# On the common zeros of $1-x\tan{x}$ and $1-\frac1x \arctan{\frac1x}$

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:

1:

• 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 $x\tan(x) = 1$. Commented Aug 25, 2015 at 9:27
• @GeoffRobinson Thanks, will check this. Modulo my errors this contradicts the plot, do you have other zero, taking any branch?
– joro
Commented Aug 25, 2015 at 9:37
• Well, I was thinking of $\arctan$ as a single-valued function above.. Commented Aug 25, 2015 at 9:39
• @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?
– joro
Commented Aug 25, 2015 at 9:40
• it vanishes for some other branch of arctan. Commented Aug 25, 2015 at 9:52