5
$\begingroup$

I am stuck at this question where I have to calculate what is big O of what,

$n!$ and $n^\sqrt n$

I tried replacing n! by it's equivalent formula but it makes everything more complicated, I can't even think about doing it by induction.

Any help would be appreciated

  • 0
    big O? $\ \ \ $ – 2017-01-19
  • 1
    Yeah, I mean I need to see what's the result of lim n-> infinity to these 2 functions so I can decide which one is greater than the other.. – 2017-01-19
  • 0
    I think you can use Stirling approximation https://en.m.wikipedia.org/wiki/Stirling's_approximation – 2017-01-19
  • 0
    It is enough to use the estimate $n!>\left(\dfrac{n}2\right)^{n/2}$. – 2017-01-19
  • 0
    @JonasMeyer I tried using the Stirling app. but it's still hard to get a result – 2017-01-19
  • 0
    Why would you use the ln ? I thought more about n! / n^ sqrt n ~ Sqrt(2 pi n ) * (n/e)^n / n^ sqrt n – 2017-01-19
  • 0
    Huh, this was less trivial than I thought it'd be :P – 2017-01-19

3 Answers 3

3

Note that $n!\ge \left(\frac{n}{2}\right)^{n/2}$. Hence, we have

$$\begin{align} \frac{n^{\sqrt{n}}}{n!}&\le \frac{n^{\sqrt n}}{(n/2)^{n/2}}\\\\ &=\left(\frac{2}{n^{1-2/\sqrt{n}}}\right)^{n/2}\\\\ &\to 0\,\,\text{as}\,\,n\to \infty \end{align}$$

  • 0
    OK this helps alot ! but do I have to prove that n! >= (n/2) ^ n/2 ? – 2017-01-19
  • 0
    @Zok: Why not? Write out the factors of $n!$, and count how many are larger than $n/2$. – 2017-01-19
  • 0
    @Zok In fact, it is a gross overestimate. Note that $$(2n)!=(2n)(2n-1)\cdots (n+1)n!\ge (n)^n\,n!\ge n^n$$ – 2017-01-19
  • 0
    @Dr.MV Sorry I don't get your last equation.. – 2017-01-19
  • 0
    Which "last equation?" Do you mean the last one in the comment? – 2017-01-19
  • 0
    @Dr.MV yes.. Because here http://math.stackexchange.com/questions/1389890/limit-of-n-nn-as-n-tends-to-infinity the answer is the opposite ( n! <= and not >= (n/2)^n/2 ) So please give me the solution how to prove that n! >= (n/2) ^ n/2 – 2017-01-20
0

A result I got

$n!/ n ^ sqrt(n) $

~ $\sqrt{2*pi * n} * (n/e)^n / n^ \sqrt n$

= $\sqrt{2 * pi } * \sqrt n * (n/e)^n / n^\sqrt n $

= $\sqrt{2 * pi} * \sqrt n * n^n / n^\sqrt n * e^n $

= $\sqrt{2 * pi} * n^{n+1/2} / n^\sqrt n * e^n $

= $\sqrt{2*pi} / n^{\sqrt n-n-1/2} *e^n$

= ?

  • 0
    A suggestion: If you don't like logarithms, try taking $n^\text{th}$ roots. – 2017-01-19
  • 0
    @JonasMeyer I edited the answer, I had error and now by modifying it I am stuck again.. – 2017-01-19
0

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \ln\pars{n^{\root{n}} \over n!} & = \root{n}\ln\pars{n} - \ln\pars{n!} \sim \root{n}\ln\pars{n} - \bracks{n\ln\pars{n} - n}\quad\mbox{as}\quad n \to \infty \end{align} such that $\ds{\lim_{n \to \infty}\ln\pars{n^{\root{n}} \over n!} = -\infty \implies \bbx{\ds{\lim_{n \to \infty}{n^{\root{n}} \over n!} = 0}}}$