6
$\begingroup$

(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
$\endgroup$
  • $\begingroup$ Have you tried to use a computer to obtain an approximate value of $\ell$ ? $\endgroup$ – Jean Marie May 31 at 19:13
  • $\begingroup$ @Somos Not really for this problem. Note that the answer will depend on $u_0$. $\endgroup$ – Robert Israel May 31 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 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 at 3:21
  • $\begingroup$ @Winther You should transform your comment into an answer. $\endgroup$ – Jean Marie Jun 2 at 7:36
4
$\begingroup$

(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.

$\endgroup$
  • $\begingroup$ I have put a Magma script to solve the functional equation directly in a new answer. $\endgroup$ – LutzL Jul 12 at 15:57
4
$\begingroup$

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

$\endgroup$
  • $\begingroup$ Is the result exact? $\endgroup$ – Szeto Jul 12 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. $\endgroup$ – LutzL Jul 12 at 10:52
  • 1
    $\begingroup$ 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 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 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 $O(4^n)$. $\endgroup$ – LutzL Jul 12 at 15:10
3
$\begingroup$

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 $\, u_n \approx y^n(x - (1-y^{2n})x^3/(3(1-y^2))).\,$

$\endgroup$
2
$\begingroup$

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
u.append(it)

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

$\endgroup$
1
$\begingroup$

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 $Ψ$ 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. This can be simplified, it is not necessary to subtract the lower powers, one can 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
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.