This is ON HOLD on MO.

Let $f(x)=1-x\tan{x}$.

Let $g(x)=1-\frac1x \arctan{\frac1x}$.

Let $r=0.8603335890193797624838\ldots$ be a real root of $f(x)=0$.

High precision numerical computations suggest $f(\pm r)=g(\pm r)=0$.

$g(x)$ doesn't appear to have other complex roots (might be wrong about this).

Q1 Is it true $f(\pm r)=g(\pm r)=0$?

Q2 If this is true, is there closed form for $r$ in terms of already named functions and constants?

On MO commenter wrote "This is just arctan(tan(x))=x" but I don't understand this.

Xray (vanishing of the real and imaginary parts of f(x) and g(x) in the complex plane), common zero is where all four curves intersect:


enter image description here

  • 3
    $\begingroup$ Aren't they the same equation? For $\frac{1}{x} \arctan(\frac{1}{x}) = 1$ if and only if $x = \arctan(\frac{1}{x})$ if and only if $\tan(x) = \frac{1}{x}$ if and only if $x\tan(x) = 1$. $\endgroup$ Commented Aug 25, 2015 at 9:27
  • $\begingroup$ @GeoffRobinson Thanks, will check this. Modulo my errors this contradicts the plot, do you have other zero, taking any branch? $\endgroup$
    – joro
    Commented Aug 25, 2015 at 9:37
  • $\begingroup$ Well, I was thinking of $\arctan$ as a single-valued function above.. $\endgroup$ Commented Aug 25, 2015 at 9:39
  • $\begingroup$ @GeoffRobinson The plot takes the principal branch. For the root of f(x)=0: r=6.43729817917195..., g(r) doesn't vanish for me or am I missing something? $\endgroup$
    – joro
    Commented Aug 25, 2015 at 9:40
  • 1
    $\begingroup$ it vanishes for some other branch of arctan. $\endgroup$
    – mercio
    Commented Aug 25, 2015 at 9:52


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