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Math Help - Continouty of a mapping, need help

  1. #1
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    Continouty of a mapping, need help

    Hi,

    I'm having trouble with an assignment of mine. I hope someone here can help me. It should be trivial - I just think I've overseen something... Well, here goes:

    I have a mapping \theta : \mathbb{R}^2 \setminus \{ (x,y) | y = 0, x \leq 0 \} \rightarrow (-\pi,\pi) defined such that \frac{x}{\sqrt{x^2 + y^2}} = \cos\theta(x,y) and \frac{y}{\sqrt{x^2 + y^2}} = \sin\theta(x,y). So, it is the angle between the x-axis and the line connecting origo and the point (x,y).

    How do I show that the mapping \theta(x,y) is not continous on the negative x-axis?
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  2. #2
    Senior Member vincisonfire's Avatar
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    This mapping is discontinuous at (0,0).
    Therefore, since you include this point in your mapping, it is discontinuous.
    Remark that if your mapping would be <br />
\theta : \mathbb{R}^2 \setminus \{ (x,y) | y = 0, x < 0 \} \rightarrow (-\pi,\pi)<br />
then it would be continuous.
    Now to prove that the function is discontinuous at (0,0), you need to show that  \lim_{(x,y)\rightarrow (0,0)} \frac{x}{\sqrt{x^2+y^2}} \neq f(0,0)
    We don't care wath function f is since we can prove that the limit doesn't exist.
    Let approach the point along the line of equation  y = mx The limit then become  \lim_{(x,y)\rightarrow (0,0)} \frac{x}{x\sqrt{1+m^2}} =\lim_{(x,y)\rightarrow (0,0)} \frac{1}{\sqrt{1+m^2}} Since the limit depends on m, by uniqueness of limit we conclude that the limit doesn't exist.
    Finally, \theta(x,y) = arccos(\frac{x}{\sqrt{x^2+y^2}}) is not continuous since  \frac{x}{\sqrt{x^2+y^2}} is not.
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  3. #3
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    No

    I'm sorry, but it was not the point (0,0) I wanted to show a discontinouity at - this point is not part of the domain of the function. The domain was \mathbb{R}^2\setminus\{ (x,y)| y=0,x \leq 0 \} - that is the whole x-y plane minus the negative x-axis, also minus the point (0,0).

    I want to show that \theta(x,y) is not continous on the negative x-axis (not including the point (0,0)). To prove that it is not possible to extend the function to a continous function on the domain \mathbb{R}^2\setminus\{ (0,0) \}.

    It is easily seen that \theta must obey that

    \lim_{y\rightarrow0^+} \theta(-1,y) \rightarrow \pi  \qquad\text{and}\qquad  \lim_{y\rightarrow0^-} \theta(-1,y) \rightarrow -\pi .

    for x < 0. But how do I show it?

    You can see the system here: http://ragelse.com/file.php/8500c2bd...b9f3fc17f5d2da
    Last edited by Bozack; November 30th 2008 at 04:30 AM.
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  4. #4
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    Doesn't matter any more - I didn't had more time to do the assignment...
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