About a function approximating the $\arctan(x)$

Question

I found in a paper this function: $$f(x)=\frac{8x}{3+\sqrt{25+\left(\frac{16x}{\pi}\right)^2}}$$ is a good approximation of the $\arctan(x)$. If we consider the difference function: $$d(x)=|\arctan(x)-f(x)|$$ its maximum is about $0.00907$ for $|x|=3.135917295$. Is there a proof about the reason why the $f(x)$ approximates so well the $\arctan(x)$? Is it possible to find function similar to f(x) which are good approximations of other trigonometric functions like $sin(x),cos(x)$, ecc.? Thanks.

Answer 1

If you are interested in the best approximation of $\arctan(x)$ by the function of the form $$\frac {ax} {b+\sqrt{c+dx^2}},$$ then the Maple global optimizer DirectSearch does the job in such way: the code $$A := proc (alpha, beta, delta, eps) DirectSearch:-GlobalSearch(abs(alpha*x/(beta+sqrt(delta+eps*x^2))-arctan(x)), {x = -infinity .. infinity}, solutions = 1) end proc: $$ $$ DirectSearch:-Search(proc (a, b, c, d) -> A(a, b, c, d)[1, 1] , {abs(a) <= 20, abs(b) <= 20, abs(c) <= 20, abs(d) <= 20});$$ produces $$ [ 0.0, \left[ \begin {array}{c} 1.76051623816363034 \\ - 0.155583793604506360\\ 1.72404758813357750\\ 1.36875140919455384 \end {array} \right] ,87] $$

Answer 2

A much better approximation than the one in the question can be obtained by defining $$f(x)=\frac{ax}{b+\sqrt{c+x^2}}$$ and simply matching the values $f'(0)=1$, $f(1)=\pi/4$, and $f(\infty)=\pi/2$. This gives the solution $$a=\frac\pi2, \qquad b=\frac{12-\pi^2}{4(4-\pi)}\approx 0.620, \qquad c=\frac{(6-\pi)^2(2-\pi)^2}{16(4-\pi)^2}\approx 0.903,$$ for which $\max|f(x)-\arctan x|$ is less than $0.00209$.

P.S. I don't understand why $b$ is negative in @user64494's answer.