That’s Unpossible! Part Deux: The Mathemagicing

So the other day I introduced you to the Koch Snowflake, and then left you hanging on two fronts. First, I stated that

$\displaystyle{\sum_{i=0}^{\infty}} \left(\frac{4}{9}\right)^{i}=\frac{9}{5},$

and asked simply for you to trust me. Second, I suggested that the Koch Snowflake existed somewhere between the 1st and 2nd dimensions. In this post, I’ll address the first of these items.

In fact, with a little bit-o-mathemagic, we’ll come up with a general statement about the sum of the powers of positive fractions that are less than 1. I know, I’m excited too!

Screen shot 2011-04-09 at 8.38.11 AM — Showing the sum of an infinite series using areas = wicked cool.

To begin, let’s simplify our fraction to something that most people can easily relate to: $\frac{1}{2}$ . So, we want to figure out the sum if we were to add up the powers of $\frac{1}{2}$ , from 0 to infinity (and beyond). Hence, we want to find

$\displaystyle{\sum_{i=0}^{\infty}} \left(\frac{1}{2}\right)^{i}.$

That is, we want to find

$1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\ldots$

Consider a square that has an area of 1 metre squared (or 1 foot squared for those imperialists among us). It’s easy to imagine dividing this area in half by cutting through the centre of the square. The result would be a rectangle with an area of $\frac{1}{2}$ metre squared. Place that half square to the right of the square (as shown in the picture). Now, let’s take half of this area as well, and place it on top. You should end up with a square with area one-fourth of the original square. If we continue this process, placing the new areas in a counter-clockwise manner, it becomes clear rather quickly that we are filling in another square that is of equal area to our first. That is, it looks that if we were to continue this process to infinity, the area on the right side of our original square would be 1 as well. Hence, we might imagine that

$\displaystyle{\sum_{i=0}^{\infty}}\left(\frac{1}{2}\right)^{i}=2.$

But is this really the case?

Let’s math this problem up a bit. Begin by replacing our fraction with the letter $p$ . That is, $p$ represents any proportion between 0 and 1. Our problem is restated as:

$S=\displaystyle{\sum_{i=0}^{\infty}}p^{i}=1+p+p^{2}+p^{3}+\ldots,$

where $S$ represents the overall sum. For fun (yes, I wrote fun), let’s find $pS$ . That is, we’re finding the $p$ portion of $S$ . Hence:

$pS=p\displaystyle{\sum_{i=0}^{\infty}}p^{i}=p+p^{2}+p^{3}+\ldots$

Notice that both $S$ and $pS$ contain all of the same powers of $p$ . The only difference is that $S$ also includes the number 1. That is, if we were to remove $pS$ from $S$ we’d have nothing left but the number 1. Check it out:

$S-pS=\left(1+p+p^{2}+p^{3}+\ldots\right)-\left(p+p^{2}+p^{3}+\ldots\right)=1.$

If I do a little factoring, we can rewrite this bad-boy as

$S(1-p)=1,$

$S=\frac{1}{1-p}.$

What the what!? We’ve actually managed to simplify our infinite summation to the following fancy-pants-formula-of-infinite-awesomeness:

$S=\displaystyle{\sum_{i=0}^{\infty}}p^{i}=\frac{1}{1-p}.$

AWESOME!

Returning to our original question, we had $p=\frac{4}{9}$ . Plugging this into our fancy-pants-formula-of-infinite-awesomeness, we get

$\displaystyle{\sum_{i=0}^{\infty}}\left(\frac{4}{9}\right)^{i}=\frac{1}{1-\frac{4}{9}}=\frac{1}{\frac{9}{9}-\frac{4}{9}}=\frac{1}{\frac{5}{9}}=\frac{9}{5},$

as I promised you it would be.

Now that you know this, you can impress all of your friends with your wicked fast ability to add up an infinite number of numbers.

Happy Mathematics Awareness Month all y’all.

consumedbywanderlust

That’s Unpossible! Part Deux: The Mathemagicing

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