1数学基础

1数学基础

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数学基础 ± 符号说明 ² 取整函数 ² 对数 ² 阶乘 ± 求和 ² 估计和式的上界 ± 递推方程求解 ± 递推方程中涉及 x x 的处理方法 1
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符号说明 2 = = = + ab n b a n ab n b a n n n n , 2 2 取整函数 x :小于等于 x 的最大整数 x :大于等于 x 的最小整数 性质 x -1 < x ⎦≤ x ≤⎡ x < x +1
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3 n c n n a n n n n n n n n l k a n k k b b log log ) log(log log log ) (log log ) log (lg , log log log log 2 2 = = = = = = 性质: 符号: 对数函数
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4 阶乘 )) 1 ( 1 ( ) ( 2 ! n Θ e n n n n + = π n ! = o ( n n ) 2 n =o ( n !), log n ! = Θ ( n log n ) Stirling 公式
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5 求和 = + = + = = n k n n k k O n k x x x 1 1 0 ) 1 ( ln 1 1 1 例1 ∑∑ = = = = = = = = + = + = + n k n k n k n k n k n k n k k k k k k k k 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 ) 1 1 1 ( ) 1 ( 1 公式
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例题 6 1 2 ) 1 ( ) 1 2 ( 2 2 2 2 2 ) 1 ( 2 2 2 ) 2 2 ( 2 1 0 1 0 1 1 0 1 1 1 1 1 1 1 1 + = = = + = = = = = = = = = = = = k k k k t t k t t k t t k t t k t t k t t k t t t k t t k t t k k t t t t t t t t 例2
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估计和式的上界 7 r a r a r a a r k r a a na a k k k k n k k k k n k k = = < = = = + = 1 1 , 0 , 0 0 0 0 0 0 1 1 max 为常数,则 对于一切 方法一:放大法
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