排序不等式：设 \(a_1 \leq a_2 \leq ... \leq a_n, b_1 \leq b_2 \leq ... \leq b_n\)，\(P\) 是 \(\{1, 2, ..., n\}\) 的一个排列，则有
\[\displaystyle\sum_{i = 1}^n a_ib_{n - i + 1} \leq \displaystyle\sum_{i = 1}^n a_ib_{P(i)} \leq \displaystyle\sum_{i = 1}^n a_ib_i \]切比雪夫单调不等式：设 \(a_1 \leq a_2 \leq ... \leq a_n, b_1 \leq b_2 \leq ... \leq b_n\)，有
\[n\displaystyle\sum_{i = 1}^n a_ib_{n - i + 1} \leq (\displaystyle\sum_{i = 1}^na_i)(\displaystyle\sum_{i = 1}^nb_i) \leq n\displaystyle\sum_{i = 1}^n a_ib_i \]即设 \(f(x), g(x)\) 在区间 \([a, b]\) 上单调不减，有
\[(b - a)\displaystyle\int_a^b f(x)g(a + b - x) \, \mathrm{d}x \leq (\displaystyle\int_a^b f(x) \, \mathrm{d}x)(\displaystyle\int_a^b g(x) \, \mathrm{d}x) \leq (b - a)\displaystyle\int_a^b f(x)g(x) \, \mathrm{d}x \]以上前两式在 \(a_1 = a_2 = ... = a_n\) 或 \(b_1 = b_2 = ... = b_n\) 时取等，后一式在 \(f(x)\) 或 \(g(x)\) 在区间 \([a, b]\) 上的每个值都相等时取等．
拉格朗日插值：对于一个 \(n\) 次函数 \(f(x)\)，若已知 \(x_i\) 与 \(f(x_i) (1 \leq i \leq n + 1\) 且 \(x_i\) 互不相同 \()\)（即已知该函数经过平面上 \(n + 1\) 个点的位置，且点两两不重合），则
\[f(x) = \displaystyle\sum_{i = 1}^{n + 1} f(x_i) \cdot \frac{\displaystyle\prod_{1 \leq k \leq n + 1, k \neq i} (x - x_k) }{ \displaystyle\prod_{1 \leq k \leq n + 1, k \neq i} (x_i - x_k) } \]例如未知 \(f(x) = 3x^2 + 2x + 5\)，而知道 \(f(1) = 10, f(3) = 38, f(7) = 166\)，则
\[f(x) = f(1) \cdot \frac{ (x - 3)(x - 7) }{(1 - 3) \times (1 - 7)} + f(3) \cdot \frac{ (x - 1)(x - 7) }{(3 - 1) \times (3 - 7)} + f(7) \cdot \frac{ (x - 1)(x - 3) }{(7 - 1) \times (7 - 3)} \]\[= 5 \times \frac{ (x - 3)(x - 7) }{6} - 19 \times \frac{ (x - 1)(x - 7) }{4} + 83 \times \frac{ (x - 1)(x - 3) }{12} \]将 \(x = 2\) 代入上式，可得
\[f(2) = 5 \times \frac{ (2 - 3) \times (2 - 7) }{6} - 19 \times \frac{ (2 - 1) \times (2 - 7) }{4} + 83 \times \frac{ (2 - 1) \times (2 - 3) }{12} \]\[= 21 \]设 \(H_k = \displaystyle\sum_{i = 1}^k \dfrac{1}{i}\)（调和级数），则 \(\displaystyle\sum_{i = 1}^{n - 1} H_i = nH_n - n\)
证明：
\[\displaystyle\sum_{i = 1}^{n - 1} H_i = (n - 1) + \frac{n - 2}{2} + \frac{n - 3}{3} + ... + \frac{1}{n - 1} \]\[= \displaystyle\sum_{i = 1}^{n - 1} \frac{n - i}{i} \]\[= \displaystyle\sum_{i = 1}^{n - 1} \frac{n}{i} - \displaystyle\sum_{i = 1}^{n - 1} 1 \]\[= n\displaystyle\sum_{i = 1}^{n - 1} \frac{1}{i} - n + 1 \]\[= n(H_{n - 1} + \frac{1}{n}) - n \]\[= nH_n - n \]证毕．
\(\mathrm{D}(x^m) = mx^{m - 1}, \mathrm{\Delta}(x^{\underline{m}}) = mx^{\underline{m - 1}}\)，其中 \(x^{\underline{m}} = \displaystyle\prod_{i = 0}^{m - 1} (x - i) = x(x - 1)(x - 2) ... (x - m + 1)\)，\(\mathrm{D}\) 为微分算子，\(\mathrm{\Delta}\) 为差分算子．
\(g(x) = \mathrm{D}f(x)\) 当且仅当 \(\displaystyle\int g(x) \, \mathrm{d}x = f(x) + C\)，则 \(\displaystyle\int_a^b g(x) \, \mathrm{d}x = f(x) \big|_a^b\)
\(g(x) = \mathrm{\Delta}f(x)\) 当且仅当 \(\displaystyle\sum g(x) \, \mathrm{\delta}x = f(x) + C\)，则 \(\displaystyle\sum_a^b g(x) \, \mathrm{\delta}x = f(x) \big|_a^b = \displaystyle\sum_{i = a}^{b - 1} g(x)\)
\(\displaystyle\int_0^n x^m \, \mathrm{d}x = \dfrac{n^{m + 1}}{ m + 1}\)，\(\displaystyle\sum_0^n x^{\underline{m}} = \dfrac{n^{\underline{m + 1}}}{m + 1} (m, n \in \N)\)．
扩展欧拉定理：
\[a^b \equiv \left\{ \begin{array}{rcl} a^{b \bmod \varphi(m)} & \gcd(a, m) = 1, \\ a^b & \gcd(a, m) \neq 1, b < \varphi(m), \\ a^{(b \bmod \varphi(m)) + \varphi(m)} & \gcd(a, m) \neq 1, b \geq \varphi(m). \end{array} \right. \]延伸：若 \(a, m\) 互质，有
\[a \cdot a^{\varphi(m) - 1} \equiv 1 \pmod m \]上式当 \(m\) 是质数时即为费马小定理．
泰勒公式：若 \(f(x)\) 在包含 \(x_0\) 的某个开区间 \((a, b)\) 上有 \((n + 1)\) 阶的导数，则对于任一 \(x \in (a, b)\)，有
\[f(x) = \displaystyle\sum_{i = 0}^{n} \frac{f^{(i)}(x_0)}{i!}(x - x_0)^i + R_n(x) \]其中 \(R_n(x) = \dfrac{f^{(n + 1)}(\varepsilon)}{(n + 1)!}(x - x_0)^{(n + 1)}\)，其中 \(\varepsilon\) 是 \(x\) 与 \(x_0\) 之间的某个数。其中 \(P_n(x) = \displaystyle\sum_{i = 0}^n \dfrac{f^{(i)}(x_0)}{i!} (x - x_0)^i\) 称为 \(n\) 阶泰勒多项式，其与 \(f(x)\) 的误差 \(R_n(x)\) 称为 \(n\) 阶泰勒余项。
组合
\[\text{阶乘展开式：} {n \choose k} = \dfrac{n!}{k!(n - k)!} \]\[\text{对称恒等式：} {n \choose k} = {n \choose {n - k}} \]\[\text{吸收恒等式：} {n \choose k} = \dfrac{n}{k} {{n - 1} \choose {k - 1}} \]\[\text{归纳恒等式：} {n \choose k} = {{n - 1} \choose k} + {{n - 1} \choose {k - 1}} \]\[\text{上指标反转：} {n \choose k} = {{k - n - 1} \choose k} \]\[\text{三项式版恒等式：} {n \choose m} {m \choose k} = {n \choose k} {{n - k} \choose {m - k}} \]\[\text{二项式定理：} (a + b)^p = \displaystyle\sum_{k = 0}^p {p \choose k} a^kb^{p - k} \]\[\text{平行求和法：} \displaystyle\sum_{k = 0}^n {{m + k} \choose k} = {{m + n + 1} \choose n} \]\[\text{上指标求和法：} \displaystyle\sum_{k = 0}^n {k \choose m} = {{n + 1} \choose {m + 1}} \]\[\text{范德蒙德卷积：} \displaystyle\sum_{k} {s \choose {m + k}}{l \choose {n - k}} = {{s + l} \choose {m + n}} \] 标签：frac,sum,leq,数学,choose,杂记,displaystyle,mathrm From： https://www.cnblogs.com/wf715/p/Mathematics.html