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 $L(E,1)=0$ ?



Vanishing of L-function of elliptic curve over $\mathbb{Q}$
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.


