1
$\begingroup$

Let $$L(C,s)=\sum_{n=1}^\infty \frac{a_n}{n^s}$$ be the Dirichlet series of the Hasse--Weil L-function of an elliptic curve $C$ over $ℚ$. The modularity theorem implies that $L(C,s)$ is the $L$-function of a holomorphic cusp form for a congruence subgroup and it is entire function and have a holomorphic continuation. Also there is a rapidly-converging series $f(s)$ expression $L(C,s)$ for any complex number $s$ given in http://modular.math.washington.edu/books/bsd/ on page 9.

I am interested on the real points $a∈ℝ$ such that the equation $$f^{(k)}(s)=a$$ have a finite number of real solutions $s$ for some $k$. Unfortunately, I have no idea to start.

$\endgroup$
2
$\begingroup$

I think that for any real number $a$ and any $k \geq 0$ there should be infinitely many solutions to $f^{(k)}(s) = a$ (where $f(s) = L(C,s)$). The idea is that (as you observed in your previous question), we have $f(m) = 0$ for any integer $m \leq 0$ and that $f(m+1/2)$ tends to infinity (in absolute value) and alternates sign. (This should be easy to justify using that $f(s) \to 1$ as $s \to \infty$ and the functional equation.)

Now, it suffices to prove the following lemma:

Suppose that $g(x)$ is an infinitely differentiable function and $a_{1}, a_{2}, \ldots$ is a sequence of real numbers with $\cdots < a_{k} < a_{k-1} < \cdots < a_{1}$ so that

$\bullet$ $g(a_{i}) = 0$ for all $i$,

$\bullet$ There is a constant $C$ so that $a_{i} - a_{i+1} \leq C$ for all $i$, and

$\bullet$ $g(x)$ is neither bounded from above nor bounded from below on $(-\infty,a_{1})$.

Then the same is true of $g'(x)$. That is there is a sequence $\cdots < b_{k} < b_{k-1} < \cdots < b_{1}$ so that $g'(b_{i}) = 0$, there is a constant $D$ so that $b_{i} - b_{i+1} \leq D$ for all $i$, and $g'(x)$ is neither bounded from above nor bounded from below on $(-\infty,b_{1})$.

This lemma should be easy to prove using the mean value theorem.

$\endgroup$
  • $\begingroup$ @ Jeremy Rouse: It is not clear how to prove the second and the third items for $g'$. $\endgroup$ – China-Hong Kong May 9 '14 at 10:27
  • 1
    $\begingroup$ For the second item, note that there must be a zero of $g'$ between $a_{i}$ and $a_{i+1}$ for all $i$. For the third, note that if $g(x) > A$ for some $x < a_{1}$, there must be an $a_{i} < x$ with $x - a_{i} < C$. Now apply the mean value theorem on the interval $[a_{i},x]$. (The same argument works to construct negative values of $g'(x)$.) $\endgroup$ – Jeremy Rouse May 9 '14 at 12:05
  • $\begingroup$ @ Jeremy Rouse: Ok and thank you very much. $\endgroup$ – China-Hong Kong May 9 '14 at 13:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.