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# System prepended metadata

title: 數學作業A
tags: [homework of calculus]

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###### tags: `homework of calculus`
# 數學作業A
## Calculus
![](https://i.imgur.com/CShQrHZ.png)
$f(x)=\left\{\begin{array}{lll} x \in\mathbb{Q}&f(x)=0 \\ x\in\mathbb{Q^c}&f(x)=1\end{array}\right.$
此函數處處不連續，應該符合9(b)的需求，至於為啥處處不連續你應該會證吧，假設他連續就行
好，我去證明，這例子好像之前作業有出現XD謝謝學長
不過題目有沒有要求$a=0$是要求她連續的呢?恩~(思索
如果有，把函數改成$f(x)=\left\{\begin{array}{lll} x \in\mathbb{Q}&f(x)=0 \\ x\in\mathbb{Q^c}&f(x)=x\end{array}\right.$
## Linear Algebra
*  prove $\small\left|\begin{array}{ccc} 1 & a_1 & a^2_1 & a^3_1&\cdots & a_1^{n-1} \\ 1 &a_2 &a_2^2 & a_2^2 &\cdots & a_2^{n-1} \\ \vdots & \vdots &\vdots &\vdots & \ddots & \vdots \\ 1 & a_{n-1} &a_{n-1}^2&a_{n-1}^2&\cdots&a_{n-1}^{n-1}\\1 & a_n & a_n^2 &a_n^2 &\cdots & a_n^{n-1}\end{array}\right|=\displaystyle\prod_{1\leq i\leq j\leq n}\left(a_j-a_i\right)$
*  solution
    1.  $n=2 ,\quad\left|\begin{array}{ccc}1&a_1\\ 1&a_2\end{array}\right|=(a_2-a_1)$, 成立
    2. 假設 $n=k$時，$\left|\begin{array}{ccc} 1 & a_1 & a^2_1 & a^3_1&\cdots & a_1^{k-1} \\ 1 &a_2 &a_2^2 & a_2^2 &\cdots & a_2^{k-1} \\ \vdots & \vdots &\vdots &\vdots & \ddots & \vdots\\ 1 & a_{k} &a_{k}^2&a_{k}^2&\cdots&a_{k}^{k-1} \end{array}\right|=\displaystyle\prod_{1\leq i\leq j\leq k}\left(a_j-a_i\right)$成立
    3. $n=k+1$時，考慮 $\large f(t)=\left|\begin{array}{ccc} 1 & a_1 & a^2_1 & a^3_1&\cdots & a_1^{k} \\ 1 &a_2 &a_2^2 & a_2^2 &\cdots & a_2^{k} \\ \vdots & \vdots &\vdots &\vdots & \ddots & \vdots\\ 1 & a_{k} &a_{k}^2&a_{k}^2&\cdots&a_{k}^{k} \\ 1 & t & t^2 &t^2 &\cdots & t^{n-1}\end{array}\right|$
    4. 我們可以發現當$t=a_1、a_2、a_3\cdots a_{k}$時，$f(t)=0$(有兩列相同)，這代表$a_1、a_2、a_3\cdots a_{k}$是$f(t)$的根，且$f(t)$可以改寫為
    \begin{gather}f(t)=C\times\displaystyle\prod_{1\leq i\leq k}\left(t-a_i\right)\quad C是常數(首項係數)
    \end{gather}
    而我們可以發現首相係數剛好是$\left|\begin{array}{ccc} 1 & a_1 & a^2_1 & a^3_1&\cdots & a_1^{k-1} \\ 1 &a_2 &a_2^2 & a_2^2 &\cdots & a_2^{k-1} \\ \vdots & \vdots &\vdots &\vdots & \ddots & \vdots\\ 1 & a_{k-1} &a_{k-1}^2&a_{k-1}^2&\cdots&a_{k-1}^{k-1} \end{array}\right|$，所以
    \begin{gather}
    f(a_{k+1})=\displaystyle\prod_{1\leq i\leq j\leq k}\left(a_j-a_i\right)\times\displaystyle\prod_{1\leq i\leq k}\left(a_{k+1}-a_i\right)=\displaystyle\prod_{1\leq i\leq j\leq k+1}\left(a_j-a_i\right)
    \end{gather}
    成立，by 數學歸納法，得證: $\small\left|\begin{array}{ccc} 1 & a_1 & a^2_1 & a^3_1&\cdots & a_1^{n-1} \\ 1 &a_2 &a_2^2 & a_2^2 &\cdots & a_2^{n-1} \\ \vdots & \vdots &\vdots &\vdots & \ddots & \vdots \\ 1 & a_{n-1} &a_{n-1}^2&a_{n-1}^2&\cdots&a_{n-1}^{n-1}\\1 & a_n & a_n^2 &a_n^2 &\cdots & a_n^{n-1}\end{array}\right|=\displaystyle\prod_{1\leq i\leq j\leq n}\left(a_j-a_i\right)$
## Introductory mathematics
### Ch-3，Mathematical Introduction
#### 16
hint: k is odd $\left(1-\frac{-1}{k}\right)\times\left(1-\frac{1}{k+1}\right)=\frac{k+1}{k}\times\frac{k}{k+1}=1$
#### 19
hint: k is even $\left(1+\frac{1}{k}\right)\times\left(1+\frac{-1}{k+1}\right)=\frac{k+1}{k}\times\frac{k}{k+1}=1$
#### 37
$C^{n}_{2}=\frac{n(n-1)}{2}$
#### 101
use the lemma:if $p|ab$ then $p|a$ or $p|b$.
#### 102
use the lemma:if $gcd(a,b)=d$ then there exist $x,y\in\mathbb{Z}$ such that $ax+by=d$
#### 106
use the lemma:if $p|ab$ then $p|a$ or $p|b$.
#### 109
輾轉相除法
#### 112
使用反證法並使用以下敘述
note : $\mathbb{Z}$ can be classified by $4n$、$4n+1$、$4n+2$、$4n+3$
## Calculus
* Prove if $f$ and $g$ are Riemann integralbe on $[a,b]$, then so is $fg$
    * Observe that $fg=\frac{(f+g)^2-f^2-g^2}{2}$. Since $f$ and $g$ is integrable, $f+g$ is integrable.Thus, we prove that if $f$ is Riemann integralbe on $[a,b]$, then so is $f^2$
    * Let $P=\{x_j|1<j<n,a=x_1<x_2<\cdots<x_{j-1}<x_{j}<x_n=b\}$ be a partition of $[a,b]$, $M_j(f)$ and $m_j(f)$ is maximum of $f$ on $[x_{j-1},x_j]$, minimum of $f$ on $[x_{j-1},x_j]$ respectively.
    * Since $M_j(f^2)=\left(M_j(|f|)\right)^2$ and $m_j(f^2)=\left(m_j(|f|)\right)^2$, it is clear that
\begin{align}
M_j(f^2)-m_j(f^2)&=\left(M_j(|f|)\right)^2-\left(m_j(|f|)\right)^2\\
                 &=\left(M_j(|f|)\right)+\left(m_j(|f|)\right)\left(M_j(|f|)\right)-\left(m_j(|f|)\right)\\
                 &=2M\left(M_j(|f|)-m_j(|f|)\right)
\end{align}
where $M=sup|f|\left([a,b]\right)$; that is, $|f(x)|\leq M$ for all $x\in[a,b]$. Multiplying the displayed inequality by $\Delta x_j$ and summing over $j=1,2,3,\cdots,n$, we have 
\begin{align}
U(f^2,P)-L(f^2,P)\leq 2M\left(U(|f|,P)-L(|f|,P)\right).
\end{align}
Hence, it follows from Lemma : if $f$ is integrable , then $|f|$ is integrable. So $f^2$ is integrable too. That is mean $fg$ is also integrable.