Does the number of roots of the modular form associated to an elliptic curve, on the positive imaginary axis, equal the analytic rank?

Question

Recently I've been playing around with elliptic curves and have seemingly come up with a conjecture that I could not find elsewhere:

Let $E$ be an elliptic curve, and $f(q)$ its associated modular form. Is it known whether along the positive imaginary axis, the number of roots of $f(q)$ is equal to the order of $L(E,1)$, i.e. the analytic rank?

I've checked a few cases on LMFDB, and the plots of $f\left(e^{-2\pi x}\right)$ along $x>0$ seem to show the same number of roots as the analytic rank of the L-function (or equivalently the algebraic rank of the elliptic curve assuming BSD). One requires several thousand terms of the q-expansion when the conductor is large to have sufficient convergence for all roots to appear. I am curious to know if this has been conjectured before (say as part of the Langlands program) or if it is a known result, or perhaps is actually false?

Answer 1

The elliptic curve $E = 990.e2$ (Cremona label $990b1$) has rank $0$, and the number of zeros of the associated newform $f$ on the imaginary axis seems to be 2 (numerically). Here is some PARI/GP code to plot $y \mapsto f(iy)$ on an interval $[y_1,y_2]$ for a given elliptic curve:

{
graph(E, y1, y2) = 
my(e = ellinit(E), [mf, f, v] = mffromell(e));
print("Analytic rank: ", ellanalyticrank(e)[1]);
ploth(y = y1, y2, real(mfeval(mf,f,I*y)));
}

graph("990b1", .01, 2)
graph("990b1", .0001, .01)

I found this curve randomly, so there may be examples with smaller conductor.

Answer 2

Let $Z$ be the number of zeroes of $y\mapsto f(iy)$ for $0<y<\infty$ and let $r$ be the analytic rank of $E$. Then $Z\geqslant r$ and $Z\equiv r \pmod{2}$ as I will explain below. I don't know of an example when $Z\neq r$ but also no reason to believe they do not exist.

Let $g(y) = 2\pi \int_{\infty}^y f(i\,t)dt$. Then $g(0) = L(f,1)$ and your zeroes are the stationary points $g'(y)=0$. Mazur and Swinnerton-Dyer call your zeros the fundamental critical point in section 2.4 in "The arithmetic of Weil curves". This question is about them.

If $L(f,1)=0$, i.e., the analytic rank is positive, then $g$ has to have at least one maximum. Instead if $L(f,1)\neq 0$, then it may well be that $g$ has no stationary point and $g$ is a decreasing function. Your function also satisfies $f(i\,t) = \varepsilon\cdot f(i\, t/N)$ where $N$ is the conductor and the root number $\varepsilon$ is $1$ if the analytic rank is even and $-1$ if it is odd. For odd analytic rank we must have a zero at $t=1/\sqrt{N}$. In general this tells us that $Z$ has the same parity as $r$. Further Mazur and Swinnerton-Dyer show that the number of zeroes of odd order is at least the analytic rank.