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







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$endgroup$












  • $begingroup$
    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









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      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

          1,059414




          1,059414























              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

                  73.6k549128




                  73.6k549128






























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