对于所有自然数m，f（n）= log（n）^ m是O（n）吗？

Question

A TA told me that this is true today but I was unable to verify this by googling. This is saying functions like log(n)^2, log(n)^3, ... , log(n)^m are all O(n).

Is this true?

Answer 1

Claim

The function f(n) = log(n)^m , for any natural number m > 2 ( m ∈ ℕ+ ) is in O(n) .

Ie there exists a set of positive constants c and n0 such that the following holds:

 log(n)^m < c · n, for all n > n0, { m > 2 | m ∈ ℕ+ } (+) 

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Proof

• Assume that (+) does not hold, and denote this assumption as (*) .

Ie, given (*) , there exists no set of positive constants c and n0 such that (+) holds for any value of m > 2 . Under this assumption, the following holds, that for all positive constants c and n0 , there exists a n > n0 such that (thanks @Oriol):

 log(n)^m ≥ c · n, { m > 2 | m ∈ ℕ+ }                       (++)

Now, if (++) holds, then the inequality in (++) will hold also after applying any monotonically increasing function to both sides of the inequality. One such function is, conveniently, the log function itself

Hence, under the assumption that (++) holds, then, for all positive constants c and n0 , there exists a n > n0 such that the following holds

 log(log(n)^m) ≥ log(c · n), { m > 2 | m ∈ ℕ+ }

 m · log(log(n)) ≥ log(c · n), { m > 2 | m ∈ ℕ+ }           (+++)

However, (+++) is obviously a contradiction: since log(n) dominates (wrt growth) over log(log(n)) ,

we can—for any given value of m > 2 —always find a set of constants c and n0 such that (+++) (and hence (++) ) is violated for all n > n0 .

Hence, assumption (*) is proven false by contradiction, and hence, (+) holds.

=> for f(n) = log(n)^m , for any finite integer m > 2 , it holds that f ∈ O(n) .

Answer 2

Yes. If the function it's f(n) , it means m is a parameter and f does not depend on it. In fact, we have a different f_m function for each m .

f_m(n) = log(n)^m

Then it's easy. Given m ∈ ℕ , use L'Hôpital's rule repeatively

        f_m(n)            log(n)^m            m * log(n)^(m-1)
limsup ──────── = limsup ────────── = limsup ────────────────── =
 n→∞      n        n→∞       n         n→∞           n

          m*(m-1) * log(n)^(m-2)                m!
= limsup ──────────────────────── = … = limsup ──── = 0
                 n                       n→∞    n

Therefore, f_m ∈ O(n) .

Of course, it would be different if we had f(m,n) = log(n)^m . For example, taking m=n ,

        f(n,n)            log(n)^n
limsup ──────── = limsup ────────── = ∞
 n→∞      n        n→∞       n

Then f ∉ O(n)

Answer 3

In many ways it is more intuitive that for any positive integer m we have:

x^m = O(e^x)

This says that exponential growth dominates polynomial growth (which is why exponential time algorithms are bad news in computer programming).

Assuming that this is true, simply take x = log(n) and use the fact that then x tends to infinity if and only if n tends to infinity and that e^x and log(x) are inverses:

log(n)^m = O(e^log(n)) = O(n)

Finally, since for any natural number m , the root function n => n^(1/m) is increasing, we can rewrite the result as

log(n) = O(n^(1/m))

This way of writing it says that log(n) grows slower than any root (square, cube, etc.) of n , which more obviously corresponds to e^n growing faster than any power of n .

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On Edit: the above showed that log(n)^m = O(n) followed from the more familiar x^m = O(e^x) . To convert it to a more self-contained proof, we can show the later somewhat directly.

Start with the Taylor series for e^x :

e^x = 1 + x/1! + x^2/2! + x^3/3! + ... + x^n/n! + ...

This is known to converge for all real numbers x . If a positive integer m is given, let K = (m+1)! . Then, if x > K we have 1/x < 1/(m+1)! , hence

x^m = x^(m+1)/x < x^(m+1)/(m+1)! < e^x

which implies x^m = O(e^x) . (The last inequality in the above is true since all terms in the expansion for e^x are strictly positive if x>0 and x^(m+1)/(m+1)! is just one of those terms.)