$∏(1-2^{-n})$

hbghlyj

OEIS: This is the limiting probability that a large random binary matrix is nonsingular
相关帖子(1 − 3⁻¹)(1 − 3⁻²)⋯(1 − 3⁻ⁿ)

Czhang271828

先算出
\[
\left(\frac{1}{2}; \frac{1}{2}\right)_\infty=\frac{2^{1/24}}{\sqrt{3}}\vartheta_2\left(\frac{1}{6}\pi,\frac{1}{2^{1/6}}\right)=0.28878809508660\ldots
\]
查一下数字, 发现文章. 看来是用 Mellin 变换做的.


直接做: $\ln(1-x)+x+x^2$ 在 $(0,\frac12)$ 单调递增, 从而
\[
(1-x)\geq e^{-x-x^2}.
\]
原式 $\geq$
\begin{align*}
\prod_{k\geq 1}e^{-2^{-k}-4^{-k}}=e^{-4/3}>\frac14.
\end{align*}
最后是因为 $8=(2\sqrt 2)^2>(2\cdot 1.414)^2>e^2$.

hbghlyj

等价于$\Bbb F_2$ n阶矩阵可逆的概率大于 25%

Czhang271828

补一个最快的方法: 注意到展开
\[
\prod_{n\geq 1}(1-q^n)=\sum_{d\in \mathbb Z}(-1)^dq^{(3d^2+d)/2}
\]
从而
\[
\prod_{n\geq 1}(1-2^{-n})\geq1-\dfrac12-\dfrac14+\dfrac1{32}-\sum_{k\geq 7}2^{-k}>\dfrac 14.
\]

江西卷的题目怀疑就是这样凑的:
\begin{align*}
\prod_{n\geq 1}(1-3^{-n})&\geq 1-\dfrac13-\dfrac19+\dfrac1{3^5}+\dfrac{1}{3^7}-\sum_{k\geq 12}\dfrac{1}{3^k}\\[8pt]
&>1-\dfrac{1}{3}-\dfrac{1}{9}+\dfrac{1}{3^5}+\dfrac{2}{3^8}+\dfrac1{3^9}\\[8pt]
&=\dfrac59+\dfrac{1}{3^5}+\dfrac{2}{3^8}+\dfrac1{3^9}\\[8pt]
&=\left(\dfrac{14}{25}-\dfrac{1}{9\cdot 25}\right)+\dfrac{1}{9\cdot 27}+\dfrac{2}{9\cdot 27^2}+\dfrac1{3^9}\\[8pt]
&=\dfrac{14}{25}-\dfrac{2}{9\cdot 25\cdot 27}+\dfrac{2}{9\cdot 27^2}+\dfrac1{3^9}\\[8pt]
&=\dfrac{14}{25}-\dfrac{4}{9\cdot 25\cdot 27^2}+\dfrac4{4\cdot27\cdot 27^2}\\[8pt]
&>\dfrac{14}{25}.
\end{align*}