Parametric Evaluations of the Rogers-Ramanujan Continued Fraction

Abstract

In this paper with the help of the inverse function of the singular moduli we evaluate the Rogers-Ranmanujan continued fraction and its first derivative.

1. Introductory Definitions and Formulas

For |𝑞|<1, the Rogers-Ramanujan continued fraction (RRCF) (see [1]) is defined as 𝑅𝑞𝑞∶=^1/5𝑞1+¹𝑞1+²𝑞1+1+³⋯. (1.1) We also define 𝑎;𝑞_𝑛∶=^𝑛−1_𝑘=01−𝑎𝑞^𝑘,𝑓−𝑞∶=^∞_𝑛=11−𝑞^𝑛=𝑞;𝑞_∞. (1.2) Ramanujan give the following relations which are very useful: 1𝑅𝑞𝑞=𝑓−1−𝑅−𝑞^1/5𝑞^1/5𝑓−𝑞⁵, (1.3)1𝑅⁵𝑞−11−𝑅⁵𝑞=𝑓⁶−𝑞𝑞𝑓⁶−𝑞⁵. (1.4) From the theory of elliptic functions (see [1–3]), 𝐾(𝑥)=₀^𝜋/211−𝑥²sin(𝑡)²𝑑𝑡, (1.5) is the elliptic integral of the first kind. It is known that the inverse elliptic nome 𝑘=𝑘_𝑟, 𝑘^′2_𝑟=1−𝑘²_𝑟 is the solution of the equation 𝐾𝑘^′_𝑟𝐾(𝑘)=√𝑟, (1.6) where 𝑟∈𝐑^∗₊. When 𝑟 is rational then the 𝑘_𝑟 are algebraic numbers.

We can also write the function 𝑓 using elliptic functions. It holds (see [3]) 𝑓−𝑞⁸=2^8/3𝜋⁴𝑞^−1/3𝑘_𝑟^2/3𝑘^′_𝑟^8/3𝐾𝑘_𝑟⁴, (1.7) and also holds 𝑓−𝑞²⁶=2𝑘_𝑟𝑘^′_𝑟𝐾𝑘_𝑟³𝜋³𝑞^1/2. (1.8) From [4] it is known that 𝑅^′𝑞=15𝑞^−5/6𝑓−𝑞⁴𝑅𝑞⁶𝑅𝑞⁻⁵−11−𝑅𝑞⁵. (1.9) Consider now for every 0<𝑥<1 the equation 𝑥=𝑘_𝑟, (1.10) which has solution 𝑟=𝑘⁽⁻¹⁾(𝑥). (1.11) Hence for example 𝑘⁽⁻¹⁾1√2=1. (1.12) With the help of 𝑘⁽⁻¹⁾ function we evaluate the Rogers Ramanujan continued fraction.

2. Propositions

The relation between 𝑘_25𝑟 and 𝑘_𝑟 is (see [1] page 280)𝑘_𝑟𝑘_25𝑟+𝑘^′_𝑟𝑘^′_25𝑟+2⋅4^1/3𝑘_𝑟𝑘_25𝑟𝑘^′_𝑟𝑘^′_25𝑟^1/3=1. (2.1) For to solve (2.1) we give the following.

Proposition 2.1. The solution of the equation 𝑥⁶+𝑥³−16+10𝑥²𝑤+15𝑥⁴𝑤²−20𝑥³𝑤³+15𝑥²𝑤⁴+𝑥10−16𝑥²𝑤⁵+𝑤⁶=0. (2.2) when one knows 𝑤 is given by 𝑦^1/2𝑤^1/2=𝑤^1/2𝑥^1/2=1224+3𝐿^1/6𝑀^1/6𝑀−4^1/6𝐿^1/62+1223𝐿^1/6𝑀^1/6𝑀−4^1/6𝐿^1/6, (2.3) where 𝑤=𝐿(18+𝐿)6(64+3𝐿)<1, (2.4)𝑀=18+𝐿64+3𝐿. (2.5) If it happens that 𝑥=𝑘_𝑟 and 𝑦=𝑘_25𝑟, then 𝑟=𝑘⁽⁻¹⁾(𝑥) and 𝑤²=𝑘_25𝑟𝑘_𝑟, (𝑤^′)²=𝑘^′_25𝑟𝑘^′_𝑟.

Proof. The relation (2.3) can be found using Mathematica. See also [5].

Proposition 2.2. If 𝑞=𝑒^{√−𝜋𝑟} and 𝑎=𝑎_𝑟=𝑘^′_𝑟𝑘^′_25𝑟²𝑘_𝑟𝑘_25𝑟𝑀₅(𝑟)⁻³, (2.6) then 𝑎_𝑟=𝑅𝑞⁻⁵−11−𝑅⁵𝑞, (2.7) where 𝑀₅(𝑟) is root of (5𝑥−1)⁵(1−𝑥)=256(𝑘_𝑟)²(𝑘^′_𝑟)²𝑥.

Proof. Suppose that 𝑁=𝑛²𝜇, where 𝑛 is positive integer and 𝜇 is positive real then it holds that 𝐾𝑛²𝜇=𝑀_𝑛𝜇𝐾𝜇, (2.8) where 𝐾[𝜇]=𝐾(𝑘_𝜇)
The following formula for 𝑀₅(𝑟) is known: 5𝑀₅(𝑟)−1⁵1−𝑀₅(𝑟)=256𝑘_𝑟²𝑘^′_𝑟²𝑀₅(𝑟). (2.9) Thus, if we use (1.4) and (1.7) and the above consequence of the theory of elliptic functions, we get: 𝑅⁻⁵𝑞−11−𝑅⁵𝑞=𝑓⁶−𝑞𝑞𝑓⁶−𝑞⁵=𝑎=𝑎_𝑟=𝑘^′_𝑟𝑘^′_25𝑟²𝑘_𝑟𝑘_25𝑟𝑀₅(𝑟)⁻³. (2.10) See also [4, 5].

3. The Main Theorem

From Proposition 2.2 and relation 𝑤²=𝑘_25𝑟𝑘_𝑟 we get 𝑤⁵−𝑘²_𝑟𝑘𝑤=^3𝑟𝑘²_𝑟−1𝑎_𝑟𝑀₅(𝑟)³. (3.1) Combining (2.2) and (3.1), we get −10𝑘⁴_𝑟+26𝑘⁶_𝑟+𝑎_𝑟𝑀₅(𝑟)³𝑘⁶_𝑟−16𝑘⁸_𝑟+−𝑘³_𝑟−6𝑎_𝑟𝑀₅(𝑟)³𝑘³_𝑟+𝑘⁵_𝑟−6𝑎_𝑟𝑀₅(𝑟)³𝑘⁵_𝑟𝑤+𝑎_𝑟𝑀₅(𝑟)³𝑘²_𝑟+15𝑎_𝑟𝑀₅(𝑟)𝑘⁴_𝑟𝑤²−20𝑎_𝑟𝑀₅(𝑟)³𝑘³_𝑟𝑤³+15𝑎_𝑟𝑀₅(𝑟)³𝑘²_𝑟𝑤⁴=0. (3.2) Solving with respect to 𝑎_𝑟𝑀₅(𝑟)³, we get 𝑎_𝑟𝑀₅(𝑟)³=16𝑘⁶_𝑟−26𝑘⁴_𝑟−𝑤𝑘³_𝑟+10𝑘²_𝑟+𝑤𝑘_𝑟𝑘⁴_𝑟−6𝑘³_𝑟𝑤−20𝑘³_𝑟𝑤+15𝑤²𝑘²_𝑟−6𝑘_𝑟𝑤+15𝑤⁴+𝑤². (3.3) Also we have 𝐾𝑘_25𝑟𝐾𝑘_𝑟=𝑀₅(𝑟)=1𝑚=𝑘_25𝑟𝑘_𝑟+𝑘^′_25𝑟𝑘^′_𝑟−𝑘_25𝑟𝑘^′_25𝑟𝑘_𝑟𝑘^′_𝑟⁻¹=𝑤𝑘_𝑟+𝑤^′𝑘^′_𝑟−𝑤𝑤^′𝑘_𝑟𝑘^′_𝑟⁻¹. (3.4) The above equalities follow from [1] page 280 Entry 13-xii and the definition of 𝑤. Note that 𝑚 is the multiplier.

Hence for given 0<𝑤<1, we find 𝐿∈𝐑 and we get the following parametric evaluation for the Rogers Ramanujan continued fraction 𝑅𝑒^{−𝜋𝑟(𝐿)−5}−11−𝑅𝑒^{−𝜋𝑟(𝐿)5}=𝑎_𝑟=16𝑘⁶_𝑟−26𝑘⁴_𝑟−𝑤𝑘³_𝑟+10𝑘²_𝑟+𝑤𝑘_𝑟𝑘⁴_𝑟−6𝑘³_𝑟𝑤−20𝑘_𝑟𝑤³+15𝑤²𝑘²_𝑟−6𝑘_𝑟𝑤+15𝑤⁴+𝑤²𝑤𝑘_𝑟+𝑤^′𝑘^′_𝑟−𝑤𝑤^′𝑘_𝑟𝑘^′_𝑟³. (3.5) Thus for a given 𝑤 we find 𝐿 and 𝑀 from (2.4) and (2.5). Setting the values of 𝑀, 𝐿, 𝑤 in (2.3) we get the values of 𝑥 and 𝑦 (see Proposition 2.1). Hence from (3.5) if we find 𝑘⁽⁻¹⁾(𝑥)=𝑟 we know 𝑅(𝑒^{√−𝜋𝑟}). The clearer result is as follows.

Main Theorem 3. When 𝑤 is a given real number, one can find 𝑥 from (2.3). Then for the Rogers-Ramanujan continued fraction the following holds: 𝑅𝑒^{−𝜋𝑘(−1)(𝑥)−5}−11−𝑅𝑒^{−𝜋𝑘(−1)(𝑥)5}=𝑎_𝑟=16𝑥⁶−26𝑥⁴−𝑤𝑥³+10𝑥²+𝑤𝑥𝑥⁴−6𝑥³𝑤−20𝑥𝑤³+15𝑤²𝑥²−6𝑥𝑤+15𝑤⁴+𝑤²×𝑤𝑥+𝑤^′√1−𝑥²−𝑤𝑤^′𝑥√1−𝑥²³. (3.6)

Theorem 3.1. (the first derivative). One has𝑅^′𝑒^{−𝜋𝑘(−1)(𝑥)}=2^4/3𝑥^1/21−𝑥²5𝑤^1/6𝑤^′2/3𝑤𝑥+𝑤^′√1−𝑥²−𝑤𝑤^′𝑥√1−𝑥^21/2×𝑅𝑒^{−𝜋𝑘(−1)(𝑥)}𝐾²(𝑥)𝑒^{𝜋𝑘(−1)(𝑥)}𝜋². (3.7)

Proof. Combining (1.7) and (1.9) and Proposition 2.2 we get the proof.

We will see now how the function 𝑘⁽⁻¹⁾(𝑥) plays the same role in other continued fractions. Here we consider also the Ramanujan's Cubic fraction (see [5]), which is completely solvable using 𝑘_𝑟.

Define the function 𝐺(𝑥)=𝑥2√𝑥−3𝑥+2𝑥^3/2−2√𝑥1−3√𝑥+4𝑥−3𝑥^3/2+𝑥². (3.8) Set for a given 0<𝑤₃<1𝑥=𝐺𝑤₃. (3.9) Then as in Main Theorem, for the Cubic continued fraction 𝑉(𝑞), the following holds (see [5]): 𝑡=𝑉𝑒^{−𝜋𝑘(−1)(𝑥)}=1−𝑥^21/3𝑤₃^1/42^1/3𝑥^1/31−√𝑤₃. (3.10) Observe here that again we only have to know 𝑘⁽⁻¹⁾(𝑥).

If 𝑥=𝑘_𝑟, for a certain 𝑟, then 𝑘_9𝑟=𝑤₃𝑘_𝑟, (3.11) and if we set 𝑇=1−8𝑉𝑞³, (3.12) then the follwing holds: 𝑘_𝑟²=𝑥²=(1−𝑇)(3+𝑇)³(1+𝑇)(3−𝑇)³, (3.13) which is solvable always in radicals quartic equation. When we know 𝑤₃ we can find 𝑘_𝑟=𝑥 from 𝑥=𝐺(𝑤₃) and hence 𝑡.

The inverse also holds: if we know 𝑡=𝑉(𝑞) we can find 𝑇 and hence 𝑘_𝑟=𝑥. The 𝑤₃ can be found by the degree 3 modular equation which is always solvable in radicals: 𝑘_𝑟𝑘^′_𝑟+𝑘_9𝑟𝑘^′_9𝑟=1. (3.14) Let now 𝑉𝑞=𝑧⟺𝑞=𝑉⁽⁻¹⁾(𝑧), (3.15) if 𝑉_𝑖(𝑡)∶=1−√1−8𝑡³1+√1−8𝑡³3+√1−8𝑡³3−√1−8𝑡³³, (3.16) then 𝑉_𝑖𝑉𝑒^{√−𝜋𝑥}=𝑘_𝑥, (3.17) or 𝑉𝑒^{√−𝜋𝑟}=𝑉_𝑖⁽⁻¹⁾𝑘_𝑟,𝑉𝑒^{−𝜋𝑘(−1)(𝑥)}=𝑉_𝑖⁽⁻¹⁾(𝑥), (3.18) or 𝑒^{−𝜋𝑘(−1)(𝑥)}=𝑉⁽⁻¹⁾𝑉_𝑖⁽⁻¹⁾(𝑥)=𝑉_𝑖∘𝑉⁽⁻¹⁾(𝑥),𝑘⁽⁻¹⁾𝑉_𝑖𝑉𝑞=1𝜋²log𝑞²=𝑟. (3.19) Setting now values into (3.19) we get values for 𝑘⁽⁻¹⁾(⋅). The function 𝑉_𝑖(⋅) is an algebraic function.

4. Evaluations of the Rogers-Ramanujan Continued Fraction

Note that if 𝑥=𝑘_𝑟, 𝑟∈𝐐^∗₊, then we have the classical evaluations with 𝑘_𝑟 and 𝑘_25𝑟.

Evaluations
(1) We have 𝑅𝑒^−2𝜋=−12−√52+5+√52,𝑅^′𝑒^−2𝜋=8259+5√5−250+22√5𝑒^2𝜋𝜋³Γ54⁴. (4.1) (2) Assume that 𝑥=1/√2, hence 𝑘⁽⁻¹⁾(1/√2)=1. From (2.5) which for this 𝑥 can be solved in radicals, with respect to 𝑤, we find √𝑤=24√5−1−127√5−15. (4.2) Hence from 𝑤^′=𝑤1−⁴𝑥²√1−𝑥², (4.3) we get 𝑤^′=1+21−30+14√5−9−150+70√5√2^1/4. (4.4) Setting these values to (3.6) we get the value of 𝑎_𝑟 and then 𝑅(𝑞) in radicals. The result is 𝑅𝑒^−𝜋−5−11−𝑅𝑒^−𝜋51=−83+√5−−30+14√5×1−√5+−30+14√5+2^3/8−3+√5−−30+14√5×1+21−30+14√5−9−150+70√5^1/4×−1574+704√5−655−30+14√5+293−150+70√5⁻¹. (4.5) (3) Set 𝑤=1/64 and 𝑎=1359863889, 𝑏=36855, then 𝑥=9√𝑎+𝑏^5/649152⋅6^1/3√𝑎+𝑏^1/6−960√𝑎+𝑏^5/6+2⋅6^2/3√𝑎+𝑏^3/2+384⋅2^2/3⋅3^1/6√−64√𝑎+𝑏+3^1/6⋅2^2/3√453287963⋅𝑏+√𝑎^2/3+8192⋅6^1/3√𝑎+𝑏^1/3+12285⋅6^2/3√𝑎+b^2/3−2⋅6^5/6√𝑎+b^1/6√4096.2^1/3⋅3^5/6√453287963√𝑎+b^1/3+36855⋅2^2/33^1/6√453287963√𝑎+b^2/3[phone]⋅6^1/3√𝑎+𝑏^1/3453025819⋅6^2/3√𝑎+b^2/3−192453025819+12285√𝑎⁻¹. (4.6) (4) For 𝑤=277+√10813385108, (4.7) we get 𝑥=277/12+13√385/124+√7. (4.8) Hence 𝑅exp−𝜋⋅𝑘⁽⁻¹⁾277/12+13√385/124+√7^1/2=−−8071+√18107555+118185(25740148−3470530√55)^1/5. (4.9) (5) Set 𝑞=𝑒^{√−𝜋𝑟0}, then from 𝑉𝑒^{√−𝜋𝑟0}=𝑉_𝑖⁽⁻¹⁾𝑘_𝑟0=𝑉₀,𝑉𝑞^1/3=³𝑉𝑞1−𝑉𝑞+𝑉𝑞²1+2𝑉𝑞+4𝑉𝑞². (4.10) We can evaluate all 𝑉𝑞₀(𝑛)=𝑏₀(𝑛)=Algebraicfunctionof𝑟₀, (4.11) where 𝑞₀(𝑛)=𝑒^{√−𝜋𝑟0/3𝑛},𝑉_𝑖𝑉𝑞₀(𝑛)=𝑉_𝑖𝑏₀(𝑛)=𝑘_𝑟0/9^𝑛, (4.12) hence 𝑘⁽⁻¹⁾𝑉_𝑖𝑏₀(𝑛)=𝑟₀9^𝑛. (4.13) An example for 𝑟₀=2 is 𝑉𝑒^√−𝜋2=−1+32,𝑉𝑒^{√−𝜋2/3}=12^1/3−1+32^1/3,𝑉𝑒^{√−𝜋2/9}=𝜌₃^1/3, (4.14) where 𝜌₃ can be evaluated in radicals but for simplicity we give the polynomial form −1−72𝑥−6408𝑥²+50048𝑥³+51264𝑥⁴−4608𝑥⁵+512𝑥⁶=0⋯. (4.15) Then, respectively, we get the values 𝑘⁽⁻¹⁾−49+35√2+4399−70√2=29,𝑘⁽⁻¹⁾𝑉_𝑖𝜌₃^1/3=2,81⋯. (4.16) Hence 𝑘⁽⁻¹⁾𝑉_𝑖𝑏₀(𝑛)=𝑟₀9^𝑛. (4.17) Also it holds that 𝑅𝑒^{√−𝜋𝑟0/3𝑛−5}−11−𝑅𝑒^{√−𝜋𝑟0/3𝑛5}=16𝑥⁶_𝑛−26𝑥⁴_𝑛−𝑤_𝑛𝑥³_𝑛+10𝑥²_𝑛+𝑤_𝑛𝑥_𝑛𝑥⁴_𝑛−6𝑥³_𝑛𝑤_𝑛−20𝑥_𝑛𝑤³_𝑛+15𝑤²_𝑛𝑥²_𝑛−6𝑥_𝑛𝑤_𝑛+15𝑤⁴_𝑛+𝑤²_𝑛×𝑤_𝑛𝑥_𝑛+𝑤^′_𝑛1−𝑥²_𝑛−𝑤_𝑛𝑤^′_𝑛𝑥_𝑛1−𝑥²_𝑛³, (4.18) where 𝑥_𝑛=𝑉_𝑖(𝑏₀(𝑛))=known. The 𝑤_𝑛 are given from (2.2) (in this case we do not find a way to evaluate 𝑤_𝑛 in radicals).

Theorem 4.1. Set 𝑤=108+144𝑎⁴+24𝑎⁸+𝑎¹²−−11664+108+144𝑎⁴+24𝑎⁸+𝑎¹²²6√3, (4.19) then 𝑥=108+𝑎⁴12+𝑎⁴²−𝑎⁴6+𝑎⁴12+𝑎⁴²36+18𝑎⁴+𝑎⁸2√33+𝑎⁴−𝑎²√6+𝑎⁴,𝑅𝑒^{−𝜋𝑘(−1)(𝑥)−5}−11−𝑅𝑒^{−𝜋𝑘(−1)(𝑥)5}=𝐴(𝑎), (4.20) where 𝑤^′=𝑤1−⁴𝑥²√1−𝑥². (4.21) The 𝐴(𝑎) is a known algebraic function of 𝑎 and can calculated from the Main Theorem.