0027c: Part 6c, Complex series for K(k)


Part 6c. Complex Series for the Complete Elliptic Integral K(kd)


By Tito Piezas III



Abstract:  Using known complex series for 1/pi, we give complex series for the complete elliptic integrals of the first kind K(k7), K(k2), K(k15).


I. Introduction

II. Dedekind eta quotients

III. The complete elliptic integral K(k7)

IV. The complete elliptic integral K(k2)

V. The complete elliptic integral K(k15)



I. Introduction


In “Divergent” Ramanujan-Type Supercongruences, J. Guillera and W. Zudilin discovered the first series for 1/π with complex coefficients,



However, it can also be expressed in a nice equivalent form. Let, 


a complex number with absolute value equal to 1.  (Note that the absolute value of 47+45√-7 is 128.)  Then, 

So it turns out the rightmost factor (negated) is a perfect twelfth power in α.  However, the fact that it is a 12th power is not an isolated result, but occurs when d = 8n+7, and d ≠ 3m. 



II. Dedekind Eta quotients


A formula for α12(τ) can be given as,



If we define the argument τ as,



then the example above used d = 7 to get,



But it is better to use two similar functions that are the 12th root of the eta quotient. Given the 6th and 12th root of unity,






Let d ≠ 3m, if d = 16n+7 for the first function, and d = 16n+15 for the second, then βn(τ) is an algebraic number of degree twice the class number h(-d).  For example, for d = {7, 23, 31} with h(-d) = 1, 3, 3, respectively, then we have the complex numbers,



where the root object follows the numbering convention of Mathematica.  Affixing 1/√2 and raising the expression to the 12th power, one easily gets α12(τ).    



III. The complete elliptic integral K(k7)

For the elliptic modulus k7, given in Mathematica as k7 = ModularLambda[√-7], then the complete elliptic integral of the first kind K(kd) is,



However, this real number can also be expressed as a complex series using α(τ). Still let,



then we have,



Compare to the two series in the reals,



where P = (x3-x-1)1 is the plastic constant.  Of course, it is possible to find complex series for K(kd) for other d, but the expressions on the LHS of the sum are slightly more complicated.  For the next sections, the formulas no longer use α(τ).



IV. The complete elliptic integral K(k2)


Similarly, given the series found by this author,



(where i is the imaginary unit), which is the first one found with only rational complex numbers.  The complex denominator on the RHS turns out to be also a cube,



and can be used to express K(k2) as,




V. The complete elliptic integral K(k15)


In “Complex Series For 1/π”, H.H. Chan, J. Wan, and W. Zudilin mentioned the formula,






(with two typos corrected by this author.)  We can also use this to find a nice complex series for K(k15).  Given the golden ratio, φ = (1+√5)/2, then,




-- End -- 


© Tito Piezas III, Jan 2012

You can email author at tpiezas@gmail.com.



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