Messing with inequalities

[LB]

I was messing with inequalities using Wolfram|Alpha's generated graph and seeing what shapes I could get. I started with a simple circle:

1 >= x² + y²
http://www.wolframalpha.com/input/?i=1+%3E%3D+x%5E2+%2B+y%5E2

Then I tried some variants in the exponents and noticed some strange things with odd-numbered exponents vs even-numbered vs mixed vs negative, etc. and I found that the higher you go with even-numbered exponents, the closer it gets to making a square:

1 >= x¹⁰⁰ + y¹⁰⁰
http://www.wolframalpha.com/input/?i=1+%3E%3D+x%5E100+%2B+y%5E100

Then I tried messing with decimal places and got a 2x1 rectangle:

1 >= x¹⁰⁰ + y^100.0001
http://www.wolframalpha.com/input/?i=1+%3E%3D+x%5E100+%2B+y%5E100.0001

If I change the number of zeroes in that 100.0001 it causes the rectangle to stretch in ways that I can't predict. Why is that 100.0001 so close to a perfect 2x1 rectangle? Is it a side-effect of using a base-10 numbering system, and I just happened to get close with that number?

In general I've noticed that things go haywire with even the slightest hint of non-whole numbers.

[Duthomhas]

Look down at the "Solutions for the variable y" and you'll see why you are getting what you get.

It isn't square, and never will be. There is always a slight offset (or curve) from the square at x=1 and y=1.

Remember, power functions also have some specific properties, which you can examine for yourself.

For example: (+5)³ will always be positive. But (-5)³ will be negative.

Here's some good reading right off the top of google (complete with pretty pictures, which I find handy)
http://wmueller.com/precalculus/families/1_41.html

Here are more pictures to look at
http://www.meipokwan.org/Figures/ga_links.htm

Hope this helps.

[LB]

Duoas wrote:

Look down at the "Solutions for the variable y" and you'll see why you are getting what you get.

I don't see that, is it a pro feature?

If we have a limit of c approaching infinity and use 1 >= x^2c + y^2c, can we not say that is a square?

[Duthomhas]

I don't have a wolfram account, so IDK. Anyway, just solve for y.

    y = ⁿ√(1-xⁿ)

The function, no matter what n>0 you choose, will never create straight lines. To us humans it may eventually like you get a square, but it will always be a curve. You can prove this by zooming in really, really close at any point along the curve and you will see that the only place it touches the x-axis is at the origin.

To see how it behaves, watch how the following plots change each time.

http://www.wolframalpha.com/input/?i=y+%3D+x%5E2
http://www.wolframalpha.com/input/?i=y+%3D+x%5E4
http://www.wolframalpha.com/input/?i=y+%3D+x%5E6
http://www.wolframalpha.com/input/?i=y+%3D+x%5E8
http://www.wolframalpha.com/input/?i=y+%3D+x%5E10
http://www.wolframalpha.com/input/?i=y+%3D+x%5E50
http://www.wolframalpha.com/input/?i=y+%3D+x%5E100

You can see how the parabola gets more and more 'square' -- the curve gets pushed closer and closer down towards (1,0) -- but it will never touch. It is always a curve.

Limits describe where a thing appears to be, not where a thing is.

For example, consider the following line:

-------o--------
   |   |
   0   x

At x, the function as a limit that produces a value on the line. Except, the function specifically excludes that point on the line.

Limits are used to show how things near a point behave.

Hope this helps. (Gotta go change a poopy diaper.)

[LB]

I understand all that, but surely since it becomes more square-like with higher even powers, we can say that as that limit approaches infinity we have a true square? There are other things in math that are defined this way, maybe I'm just not wording my conjecture correctly.

[helios]

Yes, the limit of the area as natural c goes to infinity is a square. This is because k^n for any 0 < k < 1 goes to 0 as n goes to infinity.
By the way, a generalized form of the expression would be 1 >= |x^c| + |y^c|.

[Ispil]

Well, let's take a look at the equation, then.

y = ⁿ√(1-xⁿ)

Now, we can obviously take the limit as n approaches infinity here:

L = lim_n→∞(1-xⁿ)^1/n

Which is indeterminate. Now, let's actually look at the problem. We have 1-x^∞, which will be -∞ so long as |x|>1, indeterminate when |x| = 1, and 0 when |x|<1. So, we re-write it for the two valid cases:

L = lim_n→∞(-n)^1/n
L = lim_n→∞(0)^1/n

The only case where the function is defined is when |x|<1, at which point the limit does equal 1. However, this causes another implication:

Even as the limit approaches infinity, the function is not a true square- the corners are indeterminate. In fact, because the case isn't just where the function reduces down to 1^∞ but actually is that case, one would be better off saying that the limit is undefined.

So, no. It is not a true square. Comes close, though.