スターリングの近似(Stirling's approximation)
$$\begin{array}{rcl}
\displaystyle n! & \simeq & \sqrt{2\pi n}\left(\dfrac{n}{e}\right)^n\\
\end{array}$$
階乗の対数
$$\begin{array}{rcl}
\displaystyle \log(n!)&\simeq& \log\left( \sqrt{2\pi n}\left(\dfrac{n}{e}\right)^n \right)\\
&\simeq& \log\left( \sqrt{2\pi n}\,n^n\,e^{-n} \right)\\
\displaystyle &\simeq& \log\left( \sqrt{2\pi} \right)
\displaystyle + \log\left( \sqrt{n} \right)
\displaystyle + \log\left(n^n \right)
\displaystyle + \log\left(e^{-n} \right)\\
\displaystyle &\simeq& \log\left( \sqrt{2\pi} \right)
\displaystyle + \frac{1}{2}\log\left( n \right)
\displaystyle + n\log\left(n \right)
\displaystyle - n \log\left(e \right)\\
\displaystyle &\simeq& \log\left( \sqrt{2\pi} \right)
\displaystyle + \frac{1}{2}\log\left( n \right)
\displaystyle + n\log\left(n \right)
\displaystyle - n\\
\end{array}$$
階乗の対数の微分
$$\begin{array}{rcl}
\displaystyle \left(\log(n!)\right)'
\displaystyle &\simeq& \left(\log\left( \sqrt{2\pi} \right)\right)'
\displaystyle + \left(\frac{1}{2}\log\left( n \right)\right)'
\displaystyle + \left(n\log\left(n \right)\right)'
\displaystyle - \left(n\right)'\\
\displaystyle &\simeq& \left(\log\left( \sqrt{2\pi} \right)\right)'
\displaystyle + \left(\frac{1}{2}\log\left( n \right)\right)'
\displaystyle + \left(n(\log\left(n \right))'+(n)'\log\left(n \right)\right)
\displaystyle - \left(n\right)'\\
\displaystyle &\simeq& 0
\displaystyle + \frac{1}{2}\left(\frac{1}{n}\right)
\displaystyle + \left(n\left(\frac{1}{n}\right)+1\,\log\left(n \right)\right)
\displaystyle - 1\\
\displaystyle &\simeq& \frac{1}{2n}+1+ \log\left(n \right)-1\\
\displaystyle &\simeq& \frac{1}{2n}+\log\left(n \right)
\end{array}$$
\(n\)が十分に大きい場合
$$\begin{array}{rcl}
\displaystyle \left(\log(n!)\right)'
\displaystyle &\simeq& \frac{1}{2n}+\log\left(n \right)\\
\displaystyle &\simeq& 0+\log\left(n \right)\,\dotso\,nが十分い大きい \frac{1}{2n} \to 0\\
\displaystyle &\simeq& \log\left(n \right)
\end{array}$$
0 件のコメント:
コメントを投稿