Suppose I know the Dirchlet series $$\sum_{n=1}^{\infty} \frac{f(n)}{n^s} = \frac{\zeta(s)}{\zeta(3s)},$$ where $\zeta(s)$ is the usual Riemann zeta function.

My question is - is there a way to determine $f(n)$ from this information? If so, how?

  • $\begingroup$ Check out Dirichlet series and Dirichlet convolution $\endgroup$ – Sasha Apr 20 '14 at 14:04
  • $\begingroup$ @Sasha Yes, I've checked these two pages before asking but somehow I am lacking the intuition to apply these ideas. Hence, do you mind elaborating a bit more please? $\endgroup$ – Jernej Apr 20 '14 at 14:09
  • $\begingroup$ Hint: Use the Euler product for $\zeta(s)$. (I'm assuming $f$ is multiplicative.) $\endgroup$ – André 3000 Apr 20 '14 at 15:57
  • $\begingroup$ @SpamIAm - It isn't necessary to assume that $f$ is multiplicative. Given the Dirichlet series $\sum_{n \geq 1} a_nn^{-s}$ associated with the Euler product of $\zeta(s) \zeta(3s)^{-1}$, it is identical to the Dirichlet series of $f$, which will imply that $f(n) = a_n$ for each $n$. Its a sort of Riemann-Lebesgue lemma analogue for Dirichlet series, see for example Tenenbaum's book II.1. $\endgroup$ – Dead-End Apr 22 '14 at 2:41
  • $\begingroup$ @smangerel Okay, so $f(n) = a_n$ for all $n$. Thanks for pointing that out. $\endgroup$ – André 3000 Apr 22 '14 at 3:40

Yes: \begin{align} \sum_{n = 1}^\infty \frac{f(n)}{n^s} & = \frac{\zeta(s)}{\zeta(3s)}\\ & = \prod_p \frac{1 - p^{-3s}}{1 - p^{-s}} &&\text{by the Euler product for $\zeta$ where the $p$s are all primes}\\ & = \prod_p \left(1 + \frac{1}{p^s} + \frac{1}{p^{2s}} \right) &&\text{by the identity $ 1 - x^3 = (1 - x)(1 + x + x^2)$} \end{align} Comparing with the Euler product formula, we see that $f$ is a multiplicative function and that $$f(p^\alpha) = \begin{cases} 1 & \text{if $\alpha = 1$ or $2$,} \\ 0 & \text{if $\alpha = 3, 4, 5, \ldots$.} \end{cases}$$

Because a multiplicative function is completely determined by its values at the powers of prime numbers, we can restate the function as $$f(n) = \begin{cases} 1 & \text{if 1 is the largest cube that divides $n$,} \\ 0 & \text{otherwise.} \end{cases}$$

The Dirichlet series for $\zeta(s)$ and $\zeta(3s)$ both converge absolutely for $s > 1$, so that definition of $f(n)$ is the only answer when $s > 1$ (see Theorem 4.8 at http://www.math.illinois.edu/~ajh/ant/main4.pdf).

  • $\begingroup$ (The theorem 4.8 is simply that if $H(s) = \sum_{n=1}^\infty h(n) n^{-s}$ converges then as $\Re(s) \to \infty$ : $H(s) \sim h_m m^{-s}$ where $h(m)$ is the first non-zero coef, so we can recover uniquely $h(m)$ from its Dirichlet series) $\endgroup$ – reuns Jun 13 '17 at 18:08

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