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

Definition

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

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

[math]\displaystyle{ \begin{align}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]{q\,j}\,\,_2F_1\left(-\tfrac{1}{60},\tfrac{19}{60};\tfrac{4}{5};\tfrac{1728}{j}\right)\\ &=\sqrt[60]{q\left(j-1728\right)}\,_2F_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{align} }[/math]

and,

[math]\displaystyle{ \begin{align}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}}}\,_2F_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}}}\,_2F_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{align} }[/math]

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

[math]\displaystyle{ \begin{align}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^{\infty}_{n=1}(1-q^n)^{(n|5)} \\ &= \cfrac{q^{1/5}}{1+\cfrac{q}{1+\cfrac{q^2}{1+\cfrac{q^3}{1+\ddots}}}} \end{align} }[/math]

[math]\displaystyle{ (n|m) }[/math] denotes the Jacobi symbol.

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

Special values

If q is the nome or its square, then [math]\displaystyle{ q^{-\frac{1}{60}}G(q) }[/math] and [math]\displaystyle{ q^{\frac{11}{60}}H(q) }[/math], as well as their quotient [math]\displaystyle{ R(q) }[/math], are related to modular functions of [math]\displaystyle{ \tau }[/math]. Since they have integral coefficients, the theory of complex multiplication implies that their values for [math]\displaystyle{ \tau }[/math] 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 [math]\displaystyle{ q = e^{\pi i \tau} }[/math],

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

when [math]\displaystyle{ \tau = i }[/math],

[math]\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}\phi\,(\sqrt{5}-\phi^{3/2})(\sqrt[4]{5}+\phi^{3/2}) = 0.511428\dots }[/math]

when [math]\displaystyle{ \tau = 2i }[/math],

[math]\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} = 0.284079\dots }[/math]

when [math]\displaystyle{ \tau = 4i }[/math],

[math]\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}\phi\,(\sqrt{5}-\phi^{3/2})(-\sqrt[4]{5}+\phi^{3/2}) = 0.081002\dots }[/math]

when [math]\displaystyle{ \tau = 2\sqrt{5}i }[/math],

[math]\displaystyle{ R\big(e^{-2\sqrt{5}\pi}\big) = \cfrac{e^{-\frac{2\pi}{\sqrt5}}}{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} = 0.0602094\dots }[/math]

when [math]\displaystyle{ \tau = 5i }[/math],

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

when [math]\displaystyle{ \tau = 10i }[/math],

[math]\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+\phi^2}{\phi+\big(3\sqrt{1+\phi^2}-4-\phi\big)^{1/5}} - {\phi} = 0.00186744\dots }[/math]

when [math]\displaystyle{ \tau = 20i }[/math],

[math]\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+\phi^2}{\phi+\big(\frac{1}{2}(4-\phi-3\sqrt{\phi-1})(3\phi^{3/2}+\sqrt[4]{5})\big)^{1/5}} - {\phi} = 0.00000348734\dots }[/math]

and [math]\displaystyle{ \phi=\tfrac{1+\sqrt5}{2} }[/math] is the golden ratio. Note that [math]\displaystyle{ R\big(e^{-2\pi}\big) }[/math] is a positive root of the quartic equation,

[math]\displaystyle{ x^4+2x^3-6x^2-2x+1=0 }[/math]

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

[math]\displaystyle{ y^4+2\phi^4 y^3+6\phi^2 y^2-2\phi^4 y+1=0 }[/math]

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

[math]\displaystyle{ \bigl[R(e^{-2\pi/m}) +\phi\bigr] \bigl[R(e^{-2\pi\, m}) +\phi\bigr] = \sqrt5\,\phi }[/math]

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

[math]\displaystyle{ \begin{align} G(q) &= \prod_{n=1}^\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{align} }[/math]

[math]\displaystyle{ \begin{align} H(q) &= \prod_{n=1}^\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{align} }[/math]

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

[math]\displaystyle{ \begin{align} G(e^{-2\pi}) &= (e^{-2\pi})^{1/60} \frac1{(5\,\phi)^{1/4}} \frac1{\sqrt{ R(e^{-2\pi}) }} \\ &= 1.00187093\dots\\ H(e^{-2\pi}) &= \frac1{(e^{-2\pi})^{11/60}} \frac1{(5\,\phi)^{1/4}} \sqrt{ R(e^{-2\pi}) } \\ &= 1.00000349\dots\\ \end{align} }[/math]

and,

[math]\displaystyle{ \begin{align} G(e^{-4\pi}) &= (e^{-4\pi})^{1/60} \frac1{(5\,\phi^3)^{1/4}\,(\phi+\sqrt[4]{5})^{1/4}} \frac1{\sqrt{ R(e^{-4\pi}) }} \\ &= 1.000003487354\dots\\ H(e^{-4\pi}) &= \frac1{(e^{-4\pi})^{11/60}} \frac1{(5\,\phi^3)^{1/4}\,(\phi+\sqrt[4]{5})^{1/4}} \sqrt{ R(e^{-4\pi}) } \\ &= 1.000000000012\dots\\ \end{align} }[/math]

and so on, with [math]\displaystyle{ \phi }[/math] 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:

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

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

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

[math]\displaystyle{ \ln(q) \ln(q_{1}) = \pi^2 }[/math]

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

[math]\displaystyle{ q_{1}(k) = q(k') = q(\sqrt{1 - k^2}) }[/math]

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

[math]\displaystyle{ S(q) \oplus S(q_{1})= \Phi }[/math]

[math]\displaystyle{ R(q^2) \oplus R(q_{1}^2)= \Phi^{-1} }[/math]

The letter [math]\displaystyle{ \Phi }[/math] represents the Golden number exactly:

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

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

The theorems for the squared nome are constructed as follows:

[math]\displaystyle{ R(q)^2 R(q^2)^{-1}\oplus R(q)R(q^2)^2 =1 }[/math]

[math]\displaystyle{ S(q)^2 R(q^2)^{-1}\ominus S(q)R(q^2)^2 =1 }[/math]

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

[math]\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} }[/math]

[math]\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} }[/math]

Derivation of Lemniscatic values

Into the now shown theorems certain values are inserted:

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

Therefore following identity is valid:

[math]\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] }[/math]

In an analogue pattern we get this result:

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

Therefore following identity is valid:

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

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

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

[math]\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 }[/math]

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

[math]\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] }[/math]

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

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

[math]\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] }[/math]

That equation chain leads to this tangential sum:

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

And therefore following result appears:

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

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

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

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

And a further result appears:

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

Derivation of Non-Lemniscatic values

The reflection theorem is now used for following values:

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

The Jacobi theta theorem leads to a further relation:

[math]\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] = }[/math]

[math]\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] }[/math]

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

[math]\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] }[/math]

[math]\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}\arccot(2)\bigr] }[/math]

By tangential substraction that result appears:

[math]\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] }[/math]

[math]\displaystyle{ R\bigl[\exp(-2\sqrt{2}\,\pi)\bigr] = \tan\bigl[\tfrac{1}{4}\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] }[/math]

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

[math]\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 }[/math]

[math]\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 }[/math]

Now the reflection theorem is taken again:

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

[math]\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]} }[/math]

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

[math]\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 }[/math]

Erasing the denominators gives an equation of sixth degree:

[math]\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 + }[/math]

[math]\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 }[/math]

The solution of this equation is the already mentioned solution:

[math]\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}\arccot(2)\bigr] }[/math]

Relation to modular forms

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

[math]\displaystyle{ \frac{1}{R(q)}-R(q) = \frac{\eta(\frac{\tau}{5})}{\eta(5\tau)}+1 }[/math]

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

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

[math]\displaystyle{ \begin{align} \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{align} }[/math]

The notation [math]\displaystyle{ \theta_n }[/math] is slightly easier to remember since [math]\displaystyle{ \theta_2^4+\theta_4^4 =\theta_3^4 }[/math], with even subscripts on the LHS. Thus,

[math]\displaystyle{ R(x) = \tan\biggl\{\frac{1}{2}\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\} }[/math]

[math]\displaystyle{ R(x) = \tan\biggl\{\frac{1}{2}\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\} }[/math]

[math]\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}\arccot\biggl[\frac{1}{2} - \frac{\theta_4(x)^2}{2\,\theta_4(x^5)^2}\biggr]\biggr\}^{2/5} }[/math]

[math]\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}\arccot\biggl[\frac{1}{2} - \frac{\theta_4(x^{1/2})^2}{2\,\theta_4(x^{5/2})^2}\biggr]\biggr\}^{1/5} }[/math]

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 [math]\displaystyle{ \tau = \sqrt{-1} }[/math] so [math]\displaystyle{ x = e^{-\pi} }[/math]. 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,

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

One can also define the elliptic nome,

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

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:

[math]\displaystyle{ R\big(q(k)\big) = \tan \biggl\{\frac{1}{2}\arctan y\biggr\}^{1/5} \tan \biggl\{\frac{1}{2}\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} }[/math]

with

[math]\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}. }[/math]

Relation to j-function

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

[math]\displaystyle{ j(\tau) = \frac{(x^2+10x+5)^3}{x} }[/math]

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

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

Eliminating the eta quotient [math]\displaystyle{ x }[/math] between the two equations, one can then express j(τ) in terms of [math]\displaystyle{ r =R(q) }[/math] as,

[math]\displaystyle{ \begin{align} & 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}- 10005 r^{20}- 10005 r^{10}- 522 r^5 + 1)^2}{r^5(r^{10}+ 11 r^5-1)^5} \end{align} }[/math]

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

[math]\displaystyle{ \begin{align} & 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{align} }[/math]

Let [math]\displaystyle{ z=r^5-\frac{1}{r^5} }[/math], then [math]\displaystyle{ j(5\tau) = -\frac{ \left( z^2 + 12z + 16 \right)^3}{z+11} }[/math]

where

[math]\displaystyle{ \begin{align} & 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{align} }[/math]

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

[math]\displaystyle{ y^2+(1+r^5)xy+r^5y=x^3+r^5x^2 }[/math]

parameterized by the non-cusp points of the modular curve [math]\displaystyle{ X_1(5) }[/math].

Functional equation

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

[math]\displaystyle{ j(-\tfrac{1}{\tau}) = j(\tau) }[/math]

and the Dedekind eta function has,

[math]\displaystyle{ \eta(-\tfrac{1}{\tau}) =\sqrt{-i\tau}\, \eta(\tau) }[/math]

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

[math]\displaystyle{ r(-\tfrac{1}{\tau}) = \frac{1-\phi\,r(\tau)}{\phi+r(\tau)} }[/math]

Incidentally,

[math]\displaystyle{ r(\tfrac{7+i}{10}) = i }[/math]

Modular equations

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

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

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

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

Or equivalently for [math]\displaystyle{ n = 5 }[/math], let [math]\displaystyle{ u=R(q) }[/math] and [math]\displaystyle{ v=R(q^5) }[/math] and [math]\displaystyle{ \phi=\tfrac{1+\sqrt5}2 }[/math], then [math]\displaystyle{ u^5 = \frac{v\,(v^2-\phi^2v+\phi^2)(v^2-\phi^{-2}v+\phi^{-2})}{(v^2+v+\phi^2)(v^2+v+\phi^{-2})}. }[/math]

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

Regarding [math]\displaystyle{ n = 5 }[/math], note that [math]\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). }[/math]

Other results

Ramanujan found many other interesting results regarding [math]\displaystyle{ R(q) }[/math]. Let [math]\displaystyle{ a,b\in\mathbb{R}^+ }[/math], and [math]\displaystyle{ \phi }[/math] as the golden ratio.

If [math]\displaystyle{ ab=\pi^2 }[/math] then,

[math]\displaystyle{ \bigl[R(e^{-2a})+\phi\bigl]\bigl[R(e^{-2b})+\phi\bigr]=\sqrt{5}\,\phi. }[/math]

If [math]\displaystyle{ 5ab=\pi^2 }[/math] then,

[math]\displaystyle{ \bigl[R^5(e^{-2a})+\phi^5\bigl]\bigl[R^5(e^{-2b})+\phi^5\bigr]=5\sqrt{5}\,\phi^5. }[/math]

The powers of [math]\displaystyle{ R(q) }[/math] also can be expressed in unusual ways. For its cube,

[math]\displaystyle{ R^3(q) = \frac{\alpha}{\beta} }[/math]

where,

[math]\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}} }[/math]

[math]\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}} }[/math]

For its fifth power, let [math]\displaystyle{ w=R(q)R^2(q^2) }[/math], then,

[math]\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) }[/math]

Quintic equations

The general quintic equation in Bring-Jerrard form:

[math]\displaystyle{ x^5 - 5x - 4a = 0 }[/math]

for every real value [math]\displaystyle{ a \gt 1 }[/math] can be solved in terms of Rogers-Ramanujan continued fraction [math]\displaystyle{ R(q) }[/math] and the elliptic nome:

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

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

[math]\displaystyle{ k = \tan[\tfrac{1}{4}\pi - \tfrac{1}{4}\arccsc(a^2)] }[/math]

Then the real solution is:

[math]\displaystyle{ \begin{align}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{align} }[/math]

where [math]\displaystyle{ S=R[q(k)]\,R^2[q(k)^2]. }[/math]. Recall in the previous section the 5th power of [math]\displaystyle{ R(q) }[/math] can be expressed by [math]\displaystyle{ S }[/math]:

[math]\displaystyle{ R^5[q(k)] = S\left(\frac{1-S}{1+S}\right)^2 }[/math]

Example 1

[math]\displaystyle{ x^5 - x - 1 = 0 }[/math]

Transform to,

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

thus,

[math]\displaystyle{ a = \tfrac{5}{4}\sqrt[4]{5} }[/math]

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

[math]\displaystyle{ q(k) = 0.0851414716\dots }[/math]

[math]\displaystyle{ R[q(k)] = 0.5633613184\dots }[/math]

[math]\displaystyle{ R[q(k)^2] = 0.3706122329\dots }[/math]

and the solution is:

[math]\displaystyle{ 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 }[/math]

and can not be represented by elementary root expressions.

Example 2

[math]\displaystyle{ x^5 - 5x - 4\Bigl(\sqrt[4]{\tfrac{81}{32}}\Bigr) = 0 }[/math]

thus,

[math]\displaystyle{ a = \sqrt[4]{\tfrac{81}{32}} }[/math]

Given the more familiar continued fractions with closed-forms,

[math]\displaystyle{ r_1 = R\big(e^{-\pi}\big) = \tfrac{1}{2}\phi\,(\sqrt{5}-\phi^{3/2})(\sqrt[4]{5}+\phi^{3/2}) = 0.511428\dots }[/math]

[math]\displaystyle{ r_2 = R\big(e^{-2\pi}\big) = \sqrt[4]{5}\,\phi^{1/2}-\phi = 0.284079\dots }[/math]

[math]\displaystyle{ r_4 = R\big(e^{-4\pi}\big) = \tfrac{1}{2}\phi\,(\sqrt{5}-\phi^{3/2})(-\sqrt[4]{5}+\phi^{3/2}) = 0.081002\dots }[/math]

with golden ratio [math]\displaystyle{ \phi = \tfrac{1+\sqrt5}{2} }[/math] and the solution simplifies to:

[math]\displaystyle{ \begin{align}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}}\\ &= \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}}\\ &= \sqrt[4]{8} = 1.681792\dots\end{align} }[/math]

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.