1. ## A curious function.

Marie and Fred are arguing again, this time about the function:

$g \left( x \right) =2\,\arctan \left( {\frac {1+x}{1-x}} \right) +
\arcsin \left( {\frac {1-{x}^{2}}{1+{x}^{2}}} \right)$
.

Fred says that if x > 0 then g(x) is a non-constant function whose derivative is zero. Marie says that's nonsense, and even if it weren't, for x < 0, g(x) is a smooth function whose derivative is never zero.

2. Originally Posted by jbuddenh
Marie and Fred are arguing again, this time about the function:

$g \left( x \right) =2\,\arctan \left( {\frac {1+x}{1-x}} \right) +
\arcsin \left( {\frac {1-{x}^{2}}{1+{x}^{2}}} \right)$
.

Fred says that if x > 0 then g(x) is a non-constant function whose derivative is zero. Marie says that's nonsense, and even if it weren't, for x < 0, g(x) is a smooth function whose derivative is never zero.

For $x>0$ we have $g'(x) = 0$ but $x\not =1$ since it is not defined for $x\not = 0$. Therefore, $0 < x < 1$ we have a zero derivative and for $x>1$ we have a zero derivative. But those two vertical lines are not the same. Now for $x<0$ we have a non-zero derivative.

3. Thanks to ThePerfectHacker for the repy. Yes you are basically right. I'm not sure what you mean by "those two vertical lines" though. The graph of y=g(x) looks like this:

What seems strange to me is that a function defined by a nice single line analytic formula, namely

Should be a step function, i.e. constant but for a jump discontinuity for x>0 but a smooth monotonically increasing function for x<0. The explanation comes from the fact that if you let

$\alpha=2\,\arctan \left( {\frac {1+x}{1-x}} \right)$ and $\beta=\arcsin \left( {\frac {1-{x}^{2}}{1+{x}^{2}}} \right)$

then using, for example the double angle formula for the tangent function, it is not hard to show that $\sin \left( \alpha \right) =\sin \left( \beta \right)$. This, however, can happen in different ways, and for x<0 it happens via, $\alpha=\beta$, whereas for 0<x<1 it happens as $\alpha+\beta=\pi$ and after the discontinuity at x=1 it happens as $\alpha+\beta=-\pi$.

4. Originally Posted by jbuddenh
Thanks to ThePerfectHacker for the repy. Yes you are basically right. I'm not sure what you mean by "those two vertical lines" though.
If $0 then $g'(x) = 0$ therefore the graph must be a vertical line.
If $x>1$ then $g'(x) = 0$ must be a vertical line.
Those are "those two vertical lines".

5. Originally Posted by ThePerfectHacker
If $0 then $g'(x) = 0$ therefore the graph must be a vertical line.
If $x>1$ then $g'(x) = 0$ must be a vertical line.
Those are "those two vertical lines".
One of us has vertical mixed up with horizontal. No portion of the graph of g nor of g' is vertical.

6. Originally Posted by jbuddenh
One of us has vertical mixed up with horizontal. No portion of the graph of g nor of g' is vertical.
Sorry. I meant to say "horizontal" thank you for the correction.