Definición Representación de la aproximación q 1 / 5 A 400 ( q ) / B 400 ( q ) {\displaystyle q^{1/5}A_{400}(q)/B_{400}(q)} Dadas las funcionesGRAMO ( q ) {\displaystyle G(q)} H ( q ) {\displaystyle H(q)} q = mi 2 π i τ {\displaystyle q=e^{2\pi i\tau }}
GRAMO ( q ) = ∑ norte = 0 ∞ q norte 2 ( 1 − q ) ( 1 − q 2 ) ⋯ ( 1 − q norte ) = ∑ norte = 0 ∞ q norte 2 ( q ; q ) norte = 1 ( q ; q 5 ) ∞ ( q 4 ; q 5 ) ∞ = ∏ norte = 1 ∞ 1 ( 1 − q 5 norte − 1 ) ( 1 − q 5 norte − 4 ) = q j 60 2 F 1 ( − 1 60 , 19 60 ; 4 5 ; 1728 j ) = q ( j − 1728 ) 60 2 F 1 ( − 1 60 , 29 60 ; 4 5 ; − 1728 j − 1728 ) = 1 + q + q 2 + q 3 + 2 q 4 + 2 q 5 + 3 q 6 + ⋯ {\displaystyle {\begin{aligned}G(q)&=\sum _{n=0}^{\infty }{\frac {q^{n^{2}}}{(1-q)(1-q^{2})\cdots (1-q^{n})}}=\sum _{n=0}^{\infty }{\frac {q^{n^{2}}}{(q;q)_{n}}}={\frac {1}{(q;q^{5})_{\infty }(q^{4};q^{5})_{\infty }}}\\[6pt]&=\prod _{n=1}^{\infty }{\frac {1}{(1-q^{5n-1})(1-q^{5n-4})}}\\[6pt]&={\sqrt[{60}]{q\,j}}\,\,_{2}F_{1}\left(-{\tfrac {1}{60}},{\tfrac {19}{60}};{\tfrac {4}{5}};{\tfrac {1728}{j}}\right)\\[6pt]&={\sqrt[{60}]{q\left(j-1728\right)}}\,_{2}F_{1}\left(-{\tfrac {1}{60}},{\tfrac {29}{60}};{\tfrac {4}{5}};-{\tfrac {1728}{j-1728}}\right)\\[6pt]&=1+q+q^{2}+q^{3}+2q^{4}+2q^{5}+3q^{6}+\cdots \end{aligned}}} y,
H ( q ) = ∑ norte = 0 ∞ q norte 2 + norte ( 1 − q ) ( 1 − q 2 ) ⋯ ( 1 − q norte ) = ∑ norte = 0 ∞ q norte 2 + norte ( q ; q ) norte = 1 ( q 2 ; q 5 ) ∞ ( q 3 ; q 5 ) ∞ = ∏ norte = 1 ∞ 1 ( 1 − q 5 norte − 2 ) ( 1 − q 5 norte − 3 ) = 1 q 11 j 11 60 2 F 1 ( 11 60 , 31 60 ; 6 5 ; 1728 j ) = 1 q 11 ( j − 1728 ) 11 60 2 F 1 ( 11 60 , 41 60 ; 6 5 ; − 1728 j − 1728 ) = 1 + q 2 + q 3 + q 4 + q 5 + 2 q 6 + 2 q 7 + ⋯ {\displaystyle {\begin{aligned}H(q)&=\sum _{n=0}^{\infty }{\frac {q^{n^{2}+n}}{(1-q)(1-q^{2})\cdots (1-q^{n})}}=\sum _{n=0}^{\infty }{\frac {q^{n^{2}+n}}{(q;q)_{n}}}={\frac {1}{(q^{2};q^{5})_{\infty }(q^{3};q^{5})_{\infty }}}\\[6pt]&=\prod _{n=1}^{\infty }{\frac {1}{(1-q^{5n-2})(1-q^{5n-3})}}\\[6pt]&={\frac {1}{\sqrt[{60}]{q^{11}j^{11}}}}\,_{2}F_{1}\left({\tfrac {11}{60}},{\tfrac {31}{60}};{\tfrac {6}{5}};{\tfrac {1728}{j}}\right)\\[6pt]&={\frac {1}{\sqrt[{60}]{q^{11}\left(j-1728\right)^{11}}}}\,_{2}F_{1}\left({\tfrac {11}{60}},{\tfrac {41}{60}};{\tfrac {6}{5}};-{\tfrac {1728}{j-1728}}\right)\\[6pt]&=1+q^{2}+q^{3}+q^{4}+q^{5}+2q^{6}+2q^{7}+\cdots \end{aligned}}} con los coeficientes de la expansión q siendo OEIS : A003114 OEIS : A003106 ( a ; q ) ∞ {\displaystyle (a;q)_{\infty }} denota el símbolo q-Pochhammer infinito , j es la j-función y 2 F 1 es la función hipergeométrica . La fracción continua de Rogers-Ramanujan es entonces
R ( q ) = q 11 60 H ( q ) q − 1 60 GRAMO ( q ) = q 1 5 ∏ norte = 1 ∞ ( 1 − q 5 norte − 1 ) ( 1 − q 5 norte − 4 ) ( 1 − q 5 norte − 2 ) ( 1 − q 5 norte − 3 ) = q 1 / 5 ∏ norte = 1 ∞ ( 1 − q norte ) ( norte | 5 ) = q 1 / 5 1 + q 1 + q 2 1 + q 3 1 + ⋱ {\displaystyle {\begin{aligned}R(q)&={\frac {q^{\frac {11}{60}}H(q)}{q^{-{\frac {1}{60}}}G(q)}}=q^{\frac {1}{5}}\prod _{n=1}^{\infty }{\frac {(1-q^{5n-1})(1-q^{5n-4})}{(1-q^{5n-2})(1-q^{5n-3})}}=q^{1/5}\prod _{n=1}^{\infty }(1-q^{n})^{(n|5)}\\[8pt]&={\cfrac {q^{1/5}}{1+{\cfrac {q}{1+{\cfrac {q^{2}}{1+{\cfrac {q^{3}}{1+\ddots }}}}}}}}\end{aligned}}} ( norte ∣ metro ) {\displaystyle (n\mid m)} símbolo jacobino .Hay que tener cuidado con la notación, ya que las fórmulas que emplean la función j j {\displaystyle j} q = mi 2 π i τ {\displaystyle q=e^{2\pi i\tau }} nomo ) se utiliza a lo largo de esta sección ya que la expansión q de la función j (así como la conocida función eta de Dedekind ) utilizaq = mi 2 π i τ {\displaystyle q=e^{2\pi i\tau }} nombre q = mi π i τ {\displaystyle q=e^{\pi i\tau }}
Valores especiales Si q es el nomo o su cuadrado, entoncesq − 1 60 GRAMO ( q ) {\displaystyle q^{-{\frac {1}{60}}}G(q)} q 11 60 H ( q ) {\displaystyle q^{\frac {11}{60}}H(q)} R ( q ) {\displaystyle R(q)} funciones modulares deτ {\displaystyle \tau } la multiplicación compleja implica que sus valores paraτ {\displaystyle \tau } Los números algebraicos que involucran un campo cuadrático imaginario pueden evaluarse explícitamente.
Dada la forma general en la que Ramanujan usó el nombre q = mi π i τ {\displaystyle q=e^{\pi i\tau }}
R ( q ) = q 1 / 5 1 + q 1 + q 2 1 + q 3 1 + ⋱ {\displaystyle R(q)={\cfrac {q^{1/5}}{1+{\cfrac {q}{1+{\cfrac {q^{2}}{1+{\cfrac {q^{3}}{1+\ddots }}}}}}}}} f cuandoτ = i {\displaystyle \tau =i}
R ( mi − π ) = mi − π 5 1 + mi − π 1 + mi − 2 π 1 + ⋱ = 1 2 φ ( 5 − φ 3 / 2 ) ( 5 4 + φ 3 / 2 ) = 0,511428 … {\displaystyle R{\big (}e^{-\pi }{\big )}={\cfrac {e^{-{\frac {\pi }{5}}}}{1+{\cfrac {e^{-\pi }}{1+{\cfrac {e^{-2\pi }}{1+\ddots }}}}}}={\tfrac {1}{2}}\varphi \,({\sqrt {5}}-\varphi ^{3/2})({\sqrt[{4}]{5}}+\varphi ^{3/2})=0.511428\dots } cuandoτ = 2 i {\displaystyle \tau =2i}
R ( mi − 2 π ) = mi − 2 π 5 1 + mi − 2 π 1 + mi − 4 π 1 + ⋱ = 5 4 φ 1 / 2 − φ = 0,284079 … {\displaystyle R{\big (}e^{-2\pi }{\big )}={\cfrac {e^{-{\frac {2\pi }{5}}}}{1+{\cfrac {e^{-2\pi }}{1+{\cfrac {e^{-4\pi }}{1+\ddots }}}}}}={{\sqrt[{4}]{5}}\,\varphi ^{1/2}-\varphi }=0.284079\dots } cuandoτ = 4 i {\displaystyle \tau =4i}
R ( mi − 4 π ) = mi − 4 π 5 1 + mi − 4 π 1 + mi − 8 π 1 + ⋱ = 1 2 φ ( 5 − φ 3 / 2 ) ( − 5 4 + φ 3 / 2 ) = 0,081002 … {\displaystyle R{\big (}e^{-4\pi }{\big )}={\cfrac {e^{-{\frac {4\pi }{5}}}}{1+{\cfrac {e^{-4\pi }}{1+{\cfrac {e^{-8\pi }}{1+\ddots }}}}}}={\tfrac {1}{2}}\varphi \,({\sqrt {5}}-\varphi ^{3/2})(-{\sqrt[{4}]{5}}+\varphi ^{3/2})=0.081002\dots } cuandoτ = 2 5 i {\displaystyle \tau =2{\sqrt {5}}i}
R ( mi − 2 5 π ) = mi − 2 π 5 1 + mi − 2 π 5 1 + mi − 4 π 5 1 + ⋱ = 5 1 + ( 5 3 / 4 ( φ − 1 ) 5 / 2 − 1 ) 1 / 5 − φ = 0,0602094 … {\displaystyle R{\big (}e^{-2{\sqrt {5}}\pi }{\big )}={\cfrac {e^{-{\frac {2\pi }{\sqrt {5}}}}}{1+{\cfrac {e^{-2\pi {\sqrt {5}}}}{1+{\cfrac {e^{-4\pi {\sqrt {5}}}}{1+\ddots }}}}}}={\frac {\sqrt {5}}{1+{\big (}5^{3/4}(\varphi -1)^{5/2}-1{\big )}^{1/5}}}-\varphi =0.0602094\dots } cuandoτ = 5 i {\displaystyle \tau =5i}
R ( mi − 5 π ) = mi − π 1 + mi − 5 π 1 + mi − 10 π 1 + ⋱ = 1 + φ 2 φ + ( 1 2 ( 4 − φ − 3 φ − 1 ) ( 3 φ 3 / 2 − 5 4 ) ) 1 / 5 − φ = 0,0432139 … {\displaystyle R{\big (}e^{-5\pi }{\big )}={\cfrac {e^{-\pi }}{1+{\cfrac {e^{-5\pi }}{1+{\cfrac {e^{-10\pi }}{1+\ddots }}}}}}={\frac {1+\varphi ^{2}}{\varphi +{\big (}{\frac {1}{2}}(4-\varphi -3{\sqrt {\varphi -1}})(3\varphi ^{3/2}-{\sqrt[{4}]{5}}){\big )}^{1/5}}}-\varphi =0.0432139\dots } cuandoτ = 10 i {\displaystyle \tau =10i}
R ( mi − 10 π ) = mi − 2 π 1 + mi − 10 π 1 + mi − 20 π 1 + ⋱ = 1 + φ 2 φ + ( 3 1 + φ 2 − 4 − φ ) 1 / 5 − φ = 0,00186744 … {\displaystyle R{\big (}e^{-10\pi }{\big )}={\cfrac {e^{-2\pi }}{1+{\cfrac {e^{-10\pi }}{1+{\cfrac {e^{-20\pi }}{1+\ddots }}}}}}={\frac {1+\varphi ^{2}}{\varphi +{\big (}3{\sqrt {1+\varphi ^{2}}}-4-\varphi {\big )}^{1/5}}}-\varphi =0.00186744\dots } cuandoτ = 20 i {\displaystyle \tau =20i}
R ( mi − 20 π ) = mi − 4 π 1 + mi − 20 π 1 + mi − 40 π 1 + ⋱ = 1 + φ 2 φ + ( 1 2 ( 4 − φ − 3 φ − 1 ) ( 3 φ 3 / 2 + 5 4 ) ) 1 / 5 − φ = 0,00000348734 … {\displaystyle R{\big (}e^{-20\pi }{\big )}={\cfrac {e^{-4\pi }}{1+{\cfrac {e^{-20\pi }}{1+{\cfrac {e^{-40\pi }}{1+\ddots }}}}}}={\frac {1+\varphi ^{2}}{\varphi +{\big (}{\frac {1}{2}}(4-\varphi -3{\sqrt {\varphi -1}})(3\varphi ^{3/2}+{\sqrt[{4}]{5}}){\big )}^{1/5}}}-\varphi =0.00000348734\dots } yφ = 1 + 5 2 {\displaystyle \varphi ={\tfrac {1+{\sqrt {5}}}{2}}} proporción áurea . Tenga en cuenta queR ( mi − 2 π ) {\displaystyle R{\big (}e^{-2\pi }{\big )}} ecuación cuártica ,
incógnita 4 + 2 incógnita 3 − 6 incógnita 2 − 2 incógnita + 1 = 0 {\displaystyle x^{4}+2x^{3}-6x^{2}-2x+1=0} mientrasR ( mi − π ) {\displaystyle R{\big (}e^{-\pi }{\big )}} R ( mi − 4 π ) {\displaystyle R{\big (}e^{-4\pi }{\big )}} óctico ,
y 4 + 2 φ 4 y 3 + 6 φ 2 y 2 − 2 φ 4 y + 1 = 0 {\displaystyle y^{4}+2\varphi ^{4}y^{3}+6\varphi ^{2}y^{2}-2\varphi ^{4}y+1=0} (desdeφ {\displaystyle \varphi } un entero positivo m , entoncesR ( mi − 2 π / metro ) {\displaystyle R(e^{-2\pi /m})} R ( mi − 2 π metro ) {\displaystyle R(e^{-2\pi \,m})}
[ R ( mi − 2 π / metro ) + φ ] [ R ( mi − 2 π metro ) + φ ] = 5 φ {\displaystyle {\bigl [}R(e^{-2\pi /m})+\varphi {\bigr ]}{\bigl [}R(e^{-2\pi \,m})+\varphi {\bigr ]}={\sqrt {5}}\,\varphi } El grado algebraico k deR ( mi − π norte ) {\displaystyle R(e^{-\pi \,n})} norte = 1 , 2 , 3 , 4 , … {\displaystyle n=1,2,3,4,\dots } k = 8 , 4 , 32 , 8 , … {\displaystyle k=8,4,32,8,\dots } OEIS : A082682
Por cierto, estas fracciones continuas pueden utilizarse para resolver algunas ecuaciones de quinto grado, como se muestra en una sección posterior.
G ( q ) y H ( q )Curiosamente, existen fórmulas explícitas paraGRAMO ( q ) {\displaystyle G(q)} H ( q ) {\displaystyle H(q)} función j j ( τ ) {\displaystyle j(\tau )} R ( q ) {\displaystyle R(q)} j ( τ ) {\displaystyle j(\tau )} q = mi 2 π i τ {\displaystyle q=e^{2\pi \,i\tau }} j ( τ ) , GRAMO ( q ) , H ( q ) {\displaystyle j(\tau ),\,G(q),\,H(q)} r = R ( q ) {\displaystyle r=R(q)} q {\displaystyle q}
GRAMO ( q ) = ∏ norte = 1 ∞ 1 ( 1 − q 5 norte − 1 ) ( 1 − q 5 norte − 4 ) = q 1 / 60 j ( τ ) 1 / 60 ( r 20 − 228 r 15 + 494 r 10 + 228 r 5 + 1 ) 1 / 20 {\displaystyle {\begin{aligned}G(q)&=\prod _{n=1}^{\infty }{\frac {1}{(1-q^{5n-1})(1-q^{5n-4})}}\\[6pt]&=q^{1/60}{\frac {j(\tau )^{1/60}}{(r^{20}-228r^{15}+494r^{10}+228r^{5}+1)^{1/20}}}\end{aligned}}} H ( q ) = ∏ norte = 1 ∞ 1 ( 1 − q 5 norte − 2 ) ( 1 − q 5 norte − 3 ) = − 1 q 11 / 60 ( r 20 − 228 r 15 + 494 r 10 + 228 r 5 + 1 ) 11 / 20 j ( τ ) 11 / 60 ( r 10 + 11 r 5 − 1 ) {\displaystyle {\begin{aligned}H(q)&=\prod _{n=1}^{\infty }{\frac {1}{(1-q^{5n-2})(1-q^{5n-3})}}\\[6pt]&={\frac {-1}{q^{11/60}}}{\frac {(r^{20}-228r^{15}+494r^{10}+228r^{5}+1)^{11/20}}{j(\tau )^{11/60}\,(r^{10}+11r^{5}-1)}}\end{aligned}}} Por supuesto, las fórmulas secundarias implican queq − 1 / 60 GRAMO ( q ) {\displaystyle q^{-1/60}G(q)} q 11 / 60 H ( q ) {\displaystyle q^{11/60}H(q)} números algebraicos (aunque normalmente de alto grado) paraτ {\displaystyle \tau } campo cuadrático imaginario . Por ejemplo, las fórmulas anteriores se simplifican a:
GRAMO ( mi − 2 π ) = ( mi − 2 π ) 1 / 60 1 ( 5 φ ) 1 / 4 1 R ( mi − 2 π ) = 1.00187093 … H ( mi − 2 π ) = 1 ( mi − 2 π ) 11 / 60 1 ( 5 φ ) 1 / 4 R ( mi − 2 π ) = 1.00000349 … {\displaystyle {\begin{aligned}G(e^{-2\pi })&=(e^{-2\pi })^{1/60}{\frac {1}{(5\,\varphi )^{1/4}}}{\frac {1}{\sqrt {R(e^{-2\pi })}}}\\[6pt]&=1.00187093\dots \\[6pt]H(e^{-2\pi })&={\frac {1}{(e^{-2\pi })^{11/60}}}{\frac {1}{(5\,\varphi )^{1/4}}}{\sqrt {R(e^{-2\pi })}}\\[6pt]&=1.00000349\ldots \end{aligned}}} y,
GRAMO ( mi − 4 π ) = ( mi − 4 π ) 1 / 60 1 ( 5 φ 3 ) 1 / 4 ( φ + 5 4 ) 1 / 4 1 R ( mi − 4 π ) = 1.000003487354 … H ( mi − 4 π ) = 1 ( mi − 4 π ) 11 / 60 1 ( 5 φ 3 ) 1 / 4 ( φ + 5 4 ) 1 / 4 R ( mi − 4 π ) = 1.000000000012 … {\displaystyle {\begin{aligned}G(e^{-4\pi })&=(e^{-4\pi })^{1/60}{\frac {1}{(5\,\varphi ^{3})^{1/4}\,(\varphi +{\sqrt[{4}]{5}})^{1/4}}}{\frac {1}{\sqrt {R(e^{-4\pi })}}}\\[6pt]&=1.000003487354\dots \\[6pt]H(e^{-4\pi })&={\frac {1}{(e^{-4\pi })^{11/60}}}{\frac {1}{(5\,\varphi ^{3})^{1/4}\,(\varphi +{\sqrt[{4}]{5}})^{1/4}}}{\sqrt {R(e^{-4\pi })}}\\[6pt]&=1.000000000012\dots \end{aligned}}} y así sucesivamente, conφ {\displaystyle \varphi }
Derivación de valores especiales
sumas tangenciales A continuación, expresamos los teoremas esenciales de las fracciones continuas R y S de Rogers-Ramanujan utilizando sumas tangenciales y diferencias tangenciales:
a ⊕ b = broncearse [ arctan ( a ) + arctan ( b ) ] = a + b 1 − a b {\displaystyle a\oplus b=\tan {\bigl [}\arctan(a)+\arctan(b){\bigr ]}={\frac {a+b}{1-ab}}} do ⊖ d = broncearse [ arctan ( do ) − arctan ( d ) ] = do − d 1 + do d {\displaystyle c\ominus d=\tan {\bigl [}\arctan(c)-\arctan(d){\bigr ]}={\frac {c-d}{1+cd}}} El nomo elíptico y el nomo complementario tienen esta relación entre sí:
ln ( q ) ln ( q 1 ) = π 2 {\displaystyle \ln(q)\ln(q_{1})=\pi ^{2}} El nomo complementario de un módulo k es igual al nomo del módulo complementario pitagórico:
q 1 ( k ) = q ( k ′ ) = q ( 1 − k 2 ) {\displaystyle q_{1}(k)=q(k')=q({\sqrt {1-k^{2}}})} Estos son los teoremas de reflexión para las fracciones continuas R y S:
La cartaΦ {\displaystyle \Phi } número áureo :
Φ = 1 2 ( 5 + 1 ) = cuna [ 1 2 arctan ( 2 ) ] = 2 porque ( 1 5 π ) {\displaystyle \Phi ={\tfrac {1}{2}}({\sqrt {5}}+1)=\cot[{\tfrac {1}{2}}\arctan(2)]=2\cos({\tfrac {1}{5}}{\pi })} Φ − 1 = 1 2 ( 5 − 1 ) = broncearse [ 1 2 arctan ( 2 ) ] = 2 pecado ( 1 10 π ) {\displaystyle \Phi ^{-1}={\tfrac {1}{2}}({\sqrt {5}}-1)=\tan[{\tfrac {1}{2}}\arctan(2)]=2\sin({\tfrac {1}{10}}{\pi })} Los teoremas para el nomo cuadrado se construyen de la siguiente manera:
A continuación se presentan las relaciones entre las fracciones continuas y las funciones theta de Jacobi:
Derivación de valores lemniscáticos En los teoremas que se muestran ahora se insertan ciertos valores:
S [ exp ( − π ) ] ⊕ S [ exp ( − π ) ] = Φ {\displaystyle S{\bigl [}\exp(-\pi ){\bigr ]}\oplus S{\bigl [}\exp(-\pi ){\bigr ]}=\Phi } Por lo tanto, la siguiente identidad es válida:
En un patrón análogo obtenemos este resultado:
R [ exp ( − 2 π ) ] ⊕ R [ exp ( − 2 π ) ] = Φ − 1 {\displaystyle R{\bigl [}\exp(-2\pi ){\bigr ]}\oplus R{\bigl [}\exp(-2\pi ){\bigr ]}=\Phi ^{-1}} Por lo tanto, la siguiente identidad es válida:
Además, obtenemos la misma relación utilizando el teorema mencionado anteriormente sobre las funciones theta de Jacobi:
S [ exp ( − π ) ] ⊕ R [ exp ( − 2 π ) ] = S ( q ) ⊕ R ( q 2 ) [ q = exp ( − π ) ] = {\displaystyle S{\bigl [}\exp(-\pi ){\bigr ]}\oplus R{\bigl [}\exp(-2\pi ){\bigr ]}=S(q)\oplus R(q^{2}){\bigl [}q=\exp(-\pi ){\bigr ]}=} = ϑ 00 ( q 1 / 5 ) 2 − ϑ 00 ( q ) 2 5 ϑ 00 ( q 5 ) 2 − ϑ 00 ( q ) 2 [ q = exp ( − π ) ] = 1 {\displaystyle ={\frac {\vartheta _{00}(q^{1/5})^{2}-\vartheta _{00}(q)^{2}}{5\,\vartheta _{00}(q^{5})^{2}-\vartheta _{00}(q)^{2}}}{\bigl [}q=\exp(-\pi ){\bigr ]}=1} Este resultado aparece debido a la fórmula de suma de Poisson y esta ecuación se puede resolver de esta manera:
R [ exp ( − 2 π ) ] = 1 ⊖ S [ exp ( − π ) ] = 1 ⊖ broncearse [ 1 4 π − 1 4 arctan ( 2 ) ] = broncearse [ 1 4 arctan ( 2 ) ] {\displaystyle R{\bigl [}\exp(-2\pi ){\bigr ]}=1\ominus S{\bigl [}\exp(-\pi ){\bigr ]}=1\ominus \tan {\bigl [}{\tfrac {1}{4}}\pi -{\tfrac {1}{4}}\arctan(2){\bigr ]}=\tan {\bigl [}{\tfrac {1}{4}}\arctan(2){\bigr ]}} Aplicando el otro teorema mencionado sobre las funciones theta de Jacobi, se puede determinar el siguiente valor:
R [ exp ( − π ) ] ⊖ R [ exp ( − 2 π ) ] = R ( q ) ⊖ R ( q 2 ) [ q = exp ( − π ) ] = {\displaystyle R{\bigl [}\exp(-\pi ){\bigr ]}\ominus R{\bigl [}\exp(-2\pi ){\bigr ]}=R(q)\ominus R(q^{2}){\bigl [}q=\exp(-\pi ){\bigr ]}=} = ϑ 01 ( q ) 2 − ϑ 01 ( q 1 / 5 ) 2 5 ϑ 01 ( q 5 ) 2 − ϑ 01 ( q ) 2 [ q = exp ( − π ) ] = 5 4 − 1 5 4 + 1 = 5 4 ⊖ 1 = broncearse [ arctan ( 5 4 ) − 1 4 π ] {\displaystyle ={\frac {\vartheta _{01}(q)^{2}-\vartheta _{01}(q^{1/5})^{2}}{5\,\vartheta _{01}(q^{5})^{2}-\vartheta _{01}(q)^{2}}}{\bigl [}q=\exp(-\pi ){\bigr ]}={\frac {{\sqrt[{4}]{5}}-1}{{\sqrt[{4}]{5}}+1}}={\sqrt[{4}]{5}}\ominus 1=\tan {\bigl [}\arctan({\sqrt[{4}]{5}}\,)-{\tfrac {1}{4}}\pi {\bigr ]}} Esa cadena de ecuaciones conduce a esta suma tangencial:
R [ exp ( − π ) ] = R [ exp ( − 2 π ) ] ⊕ broncearse [ arctan ( 5 4 ) − 1 4 π ] {\displaystyle R{\bigl [}\exp(-\pi ){\bigr ]}=R{\bigl [}\exp(-2\pi ){\bigr ]}\oplus \tan {\bigl [}\arctan({\sqrt[{4}]{5}}\,)-{\tfrac {1}{4}}\pi {\bigr ]}} Y por lo tanto aparece el siguiente resultado:
En el siguiente paso volvemos a usar el teorema de reflexión para la fracción continua R:
R [ exp ( − π ) ] ⊕ R [ exp ( − 4 π ) ] = Φ − 1 {\displaystyle R{\bigl [}\exp(-\pi ){\bigr ]}\oplus R{\bigl [}\exp(-4\pi ){\bigr ]}=\Phi ^{-1}} R [ exp ( − 4 π ) ] = broncearse [ 1 2 arctan ( 2 ) ] ⊖ R [ exp ( − π ) ] {\displaystyle R{\bigl [}\exp(-4\pi ){\bigr ]}=\tan {\bigl [}{\tfrac {1}{2}}\arctan(2){\bigr ]}\ominus R{\bigl [}\exp(-\pi ){\bigr ]}} Y aparece un resultado adicional:
Derivación de valores no lemniscáticos El teorema de reflexión se utiliza ahora para los siguientes valores:
R [ exp ( − 2 π ) ] ⊕ R [ exp ( − 2 2 π ) ] = Φ − 1 {\displaystyle R{\bigl [}\exp(-{\sqrt {2}}\,\pi ){\bigr ]}\oplus R{\bigl [}\exp(-2{\sqrt {2}}\,\pi ){\bigr ]}=\Phi ^{-1}} El teorema theta de Jacobi conduce a una relación adicional:
R [ exp ( − 2 π ) ] ⊖ R [ exp ( − 2 2 π ) ] = R ( q ) ⊖ R ( q 2 ) [ q = exp ( − 2 π ) ] = {\displaystyle R{\bigl [}\exp(-{\sqrt {2}}\,\pi ){\bigr ]}\ominus R{\bigl [}\exp(-2{\sqrt {2}}\,\pi ){\bigr ]}=R(q)\ominus R(q^{2}){\bigl [}q=\exp(-{\sqrt {2}}\,\pi ){\bigr ]}=} = ϑ 01 ( q ) 2 − ϑ 01 ( q 1 / 5 ) 2 5 ϑ 01 ( q 5 ) 2 − ϑ 01 ( q ) 2 [ q = exp ( − 2 π ) ] = broncearse [ 2 arctan ( 1 3 5 − 1 3 6 30 + 4 5 3 + 1 3 6 30 − 4 5 3 ) − 1 4 π ] {\displaystyle ={\frac {\vartheta _{01}(q)^{2}-\vartheta _{01}(q^{1/5})^{2}}{5\,\vartheta _{01}(q^{5})^{2}-\vartheta _{01}(q)^{2}}}{\bigl [}q=\exp(-{\sqrt {2}}\,\pi ){\bigr ]}=\tan {\bigl [}2\arctan({\tfrac {1}{3}}{\sqrt {5}}-{\tfrac {1}{3}}{\sqrt[{3}]{6{\sqrt {30}}+4{\sqrt {5}}}}+{\tfrac {1}{3}}{\sqrt[{3}]{6{\sqrt {30}}-4{\sqrt {5}}}}\,)-{\tfrac {1}{4}}\pi {\bigr ]}} Al sumar tangencialmente los dos teoremas mencionados, obtenemos este resultado:
R [ exp ( − 2 π ) ] ⊕ R [ exp ( − 2 π ) ] = Φ − 1 ⊕ broncearse [ 2 arctan ( 1 3 5 − 1 3 6 30 + 4 5 3 + 1 3 6 30 − 4 5 3 ) − 1 4 π ] {\displaystyle R{\bigl [}\exp(-{\sqrt {2}}\,\pi ){\bigr ]}\oplus R{\bigl [}\exp(-{\sqrt {2}}\,\pi ){\bigr ]}=\Phi ^{-1}\oplus \tan {\bigl [}2\arctan({\tfrac {1}{3}}{\sqrt {5}}-{\tfrac {1}{3}}{\sqrt[{3}]{6{\sqrt {30}}+4{\sqrt {5}}}}+{\tfrac {1}{3}}{\sqrt[{3}]{6{\sqrt {30}}-4{\sqrt {5}}}}\,)-{\tfrac {1}{4}}\pi {\bigr ]}} Mediante una sustracción tangencial aparece ese resultado:
R [ exp ( − 2 2 π ) ] ⊕ R [ exp ( − 2 2 π ) ] = Φ − 1 ⊖ broncearse [ 2 arctan ( 1 3 5 − 1 3 6 30 + 4 5 3 + 1 3 6 30 − 4 5 3 ) − 1 4 π ] {\displaystyle R{\bigl [}\exp(-2{\sqrt {2}}\,\pi ){\bigr ]}\oplus R{\bigl [}\exp(-2{\sqrt {2}}\,\pi ){\bigr ]}=\Phi ^{-1}\ominus \tan {\bigl [}2\arctan({\tfrac {1}{3}}{\sqrt {5}}-{\tfrac {1}{3}}{\sqrt[{3}]{6{\sqrt {30}}+4{\sqrt {5}}}}+{\tfrac {1}{3}}{\sqrt[{3}]{6{\sqrt {30}}-4{\sqrt {5}}}}\,)-{\tfrac {1}{4}}\pi {\bigr ]}} Como solución alternativa, utilizamos el teorema del nomo al cuadrado:
R [ exp ( − 2 π ) ] 2 R [ exp ( − 2 2 π ) ] − 1 ⊕ R [ exp ( − 2 π ) ] R [ exp ( − 2 2 π ) ] 2 = 1 {\displaystyle R{\bigl [}\exp(-{\sqrt {2}}\,\pi ){\bigr ]}^{2}R{\bigl [}\exp(-2{\sqrt {2}}\,\pi ){\bigr ]}^{-1}\oplus R{\bigl [}\exp(-{\sqrt {2}}\,\pi ){\bigr ]}R{\bigl [}\exp(-2{\sqrt {2}}\,\pi ){\bigr ]}^{2}=1} { R [ exp ( − 2 π ) ] 2 R [ exp ( − 2 2 π ) ] − 1 + 1 } { R [ exp ( − 2 π ) ] R [ exp ( − 2 2 π ) ] 2 + 1 } = 2 {\displaystyle {\bigl \{}R{\bigl [}\exp(-{\sqrt {2}}\,\pi ){\bigr ]}^{2}R{\bigl [}\exp(-2{\sqrt {2}}\,\pi ){\bigr ]}^{-1}+1{\bigr \}}{\bigl \{}R{\bigl [}\exp(-{\sqrt {2}}\,\pi ){\bigr ]}R{\bigl [}\exp(-2{\sqrt {2}}\,\pi ){\bigr ]}^{2}+1{\bigr \}}=2} Ahora se vuelve a aplicar el teorema de reflexión:
R [ exp ( − 2 2 π ) ] = Φ − 1 ⊖ R [ exp ( − 2 π ) ] {\displaystyle R{\bigl [}\exp(-2{\sqrt {2}}\,\pi ){\bigr ]}=\Phi ^{-1}\ominus R{\bigl [}\exp(-{\sqrt {2}}\,\pi ){\bigr ]}} R [ exp ( − 2 2 π ) ] = 1 − Φ R [ exp ( − 2 π ) ] Φ + R [ exp ( − 2 π ) ] {\displaystyle R{\bigl [}\exp(-2{\sqrt {2}}\,\pi ){\bigr ]}={\frac {1-\Phi R{\bigl [}\exp(-{\sqrt {2}}\,\pi ){\bigr ]}}{\Phi +R{\bigl [}\exp(-{\sqrt {2}}\,\pi ){\bigr ]}}}} La inserción de la última expresión mencionada en el teorema del nomo al cuadrado da como resultado la siguiente ecuación:
{ R [ exp ( − 2 π ) ] 2 Φ + R [ exp ( − 2 π ) ] 1 − Φ R [ exp ( − 2 π ) ] + 1 } ⟨ R [ exp ( − 2 π ) ] { 1 − Φ R [ exp ( − 2 π ) ] } 2 { Φ + R [ exp ( − 2 π ) ] } 2 + 1 ⟩ = 2 {\displaystyle {\biggl \{}R{\bigl [}\exp(-{\sqrt {2}}\,\pi ){\bigr ]}^{2}{\frac {\Phi +R{\bigl [}\exp(-{\sqrt {2}}\,\pi ){\bigr ]}}{1-\Phi R{\bigl [}\exp(-{\sqrt {2}}\,\pi ){\bigr ]}}}+1{\biggr \}}{\biggl \langle }R{\bigl [}\exp(-{\sqrt {2}}\,\pi ){\bigr ]}{\frac {{\bigl \{}1-\Phi R{\bigl [}\exp(-{\sqrt {2}}\,\pi ){\bigr ]}{\bigr \}}^{2}}{{\bigl \{}\Phi +R{\bigl [}\exp(-{\sqrt {2}}\,\pi ){\bigr ]}{\bigr \}}^{2}}}+1{\biggr \rangle }=2} Si se eliminan los denominadores, se obtiene una ecuación de sexto grado:
R [ exp ( − 2 π ) ] 6 + 2 Φ − 2 R [ exp ( − 2 π ) ] 5 − 5 Φ − 1 R [ exp ( − 2 π ) ] 4 + {\displaystyle R{\bigl [}\exp(-{\sqrt {2}}\,\pi ){\bigr ]}^{6}+2\,\Phi ^{-2}R{\bigl [}\exp(-{\sqrt {2}}\,\pi ){\bigr ]}^{5}-{\sqrt {5}}\,\Phi ^{-1}R{\bigl [}\exp(-{\sqrt {2}}\,\pi ){\bigr ]}^{4}+} + 2 5 Φ R [ exp ( − 2 π ) ] 3 + 5 Φ − 1 R [ exp ( − 2 π ) ] 2 + 2 Φ − 2 R [ exp ( − 2 π ) ] − 1 = 0 {\displaystyle +2\,{\sqrt {5}}\,\Phi R{\bigl [}\exp(-{\sqrt {2}}\,\pi ){\bigr ]}^{3}+{\sqrt {5}}\,\Phi ^{-1}R{\bigl [}\exp(-{\sqrt {2}}\,\pi ){\bigr ]}^{2}+2\,\Phi ^{-2}R{\bigl [}\exp(-{\sqrt {2}}\,\pi ){\bigr ]}-1=0} La solución de esta ecuación es la solución ya mencionada:
R [ exp ( − 2 π ) ] = broncearse [ arctan ( 1 3 5 − 1 3 6 30 + 4 5 3 + 1 3 6 30 − 4 5 3 ) − 1 4 arccot ( 2 ) ] {\displaystyle R{\bigl [}\exp(-{\sqrt {2}}\,\pi ){\bigr ]}=\tan {\bigl [}\arctan({\tfrac {1}{3}}{\sqrt {5}}-{\tfrac {1}{3}}{\sqrt[{3}]{6{\sqrt {30}}+4{\sqrt {5}}}}+{\tfrac {1}{3}}{\sqrt[{3}]{6{\sqrt {30}}-4{\sqrt {5}}}}\,)-{\tfrac {1}{4}}\operatorname {arccot}(2){\bigr ]}}
R ( q ) {\displaystyle R(q)} función eta de Dedekind , una forma modular de peso 1/2, como, [ 1 ]
1 R ( q ) − R ( q ) = η ( τ 5 ) η ( 5 τ ) + 1 {\displaystyle {\frac {1}{R(q)}}-R(q)={\frac {\eta ({\frac {\tau }{5}})}{\eta (5\tau )}}+1} 1 R 5 ( q ) − R 5 ( q ) = [ η ( τ ) η ( 5 τ ) ] 6 + 11 {\displaystyle {\frac {1}{R^{5}(q)}}-R^{5}(q)=\left[{\frac {\eta (\tau )}{\eta (5\tau )}}\right]^{6}+11} La fracción continua de Rogers-Ramanujan también puede expresarse en términos de las funciones theta de Jacobi . Recordemos la notación,
ϑ 10 ( 0 ; τ ) = θ 2 ( q ) = ∑ norte = − ∞ ∞ q ( norte + 1 / 2 ) 2 ϑ 00 ( 0 ; τ ) = θ 3 ( q ) = ∑ norte = − ∞ ∞ q norte 2 ϑ 01 ( 0 ; τ ) = θ 4 ( q ) = ∑ norte = − ∞ ∞ ( − 1 ) norte q norte 2 {\displaystyle {\begin{aligned}\vartheta _{10}(0;\tau )&=\theta _{2}(q)=\sum _{n=-\infty }^{\infty }q^{(n+1/2)^{2}}\\\vartheta _{00}(0;\tau )&=\theta _{3}(q)=\sum _{n=-\infty }^{\infty }q^{n^{2}}\\\vartheta _{01}(0;\tau )&=\theta _{4}(q)=\sum _{n=-\infty }^{\infty }(-1)^{n}q^{n^{2}}\end{aligned}}} La notaciónθ norte {\displaystyle \theta _{n}} θ 2 4 + θ 4 4 = θ 3 4 {\displaystyle \theta _{2}^{4}+\theta _{4}^{4}=\theta _{3}^{4}}
R ( incógnita ) = broncearse { 1 2 arccot [ 1 2 + θ 4 ( incógnita 1 / 5 ) [ 5 θ 4 ( incógnita 5 ) 2 − θ 4 ( incógnita ) 2 ] 2 θ 4 ( incógnita 5 ) [ θ 4 ( incógnita ) 2 − θ 4 ( incógnita 1 / 5 ) 2 ] ] } {\displaystyle R(x)=\tan {\biggl \{}{\frac {1}{2}}\operatorname {arccot} {\biggl [}{\frac {1}{2}}+{\frac {\theta _{4}(x^{1/5})[5\,\theta _{4}(x^{5})^{2}-\theta _{4}(x)^{2}]}{2\,\theta _{4}(x^{5})[\theta _{4}(x)^{2}-\theta _{4}(x^{1/5})^{2}]}}{\biggr ]}{\biggr \}}} R ( incógnita ) = broncearse { 1 2 arccot [ 1 2 + ( θ 2 ( incógnita 1 / 10 ) θ 3 ( incógnita 1 / 10 ) θ 4 ( incógnita 1 / 10 ) 2 3 θ 2 ( incógnita 5 / 2 ) θ 3 ( incógnita 5 / 2 ) θ 4 ( incógnita 5 / 2 ) ) 1 / 3 ] } {\displaystyle R(x)=\tan {\biggl \{}{\frac {1}{2}}\operatorname {arccot} {\biggl [}{\frac {1}{2}}+{\bigg (}{\frac {\theta _{2}(x^{1/10})\,\theta _{3}(x^{1/10})\,\theta _{4}(x^{1/10})}{2^{3}\,\theta _{2}(x^{5/2})\,\theta _{3}(x^{5/2})\,\theta _{4}(x^{5/2})}}{\bigg )}^{1/3}{\biggr ]}{\biggr \}}} R ( incógnita ) = broncearse { 1 2 arctan [ 1 2 − θ 4 ( incógnita ) 2 2 θ 4 ( incógnita 5 ) 2 ] } 1 / 5 × broncearse { 1 2 arccot [ 1 2 − θ 4 ( incógnita ) 2 2 θ 4 ( incógnita 5 ) 2 ] } 2 / 5 {\displaystyle R(x)=\tan {\biggl \{}{\frac {1}{2}}\arctan {\biggl [}{\frac {1}{2}}-{\frac {\theta _{4}(x)^{2}}{2\,\theta _{4}(x^{5})^{2}}}{\biggr ]}{\biggr \}}^{1/5}\times \tan {\biggl \{}{\frac {1}{2}}\operatorname {arccot} {\biggl [}{\frac {1}{2}}-{\frac {\theta _{4}(x)^{2}}{2\,\theta _{4}(x^{5})^{2}}}{\biggr ]}{\biggr \}}^{2/5}} R ( incógnita ) = broncearse { 1 2 arctan [ 1 2 − θ 4 ( incógnita 1 / 2 ) 2 2 θ 4 ( incógnita 5 / 2 ) 2 ] } 2 / 5 × cuna { 1 2 arccot [ 1 2 − θ 4 ( incógnita 1 / 2 ) 2 2 θ 4 ( incógnita 5 / 2 ) 2 ] } 1 / 5 {\displaystyle R(x)=\tan {\biggl \{}{\frac {1}{2}}\arctan {\biggl [}{\frac {1}{2}}-{\frac {\theta _{4}(x^{1/2})^{2}}{2\,\theta _{4}(x^{5/2})^{2}}}{\biggr ]}{\biggr \}}^{2/5}\times \cot {\biggl \{}{\frac {1}{2}}\operatorname {arccot} {\biggl [}{\frac {1}{2}}-{\frac {\theta _{4}(x^{1/2})^{2}}{2\,\theta _{4}(x^{5/2})^{2}}}{\biggr ]}{\biggr \}}^{1/5}} Sin embargo, tenga en cuenta que las funciones theta normalmente usan el nomo q = e iπτ función eta de Dedekind usa el cuadrado del nomo q = e 2iπτ x en su lugar para mantener la coherencia entre todas las funciones. Por ejemplo, seaτ = − 1 {\displaystyle \tau ={\sqrt {-1}}} incógnita = mi − π {\displaystyle x=e^{-\pi }} R ( x ), que es la evaluación correcta de la fracción continua dada anteriormente,
R ( mi − π ) = 1 2 φ ( 5 − φ 3 / 2 ) ( 5 4 + φ 3 / 2 ) = 0,511428 … {\displaystyle R{\big (}e^{-\pi }{\big )}={\frac {1}{2}}\varphi \,({\sqrt {5}}-\varphi ^{3/2})({\sqrt[{4}]{5}}+\varphi ^{3/2})=0.511428\dots } También se puede definir el noma elíptico ,
q ( k ) = exp [ − π K ( 1 − k 2 ) / K ( k ) ] {\displaystyle q(k)=\exp {\big [}-\pi K({\sqrt {1-k^{2}}})/K(k){\big ]}} La letra minúscula k describe el módulo elíptico y la letra mayúscula K describe la integral elíptica completa de primera especie. La fracción continua también puede expresarse mediante las funciones elípticas de Jacobi de la siguiente manera:
R ( q ( k ) ) = broncearse { 1 2 arctan y } 1 / 5 broncearse { 1 2 arccot y } 2 / 5 = { y 2 + 1 − 1 y } 1 / 5 { y [ 1 y 2 + 1 − 1 ] } 2 / 5 {\displaystyle R{\big (}q(k){\big )}=\tan {\biggl \{}{\frac {1}{2}}\arctan y{\biggr \}}^{1/5}\tan {\biggl \{}{\frac {1}{2}}\operatorname {arccot} y{\biggr \}}^{2/5}=\left\{{\frac {{\sqrt {y^{2}+1}}-1}{y}}\right\}^{1/5}\left\{y\left[{\sqrt {{\frac {1}{y^{2}}}+1}}-1\right]\right\}^{2/5}} con
y = 2 k 2 sn [ 2 5 K ( k ) ; k ] 2 sn [ 4 5 K ( k ) ; k ] 2 5 − k 2 sn [ 2 5 K ( k ) ; k ] 2 sn [ 4 5 K ( k ) ; k ] 2 . {\displaystyle y={\frac {2k^{2}\,{\text{sn}}[{\tfrac {2}{5}}K(k);k]^{2}\,{\text{sn}}[{\tfrac {4}{5}}K(k);k]^{2}}{5-k^{2}\,{\text{sn}}[{\tfrac {2}{5}}K(k);k]^{2}\,{\text{sn}}[{\tfrac {4}{5}}K(k);k]^{2}}}.}
Relación con la función j Una fórmula que involucra la función j y la función eta de Dedekind es la siguiente:
j ( τ ) = ( incógnita 2 + 10 incógnita + 5 ) 3 incógnita {\displaystyle j(\tau )={\frac {(x^{2}+10x+5)^{3}}{x}}} dóndeincógnita = [ 5 η ( 5 τ ) η ( τ ) ] 6 . {\displaystyle x=\left[{\frac {{\sqrt {5}}\,\eta (5\tau )}{\eta (\tau )}}\right]^{6}.\,}
1 R 5 ( q ) − R 5 ( q ) = [ η ( τ ) η ( 5 τ ) ] 6 + 11 {\displaystyle {\frac {1}{R^{5}(q)}}-R^{5}(q)=\left[{\frac {\eta (\tau )}{\eta (5\tau )}}\right]^{6}+11} Eliminando el cociente etaincógnita {\displaystyle x} j ( τ ) en términos der = R ( q ) {\displaystyle r=R(q)}
j ( τ ) = − ( r 20 − 228 r 15 + 494 r 10 + 228 r 5 + 1 ) 3 r 5 ( r 10 + 11 r 5 − 1 ) 5 j ( τ ) − 1728 = − ( r 30 + 522 r 25 − 10005 r 20 − 10005 r 10 − 522 r 5 + 1 ) 2 r 5 ( r 10 + 11 r 5 − 1 ) 5 {\displaystyle {\begin{aligned}&j(\tau )=-{\frac {(r^{20}-228r^{15}+494r^{10}+228r^{5}+1)^{3}}{r^{5}(r^{10}+11r^{5}-1)^{5}}}\\[6pt]&j(\tau )-1728=-{\frac {(r^{30}+522r^{25}-10005r^{20}-10005r^{10}-522r^{5}+1)^{2}}{r^{5}(r^{10}+11r^{5}-1)^{5}}}\end{aligned}}} donde el numerador y el denominador son invariantes polinomiales del icosaedro . Usando la ecuación modular entreR ( q ) {\displaystyle R(q)} R ( q 5 ) {\displaystyle R(q^{5})}
j ( 5 τ ) = − ( r 20 + 12 r 15 + 14 r 10 − 12 r 5 + 1 ) 3 r 25 ( r 10 + 11 r 5 − 1 ) j ( 5 τ ) − 1728 = − ( r 30 + 18 r 25 + 75 r 20 + 75 r 10 − 18 r 5 + 1 ) 2 r 25 ( r 10 + 11 r 5 − 1 ) {\displaystyle {\begin{aligned}&j(5\tau )=-{\frac {(r^{20}+12r^{15}+14r^{10}-12r^{5}+1)^{3}}{r^{25}(r^{10}+11r^{5}-1)}}\\[6pt]&j(5\tau )-1728=-{\frac {(r^{30}+18r^{25}+75r^{20}+75r^{10}-18r^{5}+1)^{2}}{r^{25}(r^{10}+11r^{5}-1)}}\end{aligned}}} Dejarz = r 5 − 1 r 5 {\displaystyle z=r^{5}-{\frac {1}{r^{5}}}} j ( 5 τ ) = − ( z 2 + 12 z + 16 ) 3 z + 11 {\displaystyle j(5\tau )=-{\frac {\left(z^{2}+12z+16\right)^{3}}{z+11}}}
dónde
z ∞ = − [ 5 η ( 25 τ ) η ( 5 τ ) ] 6 − 11 , z 0 = − [ η ( τ ) η ( 5 τ ) ] 6 − 11 , z 1 = [ η ( 5 τ + 2 5 ) η ( 5 τ ) ] 6 − 11 , z 2 = − [ η ( 5 τ + 4 5 ) η ( 5 τ ) ] 6 − 11 , z 3 = [ η ( 5 τ + 6 5 ) η ( 5 τ ) ] 6 − 11 , z 4 = − [ η ( 5 τ + 8 5 ) η ( 5 τ ) ] 6 − 11 {\displaystyle {\begin{aligned}&z_{\infty }=-\left[{\frac {{\sqrt {5}}\,\eta (25\tau )}{\eta (5\tau )}}\right]^{6}-11,\ z_{0}=-\left[{\frac {\eta (\tau )}{\eta (5\tau )}}\right]^{6}-11,\ z_{1}=\left[{\frac {\eta ({\frac {5\tau +2}{5}})}{\eta (5\tau )}}\right]^{6}-11,\\[6pt]&z_{2}=-\left[{\frac {\eta ({\frac {5\tau +4}{5}})}{\eta (5\tau )}}\right]^{6}-11,\ z_{3}=\left[{\frac {\eta ({\frac {5\tau +6}{5}})}{\eta (5\tau )}}\right]^{6}-11,\ z_{4}=-\left[{\frac {\eta ({\frac {5\tau +8}{5}})}{\eta (5\tau )}}\right]^{6}-11\end{aligned}}} que de hecho es el j-invariante de la curva elíptica ,
y 2 + ( 1 + r 5 ) incógnita y + r 5 y = incógnita 3 + r 5 incógnita 2 {\displaystyle y^{2}+(1+r^{5})xy+r^{5}y=x^{3}+r^{5}x^{2}} parametrizado por los puntos no cúspides de la curva modular incógnita 1 ( 5 ) {\displaystyle X_{1}(5)}
Ecuación funcional Para mayor comodidad, también se puede utilizar la notaciónr ( τ ) = R ( q ) {\displaystyle r(\tau )=R(q)} q = e 2πiτ . Mientras que otras funciones modulares como el invariante j satisfacen,
j ( − 1 τ ) = j ( τ ) {\displaystyle j(-{\tfrac {1}{\tau }})=j(\tau )} y la función eta de Dedekind tiene,
η ( − 1 τ ) = − i τ η ( τ ) {\displaystyle \eta (-{\tfrac {1}{\tau }})={\sqrt {-i\tau }}\,\eta (\tau )} La ecuación funcional de la fracción continua de Rogers-Ramanujan involucra [ 2 ] proporción áurea. φ {\displaystyle \varphi }
r ( − 1 τ ) = 1 − φ r ( τ ) φ + r ( τ ) {\displaystyle r(-{\tfrac {1}{\tau }})={\frac {1-\varphi \,r(\tau )}{\varphi +r(\tau )}}} De paso,
r ( 7 + i 10 ) = i {\displaystyle r({\tfrac {7+i}{10}})=i}
ecuaciones modulares Hay ecuaciones modulares entreR ( q ) {\displaystyle R(q)} R ( q norte ) {\displaystyle R(q^{n})} números primos pequeños n son las siguientes. [ 3 ]
Paranorte = 2 {\displaystyle n=2} tú = R ( q ) {\displaystyle u=R(q)} v = R ( q 2 ) {\displaystyle v=R(q^{2})} v − tú 2 = ( v + tú 2 ) tú v 2 . {\displaystyle v-u^{2}=(v+u^{2})uv^{2}.}
Paranorte = 3 {\displaystyle n=3} tú = R ( q ) {\displaystyle u=R(q)} v = R ( q 3 ) {\displaystyle v=R(q^{3})} ( v − tú 3 ) ( 1 + tú v 3 ) = 3 tú 2 v 2 . {\displaystyle (v-u^{3})(1+uv^{3})=3u^{2}v^{2}.}
Paranorte = 5 {\displaystyle n=5} tú = R ( q ) {\displaystyle u=R(q)} v = R ( q 5 ) {\displaystyle v=R(q^{5})} v ( v 4 − 3 v 3 + 4 v 2 − 2 v + 1 ) = ( v 4 + 2 v 3 + 4 v 2 + 3 v + 1 ) tú 5 . {\displaystyle v(v^{4}-3v^{3}+4v^{2}-2v+1)=(v^{4}+2v^{3}+4v^{2}+3v+1)u^{5}.}
O equivalentemente paranorte = 5 {\displaystyle n=5} tú = R ( q ) {\displaystyle u=R(q)} v = R ( q 5 ) {\displaystyle v=R(q^{5})} φ = 1 + 5 2 {\displaystyle \varphi ={\tfrac {1+{\sqrt {5}}}{2}}} tú 5 = v ( v 2 − φ 2 v + φ 2 ) ( v 2 − φ − 2 v + φ − 2 ) ( v 2 + v + φ 2 ) ( v 2 + v + φ − 2 ) . {\displaystyle u^{5}={\frac {v\,(v^{2}-\varphi ^{2}v+\varphi ^{2})(v^{2}-\varphi ^{-2}v+\varphi ^{-2})}{(v^{2}+v+\varphi ^{2})(v^{2}+v+\varphi ^{-2})}}.}
Paranorte = 11 {\displaystyle n=11} tú = R ( q ) {\displaystyle u=R(q)} v = R ( q 11 ) {\displaystyle v=R(q^{11})} tú v ( tú 10 + 11 tú 5 − 1 ) ( v 10 + 11 v 5 − 1 ) = ( tú − v ) 12 . {\displaystyle uv(u^{10}+11u^{5}-1)(v^{10}+11v^{5}-1)=(u-v)^{12}.}
Acerca denorte = 5 {\displaystyle n=5} v 10 + 11 v 5 − 1 = ( v 2 + v − 1 ) ( v 4 − 3 v 3 + 4 v 2 − 2 v + 1 ) ( v 4 + 2 v 3 + 4 v 2 + 3 v + 1 ) . {\displaystyle v^{10}+11v^{5}-1=(v^{2}+v-1)(v^{4}-3v^{3}+4v^{2}-2v+1)(v^{4}+2v^{3}+4v^{2}+3v+1).}
Otros resultados Ramanujan encontró muchos otros resultados interesantes con respecto aR ( q ) {\displaystyle R(q)} [ 4 ] a , b ∈ R + {\displaystyle a,b\in \mathbb {R} ^{+}} φ {\displaystyle \varphi } proporción áurea .
Sia b = π 2 {\displaystyle ab=\pi ^{2}}
[ R ( mi − 2 a ) + φ ] [ R ( mi − 2 b ) + φ ] = 5 φ . {\displaystyle {\bigl [}R(e^{-2a})+\varphi {\bigl ]}{\bigl [}R(e^{-2b})+\varphi {\bigr ]}={\sqrt {5}}\,\varphi .} Si5 a b = π 2 {\displaystyle 5ab=\pi ^{2}}
[ R 5 ( mi − 2 a ) + φ 5 ] [ R 5 ( mi − 2 b ) + φ 5 ] = 5 5 φ 5 . {\displaystyle {\bigl [}R^{5}(e^{-2a})+\varphi ^{5}{\bigl ]}{\bigl [}R^{5}(e^{-2b})+\varphi ^{5}{\bigr ]}=5{\sqrt {5}}\,\varphi ^{5}.} Los poderes deR ( q ) {\displaystyle R(q)} cubo ,
R 3 ( q ) = α β {\displaystyle R^{3}(q)={\frac {\alpha }{\beta }}} dónde
α = ∑ norte = 0 ∞ q 2 norte 1 − q 5 norte + 2 − ∑ norte = 0 ∞ q 3 norte + 1 1 − q 5 norte + 3 , {\displaystyle \alpha =\sum _{n=0}^{\infty }{\frac {q^{2n}}{1-q^{5n+2}}}-\sum _{n=0}^{\infty }{\frac {q^{3n+1}}{1-q^{5n+3}}},} β = ∑ norte = 0 ∞ q norte 1 − q 5 norte + 1 − ∑ norte = 0 ∞ q 4 norte + 3 1 − q 5 norte + 4 . {\displaystyle \beta =\sum _{n=0}^{\infty }{\frac {q^{n}}{1-q^{5n+1}}}-\sum _{n=0}^{\infty }{\frac {q^{4n+3}}{1-q^{5n+4}}}.} Para su quinta potencia, dejemosw = R ( q ) R 2 ( q 2 ) {\displaystyle w=R(q)R^{2}(q^{2})}
R 5 ( q ) = w ( 1 − w 1 + w ) 2 , R 5 ( q 2 ) = w 2 ( 1 + w 1 − w ) {\displaystyle R^{5}(q)=w\left({\frac {1-w}{1+w}}\right)^{2},\;\;R^{5}(q^{2})=w^{2}\left({\frac {1+w}{1-w}}\right)}
Ecuaciones quínticas La ecuación quíntica general en forma de Bring-Jerrard:
incógnita 5 − 5 incógnita − 4 a = 0 {\displaystyle x^{5}-5x-4a=0} por cada valor reala > 1 {\displaystyle a>1} R ( q ) {\displaystyle R(q)}
q ( k ) = exp [ − π K ( 1 − k 2 ) / K ( k ) ] . {\displaystyle q(k)=\exp {\big [}-\pi K({\sqrt {1-k^{2}}})/K(k){\big ]}.} Para resolver esta ecuación quíntica, primero se debe determinar el módulo elíptico como
k = broncearse [ 1 4 π − 1 4 arccsc ( a 2 ) ] . {\displaystyle k=\tan[{\tfrac {1}{4}}\pi -{\tfrac {1}{4}}\operatorname {arccsc}(a^{2})].} Entonces la solución real es
incógnita = 2 − { 1 − R [ q ( k ) ] } { 1 + R [ q ( k ) 2 ] } R [ q ( k ) ] R [ q ( k ) 2 ] 4 cuna ⟨ 4 arctan { S } ⟩ − 3 4 = 2 − { 1 − R [ q ( k ) ] } { 1 + R [ q ( k ) 2 ] } R [ q ( k ) ] R [ q ( k ) 2 ] 2 S − 1 + 2 S + 1 + 1 S − S − 3 4 . {\displaystyle {\begin{aligned}x&={\frac {2-{\bigl \{}1-R[q(k)]{\bigr \}}{\bigl \{}1+R[q(k)^{2}]{\bigr \}}}{{\sqrt {R[q(k)]\,R[q(k)^{2}]}}\,{\sqrt[{4}]{4\cot \langle 4\arctan\{S\}\rangle -3}}}}\\&={\frac {2-{\bigl \{}1-R[q(k)]{\bigr \}}{\bigl \{}1+R[q(k)^{2}]{\bigr \}}}{{\sqrt {R[q(k)]R[q(k)^{2}]}}\,{\sqrt[{4}]{{\frac {2}{S-1}}+{\frac {2}{S+1}}+{\frac {1}{S}}-S-3}}}}.\end{aligned}}} dóndeS = R [ q ( k ) ] R 2 [ q ( k ) 2 ] . {\displaystyle S=R[q(k)]\,R^{2}[q(k)^{2}].} R ( q ) {\displaystyle R(q)} S {\displaystyle S}
R 5 [ q ( k ) ] = S ( 1 − S 1 + S ) 2 {\displaystyle R^{5}[q(k)]=S\left({\frac {1-S}{1+S}}\right)^{2}}
Ejemplo 2 incógnita 5 − 5 incógnita − 4 ( 81 32 4 ) = 0 {\displaystyle x^{5}-5x-4{\Bigl (}{\sqrt[{4}]{\tfrac {81}{32}}}{\Bigr )}=0} de este modo,
a = 81 32 4 {\displaystyle a={\sqrt[{4}]{\tfrac {81}{32}}}} Dadas las fracciones continuas más familiares con formas cerradas,
r 1 = R ( mi − π ) = 1 2 φ ( 5 − φ 3 / 2 ) ( 5 4 + φ 3 / 2 ) = 0,511428 … {\displaystyle r_{1}=R{\big (}e^{-\pi }{\big )}={\tfrac {1}{2}}\varphi \,({\sqrt {5}}-\varphi ^{3/2})({\sqrt[{4}]{5}}+\varphi ^{3/2})=0.511428\dots } r 2 = R ( mi − 2 π ) = 5 4 φ 1 / 2 − φ = 0,284079 … {\displaystyle r_{2}=R{\big (}e^{-2\pi }{\big )}={\sqrt[{4}]{5}}\,\varphi ^{1/2}-\varphi =0.284079\dots } r 4 = R ( mi − 4 π ) = 1 2 φ ( 5 − φ 3 / 2 ) ( − 5 4 + φ 3 / 2 ) = 0,081002 … {\displaystyle r_{4}=R{\big (}e^{-4\pi }{\big )}={\tfrac {1}{2}}\varphi \,({\sqrt {5}}-\varphi ^{3/2})(-{\sqrt[{4}]{5}}+\varphi ^{3/2})=0.081002\dots } con la proporción áureaφ = 1 + 5 2 {\displaystyle \varphi ={\tfrac {1+{\sqrt {5}}}{2}}}
incógnita = 5 4 2 − { 1 − r 1 } { 1 + r 2 } r 1 r 2 20 cuna ⟨ 4 arctan { r 1 r 2 2 } ⟩ − 15 4 = 5 4 2 − { 1 − r 2 } { 1 + r 4 } r 2 r 4 20 cuna ⟨ 4 arctan { r 2 r 4 2 } ⟩ − 15 4 = 8 4 = 1.681792 … {\displaystyle {\begin{aligned}x&={\sqrt[{4}]{5}}\,{\frac {2-{\bigl \{}1-r_{1}{\bigr \}}{\bigl \{}1+r_{2}{\bigr \}}}{{\sqrt {r_{1}\,r_{2}}}\,{\sqrt[{4}]{20\cot \langle 4\arctan\{r_{1}\,r_{2}^{2}\}\rangle -15}}}}\\[6pt]&={\sqrt[{4}]{5}}\,{\frac {2-{\bigl \{}1-r_{2}{\bigr \}}{\bigl \{}1+r_{4}{\bigr \}}}{{\sqrt {r_{2}\,r_{4}}}\,{\sqrt[{4}]{20\cot \langle 4\arctan\{r_{2}\,r_{4}^{2}\}\rangle -15}}}}\\[6pt]&={\sqrt[{4}]{8}}=1.681792\dots \end{aligned}}}
Referencias ↑ Duke, W. "Fracciones continuas y funciones modulares", https://www.math.ucla.edu/~wdduke/preprints/bams4.pdf ↑ Duke, W. "Fracciones continuas y funciones modulares" (p. 9) ↑ Berndt, B. et al. "La fracción continua de Rogers-Ramanujan", http://www.math.uiuc.edu/~berndt/articles/rrcf.pdf ↑ Berndt, B. et al. "La fracción continua de Rogers-Ramanujan" Rogers, LJ (1894), "Segunda Memoria sobre la Expansión de ciertos Productos Infinitos" , Proc. London Math. Soc. , s1-25 (1): 318–343 , doi : 10.1112/plms/s1-25.1.318Berndt, BC; Chan, HH; Huang, SS; Kang, SY; Sohn, J.; Son, SH (1999), "La fracción continua de Rogers-Ramanujan" (PDF) , Journal of Computational and Applied Mathematics , 105 ( 1–2 ): 9–24 , doi : 10.1016/S0377-0427(99)00033-3 
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![{\displaystyle {\begin{aligned}G(q)&=\sum _{n=0}^{\infty }{\frac {q^{n^{2}}}{(1-q)(1-q^{2})\cdots (1-q^{n})}}=\sum _{n=0}^{\infty }{\frac {q^{n^{2}}}{(q;q)_{n}}}={\frac {1}{(q;q^{5})_{\infty }(q^{4};q^{5})_{\infty }}}\\[6pt]&=\prod _{n=1}^{\infty }{\frac {1}{(1-q^{5n-1})(1-q^{5n-4})}}\\[6pt]&={\sqrt[{60}]{q\,j}}\,\,_{2}F_{1}\left(-{\tfrac {1}{60}},{\tfrac {19}{60}};{\tfrac {4}{5}};{\tfrac {1728}{j}}\right)\\[6pt]&={\sqrt[{60}]{q\left(j-1728\right)}}\,_{2}F_{1}\left(-{\tfrac {1}{60}},{\tfrac {29}{60}};{\tfrac {4}{5}};-{\tfrac {1728}{j-1728}}\right)\\[6pt]&=1+q+q^{2}+q^{3}+2q^{4}+2q^{5}+3q^{6}+\cdots \end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/17f340bf6668c9c9d6b2ba43139d7d57edb2897d)
![{\displaystyle {\begin{aligned}H(q)&=\sum _{n=0}^{\infty }{\frac {q^{n^{2}+n}}{(1-q)(1-q^{2})\cdots (1-q^{n})}}=\sum _{n=0}^{\infty }{\frac {q^{n^{2}+n}}{(q;q)_{n}}}={\frac {1}{(q^{2};q^{5})_{\infty }(q^{3};q^{5})_{\infty }}}\\[6pt]&=\prod _{n=1}^{\infty }{\frac {1}{(1-q^{5n-2})(1-q^{5n-3})}}\\[6pt]&={\frac {1}{\sqrt[{60}]{q^{11}j^{11}}}}\,_{2}F_{1}\left({\tfrac {11}{60}},{\tfrac {31}{60}};{\tfrac {6}{5}};{\tfrac {1728}{j}}\right)\\[6pt]&={\frac {1}{\sqrt[{60}]{q^{11}\left(j-1728\right)^{11}}}}\,_{2}F_{1}\left({\tfrac {11}{60}},{\tfrac {41}{60}};{\tfrac {6}{5}};-{\tfrac {1728}{j-1728}}\right)\\[6pt]&=1+q^{2}+q^{3}+q^{4}+q^{5}+2q^{6}+2q^{7}+\cdots \end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/87587e00106bca786494f3e62ce429384f4c2798)
![{\displaystyle {\begin{aligned}R(q)&={\frac {q^{\frac {11}{60}}H(q)}{q^{-{\frac {1}{60}}}G(q)}}=q^{\frac {1}{5}}\prod _{n=1}^{\infty }{\frac {(1-q^{5n-1})(1-q^{5n-4})}{(1-q^{5n-2})(1-q^{5n-3})}}=q^{1/5}\prod _{n=1}^{\infty }(1-q^{n})^{(n|5)}\\[8pt]&={\cfrac {q^{1/5}}{1+{\cfrac {q}{1+{\cfrac {q^{2}}{1+{\cfrac {q^{3}}{1+\ddots }}}}}}}}\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fc22e435ee365200f68b711f8c6d5b535f315c11)
![{\displaystyle R{\big (}e^{-\pi }{\big )}={\cfrac {e^{-{\frac {\pi }{5}}}}{1+{\cfrac {e^{-\pi }}{1+{\cfrac {e^{-2\pi }}{1+\ddots }}}}}}={\tfrac {1}{2}}\varphi \,({\sqrt {5}}-\varphi ^{3/2})({\sqrt[{4}]{5}}+\varphi ^{3/2})=0,511428\dots }](https://wikimedia.org/api/rest_v1/media/math/render/svg/9a6c3405e7c30139be9ebc5883957bbe2961a640)
![{\displaystyle R{\big (}e^{-2\pi }{\big )}={\cfrac {e^{-{\frac {2\pi }{5}}}}{1+{\cfrac {e^{-2\pi }}{1+{\cfrac {e^{-4\pi }}{1+\ddots }}}}}}={{\sqrt[{4}]{5}}\,\varphi ^{1/2}-\varphi }=0,284079\dots }](https://wikimedia.org/api/rest_v1/media/math/render/svg/bf153ea27c1bdc48270a3aca1ea339ba0bde25c6)
![{\displaystyle R{\big (}e^{-4\pi }{\big )}={\cfrac {e^{-{\frac {4\pi }{5}}}}{1+{\cfrac {e^{-4\pi }}{1+{\cfrac {e^{-8\pi }}{1+\ddots }}}}}}={\tfrac {1}{2}}\varphi \,({\sqrt {5}}-\varphi ^{3/2})(-{\sqrt[{4}]{5}}+\varphi ^{3/2})=0.081002\dots }](https://wikimedia.org/api/rest_v1/media/math/render/svg/dc99d55aaf5b1eb8d359179bc54934e150593ca2)
![{\displaystyle R{\big (}e^{-5\pi }{\big )}={\cfrac {e^{-\pi }}{1+{\cfrac {e^{-5\pi }}{1+{\cfrac {e^{-10\pi }}{1+\ddots }}}}}}={\frac {1+\varphi ^{2}}{\varphi +{\big (}{\frac) {1}{2}}(4-\varphi -3{\sqrt {\varphi -1}})(3\varphi ^{3/2}-{\sqrt[{4}]{5}}){\big )}^{1/5}}}-\varphi =0.0432139\dots }](https://wikimedia.org/api/rest_v1/media/math/render/svg/60a554a375df6c825bfa296bad44ad31f95f29c0)
![{\displaystyle R{\big (}e^{-20\pi }{\big )}={\cfrac {e^{-4\pi }}{1+{\cfrac {e^{-20\pi }}{1+{\cfrac {e^{-40\pi }}{1+\ddots }}}}}}={\frac {1+\varphi ^{2}}{\varphi +{\big (}{\frac) {1}{2}}(4-\varphi -3{\sqrt {\varphi -1}})(3\varphi ^{3/2}+{\sqrt[{4}]{5}}){\big )}^{1/5}}}-\varphi =0.00000348734\dots }](https://wikimedia.org/api/rest_v1/media/math/render/svg/1c901757a378f5272b59181c76b14ea2cec9f96b)
![{\displaystyle {\bigl [}R(e^{-2\pi \pi /m})+\varphi {\bigr ]}{\bigl [}R(e^{-2\pi \,m})+\varphi {\bigr ]}={\sqrt {5}}\,\varphi }](https://wikimedia.org/api/rest_v1/media/math/render/svg/6bd65a5de209ee413c11f16d48c302e2aaa695f4)
![{\displaystyle {\begin{aligned}G(q)&=\prod _{n=1}^{\infty }{\frac {1}{(1-q^{5n-1})(1-q^{5n-4})}}\\[6pt]&=q^{1/60}{\frac {j(\tau )^{1/60}}{(r^{20}-228r^{15}+494r^{10}+228r^{5}+1)^{1/20}}}\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6bed060d9273df8e5ead63436a909dc5fdbb403f)
![{\displaystyle {\begin{aligned}H(q)&=\prod _{n=1}^{\infty }{\frac {1}{(1-q^{5n-2})(1-q^{5n-3})}}\\[6pt]&={\frac {-1}{q^{11/60}}}{\frac {(r^{20}-228r^{15}+494r^{10}+228r^{5}+1)^{11/20}}{j(\tau )^{11/60}\,(r^{10}+11r^{5}-1)}}\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/36323b5de18a9b28b21f3eea8bad32bd6979b040)
![{\displaystyle {\begin{aligned}G(e^{-2\pi })&=(e^{-2\pi })^{1/60}{\frac {1}{(5\,\varphi )^{1/4}}}{\frac {1}{\sqrt {R(e^{-2\pi })}}}\\[6pt]&=1.00187093\dots \\[6pt]H(e^{-2\pi })&={\frac {1}{(e^{-2\pi })^{11/60}}}{\frac {1}{(5\,\varphi )^{1/4}}}{\sqrt {R(e^{-2\pi })}}\\[6pt]&=1.00000349\ldots \end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0540a9b2d189ce12385c3a3eabc99832d16b8e64)
![{\displaystyle {\begin{aligned}G(e^{-4\pi })&=(e^{-4\pi })^{1/60}{\frac {1}{(5\,\varphi ^{3})^{1/4}\,(\varphi +{\sqrt[{4}]{5}})^{1/4}}}{\frac {1}{\sqrt {R(e^{-4\pi }) })}}}\\[6pt]&=1.000003487354\dots \\[6pt]H(e^{-4\pi })&={\frac {1}{(e^{-4\pi })^{11/60}}}{\frac {1}{(5\,\varphi ^{3})^{1/4}\,(\varphi +{\sqrt[{4}]{5}})^{1/4}}}{\sqrt {R(e^{-4\pi })}}\\[6pt]&=1.000000000012\dots \end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/415d12f61351a318ce5cb5565f6a3c56f62b5d0a)
![{\displaystyle a\oplus b=\tan {\bigl [}\arctan(a)+\arctan(b){\bigr ]}={\frac {a+b}{1-ab}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d4694eeba504d3db4a4c2a1812403bfc3e17f836)
![{\displaystyle c\ominus d=\tan {\bigl [}\arctan(c)-\arctan(d){\bigr ]}={\frac {cd}{1+cd}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ebd5804fe87d111f63768e8f219419efa7725e57)
![{\displaystyle \Phi ={\tfrac {1}{2}}({\sqrt {5}}+1)=\cot[{\tfrac {1}{2}}\arctan(2)]=2\cos({\tfrac {1}{5}}{\pi })}](https://wikimedia.org/api/rest_v1/media/math/render/svg/00cd707c883f6a33b179764fe529673a685d30b0)
![{\displaystyle \Phi ^{-1}={\tfrac {1}{2}}({\sqrt {5}}-1)=\tan[{\tfrac {1}{2}}\arctan(2)]=2\sin({\tfrac {1}{10}}{\pi })}](https://wikimedia.org/api/rest_v1/media/math/render/svg/00e6aa1506dfff4cd10059114065f1a571130e1b)
![{\displaystyle S{\bigl [}\exp(-\pi ){\bigr ]}\oplus S{\bigl [}\exp(-\pi ){\bigr ]}=\Phi }](https://wikimedia.org/api/rest_v1/media/math/render/svg/7d39217aba6ce97c57822a0a8fedf5d7726728d9)
![{\displaystyle R{\bigl [}\exp(-2\pi ){\bigr ]}\oplus R{\bigl [}\exp(-2\pi ){\bigr ]}=\Phi ^{-1}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/90ab593a1f3d5e881102454dbf15a36e72afc0e3)
![{\displaystyle S{\bigl [}\exp(-\pi ){\bigr ]}\oplus R{\bigl [}\exp(-2\pi ){\bigr ]}=S(q)\oplus R(q^{2}){\bigl [}q=\exp(-\pi ){\bigr ]}=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/178cde333a5b70c5ff7fbb710e602e594d66d44a)
![{\displaystyle ={\frac {\vartheta _{00}(q^{1/5})^{2}-\vartheta _{00}(q)^{2}}{5\,\vartheta _{00}(q^{5})^{2}-\vartheta _{00}(q)^{2}}}{\bigl [}q=\exp(-\pi ){\bigr ]}=1}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2fd833ed5fb7e5980d12a76241e496b9d5a7bd1b)
![{\displaystyle R{\bigl [}\exp(-2\pi ){\bigr ]}=1\ominus S{\bigl [}\exp(-\pi ){\bigr ]}=1\ominus \tan {\bigl [}{\tfrac {1}{4}}\pi -{\tfrac {1}{4}}\arctan(2){\bigr ]}=\tan {\bigl [}{\tfrac {1}{4}}\arctan(2){\bigr ]}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/93961e24edfe705f24ead2bf290ca4ed5e79f7bb)
![{\displaystyle R{\bigl [}\exp(-\pi ){\bigr ]}\ominus R{\bigl [}\exp(-2\pi ){\bigr ]}=R(q)\ominus R(q^{2}){\bigl [}q=\exp(-\pi ){\bigr ]}=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/467dc496e6b3529dd86f63acea4f98ecb42beec9)
![{\displaystyle ={\frac {\vartheta _{01}(q)^{2}-\vartheta _{01}(q^{1/5})^{2}}{5\,\vartheta _{01}(q^{5})^{2}-\vartheta _{01}(q)^{2}}}{\bigl [}q=\exp(-\pi ){\bigr ]}={\frac {{\sqrt[{4}]{5}}-1}{{\sqrt[{4}]{5}}+1}}={\sqrt[{4}]{5}}\ominus 1=\tan {\bigl [}\arctan({\sqrt[{4}]{5}}\,)-{\tfrac {1}{4}}\pi {\bigr ]}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fcf199a4e70a3743b61131ea9c2118de3aee91fd)
![{\displaystyle R{\bigl [}\exp(-\pi ){\bigr ]}=R{\bigl [}\exp(-2\pi ){\bigr ]}\oplus \tan {\bigl [}\arctan({\sqrt[{4}]{5}}\,)-{\tfrac {1}{4}}\pi {\bigr ]}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0bdb2ba5fd90f09aee88032cf729907f49c8a548)
![{\displaystyle R{\bigl [}\exp(-\pi ){\bigr ]}\oplus R{\bigl [}\exp(-4\pi ){\bigr ]}=\Phi ^{-1}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cbb4d5e961fad8c6ce68dba48f8fe696b384eef4)
![{\displaystyle R{\bigl [}\exp(-4\pi ){\bigr ]}=\tan {\bigl [}{\tfrac {1}{2}}\arctan(2){\bigr ]}\ominus R{\bigl [}\exp(-\pi ){\bigr ]}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/62349fa9ff32f7b0d2070a35ba39e36e07ef9d8d)
![{\displaystyle R{\bigl [}\exp(-{\sqrt {2}}\,\pi ){\bigr ]}\oplus R{\bigl [}\exp(-2{\sqrt {2}}\,\pi ){\bigr ]}=\Phi ^{-1}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b9d8fbc08bb4a3a852a23c7a9e1214ac28d30941)
![{\displaystyle R{\bigl [}\exp(-{\sqrt {2}}\,\pi ){\bigr ]}\ominus R{\bigl [}\exp(-2{\sqrt {2}}\,\pi ){\bigr ]}=R(q)\ominus R(q^{2}){\bigl [}q=\exp(-{\sqrt {2}}\,\pi ){\bigr ]}=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/039f770796d02ff14157dcb5ab2c7ef5c7a500bd)
![{\displaystyle ={\frac {\vartheta _{01}(q)^{2}-\vartheta _{01}(q^{1/5})^{2}}{5\,\vartheta _{01}(q^{5})^{2}-\vartheta _{01}(q)^{2}}}{\bigl [}q=\exp(-{\sqrt {2}}\,\pi ){\bigr ]}=\tan {\bigl [}2\arctan({\tfrac {1}{3}}{\sqrt {5}}-{\tfrac {1}{3}}{\sqrt[{3}]{6{\sqrt {30}}+4{\sqrt {5}}}}+{\tfrac {1}{3}}{\sqrt[{3}]{6{\sqrt {30}}-4{\sqrt {5}}}}\,)-{\tfrac {1}{4}}\pi {\bigr ]}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6225f8ac541e8bdc2e7cb17ae1798dd21bba046f)
![{\displaystyle R{\bigl [}\exp(-{\sqrt {2}}\,\pi ){\bigr ]}\oplus R{\bigl [}\exp(-{\sqrt {2}}\,\pi ){\bigr ]}=\Phi ^{-1}\oplus \tan {\bigl [}2\arctan({\tfrac {1}{3}}{\sqrt {5}}-{\tfrac {1}{3}}{\sqrt[{3}]{6{\sqrt {30}}+4{\sqrt {5}}}}+{\tfrac {1}{3}}{\sqrt[{3}]{6{\sqrt {30}}-4{\sqrt {5}}}}\,)-{\tfrac {1}{4}}\pi {\bigr ]}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6abdeb36fe2008a02304892182a7645670d38557)
![{\displaystyle R{\bigl [}\exp(-2{\sqrt {2}}\,\pi ){\bigr ]}\oplus R{\bigl [}\exp(-2{\sqrt {2}}\,\pi ){\bigr ]}=\Phi ^{-1}\ominus \tan {\bigl [}2\arctan({\tfrac {1}{3}}{\sqrt {5}}-{\tfrac {1}{3}}{\sqrt[{3}]{6{\sqrt {30}}+4{\sqrt {5}}}}+{\tfrac {1}{3}}{\sqrt[{3}]{6{\sqrt {30}}-4{\sqrt {5}}}}\,)-{\tfrac {1}{4}}\pi {\bigr ]}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/dc072e8da9259e662e23173f6da154478d0fd633)
![{\displaystyle R{\bigl [}\exp(-{\sqrt {2}}\,\pi ){\bigr ]}^{2}R{\bigl [}\exp(-2{\sqrt {2}}\,\pi ){\bigr ]}^{-1}\oplus R{\bigl [}\exp(-{\sqrt {2}}\,\pi ){\bigr ]}R{\bigl [}\exp(-2{\sqrt {2}}\,\pi ){\bigr ]}^{2}=1}](https://wikimedia.org/api/rest_v1/media/math/render/svg/21afb95996f9d394404603afcc761735ad3fad2b)
![{\displaystyle {\bigl \{}R{\bigl [}\exp(-{\sqrt {2}}\,\pi ){\bigr ]}^{2}R{\bigl [}\exp(-2{\sqrt {2}}\,\pi ){\bigr ]}^{-1}+1{\bigr \}}{\bigl \{}R{\bigl [}\exp(-{\sqrt {2}}\,\pi ){\bigr ]}R{\bigl [}\exp(-2{\sqrt {2}}\,\pi ){\bigr ]}^{2}+1{\bigr \}}=2}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4f8fd7359989247d309f916d5070b83508ca0b57)
![{\displaystyle R{\bigl [}\exp(-2{\sqrt {2}}\,\pi ){\bigr ]}=\Phi ^{-1}\ominus R{\bigl [}\exp(-{\sqrt {2}}\,\pi ){\bigr ]}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/83cecbb4dec7aa95cc6c182f11623fdb86060cde)
![{\displaystyle R{\bigl [}\exp(-2{\sqrt {2}}\,\pi ){\bigr ]}={\frac {1-\Phi R{\bigl [}\exp(-{\sqrt {2}}\,\pi ){\bigr ]}}{\Phi +R{\bigl [}\exp(-{\sqrt {2}}\,\pi ){\bigr ]}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cac30d6d5d5bf53533d3bc573c7b69a547912fa2)
![{\displaystyle {\biggl \{}R{\bigl [}\exp(-{\sqrt {2}}\,\pi ){\bigr ]}^{2}{\frac {\Phi +R{\bigl [}\exp(-{\sqrt {2}}\,\pi ){\bigr ]}}{1-\Phi R{\bigl [}\exp(-{\sqrt {2}}\,\pi ){\bigr ]}}}+1{\biggr \}}{\biggl \langle }R{\bigl [}\exp(-{\sqrt {2}}\,\pi ){\bigr ]}{\frac {{\bigl \{}1-\Phi R{\bigl [}\exp(-{\sqrt {2}}\,\pi ){\bigr ]}{\bigr \}}^{2}}{{\bigl \{}\Phi +R{\bigl [}\exp(-{\sqrt {2}}\,\pi ){\bigr ]}{\bigr \}}^{2}}}+1{\biggr \rangle }=2}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d00ea266326bcdbb8873cfc054be0e6653478012)
![{\displaystyle R{\bigl [}\exp(-{\sqrt {2}}\,\pi ){\bigr ]}^{6}+2\,\Phi ^{-2}R{\bigl [}\exp(-{\sqrt {2}}\,\pi ){\bigr ]}^{5}-{\sqrt {5}}\,\Phi ^{-1}R{\bigl [}\exp(-{\sqrt {2}}\,\pi ){\bigr ]}^{4}+}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8970c3dd6242ae8dee9d39d0ff593b54c69c90b7)
![{\displaystyle +2\,{\sqrt {5}}\,\Phi R{\bigl [}\exp(-{\sqrt {2}}\,\pi ){\bigr ]}^{3}+{\sqrt {5}}\,\Phi ^{-1}R{\bigl [}\exp(-{\sqrt {2}}\,\pi ){\bigr ]}^{2}+2\,\Phi ^{-2}R{\bigl [}\exp(-{\sqrt {2}}\,\pi ){\bigr ]}-1=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f8e9e08500522cf65d470bbaa6ff78cc728d9428)
![{\displaystyle R{\bigl [}\exp(-{\sqrt {2}}\,\pi ){\bigr ]}=\tan {\bigl [}\arctan({\tfrac {1}{3}}{\sqrt {5}}-{\tfrac {1}{3}}{\sqrt[{3}]{6{\sqrt {30}}+4{\sqrt {5}}}}+{\tfrac {1}{3}}{\sqrt[{3}]{6{\sqrt {30}}-4{\sqrt {5}}}}\,)-{\tfrac {1}{4}}\operatorname {arccot} (2){\bigr ]}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1d1ba43a88aafaece47de397366208a98b53304d)
![{\displaystyle {\frac {1}{R^{5}(q)}}-R^{5}(q)=\left[{\frac {\eta (\tau )}{\eta (5\tau )}}\right]^{6}+11}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1bd161c8522f2d0446ee11792d6f37d8f8aeb7d5)
![{\displaystyle R(x)=\tan {\biggl \{}{\frac {1}{2}}\operatorname {arccot} {\biggl [}{\frac {1}{2}}+{\frac {\theta _{4}(x^{1/5})[5\,\theta _{4}(x^{5})^{2}-\theta _{4}(x)^{2}]}{2\,\theta _{4}(x^{5})[\theta _{4}(x)^{2}-\theta _{4}(x^{1/5})^{2}]}}{\biggr ]}{\biggr \}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bc53eb3cb13875f9bf0ecc132c54d61e4ced3610)
![{\displaystyle R(x)=\tan {\biggl \{}{\frac {1}{2}}\operatorname {arccot} {\biggl [}{\frac {1}{2}}+{\bigg (}{\frac {\theta _{2}(x^{1/10})\,\theta _{3}(x^{1/10})\,\theta _{4}(x^{1/10})}{2^{3}\,\theta _{2}(x^{5/2})\,\theta _{3}(x^{5/2})\,\theta _{4}(x^{5/2})}}{\bigg )}^{1/3}{\biggr ]}{\biggr \}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/357426562078839bb726ac261902f089fe6e47a4)
![{\displaystyle R(x)=\tan {\biggl \{}{\frac {1}{2}}\arctan {\biggl [}{\frac {1}{2}}-{\frac {\theta _{4}(x)^{2}}{2\,\theta _{4}(x^{5})^{2}}}{\biggr ]}{\biggr \}}^{1/5}\times \tan {\biggl \{}{\frac {1}{2}}\operatorname {arccot} {\biggl [}{\frac {1}{2}}-{\frac {\theta _{4}(x)^{2}}{2\,\theta _{4}(x^{5})^{2}}}{\biggr ]}{\biggr \}}^{2/5}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c18d0dec82ce02b84a41c35ae5d0eb624c7b85a8)
![{\displaystyle R(x)=\tan {\biggl \{}{\frac {1}{2}}\arctan {\biggl [}{\frac {1}{2}}-{\frac {\theta _{4}(x^{1/2})^{2}}{2\,\theta _{4}(x^{5/2})^{2}}}{\biggr ]}{\biggr \}}^{2/5}\times \cot {\biggl \{}{\frac {1}{2}}\operatorname {arccot} {\biggl [}{\frac {1}{2}}-{\frac {\theta _{4}(x^{1/2})^{2}}{2\,\theta _{4}(x^{5/2})^{2}}}{\biggr ]}{\biggr \}}^{1/5}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/653ddfd03c750f090257220e797ef12f57d5a12e)
![{\displaystyle R{\big (}e^{-\pi }{\big )}={\frac {1}{2}}\varphi \,({\sqrt {5}}-\varphi ^{3/2})({\sqrt[{4}]{5}}+\varphi ^{3/2})=0,511428\dots }](https://wikimedia.org/api/rest_v1/media/math/render/svg/7d2239d945b4400f5264c067197b803342551d9d)
![{\displaystyle q(k)=\exp {\big [}-\pi K({\sqrt {1-k^{2}}})/K(k){\big ]}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/43ce6a2e5f468a1bce63fdd37a00bfecd4b335a3)
![{\displaystyle R{\big (}q(k){\big )}=\tan {\biggl \{}{\frac {1}{2}}\arctan y{\biggr \}}^{1/5}\tan {\biggl \{}{\frac {1}{2}}\operatorname {arccot} y{\biggr \}}^{2/5}=\left\{{\frac {{\sqrt {y^{2}+1}}-1}{y}}\right\}^{1/5}\left\{y\left[{\sqrt {{\frac {1}{y^{2}}}+1}}-1\right]\right\}^{2/5}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6fc64c3e2822d40b917901360429284d24cbe293)
![{\displaystyle y={\frac {2k^{2}\,{\text{sn}}[{\tfrac {2}{5}}K(k);k]^{2}\,{\text{sn}}[{\tfrac {4}{5}}K(k);k]^{2}}{5-k^{2}\,{\text{sn}}[{\tfrac {2}{5}}K(k);k]^{2}\,{\text{sn}}[{\tfrac {4}{5}}K(k);k]^{2}}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7719b360c0d190bf112e626de3f87cc19e20e34e)
![{\displaystyle x=\left[{\frac {{\sqrt {5}}\,\eta (5\tau )}{\eta (\tau )}}\right]^{6}.\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2314c008974113c9585ab1e63248c776b59c0b02)
![{\displaystyle {\begin{aligned}&j(\tau )=-{\frac {(r^{20}-228r^{15}+494r^{10}+228r^{5}+1)^{3}}{r^{5}(r^{10}+11r^{5}-1)^{5}}}\\[6pt]&j(\tau )-1728=-{\frac {(r^{30}+522r^{25}-10005r^{20}-10005r^{10}-522r^{5}+1)^{2}}{r^{5}(r^{10}+11r^{5}-1)^{5}}}\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/883827756d7f47131c78695dfb009752642db072)
![{\displaystyle {\begin{aligned}&j(5\tau )=-{\frac {(r^{20}+12r^{15}+14r^{10}-12r^{5}+1)^{3}}{r^{25}(r^{10}+11r^{5}-1)}}\\[6pt]&j(5\tau )-1728=-{\frac {(r^{30}+18r^{25}+75r^{20}+75r^{10}-18r^{5}+1)^{2}}{r^{25}(r^{10}+11r^{5}-1)}}\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b0654694c4cb81f6e887c2fe76573c7378e1202e)
![{\displaystyle {\begin{aligned}&z_{\infty }=-\left[{\frac {{\sqrt {5}}\,\eta (25\tau )}{\eta (5\tau )}}\right]^{6}-11,\ z_{0}=-\left[{\frac {\eta (\tau )}{\eta (5\tau )}}\right]^{6}-11,\ z_{1}=\left[{\frac {\eta ({\frac {5\tau +2}{5}})}{\eta (5\tau )}}\right]^{6}-11,\\[6pt]&z_{2}=-\left[{\frac {\eta ({\frac {5\tau +4}{5}})}{\eta (5\tau )}}\right]^{6}-11,\ z_{3}=\left[{\frac {\eta ({\frac {5\tau +6}{5}})}{\eta (5\tau )}}\right]^{6}-11,\ z_{4}=-\left[{\frac {\eta ({\frac {5\tau +8}{5}})}{\eta (5\tau )}}\right]^{6}-11\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bfc22fc990829510a6e4dc7225adb6abe4ad177f)
![{\displaystyle {\bigl [}R(e^{-2a})+\varphi {\bigl ]}{\bigl [}R(e^{-2b})+\varphi {\bigr ]}={\sqrt {5}}\,\varphi .}](https://wikimedia.org/api/rest_v1/media/math/render/svg/41cf03a262ca7ba345d4adb584a74b89e1f72f56)
![{\displaystyle {\bigl [}R^{5}(e^{-2a})+\varphi ^{5}{\bigl ]}{\bigl [}R^{5}(e^{-2b})+\varphi ^{5}{\bigr ]}=5{\sqrt {5}}\,\varphi ^{5}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e791a2350a718cff4bbb54d7bd2ae8f5c2fbf6d2)
![{\displaystyle q(k)=\exp {\big [}-\pi K({\sqrt {1-k^{2}}})/K(k){\big ]}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e103968699b46b8638435465792de472cc5a077f)
![{\displaystyle k=\tan[{\tfrac {1}{4}}\pi -{\tfrac {1}{4}}\operatorname {arccsc} (a^{2})].}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a7daf53c185d5126a6f836573079e44c54b13d02)
![{\displaystyle {\begin{aligned}x&={\frac {2-{\bigl \{}1-R[q(k)]{\bigr \}}{\bigl \{}1+R[q(k)^{2}]{\bigr \}}}{{\sqrt {R[q(k)]\,R[q(k)^{2}]}}\,{\sqrt[{4}]{4\cot \langle 4\arctan\{S\}\rangle -3}}}}\\&={\frac {2-{\bigl \{}1-R[q(k)]{\bigr \}}{\bigl \{}1+R[q(k)^{2}]{\bigr \}}}{{\sqrt {R[q(k)]R[q(k)^{2}]}}\,{\sqrt[{4}]{{\frac {2}{S-1}}+{\frac {2}{S+1}}+{\frac {1}{S}}-S-3}}}}.\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/685c5763932156ecc824ae9d8be60ab687407a17)
![{\displaystyle S=R[q(k)]\,R^{2}[q(k)^{2}].}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e2fcdd7b37eadcd19870e06b76bc198c952a77c8)
![{\displaystyle R^{5}[q(k)]=S\left({\frac {1-S}{1+S}}\right)^{2}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/85926c9b781d096afba0c57300edd0daaf2420a8)
![{\displaystyle ({\sqrt[{4}]{5}}x)^{5}-5({\sqrt[{4}]{5}}x)-4({\tfrac {5}{4}}{\sqrt[{4}]{5}})=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/61cbe3f71016c548429ccaa965215a70dd5684a1)
![{\displaystyle a={\tfrac {5}{4}}{\sqrt[{4}]{5}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/58854d16aba0f4a3929b92e77c60e1cf4936b709)
![{\displaystyle k=\tan[{\tfrac {1}{4}}\pi -{\tfrac {1}{4}}\operatorname {arccsc} (a^{2})]={\tfrac {5^{5/4}+{\sqrt {25{\sqrt {5}}-16}}}{5^{5/4}+{\sqrt {25{\sqrt {5}}+16}}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f3325d9072bcc7241357faa1789ccb8f610d72ca)
![{\displaystyle R[q(k)]=0.5633613184\dots }](https://wikimedia.org/api/rest_v1/media/math/render/svg/f14c09d3a3e231c86765a5558de4dc567a5a9a45)
![{\displaystyle R[q(k)^{2}]=0.3706122329\dots }](https://wikimedia.org/api/rest_v1/media/math/render/svg/953a83069f8350cffc6b99371d1aef0cfed50476)
![{\displaystyle x={\frac {2-{\bigl \{}1-R[q(k)]{\bigr \}}{\bigl \{}1+R[q(k)^{2}]{\bigr \}}}{{\sqrt {R[q(k)]\,R[q(k)^{2}]}}\,{\sqrt[{4}]{20\cot \langle 4\arctan\{R[q(k)]\,R[q(k)^{2}]^{2}\}\rangle -15}}}}=1.167303978\dots }](https://wikimedia.org/api/rest_v1/media/math/render/svg/1d376cd22fe9212abeb1d720c8a4d16f5e3b8eed)
![{\displaystyle x^{5}-5x-4{\Bigl (}{\sqrt[{4}]{\tfrac {81}{32}}}{\Bigr )}=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/26c633789bedcf0852350bfe130dea2e4b7a1962)
![{\displaystyle a={\sqrt[{4}]{\tfrac {81}{32}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2b08932766845ca177ab4d228ec02299e00664df)
![{\displaystyle r_{1}=R{\big (}e^{-\pi }{\big )}={\tfrac {1}{2}}\varphi \,({\sqrt {5}}-\varphi ^{3/2})({\sqrt[{4}]{5}}+\varphi ^{3/2})=0,511428\dots }](https://wikimedia.org/api/rest_v1/media/math/render/svg/cf43bdad2722185a9a278aa7a3d39ee5c4b9fbe0)
![{\displaystyle r_{2}=R{\big (}e^{-2\pi }{\big )}={\sqrt[{4}]{5}}\,\varphi ^{1/2}-\varphi =0.284079\dots }](https://wikimedia.org/api/rest_v1/media/math/render/svg/4a831e73343c7f98f716bab3de2dc8c20953883b)
![{\displaystyle r_{4}=R{\big (}e^{-4\pi }{\big )}={\tfrac {1}{2}}\varphi \,({\sqrt {5}}-\varphi ^{3/2})(-{\sqrt[{4}]{5}}+\varphi ^{3/2})=0.081002\dots }](https://wikimedia.org/api/rest_v1/media/math/render/svg/2d9de6c6425583a93eca75d05cd0f7fa97b907aa)
![{\displaystyle {\begin{aligned}x&={\sqrt[{4}]{5}}\,{\frac {2-{\bigl \{}1-r_{1}{\bigr \}}{\bigl \{}1+r_{2}{\bigr \}}}{{\sqrt {r_{1}\,r_{2}}}\,{\sqrt[{4}]{20\cot \langle 4\arctan\{r_{1}\,r_{2}^{2}\}\rangle -15}}}}\\[6pt]&={\sqrt[{4}]{5}}\,{\frac {2-{\bigl \{}1-r_{2}{\bigr \}}{\bigl \{}1+r_{4}{\bigr \}}}{{\sqrt {r_{2}\,r_{4}}}\,{\sqrt[{4}]{20\cot \langle 4\arctan\{r_{2}\,r_{4}^{2}\}\rangle -15}}}}\\[6pt]&={\sqrt[{4}]{8}}=1.681792\dots \end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b4f782ba7e718ea81519365e1e4bc278eb6a7c9f)