Yearly Archives: 2014

The following NPR story starts with an anecdote about Kepler. Apparently, he was trying to find himself a wife, and he had 11 candidates, but he was too thorough in interviewing them (read: slow!), and they all got impatient and rejected him. Damn.

Turns out, if you have limited options, the best strategy for a good match (though maybe not the best match) is to interview 1/e (~36.8%) of your candidates without offering the “job” to any of them. Then, if you’re interested in anyone after that point, pop the question and forget the rest of the list.

Read more at NPR:
NPR: How to marry the right girl

Golden Power Series!

As you probably know, I wear my heart on my sleeve:

Well, I took the golden opportunity (ha!) to bring the golden ratio \Phi = \frac{1+\sqrt{5}}{2} into Calc 2 this week, using it (and its little pal \Psi = \frac{1-\sqrt{5}}{2}) to find a closed formula for the n-th term of the Fibonacci sequence.

The ubiquitous Fibonacci sequence! It’s something you may have encountered out in the wild. You know, it goes a little like this:

F_0 = 1, \, \, F_1 = 1, \, \, F_n = F_{n-1} + F_{n-2},
so F_2 = 2, \, \, F_3 = 3, \, \, F_4 = 5, \, \, F_5 = 8, \, \, F_6 = 13, \, \, F_7 = 21 \, \, \ldots.

And let’s say for some reason, you need to cook up F_{108}. I hope you have some time on your hands if you’re planning to add all the way up to that. Instead, wouldn’t it be nice if we had a simple formula that we could use — i.e., a formula that was not recursive — to figure out the n-th Fibonacci number?

Luckily, such a formula exists, and there are lots of ways to find it. In this post, we’ll find it using power series. Read on, brave blogosphere traveler.