라마누잔과 파이

이 항목의 스프링노트 원문주소

• 라마누잔과 파이

 

 

간단한 소개

• 라마누잔은 1914년에 다음과 같은 공식을 발표 [RAM1914]
  \(\frac{1}{\pi}= \frac{2\sqrt2}{9801}\sum_{n=0}^{\infty}\frac{(4n)!(1103+26390n)}{(n!)^{4}396^{4n}}\)
  
• Chudnovsky 형제  [CHU88]
  

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

• 숫자 163

 

 

정의와 미리 알아야 할 것들

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

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

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

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

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

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

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

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

• 위의 함수들을 이용하여, 양수 \(r\)에 대하여 다음을 정의

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

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

• 라마누잔의 class invariants
  \(g_n:=2^{-1/4}\frac{\eta(\frac{\sqrt{-n}}{2})}{\eta(\sqrt{-n})}\)
  

 

 

singular value function 

• 타원적분이 만족시키는 르장드르 항등식
   \(E(k)K'(k)+E'(k)K(k)-K(k)K'(k)=\frac{\pi}{2}\) (AGM과 파이값의 계산)
  
• 타원적분의 성질 
  \(K'(\lambda^{*}(r))=\sqrt{r}K(\lambda^{*}(r))\)
  
• 위의 둘을 사용하여 다음을 얻는다
  \(\alpha(r)=\frac{\pi}{4K^2}-\sqrt{r}(\frac{E}{K}-1)\)
  
• 여기에 타원적분이 만족시키는 미분방정식
  \(\frac{dK}{dk}=\frac{E-k'^2K}{kk'^2}\)
  을 사용하면
  \(\alpha(r)=\frac{1}{\pi}(\frac{\pi}{2K})^2-\sqrt{r}(kk'^2\frac{\.K}{K}-k^2)\)
  를 얻게 되고, 이를 다시 쓰면
  \(\frac{1}{\pi}=\sqrt{N}k_Nk'_N^2\frac{4K\.K}{\pi^2}+[\alpha(N)-\sqrt{N}k^2_N]\frac{4K^2}{\pi^2}\)
  

 

• \([\frac{2}{\pi}K(k)]^2 =m(k)F(y(k))\) 꼴로 쓰여질때, 양변을 미분하면 다음을 얻는다
  \(\frac{4K\.K}{\pi^2}=\frac{1}{2}\.mF+\frac{1}{2}m\.y\.F(y)\)
  
• 초기하급수를 다음과 같이 쓰면
  \(F(y)=\sum_{n=0}^{\infty}a_ny^n\)
  
• \(\frac{1}{\pi}=\sum_{n=0}^\infty a_n[\frac{\sqrt{N}}{2}k{k'}^2\.m+[\alpha(N)-\sqrt{N}k^2_N]m+\frac{n\sqrt{N}}{2}m\frac{\.y}{y}kk'^2]y^n\)

 

 

라마누잔 파이 공식의 유도

• 아래의 prop, thm 번호는 [BB1998] 참조
• 초기하급수(Hypergeometric series) 항목의 Clausen 항등식이 중요하게 사용됨
• prop 5.6
  \(\frac{2}{\pi}K_s(h) = \,_2F_1(\frac{1}{4}-\frac{s}{2},\frac{1}{4}+\frac{s}{2};1;(2hh')^2)\)
  \([\frac{2}{\pi}K_s(h)]^2 = \,_2F_1(\frac{1}{2}-s,\frac{1}{2}+s,\frac{1}{2};1,1;(2hh')^2)\)
  
• prop 5.7
  \(K_{1/4}(h)=(1+k^2)^{1/2}K(k)\) if \(2hh'=[\frac{g^{12}+g^{-12}}{2}]^{-1}\)
  
• Thm 5.6
  \(\frac{2}{\pi}K(k) =(1+k^2)^{-1/2} \,_2F_1(\frac{1}{8},\frac{3}{8};1;[\frac{g^{12}+g^{-12}}{2}}]^{-2})\)
  
• Thm 5.7
  \([\frac{2}{\pi}K(k)]^2 =(1+k^2)^{-1} \,_3F_2(\frac{1}{4},\frac{3}{4},\frac{1}{2};1,1;[\frac{g^{12}+g^{-12}}{2}}]^{-2})\)
  
• (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

• \(e^{\sqrt{58}\pi}=24591257751.999999822\cdots\)
•  \(\frac{6}{5} \phi^2 = \pi\)

 

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
• 타원함수
• The modular group, j-invariant and the singular moduli
• 초기하급수(Hypergeometric series)
• 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=

 

 

사전 형태의 자료

• http://ko.wikipedia.org/wiki/

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

• http://en.wikipedia.org/wiki/
• http://www.wolframalpha.com/input/?i=
• NIST Digital Library of Mathematical Functions
• The On-Line Encyclopedia of Integer Sequences
  
  • http://www.research.att.com/~njas/sequences/?q=

 

 

 

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
• [CHU88]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
• [RAM1914]Modular equations and approximations to Pi
  
  • S. Ramanujan, Quart. J. Pure Appl. Math., (1914), vol. 45, p. 350-372

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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
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