(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
  • $\begingroup$ Have you tried to use a computer to obtain an approximate value of $\ell$ ? $\endgroup$ – Jean Marie May 31 '19 at 19:13
  • $\begingroup$ @Somos Not really for this problem. Note that the answer will depend on $u_0$. $\endgroup$ – Robert Israel May 31 '19 at 19:32
  • 1
    $\begingroup$ I can't find a closed form, but can find the first terms in a power-series $$f(\ell) = \ell +\frac{4 \ell^3}{9}+ \frac{176 \ell^5}{675}+ \frac{142144 \ell^7}{893025} + \frac{67031296\ell^9}{683164125} + \frac{777200229376 \ell^{11}}{12812743164375} + \ldots$$ which is such that $u_n = f(\ell/2^n)$ and $\ell$ is determined via $f(\ell) = u_0$. For small values of $u_0 \ll 1$ we have $\ell \approx u_0$. The general term in this power-series is $\sim (0.78\ell)^{2k+1}$. $\endgroup$ – Winther May 31 '19 at 22:04
  • $\begingroup$ @JeanMarie Yes but it looks like a bit unstable, I tried to look at $\log (u_n)$ but was not able to get any fixed value, maybe I should use convergence acceleration techniques $\endgroup$ – Raito Jun 2 '19 at 3:21
  • $\begingroup$ @Winther You should transform your comment into an answer. $\endgroup$ – Jean Marie Jun 2 '19 at 7:36

The iteration has the form $$u_{n+1}=a_1u_n+a_3u_n^3+...$$ As usual in such situations (See the answer in Convergence of $\sqrt{n}x_{n}$ where $x_{n+1} = \sin(x_{n})$ with citation of de Bruijn: "Asymptotic Methods ..."), one can try a Bernoulli-like approach and examine the dynamics of $u_n^{-2}$. There one finds $$ \frac1{u_{n+1}^2}=\frac4{u_n^2(1-\frac13u_n^2+\frac15u_n^4\mp...)^2} =\frac4{u_n^2}+\frac83-\frac4{15}u_n^2+O(u_n^4)\tag1 $$ Thus for a first approximation use $$x_{n+1}=4x_n+\frac83\iff x_{n+1}+\frac89=4(x_n+\frac89)$$ so that $$u_n^{-2}\sim x_n=4^n(x_0+\frac89)-\frac89.\tag2$$

This gives as first approximation $$ u_n\sim \frac{2^{-n}u_0}{\sqrt{1+\frac89u_0^2(1-4^{-n})}}.\tag3 $$

For the next term use $v_n=(u_n^{-2}+\frac89)^{-1}$ and express (1) in terms of $v_n$ $$ \frac1{v_{n+1}}=\frac4{v_n}-\frac4{15}\frac{v_n}{1-\frac89v_n}+O(v_n^2) \tag4 $$ so that $$ \frac1{v_{n+1}}-\frac{4}{15^2}v_{n+1}=\frac4{v_n}-\frac1{15} v_n - \frac{1}{15^2}v_n+O(v_n^2)=4\left(\frac1{v_{n}}-\frac{4}{15^2}v_{n}\right)+O(v_n^2) \tag5 $$ and consequently $$ \frac1{v_{n}}-\frac{4}{15^2}v_{n}=4^n\left(\frac1{v_{0}}-\frac{4}{15^2}v_{0}+O(v_0^2)\right) \tag6 $$ As $\frac1v-\frac{4}{15^2}v=\frac1v(1-\frac4{15^2}v^2)$, leaving out the second term adds an error $O(v_n^2)$ which is a small fraction of $O(v_0^2)$. Thus $$ \frac1{u_n^2}+\frac89=\frac1{v_n}=4^n\left(\frac1{v_{0}}-\frac{4}{15^2}v_{0}+O(v_0^2)\right)=4^n\left(\frac1{u_0^2}+\frac89-\frac{4}{15^2}\frac{u_0^2}{1+\frac89u_0^2}+O(u_0^4)\right)\tag7 $$ so that the improved approximation is $$ u_n=\frac{2^{-n}u_0}{\sqrt{1+\frac89u_0^2(1-4^{-n})-\frac{4}{25}\frac{u_0^4}{9+8u_0^2}+O(u_0^6)}} \tag8 $$

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  • $\begingroup$ Is the result exact? $\endgroup$ – Szeto Jul 12 '19 at 10:43
  • $\begingroup$ No, in an exact formula there will be additional terms $O(8^{-n})$ and smaller, and possibly additional constants $O(u_0^2)$ under the square root. Note that the $x_n$ iteration only uses the first two terms of the previous formula. The summation of $4^ku_{n-k}^2$ is not arbitrarily small. I'm working on adding the next step. $\endgroup$ – Lutz Lehmann Jul 12 '19 at 10:52
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    $\begingroup$ I have a feel that, as you work on, you will finally obtain an answer equivalent to mine. Looking forward to seeing a beautiful result, as mine is extremely ugly :( $\endgroup$ – Szeto Jul 12 '19 at 10:57
  • $\begingroup$ The equation right after the word ‘consequently’ should have an error term of $O(4^n v_0^2)$. $\endgroup$ – Szeto Jul 12 '19 at 14:40
  • $\begingroup$ Yes, obviously. The coefficients form the recursion are $1,4,...,4^{n-1}$ while the terms themselves go like $16^{1-n},16^{2-n},...,1$ which makes a sum $O(4^n)$. $\endgroup$ – Lutz Lehmann Jul 12 '19 at 15:10

(For easier discussion, I suggest you to read the introduction of Schroder's equation and the section on 'Conjugacy' of iterated function, in case you are not familiar with these topics.)

Let $f(x)=\frac12\arctan x$, and $f_n(x)$ be the $n$th iteration of $f$.

Let us reduce functional iteration to multiplication: if we can solve the corresponding Schroder's equation $$\Psi(f(x))=s\Psi(x)$$

then it is well known (and also straightforward) that $$f_n(x)=\Psi^{-1}(f'(a)^n\cdot\Psi(x))$$ where $a$ is a fixed point of $f$.

For the moment, let us focus on $\Psi(f(x))=s\Psi(x)$.

Clearly, in our case, $a=0$, and $s=f'(a)=\frac12$.

For $a = 0$, if $h$ is analytic on the unit disk, fixes $0$, and $0 < |h′(0)| < 1$, then Gabriel Koenigs showed in 1884 that there is an analytic (non-trivial) $\Psi$ satisfying Schröder's equation $\Psi(h(x))=s\Psi(x)$.

Thus, $\Psi$ is analytic.

A few more observations:

  1. $\Psi(0)=0$.
  2. $\Psi'(0)$ is up to our choice, since if a function $\psi$ is a solution to the Schroder's equation, then so is $k\cdot \psi$ for any constant $k$. For convenience, set $\Psi'(0)=1$.
  3. All other Taylor series coefficients of $\Psi$ are then uniquely determined, and can be found recursively. (The method will be illustrated below.)
  4. By Lagrange inversion theorem, $\Psi$ is invertible in a neighbourhood of $0$, and $\Psi^{-1}(z)=0+\frac1{\Psi'(0)}z+o(z)\implies \Psi^{-1}(z)\sim z\quad(z\to 0)$.
  5. Therefore, $f_n(x)=\Psi^{-1}(f'(a)^n\cdot\Psi(x))=\Psi^{-1}(2^{-n}\Psi(x))\sim 2^{-n}\Psi(x)$ as $n\to\infty$.

Hence, for the limit the OP wanted to evaluate, $$\ell:=\lim_{n\to\infty}2^nf_n(x_0)=\Psi(x_0)$$

We shall now determine all the Taylor series coefficients of $\Psi(x)$ (valid only for $|x|<1$), since it can be assumed $0\le x_0<1$.

Obviously, $\Psi$ is an odd function. Let $$\Psi(x)=x+\sum^\infty_{k=2}\phi_{2k-1} x^{2k-1}$$

The basic idea is to repeatedly differentiate both sides of $\Psi(f(x))=s\Psi(x)$ and substitute in $x=0$, then recursively solve for the coefficients.

For example, differentiating both sides three times and substitute in $x=0$, we obtain $$-\Psi'(0)+\frac18\Psi'''(0)=\frac12\Psi'''(0)\implies\phi_3=-\frac49$$

Slightly modifying the notations of our respectable MSE user @Sangchul Lee, for $\lambda=(\lambda_1,\lambda_2,\cdots,\lambda_n)$ a $n$-tuple of non-negative integers:

  • write $\lambda \vdash n$ if $\sum^n_{i=1}(2i-1)\lambda_i=2n-1$.
  • write $|\lambda| = \sum_{i=1}^{n} \lambda_i$.
  • define the tuple factorial as $\lambda !=\frac{|\lambda|!}{\lambda_1!\cdot\lambda_2!\cdots\lambda_n !}$.

I will state, without proof, Faà di Bruno's formula for odd inner function:

$$(\Psi\circ f)^{(2n-1)}=(2n-1)!\sum_{\lambda \vdash n}\lambda!\cdot\phi_{|\lambda|}\prod^n_{i=1}\left(\frac{f^{(2i-1)}(0)}{(2i-1)!}\right)^{\lambda_i}$$

$$\implies \frac12\phi_{2n-1}=\sum_{\lambda \vdash n}\lambda!\cdot\phi_{|\lambda|}\prod^n_{i=1}\left(\frac{(-1)^{i+1}}{2(2i-1)}\right)^{\lambda_i}$$

Further simplifications lead to the final result:

$$\ell=\Psi(x_0)=\sum^\infty_{k=1}\phi_{2k-1} x_0^{2k-1} \qquad{\text{where}}\qquad \phi_1=1$$

$$\phi_{2n-1}=\frac{(-1)^{n}}{2^{-1}-2^{1-2n}}\sum_{\substack{\lambda \vdash n \\ \lambda_1\ne 2n-1}}\phi_{|\lambda|}\frac{\lambda! (-1)^{(|\lambda|+1)/2}}{2^{|\lambda|}}\prod^n_{i=1}\frac1{(2i-1)^{\lambda_i}}$$

Yeah, I know it’s ugly. But that’s the best we can obtain.

If anyone have a nice math software, please help me calculate the first few Taylor coefficients.

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  • $\begingroup$ I have put a Magma script to solve the functional equation directly in a new answer. $\endgroup$ – Lutz Lehmann Jul 12 '19 at 15:57

For convenience we make a slight generalization of the problem. Let $\,u_0\,$ and $\,y\,$ be given numbers and suppose $\,u_{n+1} = y \arctan(u_n)\,$ for $\,n\ge 0\,$ where $\ y=1/2\ $ in your original recursion. Define with power series the function $$ F(x,y,z) := z\left(x + \frac{-1+z^2}{1-y^2}\frac{x^3}3 +\frac{(1-z^2)((3-2z^2)+y^2(2-3z^2)}{(1-y^2)(1-y^4)}\frac{x^5}{15} + O(x^7) \right) $$ which satisfies the equation $\,F(x,y,yz) = \arctan(F(x,y,z))y.\,$ Then we get the equation $\, u_n = F(x,y,y^n)\,$ where $\, x = \lim_{n\to\infty} u_n/y^n.\,$ I know some more terms in the power series expansion if you are interested. Thus we get the result $\, u_n \approx y^n(x - (1-y^{2n})x^3/(3(1-y^2))).\,$

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Partial answer for $u_0>0$, then $$u_{n+1}-\frac{u_n}{2}=\frac{1}{2}(\arctan{u_n}-u_n)<0$$ because $f(x)=\arctan{x}-x<0$ for positive $x$, thus $$0<u_{n+1}<\frac{u_n}{2}<u_n \tag{1}$$ Using MVT, $\exists z\in(u_{n+1},u_n)$ s.t. $$u_{n+1}-u_n=\frac{1}{2}\left(\arctan{u_n}-\arctan{u_{n-1}}\right)= \frac{1}{2}\frac{u_{n}-u_{n-1}}{z^2+1}$$ or (because $\color{red}{u_n-u_{n-1}<0}$) $$\frac{1}{2}\cdot \frac{u_{n}-u_{n-1}}{u_{n+1}^2+1}< u_{n+1}-u_n< \frac{1}{2}\cdot \frac{u_{n}-u_{n-1}}{u_{n}^2+1}$$ or $$\frac{u_{0}-u_{1}}{2^n}\prod\limits_{k=1}^n\frac{1}{u_{k+1}^2+1}< u_{n+1}-u_n< \frac{u_{0}-u_{1}}{2^n}\prod\limits_{k=1}^n\frac{1}{u_{k}^2+1}$$ Considering $u_{n+1}-u_n \sim -\frac{l}{2^{n+1}}$ then $$\frac{u_{0}-u_{1}}{2^n}\prod\limits_{k=1}^n\frac{1}{u_{k+1}^2+1}> \frac{l}{2^{n+1}}> \frac{u_{0}-u_{1}}{2^n}\prod\limits_{k=1}^n\frac{1}{u_{k}^2+1}$$ or $$\frac{2(u_{0}-u_{1})}{\prod\limits_{k=1}^n\left(u_{k+1}^2+1\right)}= \frac{2(u_{0}-u_{1})\left(u_{1}^2+1\right)}{\prod\limits_{k=1}^{n+1}\left(u_{k}^2+1\right)}> l> \frac{2(u_{0}-u_{1})}{\prod\limits_{k=1}^n\left(u_{k}^2+1\right)}$$ or $$L_1>l>L_2$$ where $$L_2=\frac{2(u_{0}-u_{1})}{\lim\limits_{n\to\infty}\prod\limits_{k=1}^n(u_{k+1}^2+1)} \text{ and } L_1=L_2\left(u_{1}^2+1\right) \tag{2}$$

So, it looks like Robert (see the comments) was right, it depends on $u_0$.

Note: the following limit exists $$\lim\limits_{n\to\infty}\prod\limits_{k=1}^n(u_{k+1}^2+1)$$ because $$0<\sum\limits_{k=1}\ln(u_{k+1}^2+1)<\sum\limits_{k=1}u_{k+1}^2<\infty$$ by ratio test from $(1)$.

The following code is computing $(2)$ but with a $\frac{1}{u_0}$ factor. You will notice a certain stability for $\frac{L_1}{2^n u_0 \cdot u_n}$ and $\frac{L_2}{2^n u_0 \cdot u_n}$ for various $u_0$

from math import atan
from math import pow

N = 300
U_0 = 190.0

u = []

it = U_0

for i in range(1, N):
    it = 0.5 * atan(it)

val = 1.0
for i in range(1, N):
    val *= (u[i] * u[i] + 1.0)

L2 = (2.0 * (u[0] - u[1]) / val) / u[0]
L1 = L2 * (u[1] * u[1] + 1.0)
MID = (L1 + L2) / 2.0

print "limit L1 =",L1
print "limit L2 =",L2
print "limit MID =",MID

for i in range(N-100, N):
    Lp1 = L1 / pow(2, i)
    Lp2 = L2 / pow(2, i)
    MIDp = MID / pow(2, i)

    r1 = Lp1 / u[i]
    r2 = Lp2 / u[i]
    rMID = MIDp / u[i]

    print Lp2," vs ",u[i]," vs ",Lp1," --- ",MIDp
    print r2," vs ",r1," --- ",rMID

Try it here.

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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 $$Ψ(x)=2Ψ(\tfrac12\arctan(x))$$ 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:


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 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

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