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Math Help - A curious function.

  1. #1
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    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) +<br />
\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.

    Please help them resolve this dispute.
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    Quote Originally Posted by jbuddenh View Post
    Marie and Fred are arguing again, this time about the function:

    g \left( x \right) =2\,\arctan \left( {\frac {1+x}{1-x}} \right) +<br />
\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.

    Please help them resolve this dispute.
    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.
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    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 .
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    Quote Originally Posted by jbuddenh View Post
    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<x<1 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".
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  5. #5
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    Quote Originally Posted by ThePerfectHacker View Post
    If 0<x<1 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.
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  6. #6
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    Quote Originally Posted by jbuddenh View Post
    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.
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