Ordering w.r.t increase in growth rate












-1












$begingroup$


What is the growth rate and how can we order given functions w.r.t increase in growth rate? How can we order the following functions ascending with respective to the rate of growth:



$n^n$,



$nlog_2n+nsqrt{n}$,



$3n^3+5n^2+6$,



$n!+2^n$,



$n+log_2(log_2n)$,



$n^{20}+2^n$,



$log_2n$,



$2^n$,



5,



$log_2(log_2n)$.






asymptotics







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  • $begingroup$
    I would guess that we are interested in $n to infty$. For example, certainly $n^n> 2n$ when $nto infty$.
    $endgroup$
    – Matti P.
    Jan 29 at 10:07
















-1












$begingroup$


What is the growth rate and how can we order given functions w.r.t increase in growth rate? How can we order the following functions ascending with respective to the rate of growth:



$n^n$,



$nlog_2n+nsqrt{n}$,



$3n^3+5n^2+6$,



$n!+2^n$,



$n+log_2(log_2n)$,



$n^{20}+2^n$,



$log_2n$,



$2^n$,



5,



$log_2(log_2n)$.






asymptotics







share|cite|improve this question











$endgroup$












  • $begingroup$
    I would guess that we are interested in $n to infty$. For example, certainly $n^n> 2n$ when $nto infty$.
    $endgroup$
    – Matti P.
    Jan 29 at 10:07














-1












-1








-1





$begingroup$


What is the growth rate and how can we order given functions w.r.t increase in growth rate? How can we order the following functions ascending with respective to the rate of growth:



$n^n$,



$nlog_2n+nsqrt{n}$,



$3n^3+5n^2+6$,



$n!+2^n$,



$n+log_2(log_2n)$,



$n^{20}+2^n$,



$log_2n$,



$2^n$,



5,



$log_2(log_2n)$.






asymptotics







share|cite|improve this question











$endgroup$




What is the growth rate and how can we order given functions w.r.t increase in growth rate? How can we order the following functions ascending with respective to the rate of growth:



$n^n$,



$nlog_2n+nsqrt{n}$,



$3n^3+5n^2+6$,



$n!+2^n$,



$n+log_2(log_2n)$,



$n^{20}+2^n$,



$log_2n$,



$2^n$,



5,



$log_2(log_2n)$.






asymptotics




asymptotics






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share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 29 at 10:17







Sail Akhil

















asked Jan 29 at 9:21









Sail AkhilSail Akhil

205




205












  • $begingroup$
    I would guess that we are interested in $n to infty$. For example, certainly $n^n> 2n$ when $nto infty$.
    $endgroup$
    – Matti P.
    Jan 29 at 10:07


















  • $begingroup$
    I would guess that we are interested in $n to infty$. For example, certainly $n^n> 2n$ when $nto infty$.
    $endgroup$
    – Matti P.
    Jan 29 at 10:07
















$begingroup$
I would guess that we are interested in $n to infty$. For example, certainly $n^n> 2n$ when $nto infty$.
$endgroup$
– Matti P.
Jan 29 at 10:07




$begingroup$
I would guess that we are interested in $n to infty$. For example, certainly $n^n> 2n$ when $nto infty$.
$endgroup$
– Matti P.
Jan 29 at 10:07










1 Answer
1






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oldest

votes


















0












$begingroup$

We say that a function $f(x)$ has a greater rate of growth than $g(x)$ if



$lim limits_{xtoinfty} frac{f(x)}{g(x)} to infty$



So, one way to order some functions with respect to the rate of growth is to compute that limit. For example, if we want to see which of the first two functions growths faster, you should compute:



$lim limits_{ntoinfty} frac{n^n}{nlog{n}+nsqrt{n}} hspace{2pt} underset{mathrm{Hôpital}}{=} hspace{2pt} lim limits_{ntoinfty} frac{n^n(log{n}+1)}{frac{3}{2}sqrt{n}+log{n}} hspace{2pt} underset{mathrm{Hôpital}}{=} hspace{2pt} lim limits_{ntoinfty} frac{n^{n-1}+n^n(log{n}+1)^2}{frac{1}{4sqrt{n}}+frac{1}{n}} to frac{infty}{0} = infty $



Meaning that $n^n succ nlog{n}+nsqrt{n} hspace{15pt}$(i.e. $n^n$ grows faster than $nlog{n}+nsqrt{n}$)
Notice that if the instead of infinity the limit was zero that would mean that $n^n$ would grow faster than $nlog{n}+nsqrt{n}$



So, one way to order the functions would be to compute some limits of quotients of functions. However, in general:



$n^n succ n! succ c^n succ n^c succ log{n} succ {log(log{n})}$



where c is a constant greater than one. You can actually check this by computing the limits of quotients. by knowing this, it should be easy to order the rest of the functions. Hope I was helpful.






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    1 Answer
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    1 Answer
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    $begingroup$

    We say that a function $f(x)$ has a greater rate of growth than $g(x)$ if



    $lim limits_{xtoinfty} frac{f(x)}{g(x)} to infty$



    So, one way to order some functions with respect to the rate of growth is to compute that limit. For example, if we want to see which of the first two functions growths faster, you should compute:



    $lim limits_{ntoinfty} frac{n^n}{nlog{n}+nsqrt{n}} hspace{2pt} underset{mathrm{Hôpital}}{=} hspace{2pt} lim limits_{ntoinfty} frac{n^n(log{n}+1)}{frac{3}{2}sqrt{n}+log{n}} hspace{2pt} underset{mathrm{Hôpital}}{=} hspace{2pt} lim limits_{ntoinfty} frac{n^{n-1}+n^n(log{n}+1)^2}{frac{1}{4sqrt{n}}+frac{1}{n}} to frac{infty}{0} = infty $



    Meaning that $n^n succ nlog{n}+nsqrt{n} hspace{15pt}$(i.e. $n^n$ grows faster than $nlog{n}+nsqrt{n}$)
    Notice that if the instead of infinity the limit was zero that would mean that $n^n$ would grow faster than $nlog{n}+nsqrt{n}$



    So, one way to order the functions would be to compute some limits of quotients of functions. However, in general:



    $n^n succ n! succ c^n succ n^c succ log{n} succ {log(log{n})}$



    where c is a constant greater than one. You can actually check this by computing the limits of quotients. by knowing this, it should be easy to order the rest of the functions. Hope I was helpful.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      We say that a function $f(x)$ has a greater rate of growth than $g(x)$ if



      $lim limits_{xtoinfty} frac{f(x)}{g(x)} to infty$



      So, one way to order some functions with respect to the rate of growth is to compute that limit. For example, if we want to see which of the first two functions growths faster, you should compute:



      $lim limits_{ntoinfty} frac{n^n}{nlog{n}+nsqrt{n}} hspace{2pt} underset{mathrm{Hôpital}}{=} hspace{2pt} lim limits_{ntoinfty} frac{n^n(log{n}+1)}{frac{3}{2}sqrt{n}+log{n}} hspace{2pt} underset{mathrm{Hôpital}}{=} hspace{2pt} lim limits_{ntoinfty} frac{n^{n-1}+n^n(log{n}+1)^2}{frac{1}{4sqrt{n}}+frac{1}{n}} to frac{infty}{0} = infty $



      Meaning that $n^n succ nlog{n}+nsqrt{n} hspace{15pt}$(i.e. $n^n$ grows faster than $nlog{n}+nsqrt{n}$)
      Notice that if the instead of infinity the limit was zero that would mean that $n^n$ would grow faster than $nlog{n}+nsqrt{n}$



      So, one way to order the functions would be to compute some limits of quotients of functions. However, in general:



      $n^n succ n! succ c^n succ n^c succ log{n} succ {log(log{n})}$



      where c is a constant greater than one. You can actually check this by computing the limits of quotients. by knowing this, it should be easy to order the rest of the functions. Hope I was helpful.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        We say that a function $f(x)$ has a greater rate of growth than $g(x)$ if



        $lim limits_{xtoinfty} frac{f(x)}{g(x)} to infty$



        So, one way to order some functions with respect to the rate of growth is to compute that limit. For example, if we want to see which of the first two functions growths faster, you should compute:



        $lim limits_{ntoinfty} frac{n^n}{nlog{n}+nsqrt{n}} hspace{2pt} underset{mathrm{Hôpital}}{=} hspace{2pt} lim limits_{ntoinfty} frac{n^n(log{n}+1)}{frac{3}{2}sqrt{n}+log{n}} hspace{2pt} underset{mathrm{Hôpital}}{=} hspace{2pt} lim limits_{ntoinfty} frac{n^{n-1}+n^n(log{n}+1)^2}{frac{1}{4sqrt{n}}+frac{1}{n}} to frac{infty}{0} = infty $



        Meaning that $n^n succ nlog{n}+nsqrt{n} hspace{15pt}$(i.e. $n^n$ grows faster than $nlog{n}+nsqrt{n}$)
        Notice that if the instead of infinity the limit was zero that would mean that $n^n$ would grow faster than $nlog{n}+nsqrt{n}$



        So, one way to order the functions would be to compute some limits of quotients of functions. However, in general:



        $n^n succ n! succ c^n succ n^c succ log{n} succ {log(log{n})}$



        where c is a constant greater than one. You can actually check this by computing the limits of quotients. by knowing this, it should be easy to order the rest of the functions. Hope I was helpful.






        share|cite|improve this answer









        $endgroup$



        We say that a function $f(x)$ has a greater rate of growth than $g(x)$ if



        $lim limits_{xtoinfty} frac{f(x)}{g(x)} to infty$



        So, one way to order some functions with respect to the rate of growth is to compute that limit. For example, if we want to see which of the first two functions growths faster, you should compute:



        $lim limits_{ntoinfty} frac{n^n}{nlog{n}+nsqrt{n}} hspace{2pt} underset{mathrm{Hôpital}}{=} hspace{2pt} lim limits_{ntoinfty} frac{n^n(log{n}+1)}{frac{3}{2}sqrt{n}+log{n}} hspace{2pt} underset{mathrm{Hôpital}}{=} hspace{2pt} lim limits_{ntoinfty} frac{n^{n-1}+n^n(log{n}+1)^2}{frac{1}{4sqrt{n}}+frac{1}{n}} to frac{infty}{0} = infty $



        Meaning that $n^n succ nlog{n}+nsqrt{n} hspace{15pt}$(i.e. $n^n$ grows faster than $nlog{n}+nsqrt{n}$)
        Notice that if the instead of infinity the limit was zero that would mean that $n^n$ would grow faster than $nlog{n}+nsqrt{n}$



        So, one way to order the functions would be to compute some limits of quotients of functions. However, in general:



        $n^n succ n! succ c^n succ n^c succ log{n} succ {log(log{n})}$



        where c is a constant greater than one. You can actually check this by computing the limits of quotients. by knowing this, it should be easy to order the rest of the functions. Hope I was helpful.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 29 at 10:27









        maxbpmaxbp

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