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<math>D(n)=xlog_{q}(x)+x</math> which resembles closely the analogous result for integers <math>\sum_{k \mathop =1}^{n}d(k)=xlogx+(2\gamma-1)x+O(\sqrt{x})</math>, where <math>\gamma</math> is [[Euler constant]]. It is a famous problem in [[elementary number theory]] to find the error term. In the polynomials case, there is no error term. This is because of the very simple nature of the zeta function <math>\zeta_{A}(s)</math>.
====Polynomial von Mangoldt function function====
The Polynomial [[von Mangoldt function]] is defined by:
<math>\Lambda_{A}(f) = \begin{cases} \log |P| & \mbox{if }f=|P|^k \text{ for some prime monic} P \text{ and integer } k \ge 1, \\ 0 & \mbox{otherwise.} \end{cases}</math>
'''Proposition.''' <math>\Lambda_{A}</math> has mean value <math>log(q)</math>.
'''Proof.'''
First, notice that,
<math>\sum_{f|m}\Lambda_{A}(f)=log|m|</math>.
Hence,
<math>\zeta_{A}*\Lambda_{A}(m)=log|m|</math>
and we get that,
<math>\zeta_{A}(s)D_{\Lambda_{A}}(s)=\sum_{m}log|m||m|^{-s}</math>. Now,
<math>\sum_{m}|m|^{s}=\sum_{n}\sum_{\text{deg}m=n}u^n=\sum_{n}q^{n}u^{n}=\sum_{n}q^{n(1-s)}</math>.
Thus,
<math>\frac{d}{ds}\sum_{m}|m|^{s}=-\sum_{n}log(q^n)q^{n(1-s)}=-\sum_{n}\sum_{deg(f)=n}log(q^n)q^{-ns}=-\sum_{f}log|f||f|^{-s}</math>.
We got that:
<math>D_{\Lambda_{A}}(s)=\frac{-\zeta'_{A}(s)}{\zeta_{A}(s)}</math>
Now,
<math>\sum_{m}\Lambda_{A}(m)|m|^{-s}=\sum_{n}(\sum_{deg(m)=n}\Lambda_{A}(m)q^{-sm})=\sum_{n}(\sum_{deg(m)=n}\Lambda_{A}(m))u^n=\frac{-\zeta'_{A}(s)}{\zeta_A(s)}=\frac{q^{1-s}log(q)}{1-q^{1-s}}=log(q)\sum_{n\mathop=1}^{\infty}q^{n}u^n</math>
Hence,
<math>\sum_{deg(m)=n}\Lambda_{A}(m)=q^{n}log(q)</math>,</br> and by dividing by <math>q^n</math> we get that,
<math>Ave_{n}\Lambda_{A}(m)=log(q)</math>.
==See also==
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