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I stumbled upon this question, among $n^{\frac{7}{4}}$ and $n(\log^9(n))$, which one is bigger?

I think that leaving out n from both leaves $n^{\frac{3}{4}}$ and $\log^9(n)$.

Now let's take $n^{\frac{3}{4}} = n$.

So, according to the concept of exponential, they grow very fast making $\log(n)$ very small. Thus I can conclude that multiplying $9$ parts of this $\log(n)$ we can never make it equal to $n$.

This is what I can imagine. I don't know even if it is correct or not. Can anyone please say if I'm on the right track? Also if you can say how to deduce the conclusion will be great. Thanks in advance.

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  • $\begingroup$ Are you familiar with l'Hopital's rule? $\endgroup$
    – Servaes
    Dec 23, 2016 at 11:27
  • $\begingroup$ Yes.Solved only a few sums only though. :p $\endgroup$
    – lu5er
    Dec 23, 2016 at 11:29
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    $\begingroup$ Try applying it to the quotient $\frac{n^{\frac74}}{n\log^9(n)}$, or if you like to $\frac{n^{\frac34}}{\log^9(n)}$. What does this tell you? $\endgroup$
    – Servaes
    Dec 23, 2016 at 11:30

3 Answers 3

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The quick fact to remember is that any power function (with a positive exponent) always wins over any power of logarithms.

We can compare $$n^{\varepsilon} \lessgtr \log^a n$$ by taking the $a$th root on both sides which gives $$ n^{\varepsilon/a} \lessgtr \log n = \frac{1}{\varepsilon/a}\log(n^{\varepsilon/a}) $$ Since $n^{\varepsilon/a}\to \infty$ is the same limit as $n\to\infty$, once we know that $\log n = o(n)$, this automatically generalizes to positive powers on both sides.

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    $\begingroup$ wow. That's cool. :) $\endgroup$
    – lu5er
    Dec 23, 2016 at 11:48
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Apply l'Hopital's rule (repeatedly) to the limit $$\lim_{n\to\infty}\frac{n^{\frac34}}{\log^9(n)}.$$ What does this tell you?

Alternatively, set $x:=\log n$ and substitute this to get the functions $$n^{\frac34}=(e^x)^{\frac34}=e^{\frac34x}\qquad\text{ and }\qquad \log^9(n)=x^9.$$ Can you tell which one grows faster then?

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    $\begingroup$ $e^{\frac{3}{4}x}$, right? $\endgroup$
    – lu5er
    Dec 23, 2016 at 11:38
  • $\begingroup$ That's right. See also Hennings answer for a broader perspective. $\endgroup$
    – Servaes
    Dec 23, 2016 at 11:41
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    $\begingroup$ And like you said, $n^\frac{3}{4}$ will go on decreasing but again I'll get $\frac{1}{n}$ on every iteration of L'Hospital Rule which will make it $\infty$ eventually. $\endgroup$
    – lu5er
    Dec 23, 2016 at 11:46
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With asymptotic analysis: $\ln^an=o(n^b)$ near $\infty$ for all $a,b>0$. So $$n\ln^9n=n\,o\Bigl(n^{3/4}\bigr)=o\Bigl(n^{1+3/4}\bigr).$$

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