问题:
证明 $$ \zeta(z) =\prod_p \frac{1}{1-p^{-z}}$$
解答: (
ID:
匿名游客[匿名]
)
$$
\zeta(z)=1+\frac{1}{2^z}+\frac{1}{3^z}+\frac{1}{4^z}-\frac{1}{5^z}+\ldots \ \ \ \star
$$
(\(\star \)) \(\times \frac{1}{2^z}\) 得
$$
\frac{1}{2^z} \zeta(z)=\frac{1}{2^z}+\frac{1}{4^z}+\frac{1}{6^z}+\frac{1}{8^z}+\frac{1}{10^z}+\ldots \ \ \ \star \star
$$
(\(\star \)) -(\(\star
\star \)) 得
$$
\left(1-\frac{1}{2^z}\right) \zeta(z)=1+\frac{1}{3^z}-\frac{1}{5^z}+\frac{1}{7^z}+\frac{1}{9^z}+\frac{1}{11^z}+\frac{1}{13^z}+\ldots \ \ \
\star\star\star
$$
(\(\star \star \star \)) \(\times \frac{1}{3^z}\) 得
$$
\frac{1}{3^z}\left(1-\frac{1}{2^z}\right) \zeta(z)=\frac{1}{3^z}+\frac{1}{9^z}+\frac{1}{15^z}+\frac{1}{21^z}+\frac{1}{27^z}+\frac{1}{33^z}+\ldots \ \ \ \ (4)
$$
(\(\star\star\star
\)) - (4)得
$$
\left(1-\frac{1}{3^z}\right)\left(1-\frac{1}{2^z}\right) \zeta(z)=1-\frac{1}{5^z}-\frac{1}{7^z}+\frac{1}{11^z}+\frac{1}{13^z}+\frac{1}{17^z}+\ldots
$$
不断重复最终可以得到
$$
\ldots\left(1-\frac{1}{11^z}\right)\left(1-\frac{1}{7^z}\right)\left(1-\frac{1}{5^z}\right)\left(1-\frac{1}{3^z}\right)\left(1-\frac{1}{2^z}\right) \zeta(z)=1
$$
则
\[
\zeta(z)=\frac{1}{\ldots\left(1-\frac{1}{11^z}\right)\left(1-\frac{1}{7^z}\right)\left(1-\frac{1}{5^z}\right)\left(1-\frac{1}{3^z}\right)\left(1-\frac{1}{2^z}\right)}
\]
\[
=\left(\frac{1}{1-2^{-z}}\right)\left(\frac{1}{1-3^{-z}}\right)\left(\frac{1}{1-5^{-z}}\right)\left(\frac{1}{1-7^{-z}}\right)\left(\frac{1}{1-11^{-z}}\right) \cdots
\]
因此
\[
\zeta (z) =\prod_p \frac{1}{1-p^{-z}}
\]
