10
$\begingroup$

Suppose I have an integral Dirichlet series $f(s) = \sum c_n n^{-s}$, $c_n \in \mathbb{Z}$, with at least one non-zero term $c_N$. Suppose furthermore that this series converges absolutely and uniformly for $\mathrm{Re}(s) > 1 + \delta$ for any $\delta > 0$ (so that $c_n$ grows slower than $n^\epsilon$ for any $\epsilon$).

I want $f(s)$ to be "small" (in terms of $N$) for fixed $\mathrm{Re}(s)$ and a range of $\mathrm{Im}(s)$. How small can I get it over any given range?

I think one can take sums of $(N + \Delta n)^{-s}$ for $\Delta n \approx O(\ln N)$ and cancel out terms in the Taylor series up to order $N^{-(s + O(\ln N))}$, all while keeping the coefficients polylog in N for $Im(s) \approx O(\ln N)$. Can we do any better? What about larger values of $s$?

  • 0
    I do not believe it is possible to keep it small for all Im($s$). I am looking for statements of the form, "for $|Im(s) - \sigma | \approx O(g(N))$ the following construction gives $|f(s)| \approx O(N^{-(Re(s) + h(N))})$". As for the construction I alluded to, consider that $N^{-s} - (N+1)^{-s} = s N^{-(s+1)} + O(s^2 N^{-(s+2)})$. Furthermore if we pick any two $k,k'$, we can find a linear combination of $N^{-s}$, $(N+k)^{-s}$ and $(N+k')^{-s}$ that will be of order $s^2 N^{-(s+2)}$. We can keep doing this approximately $ln N$ times at least before the error terms start blowing up.2011-10-06

1 Answers 1

1

I alluded to one construction that gives small values for $f(s)$ in a given range of Im(s). I will elaborate on that construction here.

Consider the Taylor expansion of $(N+x)^{-s}$. This is

$(N+x)^{-s} = N^{-s} * (1 - \frac{sx}{N} + \frac{s(s+1)x^2}{2N^2} - \ldots )$.

We are going to truncate this at some finite order $k$, and allow $x$ to take on different integer values. We then obtain a number of terms polynomial in $s$ (times $N^{-s}$) and we want a linear combination of these terms to vanish for all $s$ -- that implies $x$ takes on at least $k$ different values.

Define the $c_n$ to be the coefficients in this sum. We would like the remaining error term in the Taylor series (the terms of order $s^{(k+1)}/N^{k+1}$ and higher) to be $o(N^{-k})$. This implies that $\sum c_n (n-N)^{k+1} s^{(k+1)} / k!$ is $o(N)$. Remember that we have at least $k-1$ nonzero terms in this sum, and so there is at least one $n$ with $|n-N| \geq (k-1)/2$.

To ensure that this sum is $o(N)$, we should take $k$ to be $o(\ln N)$ and similarly for $s$. We can readily find a set of $n$ and $c_n$ which will keep the sum at $o(N)$ -- I believe the $k$th-order difference should work, although I am not certain.