### 数学代写|数论代写Number theory代考|MXB 251

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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|数论代写Number theory代考|COMPUTING THE CONTINUED FRACTION OF AN ALGEBRAIC IRRATIONAL

Let $\alpha$ be a root of a known irreducible polynomial $f$ with degree $n \geq 2$ and integral coefficients. We shall assume that $\alpha>1$, and that $f$ has no other roots $\beta>1$. In this case there is a very simple algorithm to find the zeroth partial quotient $a_{0}=\lfloor\alpha\rfloor$ : calculate $f(1), f(2), f(3), \ldots$ until a change of sign occurs; then $a_{0}$ is the last argument before the change of sign. Since $\alpha$ is irrational we have $a_{0}<\alpha1$ since $0<\alpha-a_{0}<1$; so $f_{1}$ has a real root greater than 1 . Conversely, if $\beta$ is any such root of $f_{1}$, then $a_{0}+1 / \beta$ is a root of $f$ and hence $a_{0}+1 / \beta=\alpha$. Because $f_{1}$ is a polynomial having integral coefficients and a unique real root $\alpha_{1}>1$, the procedure can be iterated to find the sequence of partial quotients of $\alpha$. Observe that the complete quotients $\alpha=\alpha_{0}, \alpha_{1}, \alpha_{2}, \ldots$ need never be calculated, so we do not have the problem of calculating decimal expansions to many places: all our calculations will be performed in terms of integer arithmetic, and so the process will be free of rounding errors.

Example. We can find the continued fraction for $\sqrt[3]{2}$ by starting with the polynomial $f(z)=f_{0}(z)=z^{3}-2$. We have
\begin{aligned} f_{0}(z)=z^{3}-2, & a_{0}=1 \ f_{1}(z)=-z^{3}+3 z^{2}+3 z+1, & a_{1}=3 \ f_{2}(z)=10 z^{3}-6 z^{2}-6 z-1, & a_{2}=1 \ f_{3}(z)=-3 z^{3}+12 z^{2}+24 z+10, & a_{3}=5 \ f_{4}(z)=55 z^{3}-81 z^{2}-33 z-3, & a_{4}=1 \ f_{5}(z)=-62 z^{3}-30 z^{2}+84 z+55, & a_{5}=1 \ f_{6}(z)=47 z^{3}-162 z^{2}-216 z-62, & a_{6}=4 \end{aligned}
and so
$$\sqrt[3]{2}=1+\frac{1}{3+} \frac{1}{1+} \frac{1}{5+} \frac{1}{1+} \frac{1}{1+} \frac{1}{4+} \cdots$$

## 数学代写|数论代写Number theory代考|THE CONTINUED FRACTION OF e

We shall determine the continued fractions for a class of numbers related to the exponential constant $e$. To do so, we first consider the functions defined by the infinite series
$$f(c ; z)=\sum_{k=0}^{\infty} \frac{1}{c(c+1) \cdots(c+k-1)} \frac{z^{k}}{k !}$$
here the parameter $c$ is any real number except $0,-1,-2, \ldots$, and it is easy to show that the series converges for all $z$. To simplify the notation we write $c^{(k)}=c(c+1) \cdots(c+k-1)$, with the understanding that $c^{(0)}=1$; thus
$$f(c ; z)=\sum_{k=0}^{\infty} \frac{1}{c^{(k)}} \frac{z^{k}}{k !} .$$
The expression $c^{(k)}$ is referred to as ” $c$ rising factorial $k$ “. It satisfies the two important recurrences
$$c^{(k+1)}=c^{(k)}(c+k)=c(c+1)^{(k)},$$
both of which are instances of the more general relation $c^{(k+m)}=c^{(k)}(c+k)^{(m)}$.
Lemma 4.18. Let $c$ be a positive real number, $z$ a non-zero real number and $k$ a non-negative integer; then
$$\frac{c}{z} \frac{f\left(c ; z^{2}\right)}{f\left(c+1 ; z^{2}\right)}=\left[\frac{c}{z}, \frac{c+1}{z}, \ldots, \frac{c+k-1}{z}, \frac{c+k}{z} \frac{f\left(c+k ; z^{2}\right)}{f\left(c+k+1 ; z^{2}\right)}\right] .$$
Proof. First, observe that under the stated conditions $f\left(c+k+1 ; z^{2}\right)$ is given by a series of positive terms, so it does not vanish and the last term in the continued fraction makes sense. From (4.14) the rising factorials satisfy
$$\frac{1}{c^{(k)}}=\frac{c+k}{c^{(k+1)}}=\frac{1}{(c+1)^{(k)}}+\frac{k}{c^{(k+1)}},$$
and hence
$$f(c ; z)=\sum_{k=0}^{\infty} \frac{1}{(c+1)^{(k)}} \frac{z^{k}}{k !}+\sum_{k=0}^{\infty} \frac{k}{c^{(k+1)}} \frac{z^{k}}{k !} .$$
The first series on the right-hand side is evidently $f(c+1 ; z)$. The second may be written
$$\sum_{k=1}^{\infty} \frac{1}{c^{(k+1)}} \frac{z^{k}}{(k-1) !}=\sum_{k=0}^{\infty} \frac{1}{c^{(k+2)}} \frac{z^{k+1}}{k !}=\frac{z}{c(c+1)} \sum_{k=0}^{\infty} \frac{1}{(c+2)^{(k)}} \frac{z^{k}}{k !},$$
and we have the second-order recurrence
$$f(c ; z)=f(c+1 ; z)+\frac{z}{c(c+1)} f(c+2 ; z)$$

## 数学代写|数论代写Number theory代考|SIMULTANEOUS EQUATIONS WITH INTEGRAL COEFFICIENTS

Let $a, b, c, d$ and $p, q$ be integers. If $a d-b c=\pm 1$, then the simultaneous equations
$$\left{\begin{array}{l} a x+b y=p \ c x+d y=q \end{array}\right.$$
have an integral solution $x, y$. Conversely, if the system has an integral solution for all integers $p, q$, then $a d-b c=\pm 1$.
Proof. The solution can be written
$$\left(\begin{array}{l} x \ y \end{array}\right)=\left(\begin{array}{ll} a & b \ c & d \end{array}\right)^{-1}\left(\begin{array}{l} p \ q \end{array}\right)=\frac{1}{a d-b c}\left(\begin{array}{cc} d & -b \ -c & a \end{array}\right)\left(\begin{array}{l} p \ q \end{array}\right)=\frac{1}{a d-b c}\left(\begin{array}{l} d p-b q \ a q-c p \end{array}\right)$$
provided that $a d-b c \neq 0$. It is clear that if $a d-b c=\pm 1$, then $x$ and $y$ are integers. Conversely, suppose that $|a d-b c|>1$ and consider the solutions when $p=1, q=0$ and when $p=0, q=1$. If these solutions are to be integers, then $a d-b c$ must be a factor of $a, b, c$ and $d$. But this leads to
$$(a d-b c)^{2} \mid a d-b c$$
which is impossible. Finally note that if $a d-b c=0$, then there exist $p, q$ for which the system has no solution at all, and therefore certainly no integral solution.

Exercise. Let $A$ be an $n \times n$ matrix with integral entries. Show that the linear equations $A \mathbf{x}=\mathbf{b}$ have a solution $\mathbf{x}$ with integral components for all integer vectors $\mathbf{b}$, if and only if $\operatorname{det}(A)=\pm 1$.

## 数学代写|数论代写Number theory代考|COMPUTING THE CONTINUED FRACTION OF AN ALGEBRAIC IRRATIONAL

F0(和)=和3−2,一个0=1 F1(和)=−和3+3和2+3和+1,一个1=3 F2(和)=10和3−6和2−6和−1,一个2=1 F3(和)=−3和3+12和2+24和+10,一个3=5 F4(和)=55和3−81和2−33和−3,一个4=1 F5(和)=−62和3−30和2+84和+55,一个5=1 F6(和)=47和3−162和2−216和−62,一个6=4

23=1+13+11+15+11+11+14+⋯

## 数学代写|数论代写Number theory代考|THE CONTINUED FRACTION OF e

F(C;和)=∑ķ=0∞1C(C+1)⋯(C+ķ−1)和ķķ!

F(C;和)=∑ķ=0∞1C(ķ)和ķķ!.

C(ķ+1)=C(ķ)(C+ķ)=C(C+1)(ķ),

C和F(C;和2)F(C+1;和2)=[C和,C+1和,…,C+ķ−1和,C+ķ和F(C+ķ;和2)F(C+ķ+1;和2)].

1C(ķ)=C+ķC(ķ+1)=1(C+1)(ķ)+ķC(ķ+1),

F(C;和)=∑ķ=0∞1(C+1)(ķ)和ķķ!+∑ķ=0∞ķC(ķ+1)和ķķ!.

∑ķ=1∞1C(ķ+1)和ķ(ķ−1)!=∑ķ=0∞1C(ķ+2)和ķ+1ķ!=和C(C+1)∑ķ=0∞1(C+2)(ķ)和ķķ!,

F(C;和)=F(C+1;和)+和C(C+1)F(C+2;和)

## 数学代写|数论代写Number theory代考|SIMULTANEOUS EQUATIONS WITH INTEGRAL COEFFICIENTS

$$\left{ 一个X+b是=p CX+d是=q\正确的。 H一个在和一个n一世n吨和Gr一个ls○l在吨一世○nX,是.C○n在和rs和l是,一世F吨H和s是s吨和米H一个s一个n一世n吨和Gr一个ls○l在吨一世○nF○r一个ll一世n吨和G和rsp,q,吨H和n一个d−bC=±1.磷r○○F.吨H和s○l在吨一世○nC一个nb和在r一世吨吨和n \剩下（ X 是\右）=\左（ 一个b Cd\right)^{-1}\left( p q\right)=\frac{1}{a db c}\left( d−b −C一个\右左（ p q\right)=\frac{1}{a db c}\left( dp−bq 一个q−Cp\正确的） pr○在一世d和d吨H一个吨一个d−bC≠0.我吨一世sCl和一个r吨H一个吨一世F一个d−bC=±1,吨H和nX一个nd是一个r和一世n吨和G和rs.C○n在和rs和l是,s在pp○s和吨H一个吨|一个d−bC|>1一个ndC○ns一世d和r吨H和s○l在吨一世○ns在H和np=1,q=0一个nd在H和np=0,q=1.我F吨H和s和s○l在吨一世○ns一个r和吨○b和一世n吨和G和rs,吨H和n一个d−bC米在s吨b和一个F一个C吨○r○F一个,b,C一个ndd.乙在吨吨H一世sl和一个ds吨○ (a db c)^{2} \mid a db c$$

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## MATLAB代写

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