Find a general solution for this recurrence: $a_n = sqrt{a_{n-1}a_{n-2}}$












2












$begingroup$


Find a general solution for this recurrence:
$$a_n = sqrt{a_{n-1}a_{n-2}}$$
when $a_1 = 2$, $a_2 = 8$.




My attempt to solve it:



This recurrence isn't a regular one. In order to solve it,
I have tried to count the first elements in this recurrence, which are $$a_1 = 2, a_2 = 8, a_3 = 4, a_4 = sqrt{32}, a_5 = sqrt{512} ...,$$ but I didn't find any way to proceed.



Any help will be very appreciated.






recurrence-relations combinations







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    Try $a_n$ = $2^{b_n}$
    $endgroup$
    – DanielV
    Jan 11 at 20:06










  • $begingroup$
    and what is $b_n$?
    $endgroup$
    – Robo Yonuomaro
    Jan 11 at 20:10










  • $begingroup$
    I mean how can I find a general solution for $b_n$?
    $endgroup$
    – Robo Yonuomaro
    Jan 11 at 20:16










  • $begingroup$
    @RoboYonuomaro using the given recurrence formula for $a_n$.
    $endgroup$
    – Scientifica
    Jan 11 at 20:18










  • $begingroup$
    First write the recurrence among $b_n, b_{n-1}, b_{n-2}$ using
    $endgroup$
    – Will Jagy
    Jan 11 at 20:18
















2












$begingroup$


Find a general solution for this recurrence:
$$a_n = sqrt{a_{n-1}a_{n-2}}$$
when $a_1 = 2$, $a_2 = 8$.




My attempt to solve it:



This recurrence isn't a regular one. In order to solve it,
I have tried to count the first elements in this recurrence, which are $$a_1 = 2, a_2 = 8, a_3 = 4, a_4 = sqrt{32}, a_5 = sqrt{512} ...,$$ but I didn't find any way to proceed.



Any help will be very appreciated.






recurrence-relations combinations







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    Try $a_n$ = $2^{b_n}$
    $endgroup$
    – DanielV
    Jan 11 at 20:06










  • $begingroup$
    and what is $b_n$?
    $endgroup$
    – Robo Yonuomaro
    Jan 11 at 20:10










  • $begingroup$
    I mean how can I find a general solution for $b_n$?
    $endgroup$
    – Robo Yonuomaro
    Jan 11 at 20:16










  • $begingroup$
    @RoboYonuomaro using the given recurrence formula for $a_n$.
    $endgroup$
    – Scientifica
    Jan 11 at 20:18










  • $begingroup$
    First write the recurrence among $b_n, b_{n-1}, b_{n-2}$ using
    $endgroup$
    – Will Jagy
    Jan 11 at 20:18














2












2








2





$begingroup$


Find a general solution for this recurrence:
$$a_n = sqrt{a_{n-1}a_{n-2}}$$
when $a_1 = 2$, $a_2 = 8$.




My attempt to solve it:



This recurrence isn't a regular one. In order to solve it,
I have tried to count the first elements in this recurrence, which are $$a_1 = 2, a_2 = 8, a_3 = 4, a_4 = sqrt{32}, a_5 = sqrt{512} ...,$$ but I didn't find any way to proceed.



Any help will be very appreciated.






recurrence-relations combinations







share|cite|improve this question











$endgroup$




Find a general solution for this recurrence:
$$a_n = sqrt{a_{n-1}a_{n-2}}$$
when $a_1 = 2$, $a_2 = 8$.




My attempt to solve it:



This recurrence isn't a regular one. In order to solve it,
I have tried to count the first elements in this recurrence, which are $$a_1 = 2, a_2 = 8, a_3 = 4, a_4 = sqrt{32}, a_5 = sqrt{512} ...,$$ but I didn't find any way to proceed.



Any help will be very appreciated.






recurrence-relations combinations




recurrence-relations combinations






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 11 at 20:23









Namaste

1




1










asked Jan 11 at 20:04









Robo YonuomaroRobo Yonuomaro

786




786








  • 3




    $begingroup$
    Try $a_n$ = $2^{b_n}$
    $endgroup$
    – DanielV
    Jan 11 at 20:06










  • $begingroup$
    and what is $b_n$?
    $endgroup$
    – Robo Yonuomaro
    Jan 11 at 20:10










  • $begingroup$
    I mean how can I find a general solution for $b_n$?
    $endgroup$
    – Robo Yonuomaro
    Jan 11 at 20:16










  • $begingroup$
    @RoboYonuomaro using the given recurrence formula for $a_n$.
    $endgroup$
    – Scientifica
    Jan 11 at 20:18










  • $begingroup$
    First write the recurrence among $b_n, b_{n-1}, b_{n-2}$ using
    $endgroup$
    – Will Jagy
    Jan 11 at 20:18














  • 3




    $begingroup$
    Try $a_n$ = $2^{b_n}$
    $endgroup$
    – DanielV
    Jan 11 at 20:06










  • $begingroup$
    and what is $b_n$?
    $endgroup$
    – Robo Yonuomaro
    Jan 11 at 20:10










  • $begingroup$
    I mean how can I find a general solution for $b_n$?
    $endgroup$
    – Robo Yonuomaro
    Jan 11 at 20:16










  • $begingroup$
    @RoboYonuomaro using the given recurrence formula for $a_n$.
    $endgroup$
    – Scientifica
    Jan 11 at 20:18










  • $begingroup$
    First write the recurrence among $b_n, b_{n-1}, b_{n-2}$ using
    $endgroup$
    – Will Jagy
    Jan 11 at 20:18








3




3




$begingroup$
Try $a_n$ = $2^{b_n}$
$endgroup$
– DanielV
Jan 11 at 20:06




$begingroup$
Try $a_n$ = $2^{b_n}$
$endgroup$
– DanielV
Jan 11 at 20:06












$begingroup$
and what is $b_n$?
$endgroup$
– Robo Yonuomaro
Jan 11 at 20:10




$begingroup$
and what is $b_n$?
$endgroup$
– Robo Yonuomaro
Jan 11 at 20:10












$begingroup$
I mean how can I find a general solution for $b_n$?
$endgroup$
– Robo Yonuomaro
Jan 11 at 20:16




$begingroup$
I mean how can I find a general solution for $b_n$?
$endgroup$
– Robo Yonuomaro
Jan 11 at 20:16












$begingroup$
@RoboYonuomaro using the given recurrence formula for $a_n$.
$endgroup$
– Scientifica
Jan 11 at 20:18




$begingroup$
@RoboYonuomaro using the given recurrence formula for $a_n$.
$endgroup$
– Scientifica
Jan 11 at 20:18












$begingroup$
First write the recurrence among $b_n, b_{n-1}, b_{n-2}$ using
$endgroup$
– Will Jagy
Jan 11 at 20:18




$begingroup$
First write the recurrence among $b_n, b_{n-1}, b_{n-2}$ using
$endgroup$
– Will Jagy
Jan 11 at 20:18










2 Answers
2






active

oldest

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11












$begingroup$

Hint: By taking the log (with respect to any base) of both sides, we can rewrite the recurrence as
$$
log(a_n) = frac 12[log(a_{n-1}) + log(a_{n-2})]
$$
So, the sequence $b_n := log(a_n)$ satisfies a linear recurrence.






share|cite|improve this answer









$endgroup$





















    3












    $begingroup$

    Taking logs,
    and letting
    $b_n = log a_n$,
    this becomes
    $b_n
    =frac12(b_{n-1}+b_{n-2})
    $    .



    The characteristic polynomial is
    $x^2-frac12 x-frac12 = 0$
    which has roots
    $x
    =dfrac{frac12pmsqrt{frac14+2}}{2}
    =dfrac{frac12pmsqrt{frac94}}{2}
    =dfrac{frac12pmfrac32}{2}
    =dfrac{2, -1}{2}
    =1, -frac12
    $
    so the solutions are
    $b_n = 1$
    and
    $b_n = (-1/2)^n
    $    .



    As a check
    $frac12(b_{n-1}+b_{n-2})
    =frac12((-1/2)^{n-1}+(-1/2)^{n-2})
    =frac12(-1/2)^{n-2}(-frac12+1)
    =frac12(-1/2)^{n-2}(frac12)
    =frac14(-1/2)^{n-2}
    =(-1/2)^{n}
    $    .



    Therefore
    $b_n = u+v(-1/2)^{n}$
    for any reals $u$ and $v$,
    so,
    $a_n = rs^{(-1/2)^{n}}$
    where $r$ and $s$ are
    positive reals
    (actually, you can let them be
    any types that you can raise to a real power).






    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









      11












      $begingroup$

      Hint: By taking the log (with respect to any base) of both sides, we can rewrite the recurrence as
      $$
      log(a_n) = frac 12[log(a_{n-1}) + log(a_{n-2})]
      $$
      So, the sequence $b_n := log(a_n)$ satisfies a linear recurrence.






      share|cite|improve this answer









      $endgroup$


















        11












        $begingroup$

        Hint: By taking the log (with respect to any base) of both sides, we can rewrite the recurrence as
        $$
        log(a_n) = frac 12[log(a_{n-1}) + log(a_{n-2})]
        $$
        So, the sequence $b_n := log(a_n)$ satisfies a linear recurrence.






        share|cite|improve this answer









        $endgroup$
















          11












          11








          11





          $begingroup$

          Hint: By taking the log (with respect to any base) of both sides, we can rewrite the recurrence as
          $$
          log(a_n) = frac 12[log(a_{n-1}) + log(a_{n-2})]
          $$
          So, the sequence $b_n := log(a_n)$ satisfies a linear recurrence.






          share|cite|improve this answer









          $endgroup$



          Hint: By taking the log (with respect to any base) of both sides, we can rewrite the recurrence as
          $$
          log(a_n) = frac 12[log(a_{n-1}) + log(a_{n-2})]
          $$
          So, the sequence $b_n := log(a_n)$ satisfies a linear recurrence.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 11 at 20:23









          OmnomnomnomOmnomnomnom

          129k794188




          129k794188























              3












              $begingroup$

              Taking logs,
              and letting
              $b_n = log a_n$,
              this becomes
              $b_n
              =frac12(b_{n-1}+b_{n-2})
              $              .



              The characteristic polynomial is
              $x^2-frac12 x-frac12 = 0$
              which has roots
              $x
              =dfrac{frac12pmsqrt{frac14+2}}{2}
              =dfrac{frac12pmsqrt{frac94}}{2}
              =dfrac{frac12pmfrac32}{2}
              =dfrac{2, -1}{2}
              =1, -frac12
              $
              so the solutions are
              $b_n = 1$
              and
              $b_n = (-1/2)^n
              $              .



              As a check
              $frac12(b_{n-1}+b_{n-2})
              =frac12((-1/2)^{n-1}+(-1/2)^{n-2})
              =frac12(-1/2)^{n-2}(-frac12+1)
              =frac12(-1/2)^{n-2}(frac12)
              =frac14(-1/2)^{n-2}
              =(-1/2)^{n}
              $              .



              Therefore
              $b_n = u+v(-1/2)^{n}$
              for any reals $u$ and $v$,
              so,
              $a_n = rs^{(-1/2)^{n}}$
              where $r$ and $s$ are
              positive reals
              (actually, you can let them be
              any types that you can raise to a real power).






              share|cite|improve this answer









              $endgroup$


















                3












                $begingroup$

                Taking logs,
                and letting
                $b_n = log a_n$,
                this becomes
                $b_n
                =frac12(b_{n-1}+b_{n-2})
                $                .



                The characteristic polynomial is
                $x^2-frac12 x-frac12 = 0$
                which has roots
                $x
                =dfrac{frac12pmsqrt{frac14+2}}{2}
                =dfrac{frac12pmsqrt{frac94}}{2}
                =dfrac{frac12pmfrac32}{2}
                =dfrac{2, -1}{2}
                =1, -frac12
                $
                so the solutions are
                $b_n = 1$
                and
                $b_n = (-1/2)^n
                $                .



                As a check
                $frac12(b_{n-1}+b_{n-2})
                =frac12((-1/2)^{n-1}+(-1/2)^{n-2})
                =frac12(-1/2)^{n-2}(-frac12+1)
                =frac12(-1/2)^{n-2}(frac12)
                =frac14(-1/2)^{n-2}
                =(-1/2)^{n}
                $                .



                Therefore
                $b_n = u+v(-1/2)^{n}$
                for any reals $u$ and $v$,
                so,
                $a_n = rs^{(-1/2)^{n}}$
                where $r$ and $s$ are
                positive reals
                (actually, you can let them be
                any types that you can raise to a real power).






                share|cite|improve this answer









                $endgroup$
















                  3












                  3








                  3





                  $begingroup$

                  Taking logs,
                  and letting
                  $b_n = log a_n$,
                  this becomes
                  $b_n
                  =frac12(b_{n-1}+b_{n-2})
                  $                  .



                  The characteristic polynomial is
                  $x^2-frac12 x-frac12 = 0$
                  which has roots
                  $x
                  =dfrac{frac12pmsqrt{frac14+2}}{2}
                  =dfrac{frac12pmsqrt{frac94}}{2}
                  =dfrac{frac12pmfrac32}{2}
                  =dfrac{2, -1}{2}
                  =1, -frac12
                  $
                  so the solutions are
                  $b_n = 1$
                  and
                  $b_n = (-1/2)^n
                  $                  .



                  As a check
                  $frac12(b_{n-1}+b_{n-2})
                  =frac12((-1/2)^{n-1}+(-1/2)^{n-2})
                  =frac12(-1/2)^{n-2}(-frac12+1)
                  =frac12(-1/2)^{n-2}(frac12)
                  =frac14(-1/2)^{n-2}
                  =(-1/2)^{n}
                  $                  .



                  Therefore
                  $b_n = u+v(-1/2)^{n}$
                  for any reals $u$ and $v$,
                  so,
                  $a_n = rs^{(-1/2)^{n}}$
                  where $r$ and $s$ are
                  positive reals
                  (actually, you can let them be
                  any types that you can raise to a real power).






                  share|cite|improve this answer









                  $endgroup$



                  Taking logs,
                  and letting
                  $b_n = log a_n$,
                  this becomes
                  $b_n
                  =frac12(b_{n-1}+b_{n-2})
                  $                  .



                  The characteristic polynomial is
                  $x^2-frac12 x-frac12 = 0$
                  which has roots
                  $x
                  =dfrac{frac12pmsqrt{frac14+2}}{2}
                  =dfrac{frac12pmsqrt{frac94}}{2}
                  =dfrac{frac12pmfrac32}{2}
                  =dfrac{2, -1}{2}
                  =1, -frac12
                  $
                  so the solutions are
                  $b_n = 1$
                  and
                  $b_n = (-1/2)^n
                  $                  .



                  As a check
                  $frac12(b_{n-1}+b_{n-2})
                  =frac12((-1/2)^{n-1}+(-1/2)^{n-2})
                  =frac12(-1/2)^{n-2}(-frac12+1)
                  =frac12(-1/2)^{n-2}(frac12)
                  =frac14(-1/2)^{n-2}
                  =(-1/2)^{n}
                  $                  .



                  Therefore
                  $b_n = u+v(-1/2)^{n}$
                  for any reals $u$ and $v$,
                  so,
                  $a_n = rs^{(-1/2)^{n}}$
                  where $r$ and $s$ are
                  positive reals
                  (actually, you can let them be
                  any types that you can raise to a real power).







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 11 at 20:35









                  marty cohenmarty cohen

                  75.4k549130




                  75.4k549130






























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