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This is copied from math.SE after a kind comment's suggestion as I am sure people here are very well knowledged in this method :)

I am currently reading Vaughan's "The Hardy-Littlewood Method", and in particular the chapter on the major arcs for Waring's Problem (2.4) - a (essentially) copy (all proofs are the same with more details) can be found here. However, I am a bit lost of the "general idea" among the many lemmas and technical calculations. It seems that it's just a lot of unrelated lemmas that happen to work together and happen to give a nice bound at the end.

Therefore, I am wondering if someone can provide a high level overview of the idea behind estimating the major arc contribution? For example, the idea behind bounding the minor arcs would be

  1. Crudely bounding $\int |f(\alpha)|^s d\alpha \ll \left(\sup_{\alpha\in\mathfrak{m}} |f(\alpha)|\right)^{s-2^k} \int_0^1 |f(\alpha)|^{2^k} d\alpha$
  2. Using Weyl's Inequality to bound $|f(\alpha)|$ over the minor arcs and hence the first term
  3. Using Hua's Lemma to reduce the constant in the exponent for the second term (over the trivial bound $\int_{\mathfrak{m}} |f(\alpha)|^{2^k} d\alpha \ll \int_0^1 \left|\sum_{m=1}^{N} e(\alpha m^k)\right|^{2^k} \ll N^{2^k}$).

Thank you and hope this helps others!

P.S. Vaughan's text seems to be quite dense, and I essentially have to think about every mathematical (and non-mathematical) statement he makes for a while. Is that normal?

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    $\begingroup$ please delete the math stack exchange version $\endgroup$
    – kodlu
    Commented Oct 20, 2022 at 14:21
  • $\begingroup$ The motivation is that the function $f(\alpha)$ has good properties when $\alpha$ is close to some rational number, and it is also true that the integral over some small neighborhoods of rationals turn out to be the constituents of the main term in the asymptotic formula. $\endgroup$
    – TravorLZH
    Commented Oct 20, 2022 at 16:22
  • $\begingroup$ As a result, the strategy is to give asymptotic estimates for integral over major arcs (collection of these small neighborhoods of rationals) and to give upper bound for integral over minor arcs (i.e. the integral over region in $[0,1]$ that are relatively distant from rationals) $\endgroup$
    – TravorLZH
    Commented Oct 20, 2022 at 16:23
  • $\begingroup$ @TravorLZH (Hi Travor I actually emailed you before haha) Yeah Travor I understand what you said so far, and in the text I am referring we will prove that $\int_{\mathfrak{M}} f^s(\alpha) e(-\alpha n) d\alpha \ll n^{s/k - 1}$ while $\int_{\mathfrak{m}} f^s(\alpha) e(-\alpha n) d\alpha = o(n^{s/k - 1})$, which will imply that number of representations is positive for large enough $n$. However, what I am asking in this question is the technicalities of the first estimate, like what are the lemmas used and what does each approximately say/do? (See what I wrote for the minor arcs). Hope you get $\endgroup$
    – Gareth Ma
    Commented Oct 20, 2022 at 21:42
  • $\begingroup$ By the way, I have written a Chinese article explaining the application of circle method to Goldbach's problem. Hope you will find that one helpful as well. $\endgroup$
    – TravorLZH
    Commented Oct 20, 2022 at 23:21

1 Answer 1

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The following explanation not only accounts for the treatments in the major arcs of Waring's problem, but also the major arcs in a general situation.

Suppose $F(\alpha)$ is some exponential sum that we wish to extract arithmetical information from. Then our task will be to estimate

$$g(n)=\int_0^1F(\alpha)e(-n\alpha)\mathrm d\alpha.\tag1 $$

Hardy and Littlewood found out that when $\alpha$ is near rational (e.g. $\alpha=a/q+\beta$ for some $(a,q)=1$), $F(\alpha)$ is well approximated by a product of two functions, first dependent on $a$ and $q$ while the other only depends on $\beta$:

$$ F\left(\frac aq+\beta\right)\sim S(q,a)u(\beta)\tag2 $$

provided that $|\beta|$ is very small (e.g. $|\beta|\le1/Q$ for some large $Q$).

Thus, we are motivated to craft the main term of $g(n)$ by summing over contribution of integrals over arcs that are near rationals (i.e. the major arcs):

$$ \mathfrak M(q,a)=\left\{0\le\alpha\le1:\left|\alpha-\frac aq\right|\le\frac1Q\right\} $$

There are infinitely many rationals in $[0,1]$, so we are not going to sum over $\int_{\mathfrak M(q,a)}$ for every rational. Instead, we only estimate the ones that have small denominators (e.g. $q\le P$):

\begin{aligned} g(n) &=\sum_{q\le P}\sum_{\substack{1\le a\le q\\(a,q)=1}}\int_{\mathfrak M(q,a)}F(\alpha)e(-n\alpha)\mathrm d\alpha+\int_{\text{minor arc}} \\ &\approx\sum_{\color{blue}{q\le P}}\sum_{\substack{1\le a\le q\\(a,q)=1}}S(q,a)e(-na/q)\int_\color{blue}{-1/Q}^\color{blue}{1/Q}u(\beta)e(-n\beta)\mathrm d\beta \\ &\approx\underbrace{\sum_{\color{red}{q\ge1}}\sum_{\substack{1\le a\le q\\(a,q)=1}}S(q,a)e(-na/q)}_{\mathfrak S(n)}\underbrace{\int_\color{red}{-1/2}^\color{red}{1/2}u(\beta)e(-n\beta)\mathrm d\beta}_{J(n)} \end{aligned}

where $\mathfrak S(n)$ is the singular series and $J(n)$ is the singular integral, which can be evaluated by extracting combinatorial properties from $u(\beta)$. Therefore, we have $g(n)\approx \mathfrak S(n)J(n)$.

This is the motivation behind all the lemmas emerging in Vaughan's book. Some lemmas are dedicated to deduce (2), and others are intended to estimate the errors emerging from replacing $q\le P$ with $q\ge1$ and replacing $\pm 1/Q$ with $\pm1/2$. I hope this answer can address your concern.

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  • $\begingroup$ Yes this helps a lot more, thanks! I will also take a look at your chinese blog (I read up to like blog 6 before). By the way, is there any connection between sieve methods and circle method? Or are they just different methods of attacking problems $\endgroup$
    – Gareth Ma
    Commented Oct 21, 2022 at 11:23
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    $\begingroup$ They are regarded as 2 different approaches to number-theoretic problems $\endgroup$
    – TravorLZH
    Commented Oct 21, 2022 at 12:03

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