comparing Asymptotic functions $ 3^{{2}^{n}}$ and $n! times n^{3}$












0












$begingroup$



$f(n)= 3^{{2}^{n}}$



$g(n)=n! times n^{3}$




$text{which is asymptotically greater} ?$




My Approach-:



$$f(n)=e^{log 4^{{3}^{n}} }$$
$$g(n)=e^{log n! times n^{3}}$$



i am comparing just the power of the functions



i.e



$$log 3^{{2}^{n}}$$ and $$log n! times n^{3}$$



$$f_1(n)=log 3^{{2}^{n}}=2^{n} log 3$$



$$f_2(n)=log (n! times n^{3})=log (n!) +log (n^{3}) =n log n+3 log n$$



Again we can express $f_1(n) $ and $f_2(n)$ as



$f_1(n)=e^{log( 2^n log 3)}$ and $f_2(n)=e^{log( n log n)}$



Again comparing the powers,



$f_{11}(n)=log (2^n log 3)=n + log log 3$



$f_{12}=log( n log n)=log n + log log n $



here $f_{11}(n) >f_{12}(n) Rightarrow f_{1}(n)>f_{12}(n) Rightarrow f(n)>g(n) $



here $>$ means asymptotically greater



Am i correct?






functions algorithms asymptotics







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  • $begingroup$
    Final result is correct, the way you got to that result might be debatable
    $endgroup$
    – Jakobian
    Dec 1 '18 at 16:37


















0












$begingroup$



$f(n)= 3^{{2}^{n}}$



$g(n)=n! times n^{3}$




$text{which is asymptotically greater} ?$




My Approach-:



$$f(n)=e^{log 4^{{3}^{n}} }$$
$$g(n)=e^{log n! times n^{3}}$$



i am comparing just the power of the functions



i.e



$$log 3^{{2}^{n}}$$ and $$log n! times n^{3}$$



$$f_1(n)=log 3^{{2}^{n}}=2^{n} log 3$$



$$f_2(n)=log (n! times n^{3})=log (n!) +log (n^{3}) =n log n+3 log n$$



Again we can express $f_1(n) $ and $f_2(n)$ as



$f_1(n)=e^{log( 2^n log 3)}$ and $f_2(n)=e^{log( n log n)}$



Again comparing the powers,



$f_{11}(n)=log (2^n log 3)=n + log log 3$



$f_{12}=log( n log n)=log n + log log n $



here $f_{11}(n) >f_{12}(n) Rightarrow f_{1}(n)>f_{12}(n) Rightarrow f(n)>g(n) $



here $>$ means asymptotically greater



Am i correct?






functions algorithms asymptotics







share|cite|improve this question









$endgroup$












  • $begingroup$
    Final result is correct, the way you got to that result might be debatable
    $endgroup$
    – Jakobian
    Dec 1 '18 at 16:37
















0












0








0





$begingroup$



$f(n)= 3^{{2}^{n}}$



$g(n)=n! times n^{3}$




$text{which is asymptotically greater} ?$




My Approach-:



$$f(n)=e^{log 4^{{3}^{n}} }$$
$$g(n)=e^{log n! times n^{3}}$$



i am comparing just the power of the functions



i.e



$$log 3^{{2}^{n}}$$ and $$log n! times n^{3}$$



$$f_1(n)=log 3^{{2}^{n}}=2^{n} log 3$$



$$f_2(n)=log (n! times n^{3})=log (n!) +log (n^{3}) =n log n+3 log n$$



Again we can express $f_1(n) $ and $f_2(n)$ as



$f_1(n)=e^{log( 2^n log 3)}$ and $f_2(n)=e^{log( n log n)}$



Again comparing the powers,



$f_{11}(n)=log (2^n log 3)=n + log log 3$



$f_{12}=log( n log n)=log n + log log n $



here $f_{11}(n) >f_{12}(n) Rightarrow f_{1}(n)>f_{12}(n) Rightarrow f(n)>g(n) $



here $>$ means asymptotically greater



Am i correct?






functions algorithms asymptotics







share|cite|improve this question









$endgroup$





$f(n)= 3^{{2}^{n}}$



$g(n)=n! times n^{3}$




$text{which is asymptotically greater} ?$




My Approach-:



$$f(n)=e^{log 4^{{3}^{n}} }$$
$$g(n)=e^{log n! times n^{3}}$$



i am comparing just the power of the functions



i.e



$$log 3^{{2}^{n}}$$ and $$log n! times n^{3}$$



$$f_1(n)=log 3^{{2}^{n}}=2^{n} log 3$$



$$f_2(n)=log (n! times n^{3})=log (n!) +log (n^{3}) =n log n+3 log n$$



Again we can express $f_1(n) $ and $f_2(n)$ as



$f_1(n)=e^{log( 2^n log 3)}$ and $f_2(n)=e^{log( n log n)}$



Again comparing the powers,



$f_{11}(n)=log (2^n log 3)=n + log log 3$



$f_{12}=log( n log n)=log n + log log n $



here $f_{11}(n) >f_{12}(n) Rightarrow f_{1}(n)>f_{12}(n) Rightarrow f(n)>g(n) $



here $>$ means asymptotically greater



Am i correct?






functions algorithms asymptotics




functions algorithms asymptotics






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asked Dec 1 '18 at 16:35









viratvirat

18118




18118












  • $begingroup$
    Final result is correct, the way you got to that result might be debatable
    $endgroup$
    – Jakobian
    Dec 1 '18 at 16:37




















  • $begingroup$
    Final result is correct, the way you got to that result might be debatable
    $endgroup$
    – Jakobian
    Dec 1 '18 at 16:37


















$begingroup$
Final result is correct, the way you got to that result might be debatable
$endgroup$
– Jakobian
Dec 1 '18 at 16:37






$begingroup$
Final result is correct, the way you got to that result might be debatable
$endgroup$
– Jakobian
Dec 1 '18 at 16:37












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$begingroup$

You can do it easily by applying $log$ function to $f(n)$ and $g(n)$.



$$log(f(n)) = 2^nlog(3), log(g(n)) = log(n!) + 3log(n)$$



As we know $log(n!) = Theta(nlog(n))$, we have:
$$g(n) = Theta(nlog(n)), f(n) = Theta(2^n)$$



Hence, $f(n)$ is greater than $g(n)$.






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    $begingroup$

    You can do it easily by applying $log$ function to $f(n)$ and $g(n)$.



    $$log(f(n)) = 2^nlog(3), log(g(n)) = log(n!) + 3log(n)$$



    As we know $log(n!) = Theta(nlog(n))$, we have:
    $$g(n) = Theta(nlog(n)), f(n) = Theta(2^n)$$



    Hence, $f(n)$ is greater than $g(n)$.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      You can do it easily by applying $log$ function to $f(n)$ and $g(n)$.



      $$log(f(n)) = 2^nlog(3), log(g(n)) = log(n!) + 3log(n)$$



      As we know $log(n!) = Theta(nlog(n))$, we have:
      $$g(n) = Theta(nlog(n)), f(n) = Theta(2^n)$$



      Hence, $f(n)$ is greater than $g(n)$.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        You can do it easily by applying $log$ function to $f(n)$ and $g(n)$.



        $$log(f(n)) = 2^nlog(3), log(g(n)) = log(n!) + 3log(n)$$



        As we know $log(n!) = Theta(nlog(n))$, we have:
        $$g(n) = Theta(nlog(n)), f(n) = Theta(2^n)$$



        Hence, $f(n)$ is greater than $g(n)$.






        share|cite|improve this answer









        $endgroup$



        You can do it easily by applying $log$ function to $f(n)$ and $g(n)$.



        $$log(f(n)) = 2^nlog(3), log(g(n)) = log(n!) + 3log(n)$$



        As we know $log(n!) = Theta(nlog(n))$, we have:
        $$g(n) = Theta(nlog(n)), f(n) = Theta(2^n)$$



        Hence, $f(n)$ is greater than $g(n)$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 1 '18 at 16:51









        OmGOmG

        2,180621




        2,180621






























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