|
|
利用解析数论的方法可以得到非常好的估计
我们来计算
$$S = \log P =\sum_{n\ge 1} \log\left(1-\frac{1}{3^n}\right).$$
定义
$$S(x) = \sum_{n\ge 1} \log\left(1-\frac{1}{3^{xn}}\right).$$
我们现在来计算 Mellin transform.
我们想起定义harmonic sum identity
$$\left(\sum_{k\ge 1} \lambda_k g(\mu_k x);s\right) =
\left(\sum_{k\ge 1} \frac{\lambda_k}{\mu_k^s} \right) g^*(s)$$
其中$g^*(s)$ 是 $g(x).$ 的 Mellin transform
则我们有
$$\lambda_k = 1, \quad \mu_k = k \quad \text{and} \quad
g(x) = \log\left(1-\frac{1}{3^x}\right).$$
显然我们有
$$\int_0^\infty \log\left(1-\frac{1}{3^x}\right) x^{s-1} dx
= \left[\log\left(1-\frac{1}{3^x}\right) \times \frac{x^s}{s} \right]_0^\infty
- \int_0^\infty \frac{\log 3}{3^x-1} \frac{x^s}{s} dx .$$
注意第一个式子有$\Re(s)\ge 1.$ 则有
$$\int_0^\infty \frac{1}{3^x-1} x^s dx
= \int_0^\infty \frac{3^{-x}}{1-3^{-x}} x^s dx
= \int_0^\infty 3^{-x}\sum_{q\ge 0} 3^{-qx} x^s dx
= \sum_{q\ge 0} \int_0^\infty 3^{-(q+1)x} x^s dx.$$
发现
$$\sum_{q\ge 0} \int_0^\infty e^{-(q+1)\log 3 \times x} x^s dx
= \sum_{q\ge 0} \frac{1}{(\log 3\times (q+1))^{s+1}}\Gamma(s+1)
= \frac{\Gamma(s+1)\zeta(s+1)}{(\log 3)^{s+1}}.$$
得到
$$g^*(s) = - \frac{\log 3}{s} \frac{\Gamma(s+1)\zeta(s+1)}{(\log 3)^{s+1}}
= - \frac{1}{s} \frac{\Gamma(s+1)\zeta(s+1)}{(\log 3)^s}.$$
对于 $Q(s)$我们利用harmonic sum 有
$$Q(s) = - \frac{1}{s (\log 3)^s} \Gamma(s+1)\zeta(s)\zeta(s+1).$$
我们下面经过简单计算有
\begin{align*}
\mathrm{Res}(Q(s)/x^s; s=1) &= -\frac{1}{6} \frac{\pi^2}{\log 3} \frac{1}{x},\\
\mathrm{Res}(Q(s)/x^s; s=0) &= \frac{1}{2} \log \frac{2\pi}{\log 3} - \frac{1}{2}\log x,
\quad\text{and}\\
\mathrm{Res}(Q(s)/x^s; s=-1) &= \frac{1}{24} x \log 3.
\end{align*}
令$x=1$ ,我们可以得到 $S(1)$ 的渐进估计,则有
$$S(1) = S \approx
-\frac{1}{6} \frac{\pi^2}{\log 3} + \frac{1}{2} \log \frac{2\pi}{\log 3}
+ \frac{1}{24} \log 3\\
\approx -0.57959338143590694$$
那么我们有
$$ P \approx \exp\left(-\frac{1}{6} \frac{\pi^2}{\log 3}\right) \times
\sqrt{\frac{2\pi}{\log 3} } \times 3^{1/24}
\approx 0.56012607792794900$$ |
|
isee
发表于 2013-10-27 22:39
kuing
发表于 2013-10-30 21:46
