NTIC From Riemann to Dirichlet and Euler

Section 24.3 From Riemann to Dirichlet and Euler

In order to see this (the convergence of the infinite product), let's instead observe our other main example of a sum over divisors equalling a product over primes working. When we compared them for

ϕ

above, we got

\sum_{d ∣ n} \frac{μ (d)}{d} = \prod_{p ∣ n} (1 - \frac{1}{p})

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@interact
def _(e=(1,[0..3]),f=(2,[0..3]),g=(0,[0..3])):
    n = 2^e*3^f*5^g
    pretty_print(html("You picked $%s=2^{%s}3^{%s}5^{%s}$"%(n,e,f,g)))
    str = '$$'+'+'.join([r'\frac{%s}{%s}'%(moebius(d),d) for d in divisors(n)])+'=%s$$'%sum([moebius(d)/d for d in divisors(n)])
    str2 = '$$'+''.join([r'\left(1-\frac{1}{%s}\right)'%p for (p,e) in factor(n)])+'=%s$$'%prod([1-1/p for (p,e) in factor(n)])
    pretty_print(html(str))
    pretty_print(html("compare to "+str2))

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We could make the powers far higher, or include more primes, and it would still work. Going to both limits, this would lead to the series

\sum_{n = 1}^{\infty} \frac{μ (n)}{n^{s}} .

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Subsection 24.3.1 Dirichlet series

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We give such series a name. The following definition is purely formal, considered without considering issues such as convergence. (See [E.2.8, Chapter 4.6] for an interesting formal viewpoint on the set of these series.)

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Definition 24.3.1. Dirichlet Series.

In general, for an arithmetic function $f (n),$ its Dirichlet series is

F (s) = \sum_{n = 1}^{\infty} \frac{f (n)}{n^{s}} .

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Answer the following three questions to see if you understand this definition. (See Exercise 24.7.1.)

For what arithmetic function is the Riemann zeta function the Dirichlet series?
What would the Dirichlet series of $N$ be?
What about the Dirichlet series of $I ?$

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Note that this already indicates some level of connection between arithmetic functions. These are connections which may not have been evident otherwise.

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Subsection 24.3.2 Euler products

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For our purposes, the very important thing to note about such series is that they often can be expanded as infinite products.

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Definition 24.3.2. Euler Products.

In general, for an arithmetic function $f (n),$ its Dirichlet series is said to have an Euler product if the series can be written as an infinite product in the following manner:

\sum_{n = 1}^{\infty} \frac{f (n)}{n^{s}} = \prod_{p} (a formula involving f (p) and p^{s}) .

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Example 24.3.3. Euler product for Riemann zeta function.

We have already suggested one for the zeta function:

ζ (s) = \sum_{n = 1}^{\infty} \frac{1}{n^{s}} = \prod_{p} (\frac{1}{1 - p^{- s}}) .

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Based on the logic of this section, we have a potential new Euler product for the Dirichlet series of the Moebius function:

\sum_{n = 1}^{\infty} \frac{μ (n)}{n^{s}} = \prod_{p} (1 - \frac{1}{p^{s}}) = \prod_{p} (1 - p^{- s})

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At least, we can consider this wherever it makes sense. See [E.4.6, Chapter 11.5] or [E.2.1, Chapter 9.8] for some criteria, or simply below at Theorem 24.5.4.

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In the next section, we justify more of this discussion, and connect our wonderful results about Dirichlet products of finite arithmetic functions to deep properties of their Dirichlet series.