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Could anyone tell me in what sense the following is an "asymptotic formula":

Theorem 1 from

Link

(this is open access, so I think I'm allowed to link it)

At the moment, I'm just trying to understand what the bit with $Q\leq x$ means. I assume the Theorem says something like: the LHS varies like the main term, in that LHS/[main term] tends to one as x tends to infinity. But it's not clear to me that [error terms]/[main terms] tend to zero. There seems to be definitely something I'm not understanding with the relationship between $Q$ and $x$.

Can anyone clarify?

Thanks very much.

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  • $\begingroup$ I answered your question, then I edited it and realized that half of it was missing. Next time please use TeX and post your question only when it appears in full as you intended it. Also, your questions are not of research level I am afraid. $\endgroup$
    – GH from MO
    Commented Dec 26, 2012 at 2:37
  • $\begingroup$ I added more detail in a comment below. $\endgroup$
    – GH from MO
    Commented Dec 26, 2012 at 19:18

1 Answer 1

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There is no Theorem 1 in this paper. The statement called "Theorem" contains an asymptotic formula: e.g. (2) has a main term $Qx\log x$ and two error terms $O(\dots)$. The error terms are $o(Qx\log x)$ for $x (\log x)^{-A} < Q < x $ and $x\to\infty$, say, hence in this case the left hand side is asymptotic to the main term.

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  • $\begingroup$ Oh ye it's just "Theorem", sorry. I will use TeX in the future. As for the level of the question, I didn't realise it had to be "research level". I thought it was ok, but maybe not then. (I see threads such as "What should be learned in an Analytic Number Theory course", which, whilst more useful than my thread, is not "research level".) Thanks for your answer. Can I ask what exactly is the limit taken over? I know we say $\rightarrow \infty $ but what exactly happens with $Q$ when we do that. I don't really understand $x(\log x)^{-A}Q\leq x$. $\endgroup$
    – tomos
    Commented Dec 26, 2012 at 11:25
  • $\begingroup$ Actually, it's ok, your answer has been enough. Thank you. $\endgroup$
    – tomos
    Commented Dec 26, 2012 at 11:31
  • $\begingroup$ @Tomos: I meant that the error terms are $o(Qx\log x)$ as $x\to\infty$, uniformly in $Q$ satisfying $x(\log x)^{-A}<Q<x$. This means that for any $\epsilon>0$ there is $x_0=x_0(\epsilon,A)$ such that for $x>x_0$ and $x(\log x)^{-A}<Q<x$ the absolute value of the error term is less than $\epsilon Qx\log x$. As for research level: it is subjective what is of that level. I guess if someone is a graduate student and asks a question that arose in her/his studies, then it is OK. $\endgroup$
    – GH from MO
    Commented Dec 26, 2012 at 19:16
  • $\begingroup$ That answer is perfect, thank you. I had thought it was this, but was not sure (I've never seen it actually written down, but perhaps it was clear from the context - my argument against this though is that it is not 100% clear for someone who is still getting used to the order notation with more variables involved. Seeing it written down gets rid of the 20-40% doubt I had reading it and allows me to move on. Thanks!) $\endgroup$
    – tomos
    Commented Dec 26, 2012 at 23:08
  • $\begingroup$ I am glad I could help. Much of analysis is about uniform bounds, and the $o(\dots)$ and $O(\dots)$ notations facilitate the work with them. $\endgroup$
    – GH from MO
    Commented Dec 27, 2012 at 0:49

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