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# 数学代写| Proof of the distribution theorem数论代考

## 数论代考

We will explain the proof of the theorem. We use a slightly different approach than the original article, bypassing the method of moments, and exploiting some simplifications that arise from the consideration of this single example.

The proof will be complete from a probabilistic point of view, but it relies on an extremely deep arithmetic result that we will only be able to view as a black box in this book. The crucial underlying result is the very general form of the Riemann Hypothesis over finite fields, and the formalism that is attached to it. This is due to Deligne, and the
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particular application we use relies extensively on the additional work of Katz. All of this builds on the algebraic-geometric foundations of Grothendieck and his school (see [59, Ch. 11] for an introduction).
In outline, the proof has three steps:

• Step 1: Show that the random Fourier series $\mathrm{K}$ exists, as a $\mathrm{C}([0,1])$-valued random variable;
• Step 2: Prove that (a small variant of the sequence of Fourier coefficients of $\mathrm{K}_{p}$ converges in law to the sequence of Fourier coefficients of $\mathrm{K}$;
• Step 3: Prove that the sequence $\left(\mathrm{K}{p}\right){p}$ is tight (Definition B.3.6), using Kolmogorov’s Tightness Criterion (Proposition B.11.10).
Once this is done, a simple probabilistic statement (Proposition B.11.8, which is a variant of Prokhorov’s Theorem B.11.4) shows that the combination of (2) and (3) implies that $\mathrm{K}_{p}$ converges to $\mathrm{K}$. Both steps (2) and (3) involve non-trivial arithmetic information; indeed, the main input in (2) is exceptionally deep, as we will explain soon.

We denote by $\mathbf{P}{p}$ and $\mathbf{E}{p}$ the probability and expectation with respect to the uniform measure on $\Omega_{p}=\mathbf{F}{p}^{\times} \times \mathbf{F}{p}^{\times}$. Before we begin the proof in earnest, it is useful to see why the limit arises, and why it is precisely this random Fourier series. The idea is to use discrete Fourier analysis to represent the partial sums of Kloosterman sums.
LEMMA 6.2.1. Let $p \geqslant 3$ be a prime and $a, b \in \mathbf{F}{p}^{\times}$. Let $t \in[0,1]$. Then we have (6.2) $\frac{1}{\sqrt{p}} \sum{1 \leqslant n \leqslant(p-1) t} e\left(\frac{a n+b \bar{n}}{p}\right)=\sum_{|h|<p / 2} \alpha_{p}(h, t) \mathrm{Kl}(a-h, b ; p)$,
where
$$\alpha_{p}(h, t)=\frac{1}{p} \sum_{1 \leqslant n \leqslant(p-1) t} e\left(\frac{n h}{p}\right) .$$
PROOF. This is a case of the discrete Plancherel formula, applied to the characteristic (indicator) function of the discrete interval of summation; to check it quickly, insert the definitions of $\alpha_{p}(h, t)$ and of $\mathrm{Kl}(a-h, b ; p)$ in the right hand-side of (6.2). This shows that it is equal to
\begin{aligned} \sum_{|h|<p / 2} \alpha_{p}(h, t) \mathrm{Kl}(a-h, b ; p) &=\frac{1}{p^{3 / 2}} \sum_{|h|<p / 2} \sum_{1 \leqslant n \leqslant(p-1) t} \sum_{m \in \mathbf{F}{p}} e\left(\frac{n h}{p}\right) e\left(\frac{(a-h) m+b \bar{m}}{p}\right) \ &=\frac{1}{\sqrt{p}} \sum{1 \leqslant n \leqslant(p-1) t} \sum_{m \in \mathbf{F}{p}} e\left(\frac{a m+b \bar{m}}{p}\right) \frac{1}{p} \sum{h \in \mathbf{F}{p}} e\left(\frac{h(n-m)}{p}\right) \ &=\frac{1}{\sqrt{p}} \sum{1 \leqslant n \leqslant(p-1) t} e\left(\frac{a n+b \bar{n}}{p}\right), \end{aligned}
as claimed, since by the orthogonality of characters we have
$$\frac{1}{p} \sum_{h \in \mathbf{F}{p}} e\left(\frac{h(n-m)}{p}\right)=\delta(n, m)$$ for any $n, m \in \mathbf{F}{p}$, where $\delta(n, m)=1$ if $n=m$ modulo $p$, and otherwise $\delta(n, m)=0$.
If we observe that $\alpha_{p}(h, t)$ is essentially a Riemann sum for the integral
$$\int_{0}^{t} e(h t) d t=\frac{e(h t)-1}{2 i \pi h}$$
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• 步骤 1：证明随机傅立叶级数 $\mathrm{K}$ 存在，作为 $\mathrm{C}([0,1])$ 值的随机变量；
• 步骤 2：证明 ($\mathrm{K}_{p}$ 的傅里叶系数序列的一个小变体在法律上收敛到 $\mathrm{K}$ 的傅里叶系数序列；
• 步骤 3：证明序列 $\left(\mathrm{K}{p}\right){p}$ 是紧的（定义 B.3.6），使用 Kolmogorov 的紧度准则（命题 B.11.10）。
一旦完成，一个简单的概率陈述（命题 B.11.8，它是 Prokhorov 定理 B.11.4 的一个变体）表明（2）和（3）的组合意味着 $\mathrm{K}_{p}$ 收敛到 $\mathrm{K}$。步骤（2）和（3）都涉及非平凡的算术信息；事实上，（2）中的主要输入非常深，我们很快就会解释。

(6.2) $\frac{1}{\sqrt{p}} \sum_{1 \leqslant n \leqslant(p-1) t} e\left(\frac{a n+b \bar{n}}{ p}\right)=\sum_{|h|<p / 2} \alpha_{p}(h, t) \mathrm{Kl}(ah, b ; p)$,

$$\alpha_{p}(h, t)=\frac{1}{p} \sum_{1 \leqslant n \leqslant(p-1) t} e\left(\frac{nh}{p}\right) .$$

$$\开始{对齐} \sum_{|h|<p / 2} \alpha_{p}(h, t) \mathrm{Kl}(ah, b ; p) &=\frac{1}{p^{3 / 2}} \ sum_{|h|<p / 2} \sum_{1 \leqslant n \leqslant(p-1) t} \sum_{m \in \mathbf{F}{p}} e\left(\frac{nh }{p}\right) e\left(\frac{(ah) m+b \bar{m}}{p}\right) \ &=\frac{1}{\sqrt{p}} \sum{1 \leqslant n \leqslant(p-1) t} \sum_{m \in \mathbf{F}{p}} e\left( \frac{a m+b \bar{m}}{p}\right) \frac{1}{p} \sum{h \in \mathbf{F}{p}} e\left(\frac{ h(nm)}{p}\right) \ &=\frac{1}{\sqrt{p}} \sum{1 \leqslant n \leqslant(p-1) t} e\left(\frac{a n+b \bar{n}}{p} \正确的）， \end{对齐}$$

$$\frac{1}{p} \sum_{h \in \mathbf{F}{p}} e\left(\frac{h(n-m)}{p}\right)=\delta(n, m)$$ 对于任何 $n，m \in \mathbf{F}{p}$，其中 $\delta(n, m)=1$ 如果 $n=m$ 模 $p$，否则 $\delta(n, m)=0$。

$$\int_{0}^{t} e(h t) d t=\frac{e(h t)-1}{2 i \pi h}$$
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## 数论代写

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## 编码理论代写

1. 数据压缩（或信源编码
2. 前向错误更正（或信道编码
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## 复分析代考

(1) 提到复变函数 ，首先需要了解复数的基本性左和四则运算规则。怎么样计算复数的平方根， 极坐标与 $x y$ 坐标的转换，复数的模之类的。这些在高中的时候囸本上都会学过。
(2) 复变函数自然是在复平面上来研究问题，此时数学分析里面的求导数之尖的运算就会很自然的 引入到复平面里面，从而引出解析函数的定义。那/研究解析函数的性贡就是关楗所在。最关键的 地方就是所谓的Cauchy一Riemann公式，这个是判断一个函数是否是解析函数的关键所在。
(3) 明白解析函数的定义以及性质之后，就会把数学分析里面的曲线积分 $a$ 的概念引入复分析中， 定义几乎是一致的。在引入了闭曲线和曲线积分之后，就会有出现复分析中的重要的定理: Cauchy 积分公式。 这个是易分析的第一个重要定理。