Section 24.4 Multiplication
Subsection 24.4.1 Some coincidences
One surprising thing about end of the previous subsection is that the Euler products for the Riemann ΞΆ function and the Dirichlet series of the MΓΆbius function are multiplicative inverses of each other. That is,xxxxxxxxxx
sum([moebius(n)/n^2 for n in [1..10000]]).n()
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1/zeta(2).n()
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1/(pi^2/6).n()
Remark 24.4.1.
Zeta has interesting values at integers, not just for s=2. Euler calculated many even values of ΞΆ, which all look like Ο2n times a rational number (see any description of the so-called Bernoulli numbers). However, it was only in 1978 that ΞΆ(3) was shown to be irrational. It was then named ApΓ©ry's constant after the man who proved this, Roger ApΓ©ry.
To compare with the situation for even n, as of this writing it is still only known that at least one of the next four odd values (ΞΆ(5),ΞΆ(7),ΞΆ(9),ΞΆ(11)) is irrationalβ1β. See Wadim Zudilin's website for many links, though this page hasn't been updated for some time.
Subsection 24.4.2 Multiplication of both kinds
Let's reinterpret this just a little bit. Assuming we can prove that all this makes sense (which we haven't, yet), we have the following two analogous facts.Fact 24.4.2.
The arithmetic functions u and ΞΌ are inverses as arithmetic functions; that is, uβΞΌ=I.
The Dirichlet series of these functions are also inverses, as ordinary functions:
Alternately, ββn=1ΞΌ(n)ns=1/ΞΆ(s)
Theorem 24.4.3.
Use the following notation:
Take f(n) and g(n) to be two arithmetic functions.
Let h=fβg be their Dirichlet product.
Let F,G,H be the corresponding Dirichlet series (in the variable s).
Then if the series F and G converge absolutely for any particular s, then H converges and H=FG for that s as well.
Proof.
First, we need there is a key fact you may or may not have seen in calculus, related to absolute convergence (see for example Active Calculus). Roughly speaking, when series converge absolutely, you can mess around with them with a lot with impunity. See, for instance, Mertens' Theorem on convergence of Cauchy products. Interestingly, neither [E.4.6] nor [E.2.1, Theorem 9.6] say much more about this in their presentation of this standard proof. See Exercise 24.7.3 if you have not encountered this!
In any case, since \(F\) and \(G\) do converge absolutely, we can and will mess around a lot with the product
In particular, we can group the products by the terms \(\frac{f(n)g(m)}{n^sm^s}\) (the same way we did in proving things about \(\star\) in Subsection 23.4.3), without loss of equality.
We can further group by when \(n\) and \(m\) are complementary divisors of the same number (I suggest using specific numbers to try this out). This gives
Notice that the inner sum is precisely the Dirichlet \(\star\) product (except divided by \(d^s\)). So we may rewrite this as
The numerators are the definition of \(h\text{,}\) so this is just \(H(s)\text{,}\) as desired. (In [E.4.6, Theorem 11.5] the additional detail that any Dirichlet series with these values must be the one for \(f\star g\) is proved, which requires a uniqueness result for the series we will omit.)