Rogers–Ramanujan continued fraction

Representation of the approximation ${\displaystyle q^{1/5}A_{400}(q)/B_{400}(q)}$ of the Rogers–Ramanujan continued fraction.

Given the functions G(q) and H(q) appearing in the Rogers–Ramanujan identities,

${\displaystyle {\begin{aligned}G(q)&=\sum _{n=0}^{\infty }{\frac {q^{n^{2}}}{(1-q)(1-q^{2})\cdots (1-q^{n})}}=\sum _{n=0}^{\infty }{\frac {q^{n^{2}}}{(q;q)_{n}}}={\frac {1}{(q;q^{5})_{\infty }(q^{4};q^{5})_{\infty }}}\\&=\prod _{n=1}^{\infty }{\frac {1}{(1-q^{5n-1})(1-q^{5n-4})}}\\&={\sqrt[{60}]{qj}}\,_{2}F_{1}\left(-{\tfrac {1}{60}},{\tfrac {19}{60}};{\tfrac {4}{5}};{\tfrac {1728}{j}}\right)\\&={\sqrt[{60}]{q\left(j-1728\right)}}\,_{2}F_{1}\left(-{\tfrac {1}{60}},{\tfrac {29}{60}};{\tfrac {4}{5}};-{\tfrac {1728}{j-1728}}\right)\\&=1+q+q^{2}+q^{3}+2q^{4}+2q^{5}+3q^{6}+\cdots \end{aligned}}}$

and,

${\displaystyle {\begin{aligned}H(q)&=\sum _{n=0}^{\infty }{\frac {q^{n^{2}+n}}{(1-q)(1-q^{2})\cdots (1-q^{n})}}=\sum _{n=0}^{\infty }{\frac {q^{n^{2}+n}}{(q;q)_{n}}}={\frac {1}{(q^{2};q^{5})_{\infty }(q^{3};q^{5})_{\infty }}}\\&=\prod _{n=1}^{\infty }{\frac {1}{(1-q^{5n-2})(1-q^{5n-3})}}\\&={\frac {1}{\sqrt[{60}]{q^{11}j^{11}}}}\,_{2}F_{1}\left({\tfrac {11}{60}},{\tfrac {31}{60}};{\tfrac {6}{5}};{\tfrac {1728}{j}}\right)\\&={\frac {1}{\sqrt[{60}]{q^{11}\left(j-1728\right)^{11}}}}\,_{2}F_{1}\left({\tfrac {11}{60}},{\tfrac {41}{60}};{\tfrac {6}{5}};-{\tfrac {1728}{j-1728}}\right)\\&=1+q^{2}+q^{3}+q^{4}+q^{5}+2q^{6}+2q^{7}+\cdots \end{aligned}}}$

OEIS: A003114 and OEIS: A003106, respectively, where ${\displaystyle (a;q)_{\infty }}$ denotes the infinite q-Pochhammer symbol, j is the j-function, and ₂F₁ is the hypergeometric function, then the Rogers–Ramanujan continued fraction is,

${\displaystyle {\begin{aligned}R(q)&={\frac {q^{\frac {11}{60}}H(q)}{q^{-{\frac {1}{60}}}G(q)}}=q^{\frac {1}{5}}\prod _{n=1}^{\infty }{\frac {(1-q^{5n-1})(1-q^{5n-4})}{(1-q^{5n-2})(1-q^{5n-3})}}\\&={\cfrac {q^{1/5}}{1+{\cfrac {q}{1+{\cfrac {q^{2}}{1+{\cfrac {q^{3}}{1+\ddots }}}}}}}}\end{aligned}}}$

If ${\displaystyle q=e^{2\pi {\rm {i}}\tau }}$, then ${\displaystyle q^{-{\frac {1}{60}}}G(q)}$ and ${\displaystyle q^{\frac {11}{60}}H(q)}$, as well as their quotient ${\displaystyle R(q)}$, are modular functions of ${\displaystyle \tau }$. Since they have integral coefficients, the theory of complex multiplication implies that their values for ${\displaystyle \tau }$ an imaginary quadratic irrational are algebraic numbers that can be evaluated explicitly.

It can be related to the Dedekind eta function, a modular form of weight 1/2, as,[1]

${\displaystyle {\frac {1}{R(q)}}-R(q)={\frac {\eta ({\frac {\tau }{5}})}{\eta (5\tau )}}+1}$

${\displaystyle {\frac {1}{R^{5}(q)}}-R^{5}(q)=\left[{\frac {\eta (\tau )}{\eta (5\tau )}}\right]^{6}+11}$

Among the many formulas of the j-function, one is,

${\displaystyle j(\tau )={\frac {(x^{2}+10x+5)^{3}}{x}}}$

where

${\displaystyle x=\left[{\frac {{\sqrt {5}}\,\eta (5\tau )}{\eta (\tau )}}\right]^{6}}$

Eliminating the eta quotient, one can then express j(τ) in terms of ${\displaystyle r=R(q)}$ as,

${\displaystyle {\begin{aligned}&j(\tau )=-{\frac {(r^{20}-228r^{15}+494r^{10}+228r^{5}+1)^{3}}{r^{5}(r^{10}+11r^{5}-1)^{5}}}\\[6pt]&j(\tau )-1728=-{\frac {(r^{30}+522r^{25}-10005r^{20}-10005r^{10}-522r^{5}+1)^{2}}{r^{5}(r^{10}+11r^{5}-1)^{5}}}\end{aligned}}}$

where the numerator and denominator are polynomial invariants of the icosahedron. Using the modular equation between ${\displaystyle R(q)}$ and ${\displaystyle R(q^{5})}$, one finds that,

${\displaystyle j(5\tau )=-{\frac {(r^{20}+12r^{15}+14r^{10}-12r^{5}+1)^{3}}{r^{25}(r^{10}+11r^{5}-1)}}}$

let ${\displaystyle z=r^{5}-{\frac {1}{r^{5}}}}$,then${\displaystyle j(5\tau )=-{\frac {\left(z^{2}+12z+16\right)^{3}}{z+11}}}$

where

${\displaystyle {\begin{aligned}&z_{\infty }=-\left[{\frac {{\sqrt {5}}\,\eta (25\tau )}{\eta (5\tau )}}\right]^{6}-11,\ z_{0}=-\left[{\frac {\eta (\tau )}{\eta (5\tau )}}\right]^{6}-11,\ z_{1}=\left[{\frac {\eta ({\frac {5\tau +2}{5}})}{\eta (5\tau )}}\right]^{6}-11,\\[6pt]&z_{2}=-\left[{\frac {\eta ({\frac {5\tau +4}{5}})}{\eta (5\tau )}}\right]^{6}-11,\ z_{3}=\left[{\frac {\eta ({\frac {5\tau +6}{5}})}{\eta (5\tau )}}\right]^{6}-11,\ z_{4}=-\left[{\frac {\eta ({\frac {5\tau +8}{5}})}{\eta (5\tau )}}\right]^{6}-11\end{aligned}}}$

which in fact is the j-invariant of the elliptic curve,

${\displaystyle y^{2}+(1+r^{5})xy+r^{5}y=x^{3}+r^{5}x^{2}}$

parameterized by the non-cusp points of the modular curve ${\displaystyle X_{1}(5)}$.

For convenience, one can also use the notation ${\displaystyle r(\tau )=R(q)}$ when q = e^2πiτ. While other modular functions like the j-invariant satisfies,

${\displaystyle j(-{\tfrac {1}{\tau }})=j(\tau )}$

and the Dedekind eta function has,

${\displaystyle \eta (-{\tfrac {1}{\tau }})={\sqrt {-i\tau }}\,\eta (\tau )}$

the functional equation of the Rogers–Ramanujan continued fraction involves[2] the golden ratio ${\displaystyle \phi }$,

${\displaystyle r(-{\tfrac {1}{\tau }})={\frac {1-\phi \,r(\tau )}{\phi +r(\tau )}}}$

Incidentally,

${\displaystyle r({\tfrac {7+i}{10}})=i}$

There are modular equations between ${\displaystyle R(q)}$ and ${\displaystyle R(q^{n})}$. Elegant ones for small prime n are as follows.[3]

For ${\displaystyle n=2}$, let ${\displaystyle u=R(q)}$ and ${\displaystyle v=R(q^{2})}$, then ${\displaystyle v-u^{2}=(v+u^{2})uv^{2}.}$

For ${\displaystyle n=3}$, let ${\displaystyle u=R(q)}$ and ${\displaystyle v=R(q^{3})}$, then ${\displaystyle (v-u^{3})(1+uv^{3})=3u^{2}v^{2}.}$

For ${\displaystyle n=5}$, let ${\displaystyle u=R(q)}$ and ${\displaystyle v=R(q^{5})}$, then ${\displaystyle (v^{4}-3v^{3}+4v^{2}-2v+1)v=(v^{4}+2v^{3}+4v^{2}+3v+1)u^{5}.}$

For ${\displaystyle n=11}$, let ${\displaystyle u=R(q)}$ and ${\displaystyle v=R(q^{11})}$, then ${\displaystyle uv(u^{10}+11u^{5}-1)(v^{10}+11v^{5}-1)=(u-v)^{12}.}$

Regarding ${\displaystyle n=5}$, note that

${\displaystyle v^{10}+11v^{5}-1=(v^{2}+v-1)(v^{4}-3v^{3}+4v^{2}-2v+1)(v^{4}+2v^{3}+4v^{2}+3v+1).}$

Ramanujan found many other interesting results regarding R(q).[4] Let ${\displaystyle u=R(q^{a})}$, ${\displaystyle v=R(q^{b})}$, and ${\displaystyle \phi }$ as the golden ratio.

If ${\displaystyle ab=4\pi ^{2}}$, then ${\displaystyle (u+\phi )(v+\phi )={\sqrt {5}}\,\phi .}$

If ${\displaystyle 5ab=4\pi ^{2}}$, then ${\displaystyle (u^{5}+\phi ^{5})(v^{5}+\phi ^{5})=5{\sqrt {5}}\,\phi ^{5}.}$

The powers of R(q) also can be expressed in unusual ways. For its cube,

${\displaystyle R^{3}(q)={\frac {\alpha }{\beta }}}$

where,

${\displaystyle \alpha =\sum _{n=0}^{\infty }{\frac {q^{2n}}{1-q^{5n+2}}}-\sum _{n=0}^{\infty }{\frac {q^{3n+1}}{1-q^{5n+3}}}}$

${\displaystyle \beta =\sum _{n=0}^{\infty }{\frac {q^{n}}{1-q^{5n+1}}}-\sum _{n=0}^{\infty }{\frac {q^{4n+3}}{1-q^{5n+4}}}}$

For its fifth power, let ${\displaystyle w=R(q)R^{2}(q^{2})}$, then,

${\displaystyle R^{5}(q)=w\left({\frac {1-w}{1+w}}\right)^{2},\;\;R^{5}(q^{2})=w^{2}\left({\frac {1+w}{1-w}}\right)}$

• Rogers, L. J. (1894), "Second Memoir on the Expansion of certain Infinite Products", Proc. London Math. Soc., s1-25 (1): 318–343, doi:10.1112/plms/s1-25.1.318
• Berndt, B. C.; Chan, H. H.; Huang, S. S.; Kang, S. Y.; Sohn, J.; Son, S. H. (1999), "The Rogers–Ramanujan continued fraction" (PDF), Journal of Computational and Applied Mathematics, 105 (1–2): 9–24, doi:10.1016/S0377-0427(99)00033-3

• Weisstein, Eric W. "Rogers-Ramanujan Identities". MathWorld.
• Weisstein, Eric W. "Rogers-Ramanujan Continued Fraction". MathWorld.