数学代写|数论作业代写number theory代考|MATH4307

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• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

数学代写|数论作业代写number theory代考|Answers to Selected Supplementary Problems

1.13. (a) True since $644=7(92)$. (b) True since $644=-7(-92)$.
(c) False since the remainder is 4 , not $0 . \quad$ (d) True since $2916=$ $243(12)$
(e) True since $16 m-12=4(4 m-3)$. (f) True since $-7 m=$ $m(-7)$.
(g) False since the remainder is $3 m$, not 0 . (h) True since $-k^{3}+$ $2 k=k\left(-k^{2}+2\right)$.
1.15. (a) $487=14(34)+11$, i.e., $q=34$ and $r=11$.
(b) $-386=27(-15)+19$, i.e., $q=-15$ and $r=19$.
1.16. Since $486=15 a+6$, we get $a=480 / 15=32$.
1.17. $44=2^{2} \cdot 11,111=3 \cdot 37$, so $\operatorname{gcd}(44,111)=1$.
1.19. (a) $F_{9}=34, F_{10}=55, F_{11}=89, F_{12}=144$.
1.20. (a) $84 . \quad$ (b) $525 .$
1.22. (a) $44=\left(2^{2}\right)(11)$ and $111=(3)(37)$, so $\operatorname{gcd}(44,111)=1$.
(b) $111=44(2)+23,44=23(1)+21,23=21(1)+2,21=2(10)+1$.
1.23. By the Euclidean Algorithm, we have $71=23(3)+2$ and $23=2(11)+1$.

Hence $1=23-2(11)=23-(71-23(3))(11)=23(34)+71(-11)$, so $x=34$ and $y=-11$.

数学代写|数论作业代写number theory代考|Listing Primes: The Sieve of Eratosthenes

A second question we now ask is, How can we list all of the prime numbers up to some positive value $n \geq 2$ ? A method to do this is known as the Sieve of Eratosthenes, named in honor of Eratosthenes $(276 \mathrm{BC}-194 \mathrm{BC})$, who appears to be the first to make use of this process. The process, described below, is quite efficient as long as $n$ isn’t too large.

We begin by listing all of the numbers from 2 to $n$. Then since 2 is prime, we leave it in the list and delete all multiples of 2 (except 2 itself) up to and including $n$. That knocks out all the even numbers in our list larger than 2 . We then leave 3 and delete all larger multiples of 3 . The next value not already deleted is 5 , so we leave it and delete all multiples of 5 . We continue this process with 7 which is yet to be deleted, then 11, ctc. The numbers remaining in the list give all primes up to $n$. A question you might have is, When can we stop this process so that we have indeed listed all the primes up to $n$ ? You are asked in Problem $2.2$ to show that we need only process primes which are less than or equal to the square root of $n$.

Example 2.2. We illustrate the Sieve of Eratosthenes by finding all primes up to $n=50$. We begin by listing all of the positive integers from 2 through 50 . By what we just stated, we need only process $2,3,5$, and 7 since $11>\sqrt{50}$.
$\begin{array}{rrrrrrrrrr} & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \ 11 & 12 & 13 & 14 & 15 & 16 & 17 & 18 & 19 & 20 \ 21 & 22 & 23 & 24 & 25 & 26 & 27 & 28 & 29 & 30 \ 31 & 32 & 33 & 34 & 35 & 36 & 37 & 38 & 39 & 40 \ 41 & 42 & 43 & 44 & 45 & 46 & 47 & 48 & 49 & 50\end{array}$
We boldface the number 2 as the first item in this list (since we know it is prime), and then cross out each multiple of 2 that is greater than 2. It is important to note that no actual arithmetic must be done here! We simply start at 2, skip by the amount of 2 (which gets us to the number 4), cross out the 4 , then skip by another 2 to get to 6 , cross out the 6 , and so on. This stage of the process is quite straightforward. This now leaves us with the following table.

数学代写|数论作业代写number theory代考|Answers to Selected Supplementary Problems

1.13。 (a) 真因为 $644=7(92)$. (b) 真因为 $644=-7(-92)$.
(c) 假，因为余数是 4 ，不是 $0 . \quad$ (d) 真因为 $2916=243(12)$
(e) 真因为 $16 m-12=4(4 m-3)$. (f) 真因为 $-7 m=m(-7)$.
(g) 假，因为余数是 $3 m$ ，而不是 0 。(h) 真因为 $-k^{3}+2 k=k\left(-k^{2}+2\right)$.
1.15。(一个) $487=14(34)+11$ ，那是， $q=34$ 和 $r=11$.
(二) $-386=27(-15)+19$ ， 那是， $q=-15$ 和 $r=19$.
1.16。自从 $486=15 a+6$ ，我们得到 $a=480 / 15=32$.
1.17. $44=2^{2} \cdot 11,111=3 \cdot 37$ ，所以 $\operatorname{gcd}(44,111)=1$.
1.19。 (-个) $F_{9}=34, F_{10}=55, F_{11}=89, F_{12}=144$.
1.20。(一个) 84 . (b) 525 .
1.22。(一个) $44=\left(2^{2}\right)(11)$ 和 $111=(3)(37)$ ，所以 $\operatorname{gcd}(44,111)=1$.
(二) $111=44(2)+23,44=23(1)+21,23=21(1)+2,21=2(10)+1$.
1.23。根据欧几里得算法，我们有 $71=23(3)+2$ 和 $23=2(11)+1$.

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MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。