Make an equation that graphs a square and all points inside with corners at (-4,4), (4,-4), (-4,-4) and (4,4). You cannot use inequalities or set notation. Just an equation.
For those who doubt the solvability of this problem - IT IS POSSIBLE, I just don't know how to get it...
For those who doubt the solvability of this problem - IT IS POSSIBLE, I just don't know how to get it...
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We can use this:
|x| / x = {-1, if x < 0
............ {1, if x > 0
to quantify regions. In fact, we can modify this a little. Let's define:
f(x) = (1/2)(|x|/x + 1)
which equals 1 when x is positive, and 0 when x is negative. This is great for decision procedures. If I want an equation to represent x < 4, then all I need is:
f(4 - x) = 1
Or, if I want to represent x > -4, then I would write:
f(4 + x) = 1
If I wanted both to be true at the same time, I would write:
f(4 - x)f(4 + x) = 1
which could only happen if both 4 - x and 4 + x are positive. Therefore, I can construct an equation for the open square (i.e. without the boundary) as follows:
f(4 - x)f(4 + x)f(4 - y)f(4 + y) = 1 ... (1)
If you want to close the box, we need to be a little more tricky. The problem is that f cannot handle an input of 0, since it would cause a division by 0. However, we can close the square by expanding (1) and multiplying through by the common denominator:
(1/2)(|4 - x|/(4 - x) + 1)(1/2)(|4 + x|/(4 + x) + 1)(1/2)(|4 - y|/(4 - y) + 1)(1/2)(|4 + y|/(4 + y) + 1) = 1
(|4 - x|/(4 - x) + 1)(|4 + x|/(4 + x) + 1)(|4 - y|/(4 - y) + 1)(|4 + y|/(4 + y) + 1) = 16
(|4 - x| + 4 - x)(|4 + x| + 4 + x)(|4 - y| + 4 - y)(|4 + y| + 4 + y) = 16(4 - x)(4 + x)(4 - y)(4 + y) ... (2)
Now, (2) is exactly what we want. It defines the square's boundary, because now equality does happen when x or y = 4 or -4. At any other point, (2) is equivalent to (1), so all points inside the open square is covered by (2), and all other points outside of the square are not covered by (2). Therefore, the graph of (2) is exactly the closed square defined.
|x| / x = {-1, if x < 0
............ {1, if x > 0
to quantify regions. In fact, we can modify this a little. Let's define:
f(x) = (1/2)(|x|/x + 1)
which equals 1 when x is positive, and 0 when x is negative. This is great for decision procedures. If I want an equation to represent x < 4, then all I need is:
f(4 - x) = 1
Or, if I want to represent x > -4, then I would write:
f(4 + x) = 1
If I wanted both to be true at the same time, I would write:
f(4 - x)f(4 + x) = 1
which could only happen if both 4 - x and 4 + x are positive. Therefore, I can construct an equation for the open square (i.e. without the boundary) as follows:
f(4 - x)f(4 + x)f(4 - y)f(4 + y) = 1 ... (1)
If you want to close the box, we need to be a little more tricky. The problem is that f cannot handle an input of 0, since it would cause a division by 0. However, we can close the square by expanding (1) and multiplying through by the common denominator:
(1/2)(|4 - x|/(4 - x) + 1)(1/2)(|4 + x|/(4 + x) + 1)(1/2)(|4 - y|/(4 - y) + 1)(1/2)(|4 + y|/(4 + y) + 1) = 1
(|4 - x|/(4 - x) + 1)(|4 + x|/(4 + x) + 1)(|4 - y|/(4 - y) + 1)(|4 + y|/(4 + y) + 1) = 16
(|4 - x| + 4 - x)(|4 + x| + 4 + x)(|4 - y| + 4 - y)(|4 + y| + 4 + y) = 16(4 - x)(4 + x)(4 - y)(4 + y) ... (2)
Now, (2) is exactly what we want. It defines the square's boundary, because now equality does happen when x or y = 4 or -4. At any other point, (2) is equivalent to (1), so all points inside the open square is covered by (2), and all other points outside of the square are not covered by (2). Therefore, the graph of (2) is exactly the closed square defined.