라마누잔과 파이

간단한 소개

\(\frac{1}{\pi}= \frac{2\sqrt2}{9801}\sum_{n=0}^{\infty}\frac{(4n)!(1103+26390n)}{(n!)^{4}396^{4n}}\)

 

\[\frac{426880 \sqrt{10005}}{\pi} = \sum_{k=0}^\infty \frac{(6k)! (13591409 + 545140134k)}{(3k)!(k!)^3 (-640320)^{3k}}\!\]

 

정의

• 타원적분 , 자코비 세타함수, 라마누잔의 class invariants 항목 참조
  \(q=e^{2\pi i \tau}\)
  \(\theta_{2}(\tau)= \sum_{n=-\infty}^\infty q^{(n+\frac{1}{2})^2/2}\)
  \(\theta(\tau)=\theta_3(\tau)=\sum_{n=-\infty}^\infty q^{n^2/2}= \sum_{n=-\infty}^\infty \exp(\pi i n^2\tau)\)
  

 

\(\theta_{4}(\tau)= \sum_{n=-\infty}^\infty (-1)^n q^{n^2/2}\)

\(k=k(\tau)=\frac{\theta_2^2(\tau)}{\theta_3^2(\tau)}\)

\(K(k) = \int_0^{\frac{\pi}{2}} \frac{d\theta}{\sqrt{1-k^2 \sin^2\theta}}\)

\(E(k) = \int_0^{\frac{\pi}{2}} \sqrt{1-k^2 \sin^2\theta}}d\theta}{\)

\(k'=\sqrt{1-k^2}=\frac{\theta_4^2(\tau)}{\theta_3^2(\tau)}\)

\(K'(k) = K(k')\)

\(E'(k) = E(k')\)

• 위의 함수들을 이용하여, 양수 \(r\)에 대하여 다음을 정의

\(\lambda^{*}(r):=k(i\sqrt{r})\)

\(\alpha(r):=\frac{E'}{K}-\frac{\pi}{4K^2}\)

 

 

• [BB1998]  (5.5.16)
  \(\frac{1}{\pi}=\sum_{n=0}^{\infty}\frac{(\frac{1}{4})_n(\frac{1}{2})_n(\frac{3}{4})_n}{(n!)^3}d_n(N)x_N^{2n+1}\)
  \(x_N=(\frac{g_N^{12}+g_N^{-12}}{2})^{-1}\)
  \(d_n(N)=[\frac{\alpha(N)x_N^{-1}}{1+k_N^2}-\frac{\sqrt{N}}{4}g_N^{-12}]+n\sqrt N(\frac{g_N^{12}-g_N^{-12}}{2})\)
  

 

• \(N=58\) 일 때
  \(x_{58}=\frac{1}{99^2}=\frac{1}{9801}\), \(d_n(58)=(1103+26390n)2\sqrt 2\) 이므로 다음을 얻는다
  \(\frac{1}{\pi}= \frac{2\sqrt2}{9801}\sum_{n=0}^{\infty}\frac{(4n)!(1103+26390n)}{(n!)^{4}396^{4n}}\)
   
  

 

 

라마누잔의 class invariants

• 라마누잔의 class invariants
• \(g_{58}^2=\frac{\sqrt{29}+5}{2}\)

 

 

재미있는 사실

• \(e^{\sqrt{58}\pi}=24591257751.999999822\cdots\)

 

 

역사

• Around 1910, the Indian mathematician Srinivasa Ramanujan discovered the formula
  

\(\frac{1}{\pi}= \frac{2\sqrt2}{9801}\sum_{n=0}^{\infty}\frac{(4n)!(1103+26390n)}{(n!)^{4}396^{4n}}\)

• William Gosper used this series in 1985 to compute the first 17 million digits of \(\pi\).
  

 

 

 

관련된 다른 주제들

• AGM과 파이값의 계산
• 타원적분
• The modular group, j-invariant and the singular moduli
• Hypergeometric functions
• 숫자 163

 

관련도서 및 추천도서

• [BB1998]Pi and the AGM
  
  • Jonathan M. Borwein, Peter B. Borwein, Wiley-Interscience (July 13, 1998)
• 도서내검색
  
  • http://books.google.com/books?q=
  • http://book.daum.net/search/contentSearch.do?query=
• 도서검색
  
  • http://www.amazon.com/s/ref=nb_ss_gw?url=search-alias%3Dstripbooks&field-keywords=
  • http://book.daum.net/search/mainSearch.do?query=

 

참고할만한 자료

• http://ko.wikipedia.org/wiki/
• http://en.wikipedia.org/wiki/Pi
• http://www.wolframalpha.com/input/?i=

 

참고할만한 자료

• 공식을 구현한 매쓰매티카 파일
  
  • ramanujan_pi.nb
• http://documents.wolfram.com/mathematica/Demos/Notebooks/CalculatingPi.html 참고
• Ramanujan's series for 1/π arising from his cubic and quartic theories of elliptic functions
  
  • Nayandeep Deka Baruaha, and Bruce C. Berndt, Journal of Mathematical Analysis and Applications, Volume 341, Issue 1, 2007
• A WZ Proof of Ramanujan's Formula for Pi 
  
  • Shalosh B. Ekhad and Doron Zeilberger,  `Geometry, Analysis, and Mechanics', ed. by J.M. Rassias, World Scientific, Singapore, 1994, 107-108.
• Class number three Ramanujan type series for 1/pi
  
  • J. M. Borwein ,P. B. Borwein, Journal of Computational and Applied Mathematics (Vol.46 NO.1 / 1993)
• Ramanujan, Modular Equations, and Approximations to Pi or How to Compute One Billion Digits of Pi
  
  • J. M. Borwein, P. B. Borwein and D. H. Bailey, The American Mathematical Monthly, Vol. 96, No. 3 (Mar., 1989), pp. 201-219
• Approximations and complex multiplication according to Ramanujan
  
  • D. V. Chudnovsky and G. V. Chudnovsky, Ramanujan Revisited, Academic Press Inc., Boston, (1988), p. 375-396 & p. 468-472.

• Explicit Ramanujan-type approximations to pi of high order 
  
  • J. M. Borwein, P. B. Borwein, 1987
• Modular equations and approximations to Pi
  
  • S. Ramanujan, Quart. J. Pure Appl. Math., (1914), vol. 45, p. 350-372

[1]

• http://ko.wikipedia.org/wiki/
• http://en.wikipedia.org/wiki/

 

관련기사

• The Mountains of Pi
  
  • The New Yorker, 1992-3-2
• 네이버 뉴스 검색 (키워드 수정)
  
  • http://news.search.naver.com/search.naver?where=news&x=0&y=0&sm=tab_hty&query=라마누잔
  • http://news.search.naver.com/search.naver?where=news&x=0&y=0&sm=tab_hty&query=파이
  • http://news.search.naver.com/search.naver?where=news&x=0&y=0&sm=tab_hty&query=
  • http://news.search.naver.com/search.naver?where=news&x=0&y=0&sm=tab_hty&query=
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