第四课。
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不等式右侧的证明：将组合数展开后取对数，再利用$\ln x \leq x - 1$放缩即可得证。
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\begin{lemma}
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\begin{lemma}\label{lemma for Bernolli's law of large numbers}
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令$0 \leq k \leq n$且$c = \binom{2n}{k} \bigg/ \binom{2n}{n}$（即$k-1$项的高度与中心高度的比值），那么
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\[
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\underbrace{\binom{2n}{0} + \binom{2n}{1} + \cdots + \binom{2n}{k-1}}_{\text{高斯曲线中从$-\infty$到$k-1$的面积}} < c \times \underbrace{2^{2n-1}}_{\text{曲线左半部分的面积}}
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218
04FibonacciNumbers.tex
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04FibonacciNumbers.tex
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\chapter{Fibonacci数列}
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\section{Fibonacci问题}
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Fibonacci是递推关系的一个典型问题，这个数列本身也有很多的应用。
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\subsection{问题的提出}
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假定初生的一对雌雄兔子，从出生的第2个月之后每个月都可以生出另外一对雌雄兔。如果第1个月只有一对初生的雌雄兔子，问$n$个月之后共有多少对兔子？
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\subsection{求递推关系}
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设满$n$个月时兔子的对数为$F_n$，那么第$n-1$个月留下的兔子数目为$F_{n-1}$对；第$n-2$个月的所有兔子到了第$n$个月都有了繁殖能力，因此当月新生兔子数目为$F_{n-2}$对。因此，我们可以对Fibonacci数列有如下定义：
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\begin{definition}
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定义Fibonacci数列$\{F_n\}$满足：
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\[F_n = F_{n-1} + F_{n-2}\eqco F_1 = F_2 = 1 \eqper\]
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\end{definition}
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于是我们可以定义$F_0 = 0$。
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\subsection{Fibonacci数列相关恒等式}
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\begin{proposition}
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设$\{F_n\}$为Fibonacci数列，那么：
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\begin{enumerate}
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\item $F_1 + F_2 + \cdots + F_n = F_{n+2} -1$；
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\item $F_1 + F_3 + \cdots + F_{2n-1} = F_{2n}$；
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\item $F_0 - F_1 + F_2 - F_3 + \cdots - F_{2n-1} + F_{2n} = F_{2n-1} - 1$；
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\item $F_1^2 + F_2^2 + \cdots + F_n^2 = F_n F_{n+1}$；
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\item $F_{n-1}F_{n+1} - F_n^2 = (-1)^n$；
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\item $F_n^2 + F_{n-1}^2 = F_{2n-1}$；
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\item $F_{n+1}F_n + F_nF_{n-1} = F_{2n}$。
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\end{enumerate}
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\end{proposition}
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下面依次证明这些性质。
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\begin{enumerate}
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\item 对等式$F_1 + F_2 + \cdots + F_n = F_{n+2} -1$：
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\begin{proof}
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\begin{equation*}
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\begin{aligned}
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F_n & = F_{n+2} - F_{n+1}\\
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F_{n-1} & = F_{n+1} - F_n\\
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&\cdots\\
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F_1 & = F_3 - F_2
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\end{aligned}
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\end{equation*}
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累加所有的式子，得到
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\[F_1 + F_2 + \cdots + F_n = F_{n+2} - F_2 = F_{n+2} -1 \eqper\]
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\end{proof}
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\item 对等式$F_1 + F_3 + \cdots + F_{2n-1} = F_{2n}$：
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\begin{proof}
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\begin{equation*}
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\begin{aligned}
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F_{2n-1} & = F_{2n} - F_{2n-2}\\
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F_{2n-3} & = F_{2n-2} - F_{2n-4}\\
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& \cdots \\
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F_1 & = F_2 - F_0
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\end{aligned}
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\end{equation*}
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累加所有的式子，得到
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\[F_1 + F_3 + \cdots + F_{2n-1} = F_{2n} - F_0 = F_{2n} \eqper\]
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\end{proof}
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\item 对等式$F_0 - F_1 + F_2 - F_3 + \cdots - F_{2n-1} + F_{2n} = F_{2n-1} - 1$：
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\begin{proof}
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将式子1减去式子2的两倍即可得证。
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\end{proof}
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\item 对等式$F_1^2 + F_2^2 + \cdots + F_n^2 = F_n F_{n+1}$：
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\begin{proof}
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\begin{equation*}
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\begin{aligned}
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F_n^2 & = F_n(F_{n+1} - F_{n-1}) = F_nF_{n+1} - F_nF_{n-1}\\
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F_{n-1}^2 & = F_{n-1}F_n - F_{n-1}F_{n-2}\\
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& \cdots \\
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F_1^2 & = F_1F_2 - F_1F_0
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\end{aligned}
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\end{equation*}
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累加即可得证。
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\end{proof}
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\item 对等式$F_{n-1}F_{n+1} - F_n^2 = (-1)^n$：
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\begin{proof}
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使用数学归纳法：
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\begin{enumerate}[label=(\arabic{*})]
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\item 当$n = 1$时，等式成立。
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\item 设当$n = k-1$时，等式成立。那么当$n = k$时，有
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\begin{equation*}
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\begin{aligned}
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F_{k-1}F_{k+1} - F_k^2 & = F_{k-1}(F_{k-1} + F_k) - F_k^2\\
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& = F_{k-1}^2 + F_{k-1}F_k - F_k^2\\
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& = F_{k-1}^2 + F_k(F_{k-1} - F_k)\\
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& = F_{k-1}^2 - F_k F_{k-2}\\
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& = -(F_k F_{k-1} - F_{k-1}^2)\\
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& = (-1)\times (-1)^{k-1}\\
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& = (-1)^k
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\end{aligned}
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\end{equation*}
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即等式对$n=k$也成立。
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\end{enumerate}
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综上，等式对所有的$n \in \mathbb{N}^\ast$成立。
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\end{proof}
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\item 对等式\[F_n^2 + F_{n-1}^2 = F_{2n-1} \tag{$\ast$} \label{fibonacci property 6}\]和\[F_{n+1}F_n + F_nF_{n-1} = F_{2n}\tag{$\ast\ast$} \label{fibonacci property 7}\]
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\begin{proof}
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使用数学归纳法：
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\begin{enumerate}[label=(\arabic{*})]
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\item 当$n=1$时，等式成立。
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\item 设当$n = k-1$时，等式成立。那么当$n = k$时，有
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\begin{equation*}
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\begin{aligned}
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F_k^2 + F_{k-1}^2 & = (F_{k-1} + F_{k-2})^2 + F_{k-1}^2\\
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& = F_{k-1}^2 + F_{k-2}^2 + 2F_{k-1}F_{k-2} + F_{k-1}^2\\
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& = F_{2k-3} + F_{k-1}(F_{k-2} + F_{k-1}) + F_{k-1}F_{k-2}\\
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& = F_{2k-3} + F_kF_{k-1} + F_{k-1}F_{k-2}\\
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& = F_{2k-3} + F_{2k-2}\\
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& = F_{2k-1}
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\end{aligned}
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\end{equation*}
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注意在这个推导过程中我们利用了\eqref{fibonacci property 6}式和\eqref{fibonacci property 7}式在$n=k-1$时成立的条件。
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\begin{equation*}
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\begin{aligned}
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F_{k+1}F_k + F_kF_{k-1} & = (F_k + F_{k-1})F_k + (F_{k-1} + F_{k-2})F_{k-1}\\
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& = F_k^2 + F_{k-1}^2 + F_kF_{k-1} + F_{k-1}F_{k-2}\\
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& = F_{2k-1} + F_{2k-1}\\
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& = F_{2k}
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\end{aligned}
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\end{equation*}
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注意在这个推导中我们用到了\eqref{fibonacci property 6}式在$n = k$时和\eqref{fibonacci property 7}在$n = k-1$时成立的条件。
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因此，通过这样先证\eqref{fibonacci property 6}、再证\eqref{fibonacci property 7}，我们可以同时证明两个等式对任意$n \in \mathbb{N}^\ast$都成立。
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\end{enumerate}
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综上，式子成立。
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\end{proof}
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\end{enumerate}
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\section{广义Fibonacci数列}
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\begin{definition}
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定义广义Fibonacci数列$\{E_n\}$满足：
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\[E_0 = A, E_1 = B, E_{n+1} = E_n + E_{n-1}\]
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\end{definition}
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写出$\{E_n\}$的前几项，有
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\begin{equation*}
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\begin{aligned}
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E_2 & = A + B\\
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E_3 & = A + 2B\\
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E_3 & = 2A + 3B\\
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E_4 & = 3A + 5B\\
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E_5 & = 5A + 8B\\
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E_6 & = 8A + 13B\\
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& \cdots \\
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E_n & = F_{n-1}A + F_nB
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\end{aligned}
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\end{equation*}
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\begin{proposition}
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若令$A = F_a\eqco B = F_{a+1}$，那么有
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\[F_{a+b+1} = E_{b+1} = F_{b}A + F_{b+1}B = F_{a+1}F_{b+1} + F_{a}F_{b}\eqper\]
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\end{proposition}
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\begin{corollary}
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由上式可得
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\begin{itemize}
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\item $F_n^2 + F_{n-1}^2 = F_{2n-1}$；
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\item $F_{n+1}F_n + F_nF_{n-1} = F_{2n}$。
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\end{itemize}
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\end{corollary}
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\section{Fibonnacci数列的通项公式}
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\begin{theorem}
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Fibonacci数列的通项公式为
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\[
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F_n = \frac{1}{\sqrt{5}}\left(\left(\frac{1 + \sqrt{5}}{2}\right)^n - \left(\frac{1 - \sqrt{5}}{2}\right)^n\right) \eqper
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\]
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\end{theorem}
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\begin{proof}
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将$\dfrac{F_{n+1}}{F_n}$进行数值试验，发现其稳定在1.618附近，猜测应为$\dfrac{\sqrt{5} + 1}{2}$。
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设$G_n = cq^n$满足$G_{n+1} = G_n + G_{n+1}$。那么
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\begin{equation*}
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\begin{aligned}
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cq^{n+1} & = cq^n + cq^{n-1}\\
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q^2 & = q + 1
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\end{aligned}
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\end{equation*}
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可以得到
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\begin{equation*}
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\left\{
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\begin{aligned}
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& q_1 = \dfrac{1 + \sqrt{5}}{2}\\
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& q_2 = \dfrac{1 - \sqrt{5}}{2}
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\end{aligned}
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\right.
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\end{equation*}
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猜测$q_2$是造成$\dfrac{F_{n+1}}{F_n}$在$\dfrac{\sqrt{5} + 1}{2}$附近波动的原因。因此我们猜测
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\[F_n = Aq_1^n + Bq_2^n \eqper\]
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由
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\begin{equation*}
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\left\{
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\begin{aligned}
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& F_0 = A + B = 0\\
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& F_1 = Aq_1 + Bq_2 = 1
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\end{aligned}
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\right.
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\end{equation*}
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可以得到
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\begin{equation*}
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\left\{
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\begin{aligned}
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& A = \frac{1}{\sqrt{5}}\\
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& B = -\frac{1}{\sqrt{5}}
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\end{aligned}
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\right.
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\end{equation*}
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于是有
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\[F_n = \frac{1}{\sqrt{5}}\left(\left(\frac{1 + \sqrt{5}}{2}\right)^n - \left(\frac{1 - \sqrt{5}}{2}\right)^n\right) \eqper\]
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\end{proof}
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\section{Fibonacci数列的性质}
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\begin{enumerate}
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\item Fibonacci数列可以作为表示任意正整数$N$的``基''：
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\[N = \sum_{i=2}^n a_iF_i \eqco\]
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其中$a_i = 0, 1$，且$a_ia_{i+1} = 0$（即连续两项至少有一个为零）。例如：
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\[11 = F_6 + F_4 = 8 + 3 \eqper\]
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这种用Fibonacci的表示可以写成一个五位数：10100。
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\item 有所谓的Fibonacci方形，即边长为$F_n$的正方形，可以由$F_i \times F_{i+1}, i = 1, 2, \cdots, n-1$的矩形拼接而成。
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\end{enumerate}
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51
05CombinatorialProbability.tex
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05CombinatorialProbability.tex
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\chapter{组合概率}
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\section{事件与概率}
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略。
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\section{独立重复试验}
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略。
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\section{大数定律}
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抛一枚硬币$n$次，正面朝上的次数记为$X$，有$X \sim B \left(n, \dfrac{1}{2}\right)$。对$k \in [0, n]$，有$P(X = k) = \dfrac{\binom{n}{k}}{2^n}$。
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\begin{theorem}[Bernolli大数定律]\label{Bernolli's law of large numbers}
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$\forall \varDelta > 0$，当$n \to \infty$有 \[P\left(\left|\dfrac{X}{n} - \dfrac{1}{2}\right| < \varepsilon\right) \to 1 \eqper\]
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\end{theorem}
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\begin{theorem}\label{corollary of Bernolli's law of large numbers}
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若$X \sim B \left(2n, \dfrac{1}{2}\right)$，则$\forall t \in [0, n]$，有
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\[P(\vert x - n \vert > t) \leq e^{-\frac{t^2}{n+t}} \eqper\]
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\end{theorem}
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我们先证明定理\ref{corollary of Bernolli's law of large numbers}。
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\begin{proof}
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\begin{equation*}
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\begin{aligned}
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P(\vert X - n \vert > t) & = P(X < n - t \text{或} X > n + t)\\
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& = P(X = 0) + P(X = 1) + \cdots + P(X = n - t - 1)\\
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& \ \ \ + P(X = n + t + 1) + \cdots + P(X = 2n)\\
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& = 2[P(X = 0) + \cdots + P(X = n - t - 1)]\\
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& = \frac{2\left[\binom{2n}{0} + \cdots + \binom{2n}{n-t-1}\right]}{2^{2n}}\\
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& \text{应用引理\ref{lemma for Bernolli's law of large numbers}}\\
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& < \binom{2n}{n-t} \bigg/ \binom{2n}{n}\\
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& \leq e^{-\frac{t^2}{n+t}}\eqper
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\end{aligned}
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\end{equation*}
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\end{proof}
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我们可利用定理\ref{corollary of Bernolli's law of large numbers}证明定理\ref{Bernolli's law of large numbers}。
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\begin{proof}
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我们需要证明
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\[n \to \infty \text{时} P\left(\left|\frac{X}{2n} - \frac{1}{2} \right| < \varepsilon \right) \to 1\]
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只需证
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\[n \to \infty \text{时} P\left(\left|\frac{X}{2n} - \frac{1}{2} \right| > \varepsilon \right) \to 0\]
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注意到
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\begin{equation*}
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P\left(\left|\frac{X}{2n} - \frac{1}{2} \right| > \varepsilon \right) = P\left(\left|X - n\right| > 2n\varepsilon \right)
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\end{equation*}
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应用\ref{corollary of Bernolli's law of large numbers}，这里的$2n\varepsilon$即为定理中的$t$，从而
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\begin{equation*}
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P\left(\left|X - n\right| > 2n\varepsilon \right) \leq e^{- \frac{4n^2\varepsilon^2}{n+2n\varepsilon}}
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\end{equation*}
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当$n \to \infty$时，$e^{- \frac{4n^2\varepsilon^2}{n+2n\varepsilon}} \to 0$，因此$P\left(\left|X - n\right| > 2n\varepsilon \right) \to 0$。
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\end{proof}
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6
离散数学.tex
6
离散数学.tex
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\newtheorem{theorem}{定理}[section]
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\newtheorem{axiom}{公理}[section]
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\newtheorem{definition}{定义}[section]
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\newtheorem{lemma}{引理}[theorem]
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\newtheorem{lemma}{引理}[section]
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\newtheorem{corollary}{推论}[theorem]
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\newtheorem*{corollary*}{推论}
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\newtheorem{example}{例}[section]
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\newtheorem{proposition}{命题}[theorem]
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\newtheorem{proposition}{命题}[section]
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% \renewcommand{\qedsymbol}{} %去掉证明结尾的方框
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@@ -53,4 +53,6 @@
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\include{01LetsCount.tex}
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\include{02CombinatorialTools.tex}
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\include{03Binome.tex}
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\include{04FibonacciNumbers.tex}
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\include{05CombinatorialProbability.tex}
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\end{document}
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