Hi Bill, While I think it is a good idea to extend double or half angle identities to the theta function, my interest is not to further entrench the Jacobi formalism, nor to endlessly list identities without saying why they would matter, or what they could be used for. In my opinion, elliptic function angle doubling formulae are not as exciting considered in and of themselves, but more exciting when paired with known function values on a small disk around the origin and a known period ratio. What is the shortest proof that Harold Edwards elliptic function restricted to a purely imaginary period ratio is essentially the same as Jacobi's sn? One common tactic is to compare poles and zeros, which can easily be done after Part III of "A normal form for elliptic curves". The other option, after Part II, is just to observe that the doubling formulae are the same, and that up to scale, the two functions agree on a small disk around the origin. Once it is well understood how the angle doubling formulae define the generic elliptic function psi, then Part III S.22 of Edward's article finally begins to make constructive sense. However, part of the algorithm is still confusing because multi-valuedness enters through use of double-angle formula. A half-angle, were it to exist, would be better. So here is a practical question for you Bill: Your half angle formulae, can they be used to simplify the algorithm for computing period ratio tau from known initial data? Cheers --Brad On Tue, Nov 19, 2019 at 9:34 PM Bill Gosper <billgosper@gmail.com> wrote:
prompts me to t(rot )out these four old 𝜗 half-angle formulas: Out[419]= {EllipticTheta[1, x/2, q] == (1/(2^( 1/4)))((-EllipticTheta[2, 0, q]^3 EllipticTheta[2, x, q] + EllipticTheta[3, 0, q]^3 EllipticTheta[3, x, q] - EllipticTheta[4, 0, q]^3 EllipticTheta[4, x, q])^(1/4)), EllipticTheta[2, x/2, q] == (1/(2^( 1/4)))((EllipticTheta[2, 0, q]^3 EllipticTheta[2, x, q] + EllipticTheta[3, 0, q]^3 EllipticTheta[3, x, q] - EllipticTheta[4, 0, q]^3 EllipticTheta[4, x, q])^(1/4)), EllipticTheta[3, x/2, q] == (1/(2^( 1/4)))((EllipticTheta[2, 0, q]^3 EllipticTheta[2, x, q] + EllipticTheta[3, 0, q]^3 EllipticTheta[3, x, q] + EllipticTheta[4, 0, q]^3 EllipticTheta[4, x, q])^(1/4)), EllipticTheta[4, x/2, q] == (1/(2^( 1/4)))((-EllipticTheta[2, 0, q]^3 EllipticTheta[2, x, q] + EllipticTheta[3, 0, q]^3 EllipticTheta[3, x, q] + EllipticTheta[4, 0, q]^3 EllipticTheta[4, x, q])^(1/4))} Check: In[432]:= Assuming[0 < x < \[Pi]/2 && 0 < q < 1, FullSimplify@Series[%419, {q, 0, 22}, {x, 0, 22}]] // tim
During evaluation of In[432]:= 419.850403,4 (*Seven minutes*).
Out[432]= {True, True, True, True} —rwg _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun