Deduce the formula of a sequence which is defined recursively Good evening to everyone. I am trying to solve the following problem. Let $ (a_n) $  be a sequence defined recursively as:
$\displaystyle a_{n+1}=\frac{2a_n^2+1}{2a_n}$ and $a_1=11$ . I know there is a theorem in order to find a closed formula if the number 2 on the numerator did not exist, but here it seems complicated. Actually this was an exercise on a problem sheet that gave to me one of my students. The ''highlight'' theorem in his course was the following:  If $\displaystyle a_{n+1}=\frac{a_n^2 + k^2}{2a_n}$ , then $\displaystyle \frac{a_n - k}{a_n +k} = \left({\frac{a_1-k}{a_1+k}}\right)^{2^{n-1}}$ . On the problem sheet there were many exercises that were exactly applications of the above theorem, nothing really difficult. But the LAST exercise was the one I mentioned in my original post,which was given,I suppose, either to show that it is in general hard to solve general recursions or there is some trick here I cannot think by myself . Does anybody know how to solve this? I am asking because it is possible there is a typo here, so I would like to know if indeed this exercise cannot be solved. Thanks a lot!
 A: Based on previous work in a related MSE question 3489102 I found a similar
approach that works in this case. There is no closed form solution
for these that I know of.
Given the function $\,f(x) := x + \frac1{2x}\,$ define the sequence
$\,a_{n+1} := f(a_n).\,$ Numerical experiments suggest to define
the sequence $\,b_n := a_n^2.\,$ Now define
$\,g(x) := x + 1 + \frac1{4x}\,$ and
$\,b_{n+1} := g(b_n) .\,$ The asymptotic expansion of $\,b_n\,$
depends on the solution to the functional equation $\,B(x+1) = g(B(x))\,$
and is given by
$$ B(x) := x + \frac{2+y}{4} + \frac{1+y}{16x} +
\frac{1-3y^2}{384x^2} + \frac{7-6y-9y^2+6y^3}{4608x^3} + \cdots $$
where $\,y:=\log(x).\,$ Then $\,b_n \approx B(n+c_0)\,$ where
$\,c_0\,$ depends only on $\,b_0.\,$
For example, if $\,b_0=2, a_0=\sqrt{2}\,$ then $\,c_0\approx 1.36157\,$ and
$\,a_{100} \approx 10.14986 \approx \sqrt{B(100+c_0)}.\,$
By the way, the "highlight" theorem you refer to is the Babylonian or Newton's method for computing square roots.
That is, define the iteration
$\,a_{n+1}=\frac{a_n^2 + N}{2a_n}.\,$ Then the sequence converges to
$\,k=\sqrt{N}\,$ and the formula with the $\,2^{n-1}\,$ exponent gives
the precise convergence of the iteration sequence.
