How to convert the solution of this recurrence like this form?

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3
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I am trying to solve the recurrence sequence. I tried



RSolve[a[n] == 1/2 (a[n - 1] + 9/a[n - 1]), a[1] == 1, a[n], 
 n] // Simplify


I got




a[n] -> 3 Coth[2^(-1 + n) ArcCoth[1/3]]




How to convert this solution to this form




(3*2^(2^(-1 + n)) + 3)/(2^(2^(-1 + n)) - 1)







output-formatting sequence







share|improve this question























  • This is not possible for these are entirely different number. One has $a(n) to 0$ for $n to infty$, but $(3 (49 + 49^n))/(-49 + 49^n) to 3$.
    – Henrik Schumacher
    1 hour ago











  • They are not equal at n=2 for instance.
    – b.gatessucks
    1 hour ago










  • I just realized that the solution $a(n)$ given by Mathematica does not fulfill the recurrence relation...
    – Henrik Schumacher
    1 hour ago










  • @b.gatessucks Edited.
    – minhthien_2016
    1 hour ago














up vote
3
down vote

favorite












I am trying to solve the recurrence sequence. I tried



RSolve[a[n] == 1/2 (a[n - 1] + 9/a[n - 1]), a[1] == 1, a[n], 
 n] // Simplify


I got




a[n] -> 3 Coth[2^(-1 + n) ArcCoth[1/3]]




How to convert this solution to this form




(3*2^(2^(-1 + n)) + 3)/(2^(2^(-1 + n)) - 1)







output-formatting sequence







share|improve this question























  • This is not possible for these are entirely different number. One has $a(n) to 0$ for $n to infty$, but $(3 (49 + 49^n))/(-49 + 49^n) to 3$.
    – Henrik Schumacher
    1 hour ago











  • They are not equal at n=2 for instance.
    – b.gatessucks
    1 hour ago










  • I just realized that the solution $a(n)$ given by Mathematica does not fulfill the recurrence relation...
    – Henrik Schumacher
    1 hour ago










  • @b.gatessucks Edited.
    – minhthien_2016
    1 hour ago












up vote
3
down vote

favorite









up vote
3
down vote

favorite











I am trying to solve the recurrence sequence. I tried



RSolve[a[n] == 1/2 (a[n - 1] + 9/a[n - 1]), a[1] == 1, a[n], 
 n] // Simplify


I got




a[n] -> 3 Coth[2^(-1 + n) ArcCoth[1/3]]




How to convert this solution to this form




(3*2^(2^(-1 + n)) + 3)/(2^(2^(-1 + n)) - 1)







output-formatting sequence







share|improve this question















I am trying to solve the recurrence sequence. I tried



RSolve[a[n] == 1/2 (a[n - 1] + 9/a[n - 1]), a[1] == 1, a[n], 
 n] // Simplify


I got




a[n] -> 3 Coth[2^(-1 + n) ArcCoth[1/3]]




How to convert this solution to this form




(3*2^(2^(-1 + n)) + 3)/(2^(2^(-1 + n)) - 1)







output-formatting sequence




output-formatting sequence






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited 1 hour ago

























asked 1 hour ago









minhthien_2016

447310




447310











  • This is not possible for these are entirely different number. One has $a(n) to 0$ for $n to infty$, but $(3 (49 + 49^n))/(-49 + 49^n) to 3$.
    – Henrik Schumacher
    1 hour ago











  • They are not equal at n=2 for instance.
    – b.gatessucks
    1 hour ago










  • I just realized that the solution $a(n)$ given by Mathematica does not fulfill the recurrence relation...
    – Henrik Schumacher
    1 hour ago










  • @b.gatessucks Edited.
    – minhthien_2016
    1 hour ago
















  • This is not possible for these are entirely different number. One has $a(n) to 0$ for $n to infty$, but $(3 (49 + 49^n))/(-49 + 49^n) to 3$.
    – Henrik Schumacher
    1 hour ago











  • They are not equal at n=2 for instance.
    – b.gatessucks
    1 hour ago










  • I just realized that the solution $a(n)$ given by Mathematica does not fulfill the recurrence relation...
    – Henrik Schumacher
    1 hour ago










  • @b.gatessucks Edited.
    – minhthien_2016
    1 hour ago















This is not possible for these are entirely different number. One has $a(n) to 0$ for $n to infty$, but $(3 (49 + 49^n))/(-49 + 49^n) to 3$.
– Henrik Schumacher
1 hour ago





This is not possible for these are entirely different number. One has $a(n) to 0$ for $n to infty$, but $(3 (49 + 49^n))/(-49 + 49^n) to 3$.
– Henrik Schumacher
1 hour ago













They are not equal at n=2 for instance.
– b.gatessucks
1 hour ago




They are not equal at n=2 for instance.
– b.gatessucks
1 hour ago












I just realized that the solution $a(n)$ given by Mathematica does not fulfill the recurrence relation...
– Henrik Schumacher
1 hour ago




I just realized that the solution $a(n)$ given by Mathematica does not fulfill the recurrence relation...
– Henrik Schumacher
1 hour ago












@b.gatessucks Edited.
– minhthien_2016
1 hour ago




@b.gatessucks Edited.
– minhthien_2016
1 hour ago










2 Answers
2






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up vote
1
down vote



accepted










A way:



sol = a[n] /. RSolve[a[n] == 1/2 (a[n - 1] + 9/a[n - 1]), a[1] == 1, a[n],n][[1, 1]] 
FullSimplify[sol // TrigToExp // ComplexExpand, Assumptions -> n > 1, n [Element] Integers] // Factor



$frac3 left(2^2^n-1+1right)2^2^n-1-1$







share|improve this answer



























    up vote
    1
    down vote













    The following verifies your result



    sol = RSolve[a[n] == 1/2 (a[n - 1] + 9/a[n - 1]), a[1] == 1, a[n],n][[1, 1]];
    Assuming[n > 1 && n ∈ Integers, (3*2^(2^(-1 + n)) + 3)/(2^(2^(-1 + n)) - 1) == a[n] /.sol // FullSimplify]


    Note that your second expression has the wrong limiting value for n=1. This may make it harder to find an automatic transformation.






    share|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








      up vote
      1
      down vote



      accepted










      A way:



      sol = a[n] /. RSolve[a[n] == 1/2 (a[n - 1] + 9/a[n - 1]), a[1] == 1, a[n],n][[1, 1]] 
      FullSimplify[sol // TrigToExp // ComplexExpand, Assumptions -> n > 1, n [Element] Integers] // Factor



      $frac3 left(2^2^n-1+1right)2^2^n-1-1$







      share|improve this answer
























        up vote
        1
        down vote



        accepted










        A way:



        sol = a[n] /. RSolve[a[n] == 1/2 (a[n - 1] + 9/a[n - 1]), a[1] == 1, a[n],n][[1, 1]] 
        FullSimplify[sol // TrigToExp // ComplexExpand, Assumptions -> n > 1, n [Element] Integers] // Factor



        $frac3 left(2^2^n-1+1right)2^2^n-1-1$







        share|improve this answer






















          up vote
          1
          down vote



          accepted







          up vote
          1
          down vote



          accepted






          A way:



          sol = a[n] /. RSolve[a[n] == 1/2 (a[n - 1] + 9/a[n - 1]), a[1] == 1, a[n],n][[1, 1]] 
          FullSimplify[sol // TrigToExp // ComplexExpand, Assumptions -> n > 1, n [Element] Integers] // Factor



          $frac3 left(2^2^n-1+1right)2^2^n-1-1$







          share|improve this answer












          A way:



          sol = a[n] /. RSolve[a[n] == 1/2 (a[n - 1] + 9/a[n - 1]), a[1] == 1, a[n],n][[1, 1]] 
          FullSimplify[sol // TrigToExp // ComplexExpand, Assumptions -> n > 1, n [Element] Integers] // Factor



          $frac3 left(2^2^n-1+1right)2^2^n-1-1$








          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered 32 mins ago









          Mariusz Iwaniuk

          6,26811027




          6,26811027




















              up vote
              1
              down vote













              The following verifies your result



              sol = RSolve[a[n] == 1/2 (a[n - 1] + 9/a[n - 1]), a[1] == 1, a[n],n][[1, 1]];
              Assuming[n > 1 && n ∈ Integers, (3*2^(2^(-1 + n)) + 3)/(2^(2^(-1 + n)) - 1) == a[n] /.sol // FullSimplify]


              Note that your second expression has the wrong limiting value for n=1. This may make it harder to find an automatic transformation.






              share|improve this answer
























                up vote
                1
                down vote













                The following verifies your result



                sol = RSolve[a[n] == 1/2 (a[n - 1] + 9/a[n - 1]), a[1] == 1, a[n],n][[1, 1]];
                Assuming[n > 1 && n ∈ Integers, (3*2^(2^(-1 + n)) + 3)/(2^(2^(-1 + n)) - 1) == a[n] /.sol // FullSimplify]


                Note that your second expression has the wrong limiting value for n=1. This may make it harder to find an automatic transformation.






                share|improve this answer






















                  up vote
                  1
                  down vote










                  up vote
                  1
                  down vote









                  The following verifies your result



                  sol = RSolve[a[n] == 1/2 (a[n - 1] + 9/a[n - 1]), a[1] == 1, a[n],n][[1, 1]];
                  Assuming[n > 1 && n ∈ Integers, (3*2^(2^(-1 + n)) + 3)/(2^(2^(-1 + n)) - 1) == a[n] /.sol // FullSimplify]


                  Note that your second expression has the wrong limiting value for n=1. This may make it harder to find an automatic transformation.






                  share|improve this answer












                  The following verifies your result



                  sol = RSolve[a[n] == 1/2 (a[n - 1] + 9/a[n - 1]), a[1] == 1, a[n],n][[1, 1]];
                  Assuming[n > 1 && n ∈ Integers, (3*2^(2^(-1 + n)) + 3)/(2^(2^(-1 + n)) - 1) == a[n] /.sol // FullSimplify]


                  Note that your second expression has the wrong limiting value for n=1. This may make it harder to find an automatic transformation.







                  share|improve this answer












                  share|improve this answer



                  share|improve this answer










                  answered 51 mins ago









                  mikado

                  6,0771829




                  6,0771829



























                       

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