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.

Domain coloring representation of the convergent ${\displaystyle A_{400}(q)/B_{400}(q)}$ of the function ${\displaystyle q^{-1/5}R(q)}$, where ${\displaystyle R(q)}$ is the Rogers–Ramanujan continued fraction.

Definition

 

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

Given the functions ${\displaystyle G(q)}$  and ${\displaystyle H(q)}$  appearing in the Rogers–Ramanujan identities, and assume ${\displaystyle q=e^{2\pi i\tau }}$ ,

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

and,

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

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

${\displaystyle {\begin{aligned}R(q)&={\frac {q^{\frac {11}{60}}H(q)}{q^{-{\frac {1}{60}}}G(q)}}=q^{\frac {1}{5}}\prod _{n=1}^{\infty }{\frac {(1-q^{5n-1})(1-q^{5n-4})}{(1-q^{5n-2})(1-q^{5n-3})}}=q^{1/5}\prod _{n=1}^{\infty }(1-q^{n})^{(n|5)}\\[8pt]&={\cfrac {q^{1/5}}{1+{\cfrac {q}{1+{\cfrac {q^{2}}{1+{\cfrac {q^{3}}{1+\ddots }}}}}}}}\end{aligned}}}$ 

${\displaystyle (n\mid m)}$  is the Jacobi symbol.

One should be careful with notation since the formulas employing the j-function ${\displaystyle j}$  will be consistent with the other formulas only if ${\displaystyle q=e^{2\pi i\tau }}$  (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 ${\displaystyle q=e^{2\pi i\tau }}$ . However, Ramanujan, in his examples to Hardy and given below, used the nome ${\displaystyle q=e^{\pi i\tau }}$ instead.^{[citation needed]}

Special values

If q is the nome or its square, 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 related to modular functions of ${\displaystyle \tau }$ . Since they have integral coefficients, the theory of complex multiplication implies that their values for ${\displaystyle \tau }$  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 ${\displaystyle q=e^{\pi i\tau }}$ ,

${\displaystyle R(q)={\cfrac {q^{1/5}}{1+{\cfrac {q}{1+{\cfrac {q^{2}}{1+{\cfrac {q^{3}}{1+\ddots }}}}}}}}}$ 

f when ${\displaystyle \tau =i}$ ,

${\displaystyle R{\big (}e^{-\pi }{\big )}={\cfrac {e^{-{\frac {\pi }{5}}}}{1+{\cfrac {e^{-\pi }}{1+{\cfrac {e^{-2\pi }}{1+\ddots }}}}}}={\tfrac {1}{2}}\varphi \,({\sqrt {5}}-\varphi ^{3/2})({\sqrt[{4}]{5}}+\varphi ^{3/2})=0.511428\dots }$ 

when ${\displaystyle \tau =2i}$ ,

${\displaystyle R{\big (}e^{-2\pi }{\big )}={\cfrac {e^{-{\frac {2\pi }{5}}}}{1+{\cfrac {e^{-2\pi }}{1+{\cfrac {e^{-4\pi }}{1+\ddots }}}}}}={{\sqrt[{4}]{5}}\,\varphi ^{1/2}-\varphi }=0.284079\dots }$ 

when ${\displaystyle \tau =4i}$ ,

${\displaystyle R{\big (}e^{-4\pi }{\big )}={\cfrac {e^{-{\frac {4\pi }{5}}}}{1+{\cfrac {e^{-4\pi }}{1+{\cfrac {e^{-8\pi }}{1+\ddots }}}}}}={\tfrac {1}{2}}\varphi \,({\sqrt {5}}-\varphi ^{3/2})(-{\sqrt[{4}]{5}}+\varphi ^{3/2})=0.081002\dots }$ 

when ${\displaystyle \tau =2{\sqrt {5}}i}$ ,

${\displaystyle R{\big (}e^{-2{\sqrt {5}}\pi }{\big )}={\cfrac {e^{-{\frac {2\pi }{\sqrt {5}}}}}{1+{\cfrac {e^{-2\pi {\sqrt {5}}}}{1+{\cfrac {e^{-4\pi {\sqrt {5}}}}{1+\ddots }}}}}}={\frac {\sqrt {5}}{1+{\big (}5^{3/4}(\varphi -1)^{5/2}-1{\big )}^{1/5}}}-\varphi =0.0602094\dots }$ 

when ${\displaystyle \tau =5i}$ ,

${\displaystyle R{\big (}e^{-5\pi }{\big )}={\cfrac {e^{-\pi }}{1+{\cfrac {e^{-5\pi }}{1+{\cfrac {e^{-10\pi }}{1+\ddots }}}}}}={\frac {1+\varphi ^{2}}{\varphi +{\big (}{\frac {1}{2}}(4-\varphi -3{\sqrt {\varphi -1}})(3\varphi ^{3/2}-{\sqrt[{4}]{5}}){\big )}^{1/5}}}-\varphi =0.0432139\dots }$ 

when ${\displaystyle \tau =10i}$ ,

${\displaystyle R{\big (}e^{-10\pi }{\big )}={\cfrac {e^{-2\pi }}{1+{\cfrac {e^{-10\pi }}{1+{\cfrac {e^{-20\pi }}{1+\ddots }}}}}}={\frac {1+\varphi ^{2}}{\varphi +{\big (}3{\sqrt {1+\varphi ^{2}}}-4-\varphi {\big )}^{1/5}}}-\varphi =0.00186744\dots }$ 

when ${\displaystyle \tau =20i}$ ,

${\displaystyle R{\big (}e^{-20\pi }{\big )}={\cfrac {e^{-4\pi }}{1+{\cfrac {e^{-20\pi }}{1+{\cfrac {e^{-40\pi }}{1+\ddots }}}}}}={\frac {1+\varphi ^{2}}{\varphi +{\big (}{\frac {1}{2}}(4-\varphi -3{\sqrt {\varphi -1}})(3\varphi ^{3/2}+{\sqrt[{4}]{5}}){\big )}^{1/5}}}-\varphi =0.00000348734\dots }$ 

and ${\displaystyle \varphi ={\tfrac {1+{\sqrt {5}}}{2}}}$  is the golden ratio. Note that ${\displaystyle R{\big (}e^{-2\pi }{\big )}}$  is a positive root of the quartic equation,

${\displaystyle x^{4}+2x^{3}-6x^{2}-2x+1=0}$ 

while ${\displaystyle R{\big (}e^{-\pi }{\big )}}$  and ${\displaystyle R{\big (}e^{-4\pi }{\big )}}$  are two positive roots of a single octic,

${\displaystyle y^{4}+2\varphi ^{4}y^{3}+6\varphi ^{2}y^{2}-2\varphi ^{4}y+1=0}$ 

(since ${\displaystyle \varphi }$  has a square root) which explains the similarity of the two closed-forms. More generally, for positive integer m, then ${\displaystyle R(e^{-2\pi /m})}$  and ${\displaystyle R(e^{-2\pi \,m})}$  are two roots of the same equation as well as,

${\displaystyle {\bigl [}R(e^{-2\pi /m})+\varphi {\bigr ]}{\bigl [}R(e^{-2\pi \,m})+\varphi {\bigr ]}={\sqrt {5}}\,\varphi }$ 

The algebraic degree k of ${\displaystyle R(e^{-\pi \,n})}$  for ${\displaystyle n=1,2,3,4,\dots }$  is ${\displaystyle k=8,4,32,8,\dots }$  (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 ${\displaystyle G(q)}$  and ${\displaystyle H(q)}$  in terms of the j-function ${\displaystyle j(\tau )}$  and the Rogers-Ramanujan continued fraction ${\displaystyle R(q)}$ . However, since ${\displaystyle j(\tau )}$  uses the nome's square ${\displaystyle q=e^{2\pi \,i\tau }}$ , then one should be careful with notation such that ${\displaystyle j(\tau ),\,G(q),\,H(q)}$  and ${\displaystyle r=R(q)}$  use the same ${\displaystyle q}$ .

${\displaystyle {\begin{aligned}G(q)&=\prod _{n=1}^{\infty }{\frac {1}{(1-q^{5n-1})(1-q^{5n-4})}}\\[6pt]&=q^{1/60}{\frac {j(\tau )^{1/60}}{(r^{20}-228r^{15}+494r^{10}+228r^{5}+1)^{1/20}}}\end{aligned}}}$ 

${\displaystyle {\begin{aligned}H(q)&=\prod _{n=1}^{\infty }{\frac {1}{(1-q^{5n-2})(1-q^{5n-3})}}\\[6pt]&={\frac {-1}{q^{11/60}}}{\frac {(r^{20}-228r^{15}+494r^{10}+228r^{5}+1)^{11/20}}{j(\tau )^{11/60}\,(r^{10}+11r^{5}-1)}}\end{aligned}}}$ 

Of course, the secondary formulas imply that ${\displaystyle q^{-1/60}G(q)}$  and ${\displaystyle q^{11/60}H(q)}$  are algebraic numbers (though normally of high degree) for ${\displaystyle \tau }$  involving an imaginary quadratic field. For example, the formulas above simplify to,

${\displaystyle {\begin{aligned}G(e^{-2\pi })&=(e^{-2\pi })^{1/60}{\frac {1}{(5\,\varphi )^{1/4}}}{\frac {1}{\sqrt {R(e^{-2\pi })}}}\\[6pt]&=1.00187093\dots \\[6pt]H(e^{-2\pi })&={\frac {1}{(e^{-2\pi })^{11/60}}}{\frac {1}{(5\,\varphi )^{1/4}}}{\sqrt {R(e^{-2\pi })}}\\[6pt]&=1.00000349\ldots \end{aligned}}}$ 

and,

${\displaystyle {\begin{aligned}G(e^{-4\pi })&=(e^{-4\pi })^{1/60}{\frac {1}{(5\,\varphi ^{3})^{1/4}\,(\varphi +{\sqrt[{4}]{5}})^{1/4}}}{\frac {1}{\sqrt {R(e^{-4\pi })}}}\\[6pt]&=1.000003487354\dots \\[6pt]H(e^{-4\pi })&={\frac {1}{(e^{-4\pi })^{11/60}}}{\frac {1}{(5\,\varphi ^{3})^{1/4}\,(\varphi +{\sqrt[{4}]{5}})^{1/4}}}{\sqrt {R(e^{-4\pi })}}\\[6pt]&=1.000000000012\dots \end{aligned}}}$ 

and so on, with ${\displaystyle \varphi }$  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:

${\displaystyle a\oplus b=\tan {\bigl [}\arctan(a)+\arctan(b){\bigr ]}={\frac {a+b}{1-ab}}}$ 

${\displaystyle c\ominus d=\tan {\bigl [}\arctan(c)-\arctan(d){\bigr ]}={\frac {c-d}{1+cd}}}$ 

The elliptic nome and the complementary nome have this relationship to each other:

${\displaystyle \ln(q)\ln(q_{1})=\pi ^{2}}$ 

The complementary nome of a modulus k is equal to the nome of the Pythagorean complementary modulus:

${\displaystyle q_{1}(k)=q(k')=q({\sqrt {1-k^{2}}})}$ 

These are the reflection theorems for the continued fractions R and S:

${\displaystyle S(q)\oplus S(q_{1})=\Phi }$

${\displaystyle R(q^{2})\oplus R(q_{1}^{2})=\Phi ^{-1}}$

The letter ${\displaystyle \Phi }$  represents the Golden number exactly:

${\displaystyle \Phi ={\tfrac {1}{2}}({\sqrt {5}}+1)=\cot[{\tfrac {1}{2}}\arctan(2)]=2\cos({\tfrac {1}{5}}{\pi })}$ 

${\displaystyle \Phi ^{-1}={\tfrac {1}{2}}({\sqrt {5}}-1)=\tan[{\tfrac {1}{2}}\arctan(2)]=2\sin({\tfrac {1}{10}}{\pi })}$ 

The theorems for the squared nome are constructed as follows:

${\displaystyle R(q)^{2}R(q^{2})^{-1}\oplus R(q)R(q^{2})^{2}=1}$

${\displaystyle S(q)^{2}R(q^{2})^{-1}\ominus S(q)R(q^{2})^{2}=1}$

Following relations between the continued fractions and the Jacobi theta functions are given:

${\displaystyle S(q)\oplus R(q^{2})={\frac {\vartheta _{00}(q^{1/5})^{2}-\vartheta _{00}(q)^{2}}{5\,\vartheta _{00}(q^{5})^{2}-\vartheta _{00}(q)^{2}}}}$

${\displaystyle R(q)\ominus R(q^{2})={\frac {\vartheta _{01}(q)^{2}-\vartheta _{01}(q^{1/5})^{2}}{5\,\vartheta _{01}(q^{5})^{2}-\vartheta _{01}(q)^{2}}}}$

Derivation of Lemniscatic values

Into the now shown theorems certain values are inserted:

${\displaystyle S{\bigl [}\exp(-\pi ){\bigr ]}\oplus S{\bigl [}\exp(-\pi ){\bigr ]}=\Phi }$ 

Therefore following identity is valid:

${\displaystyle S{\bigl [}\exp(-\pi ){\bigr ]}=\tan {\bigl [}{\tfrac {1}{2}}\arctan(\Phi ){\bigr ]}=\tan {\bigl [}{\tfrac {1}{4}}\pi -{\tfrac {1}{4}}\arctan(2){\bigr ]}}$

In an analogue pattern we get this result:

${\displaystyle R{\bigl [}\exp(-2\pi ){\bigr ]}\oplus R{\bigl [}\exp(-2\pi ){\bigr ]}=\Phi ^{-1}}$ 

Therefore following identity is valid:

${\displaystyle R{\bigl [}\exp(-2\pi ){\bigr ]}=\tan {\bigl [}{\tfrac {1}{2}}\arctan(\Phi ^{-1}){\bigr ]}=\tan {\bigl [}{\tfrac {1}{4}}\arctan(2){\bigr ]}}$

Furthermore we get the same relation by using the above mentioned theorem about the Jacobi theta functions:

${\displaystyle S{\bigl [}\exp(-\pi ){\bigr ]}\oplus R{\bigl [}\exp(-2\pi ){\bigr ]}=S(q)\oplus R(q^{2}){\bigl [}q=\exp(-\pi ){\bigr ]}=}$ 

${\displaystyle ={\frac {\vartheta _{00}(q^{1/5})^{2}-\vartheta _{00}(q)^{2}}{5\,\vartheta _{00}(q^{5})^{2}-\vartheta _{00}(q)^{2}}}{\bigl [}q=\exp(-\pi ){\bigr ]}=1}$ 

This result appears because of the Poisson summation formula and this equation can be solved in this way:

${\displaystyle R{\bigl [}\exp(-2\pi ){\bigr ]}=1\ominus S{\bigl [}\exp(-\pi ){\bigr ]}=1\ominus \tan {\bigl [}{\tfrac {1}{4}}\pi -{\tfrac {1}{4}}\arctan(2){\bigr ]}=\tan {\bigl [}{\tfrac {1}{4}}\arctan(2){\bigr ]}}$ 

By taking the other mentioned theorem about the Jacobi theta functions a next value can be determined:

${\displaystyle R{\bigl [}\exp(-\pi ){\bigr ]}\ominus R{\bigl [}\exp(-2\pi ){\bigr ]}=R(q)\ominus R(q^{2}){\bigl [}q=\exp(-\pi ){\bigr ]}=}$ 

${\displaystyle ={\frac {\vartheta _{01}(q)^{2}-\vartheta _{01}(q^{1/5})^{2}}{5\,\vartheta _{01}(q^{5})^{2}-\vartheta _{01}(q)^{2}}}{\bigl [}q=\exp(-\pi ){\bigr ]}={\frac {{\sqrt[{4}]{5}}-1}{{\sqrt[{4}]{5}}+1}}={\sqrt[{4}]{5}}\ominus 1=\tan {\bigl [}\arctan({\sqrt[{4}]{5}}\,)-{\tfrac {1}{4}}\pi {\bigr ]}}$ 

That equation chain leads to this tangential sum:

${\displaystyle R{\bigl [}\exp(-\pi ){\bigr ]}=R{\bigl [}\exp(-2\pi ){\bigr ]}\oplus \tan {\bigl [}\arctan({\sqrt[{4}]{5}}\,)-{\tfrac {1}{4}}\pi {\bigr ]}}$ 

And therefore following result appears:

${\displaystyle R{\bigl [}\exp(-\pi ){\bigr ]}=\tan {\bigl [}{\tfrac {1}{4}}\arctan(2)+\arctan({\sqrt[{4}]{5}}\,)-{\tfrac {1}{4}}\pi {\bigr ]}}$

In the next step we use the reflection theorem for the continued fraction R again:

${\displaystyle R{\bigl [}\exp(-\pi ){\bigr ]}\oplus R{\bigl [}\exp(-4\pi ){\bigr ]}=\Phi ^{-1}}$ 

${\displaystyle R{\bigl [}\exp(-4\pi ){\bigr ]}=\tan {\bigl [}{\tfrac {1}{2}}\arctan(2){\bigr ]}\ominus R{\bigl [}\exp(-\pi ){\bigr ]}}$ 

And a further result appears:

${\displaystyle R{\bigl [}\exp(-4\pi ){\bigr ]}=\tan {\bigl [}{\tfrac {1}{4}}\arctan(2)-\arctan({\sqrt[{4}]{5}}\,)+{\tfrac {1}{4}}\pi {\bigr ]}}$

Derivation of Non-Lemniscatic values

The reflection theorem is now used for following values:

${\displaystyle R{\bigl [}\exp(-{\sqrt {2}}\,\pi ){\bigr ]}\oplus R{\bigl [}\exp(-2{\sqrt {2}}\,\pi ){\bigr ]}=\Phi ^{-1}}$ 

The Jacobi theta theorem leads to a further relation:

${\displaystyle R{\bigl [}\exp(-{\sqrt {2}}\,\pi ){\bigr ]}\ominus R{\bigl [}\exp(-2{\sqrt {2}}\,\pi ){\bigr ]}=R(q)\ominus R(q^{2}){\bigl [}q=\exp(-{\sqrt {2}}\,\pi ){\bigr ]}=}$ 

${\displaystyle ={\frac {\vartheta _{01}(q)^{2}-\vartheta _{01}(q^{1/5})^{2}}{5\,\vartheta _{01}(q^{5})^{2}-\vartheta _{01}(q)^{2}}}{\bigl [}q=\exp(-{\sqrt {2}}\,\pi ){\bigr ]}=\tan {\bigl [}2\arctan({\tfrac {1}{3}}{\sqrt {5}}-{\tfrac {1}{3}}{\sqrt[{3}]{6{\sqrt {30}}+4{\sqrt {5}}}}+{\tfrac {1}{3}}{\sqrt[{3}]{6{\sqrt {30}}-4{\sqrt {5}}}}\,)-{\tfrac {1}{4}}\pi {\bigr ]}}$ 

By tangential adding the now mentioned two theorems we get this result:

${\displaystyle R{\bigl [}\exp(-{\sqrt {2}}\,\pi ){\bigr ]}\oplus R{\bigl [}\exp(-{\sqrt {2}}\,\pi ){\bigr ]}=\Phi ^{-1}\oplus \tan {\bigl [}2\arctan({\tfrac {1}{3}}{\sqrt {5}}-{\tfrac {1}{3}}{\sqrt[{3}]{6{\sqrt {30}}+4{\sqrt {5}}}}+{\tfrac {1}{3}}{\sqrt[{3}]{6{\sqrt {30}}-4{\sqrt {5}}}}\,)-{\tfrac {1}{4}}\pi {\bigr ]}}$ 

${\displaystyle R{\bigl [}\exp(-{\sqrt {2}}\,\pi ){\bigr ]}=\tan {\bigl [}\arctan({\tfrac {1}{3}}{\sqrt {5}}-{\tfrac {1}{3}}{\sqrt[{3}]{6{\sqrt {30}}+4{\sqrt {5}}}}+{\tfrac {1}{3}}{\sqrt[{3}]{6{\sqrt {30}}-4{\sqrt {5}}}}\,)-{\tfrac {1}{4}}\operatorname {arccot}(2){\bigr ]}}$

By tangential substraction that result appears:

${\displaystyle R{\bigl [}\exp(-2{\sqrt {2}}\,\pi ){\bigr ]}\oplus R{\bigl [}\exp(-2{\sqrt {2}}\,\pi ){\bigr ]}=\Phi ^{-1}\ominus \tan {\bigl [}2\arctan({\tfrac {1}{3}}{\sqrt {5}}-{\tfrac {1}{3}}{\sqrt[{3}]{6{\sqrt {30}}+4{\sqrt {5}}}}+{\tfrac {1}{3}}{\sqrt[{3}]{6{\sqrt {30}}-4{\sqrt {5}}}}\,)-{\tfrac {1}{4}}\pi {\bigr ]}}$ 

${\displaystyle R{\bigl [}\exp(-2{\sqrt {2}}\,\pi ){\bigr ]}=\tan {\bigl [}{\tfrac {1}{4}}\operatorname {arccot}(-2)-\arctan({\tfrac {1}{3}}{\sqrt {5}}-{\tfrac {1}{3}}{\sqrt[{3}]{6{\sqrt {30}}+4{\sqrt {5}}}}+{\tfrac {1}{3}}{\sqrt[{3}]{6{\sqrt {30}}-4{\sqrt {5}}}}\,){\bigr ]}}$

In an alternative solution way we use the theorem for the squared nome:

${\displaystyle R{\bigl [}\exp(-{\sqrt {2}}\,\pi ){\bigr ]}^{2}R{\bigl [}\exp(-2{\sqrt {2}}\,\pi ){\bigr ]}^{-1}\oplus R{\bigl [}\exp(-{\sqrt {2}}\,\pi ){\bigr ]}R{\bigl [}\exp(-2{\sqrt {2}}\,\pi ){\bigr ]}^{2}=1}$ 

${\displaystyle {\bigl \{}R{\bigl [}\exp(-{\sqrt {2}}\,\pi ){\bigr ]}^{2}R{\bigl [}\exp(-2{\sqrt {2}}\,\pi ){\bigr ]}^{-1}+1{\bigr \}}{\bigl \{}R{\bigl [}\exp(-{\sqrt {2}}\,\pi ){\bigr ]}R{\bigl [}\exp(-2{\sqrt {2}}\,\pi ){\bigr ]}^{2}+1{\bigr \}}=2}$ 

Now the reflection theorem is taken again:

${\displaystyle R{\bigl [}\exp(-2{\sqrt {2}}\,\pi ){\bigr ]}=\Phi ^{-1}\ominus R{\bigl [}\exp(-{\sqrt {2}}\,\pi ){\bigr ]}}$ 

${\displaystyle R{\bigl [}\exp(-2{\sqrt {2}}\,\pi ){\bigr ]}={\frac {1-\Phi R{\bigl [}\exp(-{\sqrt {2}}\,\pi ){\bigr ]}}{\Phi +R{\bigl [}\exp(-{\sqrt {2}}\,\pi ){\bigr ]}}}}$ 

The insertion of the last mentioned expression into the squared nome theorem gives that equation:

${\displaystyle {\biggl \{}R{\bigl [}\exp(-{\sqrt {2}}\,\pi ){\bigr ]}^{2}{\frac {\Phi +R{\bigl [}\exp(-{\sqrt {2}}\,\pi ){\bigr ]}}{1-\Phi R{\bigl [}\exp(-{\sqrt {2}}\,\pi ){\bigr ]}}}+1{\biggr \}}{\biggl \langle }R{\bigl [}\exp(-{\sqrt {2}}\,\pi ){\bigr ]}{\frac {{\bigl \{}1-\Phi R{\bigl [}\exp(-{\sqrt {2}}\,\pi ){\bigr ]}{\bigr \}}^{2}}{{\bigl \{}\Phi +R{\bigl [}\exp(-{\sqrt {2}}\,\pi ){\bigr ]}{\bigr \}}^{2}}}+1{\biggr \rangle }=2}$ 

Erasing the denominators gives an equation of sixth degree:

${\displaystyle R{\bigl [}\exp(-{\sqrt {2}}\,\pi ){\bigr ]}^{6}+2\,\Phi ^{-2}R{\bigl [}\exp(-{\sqrt {2}}\,\pi ){\bigr ]}^{5}-{\sqrt {5}}\,\Phi ^{-1}R{\bigl [}\exp(-{\sqrt {2}}\,\pi ){\bigr ]}^{4}+}$ 

${\displaystyle +2\,{\sqrt {5}}\,\Phi R{\bigl [}\exp(-{\sqrt {2}}\,\pi ){\bigr ]}^{3}+{\sqrt {5}}\,\Phi ^{-1}R{\bigl [}\exp(-{\sqrt {2}}\,\pi ){\bigr ]}^{2}+2\,\Phi ^{-2}R{\bigl [}\exp(-{\sqrt {2}}\,\pi ){\bigr ]}-1=0}$ 

The solution of this equation is the already mentioned solution:

${\displaystyle R{\bigl [}\exp(-{\sqrt {2}}\,\pi ){\bigr ]}=\tan {\bigl [}\arctan({\tfrac {1}{3}}{\sqrt {5}}-{\tfrac {1}{3}}{\sqrt[{3}]{6{\sqrt {30}}+4{\sqrt {5}}}}+{\tfrac {1}{3}}{\sqrt[{3}]{6{\sqrt {30}}-4{\sqrt {5}}}}\,)-{\tfrac {1}{4}}\operatorname {arccot}(2){\bigr ]}}$ 

Relation to modular forms

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

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

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

The Rogers-Ramanujan continued fraction can also be expressed in terms of the Jacobi theta functions. Recall the notation,

${\displaystyle {\begin{aligned}\vartheta _{10}(0;\tau )&=\theta _{2}(q)=\sum _{n=-\infty }^{\infty }q^{(n+1/2)^{2}}\\\vartheta _{00}(0;\tau )&=\theta _{3}(q)=\sum _{n=-\infty }^{\infty }q^{n^{2}}\\\vartheta _{01}(0;\tau )&=\theta _{4}(q)=\sum _{n=-\infty }^{\infty }(-1)^{n}q^{n^{2}}\end{aligned}}}$ 

The notation ${\displaystyle \theta _{n}}$  is slightly easier to remember since ${\displaystyle \theta _{2}^{4}+\theta _{4}^{4}=\theta _{3}^{4}}$ , with even subscripts on the LHS. Thus,

${\displaystyle R(x)=\tan {\biggl \{}{\frac {1}{2}}\operatorname {arccot} {\biggl [}{\frac {1}{2}}+{\frac {\theta _{4}(x^{1/5})[5\,\theta _{4}(x^{5})^{2}-\theta _{4}(x)^{2}]}{2\,\theta _{4}(x^{5})[\theta _{4}(x)^{2}-\theta _{4}(x^{1/5})^{2}]}}{\biggr ]}{\biggr \}}}$ 

${\displaystyle R(x)=\tan {\biggl \{}{\frac {1}{2}}\operatorname {arccot} {\biggl [}{\frac {1}{2}}+{\bigg (}{\frac {\theta _{2}(x^{1/10})\,\theta _{3}(x^{1/10})\,\theta _{4}(x^{1/10})}{2^{3}\,\theta _{2}(x^{5/2})\,\theta _{3}(x^{5/2})\,\theta _{4}(x^{5/2})}}{\bigg )}^{1/3}{\biggr ]}{\biggr \}}}$ 

${\displaystyle R(x)=\tan {\biggl \{}{\frac {1}{2}}\arctan {\biggl [}{\frac {1}{2}}-{\frac {\theta _{4}(x)^{2}}{2\,\theta _{4}(x^{5})^{2}}}{\biggr ]}{\biggr \}}^{1/5}\times \tan {\biggl \{}{\frac {1}{2}}\operatorname {arccot} {\biggl [}{\frac {1}{2}}-{\frac {\theta _{4}(x)^{2}}{2\,\theta _{4}(x^{5})^{2}}}{\biggr ]}{\biggr \}}^{2/5}}$ 

${\displaystyle R(x)=\tan {\biggl \{}{\frac {1}{2}}\arctan {\biggl [}{\frac {1}{2}}-{\frac {\theta _{4}(x^{1/2})^{2}}{2\,\theta _{4}(x^{5/2})^{2}}}{\biggr ]}{\biggr \}}^{2/5}\times \cot {\biggl \{}{\frac {1}{2}}\operatorname {arccot} {\biggl [}{\frac {1}{2}}-{\frac {\theta _{4}(x^{1/2})^{2}}{2\,\theta _{4}(x^{5/2})^{2}}}{\biggr ]}{\biggr \}}^{1/5}}$ 

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 ${\displaystyle \tau ={\sqrt {-1}}}$  so ${\displaystyle x=e^{-\pi }}$ . 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,

${\displaystyle R{\big (}e^{-\pi }{\big )}={\frac {1}{2}}\varphi \,({\sqrt {5}}-\varphi ^{3/2})({\sqrt[{4}]{5}}+\varphi ^{3/2})=0.511428\dots }$ 

One can also define the elliptic nome,

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

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:

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

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

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

Eliminating the eta quotient ${\displaystyle x}$  between the two equations, 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,

${\displaystyle {\begin{aligned}&j(5\tau )=-{\frac {(r^{20}+12r^{15}+14r^{10}-12r^{5}+1)^{3}}{r^{25}(r^{10}+11r^{5}-1)}}\\[6pt]&j(5\tau )-1728=-{\frac {(r^{30}+18r^{25}+75r^{20}+75r^{10}-18r^{5}+1)^{2}}{r^{25}(r^{10}+11r^{5}-1)}}\end{aligned}}}$ 

Let ${\displaystyle z=r^{5}-{\frac {1}{r^{5}}}}$ , 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 ${\displaystyle \varphi }$ ,

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

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 ${\displaystyle (v-u^{3})(1+uv^{3})=3u^{2}v^{2}.}$

May 11, 2024