哪对函数满足f(N)〜g(N)? [英] Which pair of functions satisfy f (N) ~g(N)?

问题描述

我刚刚开始使用算法，并且正在执行类似此问题的某些任务:

I've just started working with algorithms and I am doing some tasks like this question:

我认为正确的答案是A.由于功能相同，还是我错过了一些事情?

I think, the right answer is A. As the functions are the same, or do I miss something?

问题:

推荐答案

答案是A.请注意，在这种情况下，

The answer is A. Notice that in that case,

f(N)= N + 2N + 3N = 6N = g(N)

f(N) = N + 2N + 3N = 6N = g(N)

所以f(N)〜g(N).

so f(N) ~ g(N).

对于B中给出的功能，请注意

For the functions given in B, notice that

f(N)=(N +1)+(N + 2)+(N + 3)= 3N + 6

f(N) = (N + 1) + (N + 2) + (N + 3) = 3N + 6

所以当N趋于无穷大时，f(N)/g(N)的极限为3，因此f(N)不等于g(N)的波浪线.

so the limit of f(N) / g(N) as N tends toward infinity is 3, and therefore f(N) is not tilde of g(N).

对于C中的功能，请注意

For the functions in C, notice that

f(N)/g(N)= 1/N ⁵ +1/N ⁴ +1/N ³

f(N) / g(N) = 1 / N⁵ + 1 / N⁴ + 1 / N³

而且，在极限情况下，这不会趋向于一个.

and, in the limit, this does not tend to one.

对于D中的功能，请注意

For the functions in D, notice that

f(N)=对数N +对数2N +对数3N =对数6N ³ = 3对数6N

f(N) = log N + log 2N + log 3N = log 6N³ = 3 log 6N

因此，当N趋于无穷大时，f(N)在g(N)上的极限为3，因此这些函数不会互相代名词.

so the limit of f(N) over g(N) as N tends toward infinity is 3, so the functions aren't tilde of one another.

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