排序方式渐近增长率 [英] Ordering By Asymptotic Growth Rates
问题描述
订购下列EX pressions增加Θ顺序(如果两个函数是增长的顺序相同的,你应该 说明这一事实): N日志N,N- ^ -1,登录N,N- ^数N,10N + N ^ 3/2,π^ N,2的n次方,2 ^数N,2 ^ 2 ^数N,日志N!
Order the following expressions in increasing Θ-order (if two functions are of the same order of growth, you should state this fact): n log n, n^−1, log n, n^log n, 10n + n^3/2, π^n, 2^n, 2^log n, 2^2^log n, log n!.
答:
N R个-1«日志ñ«2 ^日志ñ«的n log N =日志N! «10N + N ^ 3/2«N R个日志ñ«2 ^ N = 2 ^ 2 ^登录n«π的n次方。
n^−1 ≪ log n ≪ 2^log n ≪ n log n = log n! ≪ 10n + n^3/2 ≪ n^log n ≪ 2^n= 2^2^log n≪ π^n.
有人能向我解释我们是如何找到这个sollution,我们是如何解决这个问题。请我需要一些解释。
Can someone please explain to me how did we find this sollution , how did we solve this. Please I need some explanation.
感谢
推荐答案
您应该用事实:
lim(n->∞) f(n)/ g(n) = 0 this gives you Θ(f(n)) < Θ(g(n))
lim(n->∞) f(n)/ g(n) = c; c > 0 this gives you Θ(f(n)) = Θ(g(n))
lim(n->∞) f(n)/ g(n) = ∞ this gives you Θ(f(n)) > Θ(g(n))
现在使用你:
lim(n->∞) n^−1 / log n = lim(n->∞) 1 / (n * log n) = 0.
这立刻让你Θ(N ^ -1)&LT; Θ(log n)的
走吧,其余。
对于一些计算,你可能会发现洛必达法则一>帮助。
For some of the calculations you might find L'Hôpital's rule helpful.
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