{{위키데이터 속성 추적}}
{{각주 부족|날짜=2023-12-15}}
{{기계 번역|날짜=2021-07-23}}

'''라마누잔 g함수와 G 함수'''는 [[모듈러 함수]]에서 유도된 함수이다. 이 함수의 이름은 인도의 수학자 [[스리니바사 라마누잔]]에서 유래됐으며, g함수와 G함수는 타원 [[모듈러 람다 함수]]와 대수적 관계를 갖는다. 모듈러 람다 함수는 합동 부분군 Γ(2)에 대하여 불변인 모듈러 함수이다. 타원 모듈러 람다 함수 λ*(x)의 [[아크탄젠트]]를 두 배로 늘리는 코탄젠트는 함수 g(x)의 12 제곱과 정확히 동일하다. 타원 모듈러 람다 함수 λ*(x)의 아크사인을 두 배로 늘리는 코시컨트는 함수 G(x)의 12 제곱과 정확히 동일하다.

== 정의 ==
라마누잔에 따른 정의:

: <math> g(x) = \frac{1}{\sqrt[4]{2}}\exp\left(\frac{\pi}{24}\sqrt{x}\right) \prod_{a=0}^{\infty}\{1-\exp[-(2a+1)\pi\sqrt{x}]\} </math>
: <math> G(x) = \frac{1}{\sqrt[4]{2}}\exp\left(\frac{\pi}{24}\sqrt{x}\right) \prod_{a=0}^{\infty}\{1+\exp[-(2a+1)\pi\sqrt{x}]\} </math>

타원 모듈러 람다 함수를 기반으로하는 함수 g(x) 및 G(x)의 정의:

: <math>g(x) = \tan\{2\arctan[\lambda^*(x)]\}^{-1/12}</math>
: <math>G(x) = \sin\{2\arcsin[\lambda^*(x)]\}^{-1/12}</math>

[[데데킨트 에타 함수]]에 기반한 정의:

: <math> g(x) = 2^{-1/4}\,\eta(\tfrac{1}{2}i\sqrt{x})\,\eta(i\sqrt{x})^{-1} = </math>
: <math> = 2^{-1/4} \biggl\{\sum_{a=-\infty}^\infty (-1)^a \exp\biggl[-(6a+1)^2\frac{\pi}{24}\sqrt{x}\biggr]\biggr\} \biggl\{\sum_{a=-\infty}^\infty (-1)^a \exp\biggl[-(6a+1)^2\frac{\pi}{12}\sqrt{x}\biggr]\biggr\}^{-1} </math>
: <math> G(x) = 2^{-1/4}\,\eta(\tfrac{1}{2}i\sqrt{x})^{-1}\,\eta(i\sqrt{x})^{2}\,\eta(2i\sqrt{x})^{-1} </math>

== 역함수 ==
역함수를 표현하려면 제 1 종의 완전한 [[타원 적분]]만 필요하다:

: <math> g^{\langle -1 \rangle}(x) = K(\sqrt{2}x^6\sqrt{\sqrt{x^{24}+1}-x^{12}})^2 K(\sqrt{x^{24}+1}-x^{12})^{-2} </math>
: <math> G^{\langle -1 \rangle}(x) = K\biggl(\frac{1}{2}\sqrt{1+x^{-12}}+\frac{1}{2}\sqrt{1-x^{-12}}\biggr)^2 K\biggl(\frac{1}{2}\sqrt{1+x^{-12}}-\frac{1}{2}\sqrt{1-x^{-12}}\biggr)^{-2} </math>

다음 문장이 유효하다:

: 면: <math>g(x_1 \in \mathbb{A}^+) \in \mathbb{A}^+ </math>, 그런: <math>g(x_1 \in \mathbb{A}^+) \in \mathbb{Q}^+ </math>
: 면: <math>G(x_2 \in \mathbb{A}^+) \in \mathbb{A}^+ </math>, 그런: <math>G(x_2 \in \mathbb{A}^+) \in \mathbb{Q}^+ </math>

== 성질 ==
라마누잔 함수 g(x) 과 G(x) 함수는 열린 상반평면 위에서 정칙함수이나. 이 초등이 아닌 초등 함수는 구체적으로 다음과 같은 항등식을 만족한다:

: <math> G(x) = 2^{-1/8} g(x)^{-1/2}[\sqrt{g(x)^{24}+1}+g(x)^{12}]^{1/8} </math>
: <math> G(x) = G(1/x) </math>
: <math> g(4x) = 2^{1/4}g(x)G(x) </math>
: <math> g(x)g(4/x) = 1 </math>
: <math> g(9x)^{12} - g(x)^{12} = 2\sqrt{2} g(x)^{9} g(9x)^{9} + 2\sqrt{2} g(x)^{3} g(9x)^{3} </math>
: <math> g(25x)^{6} - g(x)^{6} = 2 g(x)^{5} g(25x)^{5} + 2 g(x) g(25x) </math>
: <math> g(49x)^{8} + g(x)^{8} - 7 g(x)^{4} g(49x)^{4} = 2\sqrt{2} g(x)^{7} g(49x)^{7} + 2\sqrt{2} g(x) g(49x) </math>
: <math> g(121x)^{12} - g(x)^{12} = </math>
: <math> = 2\sqrt{2} g(x) g(121x) [g(x)^{2} g(121x)^{2} + 1] [g(x)^{4} g(121x)^{4} + 3 g(x)^{2} g(121x)^{2} + 1] [2 g(x)^{4} g(121x)^{4} + 3 g(x)^{2} g(121x)^{2} + 2] </math>
: <math> [g(169x)^{2} - g(x)^{2}] [g(169x)^{4} + g(x)^{4} - 7 g(x)^{2} g(169x)^{2}] \{ [g(169x)^{2} - g(x)^{2}]^4 - g(x)^{2} g(169x)^{2} [g(169x)^{2} + g(x)^{2}]^2 \} = </math>
: <math> = 8 g(x)^{13} g(169x)^{13} + 8 g(x) g(169x) </math>

== 특별한 값 ==
중요한 피솟 비자야라가브한 상수:
{| class="wikitable"
!상수의 이름
!대수 표현
!방정식
|-
|[[황금비]]
|<math> \Phi = \tfrac{1}{2}(\sqrt{5}+1) </math>
|<math> \Phi^2 - \Phi - 1 = 0 </math>
|-
|[[은 비율]]
|<math> \delta_S = \sqrt{2}+1 </math>
|<math> \delta_S^2 - 2\delta_S - 1 = 0 </math>
|-
|청동 비율
|<math> \delta_B = \tfrac{1}{2}(\sqrt{13}+3) </math>
|<math> \delta_B^2 - 3\delta_B - 1 = 0 </math>
|-
|트리보나치 상수
|<math> T_{TRI} = \tfrac{1}{3}\sqrt[3]{19+3\sqrt{33}} + \tfrac{1}{3}\sqrt[3]{19-3\sqrt{33}} + \tfrac{1}{3} </math>
|<math> T_{TRI}^3 - T_{TRI}^2 - T_{TRI} - 1 = 0 </math>
|-
|[[플라스틱 수]]
|<math> \rho = \tfrac{1}{6}\sqrt[3]{12}(\sqrt[3]{9+\sqrt{69}} + \sqrt[3]{9-\sqrt{69}}) </math>
|<math> \rho^3 - \rho - 1 = 0 </math>
|-
|[[초황금비]]
|<math> \psi = \tfrac{1}{6}\sqrt[3]{116+12\sqrt{93}} + \tfrac{1}{6}\sqrt[3]{116-12\sqrt{93}} + \tfrac{1}{3} </math>
|<math> \psi^3 - \psi^2 - 1 = 0 </math>
|}
다음 값이 정확하다:
: <math> g(1) = 2^{-1/8} </math>
: <math> g(2) = 1 </math>
: <math> g(3) = 2^{-1/6}(2+\sqrt{3})^{1/8} </math>
: <math> g(4) = 2^{1/8} </math>
: <math> g(5) = 2^{-1/8}(\sqrt{\Phi}+1)^{1/4} </math>
: <math> g(6) = \sqrt[6]{\delta_S} </math>
: <math> g(7) = 2^{-1/4}(8+3\sqrt{7})^{1/8} </math>
: <math> g(8) = 2^{1/8}\sqrt[8]{\delta_S} </math>
: <math> g(9) = 2^{-1/8}(2+\sqrt{3})^{1/24}(\sqrt{2}+\sqrt[4]{3})^{1/4} </math>
: <math> g(10) = \sqrt{\Phi} </math>
: <math> g(14) = \frac{1}{2}\sqrt{3+\sqrt{2}}+\frac{1}{2}\sqrt{\sqrt{2}-1} </math>
: <math> g(18) = (\sqrt{3}+\sqrt{2})^{1/3} </math>
: <math> g(22) = \sqrt{\delta_S} </math>
: <math> g(26) = \frac{1}{6}\sqrt{2\sqrt{13}-2}(\sqrt{\sqrt{3}+\sqrt{2}}\sqrt[6]{3\sqrt{3}+\sqrt{26}}-\sqrt{\sqrt{3}-\sqrt{2}}\sqrt[6]{3\sqrt{3}-\sqrt{26}})+\frac{1}{3}\sqrt{\sqrt{13}+2} </math>
: <math> g(30) = (\sqrt{10}+3)^{1/6}\sqrt{\Phi} </math>
: <math> g(34) = \frac{1}{4}\sqrt{14+2\sqrt{17}}+\frac{1}{4}\sqrt{2\sqrt{17}-2} </math>
: <math> g(38) = \frac{1}{6}\delta_S^{-1/2}(\sqrt{2\sqrt{57}+14}\sqrt[6]{3\sqrt{57}+16\sqrt{2}}+\sqrt{2\sqrt{57}-14}\sqrt[6]{3\sqrt{57}-16\sqrt{2}}+4+2\sqrt{2}) </math>
: <math> g(42) = 2^{-1/2}(\sqrt{7}+\sqrt{3})^{1/2}(2\sqrt{2}+\sqrt{7})^{1/6} </math>
: <math> g(46) = \frac{1}{2}\sqrt{5+\sqrt{2}}+\frac{1}{2}\sqrt{\sqrt{2}+1} </math>
: <math> g(50) = \frac{1}{2}\biggl\{\frac{4}{3}\sqrt{2}\cos\biggl(\frac{\pi}{10}\biggr)\cosh\biggl[\frac{1}{3}\operatorname{arcosh}\biggl(\frac{4}{5}\sqrt{10}\biggr)\biggr] + \frac{1}{3}\tan\biggl(\frac{\pi}{5}\biggr)\biggr\}^2 - \frac{1}{2} </math>
: <math> g(54) = (\delta_S^{3/2}+\sqrt{2}\delta_S^{7/6}+\sqrt{2}\delta_S^{5/6})^{1/3} </math>
: <math> g(58) = 2^{-1/2}(\sqrt{29}+5)^{1/2} </math>
: <math> g(62) = \cot\biggl[\frac{1}{2}\arcsin\biggl(\frac{1}{4}\sqrt{28\sqrt{2}+38}-\frac{3}{2}-\frac{1}{4}\sqrt{2}\biggr)\biggr]^{1/2} </math>
: <math> g(66) = \cot\biggl\{\frac{1}{2}\arcsin\biggl[\frac{1}{8}(8 + 3\sqrt{2} - \sqrt{66})\sqrt[4]{\sqrt{33} + 4\sqrt{2}} - \frac{1}{8}(8 - 3\sqrt{2} + \sqrt{66})\sqrt[4]{\sqrt{33} - 4\sqrt{2}}\biggr]\biggr\}^{1/3} </math>
: <math> g(70) = \Phi\sqrt{\delta_S} </math>
함수 값 g(74)를 결정하려면 [[오차 방정식]]을 풀어야 한다:

: <math> g(74)^5 + g(74)^3 + g(74) - \sqrt{\sqrt{37}+6} \left[g(74)^4 - g(74)^2 + 1\right] = 0 </math>

다음 추가 값이 정확하다:

: <math> G(1) = 1 </math>
: <math> G(2) = 2^{-1/8}\sqrt[8]{\delta_S} </math>
: <math> G(3) = 2^{1/12} </math>
: <math> G(4) = 2^{-3/16}\sqrt[4]{\delta_S} </math>
: <math> G(5) = \sqrt[4]{\Phi} </math>
: <math> G(6) = 2^{-1/8}(2+\sqrt{3})^{1/8}(\sqrt{3}+\sqrt{2})^{1/8}\delta_S^{-1/12} </math>
: <math> G(7) = 2^{1/4} </math>
: <math> G(8) = 2^{-3/16}\delta_S^{1/16}(\sqrt{\delta_S}+\sqrt{2})^{1/4} </math>
: <math> G(9) = (2+\sqrt{3})^{1/6} </math>
: <math> G(10) = 2^{-1/8}(\sqrt{10}+3)^{1/8}\Phi^{-1/4}\delta_S^{1/4} </math>
: <math> G(11) = 2^{-1/4}3^{-1}(\sqrt[3]{3\sqrt{33}+17}-\sqrt[3]{3\sqrt{33}-17}+2) = 2^{-1/4}(T_{TRI}^2-T_{TRI}) </math>
: <math> G(13) = \sqrt[4]{\delta_B} </math>
: <math> G(15) = 2^{1/4}\sqrt[3]{\Phi} </math>
: <math> G(17) = \frac{1}{4}\sqrt{10+2\sqrt{17}}+\frac{1}{4}\sqrt{2\sqrt{17}-6} </math>
: <math> G(19) = \tfrac{2}{3}\sqrt[4]{18}\cosh[\tfrac{1}{3}\operatorname{arcosh}(\tfrac{3}{4}\sqrt{6})] </math>
: <math> G(21) = 2^{-1/4}(8+3\sqrt{7})^{1/12}(\sqrt{7}+\sqrt{3})^{1/4} </math>
: <math> G(23) = \tfrac{2}{3}\sqrt[4]{18}\cosh[\tfrac{1}{3}\operatorname{arcosh}(\tfrac{3}{2}\sqrt{3})] = 2^{1/4} \rho </math>
: <math> G(25) = \Phi </math>
: <math> G(27) = 2^{1/12}3^{-1/3}(\sqrt[3]{2}+1) </math>
: <math> G(29) = 2^{-5/4}3^{-1}(\sqrt{29}-5)^{1/4}[\sqrt{29}+3+(\sqrt{3}+1)\sqrt[3]{2\sqrt{29}+6\sqrt{3}}-(\sqrt{3}-1)\sqrt[3]{2\sqrt{29}-6\sqrt{3}}] </math>
: <math> G(31) = \tfrac{1}{2}\sqrt[4]{18}\operatorname{csch}[\tfrac{1}{3}\operatorname{arsinh}(\tfrac{3}{2}\sqrt{3})] = 2^{1/4} \psi </math>
: <math> G(33) = (10+3\sqrt{11})^{1/12}(2+\sqrt{3})^{1/4} </math>
: <math> G(35) = 2^{-9/4}3^{-1}(\sqrt{5}-1)[(\sqrt{7}+\sqrt{3})\sqrt[3]{\sqrt{35}+3\sqrt{3}}+(\sqrt{7}-\sqrt{3})\sqrt[3]{\sqrt{35}-3\sqrt{3}}+6+2\sqrt{5}] </math>
: <math> G(37) = (\sqrt{37}+6)^{1/4} </math>
: <math> G(39) = 2^{-3/4}\delta_B^{-1/3}(\sqrt{4\delta_B+1}+1) </math>
: <math> G(41) = \frac{1}{8}\sqrt{6\sqrt{41}+38}+\frac{1}{8}\sqrt{14-2\sqrt{41}}+\frac{1}{8}\sqrt[4]{40\sqrt{41}+8}+\frac{1}{8}\sqrt[4]{40\sqrt{41}-248} </math>
: <math> G(43) = \tfrac{1}{4}\sqrt[4]{72}\operatorname{csch}[\tfrac{1}{3}\operatorname{arsinh}(\tfrac{3}{4}\sqrt{3})] </math>
: <math> G(45) = (4+\sqrt{15})^{1/6}\Phi^{3/4} </math>
함수 값 G(47)를 결정하려면 오차 방정식을 풀어야 한다:
: <math> [2^{-1/4}G(47)]^5 - [2^{-1/4}G(47)]^3 - 2[2^{-1/4}G(47)]^2 - 2[2^{-1/4}G(47)] - 1 = 0 </math>

== 참고 자료 ==
* Ramanujan, S. "Modular Equations and Approximations to pi." Quart. J. Pure. Appl. Math. 45, 350-372, 1913-1914.
* Borwein, J. M. and Borwein, P. B. Pi & the AGM: A Study in Analytic Number Theory and Computational Complexity. New York: Wiley, pp. 139 and 298, 1987.
*https://mathworld.wolfram.com/Ramanujang-andG-Functions.html
*https://mathworld.wolfram.com/EllipticLambdaFunction.html
*https://mathworld.wolfram.com/DedekindEtaFunction.html
*https://mathworld.wolfram.com/QuinticEquation.html
*https://oeis.org/A000073
*https://oeis.org/A000930
*https://oeis.org/A000931
*https://oeis.org/A084540
*https://arxiv.org/ftp/arxiv/papers/1601/1601.07110.pdf

[[분류:모듈러 형식]]