TITLE: Same functional equation implies same zero order at s=1/2?
QUESTION [1 upvotes]: Hello,
I would like to know whether, given two functions $F$ and $F'$ of the Selberg class sharing exactly the same functional equation (i.e. sharing the same root number), the order of $1/2$ as a zero of $F$ equals the order of $1/2$ as a zero of $F'$.
Thank you in advance.

REPLY [8 votes]: If $E$ and $E'$ are two elliptic curves over $\mathbb{Q}$ with the same conductor so that $E$ has rank $0$ but $E'$ has rank $2$, then assuming the BSD conjecture the $L$-functions $L(E,s)$ and $L(E',s)$ belong to the Selberg class, they have the same functional equation, yet a different order of vanishing at $1/2$ (the orders are $0$ and $2$). In practice, given $E$ and $E'$, one can prove this unconditionally, because one can find the modular forms producing the same $L$-functions, and then a calculation (Taylor expansion around $s=1/2$) should yield the order of vanishing predicted by BSD. I am sure plenty of such pairs can be found in published tables of elliptic curves and modular forms (I am not familiar with such tables or doing explicit calculations with software so I let others to come up with examples).
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