### 数学代写|数论代写Number theory代考|MATH2988

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• (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|数论代写Number theory代考|EXISTENCE OF TRANSCENDENTAL NUMBERS

The first question we need to address about transcendental numbers is whether or not there are any! It is clear that algebraic numbers exist: for a start, all rational numbers are algebraic, and we have also given a few examples of irrational algebraic numbers. However, it is conceivable that every complex number could be a root of a rational polynomial, in which case transcendental numbers would not exist.

Notice, by the way, that we have so far only seen algebraic numbers of degree up to 4 . It is not at all clear that algebraic numbers of arbitrarily high degree exist. If, for example, we were to consider polynomials with real (rather than rational) coefficients, then there would be no irreducible polynomials of degree greater than 2. The situation in this case would therefore be very simple: all real numbers would be algebraic (over $\mathbb{R}$ ) of degree 1 , and all nonreal complex numbers would be algebraic (over $\mathbb{R}$ ) of degree 2. Among the complex numbers there would be no algebraic numbers of higher degree, and no transcendental numbers.

The existence of transcendental numbers was first proved by Joseph Liouville, who attempted to show that $e$ is not an algebraic number. He failed in this aim but achieved enough to allow him in 1844 (and again, using different techniques, in 1851 ) to give specific examples of transcendental numbers. A completely different proof was given three decades later by Georg Cantor: a proof which is perhaps simpler, though, as it does not provide any specific examples of transcendentals, possibly somehow beside the point as far as number theory is concerned. We shall begin with Cantor’s proof.

Cantor proved the existence of transcendental numbers simply by showing that there are, in a sense, more complex numbers than algebraic numbers. Specifically, the set of complex numbers is uncountable – this follows immediately from the uncountability of the reals, proved by Cantor in 1874 – while, as we shall now show, the set of (complex) algebraic numbers is countable.

## 数学代写|数论代写Number theory代考|APPROXIMATION OF REAL NUMBERS BY RATIONALS

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
$$\left|\alpha-\frac{p}{q}\right| \leq \frac{1}{2 q} .$$
To obtain worthwhile results we need to note two more points. A single solution of either of the above inequalities is of little importance as it does not give rational numbers arbitrarily close to $\alpha$ : what we really want is that there be infinitely many solutions to such an inequality. Secondly, we would like the right-hand side of the inequality to be uniquely determined by the approximating fraction $p / q$; therefore we shall require that $q$ be the “true” denominator of the fraction, that is, that $p$ and $q$ have no common factor. As a result of these considerations, we introduce the following terminology.

## 数学代写|数论代写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|数论代写Number theory代考|APPROXIMATION OF REAL NUMBERS BY RATIONALS

(1809−1882)

|一个−pq|

|q一个−p|≤12⇒一个−pq∣≤12q<e

|一个−pq|<1q2,

|一个−pq|≤12q.

## 数学代写|数论代写Number theory代考|A SKETCH

G(3)=∑n=1∞1n3=1+123+133+143+⋯

G(3)=52∑n=1∞(−1)n−1n3(2n n)=52(113×2−123×6+133×20−143×70+⋯).
Apéry 论证的核心在于定义两个“神秘”序列一个n和bn具有商的性质一个n/bn形成一系列非常好的有理逼近G(3). 放

b0=1,b1=5,bn=34n3−51n2+27n−5n3bn−1−(n−1)3n3bn−2

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