分析学基本理论（一）

这里对分析学基础理论, 极限和无穷级数部分做一个 $\textbf{review}$
在 $\text{knuth}$ 的具体数学中, 很多理论都是建立在分析学基础上的

数列极限的补充问题

从每一个实数列中都可以选出一个收敛的或者是趋于无穷的子列
$\textbf{proof}\quad$ 其中, 收敛很好证明

无穷性的证明, $\forall k \in \mathbb{N}, \text{choose} \ n_k \in \mathbb{N}$
$|x_{n_k}| > k \ \textbf{and} \ n_k < n_{k+1}, {x_{n_k}}$ 构成无穷子列

$\textbf{example1}$

$$\mathop{\underline{\lim}}\limits_{k \rightarrow \infty} \frac{(-1)^{k}}{k}=\lim _{n \rightarrow \infty} \inf _{k \geqslant n} \frac{(-1)^{k}}{k}=\left\{\begin{gathered} \frac{(-1)^{2m +1}}{(2 m+1)}=-\frac{1}{n} \quad n=2 m+1 \\ \ \\ \frac{(-1)^{2 m} \cdot(-1)}{2 m+1}=\frac{-1}{n+1} \quad n=2m \end{gathered} \right\} =0$$

$$\mathop{\overline{\lim}}\limits _{k \rightarrow \infty} \frac{(-1)^{k}}{k}=\lim _{n \rightarrow \infty} \sup _{k \geqslant n} \frac{(-1)^{k}}{k}=\left\{\begin{gathered} \frac{(-1)^{2m}}{2 m}=\frac{1}{n} & n=2 m \\ \ \\ \frac{(-1)^{2 m+2}}{2 m+2}=\frac{1}{n+1} & n=2 m+1 \end{gathered}\right\} = 0$$

下极限证明语言
$\forall \varepsilon>0, \quad i-\varepsilon<i_n=\inf\limits_{k \geqslant n} x_{k} \leqslant x_{k}$
因为 $\varepsilon$ 是任意的, 所以对于 ${x_k}$
任何部分极限都不能小于 $i - \varepsilon$, 也就是不能小于 $i$

补充阅读：《不等式的秘密》

阅读这部分内容, 主要是为了提升数感, 让自己的代数变形能力有进一步的提升

AM-GM不等式

$\textbf{problem1}$
$a, b, c>0, abc = 1, \text{证明}$

$$\begin{gathered} (a+b)(b+c)(c+a) \geqslant (a+1)(b+1)(c+1) \\ \Longrightarrow ab(a+b)+bc(b+c)+ca(c+a) \geqslant a+b+c+ab+bc+ca \end{gathered}$$

重点在于轮换对称的构造

$$L H S_{c y c}=\left\{\begin{gathered} a^{2} b+a^{2} c \\ b^{2} a+b^{2} c \\ c^2a+c^{2} b \end{gathered}\right\} \quad \left\{\begin{gathered} a^{2} b+a b^{2} \\ b^{2} c+b c^{2} \\ c^{2} a+c a^{2} \end{gathered}\right\}$$

$$\begin{gathered} 2 L H S+\sum_{c y c} a b=\sum_{c y c}\left(a^{2} b+a^{2} b+a^{2} c+a^{2} c+b c\right) \geqslant 5 \sum_{c y c}a \\ 2 L H S+\sum_{c y c} a=\sum_{c y c}\left(a^{2} b+a^{2} b+a b^{2}+a b^{2}+c\right) \geqslant 5 \sum_{c y c} a b \end{gathered}$$

$$\begin{gathered} \text{综上, }L H S \geqslant \sum_{c y c} a+\sum_{c yc} a b \end{gathered}$$

$\textbf{problem2} \ 4n+3$ 是数论中非常常见的数, 因此构造 $\frac{3}{4}$ 也是很常见的
$x,y,z>0, \textbf{and}\ xyz=1$

$$\frac{x^{3}}{(1+y)(1+z)}+\frac{y^{3}}{(1+z)(1+x)}+\frac{z^{3}}{(1+x)(1+y)} \geq \frac{3}{4}$$

$$\frac{x^{3}}{(1+y)(1+z)}+\frac{1+y}{8}+\frac{1+z}{8} \geqslant \frac{3}{4} x , \text{写成轮换对称}$$

$$\begin{gathered} \sum_{c y c} \frac{x^{3}}{(1+y)(1+z)}+\frac{1}{4} \sum_{c y c}(1+x) \geqslant \sum_{c y c} \frac{3 x}{4} \\ \Rightarrow \sum_{c y c} \frac{x^{3}}{(1+y)(1+z)} \geq \frac{1}{4} \sum_{c y c}(2 x-1)=\frac{1}{4}(x+x+y+y+\ldots)-\frac{3}{4} \\ \geqslant \frac{6}{4}-\frac{3}{4}=\frac{3}{4} \end{gathered}$$

$\textbf{problem3轮换恒等变形}$

$$\begin{gathered} \left(a^{2}+b^{2}+c^{2}\right)(a b+b c+c a)=a b\left(a^{2}+b^{2}\right)+b c\left(b^{2}+c^{2}\right)+c a\left(a^{2}+c^{2}\right)+a b c(a+b+c) \\ \text{if} \ a , b, c \geqslant 0, a+b+c =2 \\ \left(a^{2}+b^{2}+c^{2}\right)(a b+b c+c a) \geqslant a b\left(a^{2}+b^{2}\right)+b c\left(b^{2}+c^{2}\right)+c a\left(a^{2}+c^{2}\right) \\ \geqslant 2\left(a^{2} b^{2}+b^2 c^{2}+a^{2} c^{2}\right) \end{gathered}$$

$$\begin{gathered} a^{2}+b^{2}+c^{2}+2(a b+b c+c a)=(a+b+c)^{2}=4 \\ 2\left(a^{2}+b^{2}+c^{2}\right)(a b+b c+c a) \leqslant\left(\frac{a^{2}+b^{2}+c^{2}+2(a b+b c+c a)}{2}\right)^{2}=4 \\ \begin{gathered} \left(a^{2}+b^{2}+c^{2}\right)(a b+b c+c a) \leqslant 2 \\ \Rightarrow a^{2} b^{2}+b^{2} c^{2}+a^{2} c^{2} \leqslant 1 \end{gathered} \end{gathered}$$

$\textbf{problem4不等式中的域分组}$
$a,b,c,d>0$

$$\frac{1}{a^{2}+a b}+\frac{1}{b^{2}+b c}+\frac{1}{c^{2}+c d}+\frac{1}{d^{2}+d a} \geqslant \frac{4}{a c+b d}$$

$$\begin{gathered} \frac{a c+b d}{a^{2}+a b}=\frac{a^{2}+a b+a c+b d}{a^{2}+a b}-1=\frac{a+c}{a+b}+\frac{b(d+a)}{a(a+b)}-1 \\ L H S=\sum_{c y c} \frac{a+c}{a+b}+\sum_{c y c} \frac{b(d+a)}{a(a+b)}-4 \end{gathered}$$

$$\begin{gathered} \left\{\frac{a+c}{a+b}, \frac{b(d+a)}{a(a+b)}\right\}, \quad \left\{\frac{b+d}{b+c}, \frac{c(a+b)}{b(b+c)} \right\} \\ \left\{\frac{c+a}{c+d}, \frac{d(c+b)}{c(c+d)}\right\}, \quad \left\{\frac{d+b}{d+a}, \frac{a(d+c)}{d(d+a)}\right\} \\ \ \\ \text{用AM-GM不等式消去第二个项, 得到 }\\ \ \\ LH S \geqslant \sum_{c y c} \frac{a+c}{a+b} \end{gathered}$$

$$\begin{gathered} \sum_{c y c} \frac{a+c}{a+b}=(a+c)\left(\frac{1}{a+b}+\frac{1}{c+d}\right)+(b+d)\left(\frac{1}{b+c}+\frac{1}{d+a}\right) \\ \text{Schwarz 不等式} \\ \ \\ \frac{1}{a+b}+\frac{1}{c+d} \geqslant \frac{(1+1)^{2}}{a+b+c+d}=\frac{4}{a+b+c+d} \\ \ \\ L H S \geqslant \frac{4(a+c)}{a+b+c+d}+\frac{4(b+d)}{a+b+c+d}=4 \end{gathered}$$

$\textbf{problem5}$
$a,b,c,d,e \geqslant 0, a+b+c+d+e = 5, \text{proof}$
$abc+bcd+cde+dea+eab \leqslant 5$
$\textbf{proof}$

$$\begin{gathered} a b c+b c d+c d e+d e a+e a b=e(a+c)(b+d)+b c(a+d-e) \\ \leqslant e\left(\frac{a+b+c+d}{2}\right)^{2}+\left(\frac{b+c+a+d-e}{3}\right)^{3} \\ \ \\ =\frac{e(5-e)^{2}}{4}+\frac{(5-2 e)^{2}}{27} \end{gathered}$$

$$\begin{gathered} f(x)=\frac{x(5-x)^{2}}{4}+\frac{(5-2 x)^{2}}{27} \\ \frac{d f(x)}{d x}=-\frac{5}{36}\left(x^{2}+4 x-5\right) \Rightarrow x=1, \frac{d f(x)}{d x}=0 \\ \ \\ \text{let } e = \min\{a,b,c,d,e\}, \Rightarrow e \leqslant 1 \\ \ \\ f(x) \leqslant f(1)=2 \leqslant 5 \end{gathered}$$

$\textbf{problem5}$
$a,b,c,d>0, \text{proof}$

$$\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\right)^{2} \geqslant \frac{1}{a^{2}}+\frac{4}{a^{2}+b^{2}}+\frac{9}{a^{2}+b^{2}+c^{2}}+\frac{16}{a^{2}+b^{2}+c^{2}+d^{2}}$$

$$\Leftrightarrow \frac{1}{b^{2}}+\frac{1}{c^{2}}+\frac{1}{d^{2}}+\sum_{s y m} \frac{2}{a b} \geqslant \frac{4}{a^{2}+b^{2}}+\frac{9}{a^{2}+b^{2}+c^{2}}+\frac{16}{a^{2}+b^{2}+c^{2}+d^{2}}$$

$$\begin{gathered} \text{AM-GM不等式, 有} \\ \ \\ \frac{2}{a b} \geqslant \frac{4}{a^{2}+b^{2}} \\ \ \\ \frac{2}{a c}+\frac{2}{b c} \geqslant \frac{(2 \sqrt{2})^{2}}{a c+b c} \geqslant \frac{8}{\frac{a^{2}+b^{2}}{2}+c^{2}}>\frac{8}{a^{2}+b^{2}+c^{2}} \ \text{(Cauchy-Schwarz)}\\ \ \\ \frac{1}{b^{2}}+\frac{1}{c^{2}} \geqslant \frac{4}{b^{2}+c^{2}} \geqslant \frac{1}{b^{2}+c^{2}+a^{2}} \\ \ \\ d=\min \{a, b, c, d\} \\ \ \\ \frac{2}{a d}+\frac{2}{b d}+\frac{2}{c d} \geqslant \frac{(3 \sqrt{2})^{2}}{a d+b d+c d} \\ \ \\ \geqslant \frac{18}{a d+b d+c d+d^{2}} \geqslant \frac{18}{a^{2}+b^{2}+c^{2}+d^{2}}>\frac{16}{a^{2}+b^{2}+c^{2}+d^{2}} \end{gathered}$$

$\textbf{problem6} \ \text{ask} \min {M}$

$$\left|a b\left(a^{2}-b^{2}\right)+b c\left(b^{2}-c^{2}\right)+c a\left(c^{2}-a^{2}\right)\right| \leqslant M\left(a^{2}+b^{2}+c^{2}\right)^{2}$$

$$\begin{gathered} L H S=a b\left(a^{2}-b^{2}+c^{2}-c^{2}\right)+b c\left(b^{2}-c^{2}\right)+c a\left(c^{2}-a^{2}\right) \\ \ \\ =b\left(b^{2}-c^{2}\right)(c-a)+a(c-a)(c+a)(c-b) \\ \ \\ =-(a-b)(b-c)(c-a)(a+b+c) \\ \ \\ \text{let } x=a-b, y = b-c, z=c-a, s=a+b+c \end{gathered}$$

$$\begin{gathered} \Leftrightarrow 9|sxyz| \leqslant M\left(s^{2}+x^{2}+y^{2}+z^{2}\right)^{2}, \quad x +y+z =0 \\ \ \\ \end{gathered}$$

$\textbf{i)}$

$$\begin{gathered} x^{2}+y^{2}+z^{2}=x^{2}+y^{2}+(x+y)^{2}=2\left((x+y)^{2}-x y\right) \\ \ \\ \geqslant 2\left((x+y)^{2}-\frac{(x+y)^{2}}{4}\right)=\frac{3}{2}(x+y)^{2} \\ \ \\ |s x y z|=|s x y(x+y)| \leqslant |s| \frac{(x+y)^{3}}{4} \end{gathered}$$

$\textbf{ii)}$

$$\begin{gathered} \text{let } t = x+y \\ \ \\ \Leftrightarrow \text{ask} \quad |s| t^{3} \leqslant ?\left(s^{2}+\frac{3}{2} t^{2}\right)^{2} \\ \ \\ \Leftrightarrow \text{ask} \quad s^{2} t^{6} \leqslant \ ? \frac{\left(2 s^{2}+3 t^{2}\right)^{4}}{4} \\ \ \\ \text{we have} \quad 2 s^{2} t^{6}=2 s^{2} \cdot t^{2} \cdot t^{2} \cdot t^{2} \leqslant\left(\frac{2 s^{2}+3 t^{2}}{4}\right)^{4} \\ \ \\ 4 \cdot \sqrt{2}|s| t^{3} \leqslant \frac{\left(2 s^{2}+3 t^{2}\right)^{2}}{4} \\ \ \\ |s x y z| \leqslant \frac{1}{16 \sqrt{2}}\left(s^{2}+x^{2}+y^{2}+z^{2}\right)^{2}, M=\frac{9}{16 \sqrt{2}} \\ \ \\ \left\{\begin{gathered} \sqrt{2}(a+b+c)=c-a \\ a+b+c=3 \\ c-a=3 \sqrt{2} \\ a-b=b-c \end{gathered}\right. \Rightarrow \left\{\begin{gathered} b=1 \\ c-a=3 \sqrt{2} \\ a+c=2 \end{gathered}\right. \\ \ \\ (a, b, c)=\left(1-\frac{3}{\sqrt{2}}, 1,1+\frac{3}{\sqrt{2}}\right) \end{gathered}$$

Cauchy求反技术

已知 $a+b+c+d = 4$, 证明

$$\frac{a}{1+b^{2} c}+\frac{b}{1+c^{2} d}+\frac{c}{1+d^{2} a}+\frac{d}{1+a^{2} b} \geq 2$$

$$\begin{gathered} \frac{a}{1+b^{2} c}=a-\frac{a b^{2} c}{1+b^{2} c} \geqslant a-\frac{2 b^{2} c}{2 b \sqrt{c}} \geqslant a-\frac{b(a+a c)}{4}\\ \ \\ \end{gathered}$$

$$\begin{gathered} 2 \sum_{c y c} a b=\left(\sum_{c y c} a\right)^{2}-\sum_{c y c}\left(a^{2}\right) \Rightarrow \\ \ \\ \Rightarrow \quad\left(\sum_{cyc} a\right)^{2}=2 \sum_{c y c} a b+\sum_{c y c} a^{2} \geqslant 4 \sum_{cyc}ab \\ \ \\ \Rightarrow \sum_{c y c} a b \leqslant \frac{1}{4}\left(\sum_{c y c} a\right)^{2} \\ \ \\ \left(\sum_{c yc4} a\right)^{3}=4 \sum_{c y c 4} a^{3}+\left(\begin{gathered} 3 \\ 1 \end{gathered}\right)\left(\begin{gathered} 4 \\ 1 \end{gathered}\right) \sum_{c y c 4} a b c \\ \ \\ 16 \sum_{c y c} a b c \leqslant(\sum_{cyc} a)^{3} \end{gathered}$$

$\textbf{AM-GM} \ \textbf{and} \ \textbf{Chebyshev}$
$\textbf{Chebyshev}$

$$a_{1} b_{1}+a_{2} b_{1}+\ldots+a_{n} b_{n} \geqslant \frac{1}{n}\left(a_{1}+a_{2}+\dots+a_{n}\right)\left(b_{1}+b_{2}+\ldots+b_{n}\right)$$

$a+b+c=3, \quad \text{proof}$

$$\begin{gathered} \sum_{c y c}(a b)^{\frac{2}{3}} \leqslant 3 \\ \ \\ (a+b+c)^{2} \geqslant \frac{1}{9}(a+b+c)(1+1+1)=\frac{1}{3}(a+b+c) \\ \ \\ (a+b+c)^{2}=\sum_{c y c} a^{2}+2 \sum_{c y c} a b \geqslant 3 \sum_{c y c} a b \\ \ \\ \Rightarrow \left(\sum_{c yc} a\right)^{2}=\frac{2}{3}\left(\sum_{c y c} a\right)^{2}+\frac{1}{3}\left(\sum_{c y c} a\right)^{2} \geqslant 2 \sum_{c yc} a+\sum_{c y c} a b \\ \ \\ =\sum_{c y c}(a+a+a b)=\sum_{c y c}(a+b+a b) \geqslant 3 \sum_{c y c}(a b)^{2 / 3} \end{gathered}$$

$\textbf{Rearrangement} \ \textbf{inequality}$

$$\begin{gathered} a+b+c =3, \quad \sum_{c y c} a \geqslant \sum_{c y c} a b \\ \ \\ \sum_{c y c} a=\frac{1}{3} \cdot\left(\sum_{c y c} a\right)\left(\sum_{c y c} a\right)=\frac{1}{3}\left(\sum_{c y c} a\right)^{2} \\ \ \\ \begin{gathered} &a^{2}+b^{2}+c^{2} \geqslant a b+b c+c a\\ &\sum_{c y c} a^{2} \geqslant \sum_{c y c} a b \end{gathered} \\ \sum_{c y c} a \geqslant \frac{1}{3} \cdot 3 \sum_{c y c} a b=\sum_{c y c} a b \end{gathered}$$

$\textbf{problem}$
$a+b+c+d=4, \ \text{proof}$

$$\begin{gathered} \sum_{c y c} a b-\sum_{c y c} a b^{2} c \\ \ \\ \Leftrightarrow a b+b c+c d+d a \leqslant a b^{2} c+b c^{2} d+c d^{2} a+d a^{2} b \\ \ \\ L H S=\frac{1}{4}(b+c+d+a)(a b+b c+c d+d a) \\ \ \\ \leqslant a b^{2}+b c^{2}+c d^{2}+d a^{2}=\frac{1}{4}(c+d+a+b)\left(a b^{2}+b c^{2}+c d^{2}+d a^{2}\right) \\ \ \\ \leqslant a b^{2} c+b c^{2} d+c d^{2} a+d a^{2} b \end{gathered}$$

$\textbf{Cauchy求反技术的技巧}$

$$\begin{gathered} a, b, c>0, \quad a^{2}+b^{2}+c^{2}=3 \ \text{proof} \\ \ \\ \frac{1}{a^{3}+2}+\frac{1}{b^{3}+2}+\frac{1}{c^{3}+2} \geqslant 1 \\ \ \\ \frac{1}{a^{3}+2}=\frac{1}{2}-\frac{a^{3}}{2\left(a^{3}+2\right)} \end{gathered}$$

连分数

连分数相关定义

$\textbf{表示}$

$$\begin{gathered} \left[a_{0}, a_{1}\right]=a_{0}+\frac{1}{a_{1}} \\ \ \\ \left[a_{0}, a_{1}, \ldots, a_{n-1}, a_{n}\right]=\left[a_{0}, a_{1}, \ldots, a_{n-2}, a_{n-1}+\frac{1}{a_{n}}\right] \end{gathered}$$

$$\begin{gathered} p_{0}=a_{0}, \quad p_{1}=a_{1} a_{0}+1, \quad p_{n}=a_{n} p_{n-1}+p_{n-2} \quad(2 \leqslant n \leqslant N) \\ q_{0}=1, \quad q_{1}=a_{1}, \quad q_{n}=a_{n} q_{n-1}+q_{n-2} \quad \quad \quad(2 \leqslant n \leqslant N) \\ \ \\ \left[a_{0}, a_{1}, \ldots . ., a_{n}\right]=\frac{p_{n}}{q_{n}} \end{gathered}$$

$\textbf{proof}$

$$\begin{gathered} \left[a_{0}, a_{1}, \ldots, a_{m-1}, a_{m}\right]=\frac{p_{m}}{q_{m}}=\frac{a_{m} p_{m-1}+p_{m-2}}{a_{m} q_{m-1}+q_{m-2}} \\ \ \\ \left[a_{0}, a_{1}, \ldots, a_{m-1}, a_{m}, a_{m+1}\right]=\left[a_{0}, a_{1}, \ldots ., a_{m-1}, a_{m}+\frac{1}{a_{m+1}}\right] \\ \ \\ =\frac{\left(a_{m}+\frac{1}{a_{m+1}}\right) p_{m-1}+p_{m-2}}{\left(a_{m}+\frac{1}{a_{m+1}}\right) q_{m-1}+q_{m-2}}=\frac{a_{m+1}\left(a_{m} p_{m-1}+p_{m-2}\right)+p_{m-1}}{a_{m+1}\left(a_{m} q_{m-1}+q_{m-2}\right)+q_{m-1}} \\ \ \\ = \frac{p_{m+1}}{q_{m+1}} \end{gathered}$$

$\textbf{lemma1}$

$$\frac{p_{n}}{q_{n}}=\frac{a_{n} p_{n-1}+p_{n-2}}{a_{n} q_{n-1}+q_{n-2}}$$

构造递推式如下

$$\begin{gathered} p_{n} q_{n-1}-p_{n-1} q_{n}=\left(a_{n} p_{n-1}+p_{n-2}\right) q_{n-1}-p_{n-1}\left(a_{n} q_{n-1}+q_{n-2}\right) \\ \ \\ =(-1)^{1}\left(p_{n-1} q_{n-2}-p_{n-2} q_{n-1}\right)=(-1)^{n-1}\left(p_1q_{0}-p_{0} q_{1}\right)=(-1)^{n-1} \end{gathered}$$

还有

$$\begin{gathered} p_{n} q_{n-2}-p_{n-2} q_{n}=\left(a_{n} p_{n-1}+p_{n-2}\right) q_{n-2}-p_{n-2}\left(a_{n} q_{n-1}+q_{n-2}\right) \\ \ \\ =a_{n}\left(p_{n-1} q_{n-2}-p_{n-2} q_{n-1}\right)=(-1)^{n-2} \cdot a_{n}=(-1)^{n} a_{n} \end{gathered}$$

$$\begin{gathered} p_{n} q_{n-1}-p_{n-1} q_{n}=(-1)^{n-1} \\ \left.p_{n} q_{n-2}-p_{n-2} q_{n}=(-1\right)^{n} a_{n} \\ \ \\ \frac{p_{n}}{q_{n}}-\frac{p_{n-1}}{q_{n-1}}=\frac{(-1)^{n-1}}{q_{n-1} q_{n}} \\ \ \\ \frac{p_{n}}{q_{n}}-\frac{p_{n-2}}{q_{n-2}}=\frac{(-1)^{n} a_{n}}{q_{n-2} q_{n}} \end{gathered}$$

由此可以知道分奇偶项, 得到的单调性

$\textbf{lemma2}$
$\text{if} \quad n > 3, \quad q_n \geqslant n$

$$\begin{gathered} q_{0}=1, \quad q_{1}=a_{1} \geqslant 1 \\ q_{n}=a_{n} q_{n-1}+q_{n-2} \geqslant q_{n-1}+1=n-1+1=n \end{gathered}$$

根据数学归纳法, 证毕

$\textbf{theorem1}$
$x\ \text{可以用一个奇数个渐进分数的连分数表示出来}$
$\text{那么也一定能用偶数个渐进分数的连分数表示}$

$$\begin{gathered} \left[a_{0}, a_{1}, \dots a_{n}\right]=\left[a_{0}, a_{1}, \dots a_{n}-1,1\right], \quad a_n \geqslant 2 \\ \ \\ \text{if} \quad a_n = 1, \left[a_{0}, a_{1}, \ldots, a_{n-1}, 1\right]=\left[a_{0}, a_{1}, \ldots, a_{n-2}, a_{n-1}+1\right] \end{gathered}$$

$\textbf{theorem2}$
连分数的第 $n$ 个完全商
$a_{n}^{\prime}=\left[a_{n}, a_{n+1}, \ldots, a_{N}\right] \quad(0 \leqslant n \leqslant N)$

$$\begin{gathered} x=a_{0}^{\prime}, \quad x=a_{0}+\frac{1}{a_{1}^{\prime}}=\frac{a_{0} a_{1}^{\prime}+1}{a_{1}^{\prime}} \\ \ \\ \left[a_{n}, a_{1}, \ldots, a_{N}\right]=\left[a_{0}, a_{1}, \ldots a_{n-1},\left[a_{n}, a_{n+1}, \dots, a_{N}\right]\right] \\ \ \\ =\frac{\left[a_{n}, a_{n+1} \ldots, a_{n}\right] p_{n-1}+p_{n-2}}{\left[a_{n}, a_{n+1} \dots, a_{n}\right] q_{n-1}+q_{n-2}} \\ \ \\ x=\frac{a_{n}' p_{n-1}+p_{n-2}}{a_{n}' q_{n-1}+q_{n-2}} \end{gathered}$$

$\textbf{theorem3}$
$\text{除了} \ a_N = 1, \textbf{有} a_{N-1} = [a_{N-1}']-1$
$\text{其他情况, }a_n = [a_n']$

$$\text{注意到, }a_{n}^{\prime}=a_{n}+\frac{1}{a_{n+1}'} \quad(0 \leqslant n \leqslant N-1)$$

$$\begin{gathered} \textbf{i)}\quad N = 0, a_0 = a_0'=[a_0'] \\ \ \\ \textbf{ii)}\quad N > 0, \ n = N-1, a_{n+1}^{\prime}=1, a_{N}=1, \text{我们有} \\ \ \\ a_{N-1}^{\prime}=a_{N-1}+\frac{1}{a_{N}^{\prime}}=a_{N-1}+1>1 \\ \ \\ \text{from } \ n + 1 \ \text{down to} \ n, a_{n+1}^{\prime}>1 \quad(0 \leqslant n \leqslant N-1) \\ \ \\ a_{n}<a_{n}^{\prime}<a_{n}+1 \quad(0 \leqslant n \leqslant N-1) \\ \ \\ a_{n}=\left[a_{n}^{\prime}\right] \quad(0 \leqslant n \leq N-1) \end{gathered}$$

$\textbf{theorem4} \ \textbf{连分数唯一展开}$
如果两个简单连分数
$[a_0, a_1, \ldots, a_{n-1}, a_n^{\prime}], [b_0, b_1, \ldots, b_{n-1}, b_n^{\prime}] \text{有同样的值 } x$
$a_N > 1, b_M > 1, M=N \ \text{并且这两个连分数完全相同}$

$$\begin{gathered} \textbf{i)} \quad n = m \\ \ \\ \text{根据 }\text{theorem3}, a_0 = [x] = b_0, \text{那么 }\\ \ \\ x=\left[a_{0}, a_{1}, \ldots, a_{n-1}, a_{n}^{\prime}\right]=\left[a_{0}, a_{1}, \ldots, a_{n-1}, b_{n}^{\prime}\right] \\ \ \\ \text{if} \ n = 1, a_{0}+\frac{1}{a_{1}^{\prime}}=a_{0}+\frac{1}{b_{1}^{\prime}} \Rightarrow a_{1}^{\prime}=b_{1}^{\prime} \Rightarrow a_{1}=b_1 \\ \ \\ \frac{a_{n}^{\prime} p_{n-1}+p_{n-2}}{a_{n}^{\prime} q_{n-1}+q_{n-2}}=\frac{b_{n}^{\prime} p_{n-1}+p_{n-2}}{b_{n}^{\prime} q_{n-1}+q_{n-2}} \\ \ \\ \left(a_{n}^{\prime}-b_{n}^{\prime}\right)\left(p_{n-1} q_{n-2}-p_{n-2} q_{n-1}\right)=0 \\ \ \\ p_{n-1} q_{n-2}-p_{n-2} q_{n-1}=(-1)^{n} \neq 0 \Rightarrow a_{n}^{\prime}=b_{n}^{\prime} \\ \ \\ \Rightarrow a_{n}=b_{n} \end{gathered}$$

$$\begin{gathered} \textbf{ii)} \quad N \leqslant M, \forall n \leqslant N \\ a_n = b_n \\ \ \\ M > N, \frac{p_{N}}{q_N}=x=\left[a_{0}, a_{1}, \ldots , a_{N}\right]=\left[a_{0}, a_{1}, \ldots, a_{N}, b_{N+1}, \ldots, b_{M}\right] \\ \ \\ =\frac{b_{N+1}^{\prime} p_{N}+p_{N-1}}{b_{N+1}^{\prime} q_{N}+ q_{N-1}} \Rightarrow p_{N}q_{ N-1}-p_{N-1} q_{N}=0 \\ \ \\ \text{这是不可能的 }\\ \ \\ \text{证明过程中, 我们假定了} n \leqslant N \text{的部分, 所以} \\ \ \\ b_{n}^{\prime}=\left[b_{N+1}, \ldots, b_{M}\right] \end{gathered}$$

Euclid算法

$$a_{0}=[x], \quad x=a_{0}+\xi_{0} \quad\left(0 \leqslant \xi_{0}<1\right)$$

$$\begin{gathered} \frac{1}{\xi_{0}}=a_{1}^{\prime}, \quad\left[a_{1}^{\prime}\right]=a_{1}, \quad a_{1}^{\prime}=a_{1}+\xi_{1} \quad\left(0 \leqslant \xi_{1}<1\right) \\ \ \\ \frac{1}{\xi_{1}}=a_{2}^{\prime}=a_{2}+\xi_{2} \quad\left(0 \leqslant \xi_{2}<1\right) \end{gathered}$$

$$\begin{gathered} a_{1}>0, \quad a_{2}>0, \quad \cdots \\ \ \\ \begin{gathered} &a_{1}>0, \quad a_{2}>0, \ldots\\ &x=a_{0}+\xi_{0} \quad\left(0 \leqslant \xi_{0}<1\right)\\ &\frac{1}{\xi_{0}}=a_{1}^{\prime}=a_{1}+\xi_{1} \quad\left(0 \leqslant \xi_{1}<1\right)\\ &\frac{1}{\xi_{1}}=a_{2}^{\prime}=a_{2}+\xi_{2} \quad\left(0 \leqslant \xi_{2}<1\right) \end{gathered} \end{gathered}$$

有理数可以用连分数表示

$$x=\frac{h}{k}, \quad h, k \in \mathbb{Z}, \quad k>1$$

$$\begin{gathered} \frac{h}{k}=a_{0}+\xi_{0} \Rightarrow h=a_{0} k+\xi_{0} k \\ \ \\ \text{let} \quad \xi_{0} k=k_1 \Rightarrow \\ \ \\ \frac{1}{\xi_{0}}=a_{1}^{\prime}=\frac{k}{k_{1}}=a_{1}+\xi_{1} \Rightarrow k=a_{1} k_{1}+\xi_{1} k_{1} \\ \ \\ \text{let} \quad k_{2}=\xi_{1} k_{1} \Rightarrow \cdots \end{gathered}$$

$$\begin{aligned} &h=a_{0} k+k_{1} \quad\left(0<k_{1}<k\right)\\ &k=a_{1} k_{1}+k_{2} \quad\left(0<k_{2}<k_{1}\right)\\ &k_{1}=\dots\\ &k_{N-2}=a_{N-1} k_{N-1}+k_{N} \quad\left(0<k_{N}<k_{N-1}\right)\\ &k_{N-1}=a_{N} k_{N} \end{aligned}$$

$$0<\frac{1}{a_{N}}=\frac{1}{a_{N}^{\prime}}=\frac{k_{N}}{k_{N-1}}=\xi_{n-1}<1$$