n}\exp h_{n}(z)=z^{n}\exp({\frac {z^{1}}{1}}+{\frac {z^{2}}{2}}+\cdots +{\frac {z^{n}}{n}})=z^{n}\exp(z)\exp({\frac {z^{2}}{2}})\exp({\frac {z^{3}}{3}})...\exp({\frac {z^{n}}{n}})=z^{n}(\sum _{k=0}^{\infty }{\frac {x^{k}}{k!}})(\sum _{l=0}^{\infty }{\frac {x^{2l}}{2^{l}l!}})...(\sum _{j=0}^{\infty }{\frac {x^{nj}}{n^{j}j!}})}$

iii) 基于${\displaystyle u_{n}}$ 新设一个级数${\displaystyle v_{n}(z)={\frac {u_{n}(z)}{z^{n+1}}}=\sum _{k=0}^{\infty }a_{k}z^{k}}$ 。因为极点是一个可消极点，所以这也是一个整函数。计算${\displaystyle v_{n}(1)={\frac {u_{n}(1)}{1^{n+1}}}=1-E_{n}(1)=1-[(1-1)\exp \left(h_{n}(1)\right)]=1}$

${\displaystyle \left|v_{n}(z)\right|\leq \sum _{k=0}^{\infty }\left|a_{k}\|z\right|^{k}\leq \sum _{k=0}^{\infty }a_{k}=v_{n}(1)=1{\text{ if }}|z|\leq 1}$

• Alford的《复分析》

## 参考资料

1. ^ Boas, R. P., Entire Functions, New York: Academic Press Inc., 1954, ISBN 0-8218-4505-5, OCLC 6487790, chapter 2.
2. Rudin, W., Real and Complex Analysis 3rd, Boston: McGraw Hill: 301–304, 1987, ISBN 0-07-054234-1, OCLC 13093736.