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Let $$L(C,s)=\sum_{n=1}^\infty \frac{a_n}{n^s}$$ be the Dirichlet series of the Hasse--Weil L-function of an elliptic curve $C$ over $ℚ$. The modularity theorem implies that $L(C,s)$ is the $L$-function of a holomorphic cusp form for a congruence subgroup and it is entire function and have a holomorphic continuation. Also there is a rapidly-converging series $f(s)$ expression $L(C,s)$ for any complex number $s$ given in http://modular.math.washington.edu/books/bsd/ on page 9.

My question is: I am not understood the word half (maybe mean almost the cases are zero not the exact number of those cases) in Section 1.4.1. Approximating the Rank (...Note that half of the $L(k)(E, 1)$ are automatically $0$ because of equation (1.3.3)).

DER
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    Presumably (1.3.3) is a functional equation for $L(C,s)$, where there is a sign $(-1)^{something}$, where the something depends on the elliptic curve $C$ and is even and odd half the time each. When it's odd, plugging in $s=1$ to both sides of the functional equation yields $L(C,1)=-L(C,1)$, forcing $L(C,1)=0$. – Greg Martin Oct 10 '14 at 16:49
  • @GregMartin: I am asking about the term half. – DER Oct 10 '14 at 16:50

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Aha, this isn't what I thought at first: I thought they'd be talking about half the elliptic curves $E$, but they're really talking about half the derivatives $L^{(k)}(E,1)$ as $k$ varies, for a fixed $E$.

(1.3.3) reads $$ (-1)^k \Lambda^{(k)}(E,2-s) = \epsilon\Lambda^{(k)}(E,s). $$ (I've corrected a typo on the right-hand side). Here $\epsilon\in\{1,-1\}$ is some constant. In particular, setting $s=1$ gives $$ (-1)^k \Lambda^{(k)}(E,1) = \epsilon\Lambda^{(k)}(E,1). $$ If $\epsilon=1$, then this equation forces $\Lambda^{(k)}(E,1)=0$ for all odd $k$. If $\epsilon=-1$, then it forces $\Lambda^{(k)}(E,1)=0$ for all even $k$.

Greg Martin
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