THE ELLIPTIC LOGARITHM FUNCTION

    In an earlier post I have discussed the elliptic exponential function

                                  E(iu, k) = cn(u, k) +i sn(u, k)

     By replacing u by -iu in this function we obtain the function E(u, k). When q approaches 0, 
E(iu, k)  reduces to the exponential function e^iu = cos u + i sin u and E(u, k) reduces to e^u. 
As is well-known, the inverse of the exponential function v = e^u is the logarithm function ln v =u, such that

ln(1+v) = Σ (-1)^{n-1}. v^n /n                                                                                                      (1)         

where the series runs from n=1 to infinity.

   I have found that the inverse of the elliptic exponential function v = E(u, k) is the elliptic logarithm function  LN(v, k) = u  such that

LN(1+v, k) = 

Σ [ ( β(1) δ/δβ(0) + β(2) δ/δβ(1)  + β(3) δ/δβ(2) +.... ..)^{n-1} (β(0)^n)]. v^n /n!,                      (2)

where δ/δβ(j) denotes the partial derivative with respect to the j-th elliptic Bernoulli number β(j) 
defined in an earlier post; and on the left hand side, the partial differential operator
   
 β(1) δ/δβ(0) + β(2) δ/δβ(1)  + β(3) δ/δβ(2) + .... .... 

acts n-1 times successively on β(0)^n, while the series runs from n = 1 to infinity.

  As q approaches 0, the expansion (2) of LN(1+v, k) reduces in the limit to the expansion (1) of ln(1+v). 

    Hence, LN(v, k) is the elliptic logarithm function. 

    The procedure yields the following result involving the ordinary Bernoulli numbers B(n).

( B(1) δ/δB(0) + B(2) δ/δB(1)  + B(3) δ/δB(2) + .... ..)^{n-1} (B(0)^n) = (-1)^{n-1} (n-1)!.

where δ/δB(j) denotes the partial derivative with respect to the j-th Bernoulli number B(j); and on the left hand side, the partial differential operator
   
 B(1) δ/δB(0) + B(2) δ/δB(1)  + B(3) δ/δB(2) +.... .... 

acts n-1 times successively on B(0)^n.


Somjit Datta, Ph.D
Calcutta, India
October 27, 2018