An Orgy Of Nerdiness

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So how did you celebrate the epicness that was yesterday? You know, the combination of all things awesome known as Towel Day, National Plank Day, National Wine Day (which I was only alerted to later in the day – thankfully not so late as to prevent me from partaking in the appropriate celebrations), and last, but definitely not least, Geek/Nerd Pride Day.

As you can see by the picture, I celebrated in appropriate style by merging all the days into one blessed orgy of nerdiness. That’s right folks, I said orgy of nerdiness. The picture clearly satisfies the requirements of each of these nerdidays. Note that it contains:

  1. Post plank/towel reward.

    a nerd (that would be me),

  2. a glass of wine,
  3. a plank, and
  4. a towel (yes, that is a towel I’m wearing).

To celebrate the Geek/Nerd Pride part of yesterday, I posed the following problem:

Imagine that you have impeccable penmanship. In fact, imagine your penmanship is so good that people think your writing is a perfectly formed font. Imagine further that this font is legible at any size. That is, no matter the page size you are working with, you could increase or decrease your font size such that one single letter or digit would occupy the width and height of the page.

Next, imagine that you want to write each digit of \pi. Being the nerd that you are, you would know immediately that this would be a fruitless effort because, as a mere mortal, you wouldn’t have enough time to write out each and every single one of the infinite digits after the decimal.

But what if, by writing each digit of \pi on a smaller page using our perfect almost font-like penmanship, we cut our writing time as well? I mean, it would seem reasonable to think that the time required to write a digit occupying a smaller page would be less than the time needed to write a digit occupying a larger area, especially if the speed of writing remained constant.

Specifically, let’s assume that your penmanship is so good that it would allow you to write each digit using half of the area of the previous digit, and requiring half the time to write. If the first digit takes 1 minute to write, and requires 1m^{2} of space, then the second digit would require 30 seconds to write and only 0.5m^{2} of space. Clearly the third would require half that time and space again (ad infinitum).

My question for you is this: how long and how much space would be needed to write out all of the digits of \pi?

To figure out our answer, we only need to resort to something that we discussed a few weeks back.  Recall that

\displaystyle{\sum_{i=0}^{\infty}}p^{i}=\frac{1}{1-p}, if -1\leq p\leq 1.

We simply need to use this formula for our particular problem. Consider our area. We start off with an area of 1m^{2} for the first digit (3). The second digit (or the first after the decimal) – 1 – would then be written on something that has half that area: \frac{1}{2}m^{2}. The third digit – 4 – would require \frac{1}{4}m^{2}, and so on. Generally, the n^{th} digit after the decimal place would require \left(\frac{1}{2}\right)^{n}m^{2}.  To determine the entire area required, we simply add up the individual areas required for each of the digits:

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

Similarly, we would find that the total time required to write out all of the infinite digits would be 2 minutes.

CRAZY! So despite the fact that there are an infinite number of digits in \pi, and we’d require all the time in the world to write them out, by applying some simple rules, we’ve turned a job that requires an infinite amount of time into one that requires 2 minutes.


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