# 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 $G(q)$ and $H(q)$ appearing in the Rogers–Ramanujan identities, and assume $q=e^{2\pi i\tau }$,
    ${\begin{aligned}G(q)&=\sum _{n=0}^{\infty }{\frac {q^{n^{2}}}{(1-q)(1-q^{2})\cdots (1-q^{n})}}=\sum _{n=0}^{\infty }{\frac {q^{n^{2}}}{(q;q)_{n}}}={\frac {1}{(q;q^{5})_{\infty }(q^{4};q^{5})_{\infty }}}\\[6pt]&=\prod _{n=1}^{\infty }{\frac {1}{(1-q^{5n-1})(1-q^{5n-4})}}\\[6pt]&={\sqrt[{60}]{q\,j}}\,\,_{2}F_{1}\left(-{\tfrac {1}{60}},{\tfrac {19}{60}};{\tfrac {4}{5}};{\tfrac {1728}{j}}\right)\\[6pt]&={\sqrt[{60}]{q\left(j-1728\right)}}\,_{2}F_{1}\left(-{\tfrac {1}{60}},{\tfrac {29}{60}};{\tfrac {4}{5}};-{\tfrac {1728}{j-1728}}\right)\\[6pt]&=1+q+q^{2}+q^{3}+2q^{4}+2q^{5}+3q^{6}+\cdots \end{aligned}}$
and,
    ${\begin{aligned}H(q)&=\sum _{n=0}^{\infty }{\frac {q^{n^{2}+n}}{(1-q)(1-q^{2})\cdots (1-q^{n})}}=\sum _{n=0}^{\infty }{\frac {q^{n^{2}+n}}{(q;q)_{n}}}={\frac {1}{(q^{2};q^{5})_{\infty }(q^{3};q^{5})_{\infty }}}\\[6pt]&=\prod _{n=1}^{\infty }{\frac {1}{(1-q^{5n-2})(1-q^{5n-3})}}\\[6pt]&={\frac {1}{\sqrt[{60}]{q^{11}j^{11}}}}\,_{2}F_{1}\left({\tfrac {11}{60}},{\tfrac {31}{60}};{\tfrac {6}{5}};{\tfrac {1728}{j}}\right)\\[6pt]&={\frac {1}{\sqrt[{60}]{q^{11}\left(j-1728\right)^{11}}}}\,_{2}F_{1}\left({\tfrac {11}{60}},{\tfrac {41}{60}};{\tfrac {6}{5}};-{\tfrac {1728}{j-1728}}\right)\\[6pt]&=1+q^{2}+q^{3}+q^{4}+q^{5}+2q^{6}+2q^{7}+\cdots \end{aligned}}$
with the coefficients of the q-expansion being OEIS: A003114 and OEIS: A003106, respectively, where $(a;q)_{\infty }$ denotes the infinite q-Pochhammer symbol, j is the j-function, and $_2$F$_1$ is the hypergeometric function. The Rogers–Ramanujan continued fraction is then
    ${\begin{aligned}R(q)&={\frac {q^{\frac {11}{60}}H(q)}{q^{-{\frac {1}{60}}}G(q)}}=q^{\frac {1}{5}}\prod _{n=1}^{\infty }{\frac {(1-q^{5n-1})(1-q^{5n-4})}{(1-q^{5n-2})(1-q^{5n-3})}}=q^{1/5}\prod _{n=1}^{\infty }(1-q^{n})^{(n|5)}\\[8pt]&={\cfrac {q^{1/5}}{1+{\cfrac {q}{1+{\cfrac {q^{2}}{1+{\cfrac {q^{3}}{1+\ddots }}}}}}}}\end{aligned}}$
    $(n\mid m)$ is the Jacobi symbol.
One should be careful with notation since the formulas employing the j-function $j$ will be consistent with the other formulas only if $q=e^{2\pi i\tau }$ (the square of the nome) is used throughout this section since the q-expansion of the j-function (as well as the well-known Dedekind eta function) uses $q=e^{2\pi i\tau }$. However, Ramanujan, in his examples to Hardy and given below, used the nome $q=e^{\pi i\tau }$instead.

## Special values

If q is the nome or its square, then $q^{-{\frac {1}{60}}}G(q)$ and $q^{\frac {11}{60}}H(q)$, as well as their quotient $R(q)$, are related to modular functions of $\tau$. Since they have integral coefficients, the theory of complex multiplication implies that their values for $\tau$ involving an imaginary quadratic field are algebraic numbers that can be evaluated explicitly.

### Examples of R(q)

Given the general form where Ramanujan used the nome $q=e^{\pi i\tau }$,
    $R(q)={\cfrac {q^{1/5}}{1+{\cfrac {q}{1+{\cfrac {q^{2}}{1+{\cfrac {q^{3}}{1+\ddots }}}}}}}}$
f when $\tau =i$,
    $R{\bige^{-\pi }{\big )}={\cfrac {e^{-{\frac {\pi }{5}}}}{1+{\cfrac {e^{-\pi }}{1+{\cfrac {e^{-2\pi }}{1+\ddots }}}}}}={\tfrac {1}{2}}\varphi \,({\sqrt {5}}-\varphi ^{3/2})({\sqrt[{4}]{5}}+\varphi ^{3/2})=0.511428\dots$
when $\tau =2i$,
    $R{\bige^{-2\pi }{\big )}={\cfrac {e^{-{\frac {2\pi }{5}}}}{1+{\cfrac {e^{-2\pi }}{1+{\cfrac {e^{-4\pi }}{1+\ddots }}}}}}={{\sqrt[{4}]{5}}\,\varphi ^{1/2}-\varphi }=0.284079\dots$
when $\tau =4i$,
    $R{\bige^{-4\pi }{\big )}={\cfrac {e^{-{\frac {4\pi }{5}}}}{1+{\cfrac {e^{-4\pi }}{1+{\cfrac {e^{-8\pi }}{1+\ddots }}}}}}={\tfrac {1}{2}}\varphi \,({\sqrt {5}}-\varphi ^{3/2})(-{\sqrt[{4}]{5}}+\varphi ^{3/2})=0.081002\dots$
when $\tau =2{\sqrt {5}}i$,
    $R{\bige^{-2{\sqrt {5}}\pi }{\big )}={\cfrac {e^{-{\frac {2\pi }{\sqrt {5}}}}}{1+{\cfrac {e^{-2\pi {\sqrt {5}}}}{1+{\cfrac {e^{-4\pi {\sqrt {5}}}}{1+\ddots }}}}}}={\frac {\sqrt {5}}{1+{\big5^{3/4}(\varphi -1)^{5/2}-1{\big )}^{1/5}}}-\varphi =0.0602094\dots$
when $\tau =5i$,
    $R{\bige^{-5\pi }{\big )}={\cfrac {e^{-\pi }}{1+{\cfrac {e^{-5\pi }}{1+{\cfrac {e^{-10\pi }}{1+\ddots }}}}}}={\frac {1+\varphi ^{2}}{\varphi +{\big{\frac {1}{2}}(4-\varphi -3{\sqrt {\varphi -1}})(3\varphi ^{3/2}-{\sqrt[{4}]{5}}){\big )}^{1/5}}}-\varphi =0.0432139\dots$
when $\tau =10i$,
    $R{\bige^{-10\pi }{\big )}={\cfrac {e^{-2\pi }}{1+{\cfrac {e^{-10\pi }}{1+{\cfrac {e^{-20\pi }}{1+\ddots }}}}}}={\frac {1+\varphi ^{2}}{\varphi +{\big3{\sqrt {1+\varphi ^{2}}}-4-\varphi {\big )}^{1/5}}}-\varphi =0.00186744\dots$
when $\tau =20i$,
    $R{\bige^{-20\pi }{\big )}={\cfrac {e^{-4\pi }}{1+{\cfrac {e^{-20\pi }}{1+{\cfrac {e^{-40\pi }}{1+\ddots }}}}}}={\frac {1+\varphi ^{2}}{\varphi +{\big{\frac {1}{2}}(4-\varphi -3{\sqrt {\varphi -1}})(3\varphi ^{3/2}+{\sqrt[{4}]{5}}){\big )}^{1/5}}}-\varphi =0.00000348734\dots$
and $\varphi ={\tfrac {1+{\sqrt {5}}}{2}}$ is the golden ratio. Note that $R{\bige^{-2\pi }{\big )}$ is a positive root of the quartic equation,
    $x^{4}+2x^{3}-6x^{2}-2x+1=0$
while $R{\bige^{-\pi }{\big )}$ and $R{\bige^{-4\pi }{\big )}$ are two positive roots of a single octic,
    $y^{4}+2\varphi ^{4}y^{3}+6\varphi ^{2}y^{2}-2\varphi ^{4}y+1=0$
(since $\varphi$ has a square root) which explains the similarity of the two closed-forms. More generally, for positive integer m, then $R(e^{-2\pi /m})$ and $R(e^{-2\pi \,m})$ are two roots of the same equation as well as,
    ${\biglR(e^{-2\pi /m})+\varphi {\bigr ]}{\biglR(e^{-2\pi \,m})+\varphi {\bigr ]}={\sqrt {5}}\,\varphi$
The algebraic degree k of $R(e^{-\pi \,n})$ for $n=1,2,3,4,\dots$ is $k=8,4,32,8,\dots$ (OEIS: A082682).
Incidentally, these continued fractions can be used to solve some quintic equations as shown in a later section.

### Examples of G(q) and H(q)

Interestingly, there are explicit formulas for $G(q)$ and $H(q)$ in terms of the j-function $j(\tau )$ and the Rogers-Ramanujan continued fraction $R(q)$. However, since $j(\tau )$ uses the nome's square $q=e^{2\pi \,i\tau }$, then one should be careful with notation such that $j(\tau ),\,G(q),\,H(q)$ and $r=R(q)$ use the same $q$.
    ${\begin{aligned}G(q)&=\prod _{n=1}^{\infty }{\frac {1}{(1-q^{5n-1})(1-q^{5n-4})}}\\[6pt]&=q^{1/60}{\frac {j(\tau )^{1/60}}{(r^{20}-228r^{15}+494r^{10}+228r^{5}+1)^{1/20}}}\end{aligned}}$
    ${\begin{aligned}H(q)&=\prod _{n=1}^{\infty }{\frac {1}{(1-q^{5n-2})(1-q^{5n-3})}}\\[6pt]&={\frac {-1}{q^{11/60}}}{\frac {(r^{20}-228r^{15}+494r^{10}+228r^{5}+1)^{11/20}}{j(\tau )^{11/60}\,(r^{10}+11r^{5}-1)}}\end{aligned}}$
Of course, the secondary formulas imply that $q^{-1/60}G(q)$ and $q^{11/60}H(q)$ are algebraic numbers (though normally of high degree) for $\tau$ involving an imaginary quadratic field. For example, the formulas above simplify to,
    ${\begin{aligned}G(e^{-2\pi })&=(e^{-2\pi })^{1/60}{\frac {1}{(5\,\varphi )^{1/4}}}{\frac {1}{\sqrt {R(e^{-2\pi })}}}\\[6pt]&=1.00187093\dots \\[6pt]H(e^{-2\pi })&={\frac {1}{(e^{-2\pi })^{11/60}}}{\frac {1}{(5\,\varphi )^{1/4}}}{\sqrt {R(e^{-2\pi })}}\\[6pt]&=1.00000349\ldots \end{aligned}}$
and,
    ${\begin{aligned}G(e^{-4\pi })&=(e^{-4\pi })^{1/60}{\frac {1}{(5\,\varphi ^{3})^{1/4}\,(\varphi +{\sqrt[{4}]{5}})^{1/4}}}{\frac {1}{\sqrt {R(e^{-4\pi })}}}\\[6pt]&=1.000003487354\dots \\[6pt]H(e^{-4\pi })&={\frac {1}{(e^{-4\pi })^{11/60}}}{\frac {1}{(5\,\varphi ^{3})^{1/4}\,(\varphi +{\sqrt[{4}]{5}})^{1/4}}}{\sqrt {R(e^{-4\pi })}}\\[6pt]&=1.000000000012\dots \end{aligned}}$
and so on, with $\varphi$ as the golden ratio.

## Derivation of special values
### Tangential sums

In the following we express the essential theorems of the Rogers-Ramanujan continued fractions R and S by using the tangential sums and tangential differences:
    $a\oplus b=\tan {\bigl\arctan(a)+\arctan(b){\bigr ]}={\frac {a+b}{1-ab}}$
    $c\ominus d=\tan {\bigl\arctan(c)-\arctan(d){\bigr ]}={\frac {c-d}{1+cd}}$
The elliptic nome and the complementary nome have this relationship to each other:
    $\ln(q)\ln(q_{1})=\pi ^{2}$
The complementary nome of a modulus k is equal to the nome of the Pythagorean complementary modulus:
    $q_{1}(k)=q(k')=q({\sqrt {1-k^{2}}})$
These are the reflection theorems for the continued fractions R and S:
    $S(q)\oplus S(q_{1})=\Phi$ $R(q^{2})\oplus R(q_{1}^{2})=\Phi ^{-1}$
The letter $\Phi$ represents the Golden number exactly:
    $\Phi ={\tfrac {1}{2}}({\sqrt {5}}+1)=\cot[{\tfrac {1}{2}}\arctan(2)]=2\cos({\tfrac {1}{5}}{\pi })$
    $\Phi ^{-1}={\tfrac {1}{2}}({\sqrt {5}}-1)=\tan[{\tfrac {1}{2}}\arctan(2)]=2\sin({\tfrac {1}{10}}{\pi })$
The theorems for the squared nome are constructed as follows:
    $R(q)^{2}R(q^{2})^{-1}\oplus R(q)R(q^{2})^{2}=1$ $S(q)^{2}R(q^{2})^{-1}\ominus S(q)R(q^{2})^{2}=1$
Following relations between the continued fractions and the Jacobi theta functions are given:
    $S(q)\oplus R(q^{2})={\frac {\vartheta _{00}(q^{1/5})^{2}-\vartheta _{00}(q)^{2}}{5\,\vartheta _{00}(q^{5})^{2}-\vartheta _{00}(q)^{2}}}$ $R(q)\ominus R(q^{2})={\frac {\vartheta _{01}(q)^{2}-\vartheta _{01}(q^{1/5})^{2}}{5\,\vartheta _{01}(q^{5})^{2}-\vartheta _{01}(q)^{2}}}$

### Derivation of Lemniscatic values

Into the now shown theorems certain values are inserted:
    $S{\bigl\exp(-\pi ){\bigr ]}\oplus S{\bigl\exp(-\pi ){\bigr ]}=\Phi$
Therefore following identity is valid:
    $S{\bigl\exp(-\pi ){\bigr ]}=\tan {\bigl{\tfrac {1}{2}}\arctan(\Phi ){\bigr ]}=\tan {\bigl{\tfrac {1}{4}}\pi -{\tfrac {1}{4}}\arctan(2){\bigr ]}$
In an analogue pattern we get this result:
    $R{\bigl\exp(-2\pi ){\bigr ]}\oplus R{\bigl\exp(-2\pi ){\bigr ]}=\Phi ^{-1}$
Therefore following identity is valid:
    $R{\bigl\exp(-2\pi ){\bigr ]}=\tan {\bigl{\tfrac {1}{2}}\arctan(\Phi ^{-1}){\bigr ]}=\tan {\bigl{\tfrac {1}{4}}\arctan(2){\bigr ]}$
Furthermore we get the same relation by using the above mentioned theorem about the Jacobi theta functions:
    $S{\bigl\exp(-\pi ){\bigr ]}\oplus R{\bigl\exp(-2\pi ){\bigr ]}=S(q)\oplus R(q^{2}){\biglq=\exp(-\pi ){\bigr ]}=$
    $={\frac {\vartheta _{00}(q^{1/5})^{2}-\vartheta _{00}(q)^{2}}{5\,\vartheta _{00}(q^{5})^{2}-\vartheta _{00}(q)^{2}}}{\biglq=\exp(-\pi ){\bigr ]}=1$
This result appears because of the Poisson summation formula and this equation can be solved in this way:
    $R{\bigl\exp(-2\pi ){\bigr ]}=1\ominus S{\bigl\exp(-\pi ){\bigr ]}=1\ominus \tan {\bigl{\tfrac {1}{4}}\pi -{\tfrac {1}{4}}\arctan(2){\bigr ]}=\tan {\bigl{\tfrac {1}{4}}\arctan(2){\bigr ]}$
By taking the other mentioned theorem about the Jacobi theta functions a next value can be determined:
    $R{\bigl\exp(-\pi ){\bigr ]}\ominus R{\bigl\exp(-2\pi ){\bigr ]}=R(q)\ominus R(q^{2}){\biglq=\exp(-\pi ){\bigr ]}=$
    $={\frac {\vartheta _{01}(q)^{2}-\vartheta _{01}(q^{1/5})^{2}}{5\,\vartheta _{01}(q^{5})^{2}-\vartheta _{01}(q)^{2}}}{\biglq=\exp(-\pi ){\bigr ]}={\frac {{\sqrt[{4}]{5}}-1}{{\sqrt[{4}]{5}}+1}}={\sqrt[{4}]{5}}\ominus 1=\tan {\bigl\arctan({\sqrt[{4}]{5}}\,)-{\tfrac {1}{4}}\pi {\bigr ]}$
That equation chain leads to this tangential sum:
    $R{\bigl\exp(-\pi ){\bigr ]}=R{\bigl\exp(-2\pi ){\bigr ]}\oplus \tan {\bigl\arctan({\sqrt[{4}]{5}}\,)-{\tfrac {1}{4}}\pi {\bigr ]}$
And therefore following result appears:
    $R{\bigl\exp(-\pi ){\bigr ]}=\tan {\bigl{\tfrac {1}{4}}\arctan(2)+\arctan({\sqrt[{4}]{5}}\,)-{\tfrac {1}{4}}\pi {\bigr ]}$
In the next step we use the reflection theorem for the continued fraction R again:
    $R{\bigl\exp(-\pi ){\bigr ]}\oplus R{\bigl\exp(-4\pi ){\bigr ]}=\Phi ^{-1}$
    $R{\bigl\exp(-4\pi ){\bigr ]}=\tan {\bigl{\tfrac {1}{2}}\arctan(2){\bigr ]}\ominus R{\bigl\exp(-\pi ){\bigr ]}$
And a further result appears:
    $R{\bigl\exp(-4\pi ){\bigr ]}=\tan {\bigl{\tfrac {1}{4}}\arctan(2)-\arctan({\sqrt[{4}]{5}}\,)+{\tfrac {1}{4}}\pi {\bigr ]}$

### Derivation of Non-Lemniscatic values

The reflection theorem is now used for following values:
    $R{\bigl\exp(-{\sqrt {2}}\,\pi ){\bigr ]}\oplus R{\bigl\exp(-2{\sqrt {2}}\,\pi ){\bigr ]}=\Phi ^{-1}$
The Jacobi theta theorem leads to a further relation:
    $R{\bigl\exp(-{\sqrt {2}}\,\pi ){\bigr ]}\ominus R{\bigl\exp(-2{\sqrt {2}}\,\pi ){\bigr ]}=R(q)\ominus R(q^{2}){\biglq=\exp(-{\sqrt {2}}\,\pi ){\bigr ]}=$
    $={\frac {\vartheta _{01}(q)^{2}-\vartheta _{01}(q^{1/5})^{2}}{5\,\vartheta _{01}(q^{5})^{2}-\vartheta _{01}(q)^{2}}}{\biglq=\exp(-{\sqrt {2}}\,\pi ){\bigr ]}=\tan {\bigl2\arctan({\tfrac {1}{3}}{\sqrt {5}}-{\tfrac {1}{3}}{\sqrt[{3}]{6{\sqrt {30}}+4{\sqrt {5}}}}+{\tfrac {1}{3}}{\sqrt[{3}]{6{\sqrt {30}}-4{\sqrt {5}}}}\,)-{\tfrac {1}{4}}\pi {\bigr ]}$
By tangential adding the now mentioned two theorems we get this result:
    $R{\bigl\exp(-{\sqrt {2}}\,\pi ){\bigr ]}\oplus R{\bigl\exp(-{\sqrt {2}}\,\pi ){\bigr ]}=\Phi ^{-1}\oplus \tan {\bigl2\arctan({\tfrac {1}{3}}{\sqrt {5}}-{\tfrac {1}{3}}{\sqrt[{3}]{6{\sqrt {30}}+4{\sqrt {5}}}}+{\tfrac {1}{3}}{\sqrt[{3}]{6{\sqrt {30}}-4{\sqrt {5}}}}\,)-{\tfrac {1}{4}}\pi {\bigr ]}$
    $R{\bigl\exp(-{\sqrt {2}}\,\pi ){\bigr ]}=\tan {\bigl\arctan({\tfrac {1}{3}}{\sqrt {5}}-{\tfrac {1}{3}}{\sqrt[{3}]{6{\sqrt {30}}+4{\sqrt {5}}}}+{\tfrac {1}{3}}{\sqrt[{3}]{6{\sqrt {30}}-4{\sqrt {5}}}}\,)-{\tfrac {1}{4}}\operatorname {arccot}(2){\bigr ]}$
By tangential substraction that result appears:
    $R{\bigl\exp(-2{\sqrt {2}}\,\pi ){\bigr ]}\oplus R{\bigl\exp(-2{\sqrt {2}}\,\pi ){\bigr ]}=\Phi ^{-1}\ominus \tan {\bigl2\arctan({\tfrac {1}{3}}{\sqrt {5}}-{\tfrac {1}{3}}{\sqrt[{3}]{6{\sqrt {30}}+4{\sqrt {5}}}}+{\tfrac {1}{3}}{\sqrt[{3}]{6{\sqrt {30}}-4{\sqrt {5}}}}\,)-{\tfrac {1}{4}}\pi {\bigr ]}$
    $R{\bigl\exp(-2{\sqrt {2}}\,\pi ){\bigr ]}=\tan {\bigl{\tfrac {1}{4}}\operatorname {arccot}(-2)-\arctan({\tfrac {1}{3}}{\sqrt {5}}-{\tfrac {1}{3}}{\sqrt[{3}]{6{\sqrt {30}}+4{\sqrt {5}}}}+{\tfrac {1}{3}}{\sqrt[{3}]{6{\sqrt {30}}-4{\sqrt {5}}}}\,){\bigr ]}$
In an alternative solution way we use the theorem for the squared nome:
    $R{\bigl\exp(-{\sqrt {2}}\,\pi ){\bigr ]}^{2}R{\bigl\exp(-2{\sqrt {2}}\,\pi ){\bigr ]}^{-1}\oplus R{\bigl\exp(-{\sqrt {2}}\,\pi ){\bigr ]}R{\bigl\exp(-2{\sqrt {2}}\,\pi ){\bigr ]}^{2}=1$
    ${\bigl \R{\bigl\exp(-{\sqrt {2}}\,\pi ){\bigr ]}^{2}R{\bigl\exp(-2{\sqrt {2}}\,\pi ){\bigr ]}^{-1}+1{\bigr \}}{\bigl \R{\bigl\exp(-{\sqrt {2}}\,\pi ){\bigr ]}R{\bigl\exp(-2{\sqrt {2}}\,\pi ){\bigr ]}^{2}+1{\bigr \}}=2$
Now the reflection theorem is taken again:
    $R{\bigl\exp(-2{\sqrt {2}}\,\pi ){\bigr ]}=\Phi ^{-1}\ominus R{\bigl\exp(-{\sqrt {2}}\,\pi ){\bigr ]}$
    $R{\bigl\exp(-2{\sqrt {2}}\,\pi ){\bigr ]}={\frac {1-\Phi R{\bigl\exp(-{\sqrt {2}}\,\pi ){\bigr ]}}{\Phi +R{\bigl\exp(-{\sqrt {2}}\,\pi ){\bigr ]}}}$
The insertion of the last mentioned expression into the squared nome theorem gives that equation:
    ${\biggl \R{\bigl\exp(-{\sqrt {2}}\,\pi ){\bigr ]}^{2}{\frac {\Phi +R{\bigl\exp(-{\sqrt {2}}\,\pi ){\bigr ]}}{1-\Phi R{\bigl\exp(-{\sqrt {2}}\,\pi ){\bigr ]}}}+1{\biggr \}}{\biggl \langle }R{\bigl\exp(-{\sqrt {2}}\,\pi ){\bigr ]}{\frac {{\bigl \1-\Phi R{\bigl\exp(-{\sqrt {2}}\,\pi ){\bigr ]}{\bigr \}}^{2}}{{\bigl \\Phi +R{\bigl\exp(-{\sqrt {2}}\,\pi ){\bigr ]}{\bigr \}}^{2}}}+1{\biggr \rangle }=2$
Erasing the denominators gives an equation of sixth degree:
    $R{\bigl\exp(-{\sqrt {2}}\,\pi ){\bigr ]}^{6}+2\,\Phi ^{-2}R{\bigl\exp(-{\sqrt {2}}\,\pi ){\bigr ]}^{5}-{\sqrt {5}}\,\Phi ^{-1}R{\bigl\exp(-{\sqrt {2}}\,\pi ){\bigr ]}^{4}+$
    $+2\,{\sqrt {5}}\,\Phi R{\bigl\exp(-{\sqrt {2}}\,\pi ){\bigr ]}^{3}+{\sqrt {5}}\,\Phi ^{-1}R{\bigl\exp(-{\sqrt {2}}\,\pi ){\bigr ]}^{2}+2\,\Phi ^{-2}R{\bigl\exp(-{\sqrt {2}}\,\pi ){\bigr ]}-1=0$
The solution of this equation is the already mentioned solution:
    $R{\bigl\exp(-{\sqrt {2}}\,\pi ){\bigr ]}=\tan {\bigl\arctan({\tfrac {1}{3}}{\sqrt {5}}-{\tfrac {1}{3}}{\sqrt[{3}]{6{\sqrt {30}}+4{\sqrt {5}}}}+{\tfrac {1}{3}}{\sqrt[{3}]{6{\sqrt {30}}-4{\sqrt {5}}}}\,)-{\tfrac {1}{4}}\operatorname {arccot}(2){\bigr ]}$

## Relation to modular forms

$R(q)$ can be related to the Dedekind eta function, a modular form of weight 1/2, as,
    ${\frac {1}{R(q)}}-R(q)={\frac {\eta ({\frac {\tau }{5}})}{\eta (5\tau )}}+1$
    ${\frac {1}{R^{5}(q)}}-R^{5}(q)=\left[{\frac {\eta (\tau )}{\eta (5\tau )}}\right]^{6}+11$
The Rogers-Ramanujan continued fraction can also be expressed in terms of the Jacobi theta functions. Recall the notation,
    ${\begin{aligned}\vartheta _{10}(0;\tau )&=\theta _{2}(q)=\sum _{n=-\infty }^{\infty }q^{(n+1/2)^{2}}\\\vartheta _{00}(0;\tau )&=\theta _{3}(q)=\sum _{n=-\infty }^{\infty }q^{n^{2}}\\\vartheta _{01}(0;\tau )&=\theta _{4}(q)=\sum _{n=-\infty }^{\infty }(-1)^{n}q^{n^{2}}\end{aligned}}$
The notation $\theta _{n}$ is slightly easier to remember since $\theta _{2}^{4}+\theta _{4}^{4}=\theta _{3}^{4}$, with even subscripts on the LHS. Thus,
    $R(x)=\tan {\biggl \{\frac {1}{2}}\operatorname {arccot} {\biggl{\frac {1}{2}}+{\frac {\theta _{4}(x^{1/5})[5\,\theta _{4}(x^{5})^{2}-\theta _{4}(x)^{2}]}{2\,\theta _{4}(x^{5})[\theta _{4}(x)^{2}-\theta _{4}(x^{1/5})^{2}]}}{\biggr ]}{\biggr \}}$
    $R(x)=\tan {\biggl \{\frac {1}{2}}\operatorname {arccot} {\biggl{\frac {1}{2}}+{\bigg{\frac {\theta _{2}(x^{1/10})\,\theta _{3}(x^{1/10})\,\theta _{4}(x^{1/10})}{2^{3}\,\theta _{2}(x^{5/2})\,\theta _{3}(x^{5/2})\,\theta _{4}(x^{5/2})}}{\bigg )}^{1/3}{\biggr ]}{\biggr \}}$
    $R(x)=\tan {\biggl \{\frac {1}{2}}\arctan {\biggl{\frac {1}{2}}-{\frac {\theta _{4}(x)^{2}}{2\,\theta _{4}(x^{5})^{2}}}{\biggr ]}{\biggr \}}^{1/5}\times \tan {\biggl \{\frac {1}{2}}\operatorname {arccot} {\biggl{\frac {1}{2}}-{\frac {\theta _{4}(x)^{2}}{2\,\theta _{4}(x^{5})^{2}}}{\biggr ]}{\biggr \}}^{2/5}$
    $R(x)=\tan {\biggl \{\frac {1}{2}}\arctan {\biggl{\frac {1}{2}}-{\frac {\theta _{4}(x^{1/2})^{2}}{2\,\theta _{4}(x^{5/2})^{2}}}{\biggr ]}{\biggr \}}^{2/5}\times \cot {\biggl \{\frac {1}{2}}\operatorname {arccot} {\biggl{\frac {1}{2}}-{\frac {\theta _{4}(x^{1/2})^{2}}{2\,\theta _{4}(x^{5/2})^{2}}}{\biggr ]}{\biggr \}}^{1/5}$
Note, however, that theta functions normally use the nome q = e$^{iπτ}$, while the Dedekind eta function uses the square of the nome q = e$^{2iπτ}$, thus the variable x has been employed instead to maintain consistency between all functions. For example, let $\tau ={\sqrt {-1}}$ so $x=e^{-\pi }$. Plugging this into the theta functions, one gets the same value for all three R(x) formulas which is the correct evaluation of the continued fraction given previously,
    $R{\bige^{-\pi }{\big )}={\frac {1}{2}}\varphi \,({\sqrt {5}}-\varphi ^{3/2})({\sqrt[{4}]{5}}+\varphi ^{3/2})=0.511428\dots$
One can also define the elliptic nome,
    $q(k)=\exp {\big-\pi K({\sqrt {1-k^{2}}})/K(k){\big ]}$
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 can then be also expressed by the Jacobi elliptic functions as follows:
    $R{\bigq(k){\big )}=\tan {\biggl \{\frac {1}{2}}\arctan y{\biggr \}}^{1/5}\tan {\biggl \{\frac {1}{2}}\operatorname {arccot} y{\biggr \}}^{2/5}=\left\{{\frac {{\sqrt {y^{2}+1}}-1}{y}}\right\}^{1/5}\left\{y\left[{\sqrt {{\frac {1}{y^{2}}}+1}}-1\right]\right\}^{2/5}$
with
    $y={\frac {2k^{2}\,{\text{sn}}[{\tfrac {2}{5}}K(k);k]^{2}\,{\text{sn}}[{\tfrac {4}{5}}K(k);k]^{2}}{5-k^{2}\,{\text{sn}}[{\tfrac {2}{5}}K(k);k]^{2}\,{\text{sn}}[{\tfrac {4}{5}}K(k);k]^{2}}}.$

## Relation to j-function

One formula involving the j-function and the Dedekind eta function is this:
    $j(\tau )={\frac {(x^{2}+10x+5)^{3}}{x}}$
where $x=\left[{\frac {{\sqrt {5}}\,\eta (5\tau )}{\eta (\tau )}}\right]^{6}.\,$ Since also,
    ${\frac {1}{R^{5}(q)}}-R^{5}(q)=\left[{\frac {\eta (\tau )}{\eta (5\tau )}}\right]^{6}+11$
Eliminating the eta quotient $x$ between the two equations, one can then express j(τ) in terms of $r=R(q)$ as,
    ${\begin{aligned}&j(\tau )=-{\frac {(r^{20}-228r^{15}+494r^{10}+228r^{5}+1)^{3}}{r^{5}(r^{10}+11r^{5}-1)^{5}}}\\[6pt]&j(\tau )-1728=-{\frac {(r^{30}+522r^{25}-10005r^{20}-10005r^{10}-522r^{5}+1)^{2}}{r^{5}(r^{10}+11r^{5}-1)^{5}}}\end{aligned}}$
where the numerator and denominator are polynomial invariants of the icosahedron. Using the modular equation between $R(q)$ and $R(q^{5})$, one finds that,
    ${\begin{aligned}&j(5\tau )=-{\frac {(r^{20}+12r^{15}+14r^{10}-12r^{5}+1)^{3}}{r^{25}(r^{10}+11r^{5}-1)}}\\[6pt]&j(5\tau )-1728=-{\frac {(r^{30}+18r^{25}+75r^{20}+75r^{10}-18r^{5}+1)^{2}}{r^{25}(r^{10}+11r^{5}-1)}}\end{aligned}}$
Let $z=r^{5}-{\frac {1}{r^{5}}}$, then $j(5\tau )=-{\frac {\left(z^{2}+12z+16\right)^{3}}{z+11}}$
where
    ${\begin{aligned}&z_{\infty }=-\left[{\frac {{\sqrt {5}}\,\eta (25\tau )}{\eta (5\tau )}}\right]^{6}-11,\ z_{0}=-\left[{\frac {\eta (\tau )}{\eta (5\tau )}}\right]^{6}-11,\ z_{1}=\left[{\frac {\eta ({\frac {5\tau +2}{5}})}{\eta (5\tau )}}\right]^{6}-11,\\[6pt]&z_{2}=-\left[{\frac {\eta ({\frac {5\tau +4}{5}})}{\eta (5\tau )}}\right]^{6}-11,\ z_{3}=\left[{\frac {\eta ({\frac {5\tau +6}{5}})}{\eta (5\tau )}}\right]^{6}-11,\ z_{4}=-\left[{\frac {\eta ({\frac {5\tau +8}{5}})}{\eta (5\tau )}}\right]^{6}-11\end{aligned}}$
which in fact is the j-invariant of the elliptic curve,
    $y^{2}+(1+r^{5})xy+r^{5}y=x^{3}+r^{5}x^{2}$
parameterized by the non-cusp points of the modular curve $X_{1}(5)$.

## Functional equation

For convenience, one can also use the notation $r(\tau )=R(q)$ when q = e$^{2πiτ}$. While other modular functions like the j-invariant satisfies,
    $j(-{\tfrac {1}{\tau }})=j(\tau )$
and the Dedekind eta function has,
    $\eta (-{\tfrac {1}{\tau }})={\sqrt {-i\tau }}\,\eta (\tau )$
the functional equation of the Rogers–Ramanujan continued fraction involves the golden ratio $\varphi$,
    $r(-{\tfrac {1}{\tau }})={\frac {1-\varphi \,r(\tau )}{\varphi +r(\tau )}}$
Incidentally,
    $r({\tfrac {7+i}{10}})=i$

## Modular equations

There are modular equations between $R(q)$ and $R(q^{n})$. Elegant ones for small prime n are as follows.
For $n=2$, let $u=R(q)$ and $v=R(q^{2})$, then $v-u^{2}=(v+u^{2})uv^{2}.$
For $n=3$, let $u=R(q)$ and $v=R(q^{3})$, then $(v-u^{3})(1+uv^{3})=3u^{2}v^{2}.$
For $n=5$, let $u=R(q)$ and $v=R(q^{5})$, then $v(v^{4}-3v^{3}+4v^{2}-2v+1)=(v^{4}+2v^{3}+4v^{2}+3v+1)u^{5}.$
Or equivalently for $n=5$, let $u=R(q)$ and $v=R(q^{5})$ and $\varphi ={\tfrac {1+{\sqrt {5}}}{2}}$, then $u^{5}={\frac {v\,(v^{2}-\varphi ^{2}v+\varphi ^{2})(v^{2}-\varphi ^{-2}v+\varphi ^{-2})}{(v^{2}+v+\varphi ^{2})(v^{2}+v+\varphi ^{-2})}}.$
For $n=11$, let $u=R(q)$ and $v=R(q^{11})$, then $uv(u^{10}+11u^{5}-1)(v^{10}+11v^{5}-1)=(u-v)^{12}.$
Regarding $n=5$, note that $v^{10}+11v^{5}-1=(v^{2}+v-1)(v^{4}-3v^{3}+4v^{2}-2v+1)(v^{4}+2v^{3}+4v^{2}+3v+1).$

## Other results

Ramanujan found many other interesting results regarding $R(q)$. Let $a,b\in \mathbb {R} ^{+}$, and $\varphi$ as the golden ratio.
If $ab=\pi ^{2}$ then,
    ${\biglR(e^{-2a})+\varphi {\bigl ]}{\biglR(e^{-2b})+\varphi {\bigr ]}={\sqrt {5}}\,\varphi .$
If $5ab=\pi ^{2}$ then,
    ${\biglR^{5}(e^{-2a})+\varphi ^{5}{\bigl ]}{\biglR^{5}(e^{-2b})+\varphi ^{5}{\bigr ]}=5{\sqrt {5}}\,\varphi ^{5}.$
The powers of $R(q)$ also can be expressed in unusual ways. For its cube,
    $R^{3}(q)={\frac {\alpha }{\beta }}$
where
    $\alpha =\sum _{n=0}^{\infty }{\frac {q^{2n}}{1-q^{5n+2}}}-\sum _{n=0}^{\infty }{\frac {q^{3n+1}}{1-q^{5n+3}}},$
    $\beta =\sum _{n=0}^{\infty }{\frac {q^{n}}{1-q^{5n+1}}}-\sum _{n=0}^{\infty }{\frac {q^{4n+3}}{1-q^{5n+4}}}.$
For its fifth power, let $w=R(q)R^{2}(q^{2})$, then,
    $R^{5}(q)=w\left({\frac {1-w}{1+w}}\right)^{2},\;\;R^{5}(q^{2})=w^{2}\left({\frac {1+w}{1-w}}\right)$

## Quintic equations

The general quintic equation in Bring-Jerrard form:
    $x^{5}-5x-4a=0$
for every real value $a>1$ can be solved in terms of Rogers-Ramanujan continued fraction $R(q)$ and the elliptic nome
    $q(k)=\exp {\big-\pi K({\sqrt {1-k^{2}}})/K(k){\big ]}.$
To solve this quintic, the elliptic modulus must first be determined as
    $k=\tan[{\tfrac {1}{4}}\pi -{\tfrac {1}{4}}\operatorname {arccsc}(a^{2})].$
Then the real solution is
    ${\begin{aligned}x&={\frac {2-{\bigl \1-R[q(k)]{\bigr \}}{\bigl \1+R[q(k)^{2}]{\bigr \}}}{{\sqrt {R[q(k)]\,R[q(k)^{2}]}}\,{\sqrt[{4}]{4\cot \langle 4\arctan\{S\}\rangle -3}}}}\\&={\frac {2-{\bigl \1-R[q(k)]{\bigr \}}{\bigl \1+R[q(k)^{2}]{\bigr \}}}{{\sqrt {R[q(k)]R[q(k)^{2}]}}\,{\sqrt[{4}]{{\frac {2}{S-1}}+{\frac {2}{S+1}}+{\frac {1}{S}}-S-3}}}}.\end{aligned}}$
where $S=R[q(k)]\,R^{2}[q(k)^{2}].$. Recall in the previous section the 5th power of $R(q)$ can be expressed by $S$:
    $R^{5}[q(k)]=S\left({\frac {1-S}{1+S}}\right)^{2}$

### Example 1

    $x^{5}-x-1=0$
Transform to,
    $({\sqrt[{4}]{5}}x)^{5}-5({\sqrt[{4}]{5}}x)-4({\tfrac {5}{4}}{\sqrt[{4}]{5}})=0$
thus,
    $a={\tfrac {5}{4}}{\sqrt[{4}]{5}}$
    $k=\tan[{\tfrac {1}{4}}\pi -{\tfrac {1}{4}}\operatorname {arccsc}(a^{2})]={\tfrac {5^{5/4}+{\sqrt {25{\sqrt {5}}-16}}}{5^{5/4}+{\sqrt {25{\sqrt {5}}+16}}}}$
    $q(k)=0.0851414716\dots$
    $R[q(k)]=0.5633613184\dots$
    $R[q(k)^{2}]=0.3706122329\dots$
and the solution is:
    $x={\frac {2-{\bigl \1-R[q(k)]{\bigr \}}{\bigl \1+R[q(k)^{2}]{\bigr \}}}{{\sqrt {R[q(k)]\,R[q(k)^{2}]}}\,{\sqrt[{4}]{20\cot \langle 4\arctan\{R[q(k)]\,R[q(k)^{2}]^{2}\}\rangle -15}}}}=1.167303978\dots$
and can not be represented by elementary root expressions.

### Example 2

    $x^{5}-5x-4{\Bigl{\sqrt[{4}]{\tfrac {81}{32}}}{\Bigr )}=0$
thus,
    $a={\sqrt[{4}]{\tfrac {81}{32}}}$
Given the more familiar continued fractions with closed-forms,
    $r_{1}=R{\bige^{-\pi }{\big )}={\tfrac {1}{2}}\varphi \,({\sqrt {5}}-\varphi ^{3/2})({\sqrt[{4}]{5}}+\varphi ^{3/2})=0.511428\dots$
    $r_{2}=R{\bige^{-2\pi }{\big )}={\sqrt[{4}]{5}}\,\varphi ^{1/2}-\varphi =0.284079\dots$
    $r_{4}=R{\bige^{-4\pi }{\big )}={\tfrac {1}{2}}\varphi \,({\sqrt {5}}-\varphi ^{3/2})(-{\sqrt[{4}]{5}}+\varphi ^{3/2})=0.081002\dots$
with golden ratio $\varphi ={\tfrac {1+{\sqrt {5}}}{2}}$ and the solution simplifies to
    ${\begin{aligned}x&={\sqrt[{4}]{5}}\,{\frac {2-{\bigl \1-r_{1}{\bigr \}}{\bigl \1+r_{2}{\bigr \}}}{{\sqrt {r_{1}\,r_{2}}}\,{\sqrt[{4}]{20\cot \langle 4\arctan\{r_{1}\,r_{2}^{2}\}\rangle -15}}}}\\[6pt]&={\sqrt[{4}]{5}}\,{\frac {2-{\bigl \1-r_{2}{\bigr \}}{\bigl \1+r_{4}{\bigr \}}}{{\sqrt {r_{2}\,r_{4}}}\,{\sqrt[{4}]{20\cot \langle 4\arctan\{r_{2}\,r_{4}^{2}\}\rangle -15}}}}\\[6pt]&={\sqrt[{4}]{8}}=1.681792\dots \end{aligned}}$