System of Equations

[VietDao29]

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

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

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

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.

 

[arildno]

Subtract the two equations from each other, yielding:
$$x^{2}(1-x^{2004})+y^{2}(1-y^{2004})=0$$
What can you say about the signs of each term here, and what does that further imply?

 

[VietDao29]

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?

 

[jing]

$$x ^ {2006} + y ^ {2006} & = & 1$$

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]

 

[NateTG]

Big fat stinking hint:
$$x^2+y^2=1$$
so
$$\left(x^2+y^2\right)^{1003}=1^{1003}=1$$

 

[arildno]

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?

 

[gammamcc]

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.

 

[arildno]

Why need the binomial theorem??

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

 

[VietDao29]

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

@jing: I still don't know how to use Squircle to solve the problem. The article doesn't say much about its properties. Can you explain it a little bit more? :) Thanks

 

[jing]

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

@jing: I still don't know how to use Squircle to solve the problem. The article doesn't say much about its properties. 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

 

[VietDao29]

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.