Big O Notation of $n^{0.999999}\log(n)$

Question

I'm taking the MIT Open Courseware for Introduction to Algorithms and I'm having trouble understanding the first homework problem/solution.

We are supposed to compare the asymptotic complexity (big-O) for different functions:

$f_1(n) = n^{0.999999}\log(n)$
$f_2(n) = 10000000n$

$f_2(n)$ is obviously O(n), but the big-O given for $f_1(n)$ confuses me and I don't follow the given solution.

The solution says $f_1(n)$ has less Big-O complexity than $f_2(n)$:

"recall that for any $c > 0$, $log(n)$ is $O(n^c)$.

Therefore we have: $f(n) = n^{0.999999}log(n) = O(n^{0.999999}n^{0.000001}) = O(n) = O(f_2(n))$"

I do not understand the logic underlying the solution. I may be forgetting something simple? Can someone break it down for me? Thanks!

Answer 1

If $g = O(g')$ and $h$ is nonnegative then $h \cdot g = O(h \cdot g')$ — in English, if $g$ grows slower than $g'$ then multiplying by a positive function doesn't affect that fact. (If you aren't sure about this, do the proof with quantifiers).

Since $\log(n) = O(n^{0.000001})$, $n^{0.999999} \cdot \log(n) = O(n^{0.999999} \cdot n^{0.000001})$. Since $n^{0.999999} \cdot n^{0.000001} = n$, we have $f_1(n) = O(n)$.

While it is true that $f_2(n) = O(n)$, this is not relevant here. What you need is $n = O(f_2(n))$, which is also true.

Intuitively, a higher exponent trumps a logarithm. A multiplicative constant is not significant.

Answer 2

If we consider really large numbers like $10^{1000}$ then $n\log n$ will be equal to

$$n\log 10^{1000} = 1000n\log 10 = 2302n < 1000000n.$$

Thus even for really really big $n$s, $n\log n<1000000n$.

This I guess is a layman logic, and theoretically, infinite $n$ should give $n\log n>10000000n$. But not really.

Answer 3

Just use the definition, i.e., check if $\lim_{n \to \infty} f_1(n) / f_2(n) < \infty$. Computing this, you'll see that $0 < \infty$ and you are done.