For an elliptic curve over Q that is defined with large coefficients, it can take mathematical software (such as Sage) a long to time calculate the analytic rank. However, it seems to quickly know if the rank is even or odd.
I would like to understand how they determine this so quickly.
This is hinted at in a PlanetMath article as the "root number" obtained from the sign of the functional equation:
$$\Lambda(E,s) = \pm \Lambda(E,2 - s)$$
Where $\Lambda$ is related to the $L$ function by: $$\Lambda(E,s) = N^{s/2} (2\pi)^{-s} \Gamma(s) L(E,s)$$ where $N$ is the conductor of $E$ over $\mathbb{Q}$.
(I'm still a little confused on the detailed definitions. Another reference, with slightly different definition relating $\Lambda$ and $L$, http://www.math.harvard.edu/~gross/preprints/ell2.pdf )
Anyway, the expansion definition of $L$ does not look like it would be valid for both sides of the functional equation, which would prevent just evaluating $\Lambda$ to check the sign. And due to the speed, I'm guessing Sage isn't evaluating $L$ at all here (or can immediately tell just by looking at a couple terms in the expansion?).
Is there some trick that allows extracting the sign without evaluating $L$?