Remember that great post about the golden ratio? I’ve been neck-deep in some more of the most awesome mathematics I’ve ever encountered, and I just had to share it all.

Yesterday, in an attempt to study for my Linear Algebra final, I thought I’d visit Khan Academy to view some supplemental lectures. While browsing for the Linear Algebra subject, I saw something interesting titled *Hexaflexagons*, and the remainder of my study time had almost nothing to do with Linear Algebra.

# Hexaflexagons

Hexaflexagons are cool flippy-flappy hexagon-shaped 2-dimensional paper things that have more than 2 faces and taste really good. These three videos are a must-watch, and I highly recommend checking out some of the others. Be warned, however, as these fun things—much like a favored desktoy so awesome the government thought no one should have it—can be potentially dangerous.

While watching the first of these videos, I instantly recognized the speaker as Vi Hart, the really fast-talking, smart, funny, mathy person who draws the cool snakes in math class. One of the videos linked to her website and I found a particularly interesting video there of her presentation on hyperbolic planes at the 2011 Joint Math Meeting (JMM) in New Orleans. I was *there*, I remember that talk, I’m even in that video, and I *met* her in person! I just figured all of this out, and at the time, I had no clue that this JMM speaker was also responsible for my familiarity with serpentine topology. It’s reminiscent of the time I had a conversation with President George Bush 41 or when I accidentally (and literally) walked into Nancy Pelosi in Washington D.C.—I had crossed paths with someone famous/important/etc. without realizing who they were.

I think now might be an appropriate time for a joke or mathematical musing about snakes on a (hyperbolic) plane… Vi?

# Tau vs. Pi

Pi is wrong. No, there’s nothing wrong with the numerical value \( \pi \approx 3.14159\ldots \), but just that \( \pi \) should have been \( 2\pi \). I’ll just let Michael Hartl explain. Oh, and yes, that’s another one of Vi Hart’s videos in the *smile* section.

To some, Euler’s equation for imaginary exponentials represents mathematical perfection:

\[ \mathrm{e}^{i\pi} = -1 \]

I recall telling my Calculus 3 instructor once that “negative signs make me sad”. Rightly so, because I try to be a positive, upbeat person. Euler’s equation is neat, but the negative sign is perfectly annoying, and any attempt to remove it takes away the elegance. But wait, take out \( \pi \) and put in \( \tau \) and that equation should have been

\[ \mathrm{e}^{i\tau} = 1 \]

Talk about elegance: *that* is perfection.

# Calculus of Finite Differences

Calculus works fine and dandy for continuous functions, but what about the things we computer scientists use (summations, discrete functions, and integer functions)? This week in my Algorithms class, my instructor (who has blown my mind with cool math and computer science stuff so many times already) is lecturing about **The Calculus of Finite Differences** (CFD)—the calculus that all computer scientists should know. No, I’m not taking this course from Purdue, and no, David Gleich is not my professor. That PDF is just recommended reference material for the topic.

The basis for CFD is the *difference* and *antidifference operators*, which are the discrete analogs of the derivative and antiderivative, respectively, and operate on falling powers:

\[
\Delta f(n) = f(n+1)-f(n) = g(n) \\\

\sum g(n) , \delta n = f(n)
\]

So for \( f(n) = n^{\underline{m}} \),

\[ \Delta f(n) = m , n^{\underline{m-1}} \]

Where the falling power \( n^{\underline{m}} \) is defined as

\[
n^{\underline{m}} = \begin{cases}
n(n-1)\dots(n-m+1) & m \ge 0 \\\

\frac{1}{(n+1)(n+2)\dots(n-m)} & m < 0
\end{cases}
\]

CFD makes it easy to find closed forms of complex sums without looking them up in a table. For example,

\[
\begin{align*}
\sum_{k=1}^n k^2 &= \sum_{k=1}^n k^{\underline{2}} + k^{\underline{1}} \\\

&= \left. \frac{1}{3}k^{\underline{3}} + \frac{1}{2} k^{\underline{2}} \right|_1^{n+1} \\\

&= \left. \frac{k(k-1)(2k-1)}{6} \right|_1^{n+1} \\\

&= \frac{n(n+1)(2n+1)}{6}
\end{align*}
\]