Asymptotic expansion of $u_{n + 1} = \frac12 \arctan(u_n)$

(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

Solution 1:

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 $$ u_n=\frac{2^{-n}u_0}{\sqrt{1+\frac89u_0^2(1-4^{-n})-\frac{4}{25}\frac{u_0^4}{9+8u_0^2}+O(u_0^6)}} \tag8 $$

Solution 2:

(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.