•  MATH 626 – Analytic Number Theory – Fall 2023
• UBC MATH 613D: Analytic Number Theory II
• Analytic Number Theory – OSU Math – The Ohio State University MATH 7122.01: Analytic Number Theory

If $p$ is a prime, the discussion of the congruence $x^2 \equiv a(p)$ is fairly easy. It is solvable iff $a^{(p-1) / 2} \equiv 1(p)$. With this fact in hand a complete analysis is a simple matter. However, if the question is turned around, the problem is much more difficult. Suppose that $a$ is an integer. For which primes $p$ is the congruence $x^2 \equiv a(p)$ solvable? The answer is provided by the law of quadratic reciprocity. This law was formulated by Euler and A. M. Legendre but Gauss was the first to provide a complete proof. Gauss was extremely proud of this result. He called it the Theorema Aureum, the golden theorem.

20 世纪 20 年代，范德尔科普特提出了一种估算三角和的方法，这种方法在数论中应用广泛。这一理论通常与狄利克雷的除数问题和圆问题有关–即对$(m, n) \in\left(\mathbb{Z}^{+}\right)^2$ 这样的$m^2+n^2 \leqslant x$ 的对数的评估。它更普遍地适用于给定平面轮廓内积分坐标点的计数。

We shall make use of the Poisson summation formula, which undoubtedly constitutes the most natural approach to the study of trigonometric sums. We define the Fourier transform by
$$\widehat{f}(\vartheta):=\int_{-\infty}^{+\infty} f(t) \mathrm{e}(-\vartheta t) \mathrm{d} t \quad\left(f \in L^1(\mathbb{R})\right) .$$
The following statement provides a precise framework for our usage of Poisson’s formula.

Theorem 1. Let $f \in L^1(\mathbb{R})$. Assume that the series
$$\varphi(t)=\sum_{n \in \mathbb{Z}} f(n+t)$$
converges for all $t$, and that its sum defines a function continuous at 0 and of bounded variation on $[0,1]$. Then we have
$$\lim {N \rightarrow \infty} \sum{|\nu| \leqslant N} \widehat{f}(\nu)=\sum_{n \in \mathbb{Z}} f(n) .$$

Proof. Jordan’s theorem on Fourier series asserts that any periodic function of bounded variation over a period is the sum of its Fourier series at each point of continuity. This implies $\lim {N \rightarrow \infty} \sum{|\nu| \leqslant N} \widehat{f}(\nu)=\sum_{n \in \mathbb{Z}} f(n)$. since the Fourier coefficient of order $\nu$ of $\varphi$ is $\widehat{f}(\nu)$.

bounding trigonometric integrals

Let $f \in \mathcal{C}^1(] a, b[)$ be such that $f^{\prime}(t)$ is monotonic, of constant sign on $] a, b\left[\right.$, and such that $m:=\inf _{a0}$. Then we have
$$\left|\int_a^b \mathrm{e}(f(t)) \mathrm{d} t\right| \leqslant \frac{1}{\pi m} \text {. }$$
Proof. Assuming without loss of generality that $f^{\prime}$ is non-increasing on ]$a, b[$, we can write
\begin{aligned} \left|2 \pi \int_a^b \mathrm{e}(f(t)) \mathrm{d} t\right| & =\left|\int_a^b \frac{\mathrm{d}{\mathrm{e}(f(t))}}{f^{\prime}(t)}\right|=\left|\left[\frac{\mathrm{e}(f(t))}{f^{\prime}(t)}\right]_a^b-\int_a^b \mathrm{e}(f(t)) \mathrm{d}\left{\frac{1}{f^{\prime}(t)}\right}\right| \ & \leqslant \frac{1}{\left|f^{\prime}(a)\right|}+\frac{1}{\left|f^{\prime}(b)\right|}+\frac{1}{\left|f^{\prime}(b)\right|}-\frac{1}{\left|f^{\prime}(a)\right|} \leqslant \frac{2}{m} . \end{aligned}

Theorem 2.

Let $f \in \mathcal{C}^2(] a, b[)$ be such that $f^{\prime \prime}(t)$ has constant sign on ]$a, b\left[\right.$ and set $r:=\inf {a0$. Then we have $$\left|\int_a^b \mathrm{e}(f(t)) \mathrm{d} t\right| \leqslant \frac{4}{\sqrt{\pi r}} .  Proof. Suppose, for example, that f^{\prime \prime}(t) \leqslant-r<0 for a{c-\delta}^{c+\delta}+\int_{c+\delta}^b=I_1+I_2+I_3 say, where the positive parameter \delta satisfies a+\delta \leqslant c \leqslant b-\delta. We have$$
\left|f^{\prime}(t)\right|=\left|\int_c^t f^{\prime \prime}(v) \mathrm{d} v\right| \geqslant r|t-c| \geqslant r \delta
$$when t is in [a, c-\delta] \cup[c+\delta, b]. it follows that$$
\left|I_1\right|+\left|I_3\right| \leqslant 2 / \pi r \delta \text {. }
$$Since, trivially, \left|I_2\right| \leqslant 2 \delta, we get$$
|I| \leqslant 2 \delta+2 / \pi r \delta
$$whence the required result, by selecting \delta=\sqrt{1 / \pi r}. It is clear that this upper bound remains valid if, with this choice of \delta, we have either cb-\delta. van der Corput’s inequality Theorem 3. (van der Corput’s inequality). Let a, b \in \mathbb{R} be such that a0 \quad(a<t<b) .  Then we have$$
\sum_{a<n \leqslant b} \mathrm{e}(f(n)) \ll(b-a+1) \lambda^{1 / 2}+\lambda^{-1 / 2} .
$$Proof. We can assume \lambda \leqslant 1, since otherwise it is trivially satisfied. With the notation of above Theorem, the left-hand side of it is$$
\begin{aligned}
& \ll(\beta-\alpha+1) \max _\nu\left|\int_a^b \mathrm{e}(f(t)-\nu t) \mathrm{d} t\right|+\ln (\beta-\alpha+2) \
& \ll(\beta-\alpha+1) \lambda^{-1 / 2}+\ln (\beta-\alpha+2),
\end{aligned}
$$where the second upper bound follows from above thm Now we have$$
\beta-\alpha=\left|\int_a^b f^{\prime \prime}(t) \mathrm{d} t\right| \asymp \lambda(b-a) .
$$The previous bound is therefore$$
\lambda^{1 / 2}(b-a)+\lambda^{-1 / 2}+1+\lambda(b-a) \ll \lambda^{1 / 2}(b-a)+\lambda^{-1 / 2} .
$$(Weyl-van der Corput). For all integers N \geqslant 1, Q \geqslant 1 and any complex sequence \left{z_n\right}_{n=1}^N, we have$$
\left|\sum_{1 \leqslant n \leqslant N} z_n\right|^2 \leqslant\left(1+\frac{N-1}{Q}\right) \sum_{|q|<Q}\left(1-\frac{|q|}{Q}\right) \sum_{1 \leqslant n, n+q \leqslant N} z_{n+q} \overline{z_n} .
$$Proof. Let us put z_n=0 for n \notin[1, N]. With this convention, we have$$
Q \sum_{n \in \mathbb{Z}} z_n=\sum_{1 \leqslant q \leqslant Q} \sum_{n \in \mathbb{Z}} z_{n+q}=\sum_{n \in \mathbb{Z}} \sum_{1 \leqslant q \leqslant Q} z_{n+q} .
$$Only integers n satisfying 1-Q \leqslant n \leqslant N-1 may contribute to the last inner sum. Hence there are at most N-1+Q such integers. By the Cauchy-Schwarz inequality, we obtain$$
\begin{aligned}
Q^2\left|\sum_{n \in \mathbb{Z}} z_n\right|^2 & \leqslant(N-1+Q) \sum_{n \in \mathbb{Z}}\left|\sum_{1 \leqslant q \leqslant Q} z_{n+q}\right|^2 \
& \leqslant(N-1+Q) \sum_{1 \leqslant q_1 \leqslant Q} \sum_{1 \leqslant q_2 \leqslant Q} \sum_{n \in \mathbb{Z}} z_{n+q_1} \overline{z_{n+q_2}} \
& \leqslant(N-1+Q) \sum_{1 \leqslant q_1 \leqslant Q} \sum_{1 \leqslant q_2 \leqslant Q} \sum_{m \in \mathbb{Z}} z_{m+q_1-q_2} \overline{z_m} \
& \leqslant(N-1+Q) \sum_{-Q \leqslant q \leqslant Q} r(q) \sum_{m \in \mathbb{Z}} z_{m+q} \overline{z_m}
\end{aligned}
$$where r(q)=\left|\left{\left(q_1, q_2\right): 1 \leqslant q_1 \leqslant Q, 1 \leqslant q_2 \leqslant Q, q_1-q_2=q\right}\right|. We plainly have$$
r(-q)=r(q)=\left|\left{q_1: \max (1, q+1) \leqslant q_1 \leqslant \min (Q, Q+q)\right}\right|=Q-|q| .
$$This implies on dividing by Q^2. 下面是一些经典的解析数论代考van der Corput method的题目 Problem 1. Erdös-Turán inequality. Denote by$$
F_H(t):=\sum_{|h| \leqslant H}\left(1-\frac{|h|}{H}\right) \cos (2 \pi h t)=\frac{1}{H}\left(\frac{\sin \pi H t}{\sin \pi t}\right)^2
$$the Fejér kernel of order H \in \mathbb{N}^*. Define the Fourier coefficient of index h \in \mathbb{Z} of a function f \in L^1(\mathbb{T}) by$$
c_h(f):=\int_{\mathrm{T}} \mathrm{e}(-h x) f(x) \mathrm{d} x,
$$with \mathrm{e}(u):=\exp (2 \pi i u). For f, g \in L^1(\mathrm{~T}) define the convolution product$$
f * g(x):=\int_{\mathbf{T}} f(t) g(x-t) \mathrm{d} t .
$$Finally, denote by \langle x\rangle the fractional part of a real number x. (a) Let f \in L^1(\mathrm{~T}) be a real-valued function satisfying, for a suitable integer H \geqslant 4,$$
$$Show that f * F_H=0. Deduce that, for all x \in \mathbb{T}, we have$$
\begin{aligned}
f(x)(1-I)+\int_{|t| \leqslant 2 / H} & {f(x-t \pm 2 / H)-f(x)} F_H(t) \mathrm{d} t \
& +\int_{2 / H<|t| \leqslant 1 / 2} f(x-t \pm 2 / H) F_H(t) \mathrm{d} t=0, \end{aligned} $$where I:=\int_{|t|>2 / H} F_H(t) \mathrm{d} t. (b) Show that I \leqslant \frac{1}{4}. (c) Assume that f is bounded on \mathrm{T} and set$$
K:=\sup {\substack{0 \leqslant y \leqslant 1 / H \ x \in \mathbf{T}}}{f(x+y)-f(x)} . $$Show that, under the assumption, we have$$ |f|{\infty} \leqslant 4 K(1-I) /(1-2 I) \leqslant 6 K .
$$(d) Show that, without assuming , we have$$
|f|_{\infty} \leqslant 6 K+13 \sum_{|h| \leqslant H}\left|c_h(f)\right| .
$$(e) Let \left{u_n\right}_{n=1}^N be a finite sequence of real numbers. Define$$
F(x):=\frac{1}{N} \sum_{1 \leqslant n \leqslant N} \mathbf{1}_{[0, x]}\left(\left\langle u_n\right\rangle\right) .
$$By applying the previous results to the function$$
f(x):=\langle x\rangle-\frac{1}{2}+c_0(F)-F(x),
$$show that we have, for every integer H \geqslant 1,$$
\sup {I \subset \mathbf{T}}\left|\frac{1}{N} \sum{1 \leqslant n \leqslant N} \mathbf{1}I\left(\left\langle u_n\right\rangle\right)-\right| I|| \leqslant \frac{12}{H}+9 \sum{0<h \leqslant H} \frac{\left|\sigma_N(h)\right|}{h},
$$where \sigma_N(h):=(1 / N) \sum_{1 \leqslant n \leqslant N} \mathrm{e}\left(h u_n\right). Problem 2. With the k-th derivatives. Let k be an integer \geqslant 3, let I be an interval containing N integers and let f \in \mathrm{C}^k(I, \mathbb{R}). Suppose there exist \lambda>0 and C>1 such that \lambda \leqslant\left|f^{(k)}(x)\right| \leqslant C \lambda for x \in I. Show by induction on k that$$
\left|\sum_{n \in I} \mathrm{e}(f(n))\right| \ll N \lambda^{\alpha_k}+N^{1-\alpha_k} \lambda^{-\alpha_k}

with $\alpha_k:=1 /\left{2^k-2\right}$.

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