题目1
*证明若f(n)和g(n)是单调递增的函数,则f(n)+g(n)和f(g(n))也是递增的函数。此外,若f(n)和g(n)是非负的,则f(n)g(n)是单调递增的
证明 :
- 不妨假设 n 1 ,n 2 是在函数的定义域内,且 n 1 < n 2
因为 f(n)和 g(n)都是单调递增的,则有 f(n 1) < f(n 2) g(n 1) < g(n 2)
令 T n = f(n) + g(n)
则 T n2 -T n1 = f(n 2) - f(n 1) + g(n 2) - g(n 1) > 0
即 T n2 > T n1
所以 T n 是单调递增的,也即是 f(n)+g(n)是递增的;
而 f(g(n))的证法相同,就不再赘述 <外增内增,复合为增 >
- 假设变量 m,n在函数定义域内,且 m > n,因为递增,则 f(m) > f(n),g(m) > g(n),则 :
f(m)/f(n)>1,g(m)/g(n)>1 f(m)/f(n) > 1,g(m)/g(n) > 1 f(m)/f(n)>1,g(m)/g(n)>1∴[f(m)∗g(m)]/[f(n)∗g(n)]>1 ∴ [f(m)*g(m)]/[f(n)*g(n)] > 1 ∴[f(m)∗g(m)]/[f(n)∗g(n)]>1
∴f(n)∗g(n)是单调递增的 ∴ f(n)*g(n)是单调递增的 ∴f(n)∗g(n)是单调递增的
题目2
证明等式n!=o(nn), n!=w(2n), lg(n!)=θ(n lgn)。并证明n!=w(2n) 和n!=o(nn)。
证:根据斯特林公式,n!=((2πn)(1/2))*((n/e)n)(1+θ(1/n)),存在任意的正常数c>0,n 0>0,当n 0->正无穷, ((2πn)(1/2))*((1/e)n)(1+θ(1/n)) >0,显然有 0 <= ((2πn)(1/2))*((n/e)n^)*(1+θ(1/n)) <= c (n n),得证 ;
假设要存在任意的正常数c>0,n 0>0,当n>n 0,有 0 <= c 2 n < n!成立,则 n!/(2^n) =((2πn)^(1/2))((n/(2 e))^n)(1+θ(1/n)),显然n 0->无穷,n!/(2 n)->无穷,得证。
同理,假设存在正常数c 1>0,c 2>0,n 0>0,当n>n 0,有c1nlgn< lg(n!) < c 2 nlgn。即当n 0->,则n!=lg(((2πn)(1/2))*((n/e)n))(1+θ(1/n)))/(nlgn) >1,则可知必然存在c 1、c 2,有c 1 <= 1 <=c 2,得证。
题目3
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∑i=0dai∗xn=ad∗nd+ad−1∗nd−1+ad−2∗nd−2+ad−3∗nd−3+⋅⋅⋅⋅⋅⋅+a1∗n+a0<=c∗nk \sum_{i=0}^d ai*x^n = a_d*n^d + a_{d-1}*n^{d-1} + a_{d-2}*n^{d-2} + a_{d-3}*n^{d-3} + ······+a_1*n+a_0 <=c*n^k i=0∑dai∗xn=ad∗nd+ad−1∗nd−1+ad−2∗nd−2+ad−3∗nd−3+⋅⋅⋅⋅⋅⋅+a1∗n+a0<=c∗nk
则有
ad+ad−1n+ad−2n2+⋅⋅⋅⋅⋅⋅+a0nd<=cnk a_d+\frac{a_{d-1}}{n}+\frac{a_{d-2}}{n^2}+······+\frac{a_0}{n^d}<=cn^k ad+nad−1+n2ad−2+⋅⋅⋅⋅⋅⋅+nda0<=cnk
令 c = d + 1只要:
n0=max(adk−d,ad−1k−d−1,⋅⋅⋅⋅⋅⋅,ak0) n_0 = max(\sqrt[k-d]{a_d},\sqrt[k-d-1]{a_{d-1}},······,\sqrt[k]a_0) n0=max(k−dad,k−d−1ad−1,⋅⋅⋅⋅⋅⋅,ka0)
即有 p(n) = O(n k) -
0<=cnk=ad∗nd+ad−1∗nd−1+ad−2∗nd−2+ad−3∗nd−3+⋅⋅⋅⋅⋅⋅+a1∗n+a0 0 <= cn^k = a_d*n^d + a_{d-1}*n^{d-1} + a_{d-2}*n^{d-2} + a_{d-3}*n^{d-3} + ······+a_1*n+a_0 0<=cnk=ad∗nd+ad−1∗nd−1+ad−2∗nd−2+ad−3∗nd−3+⋅⋅⋅⋅⋅⋅+a1∗n+a0
c<=ad+ad−1nd−k+ad−2nd−k−1+⋅⋅⋅⋅⋅⋅+a1n1−k+a0n−k c <= a_d + a_{d-1}n^{d-k}+a_{d-2}n^{d-k-1}+······+a_1n^{1-k}+a_0n^{-k} c<=ad+ad−1nd−k+ad−2nd−k−1+⋅⋅⋅⋅⋅⋅+a1n1−k+a0n−k
∵ d >= k
(后者)min=ad+ad−1nd−k+ad−2nd−k−1+⋅⋅⋅⋅⋅⋅+a1n1−k+a0n−k<1 (后者)_{min} = a_d +a_{d-1}n^{d-k}+a_{d-2}n^{d-k-1}+······+a_1n^{1-k}+a_0n^{-k} < 1 (后者)min=ad+ad−1nd−k+ad−2nd−k−1+⋅⋅⋅⋅⋅⋅+a1n1−k+a0n−k<1
故 c = a d -1时满足第二个条件 p(n) = Ω(n k)
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0<=c1nk=ad∗nd+ad−1∗nd−1+ad−2∗nd−2+ad−3∗nd−3+⋅⋅⋅⋅⋅⋅+a1∗n+a0<=c2nk 0 <= c_1n^k = a_d*n^d + a_{d-1}*n^{d-1} + a_{d-2}*n^{d-2} + a_{d-3}*n^{d-3} + ······+a_1*n+a_0 <= c_2n^k 0<=c1nk=ad∗nd+ad−1∗nd−1+ad−2∗nd−2+ad−3∗nd−3+⋅⋅⋅⋅⋅⋅+a1∗n+a0<=c2nk
因为 k = d
所以
c1<=ad+ad−1nd−k+ad−2nd−k−1+⋅⋅⋅⋅⋅⋅+a1n1−k+a0n−k<c2 c_1 <= a_d +a_{d-1}n^{d-k}+a_{d-2}n^{d-k-1}+······+a_1n^{1-k}+a_0n^{-k} < c_2 c1<=ad+ad−1nd−k+ad−2nd−k−1+⋅⋅⋅⋅⋅⋅+a1n1−k+a0n−k<c2
当c 1 = a d -1,c 2 = a 0 +d
n0=max(adk−d,ad−1k−d−1,⋅⋅⋅⋅⋅⋅,ak0) n_0 = max(\sqrt[k-d]{a_d},\sqrt[k-d-1]{a_{d-1}},······,\sqrt[k]a_0) n0=max(k−dad,k−d−1ad−1,⋅⋅⋅⋅⋅⋅,ka0)
得到 p(n) = θ(n k) -
0<=ad∗nd+ad−1∗nd−1+ad−2∗nd−2+ad−3∗nd−3+⋅⋅⋅⋅⋅⋅+a1∗n+a0<=cnk 0 <= a_d*n^d + a_{d-1}*n^{d-1} + a_{d-2}*n^{d-2} + a_{d-3}*n^{d-3} + ······+a_1*n+a_0 <= cn^{k} 0<=ad∗nd+ad−1∗nd−1+ad−2∗nd−2+ad−3∗nd−3+⋅⋅⋅⋅⋅⋅+a1∗n+a0<=cnk
ad+ad−1n+ad−2n2+⋅⋅⋅⋅⋅⋅+n0nd<=cnk−d a_d + \frac{a_{d-1}}{n}+\frac{a_{d-2}}{n^2}+······+\frac{n_0}{n_d} <= cn^{k-d} ad+nad−1+n2ad−2+⋅⋅⋅⋅⋅⋅+ndn0<=cnk−d
因为 k > d
则有当 n → ∞ 时
对于任意的c > 0都成立,故有 p(n) = o(n k)
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0<=cnk=ad∗nd+ad−1∗nd−1+ad−2∗nd−2+ad−3∗nd−3+⋅⋅⋅⋅⋅⋅+a1∗n+a0 0 <= cn^k = a_d*n^d + a_{d-1}*n^{d-1} + a_{d-2}*n^{d-2} + a_{d-3}*n^{d-3} + ······+a_1*n+a_0 0<=cnk=ad∗nd+ad−1∗nd−1+ad−2∗nd−2+ad−3∗nd−3+⋅⋅⋅⋅⋅⋅+a1∗n+a0
c<=adn+⋯+ai+ai−2nk−i+⋯+n0nk c <= a_dn+\cdots+a_i+\frac{a_{i-2}}{n^{k-i}}+\cdots+\frac{n_0}{n^k} c<=adn+⋯+ai+nk−iai−2+⋯+nkn0
n比较大时
c<=adn+⋯+ai+ai−2nk−i+⋯+n0nk c <= a_dn+\cdots+a_i+\frac{a_{i-2}}{n^{k-i}}+\cdots+\frac{n_0}{n^k} c<=adn+⋯+ai+nk−iai−2+⋯+nkn0
此时满足 p(n) = w(n k)