Real points $a∈ℝ$ such that the equation $f^{(k)}(s)=a$ have a finite number of real solutions $s$ for some $k$ 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.
 A: 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.
