1
$\begingroup$

Set-up

Let $f: \mathbb{N} \to \mathbb{C}$ be a multiplicative arithmetic function (meaning that $f(mn) = f(m)f(n)$ whenever $m$ and $n$ are relatively prime). Suppose that for every $\epsilon >0,$ that $$f(n) \ll_\epsilon n^\epsilon,$$ here I am using Vinogradov’s notation, this means that $|f(n)| \leq C_\epsilon n^\epsilon$ for some constant $C_\epsilon$ which depends on $\epsilon.$.

Suppose further that for each prime $p, f(p)=g $ for some $g \geq 1.$ This $g$ is thus independent of $p.$.

Consider now, for a prime p, the identity of formal power series $$1+gX+f(p^2)X^2 + \cdots = (1-X)^{-g} h_p(X).$$ It is then easy to see that $h_p(0)=1,h_p’(0)=0.$.

Question. Let now $b_p(k)$ be the coefficient of $X^k$ in $h_p(X).$ I have seen it claimed that the coefficients $b_p(k)$ are bounded by a polynomial in $k$ that is independent of $p.$
Could someone explain why this holds?

$\endgroup$
  • $\begingroup$ I don't think that $h'_p(0)=1$. Are you sure that your definition is correct? $\endgroup$ – FXV Jun 29 at 15:37
  • $\begingroup$ Let $f(p) = g,f(p^k) = \log p$ then $f(n) = \prod_{p^k \| n}f(p^k) = O(n^\epsilon)$ even if $ f(p^k)$ isn't bounded by a polynomial of $k$. If you mean a polynomial of $p^k$ then it is immediate from $f(n) = O(n)$. Also (for a multiplicative function) $f(n) = O(n^\epsilon)$ iff $f(p^k) = O((p^k)^\epsilon)$ $\endgroup$ – reuns Jun 29 at 16:00
  • $\begingroup$ @reuns I’d be careful with epsilons in the last statement. If $f(p^k) = 2p^k$ then it isn’t quite true that $f(n) = O(n)$, but it’s close. $\endgroup$ – Erick Wong Jun 29 at 17:14
  • $\begingroup$ @FXV Typo, thanks! Now fixed. $\endgroup$ – inequalitynoob2 Jun 29 at 17:14
  • $\begingroup$ @reuns I really want a polynomial in $k,$ not in $p^k.$ Are you giving a counterexample? I am not sure I understood your comment. $\endgroup$ – inequalitynoob2 Jun 29 at 17:15
-2
$\begingroup$

Here's my naive approach.

Let $a(x) =1+gX+f(p^2)X^2 + \cdots =\sum_{n-0}^{\infty} a_n x^n $ so that $(1-x)^{-g} h_p(x) =a(x) $.

Then

$\begin{array}\\ h_p(x) &=a(x)(1-x)^g\\ &=\sum_{n=0}^{\infty} a_n x^n \sum_{j=0}^g \binom{g}{j}(-1)^{g-j}x^{g-j}\\ &=\sum_{n=0}^{\infty} a_n x^n \sum_{j=0}^g \binom{g}{j}(-1)^{j}x^{j}\\ &=\sum_{n=0}^{\infty} x^n\sum_{i=0}^{\min(n, g)} a_{n-i} \binom{g}{i}(-1)^{i}\\ \end{array} $

and

$\begin{array}\\ |\sum_{i=0}^{\min(n, g)} a_{n-i} \binom{g}{i}(-1)^{i}| &\le \sum_{i=0}^{\min(n, g)} |a_{n-i} \binom{g}{i}|\\ &\le \sum_{i=0}^{\min(n, g)} g^if(p^{n-i})\\ &\le g^g\sum_{i=0}^{\min(n, g)} f(p^{n-i})\\ &\le g^g\sum_{i=0}^{\min(n, g)} C_{\epsilon}(n-i)^{\epsilon}\\ &\le g^{g+1}C_{\epsilon}n^{\epsilon}\\ \end{array} $

$\endgroup$
  • 2
    $\begingroup$ Why is $f(p^{n-i}) \leq C_\epsilon (n-i)^\epsilon?$ We only have that $f(n) \leq C_\epsilon n^\epsilon, $ so rather $f(p^{n-i}) \leq C_\epsilon (p^{n-i})^\epsilon.$ Am I missing something? $\endgroup$ – inequalitynoob2 Jun 29 at 18:46
  • $\begingroup$ Naw. Just my usual carelessness. I'll leave it up in case someone can make it right. $\endgroup$ – marty cohen Jun 29 at 21:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.