# 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.

NPR: How to marry the right girl

## It’s the little things.

Man’s mind, once stretched by a new idea, never regains its original dimensions.

— Oliver Wendell Holmes

When things get too complicated, it sometimes makes sense to stop and wonder: Have I asked the right question?

— Enrico Bombieri

## 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.