"But Kevin, it 330am. Doesn't that mean it's Monday?"
Don't be ridiculous, silly reader. Don't be ridiculous.
This week, we're going to take a little excursion. Instead of talking about some more set theory or abstract algebra, let's play a little game.
Say we start out with a square of side-length L, then inscribe a circle and cut out the circle's area. Next we inscribe a square into that circle hole and add that area back to the total. We have a square again, so let's cut out another circle. Then add another square. Then subtract yet another circle. Then add a smaller square... On and on, til we have something that looks like this:
The question is: If the process of subtracting circles and adding squares continues to itsy, bitsy, infinitesimal, microscopic mothaflankin' Planck-like lengths, what total area does our final figure assume?
Let's define an iteration of our process as "add 1 square, subtract its inscribed circle."
So, for example, iteration 0 is to begin with a square of side-length L and subtract its inscribed circle of radius r=L/2.
I hate to say it, but: just in case I lost you on that radius, check out this figure:
So the first area, let's denote is A0, we find to be:
= L2 - πr2
= L2 - π(L/2)2
= L2 - (π/4)L2
= (1 - π/4)L2
What is the area after a second iteration?
To answer this, we first must figure out what the side-length of the new inscribed square should be. Just note that its corner points are touching the circle: this gives us the length of its diagonal! It is two times the radius of the circle--and since the circle's radius is L/2, that makes the new square's diagonal equal to L.
We can now find the side length of square(1) using the good ol' Pythagorean Theorem, which is formally:
...where c2 is the hypotenuse, while a and b are the triangle's legs. In our case, c = L and a = b:
So the side length of square(1) is L/√2.
What will circle(1)'s radius be? Again, it's half the length of the square it's inscribed in, so it is a/2 = L/(2√2).
The area upon the second iteration then will be
= A0 + a2 - π(a/2)2
= A0 + (1/2)L2 - (π/8)L2
= (1 - π/4 + 1/2 - π/8)L2
For now on, let's just say that L = 1, then we don't have to worry about carrying it around.
Upon iteration 2, we would find:
= 1 - π/4 + 1/2 - π/8 + 1/4 - π/16
Similarly,
= 1 - π/4 + 1/2 - π/8 + 1/4 - π/16 + 1/8 - π/32
There's a pattern emerging here. We really don't have to do much more work to extrapolate this trend towards infinity. Let's regroup the terms of A3:
= (1 + 1/2 + 1/4 + 1/8) - (π/4)( 1 + 1/2 + 1/4 + 1/8)
= (1-π/4)(1 + 1/2 + 1/4 + 1/8)
Observe:
| (1/2)0 = 1 |
| (1/2)1 = 1/2 |
| (1/2)2 = 1/4 |
| (1/2)3 = 1/8 |
So, we can simplify the sum using "sigma summation notation":
where the sum runs from n=0 to n=3. (The capital sigma let's us know to SUM over n; n is called the summation index.)
Notice that it will now have this same form for every iteration. For example, the nth iteration will have total area given by:
Does this sum converge as n goes to infinity? The answer is yes. You can prove it using a technique from analysis called the "root test." Or, perhaps you know that this is a well-known type of series called a "geometric series," which is convergent as n goes off into the distant lands, ever-seeking that holy grail.
So what is the area of that object with all the weird holes in it? Here's a little table of some numerical values for the total area upon successive iterations:
| A0 | 0.214601837 |
| A1 | 0.321902755 |
| A2 | 0.375553214 |
| A3 | 0.402378444 |
| A4 | 0.415791058 |
| A5 | 0.422497366 |
| A6 | 0.42585052 |
| A7 | 0.427527096 |
| A8 | 0.428365385 |
We can pretty much see that it's going to end up somewhere around 0.43 --- that is, 43% the area of the original square.
Using some knowledge about the geometric series, I can tell you that the series ∑(1/2)n approaches 2, so our series (1-π/4)∑(1/2)n approaches (1-π/4)(2) = 2 - π/2, which equals 0.429203673 out to nine decimal places.
So we were pretty much right.
We rule.
