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Continuous extension

  1. Aug 11, 2010 #1
    How can the function

    f: ℝ² → ℝ : (x,y) |--> [tex]{{x^2+y^2-(x^3y^3)}\over{x^2+y^2}}[/tex] if (x,y) ≠ (0,0)

    be defined in the origin so that we get a continuous function?


    When I take 'x=y' (so (y,y)) and 'y=x' (so (x,x)) I get:

    [tex]{{2-y^4}\over{2}}[/tex]

    and

    [tex]{{2-x^4}\over{2}}[/tex]

    So for the first one I get '1' when y=0
    and for the second one I get '1' when x=0

    So can I say that if (x,y)=0 --> f=1 ??
     
  2. jcsd
  3. Aug 11, 2010 #2
    Yes, you can extend your function to a continuous one if you define f(0,0)=1.
     
  4. Aug 11, 2010 #3
    Thx!
     
  5. Aug 11, 2010 #4

    arildno

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    In order to make this general, you should switch to polar coordinates.

    Then, we can transforme the fraction to:
    [tex]\frac{r^{2}-r^{6}\sin^{3}\theta\cos^{3}\theta}{r^{2}}=1-r^{4}\sin^{3}\theta\cos^{3}\theta, r\neq{0}[/tex]
    clearly, this is readily extendable for r=0.
     
  6. Aug 11, 2010 #5
    Thank you very much!
    And how about this one?

    https://www.physicsforums.com/showthread.php?t=421612
     
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