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I played around with the equation

$$ f_{n}(x)=\frac{x}{\sqrt{1+nx^2}} $$

which is the $n$th composition of

$$ f(x)=\frac{x}{\sqrt{1+x^2}} $$

Surprising to me, it is a good approximation for $arctan(x)$ when plugging in $n=\frac{1}{2}$.

I tried comparing their Taylor-Series-Expansion but it didn't seem fruitful to me.

Is there maybe a connection to the derivative since

$$arctan'(x)=\frac{1}{1+x^2}$$

If you have any ideas why this is the case I would be glad to hear them.

Thanks in advance!

Blue
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Palehar
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    $\arctan$ is a function defined in all $\Bbb R$. Are you interested in an approximation for small values of $x$? For big values of $x$? I doubt there is a simpler function that approximate it well in the whole set of real numbers. – jjagmath Jul 21 '23 at 12:15
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    To match 3rd derivative near the origin, you need $n = 2/3\approx 0.67$. To match asymptotic value, you need $n = 4/\pi^2 \approx 0.41$. $n = 1/2$ is a compromise between the two. – eyeballfrog Jul 21 '23 at 12:26

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There is something interesting in the $\frac 1 2$.

Suppose that you want the best fit of the arctangent, say between $0$ and $t$ using such a model. To find the best $a$, consider the infinite norm $$\Phi(a,t)=\int_0^t \Big( \tan ^{-1}(x)-\frac{x}{\sqrt{1+a x^2}} \Big)^2\,dx$$ which is fully explicit (even if not very nice). This is equivalent to a non linear regression based on an infinite number of data points.

Now, for a given value of $t$, solve for $a$ the equation $$\frac {\partial \Phi(a,t)}{\partial a}=0$$

A few numbers $$\left( \begin{array}{cc} t & a_{\text{opt}} \\ 5 & 0.520880 \\ 6 & 0.509581 \\ 7 & 0.500667 \\ 8 & 0.493425 \\ 9 & 0.487405 \\ 10 & 0.482305 \\ \end{array} \right)$$

Claude Leibovici
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