### 数学代写|解析数论作业代写Analytic Number Theory代考|MATH4307

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• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

## 数学代写|解析数论作业代写Analytic Number Theory代考|The Euclidean algorithm

Iheorem $1.12$ provides a practical method for computing the $\operatorname{gcd}(a, b)$ when the prime-power factorizations of $a$ and $b$ are known. However, considerable calculation may be required to obtain these prime-power factorizations and it is desirable to have an alternative procedure that requires less computation. There is a useful process, known as Euclid’s algorithm, which does not require the factorizations of $a$ and $b$. This process is based on successive divisions and makes use of the following theorem.

Theorem 1.14 The division algorithm. Given integers $a$ and $b$ with $b>0$, there exists a unique pair of integers $q$ and $r$ such that
$$a=b q+r, \text { with } 0 \leq r<b .$$
Moreover, $r=0$ if, and only if, $b \mid a$.
Note. We say that $q$ is the quotient and $r$ the remainder obtained when $b$ is divided into $a$.

## 数学代写|解析数论作业代写Analytic Number Theory代考|A product formula for n

The sum for $\varphi(n)$ in Theorem $2.3$ can also be expressed as a product extended over the distinct prime divisors of $n$.
Theorem $2.4$ For $n \geq 1$ we have
$$\varphi(n)=n \prod_{p \mid n}\left(1-\frac{1}{p}\right) .$$
Proof. For $n=1$ the product is empty since there are no primes which divide 1 . In this case it is understood that the product is to be assigned the value $1 .$

Suppose, then, that $n>1$ and let $p_{1}, \ldots, p_{r}$ be the distinct prime divisors of $n$. The product can be written as
(4)
\begin{aligned} \prod_{p \mid n}\left(1-\frac{1}{p}\right) &=\prod_{i=1}^{r}\left(1-\frac{1}{p_{i}}\right) \ &=1-\sum \frac{1}{p_{i}}+\sum \frac{1}{p_{i} p_{j}}-\sum \frac{1}{p_{i} p_{j} p_{k}}+\cdots+\frac{(-1)^{r}}{p_{1} p_{2} \cdots p_{r}} . \end{aligned}
On the right, in a term such as $\sum 1 / p_{i} p_{j} p_{k}$ it is understood that we consider all possible products $p_{i} p_{j} p_{k}$ of distinct prime factors of $n$ taken three at a time. Note that each term on the right of (4) is of the form $\pm 1 / d$ where $d$ is a divisor of $n$ which is either 1 or a product of distinct primes. The numerator $\pm 1$ is exactly $\mu(d)$. Since $\mu(d)=0$ if $d$ is divisible by the square of any $p_{i}$ we see that the sum in (4) is exactly the same as
$$\sum_{d \mid n} \frac{\mu(d)}{d} .$$

## 数学代写|解析数论作业代写Analytic Number Theory代考|The Euclidean algorithm

$$a=b q+r, \text { with } 0 \leq r<b .$$

## 数学代写|解析数论作业代写Analytic Number Theory代考|A product formula for n

$$\varphi(n)=n \prod_{p \mid n}\left(1-\frac{1}{p}\right)$$

(4)
$$\prod_{p \mid n}\left(1-\frac{1}{p}\right)=\prod_{i=1}^{r}\left(1-\frac{1}{p_{i}}\right) \quad=1-\sum \frac{1}{p_{i}}+\sum \frac{1}{p_{i} p_{j}}-\sum \frac{1}{p_{i} p_{j} p_{k}}+\cdots+\frac{(-1)^{r}}{p_{1} p_{2} \cdots p_{r}}$$

$$\sum_{d \mid n} \frac{\mu(d)}{d} .$$

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## MATLAB代写

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