Monday Math 88

Last week, I demonstrated how the Dirichlet series generating function for a sequence consisting of a multiplicative function can be represented as a product over primes:
$F(s)=\sum_{n=1}^{\infty}\frac{f(n)}{n^s}=\prod_{p\;\text{prime}}\sum_{k=0}^{\infty}\frac{f\left(p^k\right)}{p^{ks}}$ .
Now, let us use this for a couple of multiplicative functions. First, the Möbius function μ(n). We recall that
$\mu\left(p^k\right)=\Large\left\{\begin{eqnarray}1&,\;&k={0}\\-1&,\;&k=1\\{0}&,\;&k\gt1\end{eqnarray}\right\.$
Thus, the series $\sum_{k=0}^{\infty}\frac{\mu\left(p^k\right)}{p^{ks}}$ inside the prime product has only two non-zero terms, and
$\begin{eqnarray}\sum_{n=1}^{\infty}\frac{\mu(n)}{n^s}&=&\prod_{p\;\text{prime}}\sum_{k=0}^{\infty}\frac{\mu\left(p^k\right)}{p^{ks}}\\&=&\prod_{p\;\text{prime}}\left(1-p^{-s}\right)\\&=&\left[\prod_{p\;\text{prime}}\frac{1}{1-p^{-s}}\right]^{-1}\\\sum_{n=1}^{\infty}\frac{\mu(n)}{n^s}&=&\frac{1}{\zeta(s)}\end{eqnarray}$ ,
using the Euler product for the Riemann zeta function $\zeta(s)=\prod_{p\;\text{prime}}\frac{1}{1-p^{-s}}$ .
We can similarly see:
$\begin{eqnarray}\sum_{n=1}^{\infty}\frac{\left|\mu(n)\right|}{n^s}&=&\prod_{p\;\text{prime}}\sum_{k=0}^{\infty}\frac{\left|\mu\left(p^k\right)\right|}{p^{ks}}\\&=&\prod_{p\;\text{prime}}\left(1+p^{-s}\right)\\&=&\prod_{p\;\text{prime}}\left(\frac{1-p^{-2s}}{1-p^{-s}}\right)\\\sum_{n=1}^{\infty}\frac{\left|\mu(n)\right|}{n^s}&=&\frac{\zeta(s)}{\zeta(2s)}\end{eqnarray}$ .

Now, for a more challenging example, consider Euler’s totient function φ(n). It is a multiplicative function, and I showed here that $\phi(p^k)=p^{k-1}(p-1)$ for prime p and positive k, so the Dirichlet series generating function will have prime product form:
$\sum_{n=1}^{\infty}\frac{\phi(n)}{n^s}=\prod_{p\;\text{prime}}\sum_{k=0}^{\infty}\frac{\phi\left(p^k\right)}{p^{ks}}=\prod_{p\;\text{prime}}\left[1+\sum_{k=1}^{\infty}\frac{p^{k-1}(p-1)}{p^{ks}}\right]$ ,
as when k=0, $\psi(p^k)=\psi(1)=1$ .
Now, to find a finite form for the series $\sum_{k=1}^{\infty}\frac{p^{k-1}(p-1)}{p^{ks}}$ , we note that we can factor out (p-1) from all terms, and with a little work, we can get a geometric series:
$\begin{eqnarray}\sum_{k=1}^{\infty}\frac{p^{k-1}(p-1)}{p^{ks}}&=&(p-1)\sum_{k=1}^{\infty}\frac{p^{k-1}}{p^{ks}}\\&=&\frac{p-1}{p}\sum_{k=1}^{\infty}\left(p^{1-s}\right)^k\\&=&\frac{p-1}{p}\frac{p^{1-s}}{1-p^{1-s}}\\&=&\frac{p^{-s}(p-1)}{1-p^{1-s}}\\&=&\frac{p-1}{p^s-p}\end{eqnarray}$ .
Thus
$\begin{eqnarray}1+\sum_{k=1}^{\infty}\frac{p^{k-1}(p-1)}{p^{ks}}&=&1+\frac{p-1}{p^s-p}\\&=&\frac{p^s-p}{p^s-p}+\frac{p-1}{p^s-p}\\&=&\frac{p^s-1}{p^s-p}\\&=&\frac{1-p^{-s}}{1-p^{1-s}}\end{eqnarray}$ ,
and so we find the product over primes is
$\begin{eqnarray}\sum_{n=1}^{\infty}\frac{\phi(n)}{n^s}&=&\prod_{p\;\text{prime}}\left[1+\sum_{k=1}^{\infty}\frac{p^{k-1}(p-1)}{p^{ks}}\right]\\&=&\prod_{p\;\text{prime}}\frac{1-p^{-s}}{1-p^{1-s}}\\&=&\frac{\prod_{p\;\text{prime}}\frac{1}{1-p^{1-s}}}{\prod_{p\;\text{prime}}\frac{1}{1-p^{-s}}}\\&=&\frac{\zeta(s-1)}{\zeta(s)}\end{eqnarray}$ .

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Tags: Dirichlet Series, Dirichlet Series Generating Function, Euler Product, Math, Möbius Function, Monday Math, Prime Numbers, Totient

This entry was posted on September 7, 2009 at 12:01 am and is filed under Math/Science. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.

One Response to “Monday Math 88”

Monday Math 94 « Twisted One 151’s Weblog Says:
October 26, 2009 at 12:19 am | Reply
[...] which have Dirichlet series generating functions and , respectively. You may also recall that here I showed that the Dirichlet series generating function for |μ| is . I also demonstrated here [...]

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Monday Math 88

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