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Math Help - What is the range of this function? the function is f(x,y)=2xy/(x^2+y^2)

  1. #1
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    What is the range of this function? the function is f(x,y)=2xy/(x^2+y^2)

    the function is f(x,y)=2xy/(x^2+y^2)
    the domain is R-{(x,y)=(0,0)}
    I managed to know that the lower bound of the function is -1 because:
    2xy=(x+y)^2-x^2-y^2
    so z=((x+y)^2+(x^2-y^2))/(x^2+y^2)
    which simplifies to (x+y)^2/(x^2+y^2) -1
    (x+y)^2/(x^2+y^2) >=0 so (x+y)^2/(x^2+y^2) -1>=-1
    but I didn't manage to know the upper bound of the domain. Thank you for helping me.
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  2. #2
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    Re: What is the range of this function? the function is f(x,y)=2xy/(x^2+y^2)

    This is a cool question. I think the range is within [-1, 1]. But I haven't found a good reason why. Have you graphed it? Check it out. : )What is the range of this function? the function is f(x,y)=2xy/(x^2+y^2)-weird.jpg
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    Re: What is the range of this function? the function is f(x,y)=2xy/(x^2+y^2)

    Yes you're right that the range is [-1,1], and to prove that I was trying to prove why (x+y)^2/(x^2+y^2)<=2 (you can try with different number combinations and you will always see that the results are less or equal to 2 but there's no way to prove why I think).
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  4. #4
    Super Member ILikeSerena's Avatar
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    Re: What is the range of this function? the function is f(x,y)=2xy/(x^2+y^2)

    Hi mohomad!

    The trick is to switch to polar coordinates.

    \frac {2xy} {x^2+y^2} = \frac {2 \cdot r\cos\theta \cdot r\sin \theta} { r^2} = 2\cos\theta\sin\theta

    Can you tell what its range is?
    Thanks from MooMooMoo
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    Re: What is the range of this function? the function is f(x,y)=2xy/(x^2+y^2)

    I tried switching it to polar but got stuck on 2cos(O)sin(O) but now after seeing it here I remembered that it's equal to sin(2O). Thank you very much!!
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  6. #6
    Junior Member Nehushtan's Avatar
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    Re: What is the range of this function?

    For all x,y\in\mathbb R we have (x+y)^2\geqslant0 \implies 2xy\geqslant-(x^2+y^2) and (x-y)^2\geqslant0 \implies x^2+y^2\geqslant2xy. Hence -(x^2+y^2)\leqslant2xy\leqslant x^2+y^2 \implies -1\leqslant\frac{2xy}{x^2+y^2}\leqslant1 ( (x,y)\ne(0,0)).
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    Re: What is the range of this function? the function is f(x,y)=2xy/(x^2+y^2)

    Thank you Nehushtan that what I was looking for.
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