# Thread: Coefficients of a Dirichlet series

1. ## Coefficients of a Dirichlet series

Let $\displaystyle F(s) = \zeta(s)L(s,\chi_1)L(s,\chi_2)L(s,\chi_1\chi_2)$, for two real primitive characters $\displaystyle \chi_1, \chi_2$. Multiplication of Euler products of these functions gives us

$\displaystyle F(s) = \displaystyle \sum_{n=1}^\infty a_n n^{-s}$

with $\displaystyle a_1=1$. Davenport (in p.129 of Mult. Numb. Theory) claims that taking logarithms of Euler products gives

$\displaystyle \displaystyle \log F(s) = \sum_p \sum_{m=1}^\infty \big(1+\chi_1(p^m)\big)\big(1+\chi_2(p^m)\big)m^{-1} p^{-ms},$

which is fine, but then claims that since the coefficients here are nonnegative, we have $\displaystyle a_n \geqslant 0$ for all $\displaystyle n$. Can someone show me the reasoning for this last fact? It doesn't seem obvious to me.

2. Suppose $\displaystyle f(z)=\prod_p(1-g(p)p^{-s})^{-1}$, with $\displaystyle g$ completely multiplicative.

Then $\displaystyle \log f(z) = \sum_{p}\sum_{n=1}^\infty g(p^n)n^{-1}p^{-ns}$. Now if each of the coefficients is non-negative, this implies for example that each $\displaystyle g(p)$ is non-negative. But each $\displaystyle g(n)$ is a product of terms of the form $\displaystyle g(p)$, hence is also non-negative.

3. Originally Posted by Bruno J.
Suppose $\displaystyle f(z)=\prod_p(1-g(p)p^{-s})^{-1}$, with $\displaystyle g$ completely multiplicative.

Then $\displaystyle \log f(z) = \sum_{p}\sum_{n=1}^\infty g(p^n)n^{-1}p^{-ns}$. Now if each of the coefficients is non-negative, this implies for example that each $\displaystyle g(p)$ is non-negative. But each $\displaystyle g(n)$ is a product of terms of the form $\displaystyle g(p)$, hence is also non-negative.
Okay, this would work, but it presupposes that $\displaystyle F(s)$ can be written in the form $\displaystyle \prod_p(1-a_p p^{-s})^{-1}$ where the a_p are the same as in my original post. This product only arises from the Dirichlet series if the a_n are completely multiplicative. If we knew explicitly what the a_n's were we might be able to prove this. However I think it doesn't work, as using your notation we would need to put

$\displaystyle g(p^m) = (1+\chi_1(p^m))(1+\chi_2(p^m))$

so $\displaystyle g(p) = (1+\chi_1(p))(1+\chi_2(p))$

but then $\displaystyle g(p)^m = (1+\chi_1(p))^m (1+\chi_2(p))^m \neq (1+\chi_1(p^m))(1+\chi_2(p^m)) = g(p^m).$

so the coefficients are actually not completely multiplicative.

4. Let G(s) = log F(s). Then F(s) = exp(G(s)). Express exp(G(s)) as a power series in G(s). Since the coefficients of the power series for the exponential function are positive, and the coefficients of G(s) are non-negative, then the coefficients of F(s) are non-negative. Does that work?

5. Originally Posted by Petek
Let G(s) = log F(s). Then F(s) = exp(G(s)). Express exp(G(s)) as a power series in G(s). Since the coefficients of the power series for the exponential function are positive, and the coefficients of G(s) are non-negative, then the coefficients of F(s) are non-negative. Does that work?
Damn right it does! Much better, thank you .