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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.

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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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    $\begingroup$ That's true, though the fact that a finite computation proves exact vanishing is not obvious (it comes down to identifying $L(E,1)$ with a rational multiple of the real period whose denominator is bounded by the exponent of the torsion group). $\endgroup$
    – Noam D. Elkies
    Commented Mar 21, 2012 at 21:17
  • $\begingroup$ @Noam: Thanks for your comment. I was thinking of switching to modular forms and working with them directly. Is obtaining the Taylor expansion of a modular $L$-function about its center computationally difficult? $\endgroup$
    – GH from MO
    Commented Mar 21, 2012 at 22:27
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    $\begingroup$ The problem is not computing $L$ and its derivatives, but recognizing zero. $L(E,1)$ can be computed to within $10^{-n}$ in time polynomial in $n$, but if you know $L(E,1) = .000000003$ to within $10^{-7}$, you still don't know that $E$ has positive arithmetic rank without some further argument. Here this argument is available, as it is for $L'(E,1)$ thanks to Gross-Zagier (which is much harder), but so far not beyond that. So if the analytic rank is at least $4$ we can guess but not prove it. Fortunately for your present purpose ranks $0$ and $2$ suffice. $\endgroup$
    – Noam D. Elkies
    Commented Mar 22, 2012 at 2:30
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    $\begingroup$ @GH: if $L(E,s)$ looks locally like $.00000003 + a_2 (s-1)^2 + \cdots$ then it has two zeros near $s=1$ but without more information you can't tell if they're at $s=1$ or just closer than the radius of the smallest contour you've tried. $\endgroup$
    – Noam D. Elkies
    Commented Mar 22, 2012 at 14:40
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    $\begingroup$ @GH: Indeed we have no proof that any eigenform $g$ yields an $L$-function with a central zero of order $4$ or more. [This does not depend on having some $f$ in the same space whose $L$-function has no central zero.] Averaging can yield upper bounds on the mean analytic rank, but (as far as we know) cannot prove higher-order multiplicities than we can find for individual $g$. (And whatever wisdom I've exhibited in this thread is my teachers', not my own.) $\endgroup$
    – Noam D. Elkies
    Commented Mar 24, 2012 at 0:25

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