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







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














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











    Your Answer





    StackExchange.ifUsing("editor", function () {
    return StackExchange.using("mathjaxEditing", function () {
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    });
    });
    }, "mathjax-editing");

    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "69"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3016310%2ffind-whether-fx-is-ogx-whether-fx-is-big-o-of-gx%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    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


















            draft saved

            draft discarded




















































            Thanks for contributing an answer to Mathematics Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.





            Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


            Please pay close attention to the following guidance:


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3016310%2ffind-whether-fx-is-ogx-whether-fx-is-big-o-of-gx%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            -

            Popular posts from this blog

            Plaza Victoria

            Image
            Plaza Victoria Plaza Victoria en 2017 Otros nombres Plaza de La Victoria Localización El Almendral, Valparaíso, Chile Vías adyacentes Chacabuco, Molina, Plaza Victoria y Edwards Creación 1841 Metro Bellavista Ubicación 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 Coordenadas: 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 [editar datos en Wikidata] La Plaza Victoria , también llamada Plaza de la Victoria , [ 1 ] ​ es una plaza ubicada en el sector El Almendral del plan de la ciudad de Valparaíso, Chile, de importancia patrimonial, que incluye valiosas estatuas y especies vegetales. [ 2 ] ​ La plaza se creó como tal en la primera mitad del siglo XIX, tomando previamente los nombres de Plaza Nueva y Plaza de Orrego . Si bien desde sus inicios tuvo relevancia histórica, no fue sino a partir de fines de los años 1860 cuando comenzó a cobrar mayor importancia y se convirtió en uno de los principales centros sociales de l
            Read more

            In PowerPoint, is there a keyboard shortcut for bulleted / numbered list?

            Image
            .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box; } 2 OneNote has two useful keyboard shortcuts for bulleted and numbered lists, namely Ctrl + . and Ctrl + - . Is there something similar for PowerPoint where the work with lists is also very common? Haven't found anything here: Use keyboard shortcuts in PowerPoint. keyboard-shortcuts microsoft-powerpoint share | improve this question edited Aug 14 '14 at 14:49 Linger 2,826 10 28 40
            Read more

            How to put 3 figures in Latex with 2 figures side by side and 1 below these side by side images but in...

            Image
            1 0 I want to put 3 images in LateX such as 2 figures are side by side horizotally and 3rd figure below these 2 side by side figures but in middle. I have following code: begin{figure}[H] centering begin{minipage}{.5textwidth} centering includegraphics[width=.50linewidth]{Text/Images/Genelec_8010_AP.jpg} captionof{figure}{Genelec 8010 AP} label{fig:Genelec 8010} end{minipage}% begin{minipage}{.5textwidth} centering includegraphics[width=.69linewidth]{Text/Images/Genelec_8020_CPM.jpg} captionof{figure}{Genelec 8020 CPM} label{fig:Genelec 8020} end{minipage}% end{figure} %%Here is my 3rd picture code begin{figure} centering includegraphics[width=5cm]{Text/Images/Genelec_8030_BPM.jpg} caption{Genelec 8030 BPM} label{fig:Genelec 8030} end{figure} I want to ask ho
            Read more