Ngô Quốc Anh

September 8, 2007

Một áp dụng của biểu diễn Euler cho hàm Gamma

Filed under: Các Bài Tập Nhỏ — Ngô Quốc Anh @ 19:12

Biểu diễn Euler cho hàm Gamma được phát biểu như sau

\Gamma\, :\quad\Gamma(1+z)=\prod_{n=1}^{\infty}\frac{\left(1+\frac{1}{n}\right)^{z}}{1+\frac{z}{n}}

 Đề bài. Chứng minh rằng

\prod_{n=1}^{\infty}\;\;\frac{1+n^{6}}{n^{6}}\;\;=\;\;\boxed{\frac{\sinh\,\pi\cdot\left(\cosh\,\pi\;-\;\cos\,(\sqrt 3\,\pi)\right)}{2\,\pi^{3}}}

Lời giải. Ký hiệu \xi là e^{\frac{2\pi i}{m}}. Khi đó ta có

\prod_{k=0}^{m-1}\Gamma(1-\xi^{k}z)=\prod_{n=1}^{\infty}\frac{\left(1+\frac{1}{n}\right)^{\overbrace{(-\xi^{0}-...-\xi^{m-1})}^{=0}z}}{\left(1-\frac{\xi^{0}z}{n}\right)\cdots\left(1-\frac{\xi^{m-1}z}{n}\right) }=\prod_{n=1}^{\infty}\left[1-\left(\frac{z}{n}\right)^{m}\,\right]^{-1}

 Nói riêng với m=6 và

z=e^{\frac{i\pi}{6}}\quad\prod_{n=1}^{\infty}\left(1+\frac{1}{n^{6}}\right)^{-1}=\prod_{k=0}^{5}\Gamma\left(1-e^{(2k+1)\frac{i\pi}{6}}\right) 

   =\Gamma(1-i)\,\Gamma(1+i)\;\;\,\Gamma(1-e^{\frac{i\pi}{6}})\,\Gamma(1+e^{\frac{i\pi}{6}})\;\;\,\Gamma(1-e^{\frac{5i\pi}{6}})\,\Gamma(1+e^{\frac{5i\pi}{6}})

   =\frac{i\pi}{\sin i\pi}\;\;\frac{e^{i\frac{\pi}{6}}\pi}{\sin\left(e^{i\frac{\pi}{6}}\pi\right)}\;\;\frac{e^{i\frac{5\pi}{6}}\pi}{\sin\left(e^{i\frac{5\pi}{6}}\pi\right)}=\frac{\pi^{3}}{\sinh\pi}\;\,\frac{-1}{\sin\left(e^{i\frac{\pi}{6}}\pi\right)\,\sin\left(e^{i\frac{5\pi}{6}}\pi\right) }

  =\frac{\pi^{3}}{\sinh\pi}\;\;\frac{2}{\cos\left(\underbrace{\left(e^{i\frac{\pi}{6}}+e^{i\frac{5\pi}{6}}\right)}_{=i}\pi\right)-\cos\left(\underbrace{\left(e^{i\frac{\pi}{6}}-e^{i\frac{5\pi}{6}}\right)}_{=\sqrt{3}}\pi\right) }

Từ đây có điều phải chứng minh.

Advertisements

Leave a Comment »

No comments yet.

RSS feed for comments on this post. TrackBack URI

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

Create a free website or blog at WordPress.com.

%d bloggers like this: