[基礎數學] Binomial Theorem 與其應用例

首先給出 Binomial Theorem 的陳述

Theorem: 
令 $a,b \in \mathbb{R}$ 且 $n \in \mathbb{N}$ 則
\[{(a + b)^n} = \mathop \sum \limits_{k = 0}^n \left( \begin{array}{l}
n\\
k
\end{array} \right){a^k}{b^{n - k}}\]


在證明 Binomial Theorem之前，我們會需要以下結果來幫助我們，

FACT: 給定 $n,k \in \mathbb{N}$ 且 $n \ge k$，則
\[\left( \begin{array}{l}
n + 1\\
k
\end{array} \right) = \left( \begin{array}{l}
n\\
k
\end{array} \right) + \left( \begin{array}{l}
n\\
k - 1

\end{array} \right)\]
Proof: 觀察左式，由定義可知
\[\left( \begin{array}{l}
n + 1\\
k
\end{array} \right): = \frac{{\left( {n + 1} \right)!}}{{k!\left( {n + 1 - k} \right)!}}\]
故如果我們可以證明右式等同於左式，則證明完畢。現在觀察右式
\[\begin{array}{l}
\left( \begin{array}{l}
n\\
k
\end{array} \right) + \left( \begin{array}{l}
n\\
k - 1
\end{array} \right) = \frac{{n!}}{{k!\left( {n - k} \right)!}} + \frac{{n!}}{{\left( {k - 1} \right)!\left( {n - \left( {k - 1} \right)} \right)!}}\\
\begin{array}{*{20}{c}}
{}&{}&{}&{}&{}
\end{array} = \frac{{n!\left( {n - k + 1} \right)}}{{k!\left( {n - k + 1} \right)\left( {n - k} \right)!}} + \frac{{n!k}}{{k\left( {k - 1} \right)!\left( {n + 1 - k} \right)!}}\\
\begin{array}{*{20}{c}}
{}&{}&{}&{}&{}
\end{array} = \frac{{n!\left( {n - k + 1} \right) + n!k}}{{k!\left( {n + 1 - k} \right)!}}\\
\begin{array}{*{20}{c}}
{}&{}&{}&{}&{}
\end{array} = \frac{{n!\left( {n + 1} \right)}}{{k!\left( {n + 1 - k} \right)!}} = \frac{{\left( {n + 1} \right)!}}{{k!\left( {n + 1 - k} \right)!}}
\end{array}\]注意到最後的等式與原本左式相等，故證明完畢。$\square$

現在我們可以開始證明 Binomial Theorem

Proof: (Binomial Theorem)
給定 $a,b \in \mathbb{R}$ 且 $n \in \mathbb{N}$ 利用數學歸納法，先觀察 $n=1$ 則
\[\begin{array}{l}
{(a + b)^1} = \sum\limits_{k = 0}^1 {\left( {\begin{array}{*{20}{l}}
1\\
k
\end{array}} \right)} {a^k}{b^{1 - k}}\\
\begin{array}{*{20}{c}}
{}&{}&{}
\end{array} = \left( {\begin{array}{*{20}{l}}
1\\
0
\end{array}} \right){a^0}{b^{1 - 0}} + \left( {\begin{array}{*{20}{l}}
1\\
1
\end{array}} \right){a^1}{b^{1 - 1}}\\
\begin{array}{*{20}{c}}
{}&{}&{}
\end{array} = {b^1} + {a^1}
\end{array}\]故成立。現在利用歸納法假設對 Binomial Theorem 對 $n$ 成立，我們要證明 $n+1$ 也成立：亦即假設 ${(a + b)^n} = \mathop \sum \limits_{k = 0}^n \left( \begin{array}{l}
n\\
k
\end{array} \right){a^k}{b^{n - k}}$成立，我們要證明
\[{(a + b)^{n+1}} = \mathop \sum \limits_{k = 0}^{n+1} \left( \begin{array}{l}
n+1\\
k
\end{array} \right){a^k}{b^{n+1 - k}} \;\;\;\;\; (*)
\]現在觀察
\[\begin{array}{l}
{\left( {a + b} \right)^{n + 1}} = {\left( {a + b} \right)^n}\left( {a + b} \right)\\
 = \sum\limits_{k = 0}^n {\left( {\begin{array}{*{20}{l}}
n\\
k
\end{array}} \right)} {a^k}{b^{n - k}}\left( {a + b} \right)\\
 = \sum\limits_{k = 0}^n {\left( {\begin{array}{*{20}{l}}
n\\
k
\end{array}} \right)} {a^{k + 1}}{b^{n - k}} + \sum\limits_{k = 0}^n {\left( {\begin{array}{*{20}{l}}
n\\
k
\end{array}} \right)} {a^k}{b^{n - k + 1}}
\end{array}\]為了使上述兩項和可以合成，我們使用變數變換，對第一項和令 $j=k+1$ 且對第二項和令 $j=k$ 則我們可得
\[\begin{array}{*{20}{l}}
{{{\left( {a + b} \right)}^{n + 1}} = \sum\limits_{k = 0}^n {\left( {\begin{array}{*{20}{l}}
n\\
k
\end{array}} \right)} {a^{k + 1}}{b^{n - k}} + \sum\limits_{k = 0}^n {\left( {\begin{array}{*{20}{l}}
n\\
k
\end{array}} \right)} {a^k}{b^{n - k + 1}}}\\
{\begin{array}{*{20}{c}}
{}&{}&{}&{}
\end{array} = \underbrace {\sum\limits_{j = 1}^{n + 1} {\left( {\begin{array}{*{20}{l}}
n\\
{j - 1}
\end{array}} \right)} {a^j}{b^{n - \left( {j - 1} \right)}}}_{term1} + \underbrace {\sum\limits_{j = 0}^n {\left( {\begin{array}{*{20}{l}}
n\\
j
\end{array}} \right)} {a^j}{b^{n - j + 1}}}_{term2}}\\
\begin{array}{l}
\begin{array}{*{20}{c}}
{}&{}&{}&{}
\end{array} = \underbrace {\left( {\begin{array}{*{20}{l}}
n\\
n
\end{array}} \right){a^{\left( {n + 1} \right)}}{b^{n - \left( {n + 1} \right) + 1}} + \sum\limits_{j = 1}^n {\left( {\begin{array}{*{20}{l}}
n\\
{j - 1}
\end{array}} \right)} {a^j}{b^{n - j + 1}}}_{ = term1}\\
\begin{array}{*{20}{c}}
{}&{}&{}&{}&{}&{}&{}&{}
\end{array} + \underbrace {\sum\limits_{j = 1}^n {\left( {\begin{array}{*{20}{l}}
n\\
j
\end{array}} \right)} {a^j}{b^{n - j + 1}} + \left( {\begin{array}{*{20}{l}}
n\\
0
\end{array}} \right){a^0}{b^{n - 0 + 1}}}_{ = term2}
\end{array}\\
{\begin{array}{*{20}{c}}
{}&{}&{}&{}
\end{array} = {a^{\left( {n + 1} \right)}} + \sum\limits_{j = 1}^n {\left[ {\left( {\begin{array}{*{20}{l}}
n\\
{j - 1}
\end{array}} \right) + \left( {\begin{array}{*{20}{l}}
n\\
j
\end{array}} \right)} \right]} {a^j}{b^{n - j + 1}} + {b^{n + 1}}}
\end{array}\]現在對中間項的和，使用前述的 FACT 可得
\[\begin{array}{l}
{\left( {a + b} \right)^{n + 1}} = {a^{\left( {n + 1} \right)}}{b^{n - \left( {n + 1} \right) + 1}} + \sum\limits_{j = 1}^n {\left[ {\left( {\begin{array}{*{20}{l}}
n\\
{j - 1}
\end{array}} \right) + \left( {\begin{array}{*{20}{l}}
n\\
j
\end{array}} \right)} \right]} {a^j}{b^{n - j + 1}} + {a^0}{b^{n - 0 + 1}}\\
\begin{array}{*{20}{c}}
{}&{}&{}&{}
\end{array} = {a^{\left( {n + 1} \right)}}{b^0} + \sum\limits_{j = 1}^n {\left( {\begin{array}{*{20}{l}}
{n + 1}\\
j
\end{array}} \right)} {a^j}{b^{n + 1 - j}} + {a^0}{b^{n + 1}}\\
\begin{array}{*{20}{c}}
{}&{}&{}&{}
\end{array} = \left( {\begin{array}{*{20}{l}}
{n + 1}\\
{n + 1}
\end{array}} \right){a^{\left( {n + 1} \right)}}{b^0} + \sum\limits_{j = 1}^n {\left( {\begin{array}{*{20}{l}}
{n + 1}\\
j
\end{array}} \right)} {a^j}{b^{n + 1 - j}} + \left( {\begin{array}{*{20}{l}}
{n + 1}\\
0
\end{array}} \right){a^0}{b^{n + 1}}\\
\begin{array}{*{20}{c}}
{}&{}&{}&{}
\end{array} = \sum\limits_{j = 0}^{n + 1} {\left( {\begin{array}{*{20}{l}}
{n + 1}\\
j
\end{array}} \right)} {a^j}{b^{n + 1 - j}}
\end{array}\]上式滿足 $(*)$ 故得證。 $\square$


以下我們看幾個例子：

Example 1: 
試求\[\mathop \sum \limits_{i = 0}^n \left( \begin{array}{l}
n\\
i
\end{array} \right) = ?\]
Proof: 利用 Binomial Theorem，令 $a=b=1$ 可立刻得到
\[\begin{array}{l}
{(a + b)^n} = \sum\limits_{k = 0}^n {\left( {\begin{array}{*{20}{l}}
n\\
k
\end{array}} \right)} {a^k}{b^{n - k}}\\
 \Rightarrow \sum\limits_{k = 0}^n {\left( {\begin{array}{*{20}{l}}
n\\
k
\end{array}} \right)} {1^k}{1^{n - k}} = {(1 + 1)^n} = {2^n}
\end{array}\]

Example 2:
試求
\[\mathop \sum \limits_{i = 0}^n {\left( { - 1} \right)^i}\left( \begin{array}{l}
n\\
i
\end{array} \right) = ?\]
Proof: 利用 Binomial Theorem，令 $a=-1$ 與 $b=1$ 可立刻得到
\[\begin{array}{l}
{(a + b)^n} = \sum\limits_{k = 0}^n {\left( {\begin{array}{*{20}{l}}
n\\
k
\end{array}} \right)} {a^k}{b^{n - k}}\\
 \Rightarrow \sum\limits_{k = 0}^n {\left( {\begin{array}{*{20}{l}}
n\\
k
\end{array}} \right)} {\left( { - 1} \right)^k}{1^{n - k}} = {( - 1 + 1)^n} = {0^n} = 0
\end{array}\]

[基礎數學] 數學歸納法應用例

以下我們簡介證明手段中一個重要的工具：數學歸納法 (Mathematical Induction) 以及一些 應用例子，以下我們給出定義

Theorem: (Mathematical Induction)
給定 $n, n_0 \in \mathbb{N}$ 滿足  $n \ge n_0$，則命題句  $P(n)$ 對任意 $n \ge n_0$ 為真若下列兩個條件成立：

1. $P(n_0)$ 為真
2. 對任意 $k \ge n_0$，若 $P(k)$ 為真，則 $P(k+1)$ 為真。

Comments:
1. 對於條件 1，我們稱其為 base case， 對於條件 2，我們稱其為 induction step
2. 想法：關於數學歸納法可以想成推倒骨牌的遊戲，條件一可以想像成推倒第一片骨牌，然後條件二假設如果第 $n$ 片骨牌被推倒，則 $n+1$ 片骨牌必定要倒。


以下我們看幾個例子：

================
Claim: 對任意 $p > -1$ 與 $n \in \mathbb{N}$，下列結果恆成立：
\[
(1+p)^n \ge 1+np
\]===============
Proof:
利用歸納法證明，先使用 $(*)$
考慮 $n=1$ 與 $p = 0$，該結果可簡化為
\[
1 \ge 1
\]故可馬上得知 $n=1$ 時候上述命題成立。接著我們使用 $(**)$，故現在假設
\[
(1+p)^n \ge 1+np
\]我們要證明
\[
(1+p)^{n+1} \ge 1+(n+1)p
\]
觀察
\[
(1+p)^{n+1} = (1+p)^n (1+p) \ge (1+np) (1+p) = 1+n p + p + n p^2
\]
注意到 $np^2 \ge 0$ 故我們有
\[
(1+p)^{n+1} \ge 1 + (n + 1)p\;\;\;\;\;\; \square
\]

Claim: 令
\[A = \left[ {\begin{array}{*{20}{c}}
1&b\\
0&1
\end{array}} \right]\]試證明
\[{A^n} = \left[ {\begin{array}{*{20}{c}}
1&{nb}\\
0&1
\end{array}} \right]\]
Proof: 首先觀察 $n=1$，原式成立
\[{A^1} = \left[ {\begin{array}{*{20}{c}}
1&b\\
0&1
\end{array}} \right]\]現在利用歸納法，假設
\[{A^n} = \left[ {\begin{array}{*{20}{c}}
1&{nb}\\
0&1
\end{array}} \right]\]我們要證明 $n+1$ 成立，亦即我們要證明
\[{A^{n + 1}} = \left[ {\begin{array}{*{20}{c}}
1&{\left( {n + 1} \right)b}\\
0&1
\end{array}} \right]\]現在觀察上式左方，
\[\begin{array}{l}
{A^{n + 1}} = {A^n}A\\
\begin{array}{*{20}{c}}
{}&{}&{}
\end{array} = \left[ {\begin{array}{*{20}{c}}
1&{nb}\\
0&1
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
1&b\\
0&1
\end{array}} \right]\\
\begin{array}{*{20}{c}}
{}&{}&{}
\end{array} = \left[ {\begin{array}{*{20}{c}}
1&{b + nb}\\
0&1
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
1&{\left( {1 + n} \right)b}\\
0&1
\end{array}} \right]
\end{array}\]即為所求。$\square$


讀者不妨練習幾題：
Exercise 1: Show that $1+2+...+n = \frac{n(n+1)}{2}$ for all $n \in \mathbb{N}$
Exercise 2: Show that $n < 2^n$ for all $n \in \mathbb{N}$
Exercise 3: Show that $2^n<n!$ for all $n\geq 4$

Claim: 最小整數原理：
對任意非空集合 $S \neq \emptyset$ 且 $S \subset \mathbb{N}$，則 $S$ 必有最小元素
Proof:
給定任意非空集合 $S := \{n: n \in \mathbb{N}\} $， 首先觀察若 $n=1$，我們有 $S = \{1\}$，且有最小值 $1$ 為其最小元素。

現在用歸納法，假設非空集合 $S_n = \{i: i \le n, i \in \mathbb{N}\} $有最小元素。我們證明 $S_{n+1} := \{i: i \le n+1, i \in \mathbb{N} \}$ 有最小元素。

注意到
$S_{n+1} = S_n \cup \{i: i= n+1\}$ 又因為 $S_n$ 有最小元素，故 $S_{n+1}$ 亦存在最小元素。$\square$

Comment:
上述最小整數原理不適用於任意實數 或者 有理數，舉例而言，若觀察 $E :=\{1, 1/2, 1/3, 1/4, ...\}$ 則此集合不存在最小元素。因為若  $x \in E$ 為最小值 則 必存在 $y \in E$ 使得 $y \leq x$，此時 $y$ 成為新的最小值。