数学公式(一)

数学公式(一)

不等式

  • 基本不等式拓展: 2 a b a + b a b a + b 2 a 2 + b 2 2 ( 当且仅当 a = b 时取“ = ”号 ) \frac{{2ab}}{{a+b}} \le \sqrt{{ab}} \le \frac{{a+b}}{{2}} \le \sqrt{{\frac{{a\mathop{{}}\nolimits^{{2}}+b\mathop{{}}\nolimits^{{2}}}}{{2}}}} \left( \text{当}\text{且}\text{仅}\text{当}a=b\text{时}\text{取}\text{“}=\text{”}\text{号} \right)
  • 均值不等式: H n = n i = 1 n 1 x i = n 1 x 1 + 1 x 2 + + 1 x n ( ) G n = i = 1 n x i n = x 1 x 2 x n n ( ) A n = 1 n i = 1 n x i = x 1 + x 2 + + x n n ( ) Q n = i = 1 n x i 2 = x 1 2 + x 2 2 + + x n 2 n ( ) H n G n A n Q n H_{n}=\frac{n}{\sum \limits_{i=1}^{n}\frac{1}{x_{i}}}= \frac{n}{\frac{1}{x_{1}}+ \frac{1}{x_{2}}+ \cdots + \frac{1}{x_{n}}}(调和平均数)\\ G_{n}=\sqrt[n]{\prod \limits_{i=1}^{n}x_{i}}= \sqrt[n]{x_{1}x_{2}\cdots x_{n}}(几何平均数)\\ A_{n}=\frac{1}{n}\sum \limits_{i=1}^{n}x_{i}=\frac{x_{1}+ x_{2}+ \cdots + x_{n}}{n}(算术平均数)\\ Q_{n}=\sqrt{\sum \limits_{i=1}^{n}x_{i}^{2}}= \sqrt{\frac{x_{1}^{2}+ x_{2}^{2}+ \cdots + x_{n}^{2}}{n}}(平方平均数)\\ \\ H_{n}\leq G_{n}\leq A_{n}\leq Q_{n}
    参考调和平均数-几何平均数-算术平均数-平方平均数的4种经典证明
  • 柯西不等式: ( k = 1 n a k b k )  ⁣ ⁣ 2 ( k = 1 n a k 2 ) ( k = 1 n b k 2 ) \left( \sum_{k=1}^n a_k b_k \right)^{\!\!2}\leq \left( \sum_{k=1}^n a_k^2 \right) \left( \sum_{k=1}^n b_k^2 \right)
  • 绝对值不等式: a b a + b a + b \left| a \left| - \left| b \left| \le \left| a+b \left| \le \left| a \left| + \left| b \right| \right. \right. \right. \right. \right. \right. \right. \right.
  • 排序不等式:设 a 1 a 2 a n a_1\leq a_2 \leq \cdots \leq a_n b 1 b 2 b n b_1\leq b_2 \leq \cdots \leq b_n , c 1 , c 2 c n c_1,c_2 \cdots c_n 为b的任意序列,则有: a 1 b n + + a n b 1 a 1 c 1 + + a n c n a 1 b 1 + + a n b n a_1b_n+\cdots +a_nb_1 \leq a_1c_1+\cdots +a_nc_n \leq a_1b_1+\cdots + a_nb_n
    即逆序和 \leq 乱序和 \leq 正序和
  • 伯努利不等式 :
    a > 0 , n > 1 , 则有 \text{若}a > 0,n > 1,\text{则}\text{有}
    a n > 1 + n ( a 1 ) (1) \mathop{{a}}\nolimits^{{n}} > 1+n{ \left( {a-1} \right) } \tag {1}
    a = b 1 n , b > 1 时,有 \text{当}a=\mathop{{b}}\nolimits^{{\frac{{1}}{{n}}}},b > 1\text{时}\text{,}\text{有}
    b 1 n 1 < b 1 n (2) {\mathop{{b}}\nolimits^{{\frac{{1}}{{n}}}}-1 < \frac{{b-1}}{{n}}} \tag{2}

级数

  • 欧拉公式: e i θ = cos θ + i sin θ e^{i\theta }=\cos \theta + i \sin \theta
    特别的当 θ = π \theta = \pi 时: e i π + 1 = 0 e^{i\pi} + 1=0
    证明:
    在这里插入图片描述

级数

  • f ( x ) , g ( x ) [ a , b ] f(x),g(x)在[a,b] 连续: [ a b f ( x ) g ( x ) d x ] 2 a b f 2 ( x ) d x a b g 2 ( x ) d x \left[\int_{a}^{b} f(x) g(x) \mathrm{d} x\right]^{2} \leqslant \int_{a}^{b} f^{2}(x) \mathrm{d} x \int_{a}^{b} g^{2}(x) \mathrm{d} x
    证明:若 f ( x ) 0 f(x)\equiv 0 显然成立
    f ( x ) ≢ 0 f(x)\not\equiv 0 a b f 2 ( x ) d x > 0 \int_{a}^{b}f^2(x)dx>0 ,令 φ ( t ) = a b [ t f ( x ) + g ( x ) ] 2 d x = t 2 a b f 2 ( x ) d x + 2 t a b f ( x ) g ( x ) d x + a b g 2 ( x ) d x \varphi(t)=\int_{a}^{b}[t f(x)+g(x)]^{2} \mathrm{d} x=t^{2} \int_{a}^{b} f^{2}(x) \mathrm{d} x+2 t \int_{a}^{b} f(x) g(x) \mathrm{d} x+\int_{a}^{b} g^{2}(x) \mathrm{d} x
    φ ( t ) \varphi(t) 为二次多项式,在R上满足 φ ( t ) 0 \varphi(t) \geqslant 0 Δ = b 2 4 a c = [ 2 a b f ( x ) g ( x ) d x ] 2 4 a b f 2 ( x ) d x a b g 2 ( x ) d x 0 \Delta =\mathop{{b}}\nolimits^{{2}}-4ac=\left[2\int_{a}^{b} f(x) g(x) \mathrm{d} x\right]^{2} -4 \int_{a}^{b} f^{2}(x) \mathrm{d} x \int_{a}^{b} g^{2}(x) \mathrm{d} x\leqslant 0
  • a b f ( x ) d x = 1 \int_{a}^{b}f(x)dx=1 则有: [ a l x f ( x ) d x ] 2 = [ a b x f ( x ) f ( x ) d x ] 2 a b x 2 f ( x ) d x a b f ( x ) d x = a b x 2 f ( x ) d x \left[\int_{a}^{l} x f(x) \mathrm{d} x\right]^{2}=\left[\int_{a}^{b} x \sqrt{f(x)} \cdot \sqrt{f(x)} \mathrm{d} x\right]^{2} \leqslant \int_{a}^{b} x^{2} f(x) \mathrm{d} x \cdot \int_{a}^{b} f(x) \mathrm{d} x=\int_{a}^{b} x^{2} f(x) \mathrm{d} x

参考

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