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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|APPROXIMATION OF REAL NUMBERS BY RATIONALS

Instead of relying on Cantor’s countability argument we can go back to Liouville’s earlier proofs. These make it possible to explicitly construct transcendental numbers and are therefore, from the number-theoretic point of view, more interesting than Cantor’s proof. They also avoid the intuitionist objections mentioned above – though perhaps not completely so, as they do make use of the Mean Value Theorem from elementary calculus.
Joseph Liouville
Liouville’s methods derive from an investigation of
$(1809-1882)$
the problem of approximating real numbers by rationals. Let $\alpha \in \mathbb{R}$; we wish to ask how closely $\alpha$ can be approximated by rational numbers $p / q$. That is, we want to know how small
$$\left|\alpha-\frac{p}{q}\right|$$
can be made by a suitable choice of the rational $p / q$. Unfortunately, this problem is too easy to be of any interest: as the rationals are dense in $\mathbb{R}$, the difference (3.4), for any $\alpha$, can be made as small as desired by choosing a large value of $q$ and an appropriate $p$. Specifically, if we want the difference to be smaller than a positive number $\varepsilon$, we choose $q>1 / 2 \varepsilon$ and let $p$ be the closest integer to $q \alpha$. Then
$$|q \alpha-p| \leq \frac{1}{2} \quad \Rightarrow \quad \alpha-\frac{p}{q} \mid \leq \frac{1}{2 q}<\varepsilon .$$
This observation, though not very interesting in itself, may suggest a more significant approach, namely, to insist that the closeness of approximation should depend on the denominator of the approximating fraction. In other words, we shall be interested in a fairly weak approximation if it is given by a fraction with very small denominator, whereas if the denominator is large we shall expect the approximation to be exceptionally close. One way to achieve this is to try to solve an inequality such as
$$\left|\alpha-\frac{p}{q}\right|<\frac{1}{q^{2}}$$
where $\alpha$ is a given real number and we seek rational $p / q$. In this case, if we are forced to choose a large value of $q$, we do at least know that the approximation is much closer than we had previously with
$$\left|\alpha-\frac{p}{q}\right| \leq \frac{1}{2 q} .$$

## 数学代写|数论作业代写number theory代考|A SKETCH

We conclude this chapter with a very brief summary of Apéry’s notoriously complex irrationality proof for $\zeta(3)$. Our only aim is to show how the argument is based fundamentally on the approximation ideas introduced in the previous section: specifically, Apéry showed that $\zeta(3)$ is approximable to order (just slightly) greater than 1 , and therefore cannot be rational. The reader should not be deluded into believing that the arguments we have omitted are easy! they most assuredly are not. More details (as well as an engaging account of the circumstances surrounding Apéry’s announcement of his result) may be found in [66]. Another, and possibly simpler, irrationality proof for $\zeta(3)$ was given by Beukers [14]. Although superficially Beukers’ approach appears quite different from Apéry’s, the author acknowledges a close connection between the two.
So, we begin by recalling the definition
$$\zeta(3)=\sum_{n=1}^{\infty} \frac{1}{n^{3}}=1+\frac{1}{2^{3}}+\frac{1}{3^{3}}+\frac{1}{4^{3}}+\cdots$$
By intricate but essentially straightforward algebra we may obtain an alternative summation formula
$$\zeta(3)=\frac{5}{2} \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^{3}\left(\begin{array}{c} 2 n \ n \end{array}\right)}=\frac{5}{2}\left(\frac{1}{1^{3} \times 2}-\frac{1}{2^{3} \times 6}+\frac{1}{3^{3} \times 20}-\frac{1}{4^{3} \times 70}+\cdots\right) .$$
The heart of Apéry’s argument consists of defining two “mysterious” sequences $a_{n}$ and $b_{n}$ with the property that the quotients $a_{n} / b_{n}$ form a sequence of very good rational approximations to $\zeta(3)$. Set
$$a_{0}=0, a_{1}=6, a_{n}=\frac{34 n^{3}-51 n^{2}+27 n-5}{n^{3}} a_{n-1}-\frac{(n-1)^{3}}{n^{3}} a_{n-2}$$
for $n \geq 2$, and
$$b_{0}=1, b_{1}=5, b_{n}=\frac{34 n^{3}-51 n^{2}+27 n-5}{n^{3}} b_{n-1}-\frac{(n-1)^{3}}{n^{3}} b_{n-2}$$
for $n \geq 2$. One observes that $a_{n}$ and $b_{n}$ satisfy the same recurrence, and differ only in their respective initial conditions. Amazingly, despite the fractional coefficients in its recurrence, it can be shown that $b_{n}$ is always an integer! This is not the case for $a_{n}$; however, it turns out that $a_{n}$ is a rational number whose denominator is a factor of $2 L_{n}^{3}$, where $L_{n}$ is the least common multiple of the integers $1,2, \ldots, n$. Apéry also proved that
$$\lim {n \rightarrow \infty} \frac{a{n}}{b_{n}}=\zeta(3) .$$

## 数学代写|数论作业代写number theory代考|APPROXIMATION OF REAL NUMBERS BY RATIONALS

Joseph Liouville 刘

$(1809-1882)$

$$\left|\alpha-\frac{p}{q}\right|$$

$$|q \alpha-p| \leq \frac{1}{2} \quad \Rightarrow \quad \alpha-\frac{p}{q} \mid \leq \frac{1}{2 q}<\varepsilon .$$

$$\left|\alpha-\frac{p}{q}\right|<\frac{1}{q^{2}}$$

$$\left|\alpha-\frac{p}{q}\right| \leq \frac{1}{2 q}$$

## 数学代写|数论作业代写number theory代考|A SKETCH

$$\zeta(3)=\sum_{n=1}^{\infty} \frac{1}{n^{3}}=1+\frac{1}{2^{3}}+\frac{1}{3^{3}}+\frac{1}{4^{3}}+\cdots$$

$$\zeta(3)=\frac{5}{2} \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^{3}(2 n n)}=\frac{5}{2}\left(\frac{1}{1^{3} \times 2}-\frac{1}{2^{3} \times 6}+\frac{1}{3^{3} \times 20}-\frac{1}{4^{3} \times 70}+\cdots\right)$$
Apéry 论证的核心在于定义两个”神秘”序列 $a_{n}$ 和 $b_{n}$ 具有商的性质 $a_{n} / b_{n}$ 形成一系列非常好的有理逼近 $\zeta(3)$. 放
$$a_{0}=0, a_{1}=6, a_{n}=\frac{34 n^{3}-51 n^{2}+27 n-5}{n^{3}} a_{n-1}-\frac{(n-1)^{3}}{n^{3}} a_{n-2}$$

$$b_{0}=1, b_{1}=5, b_{n}=\frac{34 n^{3}-51 n^{2}+27 n-5}{n^{3}} b_{n-1}-\frac{(n-1)^{3}}{n^{3}} b_{n-2}$$

$$\lim n \rightarrow \infty \frac{a n}{b_{n}}=\zeta(3)$$

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assignmentutor™您的专属作业导师
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