라마누잔 g함수와 G함수

수학에서 라마누잔 g함수와 G 함수는 모듈러 함수에서 유도된 함수이다. 이 함수의 이름은 인도의 수학자 스리니바사 라마누잔에서 유래됐으며, g함수와 G함수는 타원 모듈러 람다 함수와 대수적 관계를 갖는다. 모듈러 람다 함수는 합동 부분군 Γ(2)에 대하여 불변인 모듈러 함수이다. 타원 모듈러 람다 함수 λ*(x)의 아크탄젠트를 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. 라마누잔에 따른 정의: g ( x ) = 1 2 4 exp ⁡ ( π 24 x ) ∏ a = 0 ∞ { 1 − exp ⁡ [ − ( 2 a + 1 ) π x ] } {\displaystyle 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}}]\}} G ( x ) = 1 2 4 exp ⁡ ( π 24 x ) ∏ a = 0 ∞ { 1 + exp ⁡ [ − ( 2 a + 1 ) π x ] } {\displaystyle 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}}]\}} 타원 모듈러 람다 함수를 기반으로하는 함수 g(x) 및 G(x)의 정의: g ( x ) = tan ⁡ { 2 arctan ⁡ [ λ ∗ ( x ) ] } − 1 / 12 {\displaystyle g(x)=\tan\{2\arctan[\lambda ^{*}(x)]\}^{-1/12}} G ( x ) = sin ⁡ { 2 arcsin ⁡ [ λ ∗ ( x ) ] } − 1 / 12 {\displaystyle G(x)=\sin\{2\arcsin[\lambda ^{*}(x)]\}^{-1/12}} 데데킨트 에타 함수에 기반한 정의: g ( x ) = 2 − 1 / 4 η ( 1 2 i x ) η ( i x ) − 1 = {\displaystyle g(x)=2^{-1/4}\,\eta ({\tfrac {1}{2}}i{\sqrt {x}})\,\eta (i{\sqrt {x}})^{-1}=} = 2 − 1 / 4 { ∑ a = − ∞ ∞ ( − 1 ) a exp ⁡ [ − ( 6 a + 1 ) 2 π 24 x ] } { ∑ a = − ∞ ∞ ( − 1 ) a exp ⁡ [ − ( 6 a + 1 ) 2 π 12 x ] } − 1 {\displaystyle =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}} G ( x ) = 2 − 1 / 4 η ( 1 2 i x ) − 1 η ( i x ) 2 η ( 2 i x ) − 1 {\displaystyle G(x)=2^{-1/4}\,\eta ({\tfrac {1}{2}}i{\sqrt {x}})^{-1}\,\eta (i{\sqrt {x}})^{2}\,\eta (2i{\sqrt {x}})^{-1}} 역함수를 표현하려면 제 1 종의 완전한 타원 적분만 필요하다: g ⟨ − 1 ⟩ ( x ) = K ( 2 x 6 x 24 + 1 − x 12 ) 2 K ( x 24 + 1 − x 12 ) − 2 {\displaystyle 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}} G ⟨ − 1 ⟩ ( x ) = K ( 1 2 1 + x − 12 + 1 2 1 − x − 12 ) 2 K ( 1 2 1 + x − 12 − 1 2 1 − x − 12 ) − 2 {\displaystyle 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}} 다음 문장이 유효하다: 면: g ( x 1 ∈ A + ) ∈ A + {\displaystyle g(x_{1}\in \mathbb {A} ^{+})\in \mathbb {A} ^{+}} , 그런: g ( x 1 ∈ A + ) ∈ Q + {\displaystyle g(x_{1}\in \mathbb {A} ^{+})\in \mathbb {Q} ^{+}} 면: G ( x 2 ∈ A + ) ∈ A + {\displaystyle G(x_{2}\in \mathbb {A} ^{+})\in \mathbb {A} ^{+}} , 그런: G ( x 2 ∈ A + ) ∈ Q + {\displaystyle G(x_{2}\in \mathbb {A} ^{+})\in \mathbb {Q} ^{+}} 라마누잔 함수 g(x) 과 G(x) 함수는 열린 상반평면 위에서 정칙함수이나. 이 초등이 아닌 초등 함수는 구체적으로 다음과 같은 항등식을 만족한다: G ( x ) = 2 − 1 / 8 g ( x ) − 1 / 2 [ g ( x ) 24 + 1 + g ( x ) 12 ] 1 / 8 {\displaystyle G(x)=2^{-1/8}g(x)^{-1/2}[{\sqrt {g(x)^{24}+1}}+g(x)^{12}]^{1/8}} G ( x ) = G ( 1 / x ) {\displaystyle G(x)=G(1/x)} g ( 4 x ) = 2 1 / 4 g ( x ) G ( x ) {\displaystyle g(4x)=2^{1/4}g(x)G(x)} g ( x ) g ( 4 / x ) = 1 {\displaystyle g(x)g(4/x)=1} g ( 9 x ) 12 − g ( x ) 12 = 2 2 g ( x ) 9 g ( 9 x ) 9 + 2 2 g ( x ) 3 g ( 9 x ) 3 {\displaystyle g(9x)^{12}-g(x)^{12}=2{\sqrt {2}}g(x)^{9}g(9x)^{9}+2{\sqrt {2}}g(x)^{3}g(9x)^{3}} g ( 25 x ) 6 − g ( x ) 6 = 2 g ( x ) 5 g ( 25 x ) 5 + 2 g ( x ) g ( 25 x ) {\displaystyle g(25x)^{6}-g(x)^{6}=2g(x)^{5}g(25x)^{5}+2g(x)g(25x)} g ( 49 x ) 8 + g ( x ) 8 − 7 g ( x ) 4 g ( 49 x ) 4 = 2 2 g ( x ) 7 g ( 49 x ) 7 + 2 2 g ( x ) g ( 49 x ) {\displaystyle g(49x)^{8}+g(x)^{8}-7g(x)^{4}g(49x)^{4}=2{\sqrt {2}}g(x)^{7}g(49x)^{7}+2{\sqrt {2}}g(x)g(49x)} g ( 121 x ) 12 − g ( x ) 12 = {\displaystyle g(121x)^{12}-g(x)^{12}=} = 2 2 g ( x ) g ( 121 x ) [ g ( x ) 2 g ( 121 x ) 2 + 1 ] [ g ( x ) 4 g ( 121 x ) 4 + 3 g ( x ) 2 g ( 121 x ) 2 + 1 ] [ 2 g ( x ) 4 g ( 121 x ) 4 + 3 g ( x ) 2 g ( 121 x ) 2 + 2 ] {\displaystyle =2{\sqrt {2}}g(x)g(121x)[g(x)^{2}g(121x)^{2}+1][g(x)^{4}g(121x)^{4}+3g(x)^{2}g(121x)^{2}+1][2g(x)^{4}g(121x)^{4}+3g(x)^{2}g(121x)^{2}+2]} [ g ( 169 x ) 2 − g ( x ) 2 ] [ g ( 169 x ) 4 + g ( x ) 4 − 7 g ( x ) 2 g ( 169 x ) 2 ] { [ g ( 169 x ) 2 − g ( x ) 2 ] 4 − g ( x ) 2 g ( 169 x ) 2 [ g ( 169 x ) 2 + g ( x ) 2 ] 2 } = {\displaystyle [g(169x)^{2}-g(x)^{2}][g(169x)^{4}+g(x)^{4}-7g(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}\}=} = 8 g ( x ) 13 g ( 169 x ) 13 + 8 g ( x ) g ( 169 x ) {\displaystyle =8g(x)^{13}g(169x)^{13}+8g(x)g(169x)} 중요한 피솟 비자야라가브한 상수: 다음 값이 정확하다: g ( 1 ) = 2 − 1 / 8 {\displaystyle g(1)=2^{-1/8}} g ( 2 ) = 1 {\displaystyle g(2)=1} g ( 3 ) = 2 − 1 / 6 ( 2 + 3 ) 1 / 8 {\displaystyle g(3)=2^{-1/6}(2+{\sqrt {3}})^{1/8}} g ( 4 ) = 2 1 / 8 {\displaystyle g(4)=2^{1/8}} g ( 5 ) = 2 − 1 / 8 ( Φ + 1 ) 1 / 4 {\displaystyle g(5)=2^{-1/8}({\sqrt {\Phi }}+1)^{1/4}} g ( 6 ) = δ S 6 {\displaystyle g(6)={\sqrt[{6}]{\delta _{S}}}} g ( 7 ) = 2 − 1 / 4 ( 8 + 3 7 ) 1 / 8 {\displaystyle g(7)=2^{-1/4}(8+3{\sqrt {7}})^{1/8}} g ( 8 ) = 2 1 / 8 δ S 8 {\displaystyle g(8)=2^{1/8}{\sqrt[{8}]{\delta _{S}}}} g ( 9 ) = 2 − 1 / 8 ( 2 + 3 ) 1 / 24 ( 2 + 3 4 ) 1 / 4 {\displaystyle g(9)=2^{-1/8}(2+{\sqrt {3}})^{1/24}({\sqrt {2}}+{\sqrt[{4}]{3}})^{1/4}} g ( 10 ) = Φ {\displaystyle g(10)={\sqrt {\Phi }}} g ( 14 ) = 1 2 3 + 2 + 1 2 2 − 1 {\displaystyle g(14)={\frac {1}{2}}{\sqrt {3+{\sqrt {2}}}}+{\frac {1}{2}}{\sqrt {{\sqrt {2}}-1}}} g ( 18 ) = ( 3 + 2 ) 1 / 3 {\displaystyle g(18)=({\sqrt {3}}+{\sqrt {2}})^{1/3}} g ( 22 ) = δ S {\displaystyle g(22)={\sqrt {\delta _{S}}}} g ( 26 ) = 1 6 2 13 − 2 ( 3 + 2 3 3 + 26 6 − 3 − 2 3 3 − 26 6 ) + 1 3 13 + 2 {\displaystyle 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}}} g ( 30 ) = ( 10 + 3 ) 1 / 6 Φ {\displaystyle g(30)=({\sqrt {10}}+3)^{1/6}{\sqrt {\Phi }}} g ( 34 ) = 1 4 14 + 2 17 + 1 4 2 17 − 2 {\displaystyle g(34)={\frac {1}{4}}{\sqrt {14+2{\sqrt {17}}}}+{\frac {1}{4}}{\sqrt {2{\sqrt {17}}-2}}} g ( 38 ) = 1 6 δ S − 1 / 2 ( 2 57 + 14 3 57 + 16 2 6 + 2 57 − 14 3 57 − 16 2 6 + 4 + 2 2 ) {\displaystyle 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}})} g ( 42 ) = 2 − 1 / 2 ( 7 + 3 ) 1 / 2 ( 2 2 + 7 ) 1 / 6 {\displaystyle g(42)=2^{-1/2}({\sqrt {7}}+{\sqrt {3}})^{1/2}(2{\sqrt {2}}+{\sqrt {7}})^{1/6}} g ( 46 ) = 1 2 5 + 2 + 1 2 2 + 1 {\displaystyle g(46)={\frac {1}{2}}{\sqrt {5+{\sqrt {2}}}}+{\frac {1}{2}}{\sqrt {{\sqrt {2}}+1}}} g ( 50 ) = 1 2 { 4 3 2 cos ⁡ ( π 10 ) cosh ⁡ [ 1 3 arcosh ⁡ ( 4 5 10 ) ] + 1 3 tan ⁡ ( π 5 ) } 2 − 1 2 {\displaystyle 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}}} g ( 54 ) = ( δ S 3 / 2 + 2 δ S 7 / 6 + 2 δ S 5 / 6 ) 1 / 3 {\displaystyle g(54)=(\delta _{S}^{3/2}+{\sqrt {2}}\delta _{S}^{7/6}+{\sqrt {2}}\delta _{S}^{5/6})^{1/3}} g ( 58 ) = 2 − 1 / 2 ( 29 + 5 ) 1 / 2 {\displaystyle g(58)=2^{-1/2}({\sqrt {29}}+5)^{1/2}} g ( 62 ) = cot ⁡ [ 1 2 arcsin ⁡ ( 1 4 28 2 + 38 − 3 2 − 1 4 2 ) ] 1 / 2 {\displaystyle 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}} g ( 66 ) = cot ⁡ { 1 2 arcsin ⁡ [ 1 8 ( 8 + 3 2 − 66 ) 33 + 4 2 4 − 1 8 ( 8 − 3 2 + 66 ) 33 − 4 2 4 ] } 1 / 3 {\displaystyle 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}} g ( 70 ) = Φ δ S {\displaystyle g(70)=\Phi {\sqrt {\delta _{S}}}} 함수 값 g(74)를 결정하려면 오차 방정식을 풀어야 한다: g ( 74 ) 5 + g ( 74 ) 3 + g ( 74 ) − 37 + 6 [ g ( 74 ) 4 − g ( 74 ) 2 + 1 ] = 0 {\displaystyle g(74)^{5}+g(74)^{3}+g(74)-{\sqrt {{\sqrt {37}}+6}}\left[g(74)^{4}-g(74)^{2}+1\right]=0} 다음 추가 값이 정확하다: G ( 1 ) = 1 {\displaystyle G(1)=1} G ( 2 ) = 2 − 1 / 8 δ S 8 {\displaystyle G(2)=2^{-1/8}{\sqrt[{8}]{\delta _{S}}}} G ( 3 ) = 2 1 / 12 {\displaystyle G(3)=2^{1/12}} G ( 4 ) = 2 − 3 / 16 δ S 4 {\displaystyle G(4)=2^{-3/16}{\sqrt[{4}]{\delta _{S}}}} G ( 5 ) = Φ 4 {\displaystyle G(5)={\sqrt[{4}]{\Phi }}} G ( 6 ) = 2 − 1 / 8 ( 2 + 3 ) 1 / 8 ( 3 + 2 ) 1 / 8 δ S − 1 / 12 {\displaystyle G(6)=2^{-1/8}(2+{\sqrt {3}})^{1/8}({\sqrt {3}}+{\sqrt {2}})^{1/8}\delta _{S}^{-1/12}} G ( 7 ) = 2 1 / 4 {\displaystyle G(7)=2^{1/4}} G ( 8 ) = 2 − 3 / 16 δ S 1 / 16 ( δ S + 2 ) 1 / 4 {\displaystyle G(8)=2^{-3/16}\delta _{S}^{1/16}({\sqrt {\delta _{S}}}+{\sqrt {2}})^{1/4}} G ( 9 ) = ( 2 + 3 ) 1 / 6 {\displaystyle G(9)=(2+{\sqrt {3}})^{1/6}} G ( 10 ) = 2 − 1 / 8 ( 10 + 3 ) 1 / 8 Φ − 1 / 4 δ S 1 / 4 {\displaystyle G(10)=2^{-1/8}({\sqrt {10}}+3)^{1/8}\Phi ^{-1/4}\delta _{S}^{1/4}} G ( 11 ) = 2 − 1 / 4 3 − 1 ( 3 33 + 17 3 − 3 33 − 17 3 + 2 ) = 2 − 1 / 4 ( T T R I 2 − T T R I ) {\displaystyle 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})} G ( 13 ) = δ B 4 {\displaystyle G(13)={\sqrt[{4}]{\delta _{B}}}} G ( 15 ) = 2 1 / 4 Φ 3 {\displaystyle G(15)=2^{1/4}{\sqrt[{3}]{\Phi }}} G ( 17 ) = 1 4 10 + 2 17 + 1 4 2 17 − 6 {\displaystyle G(17)={\frac {1}{4}}{\sqrt {10+2{\sqrt {17}}}}+{\frac {1}{4}}{\sqrt {2{\sqrt {17}}-6}}} G ( 19 ) = 2 3 18 4 cosh ⁡ [ 1 3 arcosh ⁡ ( 3 4 6 ) ] {\displaystyle G(19)={\tfrac {2}{3}}{\sqrt[{4}]{18}}\cosh[{\tfrac {1}{3}}\operatorname {arcosh} ({\tfrac {3}{4}}{\sqrt {6}})]} G ( 21 ) = 2 − 1 / 4 ( 8 + 3 7 ) 1 / 12 ( 7 + 3 ) 1 / 4 {\displaystyle G(21)=2^{-1/4}(8+3{\sqrt {7}})^{1/12}({\sqrt {7}}+{\sqrt {3}})^{1/4}} G ( 23 ) = 2 3 18 4 cosh ⁡ [ 1 3 arcosh ⁡ ( 3 2 3 ) ] = 2 1 / 4 ρ {\displaystyle G(23)={\tfrac {2}{3}}{\sqrt[{4}]{18}}\cosh[{\tfrac {1}{3}}\operatorname {arcosh} ({\tfrac {3}{2}}{\sqrt {3}})]=2^{1/4}\rho } G ( 25 ) = Φ {\displaystyle G(25)=\Phi } G ( 27 ) = 2 1 / 12 3 − 1 / 3 ( 2 3 + 1 ) {\displaystyle G(27)=2^{1/12}3^{-1/3}({\sqrt[{3}]{2}}+1)} G ( 29 ) = 2 − 5 / 4 3 − 1 ( 29 − 5 ) 1 / 4 [ 29 + 3 + ( 3 + 1 ) 2 29 + 6 3 3 − ( 3 − 1 ) 2 29 − 6 3 3 ] {\displaystyle 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}}}}]} G ( 31 ) = 1 2 18 4 csch ⁡ [ 1 3 arsinh ⁡ ( 3 2 3 ) ] = 2 1 / 4 ψ {\displaystyle G(31)={\tfrac {1}{2}}{\sqrt[{4}]{18}}\operatorname {csch} [{\tfrac {1}{3}}\operatorname {arsinh} ({\tfrac {3}{2}}{\sqrt {3}})]=2^{1/4}\psi } G ( 33 ) = ( 10 + 3 11 ) 1 / 12 ( 2 + 3 ) 1 / 4 {\displaystyle G(33)=(10+3{\sqrt {11}})^{1/12}(2+{\sqrt {3}})^{1/4}} G ( 35 ) = 2 − 9 / 4 3 − 1 ( 5 − 1 ) [ ( 7 + 3 ) 35 + 3 3 3 + ( 7 − 3 ) 35 − 3 3 3 + 6 + 2 5 ] {\displaystyle 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}}]} G ( 37 ) = ( 37 + 6 ) 1 / 4 {\displaystyle G(37)=({\sqrt {37}}+6)^{1/4}} G ( 39 ) = 2 − 3 / 4 δ B − 1 / 3 ( 4 δ B + 1 + 1 ) {\displaystyle G(39)=2^{-3/4}\delta _{B}^{-1/3}({\sqrt {4\delta _{B}+1}}+1)} G ( 41 ) = 1 8 6 41 + 38 + 1 8 14 − 2 41 + 1 8 40 41 + 8 4 + 1 8 40 41 − 248 4 {\displaystyle 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}}} G ( 43 ) = 1 4 72 4 csch ⁡ [ 1 3 arsinh ⁡ ( 3 4 3 ) ] {\displaystyle G(43)={\tfrac {1}{4}}{\sqrt[{4}]{72}}\operatorname {csch} [{\tfrac {1}{3}}\operatorname {arsinh} ({\tfrac {3}{4}}{\sqrt {3}})]} G ( 45 ) = ( 4 + 15 ) 1 / 6 Φ 3 / 4 {\displaystyle G(45)=(4+{\sqrt {15}})^{1/6}\Phi ^{3/4}} 함수 값 G(47)를 결정하려면 오차 방정식을 풀어야 한다: [ 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 {\displaystyle [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} 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