"로저스-라마누잔 항등식"의 두 판 사이의 차이

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34번째 줄: 34번째 줄:
  
 
<math> \frac{G(q)}{H(q)} = 1+\frac{q}{1+\frac{q^2}{1+\frac{q^3}{1+\cdots}}}</math>
 
<math> \frac{G(q)}{H(q)} = 1+\frac{q}{1+\frac{q^2}{1+\frac{q^3}{1+\cdots}}}</math>
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175번째 줄: 177번째 줄:
 
* [http://www.ams.org/bull/2005-42-02/S0273-0979-05-01047-5/home.html#References Continued fractions and modular functions]<br>
 
* [http://www.ams.org/bull/2005-42-02/S0273-0979-05-01047-5/home.html#References Continued fractions and modular functions]<br>
 
** W. Duke, Bull. Amer. Math. Soc. 42 (2005), 137-162.
 
** W. Duke, Bull. Amer. Math. Soc. 42 (2005), 137-162.
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* [http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.31.5875 Ramanujan’s formulas for the explicit evaluation of the Rogers–Ramanujan continued fraction and theta-functions]<br>
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** Soon-Yi Kang, ACTA ARITHMETICA XC.1 (1999)
 
* [http://www.digizeitschriften.de/index.php?id=loader&tx_jkDigiTools_pi1%5BIDDOC%5D=503543 Explicit evaluations of the Rogers-Ramanujan continued fraction.]<br>
 
* [http://www.digizeitschriften.de/index.php?id=loader&tx_jkDigiTools_pi1%5BIDDOC%5D=503543 Explicit evaluations of the Rogers-Ramanujan continued fraction.]<br>
 
** Berndt, B.C,Chan, H.H.,Zhang, L.-C., Journal für die reine und angewandte Mathematik 480, 1996
 
** Berndt, B.C,Chan, H.H.,Zhang, L.-C., Journal für die reine und angewandte Mathematik 480, 1996
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* [http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.31.5875 Ramanujan’s formulas for the explicit evaluation of the Rogers–Ramanujan continued fraction and theta-functions]<br>
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** Soon-Yi Kang, ACTA ARITHMETICA XC.1 (1999)
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* [http://www.jstor.org/stable/2325145 A Motivated Proof of the Rogers-Ramanujan Identities]<br>
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** George E. Andrews and R. J. Baxter, <cite style="line-height: 2em;">The American Mathematical Monthly</cite>, Vol. 96, No. 5 (May, 1989), pp. 401-409
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*  Watson, G. N.<br>
 
*  Watson, G. N.<br>
 
** [http://www.google.com/url?sa=t&ct=res&cd=1&url=http%3A%2F%2Fjlms.oxfordjournals.org%2Fcgi%2Freprint%2Fs1-4%2F3%2F231&ei=JY1hSLWRLpSY8gSI7JSiBQ&usg=AFQjCNElhd9FwCl3m3Qcb3hW7j87K1P5FQ&sig2=4OhMIB56amm8h4EOGNSk6g Theorems Stated by Ramanujan (IX): Two Continued Fractions.]
 
** [http://www.google.com/url?sa=t&ct=res&cd=1&url=http%3A%2F%2Fjlms.oxfordjournals.org%2Fcgi%2Freprint%2Fs1-4%2F3%2F231&ei=JY1hSLWRLpSY8gSI7JSiBQ&usg=AFQjCNElhd9FwCl3m3Qcb3hW7j87K1P5FQ&sig2=4OhMIB56amm8h4EOGNSk6g Theorems Stated by Ramanujan (IX): Two Continued Fractions.]
 
** [http://www.google.com/url?sa=t&ct=res&cd=1&url=http%3A%2F%2Fjlms.oxfordjournals.org%2Fcgi%2Freprint%2Fs1-4%2F13%2F39&ei=HY5hSNa6E5ym8ASu_biqBQ&usg=AFQjCNGfZ9Hu3vXz6bawkdnRZ2UU6jDUPA&sig2=dEC2KNSntm2J6L5GwTii3A Theorems Stated by Ramanujan (VII): Theorems on a Continued Fraction.]
 
** [http://www.google.com/url?sa=t&ct=res&cd=1&url=http%3A%2F%2Fjlms.oxfordjournals.org%2Fcgi%2Freprint%2Fs1-4%2F13%2F39&ei=HY5hSNa6E5ym8ASu_biqBQ&usg=AFQjCNGfZ9Hu3vXz6bawkdnRZ2UU6jDUPA&sig2=dEC2KNSntm2J6L5GwTii3A Theorems Stated by Ramanujan (VII): Theorems on a Continued Fraction.]
  
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* [http://www.jstor.org/stable/2325145 A Motivated Proof of the Rogers-Ramanujan Identities]<br>
 
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** George E. Andrews and R. J. Baxter, <cite style="line-height: 2em;">The American Mathematical Monthly</cite>, Vol. 96, No. 5 (May, 1989), pp. 401-409
 
 
* http://ko.wikipedia.org/wiki/
 
* http://ko.wikipedia.org/wiki/
  

2009년 8월 17일 (월) 17:35 판

간단한 소개

라마누잔이 하디에게 보낸 편지에는 다음과 같은 공식이 포함되어 있음

\(\cfrac{1}{1 + \cfrac{e^{-2\pi}}{1 + \cfrac{e^{-4\pi}}{1+\dots}}} = \left({\sqrt{5+\sqrt{5}\over 2}-{\sqrt{5}+1\over 2}}\right)e^{2\pi/5} = e^{2\pi/5}\left({\sqrt{\varphi\sqrt{5}}-\varphi}\right) = 0.9981360\dots\)

 

\(\varphi\) 는 황금비

 

 

로저스-라마누잔 항등식

\(G(q) = \sum_{n=0}^\infty \frac {q^{n^2}} {(q;q)_n} = \frac {1}{(q;q^5)_\infty (q^4; q^5)_\infty} =1+ q +q^2 +q^3 +2q^4+2q^5 +3q^6+\cdots\)

 

 \(H(q) =\sum_{n=0}^\infty \frac {q^{n^2+n}} {(q;q)_n} = \frac {1}{(q^2;q^5)_\infty (q^3; q^5)_\infty} =1+q^2 +q^3 +q^4+q^5 +2q^6+\cdots\)

\((a;q)_n = \prod_{k=0}^{n-1} (1-aq^k)=(1-a)(1-aq)(1-aq^2)\cdots(1-aq^{n-1})\)

 

로저스-라마누잔 연분수

\( \frac{G(q)}{H(q)} = 1+\frac{q}{1+\frac{q^2}{1+\frac{q^3}{1+\cdots}}}\)

 

 

 

\(r(\tau)=\cfrac{q^{\frac{1}{5}}}{1+\cfrac{q}{1+\cfrac{q^2}{1+\cfrac{q^3}{1+\cdots}}}}, q=e^{2\pi i \tau}\)

\(\tau=i\) 인 경우에 값을 계산할 수 있다면, 위의 값을 얻을 수 있다.

\(r(i)=\cfrac{e^{\frac{-2\pi}{5}}}{1+\cfrac{e^{-2\pi}}{1+\cfrac{e^{-4\pi}}{1+\cfrac{e^{-6\pi}}{1+\cdots}}}}\)

 

 

modularity
  • 로저스-라마누잔 함수들은 약간의 수정을 통해 modularity를 가짐
    \(q^{-1/60}G(q) = q^{-1/60}\sum_{n=0}^\infty \frac {q^{n^2}} {(q;q)_n} = \frac {q^{-1/60}}{(q;q^5)_\infty (q^4; q^5)_\infty}\)
    \(q^{11/60}H(q) =q^{11/60}\sum_{n=0}^\infty \frac {q^{n^2+n}} {(q;q)_n} = q^{11/60}\frac {1}{(q^2;q^5)_\infty (q^3; q^5)_\infty} \)
  • 데데킨트 에타함수가 갖는 modularity와의 유사성
    \(\eta(\tau) = q^{1/24} \prod_{n=1}^{\infty} (1-q^{n})\)

 

 

j-invariant 와의 관계

\((r^{20}-228r^{15}+494r^{10}+228r^{5}+1)^3+j(\tau)r^{5}(r^{10}+11r^{5}-1)^5=0\)

여기서, \(j(\tau)\) 는 j-invariant

\( j(\sqrt{-1})=1728\)를 이용하고 방정식의 해를 계산.

 

 

 

데데킨트 \(\eta\) 함수와의 관계

\(\frac{1}{r(5\tau)}-r(5\tau)-1=\frac{\eta(\tau)}{\eta(25\tau)}\)

\(\frac{1}{r(-\frac{1}{5\tau})}-r(-\frac{1}{5\tau})-1=\frac{\eta(-\frac{1}{25\tau})}{\eta(-\frac{1}{\tau})}\)

 

 

  • 에타함수의 modularity

\(\eta(-\frac{1}{\tau}) =\sqrt{\frac{\tau}{i}}\eta(\tau)\)

\(\eta(-\frac{1}{25\tau}) =\sqrt{\frac{25\tau}{i}}\eta(25\tau)\)

\(\frac{\eta(\tau)}{\eta(25\tau)}\frac{\eta(-\frac{1}{25\tau})}{\eta(-\frac{1}{\tau})}=5\)

  • 양변을 곱하여 다음 식을 얻는다.

\((\frac{1}{r(5\tau)}-r(5\tau)-1)(\frac{1}{r(-\frac{1}{5\tau})}-r(-\frac{1}{5\tau})-1)=5\)

 

\(\tau=\frac{i}{5}\) 인 경우, \((\frac{1}{r(i)}-r(i)-1)^2=5\) 를 얻고, 방정식을 풀 수 있음.

 


 

하위주제들

 

 

 

하위페이지

 

 

재미있는 사실
  • 이 항등식은 통계물리의 Lee-Yang 모델과 밀접하게 관련되어 있음

 

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