Prove that $O(3^{2n}) subseteq O(2^{3n})$












-1












$begingroup$


I need to prove that $O(3^{2n}) subseteq O(2^{3n})$. So far I have made this solution:



I)Lets assume that this is true, and that there $ exists space c in mathbb{R}^+ $ such as $$3^{2n} leq 2^{3n} * c$$



II) therefore this fraction must be greater or equal as $1$



$$frac{c*2^{3n}}{3^{2n}} geq 1$$



III) lets look on $$lim_{ntoinfty} frac{c*2^{3n}}{3^{2n}}=(frac{8}{9})^n=0$$ since the limit is equal to $0$ there occurs an conflict that shows us that II) can not ever be greater or equal as $1$. Is my solution correct? Any help is appreciated.






computational-complexity







share|cite|improve this question











$endgroup$












  • $begingroup$
    I'm pretty sure this is false.
    $endgroup$
    – Lucas Henrique
    Dec 16 '18 at 22:19










  • $begingroup$
    The statement in your question's title is the other way round to the statement in the body.
    $endgroup$
    – Patrick Stevens
    Dec 16 '18 at 22:25










  • $begingroup$
    Edited the title, thanks
    $endgroup$
    – nocturne
    Dec 16 '18 at 22:53






  • 1




    $begingroup$
    Now the title and the body are different (small-o vs big-O) and both wrong.
    $endgroup$
    – Did
    Dec 17 '18 at 0:04










  • $begingroup$
    Edited again, it should be fine now.
    $endgroup$
    – nocturne
    Dec 17 '18 at 8:44
















-1












$begingroup$


I need to prove that $O(3^{2n}) subseteq O(2^{3n})$. So far I have made this solution:



I)Lets assume that this is true, and that there $ exists space c in mathbb{R}^+ $ such as $$3^{2n} leq 2^{3n} * c$$



II) therefore this fraction must be greater or equal as $1$



$$frac{c*2^{3n}}{3^{2n}} geq 1$$



III) lets look on $$lim_{ntoinfty} frac{c*2^{3n}}{3^{2n}}=(frac{8}{9})^n=0$$ since the limit is equal to $0$ there occurs an conflict that shows us that II) can not ever be greater or equal as $1$. Is my solution correct? Any help is appreciated.






computational-complexity







share|cite|improve this question











$endgroup$












  • $begingroup$
    I'm pretty sure this is false.
    $endgroup$
    – Lucas Henrique
    Dec 16 '18 at 22:19










  • $begingroup$
    The statement in your question's title is the other way round to the statement in the body.
    $endgroup$
    – Patrick Stevens
    Dec 16 '18 at 22:25










  • $begingroup$
    Edited the title, thanks
    $endgroup$
    – nocturne
    Dec 16 '18 at 22:53






  • 1




    $begingroup$
    Now the title and the body are different (small-o vs big-O) and both wrong.
    $endgroup$
    – Did
    Dec 17 '18 at 0:04










  • $begingroup$
    Edited again, it should be fine now.
    $endgroup$
    – nocturne
    Dec 17 '18 at 8:44














-1












-1








-1





$begingroup$


I need to prove that $O(3^{2n}) subseteq O(2^{3n})$. So far I have made this solution:



I)Lets assume that this is true, and that there $ exists space c in mathbb{R}^+ $ such as $$3^{2n} leq 2^{3n} * c$$



II) therefore this fraction must be greater or equal as $1$



$$frac{c*2^{3n}}{3^{2n}} geq 1$$



III) lets look on $$lim_{ntoinfty} frac{c*2^{3n}}{3^{2n}}=(frac{8}{9})^n=0$$ since the limit is equal to $0$ there occurs an conflict that shows us that II) can not ever be greater or equal as $1$. Is my solution correct? Any help is appreciated.






computational-complexity







share|cite|improve this question











$endgroup$




I need to prove that $O(3^{2n}) subseteq O(2^{3n})$. So far I have made this solution:



I)Lets assume that this is true, and that there $ exists space c in mathbb{R}^+ $ such as $$3^{2n} leq 2^{3n} * c$$



II) therefore this fraction must be greater or equal as $1$



$$frac{c*2^{3n}}{3^{2n}} geq 1$$



III) lets look on $$lim_{ntoinfty} frac{c*2^{3n}}{3^{2n}}=(frac{8}{9})^n=0$$ since the limit is equal to $0$ there occurs an conflict that shows us that II) can not ever be greater or equal as $1$. Is my solution correct? Any help is appreciated.






computational-complexity




computational-complexity






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 17 '18 at 8:44







nocturne

















asked Dec 16 '18 at 20:06









nocturnenocturne

689




689












  • $begingroup$
    I'm pretty sure this is false.
    $endgroup$
    – Lucas Henrique
    Dec 16 '18 at 22:19










  • $begingroup$
    The statement in your question's title is the other way round to the statement in the body.
    $endgroup$
    – Patrick Stevens
    Dec 16 '18 at 22:25










  • $begingroup$
    Edited the title, thanks
    $endgroup$
    – nocturne
    Dec 16 '18 at 22:53






  • 1




    $begingroup$
    Now the title and the body are different (small-o vs big-O) and both wrong.
    $endgroup$
    – Did
    Dec 17 '18 at 0:04










  • $begingroup$
    Edited again, it should be fine now.
    $endgroup$
    – nocturne
    Dec 17 '18 at 8:44


















  • $begingroup$
    I'm pretty sure this is false.
    $endgroup$
    – Lucas Henrique
    Dec 16 '18 at 22:19










  • $begingroup$
    The statement in your question's title is the other way round to the statement in the body.
    $endgroup$
    – Patrick Stevens
    Dec 16 '18 at 22:25










  • $begingroup$
    Edited the title, thanks
    $endgroup$
    – nocturne
    Dec 16 '18 at 22:53






  • 1




    $begingroup$
    Now the title and the body are different (small-o vs big-O) and both wrong.
    $endgroup$
    – Did
    Dec 17 '18 at 0:04










  • $begingroup$
    Edited again, it should be fine now.
    $endgroup$
    – nocturne
    Dec 17 '18 at 8:44
















$begingroup$
I'm pretty sure this is false.
$endgroup$
– Lucas Henrique
Dec 16 '18 at 22:19




$begingroup$
I'm pretty sure this is false.
$endgroup$
– Lucas Henrique
Dec 16 '18 at 22:19












$begingroup$
The statement in your question's title is the other way round to the statement in the body.
$endgroup$
– Patrick Stevens
Dec 16 '18 at 22:25




$begingroup$
The statement in your question's title is the other way round to the statement in the body.
$endgroup$
– Patrick Stevens
Dec 16 '18 at 22:25












$begingroup$
Edited the title, thanks
$endgroup$
– nocturne
Dec 16 '18 at 22:53




$begingroup$
Edited the title, thanks
$endgroup$
– nocturne
Dec 16 '18 at 22:53




1




1




$begingroup$
Now the title and the body are different (small-o vs big-O) and both wrong.
$endgroup$
– Did
Dec 17 '18 at 0:04




$begingroup$
Now the title and the body are different (small-o vs big-O) and both wrong.
$endgroup$
– Did
Dec 17 '18 at 0:04












$begingroup$
Edited again, it should be fine now.
$endgroup$
– nocturne
Dec 17 '18 at 8:44




$begingroup$
Edited again, it should be fine now.
$endgroup$
– nocturne
Dec 17 '18 at 8:44










2 Answers
2






active

oldest

votes


















1












$begingroup$

This is false.



Suppose it holds: then, since obviously $3^{2n} in O(3^{2n}) subseteq O(2^{3n}) implies 3^{2n} in O(2^{3n})$, $lim_{n to infty}, sup |frac{9}{8}|^n < infty$, which obviously does not hold.






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    More generally,
    when is
    $a^{bn} subseteq c^{dn}$?



    $a^{bn}
    =(e^{ln(a)})^{bn}
    =e^{ln(a)b n}
    =(e^n)^{ln(a)b }
    =(e^n)^{ln(a^b) }
    $    .



    Therefore
    $a^{bn} subseteq c^{dn}
    iff ln(a^b) le ln(c^d)
    iff a^b le c^d
    $    .



    Putting
    $a=3, b=2, c=2, d=3$,
    this is true
    iff
    $3^2 le 2^3$,
    which is false.



    For the case
    $c=b, d=a$,
    i.e.,
    when is
    $a^{bn} subseteq b^{an}
    $    ,
    this is the old problem
    of when
    $a^b le b^a$,
    or
    $a^{1/a} le b^{1/b}$.



    It is well known that
    this is true when
    $e le b lt a$.



    The case
    $a < e < b$
    is trickier.
    I have a result that shows
    this is decided if
    $ab > e^2$
    but I can't find it right now.
    I'll add to this if
    I find it.






    share|cite|improve this answer









    $endgroup$













      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%2f3043089%2fprove-that-o32n-subseteq-o23n%23new-answer', 'question_page');
      }
      );

      Post as a guest















      Required, but never shown

























      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      1












      $begingroup$

      This is false.



      Suppose it holds: then, since obviously $3^{2n} in O(3^{2n}) subseteq O(2^{3n}) implies 3^{2n} in O(2^{3n})$, $lim_{n to infty}, sup |frac{9}{8}|^n < infty$, which obviously does not hold.






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        This is false.



        Suppose it holds: then, since obviously $3^{2n} in O(3^{2n}) subseteq O(2^{3n}) implies 3^{2n} in O(2^{3n})$, $lim_{n to infty}, sup |frac{9}{8}|^n < infty$, which obviously does not hold.






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          This is false.



          Suppose it holds: then, since obviously $3^{2n} in O(3^{2n}) subseteq O(2^{3n}) implies 3^{2n} in O(2^{3n})$, $lim_{n to infty}, sup |frac{9}{8}|^n < infty$, which obviously does not hold.






          share|cite|improve this answer









          $endgroup$



          This is false.



          Suppose it holds: then, since obviously $3^{2n} in O(3^{2n}) subseteq O(2^{3n}) implies 3^{2n} in O(2^{3n})$, $lim_{n to infty}, sup |frac{9}{8}|^n < infty$, which obviously does not hold.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 16 '18 at 22:24









          Lucas HenriqueLucas Henrique

          1,026414




          1,026414























              1












              $begingroup$

              More generally,
              when is
              $a^{bn} subseteq c^{dn}$?



              $a^{bn}
              =(e^{ln(a)})^{bn}
              =e^{ln(a)b n}
              =(e^n)^{ln(a)b }
              =(e^n)^{ln(a^b) }
              $              .



              Therefore
              $a^{bn} subseteq c^{dn}
              iff ln(a^b) le ln(c^d)
              iff a^b le c^d
              $              .



              Putting
              $a=3, b=2, c=2, d=3$,
              this is true
              iff
              $3^2 le 2^3$,
              which is false.



              For the case
              $c=b, d=a$,
              i.e.,
              when is
              $a^{bn} subseteq b^{an}
              $              ,
              this is the old problem
              of when
              $a^b le b^a$,
              or
              $a^{1/a} le b^{1/b}$.



              It is well known that
              this is true when
              $e le b lt a$.



              The case
              $a < e < b$
              is trickier.
              I have a result that shows
              this is decided if
              $ab > e^2$
              but I can't find it right now.
              I'll add to this if
              I find it.






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                More generally,
                when is
                $a^{bn} subseteq c^{dn}$?



                $a^{bn}
                =(e^{ln(a)})^{bn}
                =e^{ln(a)b n}
                =(e^n)^{ln(a)b }
                =(e^n)^{ln(a^b) }
                $                .



                Therefore
                $a^{bn} subseteq c^{dn}
                iff ln(a^b) le ln(c^d)
                iff a^b le c^d
                $                .



                Putting
                $a=3, b=2, c=2, d=3$,
                this is true
                iff
                $3^2 le 2^3$,
                which is false.



                For the case
                $c=b, d=a$,
                i.e.,
                when is
                $a^{bn} subseteq b^{an}
                $                ,
                this is the old problem
                of when
                $a^b le b^a$,
                or
                $a^{1/a} le b^{1/b}$.



                It is well known that
                this is true when
                $e le b lt a$.



                The case
                $a < e < b$
                is trickier.
                I have a result that shows
                this is decided if
                $ab > e^2$
                but I can't find it right now.
                I'll add to this if
                I find it.






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  More generally,
                  when is
                  $a^{bn} subseteq c^{dn}$?



                  $a^{bn}
                  =(e^{ln(a)})^{bn}
                  =e^{ln(a)b n}
                  =(e^n)^{ln(a)b }
                  =(e^n)^{ln(a^b) }
                  $                  .



                  Therefore
                  $a^{bn} subseteq c^{dn}
                  iff ln(a^b) le ln(c^d)
                  iff a^b le c^d
                  $                  .



                  Putting
                  $a=3, b=2, c=2, d=3$,
                  this is true
                  iff
                  $3^2 le 2^3$,
                  which is false.



                  For the case
                  $c=b, d=a$,
                  i.e.,
                  when is
                  $a^{bn} subseteq b^{an}
                  $                  ,
                  this is the old problem
                  of when
                  $a^b le b^a$,
                  or
                  $a^{1/a} le b^{1/b}$.



                  It is well known that
                  this is true when
                  $e le b lt a$.



                  The case
                  $a < e < b$
                  is trickier.
                  I have a result that shows
                  this is decided if
                  $ab > e^2$
                  but I can't find it right now.
                  I'll add to this if
                  I find it.






                  share|cite|improve this answer









                  $endgroup$



                  More generally,
                  when is
                  $a^{bn} subseteq c^{dn}$?



                  $a^{bn}
                  =(e^{ln(a)})^{bn}
                  =e^{ln(a)b n}
                  =(e^n)^{ln(a)b }
                  =(e^n)^{ln(a^b) }
                  $                  .



                  Therefore
                  $a^{bn} subseteq c^{dn}
                  iff ln(a^b) le ln(c^d)
                  iff a^b le c^d
                  $                  .



                  Putting
                  $a=3, b=2, c=2, d=3$,
                  this is true
                  iff
                  $3^2 le 2^3$,
                  which is false.



                  For the case
                  $c=b, d=a$,
                  i.e.,
                  when is
                  $a^{bn} subseteq b^{an}
                  $                  ,
                  this is the old problem
                  of when
                  $a^b le b^a$,
                  or
                  $a^{1/a} le b^{1/b}$.



                  It is well known that
                  this is true when
                  $e le b lt a$.



                  The case
                  $a < e < b$
                  is trickier.
                  I have a result that shows
                  this is decided if
                  $ab > e^2$
                  but I can't find it right now.
                  I'll add to this if
                  I find it.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 17 '18 at 0:17









                  marty cohenmarty cohen

                  74.2k549128




                  74.2k549128






























                      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.




                      draft saved


                      draft discarded














                      StackExchange.ready(
                      function () {
                      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3043089%2fprove-that-o32n-subseteq-o23n%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

                      Bundesstraße 106

                      Image
                      Dieser Artikel beschreibt die Bundesstraße 106 in Deutschland. Zur gleichnamigen Straße in Österreich siehe Mölltal Straße. Vorlage:Infobox hochrangige Straße/Wartung/DE-B Bundesstraße 106 in Deutschland Karte Basisdaten Betreiber: Deutschland  Bundesrepublik Deutschland Straßenbeginn: Wismar ( 53° 54′  N , 11° 24′  O 53.900536 11.396875 ) Straßenende: Schwerin ( 53° 36′  N , 11° 24′  O 53.592785 11.404892 ) Gesamtlänge: 40,7 km Bundesland: Mecklenburg-Vorpommern Ausbauzustand: zweistreifig [1] Bundesstraße 106 in Schwerin (Umgehungsstraße) Straßenverlauf Land Mecklenburg-Vorpommern Landkreis Nordwestmecklenburg Ortsumgehung Wismar Wallensteingraben über   Dorf Mecklenburg Wallensteingraben Groß Stieten Bad Kleinen OT Niendorf Bahnstrecke Lübeck–Bad Kleinen Zickhusen Lübstorf Klein Trebbow OT Kirch Stück
                      Read more

                      Verónica Boquete

                      Image
                      Verónica Boquete im Trikot der Nationalmannschaft (2013) Personalia Name Verónica Boquete Giadans Geburtstag 9. April 1987 Geburtsort Santiago de Compostela, Spanien Größe 159 cm Position Mittelfeld / Sturm Juniorinnen Jahre Station 2001–2005 Xuventú Aguiño Frauen Jahre Station Spiele (Tore) 1 2005–2008 Prainsa Zaragoza 83 (40) 2008–2010 Espanyol Barcelona 53 (30) 2010 Buffalo Flash 9 0 (9) 2010 Chicago Red Stars 3 0 (1) 2010–2011 Espanyol Barcelona 26 (41) 2011 Philadelphia Independence 13 0 (5) 2011 Energija Woronesch 5 0 (1) 2012–2014 Tyresö FF 39 (14) 2014 Portland Thorns FC 15 0 (4) 2014–2015 1. FFC Frankfurt 23 0 (7) 2015–2016 FC Bayern München 9 0 (0) 2016–2018 Paris Saint-Germain 31 0 (8) 2018 Peking BG Phoenix 2019– Utah Royals FC 0 0 (0) Nationalmannschaft Jahre Auswahl
                      Read more

                      Ida-Boy-Ed-Garten

                      Image
                      Die Lage des Ida-Boy-Ed-Gartens, rot markiert Der Ida-Boy-Ed-Garten Der Ida-Boy-Ed-Garten, Blick von der Wakenitzmauer durch das Tor in der Stadtmauer Der Ida-Boy-Ed-Garten ist eine Straße der Lübecker Altstadt. Inhaltsverzeichnis 1 Lage 2 Geschichte 3 Literatur 4 Weblinks Lage | Der etwa 90 Meter lange Ida-Boy-Ed-Garten befindet sich an der nördlichen Spitze der Altstadtinsel, vor dem Burgtor. Die in einem Bogen durch eine Grünanlage hangaufwärts führende Straße verbindet die Wakenitzmauer mit der Burgtorbrücke. Geschichte | Der heutige Ida-Boy-Ed-Garten wurde ursprünglich als Verlängerung der Wakenitzmauer angelegt. 1903 wurde in den erhaltenen Teil der mittelalterlichen Stadtmauer, an dem die Wakenitzmauer bei der Einmündung der Kaiserstraße begann, ein Tor mit einem großen Bogen für die Durchführung der Straße und zwei kleineren seitlichen Öffnungen für Fußgänger gebrochen, um für die neu errichteten Wohnhäuser in diesem Bereich
                      Read more