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I wonder if the following result is already known and may be considered as interesting : let $\mathcal{C}$ be the real algebra of continuous functions $f:\left[0,1\right]\to\mathbf{R}$, $T:\mathcal{C}\to \mathcal{C}$ the map defined by $$ T\left(f\right)\left(x\right)=f\left(x\right)\left(2-\left(1+x^2\right)f\left(x\right)\right) $$ $\left(f_n\right)$ the sequence of $\mathcal{C}$ defined by $$ f_n= \begin{cases} T\left(f_{n-1}\right) & \text{if } n>0\\ 1/2 & \text{if } n=0 \end{cases} $$ and $\left(I_n\right)$ the real sequence defined by $I_n=\int_0^1 f_n\left(x\right) \, dx$. One has $$ \forall n\in\mathbf{N} \qquad 0 \leq \frac{\pi}{4} - I_n \leq 2^{-2^n} $$ and $$ \forall n\in\mathbf{N} \qquad I_n = \frac{1}{2} \sum_{k=0}^{2^n-1} \frac{2^k}{\left(2k+1\right)\binom{2k}{k}} $$

vdespax
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  • See the Arctangent part on the Wikipedia page for approximations of $\pi$. – Jakob Streipel Dec 20 '22 at 17:15
  • @prets: thank you. This recursive approach yields $S_{2^n-1}$, where $S_n$ is the partial sum of the expansion in the link. I wonder if this is remarkable/intersting or not. – vdespax Dec 20 '22 at 17:23
  • The sequence $f_n$ is an approximation from below of $\dfrac{1}{1 + x^2}$ which converges termwise to its power series around $x = 0$, so this is really the exact same thing as $\arctan(1) = \pi/4$ and $\dfrac{d}{d x} \arctan(x) = \dfrac{1}{1 + x^2}$. – Jakob Streipel Dec 20 '22 at 17:47
  • But here, the distance between the limit and $f_n$ is dominated by $2^{-2^n}$. Maybe I don't understand your remark. – vdespax Dec 20 '22 at 17:53
  • The constant term of $f_n$ is $1 - 2^{-2^n}$, which is where that bound comes from (the constant term of the power series of $\dfrac{1}{1 + x^2}$ around $x = 0$ is $1$, so the difference is the $2^{-2^n}$ you're observing). – Jakob Streipel Dec 20 '22 at 18:04
  • More generally, one can show : $$\forall n\in\mathbf{N},\forall x\in\left[0,1\right] \qquad f_n \left(x\right)=\frac{1}{2} \sum_{k=0}^{2^n-1} \left(\frac{1-x^2}{2}\right)^k$$ – vdespax Dec 20 '22 at 18:48
  • Take $x=1$ in the first formula given in this answer by Ron Gordon and the reference herein (reference found using the formula searching tool https://approach0.xyz/search/) – Jean Marie Dec 20 '22 at 19:00
  • @Jean Marie (Monier ? I'm french) : thank you for the reference. I am aware of these identities. What I would like to underline here is : 1. the speed of convergence and 2. the simple aspect of this recursion. As a non-specialist, I ask myself if this would have any interest in a computanional point of view. I am also aware that they are more efficient algorithms to approach $\pi$. But maybe this is nothing but a "gourmandise". – vdespax Dec 20 '22 at 19:59
  • Happy to meet fellow citizens ! (from which region ?) 2) I am not Jean-Marie Monnier, though I know him. My family name is Becker 3) Your question is interesting.
    • – Jean Marie Dec 20 '22 at 20:06
  • As a new user, I don't know how to chat with other users. – vdespax Dec 20 '22 at 20:28
  • If I may ask, where are you located ? I am in Pau. Cheers :-) – Claude Leibovici Dec 22 '22 at 07:53
  • Thank you for your answer. I live in the "Loire Valley" : Tours. – vdespax Dec 22 '22 at 07:59