Archive of posts filed under the Application category.

When a Dollar is not a Dollar

There must be some way to capitalize on this through arbitrage, right?

Math Photo: Circle in the Sky

I’m not sure exactly what this is, but I know that a lot of geometry went into designing it!

Mona Lisa and Information Theory

This pointilist images puts me in mind of some fundamental ideas of information theory.

Even though you are seeing a limited set of data here, 140 dots, each of uniform color, it is fairly clear what this image represents.  (Hint:  standing a bit farther back might help!)

A fundamental concept in information theory is the compression of information.  Suppose you want to communicate something, like a painting:  a mathematical way to do this would be to describe the color of each pixel, in some order.  That would require the transmission, reception, and translation of millions, or even billions, of numbers.

Here we see a very complicated and intricate image being communicated with a very small set of data.  Finding more efficient ways to pass along (and store) information is more important than ever before, and is a focus of much modern mathematics.

Of course, recognition of this image relies on the experience of the viewer, who must bring the appropriate context to this information in order to decode it.  But  that is required in all successful communication.

A Trigonometric Warning

I’ve never seen a trigonometric warning sign before, but it makes a good point!

The Algebra of Coffee Consumption

Upon discovering that I had been paying much more for my coffee than I realized, I was faced with a dilemma only algebra could solve.   Or, at least help analyze.

You see, my local coffee shop offers a simple reward program for regular customers:  purchase 10 bags of coffee, and your eleventh bag is free.  At the time of my discovery, I had credit for four bags of coffee.  Thus, the dilemma:  do I continue to buy the over-priced coffee six more times in order to get the free bag?  Or do I just start buying cheaper coffee somewhere else?

The coffee at my local shop was costing me around $12 for 12 ounces. I knew I could get comparable coffee at another shop for around$10 a pound.  So should I stay or switch?

If I bought 6 more bags from the local shop, I’d spend $72 and get seven 12-ounce bags of coffee (the six I bought, and the free one). Thus, I would get 84 ounces of coffee for$72.

Now $72 dollars at the new place would get me 7.2 pounds of coffee, or around 115 ounces. So I’d be getting an additional 31 ounces, or almost two pounds of coffee, by switching! Needless to say, I swtiched. But it got me thinking: at what point would it have been better to continue buying the over-priced coffee? I turned to algebra for the answer. Let x = the number of bags of coffee that have already been purchased. In order to get a free bag of coffee from my local shop, I’d need to buy ten total bags; thus, I’d need to purchase (10 - x) more bags of coffee. After that, I’d receive one free bag, so in the end I’d get (10 – x + 1) = (11 – x) total bags of coffee. Each 12-ounce bag of coffee costs$12, so I’d spend a total of $12 * (10 – x) to get (11 - x) * 12 ounces of coffee. This works out to a price of $\frac{10 - x}{11 - x}$ dollars per ounce for the coffee I’d get from the old place. I know I can get coffee at the other place for$10 per pound, or $\frac{10}{16} = \frac{5}{8}$ dollars per ounce.  Thus, I want to find x so that

$\frac{10-x}{11-x} < \frac{5}{8}$

which would mean that the coffee from the old place would be cheaper per ounce than the coffee from the new place.

Now we simplify our inequality:

$8 * (10 - x) < 5 * (11 - x)$,

$\newline 80 - 8x < 55 - 5x$,

$25 < 3x$,

$\frac{25}{3} < x$

Thus, I should continue buying coffee at the old place if I’d already bought more than eight bags of coffee there.  Otherwise, I’d be better off switching.

I probably didn’t need algebra to tell me to stick with the old place if I’d already bought 9 bags, but algebra did show me just how hopeless my situation was!