Find whether $f(x)$ is $O(g(x))$ [whether $f(x)$ is Big-O of $g(x)$]












0














Given: $f(x)=3^{sqrt{x}}, g(x) = 2^x$, find whether $f(x)$ is Big-O of $g(x)$, and vice-versa.



I want to use the following fact: $$lim_{xtoinfty}(ln|f(x)|-ln|g(x)|) leq ln(C) implies f(x)=O[g(x)]$$



I have done the following: $$lim_{xtoinfty}(ln|f(x)|-ln|g(x)|)$$
$$ = lim_{xtoinfty}(sqrt{x}ln(3)-xln(2))$$
$$ = lim_{xtoinfty}[ln(dfrac{3^sqrt{x}}{2^x})] $$
$$= ln[lim_{xtoinfty}(dfrac{3^sqrt{x}}{2^x})] $$



Where do I go from here?






real-analysis limits logarithms asymptotics







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  • 1




    $sqrt{x} ne x/2$
    – angryavian
    Nov 27 '18 at 21:23










  • @angryavian my bad - edited the question
    – bob
    Nov 27 '18 at 21:26










  • Let use $log_2$ for $3^{sqrt x}$ in order to have all in base $2$ and compare the terms.
    – gimusi
    Nov 27 '18 at 21:32
















0














Given: $f(x)=3^{sqrt{x}}, g(x) = 2^x$, find whether $f(x)$ is Big-O of $g(x)$, and vice-versa.



I want to use the following fact: $$lim_{xtoinfty}(ln|f(x)|-ln|g(x)|) leq ln(C) implies f(x)=O[g(x)]$$



I have done the following: $$lim_{xtoinfty}(ln|f(x)|-ln|g(x)|)$$
$$ = lim_{xtoinfty}(sqrt{x}ln(3)-xln(2))$$
$$ = lim_{xtoinfty}[ln(dfrac{3^sqrt{x}}{2^x})] $$
$$= ln[lim_{xtoinfty}(dfrac{3^sqrt{x}}{2^x})] $$



Where do I go from here?






real-analysis limits logarithms asymptotics







share|cite|improve this question




















  • 1




    $sqrt{x} ne x/2$
    – angryavian
    Nov 27 '18 at 21:23










  • @angryavian my bad - edited the question
    – bob
    Nov 27 '18 at 21:26










  • Let use $log_2$ for $3^{sqrt x}$ in order to have all in base $2$ and compare the terms.
    – gimusi
    Nov 27 '18 at 21:32














0












0








0


0





Given: $f(x)=3^{sqrt{x}}, g(x) = 2^x$, find whether $f(x)$ is Big-O of $g(x)$, and vice-versa.



I want to use the following fact: $$lim_{xtoinfty}(ln|f(x)|-ln|g(x)|) leq ln(C) implies f(x)=O[g(x)]$$



I have done the following: $$lim_{xtoinfty}(ln|f(x)|-ln|g(x)|)$$
$$ = lim_{xtoinfty}(sqrt{x}ln(3)-xln(2))$$
$$ = lim_{xtoinfty}[ln(dfrac{3^sqrt{x}}{2^x})] $$
$$= ln[lim_{xtoinfty}(dfrac{3^sqrt{x}}{2^x})] $$



Where do I go from here?






real-analysis limits logarithms asymptotics







share|cite|improve this question















Given: $f(x)=3^{sqrt{x}}, g(x) = 2^x$, find whether $f(x)$ is Big-O of $g(x)$, and vice-versa.



I want to use the following fact: $$lim_{xtoinfty}(ln|f(x)|-ln|g(x)|) leq ln(C) implies f(x)=O[g(x)]$$



I have done the following: $$lim_{xtoinfty}(ln|f(x)|-ln|g(x)|)$$
$$ = lim_{xtoinfty}(sqrt{x}ln(3)-xln(2))$$
$$ = lim_{xtoinfty}[ln(dfrac{3^sqrt{x}}{2^x})] $$
$$= ln[lim_{xtoinfty}(dfrac{3^sqrt{x}}{2^x})] $$



Where do I go from here?






real-analysis limits logarithms asymptotics




real-analysis limits logarithms asymptotics






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 27 '18 at 21:29







bob

















asked Nov 27 '18 at 21:17









bobbob

1089




1089








  • 1




    $sqrt{x} ne x/2$
    – angryavian
    Nov 27 '18 at 21:23










  • @angryavian my bad - edited the question
    – bob
    Nov 27 '18 at 21:26










  • Let use $log_2$ for $3^{sqrt x}$ in order to have all in base $2$ and compare the terms.
    – gimusi
    Nov 27 '18 at 21:32














  • 1




    $sqrt{x} ne x/2$
    – angryavian
    Nov 27 '18 at 21:23










  • @angryavian my bad - edited the question
    – bob
    Nov 27 '18 at 21:26










  • Let use $log_2$ for $3^{sqrt x}$ in order to have all in base $2$ and compare the terms.
    – gimusi
    Nov 27 '18 at 21:32








1




1




$sqrt{x} ne x/2$
– angryavian
Nov 27 '18 at 21:23




$sqrt{x} ne x/2$
– angryavian
Nov 27 '18 at 21:23












@angryavian my bad - edited the question
– bob
Nov 27 '18 at 21:26




@angryavian my bad - edited the question
– bob
Nov 27 '18 at 21:26












Let use $log_2$ for $3^{sqrt x}$ in order to have all in base $2$ and compare the terms.
– gimusi
Nov 27 '18 at 21:32




Let use $log_2$ for $3^{sqrt x}$ in order to have all in base $2$ and compare the terms.
– gimusi
Nov 27 '18 at 21:32










3 Answers
3






active

oldest

votes


















1














You want to compare $f(x)$ and $g(x)$



Let's take the logarithm of their ratio



$lnleft(dfrac{f(x)}{g(x)}right)=ln(f(x))-ln(g(x))=sqrt{x}ln(3)-xln(2)sim-xln(2)to-infty$



Since the square root of $x$ is negligible compared to $x$ at infinity: $sqrt{x}ll x$



Thus $dfrac{f(x)}{g(x)}to 0$ and $f(x)=o(g(x))$ which is even stronger than just a big-O.



Anyway it implies $f(x)=O(g(x))$






share|cite|improve this answer





















  • Thank you, this helps.
    – bob
    Nov 27 '18 at 21:46



















1














You can write $f(x)=3^{sqrt x}=2^{sqrt xlog_23}$.



Now $ln(limlimits_{xto infty}(3^sqrt x/2^x) )=log_2(limlimits_{xto infty}(2^{sqrt xlog_23}/2^x) )leq log_2(C)$



So ${limlimits_{xto infty}sqrt xlog_23-xleq C}$



$sqrt xlog_23leq x+C$ as $x to infty$



$2^{sqrt xlog_23}leq 2^{x+C}$ as $xto infty$



$implies f(x)leq 2^Cg(x)$



Hence $f(x)=O(g(x))$ by definition.






share|cite|improve this answer





























    0














    We have that



    $$large 3^{sqrt{x}}=2^{sqrt x log_2 3} implies frac{3^{sqrt{x}}}{2^x}=2^{sqrt x log_2 3-x}to 0$$






    share|cite|improve this answer























    • Could you take this a step further?
      – bob
      Nov 27 '18 at 21:36








    • 1




      @bob What is stronger $ 2^x $or $2^{sqrt x log_2 3}$?
      – gimusi
      Nov 27 '18 at 21:38











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    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1














    You want to compare $f(x)$ and $g(x)$



    Let's take the logarithm of their ratio



    $lnleft(dfrac{f(x)}{g(x)}right)=ln(f(x))-ln(g(x))=sqrt{x}ln(3)-xln(2)sim-xln(2)to-infty$



    Since the square root of $x$ is negligible compared to $x$ at infinity: $sqrt{x}ll x$



    Thus $dfrac{f(x)}{g(x)}to 0$ and $f(x)=o(g(x))$ which is even stronger than just a big-O.



    Anyway it implies $f(x)=O(g(x))$






    share|cite|improve this answer





















    • Thank you, this helps.
      – bob
      Nov 27 '18 at 21:46
















    1














    You want to compare $f(x)$ and $g(x)$



    Let's take the logarithm of their ratio



    $lnleft(dfrac{f(x)}{g(x)}right)=ln(f(x))-ln(g(x))=sqrt{x}ln(3)-xln(2)sim-xln(2)to-infty$



    Since the square root of $x$ is negligible compared to $x$ at infinity: $sqrt{x}ll x$



    Thus $dfrac{f(x)}{g(x)}to 0$ and $f(x)=o(g(x))$ which is even stronger than just a big-O.



    Anyway it implies $f(x)=O(g(x))$






    share|cite|improve this answer





















    • Thank you, this helps.
      – bob
      Nov 27 '18 at 21:46














    1












    1








    1






    You want to compare $f(x)$ and $g(x)$



    Let's take the logarithm of their ratio



    $lnleft(dfrac{f(x)}{g(x)}right)=ln(f(x))-ln(g(x))=sqrt{x}ln(3)-xln(2)sim-xln(2)to-infty$



    Since the square root of $x$ is negligible compared to $x$ at infinity: $sqrt{x}ll x$



    Thus $dfrac{f(x)}{g(x)}to 0$ and $f(x)=o(g(x))$ which is even stronger than just a big-O.



    Anyway it implies $f(x)=O(g(x))$






    share|cite|improve this answer












    You want to compare $f(x)$ and $g(x)$



    Let's take the logarithm of their ratio



    $lnleft(dfrac{f(x)}{g(x)}right)=ln(f(x))-ln(g(x))=sqrt{x}ln(3)-xln(2)sim-xln(2)to-infty$



    Since the square root of $x$ is negligible compared to $x$ at infinity: $sqrt{x}ll x$



    Thus $dfrac{f(x)}{g(x)}to 0$ and $f(x)=o(g(x))$ which is even stronger than just a big-O.



    Anyway it implies $f(x)=O(g(x))$







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Nov 27 '18 at 21:39









    zwimzwim

    11.7k729




    11.7k729












    • Thank you, this helps.
      – bob
      Nov 27 '18 at 21:46


















    • Thank you, this helps.
      – bob
      Nov 27 '18 at 21:46
















    Thank you, this helps.
    – bob
    Nov 27 '18 at 21:46




    Thank you, this helps.
    – bob
    Nov 27 '18 at 21:46











    1














    You can write $f(x)=3^{sqrt x}=2^{sqrt xlog_23}$.



    Now $ln(limlimits_{xto infty}(3^sqrt x/2^x) )=log_2(limlimits_{xto infty}(2^{sqrt xlog_23}/2^x) )leq log_2(C)$



    So ${limlimits_{xto infty}sqrt xlog_23-xleq C}$



    $sqrt xlog_23leq x+C$ as $x to infty$



    $2^{sqrt xlog_23}leq 2^{x+C}$ as $xto infty$



    $implies f(x)leq 2^Cg(x)$



    Hence $f(x)=O(g(x))$ by definition.






    share|cite|improve this answer


























      1














      You can write $f(x)=3^{sqrt x}=2^{sqrt xlog_23}$.



      Now $ln(limlimits_{xto infty}(3^sqrt x/2^x) )=log_2(limlimits_{xto infty}(2^{sqrt xlog_23}/2^x) )leq log_2(C)$



      So ${limlimits_{xto infty}sqrt xlog_23-xleq C}$



      $sqrt xlog_23leq x+C$ as $x to infty$



      $2^{sqrt xlog_23}leq 2^{x+C}$ as $xto infty$



      $implies f(x)leq 2^Cg(x)$



      Hence $f(x)=O(g(x))$ by definition.






      share|cite|improve this answer
























        1












        1








        1






        You can write $f(x)=3^{sqrt x}=2^{sqrt xlog_23}$.



        Now $ln(limlimits_{xto infty}(3^sqrt x/2^x) )=log_2(limlimits_{xto infty}(2^{sqrt xlog_23}/2^x) )leq log_2(C)$



        So ${limlimits_{xto infty}sqrt xlog_23-xleq C}$



        $sqrt xlog_23leq x+C$ as $x to infty$



        $2^{sqrt xlog_23}leq 2^{x+C}$ as $xto infty$



        $implies f(x)leq 2^Cg(x)$



        Hence $f(x)=O(g(x))$ by definition.






        share|cite|improve this answer












        You can write $f(x)=3^{sqrt x}=2^{sqrt xlog_23}$.



        Now $ln(limlimits_{xto infty}(3^sqrt x/2^x) )=log_2(limlimits_{xto infty}(2^{sqrt xlog_23}/2^x) )leq log_2(C)$



        So ${limlimits_{xto infty}sqrt xlog_23-xleq C}$



        $sqrt xlog_23leq x+C$ as $x to infty$



        $2^{sqrt xlog_23}leq 2^{x+C}$ as $xto infty$



        $implies f(x)leq 2^Cg(x)$



        Hence $f(x)=O(g(x))$ by definition.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 27 '18 at 21:45









        JimmyJimmy

        17212




        17212























            0














            We have that



            $$large 3^{sqrt{x}}=2^{sqrt x log_2 3} implies frac{3^{sqrt{x}}}{2^x}=2^{sqrt x log_2 3-x}to 0$$






            share|cite|improve this answer























            • Could you take this a step further?
              – bob
              Nov 27 '18 at 21:36








            • 1




              @bob What is stronger $ 2^x $or $2^{sqrt x log_2 3}$?
              – gimusi
              Nov 27 '18 at 21:38
















            0














            We have that



            $$large 3^{sqrt{x}}=2^{sqrt x log_2 3} implies frac{3^{sqrt{x}}}{2^x}=2^{sqrt x log_2 3-x}to 0$$






            share|cite|improve this answer























            • Could you take this a step further?
              – bob
              Nov 27 '18 at 21:36








            • 1




              @bob What is stronger $ 2^x $or $2^{sqrt x log_2 3}$?
              – gimusi
              Nov 27 '18 at 21:38














            0












            0








            0






            We have that



            $$large 3^{sqrt{x}}=2^{sqrt x log_2 3} implies frac{3^{sqrt{x}}}{2^x}=2^{sqrt x log_2 3-x}to 0$$






            share|cite|improve this answer














            We have that



            $$large 3^{sqrt{x}}=2^{sqrt x log_2 3} implies frac{3^{sqrt{x}}}{2^x}=2^{sqrt x log_2 3-x}to 0$$







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Nov 27 '18 at 21:41

























            answered Nov 27 '18 at 21:24









            gimusigimusi

            1




            1












            • Could you take this a step further?
              – bob
              Nov 27 '18 at 21:36








            • 1




              @bob What is stronger $ 2^x $or $2^{sqrt x log_2 3}$?
              – gimusi
              Nov 27 '18 at 21:38


















            • Could you take this a step further?
              – bob
              Nov 27 '18 at 21:36








            • 1




              @bob What is stronger $ 2^x $or $2^{sqrt x log_2 3}$?
              – gimusi
              Nov 27 '18 at 21:38
















            Could you take this a step further?
            – bob
            Nov 27 '18 at 21:36






            Could you take this a step further?
            – bob
            Nov 27 '18 at 21:36






            1




            1




            @bob What is stronger $ 2^x $or $2^{sqrt x log_2 3}$?
            – gimusi
            Nov 27 '18 at 21:38




            @bob What is stronger $ 2^x $or $2^{sqrt x log_2 3}$?
            – gimusi
            Nov 27 '18 at 21:38


















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