Coefficients of a Dirichlet series

**Capillarian** · March 11th 2011, 11:14 AM

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

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

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

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

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

**Bruno J.** · March 11th 2011, 05:21 PM

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

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

**Capillarian** · March 12th 2011, 01:15 AM

Originally Posted by Bruno J.

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

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

Okay, this would work, but it presupposes that $F(s)$ can be written in the form $\prod_p(1-a_p p^{-s})^{-1}$ 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

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

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

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

so the coefficients are actually not completely multiplicative.

**Petek** · March 12th 2011, 04:23 AM

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?

**Capillarian** · March 12th 2011, 05:22 AM

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

.

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