数学代写|数论作业代写number theory代考|MATH3170

Doug I. Jones

Doug I. Jones

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  • Statistical Inference 统计推断
  • Statistical Computing 统计计算
  • Advanced Probability Theory 高等概率论
  • Advanced Mathematical Statistics 高等数理统计学
  • (Generalized) Linear Models 广义线性模型
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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|数论作业代写number theory代考|MATH3170


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.
First, a slightly informal proof. Recall that an algebraic number is, (almost) by definition, a root of a non-zero polynomial with integral coefficients. Define the height of any such polynomial to be the maximum of the absolute values of its coefficients: that is, if $f(z)=a_n z^n+a_{n-1} z^{n-1}+\cdots+a_1 z+a_0$ with all $a_k$ integers and $a_n \neq 0$, then
H(f)=\max \left(\left|a_n\right|,\left|a_{n-1}\right|, \ldots,\left|a_1\right|,\left|a_0\right|\right) .
The height of any polynomial $f \neq 0$ is a positive integer. Let $m$ be a positive integer and consider all polynomials $f$ with $\operatorname{deg} f+H(f)=m$. Any such polynomial has degree less than $m$ and therefore at most $m$ non-zero coefficients; each of these coefficients is an integer from $-m$ to $m$. Therefore, the number of $f$ satisfying $\operatorname{deg} f+H(f)=m$ is at most $(2 m+1)^m$ and is hence finite. So we can construct a list of all non-zero polynomials over $\mathbb{Z}$ by writing down those with degree plus height equal to 1 , followed by those with degree plus height equal to 2 , and so on. We can now write down all the roots of the first polynomial in our list, followed by all the roots of the second, and so forth; deleting any repetitions in this list, we have listed all (real and complex) algebraic numbers once each. Since the set of algebraic numbers can be arranged in a list it is countable, and therefore cannot include all complex numbers. Thus transcendental numbers exist.


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.

Liouville’s methods derive from an investigation of
Joseph Liouville
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
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 chonsing 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\left|\alpha-\frac{p}{q}\right| \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
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|MATH3170




请注意,到目前为止,我们只看到了 4 次以内的代数数。是否存在任意高次的代数数这一点尚不清楚。例如,如果我们要考虑具有实数(而不是有理数)系数的多项式,那么就不会有次数大于 2 的不可约多项式。因此,这种情况下的情况非常简单:所有实数都是代数的(超过R) 的次数为 1 ,并且所有非实复数都是代数的(超过R) 度 2 。复数中不会有更高次的代数数,也不会有超越数。

超越数的存在首先由 Joseph Liouville 证明,他试图证明和不是代数数。他未能实现这一目标,但取得的成就足以让他在 1844 年(并在 1851 年再次使用不同的技术)给出超越数的具体例子。三十年后,Georg Cantor 给出了一个完全不同的证明:一个可能更简单的证明,因为它没有提供任何超越数的具体例子,就数论而言可能不知何故与重点无关。我们将从 Cantor 的证明开始。

康托尔简单地通过证明在某种意义上存在比代数数更复杂的数来证明超越数的存在。具体来说,复数集是不可数的——这直接由康托尔在 1874 年证明的实数的不可数性得出——而正如我们现在将要证明的那样,(复数)代数数集是可数的。
首先,一个稍微非正式的证明。回想一下,根据定义,代数数(几乎)是具有整数系数的非零多项式的根。将任何此类多项式的高度定义为其系数的绝对值的最大值:也就是说,如果F(和)=一种n和n+一种n−1和n−1+⋯+一种1和+一种0所有一种k整数和一种n≠0, 然后

任何多项式的高度F≠0是一个正整数。让米是一个正整数并考虑所有多项式F和你⁡F+H(F)=米. 任何此类多项式的次数都小于米因此至多米非零系数;这些系数中的每一个都是来自−米到米. 因此,数量F令人满意你⁡F+H(F)=米最多是(2米+1)米因此是有限的。所以我们可以构造一个所有非零多项式的列表从写下度加高度等于 1 的那些,然后是度加高度等于 2 的那些,依此类推。我们现在可以写下列表中第一个多项式的所有根,然后是第二个多项式的所有根,依此类推;删除此列表中的任何重复项,我们列出了所有(实数和复数)代数数各一次。由于代数数集可以排列在列表中,因此它是可数的,因此不能包括所有复数。因此超越数存在。


我们可以回到 Liouville 的早期证明,而不是依赖 Cantor 的可数性论 证。这些使得显式构造超越数成为可能,因此,从数论的角度来看, 比康托尔的证明更有趣。他们还避免了上面提到的直觉主义者的反对 意见一一㞔管可能不完全如此,因为他们确实利用了初等微积分中的 中值定理。
Liouville 的方法源自对 Joseph Liouville的调查 $(1809-1882)$
用有理数逼近实数的问题。让 $\alpha \in \mathbb{R}$; 我们想问问有多近 $\alpha$ 可以用有理 数来近似 $p / q$. 也就是说,我们想知道有多小
可以通过合理的理性选择来做出 $p / q$. 不幸的是,这个问题太简单了, 没有人感兴趣: 因为有理数密集 $\mathbb{R}$, 差值 (3.4),对于任何 $\alpha$ ,可以通过选 择较大的值来尼可能小 $q$ 和一个适当的 $p$. 具体来说,如果我们㳍望差值 小于正数 $\varepsilon$ ,我们选择 $q>1 / 2 \varepsilon$ 然后让 $p$ 是最接近的整数 $q \alpha$. 然后
|q \alpha-p| \leq \frac{1}{2} \Rightarrow\left|\alpha-\frac{p}{q}\right| \leq \frac{1}{2 q}<\varepsilon .
这个观察本身虽然不是很有趣,但可能暗示了一种更重要的方法,即 坚持近似的接近程度应取决于近似分数的分母。换句话说,如果分母 非常小的分数给出了一个相当弱的近似值,我们将对其感兴趣,而如 果分母很大,我们将期望近似值非常接近。实现这一目标的一种方法 是尝试解决不等式,例如
在哪里 $\alpha$ 是一个给定的实数,我们寻求理性 $p / q$. 在这种情况下,如果 我们被迫选择较大的值 $q$ ,我们至少知道近似值比我们之前用的更接近
\left|\alpha-\frac{p}{q}\right| \leq \frac{1}{2 q}

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术语 广义线性模型(GLM)通常是指给定连续和/或分类预测因素的连续响应变量的常规线性回归模型。它包括多元线性回归,以及方差分析和方差分析(仅含固定效应)。



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