System of Equations

  • Thread starter VietDao29
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  • #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:
 
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Answers and Replies

  • #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?
 
  • #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?
 
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  • #4
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[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

http://en.wikipedia.org/wiki/Squircle" [Broken]
 
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  • #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]
 
  • #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?
 
  • #7
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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.
 
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  • #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.
 
  • #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
 
  • #10
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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
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
 
  • #11
VietDao29
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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
Thanks, jing. :smile:
 

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