Calculates e1,e2,e3 from the invariants using
polyroot or Cardano's method.
Two-element vector with
Boolean, with default
Boolean, with default
Also note that setting
A pair of primitive half periods, if known. If supplied, the
function uses them to calculate approximate values for the three
es (but supplies values calculated by
Real, relative tolerance criterion for terminating Laurent summation
Returns a three-element vector.
e1e2e3(), so do not
parameters() to determine argument
doing so will result in a recursive loop.
Just to be specific:
e1e2e3(g=parameters(...)) will fail. It
would be pointless anyway, because
(inter alia) e1, e2, e3.
There is considerable confusion about the order of e1, e2 and e3, essentially due to Abramowitz and Stegun's definition of the half periods being inconsistent with that of Chandrasekharan's, and Mathematica's. It is not possible to reconcile A and S's notation for theta functions with Chandrasekharan's definition of a primitive pair. Thus, the convention adopted here is the rather strange-seeming choice of e1=P(omega_1/2), e2=P(omega_3/2), e3=P(omega_2/2). This has the advantage of making equation 18.10.5 (p650, ams55), and equation 09.13.27.0011.01, return three identical values.
The other scheme to rescue 18.10.5 would be to define
(omega1,omega3) as a primitive pair, and
omega2=-omega1-omega3. This is
the method adopted by Mathematica; it is no more inconsistent with
ams55 than the solution used in package elliptic. However,
this scheme suffers from the
disadvantage that the independent elements of
have to be supplied as
c(omega1,NA,omega3), and this is
inimical to the precepts of R.
One can realize the above in practice by
considering what this package calls
“omega2” to be really
omega3, and what this package calls
“omega1+omega2” to be
really omega2. Making function
half.periods() return a three element vector with names
omega2 might work on some
levels, and indeed might be the correct solution for a user
somewhere; but it would be confusing. This confusion would
dog my weary steps for ever more.
Robin K. S. Hankin
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