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System of Equations

  1. Feb 23, 2007 #1

    VietDao29

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    Hi,
    I just came across this problem some days ago, and I am tearing all my hair to solve it. :cry: :cry: :cry: I am nearly bald now... :cry: :cry:

    The problem asks to solve for 2 unknowns x, and y:
    [tex]\left\{ \begin{array}{ccc} x ^ 2 + y ^ 2 & = & 1 \\ x ^ {2006} + y ^ {2006} & = & 1 \end{array} \right.[/tex]

    I know there should be two sets of roots:
    [tex]\left\{ \begin{array}{ccc} x & = & 1 \\ y & = & 0 \end{array} \right. \quad \mbox{or} \quad \left\{ \begin{array}{ccc} x & = & 0 \\ y & = & 1 \end{array} \right.[/tex]

    So, I rearranged the first equation to: x2 = 1 - y2, and substituted this to the second equation, leaving me an equation of power 2004. :yuck: :yuck: I must be missing something really big. :grumpy:

    I think there should be another easy way to solve the problem. Can anyone just give me a small crack at it?

    Thanks in advance. :smile:
     
    Last edited: Feb 23, 2007
  2. jcsd
  3. Feb 23, 2007 #2

    arildno

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    Subtract the two equations from each other, yielding:
    [tex]x^{2}(1-x^{2004})+y^{2}(1-y^{2004})=0[/tex]
    What can you say about the signs of each term here, and what does that further imply?
     
  4. Feb 23, 2007 #3

    VietDao29

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    Uhmmm... Let's see if I can get it correct
    x2(1 - x2004) + y2(1 - y2004) = 0
    The sign of each term depends on (1 - x2004), and (1 - y2004) respectively.
    Both can be 0, that's the case mentioned above.
    The other case is one must be negative and one must be positive. Say (1 - x2004) < 0, and (1 - y2004) > 0
    That means x > 1, or x < -1 (cannot satisfy either equation), and -1 < y < 1. Hence, this case cannot happen.

    Am I doing it correctly?
     
    Last edited: Feb 23, 2007
  5. Feb 23, 2007 #4
    [tex] x ^ {2006} + y ^ {2006} & = & 1 [/tex]

    is an example of a supercircle.

    Examing it's properties may help you see the values of x and y

    link here to squircle
     
  6. Feb 23, 2007 #5

    NateTG

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    Big fat stinking hint:
    [tex]x^2+y^2=1[/tex]
    so
    [tex]\left(x^2+y^2\right)^{1003}=1^{1003}=1[/tex]
     
  7. Feb 23, 2007 #6

    arildno

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    You're on the right track!

    However, from your original equations, what can you say about |x| and |y|, and what will this info provide you with clues as to the signs of the terms in the derived equation?
     
  8. Feb 23, 2007 #7
    Binomial theorem helps. (x^2+y^2)^{1003} = x^2006+y^2006 + nonnegative stuff.

    Stuff is zero for certain x and y.

    I think you get more solutions than what you wrote.
     
    Last edited: Feb 23, 2007
  9. Feb 23, 2007 #8

    arildno

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    Why need the binomial theorem??

    From the first original equations, we know that |x|,|y|<=1, and therefore, we have:
    [tex]0\leq{x}^{2}(1-x^{2004}),0\leq{y}^{2}(1-y^{2004})[/tex]
    since the sum of these terms equals 0, we must have:
    [tex]x^{2}(1-x^{2004})=0[/tex]
    And:
    [tex]y^{2}(1-y^{2004})=0[/tex]
    This system of equations is readily solved, and by insertion of the solutions of this sytem into the first system, we find which solutions that system has.
     
  10. Feb 24, 2007 #9

    VietDao29

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    Yeah, thanks everyone for their help. I got it now. :smile:

    @jing: I still don't know how to use Squircle to solve the problem. The article doesn't say much about its properties. :frown: Can you explain it a little bit more? :) Thanks
     
  11. Feb 24, 2007 #10
    Though that looking at the pictures may help you see that the super circle lies within the circle and hence help you see |x| <=1 and |y| <=1
     
  12. Feb 25, 2007 #11

    VietDao29

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    Thanks, jing. :smile:
     
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