\documentclass[12pt]{amsart}
\title[Elliptic Curves]{Computing the Lowest Zero of L(E,s)}
\author{Michael Comerford}
\begin{document}
\begin{abstract}
This paper deals with a method of finding zeros of the L-function of an
elliptic curve on the critical line $s=\frac{1}{2}+it$.  It is
conjectured, due to the Reimann Hypothesis, that all zeros have real part
$\frac{1}{2}$ after normalization.  We find zeros by looking for sign
changes of $L(E,s)$.  However, we will use a different function, derived
to have the same zeros as $L(E,s)$, which is easier to calculate. This
paper will also discuss briefly the computer program I wrote to find sign
changes of this function.
\end{abstract}
\maketitle

The definition for the L-function of an elliptic curve is an Euler
product, as follows.

\begin{equation*}
L(E,s) := \prod_{p|\Delta} \left(1-a_p p^{-\frac{1}{2}-s}\right)^{-1}
\prod_{p ~not \vert \Delta} \left(1-a_p p^{-\frac{1}{2} -s} +
p^{-2s}\right)^{-1}
\end{equation*}

where $a_p = p+1-\#E_p({\bf Z}_p)$ with $p$ prime and where $\#E_p({\bf
Z}_p)$ is the number of solutions of the curve modulo $p$.

The functional equation, as asserted by the Hasse-Weil conjecture, is
defined as follows.

\begin{equation*}
\Lambda(E,s) := \left(\frac{N^{\frac{1}{2}}}{2\pi}\right)^s
\Gamma\left(s+\frac{1}{2}\right) L(E,s) = -\epsilon \Lambda(E,1-s)
\end{equation*}

where $N$ is the conductor of the elliptic curve and $\epsilon$, which is
the sign of the functional equation, is $\pm 1$.  $\Gamma(z)$ is a well
known function and is defined as follows.

\begin{equation*}
\Gamma(z) = \int_0^{\infty}e^{-x}x^{z-1}dx
\end{equation*}

Finding zeros of $\Lambda(E,s)$ is equivalent to finding zeros of
$L(E,s)$, as we will see, since
$\left(\frac{N^{\frac{1}{2}}}{2\pi}\right)^s$ and $\Gamma(s+\frac{1}{2})$
are never zero.  Therefore, zeros of $\Lambda(E,s)$ are exactly the zeros
of $L(E,s)$.

It is clear that $(\frac{N^{\frac{1}{2}}}{2\pi})^s$ is never zero since
$\frac{N^{\frac{1}{2}}}{2\pi}$ is nonzero.  It is well known that
$\Gamma(z)$ is nonzero.  This follows from the fact that
$\Gamma(z)\Gamma(1-z) = \frac{\pi}{\sin (\pi z)}$ since $\frac{\pi}{\sin
(\pi z)}$ is nonzero.

We will now show that $\Lambda(E,s)$ (with $s=\frac{1}{2} + it$) is real
or purely imaginary.  To do this, we will show that
$\overline{\Lambda(E,s)} = \Lambda(E,1-s)$.  We know that $\Lambda(E,s) =
\pm \Lambda(E,1-s)$, so if $\Lambda(E,s) = \pm \overline{\Lambda(E,s)}$,
it is clear that $\Lambda(E,s)$ is real or purely imaginary.

First of all, it is true that the complex conjugate of a product is the
product of the complex conjegates.  Therefore,

\begin{equation*}
\overline{\Lambda(E,s)} =
\overline{\left(\frac{N^{\frac{1}{2}}}{2\pi}\right)^s}
\overline{\Gamma\left(s+\frac{1}{2}\right)} \overline{L(E,s)}
\end{equation*}

\begin{equation*}
\overline{\left(\frac{N^[\frac{1}{2}]}{2\pi}\right)^s} =
\left(\frac{N^{\frac{1}{2}}}{2\pi}\right)^{\overline{s}} =
\left(\frac{N^{\frac{1}{2}}}{2\pi}\right)^{1-s}
\end{equation*}

\begin{equation*}
\overline{\Gamma\left(s+\frac{1}{2}\right)} =
\int_0^{\infty}e^{-x}x^{\overline{s-\frac{1}{2}}}dx =
\int_0^{\infty}e^{-x}x^{1-s+\frac{1}{2}}dx =
\Gamma\left(1-s+\frac{1}{2}\right)
\end{equation*}

\begin{equation*}
\overline{L(E,s)} = \prod_{p|\Delta} \left(1-a_p
p^{\overline{-\frac{1}{2}-s}}\right)^{-1} \prod_{p ~ not | \Delta}
\left(1-a_p p^{\overline{-\frac{1}{2} -s}} +
p^{\overline{-2s}}\right)^{-1} =
\end{equation*}
\begin{equation*}
\prod_{p|\Delta} \left(1-a_p p^{-\frac{3}{2} + s}\right)^{-1} \prod_{p ~
not |\Delta} \left(1-a_p p^{-\frac{3}{2}+s} + p^{-2+2s}\right)^{-1} =
L(E,1-s)
\end{equation*}

Therefore, we have shown that since the complex conjugate of each term of
$\Lambda(E,s)$ simply replaces $s$ by $1-s$, $\overline\Lambda(E,s) =
\Lambda(E,1-s) = \pm\Lambda(E,s)$.  Assuming the Shimura-Taniyama-Weil
conjecture holds for $L(E,s)$ (it has been proven for curves whose c, $N$,
is not divisible by 27) we have the following.

\begin{equation*}
\left(\frac{N^{\frac{1}{2}}}{2\pi}\right)^s
\Gamma\left(s+\frac{1}{2}\right) L(E,s) =
\left(\frac{2\pi}{N^{\frac{1}{2}}}\right)^{\frac{1}{2}}\sum_{n=1}^{\infty}a_nG\left(s+\frac{1}{2},\frac{2\pi n}{N^{\frac{1}{2}}}\right)
\end{equation*}
\begin{equation*}
-\epsilon\left(\frac{2\pi}{N^{\frac{1}{2}}}\right)^{\frac{1}{2}}\sum_{n=1}^{\infty}a_nG\left(1-s+\frac{1}{2},\frac{2\pi n}{ N^{\frac{1}{2}}}\right)
\end{equation*}

Where G(z,w) is defined as follows.

\begin{equation*}
G(z,w) = \int_1^{\infty}e^{-wx}x^{z-1}dx
\end{equation*}

We see that the left hand side is $\Lambda(E,s)$ and a formula on the
right hand side which is easier to calculate.  Since we are looking for
zeros, we can eliminate the constant terms in front of each sum, so that
we arrive at the following formula, which I use to compute zeros of
$L(E,s)$.

\begin{equation*}
\Lambda(E,s) = \sum_{n=1}^{\infty}a_n\left(G\left(s+\frac{1}{2},\frac{2\pi
n}{N^{\frac{1}{2}}}\right)-\epsilon~ G\left(1-s+\frac{1}{2},\frac{2\pi
n}{N^{\frac{1}{2}}}\right)\right)
\end{equation*}

Setting $w=\frac{2\pi n}{N^{\frac{1}{2}}}$, and $s=\frac{1}{2}+it$, we see
that we need to compute $G(1+it,w)$ and $G(1-it,w)$ as part of the above
equation.

\begin{equation*}
G(1+it,w) = \int_1^{\infty}e^{-wx}x^{it}dx =
\int_1^{\infty}e^{-wx}(\cos(t\ln x) + i\sin(t\ln x)) dx
\end{equation*}

\begin{equation*}
G(1-it,w) = \int_1^{\infty}e^{-wx}x^{-it}dx =
\int_1^{\infty}e^{-wx}(\cos(t\ln x) - i\sin(t\ln x)) dx
\end{equation*}

Therefore, combining these values of $G(z,w)$ in the sum, depending on the
value of $\epsilon$, we get the final two formulas which we use to compute
the zeros of $L(E,s)$ depending on the value of the functional equation
($\epsilon$).

\begin{equation*}
\epsilon = 1: \sum_1^{\infty}a_n\int_1^{\infty}e^{-wx}2i\sin(t\ln x)dx
\end{equation*}

\begin{equation*}
\epsilon = -1: \sum_1^{\infty}a_n\int_1^{\infty}e^{-wx}2\cos(t\ln x)dx
\end{equation*}

We can see the result is either real or purely imaginary, as suspected.
In my computer program, I calculate the integral using simpson's rule with
small step sizes up to a certain bound.  Simpson's rule is used as a
better approximation, with a comparable amount of computation, as compared
to the usual rectangle technique.  Problems arose when I originally
calculated the integral using the rectangle method since values like $x=1$
blow up.   Also, I have a calculated bound for the highest value of $n$ I
sum up.  In order to pick an appriate bound for $x$ in the integral, we
look for when $x$ will make the integral smaller than some $\epsilon>0$.
Since both $\sin(x)$ and $\cos(x)$ are less than or equal to $1$, we only
need to make $e^{-wx} < \epsilon$.  So we pick an $x$-bound $> -\frac{\ln
\epsilon}{w}$.  For a bound on $n$, it seems that a value on the order of
$N^{\frac{1}{2}}\ln(N)$ is sufficient.  It took a lot of time trying
bounds and step sizes, checking results with already published results for
zeros of certain curves, to see what works well in a reasonable amount of
time.

In order to find zeros, the program scans values of $t$ looking for sign
changes.  It starts with a large step size for $t$, usually $0.1$, and
when it finds a sign change, reduces the step size by a factor of $10$ and
scans the region where it was found.  Doing this, I have been able to
acheive $2$ or $3$ decimal point accuracy for most small zeros, especially
if the conductor is reasonably small. The problem with using this method
of finding sign changes of $L(E,s)$ is the possibility of missing zeros.
For example, if a root has even degree, the curve would have a zero but
never change signs.

In general, however, this method seems to work quite well, missing very
few zeros.  I am studying the distribution of the first zero above the
origin on the critical line $s=\frac{1}{2}$ and how it might be affected
to rank and within families.  In a particular family, I will look at
curves with forced rank and see how the distibution of the first zero
relates to other curves in the family.  I expect these zeros to be higher
as the rank grows.  Hopefully, I will find some results that clearly
support or contradict certain conjectures about such problems.
\end{document}





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