A good approximation for $\arctan(u \tan (v x))$

Question

I am trying to find an approximate solution for the pole of a transfer function that I need to analyze in electronics. I could simplify greatly my problem untill the point where I have to solve a transcendental equation of the form $$ x - a \arctan(u \tan (v x)) = 0 $$ Where $a$, $u$ and $v$ are complicated expressions. I want to find a good approximation for the first two solutions of this equation (besides $x=0$) as a function of the parameters $a$, $u$ and $v$. So far I only could get a mediocre one by using the Pade approximation technique. Do you think about a better strategy?

Answer 1

Using $\tan y\approx y+\frac{1}{3}y^3$ and $\arctan y\approx y-\frac{1}{3}y^3$, $u\tan vx\approx uvx+\frac{1}{3}uv^3x^3$ and $$\arctan(u\tan vx)\approx uvx+\frac{1}{3}uv^3x^3-\frac{1}{3}(uvx+\frac{1}{3}uv^3x^3)^3\approx uvx+\frac{1}{3}u(1-u^2)v^3x^3.$$The above calculations have all worked to order $x^3$, which is fine for small $x$. A small $x$ solves your equation provided it approximates a non-zero root of the cubic$$0=x-a(uvx+\frac{1}{3}u(1-u^2)v^3x^3),$$or equivalently of the quadratic$$0=1-auv-\frac{1}{3}au(1-u^2)v^3x^2.$$This gives$$x\approx\pm\sqrt{\frac{3(1-auv)}{au(1-u^2)v^3}}.$$I advise you to double-check my calculations.