TITLE: Asymptotic expansion of $u_{n + 1} = \frac12 \arctan(u_n)$
QUESTION [7 upvotes]: (I'm aware of Asymptotic expansion of $v_n = 2^nu_n$ where $u_{n+1} = \dfrac{1}{2}\arctan(u_n)$ but it has no answers…)
Let be $u_0 \in \mathbb{R}$ and the sequence $(u_n)_n$ defined by: $u_{n + 1} = \frac12 \arctan(u_n)$.
I define also: $v_n = 2^n u_n$, so I can show that: $\lim (u_n)_n = 0$ (by studying $x \mapsto \frac12 \arctan(x)$), thus, I can show that $(v_n)_n$ is monotone and converges because it is bound.
Now, I conclude: $u_n \sim \dfrac{l}{2^n}$, I'd like to determine $l$ more precisely.
Here is what I tried, I suspect $l$ to be something like $f(\pi)$ for some $f$ :

push the asymptotic expansion of $\arctan$ to the 2nd order and reinject it ;
use $\arctan(u_n) + \arctan(1/u_n) = \dfrac{\pi}{2}$ ;
use series techniques to look for $\sum v_{n + 1} - v_n$, maybe conclude using Cesaro summation

REPLY [2 votes]: Complement to the answer of @Szeto
In many cases when you start tinkering with the Faà di Bruno formula, you will be better served computing with truncated Taylor series.
So we want to solve
$$
Ψ(x)=2Ψ(\tfrac12\arctan(x)),
~~\text{ so that }~~ x_n=Ψ^{-1}(2^{-n}Ψ(x_0)),$$
where $Ψ(x)\sim x$ for $x\approx 0$ by the scaling normalization. As $\arctan(x)\sim x$ for $x\approx 0$, the coefficient determination for $Ψ(x)=x+c_2x^2+c_3x^3+...$ is a finite problem for each coefficient, it is only influenced by lower degree coefficients. Thus assuming that the coefficients $c_0=0,c_1=1,c_2,..c_{k-1}$ are already determined, one gets the next coefficient from
$$
(1-2^{1-k})c_kx^k=A_k(x)-x+c_2(2^{-1}A_{k-1}(x)^2-x^2)+c_3(2^{-2}A_{k-2}(x)^3-x^3)+...+c_{k-1}(2^{2-k}A_{2}(x)^{k-1}-x^{k-1})
$$
by comparing the coefficients of $x^k$ on both sides. The $A_k(x)$ are the $k$-th partial sums of the arcus tangent series, $\arctan(x)=A_k(x)+O(x^{k+1})$. This can be simplified, it is not necessary to subtract the lower powers, one can also take the odd nature of the series into account.
Using a CAS like Magma (online calculator one can extract the equation for the next coefficient directly from the unmodified equation with the following script:
A<a>:=FunctionField(Rationals());
PS<x>:=PowerSeriesRing(A);
Pol<z>:=PolynomialRing(Rationals());

Psi := x;
for k in [2..20] do
    Psia := Psi+(a+O(x))*x^k;
    eqn := Coefficient( Psia-2*Evaluate(Psia, 1/2*Arctan(x+O(x^(k+1))) ), k );
    c := Roots(Pol!eqn)[1,1]; k,c;
    Psi +:= c*x^k;
end for;

which when executed gives in the end for $\Psi(x)+O(x^{21})$
x - 4/9*x^3 + 224/675*x^5 - 51008/178605*x^7 + 25619968/97594875*x^9 
  - 91726170112/366078376125*x^11 + 45580629370863616/186023558824228125*x^13 
  - 171377650156414910464/703297837896306778125*x^15 
  + 56540215172481124229054464/230453119032672323522109375*x^17 
  - 353563937806248194123298285027328/1417897477708832149477498284609375*x^19

The inverse function $\Psi^{-1}$ is then similarly obtained as
x + 4/9*x^3+176/675*x^5 + 142144/893025*x^7 + 67031296/683164125*x^9
    + 777200229376/12812743164375*x^11 
    + 76806067707244544/2046259147066509375*x^13 
    + 7434789485314586820608/320000516242819584046875*x^15 
    + 3317928226689969972317978624/230683572151704995845631484375*x^17
    + 30692357195871908183846360294096896/3446908768310170955379798329885390625*x^19

which is the series in the comment to the question by @Winther