2016年3月15日 · O(2^(n+1)) is the same as O(2 * 2^n), and you can always pull out constant factors, so it is the same as O(2^n). However, constant factors are the only thing you can pull out. 2^(2n) can be expressed as (2^n)(2^n), and 2^n isn't a constant. So, the answer to your questions are yes and no.

2005年1月11日 · I'm to prove that for n>=4, 2^n < n! holds, but I don't know where to go after the inductive hypothesis that it holds for n>= 4 after showing it works for the base case (n = 4). Here are my steps so far: 2^(n+1) < (n+1)! 2*(2^n) < (n+1)(n!) but I …

2014年2月14日 · n 2^n grows asymptotically faster than 2^n. That's that question answered. But you could ask "if algorithm A takes 2^n nanoseconds, and algorithm B takes n 2^n nanoseconds, what is the biggest n where I can find a solution in a second / minute / hour / day / month / year? And the answers are n = 29/35/41/46/51/54 vs. 25/30/36/40/45/49.

You are right that for n = 3, n^100 is greater than 2^n but once n > 1000, 2^n is always greater than n^100 so we can disregard n^100 in O(2^n + n^100) for n much greater than 1000. For a formal mathematical description of Big O notation the wikipedia article does a good job

2011年4月1日 · I need to implement and test an algorithm with a 2^n complexity. I have been trying to find one for a while. If there is any way I can acheive this by implementation -- with a exact complexity of 2^n that would be optimal. If anyone knows of a location I can find an example, or could help me implement one, that would be awesome :-).

2017年2月8日 · ((n+2)!)/(n!) = (n+2)(n+1) Remember that: n! =n(n-1)(n-2)...1 And so (n+2)! =(n+2)(n+1)(n)(n-1) ... 1 \ \ \ \ \ \ \ \ \ \ \ \ \ \=(n+2)(n+1)n!

2015年12月10日 · While there isn't a simplification of ((2n)!)/(n!), there are other ways of expressing it. For example ((2n)!)/(n!) = prod_(k=0)^(n-1)(2n-k) = (2n)(2n-1)...(n+1) This follows directly from the definition of the factorial function and canceling common factors from the numerator and denominator.

n^2+3n+2 We can rewrite the numerator as: ((n+2) * (n+2-1) * (n+2-2)!)/((n)!) =((n+2) * (n+1) * (n)!)/((n)!) We can cancel (n)! and (n)! out: =((n+2) * (n+1) * 1)/1 ...

2017年8月20日 · The answer is =2(2^n-1) This is a geometric series, the common ratio is r=2 S=2+2^2+2^3+.....2^n 2S=2^2+2 ...

2012年8月25日 · Using Extended Masters Theorem T(n)=2T(n/2)+n/logn can be solved easily as follows. Here n/log n part can be rewritten as n * (logn)^-1, Effictively maaking value of p=-1. Now Extended Masters Theorem can be applied easily, it will relate to case 2b of Extended Masters Theorem . T(n)= O(nloglogn)