0
$\begingroup$

I am seeking a rational mediant for arcus tangent on $[0,\infty)$. The idea is to find two functions $f\colon \mathbb{R}^2 \to \mathbb{R}$ and $g\colon \mathbb{R}^2 \to \mathbb{R}$, such that:

$$\tan^{-1}(f(a,b)) = g(\tan^{-1}(a),\tan^{-1}(b))$$

Where both $f$ and $g$ are mediants, i.e. $a < f(a,b) < b$ for $a < b$ and $x < g(x,y) < y$ for $x < y$. And a further requirement is that $f$ and $g$ are algebraic.

  • 0
    What is the motivation? By algebraic you mean rational? – 2017-02-10
  • 0
    I am not sure what is possible, would be happy with either. – 2017-02-10
  • 0
    Can you give us examples of "mediants" you have been looking at ? – 2017-02-10

1 Answers 1

0

An algebraic solution can be derived from the following identity:

$$\tan^{-1}(a) + \tan^{-1}(b) = \tan^{-1}(\frac{a+b}{1-a b})$$

$$= \tan^{-1}(H(a,b))$$

Solving $H(a,a)=b$ for $a$ we get the following further identity:

$$\tan^{-1}(a)/2 = \tan^{-1}(\frac{\sqrt{1+a^2}-1}{a})$$

$$= \tan^{-1}(J(a))$$

We can now use $f(a,b) = J(H(a,b))$ and $g(x,y)=(x+y)/2$.