Average order of an arithmetic function: Difference between revisions

Revision as of 06:23, 10 August 2014

Revision as of 06:28, 10 August 2014

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Define <math>\delta(f)</math> to be 1 if <math>f</math> is k-th power free and 0 otherwise.

We calculate the average value of δ<math>\delta</math>, which is the density of the k-th power free polynomials in {{math|'''F'''_'''q'''[X]}}, in the same fashion as in the integers.

By multiplicativity of <math>\delta</math>:

<math>\sum_{f}\frac{\delta(f)}{|f|^{s}}=\prod_{P}(\sum_{j \mathop =0}^{k-1}(|P|^{-js}))=\prod_{P}\frac{1-|P|^{-sk}}{1-|P|^{-s}}=\frac{\zeta_{A}(s)}{\zeta_{A}(sk)}=\frac{1-q^{1-ks}}{1-q^{1-s}}=\frac{\zeta_A(s)}{\zeta_A(ks)}</math>

Denote <math>b_{n}</math> the number of k-th power monic polynomials of degree ''n'', we get

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<math>\sum_{f}\frac{\delta(f)}{|f|^{s}}=\sum_{n}\sum_{\text{def}f=n}\delta(f)|f|^{-s}=\sum_{n}b_{n}q^{-sn}</math>.

Making the substitution <math>u=q^{-s}</math> we get:

<math>\frac{1-qu^{k}}{1-qu}=\sum_{n \mathop =0}^{\infty}b_{n}u^{n}</math>.

Finally, expand the left-hand side in a geometric series and compare the coefficients on <math>u^{n}</math> on both sides, weto getconclude that

<math>b_{n}=\begin{cases}