Approximating the Rank

Fix an elliptic curve $E$ over $\mathbb{Q}$. The usual method to approximate the rank is to find a series that rapidly converges to $L^{(r)}(E,1)$ for $r=0,1,2,3,\ldots$, then compute $L(E,1)$, $L'(E,1)$, $L^{(2)}(E,1)$, etc., until one appears to be nonzero. Note that half of the $L^{(k)}(E,1)$ are automatically $0$ because of equation (1.3.3). For more details, see [Cre97, §2.13] and [Dok04].

In this section, we describe a slightly different method, which only uses Theorem 1.10 and the definition of the derivative.

Proposition 1.11   Write

\begin{displaymath}
L(E,s) = c_r(s-1)^r + c_{r+1}(s-1)^{r+1} + \cdots.
\end{displaymath}

with $c_r\neq 0$. Then

\begin{displaymath}
\lim_{s\rightarrow 1} 
(s-1)\cdot \frac{L'(E,s)}{L(E,s)} = r.
\end{displaymath}


\begin{proof}
Setting $L(s) = L(E,s)$, we have
\begin{align*}
\lim_{s\rightarro...
...s}
{c_r(s-1)^r + c_{r+1}(s-1)^{r+1} + \cdots}\\
&= r.
\end{align*}\end{proof}

Thus the rank $r$ is the limit as $s\rightarrow 1$ of a certain (smooth) function. We know this limit is an integer. But, for example, for the rank $4$ curve

\begin{displaymath}
y^2 +xy = x^3 - x^2 - 79x + 289
\end{displaymath} (1.4.1)

of conductor 234446 nobody has succeeded in proving that this integer limit is $4$. (We can prove that the limit is either $2$ or $4$ by using the functionality equation (1.3.2) to show that the order of vanishing is even, then verifying by computation that $L^{(4)}(E,1)= 214.65233\ldots \neq 0$.)

Using the definition of derivative, we approximate $(s-1)\frac{L'(s)}{L(s)}$ as follows. For $\vert s-1\vert$ small, we have
\begin{align*}
(s-1)\frac{L'(s)}{L(s)} &=
\frac{s-1}{L(s)}\cdot \lim_{h\righta...
...1)^2) - L(s)}{(s-1)^2}\\
&= \frac{L(s^2 - s+1) - L(s)}{(s-1)L(s)}
\end{align*}

In fact, we have

\begin{displaymath}
\lim_{s\rightarrow 1}  (s-1)\cdot \frac{L'(s)}{L(s)}
= \lim_{s\rightarrow 1} \frac{L(s^2 - s+1) - L(s)}{(s-1)L(s)}.
\end{displaymath}

We can use this formula in SAGE to ``approximate'' $r$. First we start with a curve of rank $2$.

sage: e = EllipticCurve('389a'); e.rank()
2
sage: L = e.Lseries_dokchitser()
sage: def r(e,s): L1=L(s); L2=L(s^2-s+1); return (L2-L1)/((s-1)*L1)
sage: r(e,1.01)
2.00413534247395
sage: r(e,1.001)
2.00043133754756
sage: r(e,1.00001)
2.00000433133371

Next consider the curve $y^2 +xy = x^3 - x^2 - 79x + 289$ of rank $4$:

sage: e =  EllipticCurve([1, -1, 0, -79, 289])
sage: e.rank()
4
sage: L = e.Lseries_dokchitser(100)
sage: def r(e,s): L1=L(s); L2=L(s^2-s+1); return (L2-L1)/((s-1)*L1)
sage: R = RealField(100)
sage: r(e,R('1.01'))
4.0212949184444018810727106489
sage: r(e,R('1.001'))
4.0022223745190806421850637523
sage: r(e,R('1.00001'))
4.0000223250026401574120263050
sage: r(e,R('1.000001'))
4.0000022325922257758141597819

It certainly looks like $\lim_{s\rightarrow 1} r(s) = 4$. We know that $\lim_{s\rightarrow 1} r(s)\in\mathbb{Z}$, and if only there were a good way to bound the error we could conclude that the limit is $4$. But this has stumped people for years, and probably it is nearly impossible without a deep result that somehow interprets $L''(E,1)$ in a completely different way.

William 2007-05-25