EllipticFunctions.jl

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I found a value for which the 3000 iterations are attained when calculating jtheta2: $$z=0 \quad \text{and} \quad \tau = 0.7792256 +10^{-7}i.$$

This is really a edge case: with tau = 0.78 + 1e-7, this works.

There's a formula linking theta(z, tau) and theta(z/tau, -1/tau), and one can get the result with this formula. One can see the presence of the edge case once again, because the imaginary part of -1/tau is 1.646924e-07.

@simonp0420 do you have any comment ?

See this issue for a discussion about this problem, using a different implementation (the one I used before @simonp0420 proposed the new implementation).