对任意实数 \(x\)
\[x - 1 < \lfloor x \rfloor \le x \le \lceil x \rceil < x + 1
\]
对任意整数 \(n\)
\[\lceil n / 2 \rceil + \lfloor n / 2 \rfloor = n
\]
对任意实数 \(x \ge 0\) 和整数 \(a, b > 0\)
\[\left\lceil \dfrac{\lceil x / a \rceil}b \right\rceil = \left\lceil \dfrac x {ab} \right\rceil
\]
\[\left\lfloor \dfrac{\lfloor x / a \rfloor}b \right\rfloor = \left\lfloor \dfrac x {ab} \right\rfloor
\]
\[\left\lceil \dfrac a b \right\rceil \le \dfrac {a + (b - 1)} b
\]
\[\left\lfloor \dfrac a b \right\rfloor \ge \dfrac {a - (b - 1)} b
\]
对于 \(e\)
\[e^x = \sum_{i = 1}^{\infty} \frac {x^i} {i!}
\]
由此知
\[e^x \ge 1 + x
\]
当 \(|x| \le 1\) 时，我们有近似估计
\[1 + x \le e^x \le 1 + x + x^2
\]
对于 \(x \to 0\)
\[\lim_{n \to \infty} (1 + \frac x n) = e^x
\]
对数
\[a^{log_b c} = c^{log_b a}
\]
(两边取 \(ln\) 证明）
当 \(|x| < 1\) 时
\[ln(1 + x) = \sum_{i = 1} ^ \infty (-1)^{i - 1} \frac{x ^ i} i
\]
对于 \(x > -1\)
\[\frac x {1 + x} \le ln(1 + x) \le x
\]
\[lg(n!) = \Theta(nlgn)
\]
斐波那契
\[F_0 = 0, F_1 = 1
\]
\[F_i = F_{i - 1} + F_{i - 2}
\]
黄金分割率 \(\phi\) 是 \(x^2 = x + 1\) 的两个根
\[\phi = \frac{1 + \sqrt 5} 2 = 1.61803...
\]
\[\hat{\phi} = \frac{1 - \sqrt 5} 2 = -0.61803...
\]
\[F_i = \frac{\phi^i - \hat{\phi}^i} {\sqrt 5}
\]