Vanishing of L-function of elliptic curve over $\mathbb{Q}$ For an elliptic curve $E$ over $\mathbb{Q}$, it is not very difficult to show $L(E,1)\not=0$ (when the analytic rank$=0$) by computing the several Fourier coefficients but seem to be difficult to determine whether one has $L(E,1)=0$ (when the analytic rank$\not=0$). Is there a small constant $c$ such that, if we have $\mid L(E,1)\mid <c$, one can obtain $L(E,1)=0$ ?
 A: There is not a "constant" $c$ per se, but rather an expression in terms of the real period. This follows for instance from Manin's modular symbols theory, that the central $L$-value is a rational multiple (of bounded height) of the real period. More explicitly, you need to include the torsion and Tamagawa number contributions.
Using Birch and Swinnerton-Dyer as a guide, one has
$${L(E,1)\over\Omega}{|T|^2\over \prod c_p}$$
is an integral square (possibly zero), and if it is less than 1 it indeed must vanish (one might need extra factors from Manin constants, I'm not sure).
So there is an implication:
$$L(E,1)<{\prod_p c_p\over |T|^2}\Omega_{\rm re}\implies L(E,1)=0.$$
A: Just to pick up on something mentioned in @MyNinthAccount's answer: 
If you just want to determine whether or not $L(E, 1) = 0$, then there is another approach, which doesn't involve computing $L(E, 1)$ numerically to high precision and then arguing with inequalities. The point is that the ratio 
$$\frac{L(E, 1)}{\Omega^+_E}$$
is not only rational, but is exactly computable as a rational number (using modular symbols). So there is an algorithm which will determine, rigorously and in finitely many steps, whether the analytic rank is 0 or not. The standard number theory software packages (Sage, Magma, maybe also Pari) have built-in functionality to do this.
If you find by this method that $L(E, 1) = 0$, then determining what the analytic rank actually is is a much harder problem.
