Waring’s Problem: Lecture 4

In which we show that the minor arcs contribute a negligible amount, and study the behaviour at a point in the major arcs.

  • Proposition 5 (The contribution from the minor arcs): Suppose that Q \gg N^{\gamma} and T \gg N^{\gamma} for some \gamma > 0.  Then for sufficiently large s (depending on k and \gamma), we have \displaystyle \int_{\mathfrak{m}} \widehat{1_S}(\theta)^s e(-N \theta) \mathrm{d}\theta = o(N^{\frac{s}{k}-1})We proved this using our bound from last time on |\widehat{1_S}(\theta)|, together with Parseval’s identity (which here comes down to being the statement that \displaystyle \int_0^1 |\widehat{1_S}(\theta)|^2 \mathrm{d}\theta = N^{\frac{1}{k}}).  This result does not give us a good value of s; later in the course, we shall see Hua’s lemma, and this result will enable us to get a much better value of s.
  • Definition of S(a,q) := \mathbb{E}_{n \pmod{q}} e(\frac{a}{q}n^k) = \frac{1}{q} \sum_{n=0}^{q-1} e(\frac{a}{q}n^k).
  • Lemma 6: Let I \subseteq [N^{\frac{1}{k}}] be an interval.  Then \displaystyle \sum_{n \in I} e(\frac{a}{q}n^k) = |I| S(a,q) + O(q). We saw that the key idea here is that the value of e(\frac{a}{q}n^k) depends only on the congruence class of n modulo q, so we partitioned I accordingly.
  • Lemma 7: Let f : \mathbb{R} \to \mathbb{C} be a differentiable function.  Then \displaystyle |\sum_{j=1}^J f(j) - \int_0^J f(x) \mathrm{d}x| \leq J \max_{y \in [0,J]} |f'(y)| + \max_{y \in [0,J]} |f(y)|.  This was quite a crude bound; we did not do much work to prove it (we just split [0,J] into intervals of length 1 and used an easy bound on each one).  Later in the course, we shall see a better way to approximate a sum by an integral, but this will do for now.
  • Lemma 8: Let \theta = \frac{a}{q} + t be a point in the major arc \mathfrak{M}_{a,q}.  Then \displaystyle \widehat{1_S}(\theta) = S(a,q) \int_0^{N^{\frac{1}{k}}} e(ty^k) \mathrm{d}y + O(N^{\frac{1}{2k}} Q^{\frac{1}{2}} T^{\frac{1}{2}}).  We partitioned [N^{\frac{1}{k}}] into shorter intervals on which we could treat tn^k as approximately constant, and then used Lemma 7 to replace the resulting sum by an integral.

Further reading

There’s a colourful picture showing major arcs on the Wikipedia page for the Hardy-Littlewood circle method.

Preparation for Lecture 5

Next time, we shall put together the contributions from individual points on the major arcs, by integrating.  You might like to try this for yourself before the lecture.

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4 Responses to “Waring’s Problem: Lecture 4”

  1. Waring’s Problem: Lecture 5 « Theorem of the week Says:

    […] Theorem of the week Expositions of interesting mathematical results « Waring’s Problem: Lecture 4 […]

  2. Waring’s Problem: Lecture 6 « Theorem of the week Says:

    […] 13: For sufficiently large (depending on ), we have We saw that this follows immediately from Proposition 5 and Proposition […]

  3. Waring’s Problem: Lecture 11 « Theorem of the week Says:

    […] saw that this enables us to improve some earlier bounds.  In particular, in Proposition 5 we could show that the contribution from the minor arcs is negligible if , and in Theorem 24 and […]

  4. Waring’s Problem: Lecture 11 « Theorem of the week Says:

    […] saw that this enables us to improve some earlier bounds.  In particular, in Proposition 5 we could show that the contribution from the minor arcs is negligible if , and in Theorem 24 and […]

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