Rogers–Ramanujan continued fraction

Short description: Continued fraction closely related to the Rogers–Ramanujan identities

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

Definition

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

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

$\begin{array}{rl}G(q)& =\sum _{n=0}^{\mathrm{\infty }}\frac{q^{n^2}}{(1-q)(1-q^2)\cdots (1-q^n)}=\sum _{n=0}^{\mathrm{\infty }}\frac{q^{n^2}}{(q;q{)}_n}=\frac{1}{(q;q^5{)}_{\mathrm{\infty }}(q^4;q^5{)}_{\mathrm{\infty }}}\\ & =\prod _{n=1}^{\mathrm{\infty }}\frac{1}{(1-q^{5n-1})(1-q^{5n-4})}\\ & =\sqrt[60]{qj}{}_2{F}_1(-\frac{1}{60},\frac{19}{60};\frac{4}{5};\frac{1728}{j})\\ & =\sqrt[60]{q(j-1728)}{}_2{F}_1(-\frac{1}{60},\frac{29}{60};\frac{4}{5};-\frac{1728}{j-1728})\\ & =1+q+q^2+q^3+2q^4+2q^5+3q^6+\cdots \end{array}$

and,

$\begin{array}{rl}H(q)& =\sum _{n=0}^{\mathrm{\infty }}\frac{q^{n^2+n}}{(1-q)(1-q^2)\cdots (1-q^n)}=\sum _{n=0}^{\mathrm{\infty }}\frac{q^{n^2+n}}{(q;q{)}_n}=\frac{1}{(q^2;q^5{)}_{\mathrm{\infty }}(q^3;q^5{)}_{\mathrm{\infty }}}\\ & =\prod _{n=1}^{\mathrm{\infty }}\frac{1}{(1-q^{5n-2})(1-q^{5n-3})}\\ & =\frac{1}{\sqrt[60]{q^{11}{j}^{11}}}{}_2{F}_1(\frac{11}{60},\frac{31}{60};\frac{6}{5};\frac{1728}{j})\\ & =\frac{1}{\sqrt[60]{q^{11}{(j-1728)}^{11}}}{}_2{F}_1(\frac{11}{60},\frac{41}{60};\frac{6}{5};-\frac{1728}{j-1728})\\ & =1+q^2+q^3+q^4+q^5+2q^6+2q^7+\cdots \end{array}$

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

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

$(n\mid m)$ 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 $q=e^{\pi i\tau }$,

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

f when $\tau =i$,

$R\left(e^{-\pi }\right)=\frac{e^{-\frac{\pi }{5}}}{1+\frac{e^{-\pi }}{1+\frac{e^{-2\pi }}{1+\ddots }}}=\frac{1}{2}\phi (\sqrt{5}-{\phi }^{3/2})(\sqrt[4]{5}+{\phi }^{3/2})=0.511428\dots$

when $\tau =2i$,

$R\left(e^{-2\pi }\right)=\frac{e^{-\frac{2\pi }{5}}}{1+\frac{e^{-2\pi }}{1+\frac{e^{-4\pi }}{1+\ddots }}}=\sqrt[4]{5}{\phi }^{1/2}-\phi =0.284079\dots$

when $\tau =4i$,

$R\left(e^{-4\pi }\right)=\frac{e^{-\frac{4\pi }{5}}}{1+\frac{e^{-4\pi }}{1+\frac{e^{-8\pi }}{1+\ddots }}}=\frac{1}{2}\phi (\sqrt{5}-{\phi }^{3/2})(-\sqrt[4]{5}+{\phi }^{3/2})=0.081002\dots$

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

$R\left(e^{-2\sqrt{5}\pi }\right)=\frac{e^{-\frac{2\pi }{\sqrt{5}}}}{1+\frac{e^{-2\pi \sqrt{5}}}{1+\frac{e^{-4\pi \sqrt{5}}}{1+\ddots }}}=\frac{\sqrt{5}}{1+(5^{3/4}(\phi -1{)}^{5/2}-1{)}^{1/5}}-\phi =0.0602094\dots$

when $\tau =5i$,

$R\left(e^{-5\pi }\right)=\frac{e^{-\pi }}{1+\frac{e^{-5\pi }}{1+\frac{e^{-10\pi }}{1+\ddots }}}=\frac{1+{\phi }^2}{\phi +(\frac{1}{2}(4-\phi -3\sqrt{\phi -1})(3{\phi }^{3/2}-\sqrt[4]{5}){)}^{1/5}}-\phi =0.0432139\dots$

when $\tau =10i$,

$R\left(e^{-10\pi }\right)=\frac{e^{-2\pi }}{1+\frac{e^{-10\pi }}{1+\frac{e^{-20\pi }}{1+\ddots }}}=\frac{1+{\phi }^2}{\phi +(3\sqrt{1+{\phi }^2}-4-\phi {)}^{1/5}}-\phi =0.00186744\dots$

when $\tau =20i$,

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

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

$x^4+2x^3-6x^2-2x+1=0$

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

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

(since $\phi$ 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,

$[R(e^{-2\pi /m})+\phi ][R(e^{-2\pi m})+\phi ]=\sqrt{5}\phi$

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 $q$.

$\begin{array}{rl}G(q)& =\prod _{n=1}^{\mathrm{\infty }}\frac{1}{(1-q^{5n-1})(1-q^{5n-4})}\\ & =q^{1/60}\frac{j(\tau {)}^{1/60}}{(r^{20}-228r^{15}+494r^{10}+228r^5+1{)}^{1/20}}\end{array}$

$\begin{array}{rl}H(q)& =\prod _{n=1}^{\mathrm{\infty }}\frac{1}{(1-q^{5n-2})(1-q^{5n-3})}\\ & =\frac{-1}{q^{11/60}}\frac{(r^{20}-228r^{15}+494r^{10}+228r^5+1{)}^{11/20}}{j(\tau {)}^{11/60}(r^{10}+11r^5-1)}\end{array}$

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,

$\begin{array}{rl}G(e^{-2\pi })& =(e^{-2\pi }{)}^{1/60}\frac{1}{(5\phi {)}^{1/4}}\frac{1}{\sqrt{R(e^{-2\pi })}}\\ & =1.00187093\dots \\ H(e^{-2\pi })& =\frac{1}{(e^{-2\pi }{)}^{11/60}}\frac{1}{(5\phi {)}^{1/4}}\sqrt{R(e^{-2\pi })}\\ & =1.00000349\dots \end{array}$

and,

$\begin{array}{rl}G(e^{-4\pi })& =(e^{-4\pi }{)}^{1/60}\frac{1}{(5{\phi }^3{)}^{1/4}(\phi +\sqrt[4]{5}{)}^{1/4}}\frac{1}{\sqrt{R(e^{-4\pi })}}\\ & =1.000003487354\dots \\ H(e^{-4\pi })& =\frac{1}{(e^{-4\pi }{)}^{11/60}}\frac{1}{(5{\phi }^3{)}^{1/4}(\phi +\sqrt[4]{5}{)}^{1/4}}\sqrt{R(e^{-4\pi })}\\ & =1.000000000012\dots \end{array}$

and so on, with $\phi$ 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:

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

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

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:

$q_1(k)=q(k^{\prime })=q(\sqrt{1-k^2})$

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

$S(q)\oplus S(q_1)=\mathrm{\Phi }$

$R(q^2)\oplus R(q_1^2)={\mathrm{\Phi }}^{-1}$

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

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

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

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:

$S[\exp (-\pi )]\oplus S[\exp (-\pi )]=\mathrm{\Phi }$

Therefore following identity is valid:

$S[\exp (-\pi )]=\tan [\frac{1}{2}\arctan (\mathrm{\Phi })]=\tan [\frac{1}{4}\pi -\frac{1}{4}\arctan (2)]$

In an analogue pattern we get this result:

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

Therefore following identity is valid:

$R[\exp (-2\pi )]=\tan [\frac{1}{2}\arctan ({\mathrm{\Phi }}^{-1})]=\tan [\frac{1}{4}\arctan (2)]$

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

$S[\exp (-\pi )]\oplus R[\exp (-2\pi )]=S(q)\oplus R(q^2)[q=\exp (-\pi )]=$

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

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

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

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

$R[\exp (-\pi )]\ominus R[\exp (-2\pi )]=R(q)\ominus R(q^2)[q=\exp (-\pi )]=$

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

That equation chain leads to this tangential sum:

$R[\exp (-\pi )]=R[\exp (-2\pi )]\oplus \tan [\arctan (\sqrt[4]{5})-\frac{1}{4}\pi ]$

And therefore following result appears:

$R[\exp (-\pi )]=\tan [\frac{1}{4}\arctan (2)+\arctan (\sqrt[4]{5})-\frac{1}{4}\pi ]$

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

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

$R[\exp (-4\pi )]=\tan [\frac{1}{2}\arctan (2)]\ominus R[\exp (-\pi )]$

And a further result appears:

$R[\exp (-4\pi )]=\tan [\frac{1}{4}\arctan (2)-\arctan (\sqrt[4]{5})+\frac{1}{4}\pi ]$

Derivation of Non-Lemniscatic values

The reflection theorem is now used for following values:

$R[\exp (-\sqrt{2}\pi )]\oplus R[\exp (-2\sqrt{2}\pi )]={\mathrm{\Phi }}^{-1}$

The Jacobi theta theorem leads to a further relation:

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

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

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

$R[\exp (-\sqrt{2}\pi )]\oplus R[\exp (-\sqrt{2}\pi )]={\mathrm{\Phi }}^{-1}\oplus \tan [2\arctan (\frac{1}{3}\sqrt{5}-\frac{1}{3}\sqrt[3]{6\sqrt{30}+4\sqrt{5}}+\frac{1}{3}\sqrt[3]{6\sqrt{30}-4\sqrt{5}})-\frac{1}{4}\pi ]$

$R[\exp (-\sqrt{2}\pi )]=\tan [\arctan (\frac{1}{3}\sqrt{5}-\frac{1}{3}\sqrt[3]{6\sqrt{30}+4\sqrt{5}}+\frac{1}{3}\sqrt[3]{6\sqrt{30}-4\sqrt{5}})-\frac{1}{4}\mathrm{arccot}(2)]$

By tangential substraction that result appears:

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

$R[\exp (-2\sqrt{2}\pi )]=\tan [\frac{1}{4}\mathrm{arccot}(-2)-\arctan (\frac{1}{3}\sqrt{5}-\frac{1}{3}\sqrt[3]{6\sqrt{30}+4\sqrt{5}}+\frac{1}{3}\sqrt[3]{6\sqrt{30}-4\sqrt{5}})]$

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

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

$\left\{R\right[\exp (-\sqrt{2}\pi ){]}^{2}R[\exp (-2\sqrt{2}\pi ){]}^{-1}+1\}\left\{R\right[\exp (-\sqrt{2}\pi )\left]R\right[\exp (-2\sqrt{2}\pi ){]}^2+1\}=2$

Now the reflection theorem is taken again:

$R[\exp (-2\sqrt{2}\pi )]={\mathrm{\Phi }}^{-1}\ominus R[\exp (-\sqrt{2}\pi )]$

$R[\exp (-2\sqrt{2}\pi )]=\frac{1-\mathrm{\Phi }R[\exp (-\sqrt{2}\pi )]}{\mathrm{\Phi }+R[\exp (-\sqrt{2}\pi )]}$

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

$\left\{R\right[\exp (-\sqrt{2}\pi ){]}^2\frac{\mathrm{\Phi }+R[\exp (-\sqrt{2}\pi )]}{1-\mathrm{\Phi }R[\exp (-\sqrt{2}\pi )]}+1\}⟨R[\exp (-\sqrt{2}\pi )]\frac{\{1-\mathrm{\Phi }R[\exp (-\sqrt{2}\pi )]{\}}^2}{\{\mathrm{\Phi }+R[\exp (-\sqrt{2}\pi )]{\}}^2}+1⟩=2$

Erasing the denominators gives an equation of sixth degree:

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

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

The solution of this equation is the already mentioned solution:

$R[\exp (-\sqrt{2}\pi )]=\tan [\arctan (\frac{1}{3}\sqrt{5}-\frac{1}{3}\sqrt[3]{6\sqrt{30}+4\sqrt{5}}+\frac{1}{3}\sqrt[3]{6\sqrt{30}-4\sqrt{5}})-\frac{1}{4}\mathrm{arccot}(2)]$

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,

$\begin{array}{rl}{\vartheta }_{10}(0;\tau )& ={\theta }_2(q)=\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}{q}^{(n+1/2{)}^2}\\ {\vartheta }_{00}(0;\tau )& ={\theta }_3(q)=\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}{q}^{n^2}\\ {\vartheta }_{01}(0;\tau )& ={\theta }_4(q)=\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}(-1{)}^n{q}^{n^2}\end{array}$

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,

$R(x)=\tan \{\frac{1}{2}\mathrm{arccot}[\frac{1}{2}+\frac{{\theta }_4(x^{1/5})[5{\theta }_4(x^5{)}^2-{\theta }_4(x{)}^2]}{2{\theta }_4(x^5)[{\theta }_4(x{)}^2-{\theta }_4(x^{1/5}{)}^2]}\left]\right\}$

$R(x)=\tan \{\frac{1}{2}\mathrm{arccot}[\frac{1}{2}+\left(\frac{{\theta }_2(x^{1/10}){\theta }_3(x^{1/10}){\theta }_4(x^{1/10})}{2^3{\theta }_2(x^{5/2}){\theta }_3(x^{5/2}){\theta }_4(x^{5/2})}{)}^{1/3}\right]\}$

$R(x)=\tan \{\frac{1}{2}\arctan [\frac{1}{2}-\frac{{\theta }_4(x{)}^2}{2{\theta }_4(x^5{)}^2}]{\}}^{1/5}\times \tan \{\frac{1}{2}\mathrm{arccot}[\frac{1}{2}-\frac{{\theta }_4(x{)}^2}{2{\theta }_4(x^5{)}^2}]{\}}^{2/5}$

$R(x)=\tan \{\frac{1}{2}\arctan [\frac{1}{2}-\frac{{\theta }_4(x^{1/2}{)}^2}{2{\theta }_4(x^{5/2}{)}^2}]{\}}^{2/5}\times \cot \{\frac{1}{2}\mathrm{arccot}[\frac{1}{2}-\frac{{\theta }_4(x^{1/2}{)}^2}{2{\theta }_4(x^{5/2}{)}^2}]{\}}^{1/5}$

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,

$R\left(e^{-\pi }\right)=\frac{1}{2}\phi (\sqrt{5}-{\phi }^{3/2})(\sqrt[4]{5}+{\phi }^{3/2})=0.511428\dots$

One can also define the elliptic nome,

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

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:

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

with

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

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,

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

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,

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

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

where

$\begin{array}{rl}& z_{\mathrm{\infty }}=-{\left[\frac{\sqrt{5}\eta (25\tau )}{\eta (5\tau )}\right]}^6-11,z_0=-{\left[\frac{\eta (\tau )}{\eta (5\tau )}\right]}^6-11,z_1={\left[\frac{\eta (\frac{5\tau +2}{5})}{\eta (5\tau )}\right]}^6-11,\\ & z_2=-{\left[\frac{\eta (\frac{5\tau +4}{5})}{\eta (5\tau )}\right]}^6-11,z_3={\left[\frac{\eta (\frac{5\tau +6}{5})}{\eta (5\tau )}\right]}^6-11,z_4=-{\left[\frac{\eta (\frac{5\tau +8}{5})}{\eta (5\tau )}\right]}^6-11\end{array}$

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,

$j(-\frac{1}{\tau })=j(\tau )$

and the Dedekind eta function has,

$\eta (-\frac{1}{\tau })=\sqrt{-i\tau }\eta (\tau )$

the functional equation of the Rogers–Ramanujan continued fraction involves the golden ratio $\phi$,

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

Incidentally,

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

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 $v-u^2=(v+u^2)u{v}^2.$

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

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 $\phi =\frac{1+\sqrt{5}}{2}$, then $u^5=\frac{v(v^2-{\phi }^{2}v+{\phi }^2)(v^2-{\phi }^{-2}v+{\phi }^{-2})}{(v^2+v+{\phi }^2)(v^2+v+{\phi }^{-2})}.$

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 $\phi$ as the golden ratio.

If $ab={\pi }^2$ then,

$[R(e^{-2a})+\phi ][R(e^{-2b})+\phi ]=\sqrt{5}\phi .$

If $5ab={\pi }^2$ then,

$[R^5(e^{-2a})+{\phi }^5][R^5(e^{-2b})+{\phi }^5]=5\sqrt{5}{\phi }^5.$

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

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

where

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

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

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

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

To solve this quintic, the elliptic modulus must first be determined as

$k=\tan [\frac{1}{4}\pi -\frac{1}{4}\mathrm{arccsc}(a^2)].$

Then the real solution is

$\begin{array}{rl}x& =\frac{2-\{1-R[q(k)]\}\{1+R[q(k{)}^2]\}}{\sqrt{R[q(k)]R[q(k{)}^2]}\sqrt[4]{4\cot ⟨4\arctan \{S\}⟩-3}}\\ & =\frac{2-\{1-R[q(k)]\}\{1+R[q(k{)}^2]\}}{\sqrt{R[q(k)]R[q(k{)}^2]}\sqrt[4]{\frac{2}{S-1}+\frac{2}{S+1}+\frac{1}{S}-S-3}}.\end{array}$

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,

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

thus,

$a=\frac{5}{4}\sqrt[4]{5}$

$k=\tan [\frac{1}{4}\pi -\frac{1}{4}\mathrm{arccsc}(a^2)]=\frac{5^{5/4}+\sqrt{25\sqrt{5}-16}}{5^{5/4}+\sqrt{25\sqrt{5}+16}}$

$q(k)=0.0851414716\dots$

$R[q(k)]=0.5633613184\dots$

$R[q(k{)}^2]=0.3706122329\dots$

and the solution is:

$x=\frac{2-\{1-R[q(k)]\}\{1+R[q(k{)}^2]\}}{\sqrt{R[q(k)]R[q(k{)}^2]}\sqrt[4]{20\cot ⟨4\arctan \{R[q(k)]R[q(k{)}^2{]}^2\}⟩-15}}=1.167303978\dots$

and can not be represented by elementary root expressions.

Example 2

$x^5-5x-4\left(\sqrt[4]{\frac{81}{32}}\right)=0$

thus,

$a=\sqrt[4]{\frac{81}{32}}$

Given the more familiar continued fractions with closed-forms,

$r_1=R\left(e^{-\pi }\right)=\frac{1}{2}\phi (\sqrt{5}-{\phi }^{3/2})(\sqrt[4]{5}+{\phi }^{3/2})=0.511428\dots$

$r_2=R\left(e^{-2\pi }\right)=\sqrt[4]{5}{\phi }^{1/2}-\phi =0.284079\dots$

$r_4=R\left(e^{-4\pi }\right)=\frac{1}{2}\phi (\sqrt{5}-{\phi }^{3/2})(-\sqrt[4]{5}+{\phi }^{3/2})=0.081002\dots$

with golden ratio $\phi =\frac{1+\sqrt{5}}{2}$ and the solution simplifies to

$\begin{array}{rl}x& =\sqrt[4]{5}\frac{2-\{1-r_1\}\{1+r_2\}}{\sqrt{r_1{r}_2}\sqrt[4]{20\cot ⟨4\arctan \{r_1{r}_2^2\}⟩-15}}\\ & =\sqrt[4]{5}\frac{2-\{1-r_2\}\{1+r_4\}}{\sqrt{r_2{r}_4}\sqrt[4]{20\cot ⟨4\arctan \{r_2{r}_4^2\}⟩-15}}\\ & =\sqrt[4]{8}=1.681792\dots \end{array}$

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, https://zenodo.org/record/1447732 
• Berndt, B. C.; Chan, H. H.; Huang, S. S.; Kang, S. Y.; Sohn, J.; Son, S. H. (1999), "The Rogers–Ramanujan continued fraction", Journal of Computational and Applied Mathematics 105 (1–2): 9–24, doi:10.1016/S0377-0427(99)00033-3, http://www.math.uiuc.edu/~berndt/articles/rrcf.pdf 

External links

• Weisstein, Eric W.. "Rogers-Ramanujan Identities". http://mathworld.wolfram.com/Rogers-RamanujanIdentities.html. 
• Weisstein, Eric W.. "Rogers-Ramanujan Continued Fraction". http://mathworld.wolfram.com/Rogers-RamanujanContinuedFraction.html.