Rogers–Ramanujan Continued Fraction

The Rogers–Ramanujan continued fraction is a continued fraction discovered by 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 $q=e^$,

:\beginG(q)
&= \sum_^\infty \frac =\sum_^\infty \frac = \frac \\ pt&= \prod_^\infty \frac\\ pt&=\sqrt 0,\,_2F_1\left(-\tfrac,\tfrac;\tfrac;\tfrac\right)\\ pt&=\sqrt 0,_2F_1\left(-\tfrac,\tfrac;\tfrac;-\tfrac\right)\\ pt&= 1+ q +q^2 +q^3 +2q^4+2q^5 +3q^6+\cdots
\end

and,

:\beginH(q)
&= \sum_^\infty \frac =\sum_^\infty \frac = \frac \\ pt&= \prod_^\infty \frac\\ pt&=\frac\,_2F_1\left(\tfrac,\tfrac;\tfrac;\tfrac\right)\\ pt&=\frac\,_2F_1\left(\tfrac,\tfrac;\tfrac;-\tfrac\right)\\ pt&= 1+q^2 +q^3 +q^4+q^5 +2q^6+2q^7+\cdots
\end

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

:\beginR(q)
&= \frac = q^\prod_^\infty \frac=q^\prod^_(1-q^n)^ \\ pt
&= \cfrac
\end
:$(n\mid m)$ 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 $q=e^$ (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 $q=e^$. However, Ramanujan, in his examples to Hardy and given below, used the ''nome'' $q=e^$instead.

Special values

If ''q'' is the nome or its square, then $q^G(q)$ and $q^H(q)$, 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 $q = e^$,

:$R(q) = \cfrac$
f
when $\tau = i$,

:R\big(e^\big) = \cfrac = \tfrac\varphi\,(\sqrt-\varphi^)(\sqrt \varphi^) = 0.511428\dots

when $\tau = 2i$,

:$R\big(e^\big) = \cfrac = = 0.284079\dots$

when $\tau = 4i$,

:R\big(e^\big) = \cfrac = \tfrac\varphi\,(\sqrt-\varphi^)(-\sqrt \varphi^) = 0.081002\dots

when $\tau = 2\sqrti$,

:$R\big(e^\big) = \cfrac = \frac - \varphi = 0.0602094\dots$

when $\tau = 5i$,

:$R\big(e^\big) = \cfrac = \frac - \varphi = 0.0432139\dots$

when $\tau = 10i$,

:$R\big(e^\big) = \cfrac = \frac - \varphi = 0.00186744\dots$

when $\tau = 20i$,

:$R\big(e^\big) = \cfrac = \frac - \varphi = 0.00000348734\dots$

and $\varphi=\tfrac$ is the golden ratio. Note that $R\big(e^\big)$ is a positive root of the quartic equation,

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

while $R\big(e^\big)$ and $R\big(e^\big)$ 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 $R(e^)$ and $R(e^)$ are two roots of the same equation as well as,

:\bigl (e^) +\varphi\bigr\bigl (e^) +\varphi\bigr= \sqrt5\,\varphi

The algebraic degree ''k'' of $R(e^)$ for $n = 1,2,3,4,\dots$ is $k = 8,4,32,8,\dots$ ().

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 $q = e^$, then one should be careful with notation such that $j(\tau),\,G(q),\, H(q)$ and $r = R(q)$ use the same $q$.

:\begin
G(q) &= \prod_^\infty \frac\\ pt&= q^ \frac
\end

:\begin
H(q) &= \prod_^\infty \frac\\ pt&= \frac \frac
\end

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

:\begin
G(e^)
&= (e^)^ \frac1 \frac1 \\ pt&= 1.00187093\dots\\ ptH(e^)
&= \frac1 \frac1 \sqrt \\ pt&= 1.00000349\ldots
\end

and,

:\begin
G(e^)
&= (e^)^ \frac1 \frac1 \\ pt&= 1.000003487354\dots\\ ptH(e^)
&= \frac1 \frac1 \sqrt \\ pt&= 1.000000000012\dots
\end

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

: c \ominus d = \tan\bigl arctan(c) - \arctan(d)\bigr= \frac

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

:$\ln(q) \ln(q_) = \pi^2$

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

:$q_(k) = q(k') = q(\sqrt)$

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

:

The letter $\Phi$ represents the Golden number exactly:

: \Phi = \tfrac(\sqrt + 1) = \cot tfrac\arctan(2)= 2\cos(\tfrac)

: \Phi^ = \tfrac(\sqrt - 1) = \tan tfrac\arctan(2)= 2\sin(\tfrac)

The theorems for the squared nome are constructed as follows:

:

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

:

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:

:

In an analogue pattern we get this result:

: R\bigl exp(-2\pi)\bigr\oplus R\bigl exp(-2\pi)\bigr= \Phi^

Therefore following identity is valid:

:

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) \bigl = \exp(-\pi)\bigr=

: = \frac \bigl = \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\pi - \tfrac\arctan(2)\bigr= \tan\bigl tfrac\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) \bigl = \exp(-\pi)\bigr=

: = \frac \bigl = \exp(-\pi)\bigr= \frac = \sqrt \ominus 1 = \tan\bigl arctan(\sqrt[4,) - \tfrac\pi \bigr">.html" ;"title="arctan(\sqrt[4">arctan(\sqrt[4,) - \tfrac\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,) - \tfrac\pi \bigr">.html" ;"title="arctan(\sqrt[4">arctan(\sqrt[4,) - \tfrac\pi \bigr

And therefore following result appears:

:

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^

: R\bigl[\exp(-4\pi)\bigr] = \tan\bigl tfrac\arctan(2)\bigr\ominus R\bigl exp(-\pi)\bigr

And a further result appears:

:

Derivation of Non-Lemniscatic values

The reflection theorem is now used for following values:

: R\bigl exp(-\sqrt\,\pi)\bigr\oplus R\bigl exp(-2\sqrt\,\pi)\bigr= \Phi^

The Jacobi theta theorem leads to a further relation:

: R\bigl exp(-\sqrt\,\pi)\bigr\ominus R\bigl exp(-2\sqrt\,\pi)\bigr= R(q) \ominus R(q^2) \bigl = \exp(-\sqrt\,\pi)\bigr=

: = \frac \bigl = \exp(-\sqrt\,\pi)\bigr= \tan\bigl \arctan(\tfrac\sqrt - \tfrac\sqrt[3+ \tfrac\sqrt[3">.html" ;"title="\arctan(\tfrac\sqrt - \tfrac\sqrt[3">\arctan(\tfrac\sqrt - \tfrac\sqrt[3+ \tfrac\sqrt[3,) - \tfrac\pi \bigr">">\arctan(\tfrac\sqrt - \tfrac\sqrt[3+ \tfrac\sqrt[3">.html" ;"title="\arctan(\tfrac\sqrt - \tfrac\sqrt[3">\arctan(\tfrac\sqrt - \tfrac\sqrt[3+ \tfrac\sqrt[3,) - \tfrac\pi \bigr

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

: R\bigl exp(-\sqrt\,\pi)\bigr\oplus R\bigl exp(-\sqrt\,\pi)\bigr= \Phi^ \oplus \tan\bigl \arctan(\tfrac\sqrt - \tfrac\sqrt[3+ \tfrac\sqrt[3">.html" ;"title="\arctan(\tfrac\sqrt - \tfrac\sqrt[3">\arctan(\tfrac\sqrt - \tfrac\sqrt[3+ \tfrac\sqrt[3,) - \tfrac\pi \bigr">">\arctan(\tfrac\sqrt - \tfrac\sqrt[3+ \tfrac\sqrt[3">.html" ;"title="\arctan(\tfrac\sqrt - \tfrac\sqrt[3">\arctan(\tfrac\sqrt - \tfrac\sqrt[3+ \tfrac\sqrt[3,) - \tfrac\pi \bigr

:

By tangential substraction that result appears:

: R\bigl exp(-2\sqrt\,\pi)\bigr\oplus R\bigl exp(-2\sqrt\,\pi)\bigr= \Phi^ \ominus \tan\bigl \arctan(\tfrac\sqrt - \tfrac\sqrt[3+ \tfrac\sqrt[3">.html" ;"title="\arctan(\tfrac\sqrt - \tfrac\sqrt[3">\arctan(\tfrac\sqrt - \tfrac\sqrt[3+ \tfrac\sqrt[3,) - \tfrac\pi \bigr">">\arctan(\tfrac\sqrt - \tfrac\sqrt[3+ \tfrac\sqrt[3">.html" ;"title="\arctan(\tfrac\sqrt - \tfrac\sqrt[3">\arctan(\tfrac\sqrt - \tfrac\sqrt[3+ \tfrac\sqrt[3,) - \tfrac\pi \bigr

:

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

: R\bigl exp(-\sqrt\,\pi)\bigr2 R\bigl[\exp(-2\sqrt\,\pi)\bigr]^\oplus R\bigl[\exp(-\sqrt\,\pi)\bigr]R\bigl[\exp(-2\sqrt\,\pi)\bigr]^2 = 1

:$\bigl\\bigl\ = 2$

Now the reflection theorem is taken again:

: R\bigl exp(-2\sqrt\,\pi)\bigr= \Phi^ \ominus R\bigl exp(-\sqrt\,\pi)\bigr

: R\bigl exp(-2\sqrt\,\pi)\bigr= \frac

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

: \biggl\\biggl\langle R\bigl exp(-\sqrt\,\pi)\bigr\frac + 1\biggr\rangle = 2

Erasing the denominators gives an equation of sixth degree:

: R\bigl exp(-\sqrt\,\pi)\bigr6 + 2\,\Phi^ R\bigl exp(-\sqrt\,\pi)\bigr5 - \sqrt\,\Phi^ R\bigl exp(-\sqrt\,\pi)\bigr4 +

: + 2\,\sqrt\,\Phi R\bigl exp(-\sqrt\,\pi)\bigr3 + \sqrt\,\Phi^ R\bigl exp(-\sqrt\,\pi)\bigr2 + 2\,\Phi^ R\bigl exp(-\sqrt\,\pi)\bigr- 1 = 0

The solution of this equation is the already mentioned solution:

:R\bigl exp(-\sqrt\,\pi)\bigr= \tan\bigl arctan(\tfrac\sqrt - \tfrac\sqrt[3+ \tfrac\sqrt[3">.html" ;"title="arctan(\tfrac\sqrt - \tfrac\sqrt[3">arctan(\tfrac\sqrt - \tfrac\sqrt[3+ \tfrac\sqrt[3,) - \tfrac\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,

:$\frac-R(q) = \frac+1$

:\frac-R^5(q) = \left frac\right6+11

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

:$\begin \vartheta_(0;\tau)&=\theta_2(q)=\sum_^\infty q^\\ \vartheta_(0;\tau)&=\theta_3(q)=\sum_^\infty q^\\ \vartheta_(0;\tau)&=\theta_4(q)=\sum_^\infty (-1)^n q^ \end$

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\$
: $R(x) = \tan\biggl\$
: $R(x) = \tan\biggl\^\times\tan\biggl\^$
: $R(x) = \tan\biggl\^\times\cot\biggl\^$

Note, however, that theta functions normally use the nome , while the Dedekind eta function uses the ''square'' of the nome , thus the variable ''x'' has been employed instead to maintain consistency between all functions. For example, let $\tau = \sqrt$ so $x = e^$. 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\big(e^\big) = \frac\varphi\,(\sqrt-\varphi^)(\sqrt \varphi^) = 0.511428\dots

One can also define th
elliptic nome

:q(k) = \exp\big \pi K(\sqrt)/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:

:$R\big(q(k)\big) = \tan \biggl\^ \tan \biggl\^ =\left\^ \left\^$

with

:$y=\frac.$

Relation to j-function

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

:$j(\tau) = \frac$

where x = \left frac\right6.\, Since also,

:\frac-R^5(q) = \left frac\right6+11

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

:
\begin
& j(\tau) = -\frac \\ pt& j(\tau)-1728 = -\frac
\end

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
& j(5\tau) = -\frac \\ pt& j(5\tau)-1728 = -\frac
\end

Let $z=r^5-\frac$, then $j(5\tau) = -\frac$

where

:
\begin
& z_\infty = -\left frac\right6-11,\ z_0=-\left \frac \right6-11,\ z_1=\left frac\right6-11, \\ pt& z_2=-\left frac\right6-11,\ z_3=\left frac\right6-11,\ z_4=-\left \frac\right6-11
\end

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,

:$j(-\tfrac) = j(\tau)$

and the Dedekind eta function has,

:$\eta(-\tfrac) =\sqrt\, \eta(\tau)$

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

:$r(-\tfrac) = \frac$

Incidentally,

:$r(\tfrac) = i$

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

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 $\varphi=\tfrac2$, then $u^5 = \frac.$

For $n = 11$, let $u=R(q)$ and $v=R(q^)$, then $uv(u^+11u^5-1)(v^+11v^5-1) = (u-v)^.$

Regarding $n = 5$, note that $v^+11v^5-1=(v^2+v-1)(v^4-3v^3+4v^2-2v+1)(v^4+2v^3+4v^2+3v+1).$

Other results

Ramanujan found many other interesting results regarding $R(q)$.Berndt, B. et al. "The Rogers–Ramanujan Continued Fraction" Let $a,b\in\mathbb^+$, and $\varphi$ as the golden ratio.

If $ab=\pi^2$ then,

:\bigl (e^)+\varphi\biglbigl (e^)+\varphi\bigr\sqrt\,\varphi.

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

:\bigl ^5(e^)+\varphi^5\biglbigl ^5(e^)+\varphi^5\bigr5\sqrt\,\varphi^5.

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

:$R^3(q) = \frac$

where

:$\alpha=\sum_^\infty\frac-\sum_^\infty \frac,$
:$\beta=\sum_^\infty\frac-\sum_^\infty \frac.$

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

:$R^5(q) = w\left(\frac\right)^2,\;\; R^5(q^2) = w^2\left(\frac \right)$

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)/K(k)\big

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

:k = \tan tfrac\pi - \tfrac\arccsc(a^2)

Then the real solution is

:$\beginx&= \frac\\&= \frac. \end$

where S=R (k),R^2 (k)^2. Recall in the previous section the 5th power of $R(q)$ can be expressed by $S$:

:R^5 (k)= S\left(\frac\right)^2

Example 1

:$x^5 - x - 1 = 0$

Transform to,

:(\sqrt )^5 - 5(\sqrt ) - 4(\tfrac\sqrt = 0

thus,

:a = \tfrac\sqrt

:k = \tan tfrac\pi - \tfrac\arccsc(a^2)= \tfrac

:$q(k) = 0.0851414716\dots$

:R (k)= 0.5633613184\dots

:R (k)^2= 0.3706122329\dots

and the solution is:

:$x = \frac = 1.167303978\dots$

and can not be represented by elementary root expressions.

Example 2

:x^5 - 5x - 4\Bigl(\sqrt Bigr) = 0

thus,

:a = \sqrt

Given the more familiar continued fractions with closed-forms,

:r_1 = R\big(e^\big) = \tfrac\varphi\,(\sqrt-\varphi^)(\sqrt \varphi^) = 0.511428\dots

:r_2 = R\big(e^\big) = \sqrt ,\varphi^-\varphi = 0.284079\dots

:r_4 = R\big(e^\big) = \tfrac\varphi\,(\sqrt-\varphi^)(-\sqrt \varphi^) = 0.081002\dots

with golden ratio $\varphi = \tfrac$ and the solution simplifies to

:\beginx
&= \sqrt ,\frac\\ pt&= \sqrt ,\frac\\ pt&= \sqrt = 1.681792\dots\end

References

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External links

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{{DEFAULTSORT:Rogers-Ramanujan continued fraction
Mathematical identities
Q-analogs
Modular forms
Continued fractions
Srinivasa Ramanujan