Reached 3000 iterations

[stla]

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).

[simonp0420]

Since q = exp(iπτ) = 0.999999685... in this case and the theta functions are only defined for |q| < 1, it isn't surprising that there will be some convergence problems if one approaches this boundary too closely. Of course, Wolfram can do it, but maybe they need a lot of iterations there too.

[stla]

@simonp0420 I've ported the new Fortran implementation by @fremling to Haskell and with this implementation the result is immediately obtained! That's impressive. I think I will port it to R/C++ now, and if all unit tests succeed, maybe in Julia later.

[simonp0420]

Could you provide a link to the Fortran implementation so I could take a look?

[stla]

Here. He uses the modular transformations to bring $\tau$ to a certain region where the evaluation works.

[stla]

@simonp0420 I ported the code to R. It works, and better than the previous one!

[stla]

I ported the code to Julia now: gist. It works!

[simonp0420]

I added some comments to the gist.