对任意实数 (x)
[x - 1 < lfloor x
floor le x le lceil x
ceil < x + 1
]
对任意整数 (n)
[lceil n / 2
ceil + lfloor n / 2
floor = n
]
对任意实数 (x ge 0) 和整数 (a, b > 0)
[leftlceil dfrac{lceil x / a
ceil}b
ight
ceil = leftlceil dfrac x {ab}
ight
ceil
]
[leftlfloor dfrac{lfloor x / a
floor}b
ight
floor = leftlfloor dfrac x {ab}
ight
floor
]
[leftlceil dfrac a b
ight
ceil le dfrac {a + (b - 1)} b
]
[leftlfloor dfrac a b
ight
floor 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 o 0)
[lim_{n o 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}
]