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Proving exp function is bounded and not extended continuously.

  1. Oct 20, 2009 #1
    1. The problem statement, all variables and given/known data

    The function

    exp[ (x2 + y2 - xy)/(x2 + y2) ] = f(x,y)

    is continuous on the open first quadrant.

    Prove it is bounded there.

    Prove f cannot be extended continuously to the closed first quadrant.

    3. The attempt at a solution

    Since f is a real-valued function defined and continuous on a compact set (open first quadrant...I think...), then f is bounded..? So I have to show that the open first quadrant is a compact set maybe?

    I think I have to show that f is not uniformly continuous on the closed first quadrant. So, there exists an epsilon > 0 where a number d >0 cannot be found such that |f(p) - f(q)| < epsilon whenever p and q are in the closed first quadrant and |p-q|<d.

    I have so much trouble starting my proofs properly. I don't know how to set them up at the beginning, even when I know the rough outline.
     
  2. jcsd
  3. Oct 20, 2009 #2
    You need to check your definition of compact.

    can you bound the rational function on the open first quadrant?
     
  4. Oct 21, 2009 #3

    Mark44

    Staff: Mentor

    Try working with this expression:
    [tex]\frac{x^2 + y^2 - xy}{x^2 + y^2}[/tex]
    I can't give you much more of a hint than that without essentially doing the problem.
     
  5. Oct 22, 2009 #4
    |(x2 + y2 - xy)/(x2 + y2)| = |(r2 - rcosArsinA)/r2)| = |1-cosAsinA| which is bounded by 0 and 1

    So f(x,y) is bounded by 1 and e.

    Is it enough to show that? Or do I need to use definition of limits...

    Also, for the second part, it cannot be extended because x=0=y is a singularity. Does it suffice just to say that or does that not count as a full proof?
     
  6. Oct 22, 2009 #5

    Mark44

    Staff: Mentor

    You don't need to convert to polar form. How did you conclude that f(x, y) is bounded by 1 and e? Are you saying that all values of f(x, y) on the open first quadrant are between 1 and e? That's not true.

    Think division.
     
  7. Oct 22, 2009 #6
    Ok let me try this again...

    |(x2 + y2 - xy)/(x2 + y2)|<or= 1 because you are subtracting xy from the top so something smaller divided by something larger is a fraction less than one.

    Then, since the exponent is bounded by m=1, can you say f(x,y) is bounded by 0 and e, i.e. when you plug in 1 in the exponent you get e^1 = e?

    For the second part, I THOUGHT I could use the discontinuity x=y=0 but since S is the open first quadrant, can I do that?
     
  8. Oct 22, 2009 #7

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    I would be inclined to change to polar coordinates.
     
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