Monday Math 87

For a sequence a_n, n=1,2,3,…, the series $f(s)=\sum_{n=1}^{\infty}\frac{a_n}{n^s}$ is known as the Dirichlet series generating function (or sometimes simply the Dirichlet series) for the sequence. For example, if we take the trivial sequence 1,1,1,… (a_n=1 for all n), then the Dirichlet series is the Riemann zeta function
$\zeta(s)=\sum_{n=1}^{\infty}\frac{1}{n^s}$ .
For the alternating sequence 1,-1,1,-1,… (a_n=(-1)^n-1), the Dirichlet series is the Dirichlet eta function
$\eta(s)=\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^s}$ .

Now, consider the case where a_n=f(n), where f(n) is a multiplicative function.
Then we have Dirichlet series generating function:
$F(s)=\sum_{n=1}^{\infty}\frac{f(n)}{n^s}$ .
Using the prime factorization of n, $n=p_1^{k_1}p_2^{k_2}\cdots{p}_r^{k_r}$ , we see:
$\begin{eqnarray}F(s)&=&\sum_{n=1}^{\infty}\frac{f(n)}{n^s}\\&=&\sum_{n=1}^{\infty}\frac{f\left(p_1^{k_1}p_2^{k_2}\cdots{p}_r^{k_r}\right)}{\left(p_1^{k_1}p_2^{k_2}\cdots{p}_r^{k_r}\right)^s}\\&=&\sum_{n=1}^{\infty}\frac{f\left(p_1^{k_1}\right)f\left(p_2^{k_2}\right)\cdots{f}\left({p}_r^{k_r}\right)}{\left(p_1^{k_1}p_2^{k_2}\cdots{p}_r^{k_r}\right)^s}\\&=&\sum_{n=1}^{\infty}\frac{f\left(p_1^{k_1}\right)f\left(p_2^{k_2}\right)\cdots{f}\left({p}_r^{k_r}\right)}{p_1^{k_1\cdot{s}}p_2^{k_2\cdot{s}}\cdots{p}_r^{k_r\cdot{s}}}\end{eqnarray}$ .
Now, we procede as we did in this post, which is just the above with multiplicative function $f(n)=2^{\omega(n)}$ . The above can be expressed as a product over the primes:
$\begin{eqnarray}F(s)&=&\sum_{n=1}^{\infty}\frac{f\left(p_1^{k_1}\right)f\left(p_2^{k_2}\right)\cdots{f}\left({p}_r^{k_r}\right)}{p_1^{k_1\cdot{s}}p_2^{k_2\cdot{s}}\cdots{p}_r^{k_r\cdot{s}}}\\&=&\sum_{n=1}^{\infty}\prod_i\frac{f\left(p_i^{k_i}\right)}{p_i^{k_i\cdot{s}}}\\&=&\prod_{p\;\text{prime}}\sum_{k=0}^{\infty}\frac{f\left(p^k\right)}{p^{ks}}\end{eqnarray}$ .
When f(n) is a completely multiplicative function, $f\left(p^k)\right)=\left(f(p)\right)^k$ , so then the series within the summation becomes a geometric series:
$\begin{eqnarray}F(s)&=&\prod_{p\;\text{prime}}\sum_{k=0}^{\infty}\frac{f\left(p^k)\right)}{p^{ks}}\\&=&\prod_{p\;\text{prime}}\sum_{k=0}^{\infty}\left(\frac{f(p)}{p^{s}}\right)^k\\&=&\prod_{p\;\text{prime}}\frac{1}{1-f(p)p^{-s}}\end{eqnarray}$ .
Note that if our function is the constant function f(n)=1 (it is completely multiplicative), then the above gives us the Euler product for the Riemann zeta function
$\zeta(s)=\prod_{p\;\text{prime}}\frac{1}{1-p^{-s}}$
which we found here.

Note that even if the function is just multiplicative, and not completely multiplicative, it may still be possible to find a non-series expression for the series $\sum_{k=0}^{\infty}\frac{f\left(p^k\right)}{p^{ks}}$ found in the product over primes. One example can be seen this past post.

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Tags: Dirichlet Series, Dirichlet Series Generating Function, Euler Product, Math, Monday Math, Multiplicative Function, Riemann Zeta Function

This entry was posted on August 31, 2009 at 12:18 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.

3 Responses to “Monday Math 87”

Monday Math 88 « Twisted One 151’s Weblog Says:
September 7, 2009 at 12:07 am | Reply
[...] Math 88 By twistedone151 Last week, I demonstrated how the Dirichlet series generating function for a sequence consisting of a [...]
Monday Math 89 « Twisted One 151’s Weblog Says:
September 14, 2009 at 12:18 am | Reply
[...] , we get . [Note that , so that the x=0 case is the limit of the x≠0 general case]. Recall previouly that the Dirichlet series generating function for a sequence consisting of a multiplicative [...]
Monday Math 94 « Twisted One 151's Weblog Says:
September 21, 2010 at 9:51 am | Reply
[...] we can see that λ(n) has Dirichlet series generating function ; we can confirm this via the Euler product for Dirichlet series generating functions of completely multiplicative functions. One should also note that if a function f(n) has Dirichlet series generating function F(s), then [...]

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Monday Math 87

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