The Big $O$

I need to teach myself this stuff, so I'm going to make this post as a place where I can prove a bunch of elementary facts about Big O notation, (& maybe some related topics).

Starring Rarity as Roger Smith

Definition:

$f(x) = O(g(x)) \mbox{ as  } x \rightarrow \infty$ 

if and only if

$\exists \, M, x_0 \in \mathbb{R}^+ : |f(x)| \leq M|g(x)| \, \forall x > x_0$

Some basic observations:

-The equality sign here is an odd convention, because this is more of an order relation.

-This isn't a 'total ordering' when we look at all $f$ and $g$; $e^xsin(x)$ and $x^2$ are incomparable.

-There's an obvious analogue of this definition for sequences (those discrete things) and all the results discussed here apply equally to them.

Basic result(s) come after the break.

2-Group Madness

So, beyond mathematicians' categorizing all the finite simple groups and enumerating all the ways you can get the rest of the finite groups out there by gluing smaller groups together, it turns out that that we have fairly good algorithms for enumerating all the groups of a given order, and of all the groups whose order is less than 2000, most of them have order 1024.
As this MathOverflow post and this blogpost point out, this sort of behavior may quite likely to be a general phenomenon.
Old news, but it's pretty damn neat.

Robert Ghrist!

This is what turned me on to higher math.  I had no idea these lectures were floating around the internet until today.  Enjoy them!

One

Two

Three

Let's do some basic math problems!

Maybe this is a side-effect of playing those silly "Math Blaster" games when I was growing up, but I've found this to be pretty entertaining.
On the strikeout: math is fun, and being handed problems you know how to solve is also fun.  How long has it been since you did long division in your head?  This isn't teaching me anything new, but it makes my mind feel more limber.

ABC Conjecture

Heard the news?  It's a big deal if the proof goes through.
Shinichi Mochizuki recently released a 500-page proof (split up in four sections) that claims to have solved the ABC conjecture:

 $\, \forall \, \epsilon > 1 \, \exists \,$ only finitely many coprime $a, b$ such that $d^{\epsilon} < c$ where $c=a+b$ and $d$ is the product of the unique prime factors of $abc$ (the "square free" part).

Apparently he needed to use non-well-founded sets to get this result. Neat stuff.
Paper 1     Paper 2     Paper 3     Paper 4
Updated to add better-educated folks' commentary on the paper:
Take one  Take Two  Take Three  MathOverflow

Real Analysis $\mathbb{R}$e-view

On going through a layperson-readable take on Skolem's paradox here I noticed that the author (Steven Landsburg) talked about producing different models of $\mathbb{R}$ from different 'models' of $\mathcal{P}(\mathbb{Q})$ (powersets of the rationals). Why didn't he talk about Cauchy sequences? (because that's how we construct $\mathbb{R}$, right?)
As long as we don't mind throwing out the really boring sequences (like $\{1,1,1,1...\}$) with only finitely many unique elements, no emphasis on sequences is necessary. Because $\mathbb{Q}$ is countable, any subset is at most countable, and can be ordered and considered as a sequence. Furthermore, actually assigning an ordering is unnecessary...

Call a subset $S \subset \mathbb{Q}$ a Cauchy subset iif
1) $S$ is infinite
2) $\, \forall \, \epsilon > 0 \, \exists \, x \in S$ such that $\{ y \in S : |x-y|\geq \epsilon \}$ is finite

As an exercise, show (because my probably-nonexistent readership loves exercises, right?)
-$S$ is a Cauchy sequence under a particular ordering $\Leftrightarrow$ $S$ is a Cauchy set $\Leftrightarrow$ $S$ is a Cauchy sequence under every ordering.
-All the above statements/definitions work equally well with subsets of $\mathbb{R}$, not just $\mathbb{Q}$. (hint: show that condition 2 forces $S$ to be at most countable)

(AG) Reading Rainbow: Get Your Books!

So there's an Algebraic Geometry reading group at this subreddit.

If you're up for this, it looks like you'll have to buy the Beltrametti et al book on your own.  But if you're looking to pick up the other books, and maybe you also want to score an Introduction to Commutative Algebra with some solution manuals on the side... I know a guy... (the blogger gestures down a nearby alley)

Follow the link after the break...