"라마누잔과 파이"의 두 판 사이의 차이

수학노트
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(section '관련논문' updated)
(같은 사용자의 중간 판 7개는 보이지 않습니다)
1번째 줄: 1번째 줄:
 
==개요==
 
==개요==
  
−
* 라마누잔은 1914년에 다음과 같은 공식을 발표 '''[RAM1914]''':<math>\frac{1}{\pi}= \frac{2\sqrt2}{9801}\sum_{n=0}^{\infty}\frac{(4n)!(1103+26390n)}{(n!)^{4}396^{4n}}</math><br>
+
* 라마누잔은 1914년에 다음과 같은 공식을 발표 '''[RAM1914]''':<math>\frac{1}{\pi}= \frac{2\sqrt2}{9801}\sum_{n=0}^{\infty}\frac{(4n)!(1103+26390n)}{(n!)^{4}396^{4n}}</math>
 
* Chudnovsky 형제  '''[CHU88]'''
 
* Chudnovsky 형제  '''[CHU88]'''
 
:<math>\frac{426880 \sqrt{10005}}{\pi} = \sum_{k=0}^\infty \frac{(6k)! (13591409 + 545140134k)}{(3k)!(k!)^3 (-640320)^{3k}}\!</math>
 
:<math>\frac{426880 \sqrt{10005}}{\pi} = \sum_{k=0}^\infty \frac{(6k)! (13591409 + 545140134k)}{(3k)!(k!)^3 (-640320)^{3k}}\!</math>
13번째 줄: 13번째 줄:
 
==정의와 미리 알아야 할 것들==
 
==정의와 미리 알아야 할 것들==
  
−
* [[자코비 세타함수]], [[라마누잔의 class invariants]], [[타원적분]] 참조:<math>q=e^{2\pi i \tau}</math>:<math>\theta_{2}(\tau)= \sum_{n=-\infty}^\infty q^{(n+\frac{1}{2})^2/2}</math>:<math>\theta_3(\tau)=\sum_{n=-\infty}^\infty q^{n^2/2}</math><br>
+
* [[자코비 세타함수]], [[라마누잔의 class invariants]], [[타원적분]] 참조
 +
 
 +
<math>q=e^{2\pi i \tau}</math>
 +
 
 +
<math>\theta_{2}(\tau)= \sum_{n=-\infty}^\infty q^{(n+\frac{1}{2})^2/2}</math>
 +
 
 +
<math>\theta_3(\tau)=\sum_{n=-\infty}^\infty q^{n^2/2}</math>
  
 
<math>\theta_{4}(\tau)= \sum_{n=-\infty}^\infty (-1)^n q^{n^2/2}</math>
 
<math>\theta_{4}(\tau)= \sum_{n=-\infty}^\infty (-1)^n q^{n^2/2}</math>
35번째 줄: 41번째 줄:
 
<math>\alpha(r):=\frac{E'}{K}-\frac{\pi}{4K^2}</math>
 
<math>\alpha(r):=\frac{E'}{K}-\frac{\pi}{4K^2}</math>
  
−
* [[라마누잔의 class invariants]]:<math>g_n:=2^{-1/4}\frac{\eta(\frac{\sqrt{-n}}{2})}{\eta(\sqrt{-n})}</math><br>
+
* [[라마누잔의 class invariants]]:<math>g_n:=2^{-1/4}\frac{\eta(\frac{\sqrt{-n}}{2})}{\eta(\sqrt{-n})}</math>
  
 
   
 
   
43번째 줄: 49번째 줄:
 
==singular value function ==
 
==singular value function ==
  
−
*  타원적분이 만족시키는 르장드르 항등식<br>  <math>E(k)K'(k)+E'(k)K(k)-K(k)K'(k)=\frac{\pi}{2}</math> ([[산술기하평균함수(AGM)와 파이값의 계산|AGM과 파이값의 계산]])<br>
+
*  타원적분이 만족시키는 르장드르 항등식 ([[산술기하평균함수(AGM)와 파이값의 계산|AGM과 파이값의 계산]])
−
*  타원적분의 성질 :<math>K'(\lambda^{*}(r))=\sqrt{r}K(\lambda^{*}(r))</math><br>
+
:<math>E(k)K'(k)+E'(k)K(k)-K(k)K'(k)=\frac{\pi}{2}\label{leg}</math>  
−
*  위의 둘을 사용하여 다음을 얻는다:<math>\alpha(r)=\frac{\pi}{4K^2}-\sqrt{r}(\frac{E}{K}-1)</math><br>
+
*  타원적분의 성질
−
*  여기에 타원적분이 만족시키는 미분방정식:<math>\frac{dK}{dk}=\frac{E-k'^2K}{kk'^2}</math><br> 을 사용하면:<math>\alpha(r)=\frac{1}{\pi}(\frac{\pi}{2K})^2-\sqrt{r}(kk'^2\frac{\dot{K}}{K}-k^2)</math><br> 를 얻게 되고, 이를 다시 쓰면:<math>\frac{1}{\pi}=\sqrt{N}k_Nk'^2_N\frac{4K\dot{K}}{\pi^2}+[\alpha(N)-\sqrt{N}k^2_N]\frac{4K^2}{\pi^2}</math><br>
+
:<math>K'(\lambda^{*}(r))=\sqrt{r}K(\lambda^{*}(r))\label{ell}</math>
 +
*  \ref{leg}와 \ref{ell}로부터 다음을 얻는다
 +
:<math>\alpha(r)=\frac{\pi}{4K^2}-\sqrt{r}(\frac{E}{K}-1)</math>
 +
*  여기에 타원적분이 만족시키는 미분방정식
 +
:<math>\frac{dK}{dk}=\frac{E-k'^2K}{kk'^2}</math> 을 사용하면
 +
:<math>\alpha(r)=\frac{1}{\pi}(\frac{\pi}{2K})^2-\sqrt{r}(kk'^2\frac{\dot{K}}{K}-k^2)</math> 를 얻게 되고, 이를 다시 쓰면
 +
:<math>\frac{1}{\pi}=\sqrt{N}k_Nk'^2_N\frac{4K\dot{K}}{\pi^2}+[\alpha(N)-\sqrt{N}k^2_N]\frac{4K^2}{\pi^2}</math>
  
 
   
 
   
  
−
* <math>[\frac{2}{\pi}K(k)]^2 =m(k)F(y(k))</math> 꼴로 쓰여질때, 양변을 미분하면 다음을 얻는다:<math>\frac{4K\dot{K}}{\pi^2}=\frac{1}{2}\dot{m}F+\frac{1}{2}m\dot{y}\dot{F}(y)</math><br>
+
* <math>[\frac{2}{\pi}K(k)]^2 =m(k)F(y(k))</math> 꼴로 쓰여질때, 양변을 미분하면 다음을 얻는다:<math>\frac{4K\dot{K}}{\pi^2}=\frac{1}{2}\dot{m}F+\frac{1}{2}m\dot{y}\dot{F}(y)</math>
−
*  초기하급수를 다음과 같이 쓰면:<math>F(y)=\sum_{n=0}^{\infty}a_ny^n</math><br>
+
*  초기하급수를 다음과 같이 쓰면
−
* <math>\frac{1}{\pi}=\sum_{n=0}^\infty a_n[\frac{\sqrt{N}}{2}k{k'}^2\dot{m}+[\alpha(N)-\sqrt{N}k^2_N]m+\frac{n\sqrt{N}}{2}m\frac{\dot{y}}{y}kk'^2]y^n</math>
+
:<math>F(y)=\sum_{n=0}^{\infty}a_ny^n</math>
 +
다음을 얻는다
 +
:<math>\frac{1}{\pi}=\sum_{n=0}^\infty a_n[\frac{\sqrt{N}}{2}k{k'}^2\dot{m}+[\alpha(N)-\sqrt{N}k^2_N]m+\frac{n\sqrt{N}}{2}m\frac{\dot{y}}{y}kk'^2]y^n</math>
  
 
   
 
   
62번째 줄: 76번째 줄:
 
* 아래의 prop, thm 번호는 '''[BB1998] '''참조
 
* 아래의 prop, thm 번호는 '''[BB1998] '''참조
 
* [[초기하급수(Hypergeometric series)]] 항목의 Clausen 항등식이 중요하게 사용됨
 
* [[초기하급수(Hypergeometric series)]] 항목의 Clausen 항등식이 중요하게 사용됨
−
*  prop 5.6:<math>\frac{2}{\pi}K_s(h) = \,_2F_1(\frac{1}{4}-\frac{s}{2},\frac{1}{4}+\frac{s}{2};1;(2hh')^2)</math>:<math>[\frac{2}{\pi}K_s(h)]^2 = \,_2F_1(\frac{1}{2}-s,\frac{1}{2}+s,\frac{1}{2};1,1;(2hh')^2)</math><br>
+
*  prop 5.6:<math>\frac{2}{\pi}K_s(h) = \,_2F_1(\frac{1}{4}-\frac{s}{2},\frac{1}{4}+\frac{s}{2};1;(2hh')^2)</math>:<math>[\frac{2}{\pi}K_s(h)]^2 = \,_2F_1(\frac{1}{2}-s,\frac{1}{2}+s,\frac{1}{2};1,1;(2hh')^2)</math>
−
*  prop 5.7:<math>K_{1/4}(h)=(1+k^2)^{1/2}K(k)</math> if <math>2hh'=[\frac{g^{12}+g^{-12}}{2}]^{-1}</math><br>
+
*  prop 5.7:<math>K_{1/4}(h)=(1+k^2)^{1/2}K(k)</math> if <math>2hh'=[\frac{g^{12}+g^{-12}}{2}]^{-1}</math>
−
*  Thm 5.6:<math>\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})</math><br>
+
*  Thm 5.6:<math>\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})</math>
−
*  Thm 5.7:<math>[\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})</math><br>
+
*  Thm 5.7:<math>[\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})</math>
−
*  (5.5.16):<math>\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}</math>:<math>x_N=(\frac{g_N^{12}+g_N^{-12}}{2})^{-1}</math>:<math>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})</math><br>
+
*  (5.5.16):<math>\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}</math>:<math>x_N=(\frac{g_N^{12}+g_N^{-12}}{2})^{-1}</math>:<math>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})</math>
  
 
   
 
   
  
−
* <math>N=58</math> 일 때:<math>x_{58}=\frac{1}{99^2}=\frac{1}{9801}</math>, <math>d_n(58)=(1103+26390n)2\sqrt 2</math> 이므로 다음을 얻는다:<math>\frac{1}{\pi}= \frac{2\sqrt2}{9801}\sum_{n=0}^{\infty}\frac{(4n)!(1103+26390n)}{(n!)^{4}396^{4n}}</math><br>  <br>
+
* <math>N=58</math> 일 때
 +
:<math>x_{58}=\frac{1}{99^2}=\frac{1}{9801},</math>
 +
:<math>d_n(58)=(1103+26390n)2\sqrt 2</math> 에서 다음을 얻는다
 +
:<math>\frac{1}{\pi}= \frac{2\sqrt2}{9801}\sum_{n=0}^{\infty}\frac{(4n)!(1103+26390n)}{(n!)^{4}396^{4n}}</math>   
  
 
   
 
   
79번째 줄: 96번째 줄:
  
 
* [[라마누잔의 class invariants]]
 
* [[라마누잔의 class invariants]]
−
* <math>g_{58}^2=\frac{\sqrt{29}+5}{2}</math>
+
:<math>g_{58}^2=\frac{\sqrt{29}+5}{2}</math>
  
 
   
 
   
85번째 줄: 102번째 줄:
 
   
 
   
  
−
==재미있는 사실==
+
==메모==
−
 
 
 
* <math>e^{\sqrt{58}\pi}=24591257751.999999822\cdots</math>
 
* <math>e^{\sqrt{58}\pi}=24591257751.999999822\cdots</math>
−
* <math>\frac{6}{5}{\phi}^2\approx{\pi}</math>
+
* <math>\frac{6}{5}{\phi}^2\approx{\pi}</math>
−
 
+
* http://mathoverflow.net/questions/161836/ramanujans-pi-formulas-with-a-twist
 
   
 
   
  
95번째 줄: 111번째 줄:
  
 
==역사==
 
==역사==
−
 
+
* 1910 - 라마누잔이 다음의 공식을 발견
−
*  Around 1910, the Indian mathematician Srinivasa Ramanujan discovered the formula<br>
+
:<math>\frac{1}{\pi}= \frac{2\sqrt2}{9801}\sum_{n=0}^{\infty}\frac{(4n)!(1103+26390n)}{(n!)^{4}396^{4n}}</math>
−
 
+
* 1985 William Gosper가 이 급수를 이용하여 <math>\pi</math>값을 1700만 자리까지 계산
−
; <math>\frac{1}{\pi}= \frac{2\sqrt2}{9801}\sum_{n=0}^{\infty}\frac{(4n)!(1103+26390n)}{(n!)^{4}396^{4n}}</math>
 
−
 
 
−
* William Gosper used this series in 1985 to compute the first 17 million digits of <math>\pi</math>.<br>
 
 
* [[수학사 연표]]
 
* [[수학사 연표]]
  
 
   
 
   
−
 
 
   
 
   
  
125번째 줄: 137번째 줄:
 
* http://documents.wolfram.com/mathematica/Demos/Notebooks/CalculatingPi.html
 
* http://documents.wolfram.com/mathematica/Demos/Notebooks/CalculatingPi.html
 
* http://functions.wolfram.com/Constants/Pi/06/01/02/0001/
 
* http://functions.wolfram.com/Constants/Pi/06/01/02/0001/
−
* http://www.wolframalpha.com/input/?i=series+representation+of+pi
 
−
* [http://www.research.att.com/%7Enjas/sequences/index.html The On-Line Encyclopedia of Integer Sequences]
 
  
−
* [[매스매티카 파일 목록]]
 
  
−
+
==사전 형태의 자료==
−
 
+
* http://en.wikipedia.org/wiki/Ramanujan-Sato_series
−
 
  
 
==관련도서==
 
==관련도서==
  
−
* '''[BB1998]'''[http://www.amazon.com/PI-AGM-Analytic-Computational-Complexity/dp/047131515X Pi and the AGM]<br>
+
* '''[BB1998]'''[http://www.amazon.com/PI-AGM-Analytic-Computational-Complexity/dp/047131515X Pi and the AGM]
 
** Jonathan M. Borwein, Peter B. Borwein, Wiley-Interscience (July 13, 1998)
 
** Jonathan M. Borwein, Peter B. Borwein, Wiley-Interscience (July 13, 1998)
−
*  도서내검색<br>
 
−
** http://books.google.com/books?q=
 
−
** http://book.daum.net/search/contentSearch.do?query=
 
−
*  도서검색<br>
 
−
** 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=
 
−
* [http://dlmf.nist.gov/ NIST Digital Library of Mathematical Functions]
 
  
 
   
 
   
−
 
 
   
 
   
  
172번째 줄: 158번째 줄:
 
   
 
   
  
−
==관련논문[http://documents.wolfram.com/mathematica/Demos/Notebooks/CalculatingPi.html ]==
+
==관련논문==
−
 
+
* Jesús Guillera, New proofs of Borwein-type algorithms for Pi, arXiv:1604.00193[math.NT], April 01 2016, http://arxiv.org/abs/1604.00193v1
−
* [http://www.sciencedirect.com/science?_ob=ArticleURL&_udi=B6WK2-4PW5XTP-8&_user=4420&_rdoc=1&_fmt=&_orig=search&_sort=d&view=c&_acct=C000059607&_version=1&_urlVersion=0&_userid=4420&md5=07a10c67e340156fe912e39d39c0330a Ramanujan's series for 1/π arising from his cubic and quartic theories of elliptic functions]<br>
+
* Cooper, Shaun, and Wadim Zudilin. “Holonomic Alchemy and Series for $1/\pi$.” arXiv:1512.04608 [math], December 14, 2015. http://arxiv.org/abs/1512.04608.
 +
* Osburn, Robert, and Wadim Zudilin. ‘On the (K.2) Supercongruence of Van Hamme’. arXiv:1504.01976 [math], 8 April 2015. http://arxiv.org/abs/1504.01976.
 +
* Swisher, Holly. “On the Supercongruence Conjectures of van Hamme.” arXiv:1504.01028 [math], April 4, 2015. http://arxiv.org/abs/1504.01028.
 +
* [http://www.sciencedirect.com/science?_ob=ArticleURL&_udi=B6WK2-4PW5XTP-8&_user=4420&_rdoc=1&_fmt=&_orig=search&_sort=d&view=c&_acct=C000059607&_version=1&_urlVersion=0&_userid=4420&md5=07a10c67e340156fe912e39d39c0330a 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
 
** Nayandeep Deka Baruaha, and Bruce C. Berndt, Journal of Mathematical Analysis and Applications, Volume 341, Issue 1, 2007
−
* [http://www.math.rutgers.edu/%7Ezeilberg/mamarim/mamarimhtml/ramapi.html A WZ Proof of Ramanujan's Formula for Pi ]<br>
+
* [http://www.math.rutgers.edu/%7Ezeilberg/mamarim/mamarimhtml/ramapi.html 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.
 
** Shalosh B. Ekhad and Doron Zeilberger,  `Geometry, Analysis, and Mechanics', ed. by J.M. Rassias, World Scientific, Singapore, 1994, 107-108.
−
* [http://www.cecm.sfu.ca/personal/pborwein/PAPERS/P62.pdf Class number three Ramanujan type series for 1/pi]<br>
+
* [http://www.cecm.sfu.ca/personal/pborwein/PAPERS/P62.pdf 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)
 
** J. M. Borwein ,P. B. Borwein, Journal of Computational and Applied Mathematics (Vol.46 NO.1 / 1993)
−
* [http://www.jstor.org/stable/2325206 Ramanujan, Modular Equations, and Approximations to Pi or How to Compute One Billion Digits of Pi]<br>
+
* [http://www.jstor.org/stable/2325206 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, <cite style="line-height: 2em;">The American Mathematical Monthly</cite>, Vol. 96, No. 3 (Mar., 1989), pp. 201-219
 
** J. M. Borwein, P. B. Borwein and D. H. Bailey, <cite style="line-height: 2em;">The American Mathematical Monthly</cite>, Vol. 96, No. 3 (Mar., 1989), pp. 201-219
−
* '''[CHU88]'''Approximations and complex multiplication according to Ramanujan<br>
+
* '''[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.
+
** D. V. Chudnovsky and G. V. Chudnovsky, Ramanujan Revisited, Academic Press Inc., Boston, (1988), p. 375-396 & p. 468-472
−
 
+
* [http://www.cecm.sfu.ca/personal/pborwein/PAPERS/P35.pdf Explicit Ramanujan-type approximations to pi of high order ]
−
* [http://www.cecm.sfu.ca/personal/pborwein/PAPERS/P35.pdf Explicit Ramanujan-type approximations to pi of high order ]<br>
 
 
** J. M. Borwein, P. B. Borwein, 1987
 
** J. M. Borwein, P. B. Borwein, 1987
−
* '''[RAM1914]'''[http://www.imsc.res.in/%7Erao/ramanujan/CamUnivCpapers/Cpaper6/page1.htm Modular equations and approximations to Pi]<br>
+
* '''[RAM1914]'''[http://www.imsc.res.in/%7Erao/ramanujan/CamUnivCpapers/Cpaper6/page1.htm Modular equations and approximations to Pi]
 
** S. Ramanujan, Quart. J. Pure Appl. Math., (1914), vol. 45, p. 350-372
 
** S. Ramanujan, Quart. J. Pure Appl. Math., (1914), vol. 45, p. 350-372
−
 
−
 
−
 
  
 
==관련기사==
 
==관련기사==
  
−
*  The Mountains of Pi<br>
+
*  The Mountains of Pi
 
** The New Yorker, 1992-3-2
 
** The New Yorker, 1992-3-2
 
[[분류:원주율]]
 
[[분류:원주율]]

2016년 4월 3일 (일) 23:30 판

개요

  • 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]

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



정의와 미리 알아야 할 것들

\(q=e^{2\pi i \tau}\)

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

\(\theta_3(\tau)=\sum_{n=-\infty}^\infty q^{n^2/2}\)

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

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

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

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

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

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

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

  • 위의 함수들을 이용하여, 양수 \(r\)에 대하여 다음을 정의

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

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



singular value function

\[E(k)K'(k)+E'(k)K(k)-K(k)K'(k)=\frac{\pi}{2}\label{leg}\]

  • 타원적분의 성질

\[K'(\lambda^{*}(r))=\sqrt{r}K(\lambda^{*}(r))\label{ell}\]

  • \ref{leg}와 \ref{ell}로부터 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 \[\alpha(r)=\frac{1}{\pi}(\frac{\pi}{2K})^2-\sqrt{r}(kk'^2\frac{\dot{K}}{K}-k^2)\] 를 얻게 되고, 이를 다시 쓰면 \[\frac{1}{\pi}=\sqrt{N}k_Nk'^2_N\frac{4K\dot{K}}{\pi^2}+[\alpha(N)-\sqrt{N}k^2_N]\frac{4K^2}{\pi^2}\]


  • \([\frac{2}{\pi}K(k)]^2 =m(k)F(y(k))\) 꼴로 쓰여질때, 양변을 미분하면 다음을 얻는다\[\frac{4K\dot{K}}{\pi^2}=\frac{1}{2}\dot{m}F+\frac{1}{2}m\dot{y}\dot{F}(y)\]
  • 초기하급수를 다음과 같이 쓰면

\[F(y)=\sum_{n=0}^{\infty}a_ny^n\] 다음을 얻는다 \[\frac{1}{\pi}=\sum_{n=0}^\infty a_n[\frac{\sqrt{N}}{2}k{k'}^2\dot{m}+[\alpha(N)-\sqrt{N}k^2_N]m+\frac{n\sqrt{N}}{2}m\frac{\dot{y}}{y}kk'^2]y^n\]



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

\[x_{58}=\frac{1}{99^2}=\frac{1}{9801},\] \[d_n(58)=(1103+26390n)2\sqrt 2\] 에서 다음을 얻는다 \[\frac{1}{\pi}= \frac{2\sqrt2}{9801}\sum_{n=0}^{\infty}\frac{(4n)!(1103+26390n)}{(n!)^{4}396^{4n}}\]



라마누잔의 class invariants

\[g_{58}^2=\frac{\sqrt{29}+5}{2}\]



메모



역사

  • 1910 - 라마누잔이 다음의 공식을 발견

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

  • 1985 William Gosper가 이 급수를 이용하여 \(\pi\)값을 1700만 자리까지 계산
  • 수학사 연표



관련된 항목들



매스매티카 파일 및 계산 리소스


사전 형태의 자료

관련도서

  • [BB1998]Pi and the AGM
    • Jonathan M. Borwein, Peter B. Borwein, Wiley-Interscience (July 13, 1998)



리뷰논문, 에세이, 강의노트



관련논문

관련기사

  • The Mountains of Pi
    • The New Yorker, 1992-3-2
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