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)

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  • $\begingroup$ This is not possible for these are entirely different number. One has $a(n) \to 0$ for $n \to \infty$, but $(3 (49 + 49^n))/(-49 + 49^n) \to 3$. $\endgroup$ – Henrik Schumacher Sep 30 '18 at 10:33
  • $\begingroup$ They are not equal at n=2 for instance. $\endgroup$ – b.gates.you.know.what Sep 30 '18 at 10:35
  • $\begingroup$ I just realized that the solution $a(n)$ given by Mathematica does not fulfill the recurrence relation... $\endgroup$ – Henrik Schumacher Sep 30 '18 at 10:36
  • $\begingroup$ @b.gatessucks Edited. $\endgroup$ – minhthien_2016 Sep 30 '18 at 10:42
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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

$\frac{3 \left(2^{2^{n-1}}+1\right)}{2^{2^{n-1}}-1}$

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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.

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