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My algorithm textbook has a theorem that says

'For every $r > 1$ and every $d > 0$, we have $n^d = O(r^n)$.'

However, it does not provide proof.

Of course I know exponential grows faster than polynomial in most cases, but is it true for all case?

What if the polynomial function is something like $n^{100^{100}}$ and exponential is $2^n$? Will the latter outgrow the former at some point?

Eric
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    Yes. Exponential of base $> 1$ will eventually grow faster than any polynomial. – Tunococ Sep 20 '13 at 09:20
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    For intuition, take logs on both sides and you have $100^{100}\log n$ versus $n\log 2$. Hopefully you believe that $\log n$ grows much slower than $n$... –  Sep 20 '13 at 09:38
  • The result is correct. In the very long run, $f(n)=(1.00000001)^n$ "beats" $g(n)=n^{9999999999}$. However, for all practical purposes, an "exponential" algorithm that takes time $f(n)$ on input of size $n$ may be much more practical than a "polynomial" algorithm that takes time $g(n)$. – André Nicolas Sep 20 '13 at 09:52

3 Answers3

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Yes, it is true for all cases. This can be seen by noting that

$$\lim_{n\to\infty} \frac{n^k}{e^n} = 0$$

for any $k$. This can be seen by an application of L'Hospital's rule a number of times, or by using induction as here.

Community
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Let $n = k^2$.

Then $n^c = k^{2c}$ and $2^n = (2^k)^k$.

Clearly $2^k \ge k+1 > k$ so all we need is $k > 2c$.

In general if we want to find $n \in \mathbb{N}$ such that $r^n > n^c$ where $r > 1$, we can do essentially the same:

Let $n = ak^2$ where $a > \log_r(2)$.

Then $r^n > (2^k)^k$ and $n^c = a^c k^{2c}$.

It is then enough to choose $k$ such that $k > a$ and $k > 3c$, so that $n^c < k^{3c} < (2^k)^k < 2^n$.

user21820
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  • I don't understand... – linear_combinatori_probabi Mar 12 '18 at 17:36
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    @Niing: My answer may have been a bit cryptic, but it should have been decipherable after some effort. Given constant $c$, to prove $2^n > n^c$ as $n \to \infty$, let $n = k^2$ and observe that $2^n = (2^k)^k > k^k$ and $n^c = k^{2c}$, and so it should be clear that the desired inequality holds because $k > 2c$ as $n \to \infty$. Similarly, given constants $c$ and $r>1$, to prove $r^n > n^c$ as $n \to \infty$, let $n = ak^2$ where $a > \log_r(2)$, so $r^n = (2^{ak})^k > (2^k)^k$ and $n^c = a^c k^{2c} < k^{3c} < (2^k)^k$ because $a < k$ and $3c < k$ as $n \to \infty$. – user21820 Mar 14 '18 at 11:55
  • @Niing: If you still do not get it, specify exactly which part of my comment you cannot understand, and I will explain. – user21820 Mar 14 '18 at 11:56
  • I've asked a question about how to prove what you just did... And I've read, at least for me, a lot of related posts. Your explanation is the best..., since you didn't use any advanced concept like the limit of a quotient. TLDR: thank you... – linear_combinatori_probabi Mar 14 '18 at 16:12
  • @Niing: Yes that is precisely the reason I posted my answer a year after the others, to demonstrate that we can in fact do a lot by purely elementary means. Of course, once we learn asymptotic expansion, we ought to always just use it, and in this case we can use $exp(n/c) > 1+n/c+n^2/2c^2 > n$ as $n \to \infty$. But it is good to know what we can do with simpler tools. =) – user21820 Mar 14 '18 at 16:53
  • Actually, I'm really afraid of that it's, maybe, essential for math that I would always need some advanced tools to construct some idea before I could actually understand it... If that's the case, people are both creating and discovering math. – linear_combinatori_probabi Mar 14 '18 at 17:05
  • @Niing: That is true in general, but in this case it wasn't necessary; you just need to play around enough with what you have and then suddenly you see how to use it to do interesting things. Just like how the fact that $(x-y)^2 \ge 0$ for every reals $x,y$ can give you the fact that $(x+y)/2 \ge sqrt(x·y)$ for every positive reals $x,y$. And I posted an answer to your new question. =) – user21820 Mar 14 '18 at 17:18
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Hint: Yes. See Taylor expansion of exponential function.

Arash
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