Question about the Hardy-Littlewood method (quite basic) - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T00:08:49Z http://mathoverflow.net/feeds/question/119937 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/119937/question-about-the-hardy-littlewood-method-quite-basic Question about the Hardy-Littlewood method (quite basic) Tomos Parry 2013-01-26T14:53:38Z 2013-01-26T23:23:47Z <p>Hi, I have a question about the Hardy-Littlewood method. </p> <p>Writing $R_s(n)$ for the number of ways to write $n$ as a sum of $s$ $k$-th powers and $f(\alpha )$ for the sum $\sum _{m=1}^Ne(\alpha m^k)$, we have </p> <p>$R_s(n)=\int _\mathfrak Uf(\alpha )^se(-\alpha n)d\alpha ,$</p> <p>where $\mathfrak U$ is some unit interval. The aim is to work out an asymptotic expression for $R_s(n)$ by studying this integral. We split the domain of integration into the major and minor arcs:</p> <p>$R_s(n)=\int _\mathfrak Mf(\alpha )^se(-\alpha n)d\alpha +\int _\mathfrak mf(\alpha )^se(-\alpha n)d\alpha$</p> <p>and aim to work out an asymptotic expressions for the integral over the major arcs whilst making sure the minor arc contributions are "not too big".</p> <p>Now, this rests on the fact that $f(\alpha )$ gets "big" near a rational number, and moreover gets "more big" the smaller the denominator of the rational number, so that the major arcs, being intervals around rationals with "small" denominators, give the biggest contribution. My problem essentially is, I think, that I don't understand why this should be true; why does $f$ get "big" near rationals, that is.</p> <p>In Vaughan's book, we approximate $f(\alpha )$ near these rationals through the function $v(\beta )$ (page 14). I think Lemma 2.7 contains what I don't understand. For example, does this lemma say that</p> <p>$f(\alpha )\sim \frac {S(q,a)}{q}v(\alpha -a/q)$, as $n\rightarrow \infty$?</p> <p>And how exactly does it influence our definition of the major arcs and our choice of parameter $v$ in the definition of the major arcs? I assume that $v$ is chosen small enough to gain a saving on the estimate $n^{s/k-1}$ in (2.13) on page 16, and that having $q\leq N^v,\alpha \in \mathfrak M(q,a)$ ensures we get an error no larger than $N^{2v}$ in the lemma, but I still feel I'm missing the big picture in the analysis somehow.</p> <p>Perhaps I don't have a specific problem as such but want to clarify the situation a little bit; perhaps also it is a problem in that I'm simply not quite settled in the fact that $f(\alpha )$ is large at rationals with small denominators.</p> <p>In any case, I'd appreciate any thoughts/clarifications/things to think about. Thanks very much.</p> http://mathoverflow.net/questions/119937/question-about-the-hardy-littlewood-method-quite-basic/119971#119971 Answer by Freddie Manners for Question about the Hardy-Littlewood method (quite basic) Freddie Manners 2013-01-26T23:16:47Z 2013-01-26T23:23:47Z <p>I'll split this into two questions:</p> <ol> <li>Why can $f$ be large near rationals with small denominator?</li> <li>Why is $f$ small away from these points (i.e. on the minor arcs)?</li> </ol> <p>For (1), I should first clarify that $f$ isn't <em>always</em> large at these rationals; it could well be zero. But it is large some of the time.</p> <p>Let's look at the example of <code>$f(a/8)$</code> for <code>$a \in \{1, 3, \dots, 7\}$</code>, when (say) $k=2$. We have:</p> <p><code>$f(a/8) = \sum_{n=1}^N e(a\, n^2 / 8)$</code></p> <p>Since $e(an/8)$ only detects the residue of $n \bmod{8}$, we have:</p> <p><code>$f(a/8) = \frac{N}{8} \sum_{n \in \mathbb{Z} / 8 \mathbb{Z}} e(a\, n^2 / 8) + O(1)$</code></p> <p>But $n^2$ is congruent to $0 \bmod{8}$ a quarter of the time, $4 \bmod{8}$ a quarter of the time and $1 \bmod{8}$ half the time, so:</p> <p><code>$f(a/8) = N \left(\frac{1}{4} + \frac{1}{4} e(a/2) + \frac{1}{2} e(a/8) \right) + O(1) = \frac{N}{2} e(a/8) + O(1)$</code></p> <p>(recalling $a$ was odd). That's going to be pretty large (absolute value about $N/2$). The reason is essentially that $f(a/8)$ is detecting a huge bias in the squares, namely that many more than average are $1 \bmod 8$.</p> <p>The same's going to happen for other denominators: only about half of all residues modulo $q$ get to be squares (for $q$ a prime power), and this bias causes large values of $f(a/q)$.</p> <p>Another way of saying exactly the same is that $(n^2 / 8) \pmod{1}$ has a very non-uniform distribution, with a lot of the mass bunched at $1/8$.</p> <p>This brings us to (2). The question is: might the squares have similar bias "<code>$(\bmod{\sqrt{2}})$</code>"? I have no idea where such a bias might plausibly come from, and fortunately in any case the answer is no. That is, if you were to draw the set:</p> <p><code>$\left\{ n^2 / \sqrt{2} \pmod{1}\,:\, n \in \{1 \dots N\} \right\}$</code></p> <p>as a subset of $[0, 1)$, you'd find they were smeared out all over the place. More formally, the proportion of these points contained in any interval $[a, b]$ is roughly $(b-a)$, and we say the set is <em>equidistributed</em> modulo 1. As $N$ becomes large, this equidistribution becomes quantitatively better.</p> <p>Now, by Weyl's Equidistribution Criterion (many good references on the web), my previous statement is essentially equivalent to $f(a / b \sqrt{2})$ being small, whenever $a, b \in \mathbb{Z}$ are smallish. (Intuitively, you get a lot of cancellation in the definition of $f$.) In practice, it's usual to do the converse to this, i.e. deduce equidistribution from bounds on $f(a / b \sqrt{2})$; but I think equidistribution shows more intuitively what is going on in the minor arc bounds.</p> <p>How do you prove it? Well, there's some work to be done at some point. Proving Weyl's Inequality gives quite a general bound of this form. Another (essentially equivalent) statement is Van der Corput's lemma (see e.g. the first half of <a href="http://terrytao.wordpress.com/2008/06/14/the-van-der-corputs-trick-and-equidistribution-on-nilmanifolds/" rel="nofollow">Terry Tao's post on this</a>):</p> <blockquote> <p>If $a_n$ is a real sequence such that $(a_{n+h} - a_n)$ equidistributes $\bmod{1}$ for every non-zero $h \in \mathbb{Z}$, then $a_n$ equidistributes $\bmod{1}$.</p> </blockquote> <p>In our case, $((n + h)^2 - n^2) / \sqrt{2} = (2 h n + h^2) / \sqrt{2}$ is reasonably easily shown to equidistribute $\bmod{1}$, for fixed $h$.</p> <p>Hope some of that helps.</p>