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 $A_{400}(q)/B_{400}(q)$ 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,

$\begin{align}G(q) &= \sum_{n=0}^\infty \frac {q^{n^2}} {(1-q)(1-q^2)\dots(1-q^k)} =\sum_{n=0}^\infty \frac {q^{n^2}} {(q;q)_n} = \frac {1}{(q;q^5)_\infty (q^4; q^5)_\infty}\\ &= \prod_{n=1}^\infty \frac{1}{(1-q^{5n-1})(1-q^{5n-4})}\\ &=\sqrt[60]{qj}\,_2F_1\left(-\tfrac{1}{60},\tfrac{19}{60};\tfrac{4}{5};\tfrac{1728}{j}\right)\\ &=\sqrt[60]{q\left(j-1728\right)}\,_2F_1\left(-\tfrac{1}{60},\tfrac{29}{60};\tfrac{4}{5};-\tfrac{1728}{j-1728}\right)\\ &= 1+ q +q^2 +q^3 +2q^4+2q^5 +3q^6+\cdots \end{align}$

and,

$\begin{align}H(q) &= \sum_{n=0}^\infty \frac {q^{n^2+n}} {(1-q)(1-q^2)\dots(1-q^k)} =\sum_{n=0}^\infty \frac {q^{n^2+n}} {(q;q)_n} = \frac {1}{(q^2;q^5)_\infty (q^3; q^5)_\infty}\\ &= \prod_{n=1}^\infty \frac{1}{(1-q^{5n-2})(1-q^{5n-3})}\\ &=\frac{1}{\sqrt[60]{q^{11}j^{11}}}\,_2F_1\left(\tfrac{11}{60},\tfrac{31}{60};\tfrac{6}{5};\tfrac{1728}{j}\right)\\ &=\frac{1}{\sqrt[60]{q^{11}\left(j-1728\right)^{11}}}\,_2F_1\left(\tfrac{11}{60},\tfrac{41}{60};\tfrac{6}{5};-\tfrac{1728}{j-1728}\right)\\ &= 1+q^2 +q^3 +q^4+q^5 +2q^6+2q^7+\cdots \end{align}$

 A003114 and  A003106, respectively, where $(a;q)_\infty$ denotes the infinite q-Pochhammer symbol, j is the j-function, and ₂F₁ is the hypergeometric function, then the Rogers–Ramanujan continued fraction is,

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

Modular functions

If $q=e^{2\pi{\rm{i}}\tau}$, then $q^{-\frac{1}{60}}G(q)$ and $q^{\frac{11}{60}}H(q)$, as well as their quotient $R(q)$, are modular functions of $\tau$. Since they have integral coefficients, the theory of complex multiplication implies that their values for $\tau$ an imaginary quadratic irrational are algebraic numbers that can be evaluated explicitly.

Examples

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

$R\big(e^{-2\sqrt{5}\pi}\big) = \cfrac{e^{-\frac{2\pi}{\sqrt5}}}{1+\cfrac{e^{-2\pi\sqrt{5}}}{1 + \cfrac{e^{-4\pi\sqrt{5}}}{1+\ddots}}} = \frac{\sqrt{5}}{1+\big(5^{3/4} (\phi-1)^{5/2}-1\big)^{1/5}} - {\phi}$

where $\phi=\frac{1+\sqrt5}{2}$ is the golden ratio.

Relation to modular forms

It can be related to the Dedekind eta function, a modular form of weight 1/2, as,^[1]

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

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

Relation to j-function

Among the many formulas of the j-function, one is,

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

where,

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

Eliminating the eta quotient, one can then express j(τ) in terms of $r =R(q)$ as,

$j(\tau) = -\frac{(r^{20}-228r^{15}+494r^{10}+228r^5+1)^3}{r^5(r^{10}+11r^5-1)^5}$

$j(\tau)-1728 = -\frac{(r^{30}+ 522r^{25}- 10005 r^{20}- 10005 r^{10}- 522 r^{5}+1)^2}{r^5(r^{10}+ 11 r^5-1)^5}$

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,

$j(5\tau) = -\frac{(r^{20}+12r^{15}+14r^{10}-12r^5+1)^3}{r^{25}(r^{10}+11r^5-1)}$

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

where,

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

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

$y^2+(1+r^5)xy+r^5y=x^3+r^5x^2$

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^[2] the golden ratio $\phi$,

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

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.^[3]

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 $(v-u^3)(1+uv^3) = 3u^2v^2.$

For $n = 5$, let $u=R(q)$ and $v=R(q^5)$, then $(v^4-3v^3+4v^2-2v+1)v=(v^4+2v^3+4v^2+3v+1)u^5.$

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).^[4] Let $u=R(q^a)$, $v=R(q^b)$, and $\phi$ as the golden ratio.

If $ab=4\pi^2$, then $(u+\phi)(v+\phi)=\sqrt{5}\,\phi.$

If $5ab=4\pi^2$, then $(u^5+\phi^5)(v^5+\phi^5)=5\sqrt{5}\,\phi^5.$

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

$R^3(q) = \frac{\sum_{n=0}^\infty\frac{q^{2n}}{1-q^{5n+2}}-\sum_{n=0}^\infty\frac{q^{3n+1}}{1-q^{5n+3}} }{\sum_{n=0}^\infty\frac{q^{n}}{1-q^{5n+1}}-\sum_{n=0}^\infty\frac{q^{4n+3}}{1-q^{5n+4}} }$

For its fifth power, let $w=R(q)R^2(q^2)$, then,

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

References

1. Duke, W. "Continued Fractions and Modular Functions", http://www.math.ucla.edu/~wdduke/preprints/bams4.pdf
2. Duke, W. "Continued Fractions and Modular Functions" (p.9)
3. Berndt, B. et al. "The Rogers–Ramanujan Continued Fraction", http://www.math.uiuc.edu/~berndt/articles/rrcf.pdf
4. Berndt, B. et al. "The Rogers–Ramanujan Continued Fraction"

• Rogers, L. J. (1894), "Second Memoir on the Expansion of certain Infinite Products", Proc. London Math. Soc., s1-25 (1): 318–343, doi:10.1112/plms/s1-25.1.318
• Berndt, B. C.; Chan, H. H.; Huang, S. S.; Kang, S. Y.; Sohn, J.; Son, S. H. (1999), "The Rogers–Ramanujan continued fraction" (PDF), Journal of Computational and Applied Mathematics 105: 9, doi:10.1016/S0377-0427(99)00033-3

External links

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