Rogers–Ramanujan continued fraction

The Rogers–Ramanujan continued fraction is a continued fraction discovered by Rogers (1894) and independently by Srinivasa Ramanujan, and closely related to the Rogers–Ramanujan identities. It can be evaluated explicitly for a broad class of values of its argument.

Definition[]

Given the functions ${\displaystyle G(q)}$ and ${\displaystyle H(q)}$ 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,

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

${\displaystyle (n|m)}$ denotes the Jacobi symbol.

Modular functions[]

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.

Examples[]

${\displaystyle R{\big (}e^{-\pi }{\big )}={\cfrac {e^{-{\frac {\pi }{5}}}}{1+{\cfrac {e^{-\pi }}{1+{\cfrac {e^{-2\pi }}{1+\ddots }}}}}}={\frac {1}{2}}\phi ({\sqrt {5}}-\phi ^{3/2})(\phi ^{3/2}+{\sqrt[{4}]{5}})}$

${\displaystyle R{\big (}e^{-2\pi }{\big )}={\cfrac {e^{-{\frac {2\pi }{5}}}}{1+{\cfrac {e^{-2\pi }}{1+{\cfrac {e^{-4\pi }}{1+\ddots }}}}}}={{\sqrt[{4}]{5}}\phi ^{1/2}-\phi }}$

${\displaystyle R{\big (}e^{-2{\sqrt {5}}\pi }{\big )}={\cfrac {e^{-{\frac {2\pi }{\sqrt {5}}}}}{1+{\cfrac {e^{-2\pi {\sqrt {5}}}}{1+{\cfrac {e^{-4\pi {\sqrt {5}}}}{1+\ddots }}}}}}={\frac {\sqrt {5}}{1+{\big (}5^{3/4}(\phi -1)^{5/2}-1{\big )}^{1/5}}}-{\phi }}$

where ${\displaystyle \phi ={\frac {1+{\sqrt {5}}}{2}}}$ is the golden ratio.

Relation to modular forms[]

${\displaystyle R(q)}$ can be related to the Dedekind eta function, a modular form of weight 1/2, as,^[1]

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

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

Therefore the Rogers-Ramanujan continued fraction can be expressed in terms of Jacobi theta function this way:

${\displaystyle R(x)=\tan {\biggl \langle }{\frac {1}{2}}\operatorname {arccot} {\biggl \{}{\frac {1}{2}}{\biggl [}{\frac {\vartheta _{00}(x^{1/10})\vartheta _{01}(x^{1/10})\vartheta _{10}(x^{1/10})}{\vartheta _{00}(x^{5/2})\vartheta _{01}(x^{5/2})\vartheta _{10}(x^{5/2})}}{\biggr ]}^{1/3}+{\frac {1}{2}}{\biggr \}}{\biggr \rangle }}$

${\displaystyle R(x)=\tan {\biggl \{}{\frac {1}{2}}\arctan {\biggl [}{\frac {1}{2}}-{\frac {\vartheta _{01}(x)^{2}}{2\vartheta _{01}(x^{5})^{2}}}{\biggr ]}{\biggr \}}^{1/5}\tan {\biggl \{}{\frac {1}{2}}\operatorname {arccot} {\biggl [}{\frac {1}{2}}-{\frac {\vartheta _{01}(x)^{2}}{2\vartheta _{01}(x^{5})^{2}}}{\biggr ]}{\biggr \}}^{2/5}}$

${\displaystyle R(x)=\tan {\biggl \{}{\frac {1}{2}}\arctan {\biggl [}{\frac {1}{2}}-{\frac {\vartheta _{01}(x^{1/2})^{2}}{2\vartheta _{01}(x^{5/2})^{2}}}{\biggr ]}{\biggr \}}^{2/5}\cot {\biggl \{}{\frac {1}{2}}\operatorname {arccot} {\biggl [}{\frac {1}{2}}-{\frac {\vartheta _{01}(x^{1/2})^{2}}{2\vartheta _{01}(x^{5/2})^{2}}}{\biggr ]}{\biggr \}}^{1/5}}$

Definition of the nome function:

${\displaystyle q(k)=\exp[-\pi K({\sqrt {1-k^{2}}})K(k)^{-1}]}$

The small letter k describes the elliptic modulus and the big letter K describes the complete elliptic integral of the first kind.

The continued fraction is related to the Jacobi elliptic functions as follows:

${\displaystyle R[q(k)]=\tan ^{1/5}{\biggl \{}{\frac {1}{2}}\arctan y{\biggr \}}\tan ^{2/5}{\biggl \{}{\frac {1}{2}}\operatorname {arccot} y{\biggr \}}=\left\{{\frac {{\sqrt {y^{2}+1}}-1}{y}}\right\}^{1/5}\left\{y\left[{\sqrt {{\frac {1}{y^{2}}}+1}}-1\right]\right\}^{2/5}}$

with

${\displaystyle y={\frac {2k^{2}{\text{sn}}[{\tfrac {2}{5}}K(k);k]^{2}{\text{sn}}[{\tfrac {4}{5}}K(k);k]^{2}}{5-k^{2}{\text{sn}}[{\tfrac {2}{5}}K(k);k]^{2}{\text{sn}}[{\tfrac {4}{5}}K(k);k]^{2}}}.}$

Relation to j-function[]

One formula involving the j-function and the Dedekind eta function is this:

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

where

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

Eliminating the eta quotient ${\displaystyle x}$, 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 ${\displaystyle X_{1}(5)}$.

Functional equation[]

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,

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

Modular equations[]

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

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

Derivatives[]

For ${\displaystyle |q|<1}$, the first order derivative of ${\displaystyle R(q)}$ given in terms of the Euler function ${\displaystyle f(-q):=\prod _{n=1}^{\infty }(1-q^{n})}$ is ^[4]

${\displaystyle R'(q)=5^{-1}q^{-5/6}f(-q)^{4}R(q){\sqrt[{6}]{R(q)^{-5}-11-R(q)^{5}}}.}$

Setting ${\displaystyle q=\exp(2\pi iz)}$, where ${\displaystyle \operatorname {Im} (z)>0}$, the function ${\displaystyle f(-q)}$ is related with the Dedekind eta function

${\displaystyle \eta (z)=q^{1/24}f(-q).}$

Ramanujan has stated that, for real ${\displaystyle q}$, with ${\displaystyle |q|<1}$, we have ^[5]

${\displaystyle R(q)={\frac {{\sqrt {5}}-1}{2}}\exp \left({\frac {1}{5}}\int _{1}^{q}{\frac {f(-t)^{5}}{tf(-t^{5})}}dt\right).}$

By taking the logarithmic derivative of the above relation and applying analytic continuation we get

${\displaystyle R'(q)={\frac {f(-q)^{5}}{5qf(-q^{5})}}R(q),}$

with validity to all complex ${\displaystyle q}$ with ${\displaystyle |q|<1}$.

Evaluations[]

Assume ${\displaystyle q=e^{-\pi {\sqrt {r}}}}$, ${\displaystyle r>0}$ and ${\displaystyle 0<k_{r}<1}$ is the root of the equation ${\displaystyle K\left(k'_{r}\right)/K(k_{r})={\sqrt {r}}}$, ${\displaystyle k'_{r}={\sqrt {1-k_{r}^{2}}}}$ (here ${\displaystyle k_{r}}$ is the elliptic singular modulus associated with the nome (mathematics) and ${\displaystyle K(x)=\pi /2\cdot {}_{2}F_{1}(1/2,1/2;1;x^{2})}$ is the complete elliptic integral of the first kind also see ^[6],^[7],^[8]). Then if

${\displaystyle a_{r}=\left({\frac {k'_{r}}{k'_{25r}}}\right)^{2}{\sqrt {\frac {k_{r}}{k_{25r}}}}M_{5}(r)^{-3},}$ we get ^[9]

${\displaystyle R(q)=\left(-{\frac {11}{2}}-{\frac {a_{r}}{2}}+{\frac {1}{2}}{\sqrt {125+22a_{r}+a_{r}^{2}}}\right)^{1/5},}$

where ${\displaystyle M_{5}(r)}$ is root of

${\displaystyle (5x-1)^{5}(1-x)=256(k_{r}k'_{r})^{2}x.}$

Also

${\displaystyle R'(q)={\frac {2^{4/3}(k_{r})^{5/12}(k'_{r})^{5/3}}{5(k_{25r})^{1/12}(k'_{25r})^{1/3}{\sqrt {M_{5}(r)}}}}R(q){\frac {K(k_{r})^{2}}{\pi ^{2}q}}.}$

Examples of evaluations of ${\displaystyle R(e^{-\pi {\sqrt {r}}})}$ and ${\displaystyle R'(e^{-\pi {\sqrt {r}}})}$, can be found in tables in the literature. For instance, if we set ${\displaystyle r=1}$, then ${\displaystyle k_{1}=1/{\sqrt {2}}}$, ${\displaystyle k'_{1}=1/{\sqrt {2}}}$, ${\displaystyle M_{5}(1)=5^{-1}(2+{\sqrt {5}})}$ and [8] pg.334: ${\displaystyle k_{25}^{2}=2^{-1}({\sqrt {5}}-2)^{2}(3-2\cdot 5^{1/4})^{2}}$, [8] pg.331: ${\displaystyle K(k_{1})=\Gamma (1/4)^{2}/(4{\sqrt {\pi }})}$, where ${\displaystyle \Gamma (x)}$ is the Euler gamma function.

Other results[]

Ramanujan found many other interesting results regarding ${\displaystyle R(q)}$.^[10] Let ${\displaystyle a,b\in \mathbb {R} ^{+}}$, ${\displaystyle u=R(e^{-2a})}$, ${\displaystyle v=R(e^{-2b})}$, and ${\displaystyle \phi }$ as the golden ratio.

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

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

The powers of ${\displaystyle R(q)}$ 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,

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

Quintic equations[]

The general quintic equation in Bring-Jerrard Form can be solved in terms of Rogers-Ramanujan continued fraction. For every real value a > 1 this equation can be solved with the function R(q) and the elliptic nome q(k):

${\displaystyle x^{5}-5x-4a=0}$

To solve this equation, the elliptic modulus must be determined according to the following pattern:

${\displaystyle k=\tan[{\tfrac {1}{4}}\pi -{\tfrac {1}{4}}\operatorname {arccsc}(a^{2})]=({\sqrt {a^{2}+1}}-|a|)(|a|+{\sqrt {a^{2}-1}})}$

Then this is the real solution of this quintic equation:

${\displaystyle {\begin{aligned}x&={\frac {2-{\bigl \{}1-R[q(k)]{\bigr \}}{\bigl \{}1+R[q(k)^{2}]{\bigr \}}}{{\sqrt {R[q(k)]R[q(k)^{2}]}}{\sqrt[{4}]{4\cot \langle 4\arctan\{R[q(k)]R[q(k)^{2}]^{2}\}\rangle -3}}}}\\&={\frac {2-{\bigl \{}1-R[q(k)]{\bigr \}}{\bigl \{}1+R[q(k)^{2}]{\bigr \}}}{{\sqrt {R[q(k)]R[q(k)^{2}]}}{\sqrt[{4}]{{\frac {2}{R[q(k)]R[q(k)^{2}]^{2}-1}}+{\frac {2}{R[q(k)]R[q(k)^{2}]^{2}+1}}+{\frac {1}{R[q(k)]R[q(k)^{2}]^{2}}}-R[q(k)]R[q(k)^{2}]^{2}-3}}}}\end{aligned}}}$

For example, the following equation has the following real solution:

${\displaystyle x^{5}-x-1=0}$

${\displaystyle ({\sqrt[{4}]{5}}x)^{5}-5({\sqrt[{4}]{5}}x)-4({\tfrac {5}{4}}{\sqrt[{4}]{5}})=0}$

${\displaystyle x={\frac {2-{\biggl \{}1-R{\biggl [}q{\biggl (}{\tfrac {5^{5/4}+{\sqrt {25{\sqrt {5}}-16}}}{{\sqrt {25{\sqrt {5}}+16}}+5^{5/4}}}{\biggr )}{\biggr ]}{\biggr \}}{\biggl \{}1+R{\biggl [}q{\biggl (}{\tfrac {5^{5/4}+{\sqrt {25{\sqrt {5}}-16}}}{{\sqrt {25{\sqrt {5}}+16}}+5^{5/4}}}{\biggr )}^{2}{\biggr ]}{\biggr \}}}{{\sqrt {R{\biggl [}q{\biggl (}{\tfrac {5^{5/4}+{\sqrt {25{\sqrt {5}}-16}}}{{\sqrt {25{\sqrt {5}}+16}}+5^{5/4}}}{\biggr )}{\biggr ]}R{\biggl [}q{\biggl (}{\tfrac {5^{5/4}+{\sqrt {25{\sqrt {5}}-16}}}{{\sqrt {25{\sqrt {5}}+16}}+5^{5/4}}}{\biggr )}^{2}{\biggr ]}}}{\sqrt[{4}]{20\cot {\biggl \langle }4\arctan {\biggl \{}R{\biggl [}q{\biggl (}{\tfrac {5^{5/4}+{\sqrt {25{\sqrt {5}}-16}}}{{\sqrt {25{\sqrt {5}}+16}}+5^{5/4}}}{\biggr )}{\biggr ]}R{\biggl [}q{\biggl (}{\tfrac {5^{5/4}+{\sqrt {25{\sqrt {5}}-16}}}{{\sqrt {25{\sqrt {5}}+16}}+5^{5/4}}}{\biggr )}^{2}{\biggr ]}^{2}{\biggr \}}{\biggr \rangle }-15}}}}}$

This constant is roughly 1.1673 and it can not be represented by elementary root expressions.

References[]

• 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

External links[]

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