Does for all k, n^k is O(2^n)?

Question

Is it true that for all k, n^k is O(2^n)?

What I actually want to know whether this upper bound is correct. Like we can say n^2 is O(n^3) since it's true that n^2 < c * n^3, where c is a constant. SO similarly can I say that n^k < c * 2^n, for all value of k?

Are you asking if *calculating* n^k is an O(2^n) time complexity? — paxdiablo, Sep 08 '16 at 02:25
I'm voting to close this question as off-topic because it is a math question. — Raymond Chen, Sep 08 '16 at 02:25
@paxdiablo, yes, whether calculating n^k takes less than O(2^n) time for any value of k? — russell, Sep 08 '16 at 02:28
@RaymondChen, so you think it's not algorithm, big-o, asymptotic-complexity question!!!! — russell, Sep 08 '16 at 02:29
It is not a programming question, which is what this site is for. — Raymond Chen, Sep 08 '16 at 02:32
Russel, then it depends on how you define an operation. If multiplication is O(1) the exponentiation is at most O(n) [since `x^7` can be done as `x*x*x*x*x*x*x`]. I think you're going to need to add some detail to the question (if it survives). — paxdiablo, Sep 08 '16 at 02:32
The answer is no. Many powers can be calculated in *O(1)* shifts, or in *O(log N)* multiplication steps. — user207421, Sep 08 '16 at 02:33
@paxdiablo , I added some details about the question. Please help me if possible. Thanks. — russell, Sep 08 '16 at 02:47
for any k, there exists a corresponding constant c such that c * 2 ^ n is larger. you should find a algorithm book and read it first. O is actually a fairly weak bound — Jason Hu, Sep 08 '16 at 02:50
to directly answer your question, O(1) == O(log n) == O(n) == O(2 ^ n) so yes, you are right. but it doesn't make sense to relax O notation to your extent. — Jason Hu, Sep 08 '16 at 02:53
Read the defiition of O(f(n)) again. Does it say "for all n" at any point? Maybe "for all n that satisfy some condition"? What is that condition? — n. m. could be an AI, Sep 08 '16 at 03:10
No, it doesn't say anything about n, only says for any value of k. — russell, Sep 08 '16 at 03:11

score 0 · Accepted Answer · answered Sep 08 '16 at 02:56

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To show that there is always a constant c for any constant k such that n^k < c * 2^n, consider this: (n+1)^k / n^k = ((n+1)/n)^k. As n increases, (n+1)/n tends to 1, thus ((n+1)/n)^k tends to 1. This implies that the relative increase between values decreases as n increases.

Now consider 2^(n+1) / 2^n. This is clearly 2. Thus the relative increase stays the same as n increases. Thus there will be a c for every k such that n^k < c * 2^n

answered Sep 08 '16 at 02:56

Benson Lin

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Thanks @Benson Lin, now it's bit clear . – russell Sep 08 '16 at 03:07
not even remotely a proof of the concept – Yerken Sep 08 '16 at 03:27

Does for all k, n^k is O(2^n)?

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