Too Many Mornings

In an effort to balance out that last post, I'll talk about something that is probably of marginally more interest to people but still sort of galvanizing: Bob Dylan.

He's famous among his fans or rock fans or music fans or something for recording many (at times very) different versions of his songs, and I've had one of those songs that keeps popping up in his repertoire over the decades stuck in my head, so I thought I'd use that as an example. Here's the original, from The Times They Are A-Changin' (1964). It's got a nice simple melancholy to it, which is pretty much how most of the album feels. It's probably only his second best acoustic album, after The Freewheelin' Bob Dylan, but still leagues better than Another Side, which is pretty weak in my opinion. This is probably one of the best tracks from the album, which says something for an album that contains THE LONESOME DEATH OF HATTIE CARROLL - a favorite track of IWU roommates.

Folk fans everywhere were saddened when Dylan abandoned this sound and started hanging out with the guys who would become The Band, but he didn't forget this number and mixed it up to fit their "thin, wild mercury" sound. You have to fast forward a bit to get to the song. I would have thought this would have been an easier version to find, since it is on The Bootleg Series, Vol. 4, but I guess not.

After the whole electric thing, he went back to recording acoustic stuff, famously making a short but sweet country album called Nashville Skyline, a favorite of Dan's if I recall correctly. While recording for that album, I believe, he ran into Johnny Cash, and they recorded duets of some of each of their songs as well as some others. This is one of them. It's got a very simple Johnny Cash sound to it, with Dylan using his country crooner voice. Not my favorite, but not terrible.

Just a few years later, we get a collaboration with another famous Dylan friend, George Harrison. I believe this was recorded around the time of New Morning, but I'm not sure. I kind of like it, but I've got a soft spot for that album. Overall it's not that good a version and seems highly out of character for the lyrics, but it's worth a listen, at least.

A few years later we get to painted face, cowboy hat wearing, going a bit crazy Rolling Thunder Revue Dylan, which is one of my favorite eras. This version is from Hard Rain, a live album much maligned for being listless and not as good as the first half of his tour with the RTR. I think that is part of the charm of it; the RTR was a big, energetic last push by Bob before he seems to have lost it all for a decade or so, and here we get to see Dylan rough around the edges, basically yelling into mic, breathing new life into the song.

Quite a far cry from this. Leather pants?

Alright, well, that's about it. You can dig up some more versions of it on YouTube if you want.

I Review a Movie I Have Not Seen

The Blind Side is terrible. My roommates are in love with it. I was so bored within the first fifteen minutes of it that I gave up and went to surf the internet instead. How did this movie succeed? It seems like it's just some sort of Hallmark Channel movie of the week. Does it make white people feel good about themselves, thinking that in the same situation they would do the same thing when they no doubt wouldn't? I don't get it.

Hidden Post

I had a new post, but it was rather lengthy, and due to extenuating circumstances, it didn't get done until after my last couple short posts. So, I put it up today, but I guess blogger puts things in order by when you start them, not when you finish them, so you'll have to scroll down if you want to get to it. WARNING: LOTS OF MATH.

Finally Done

To slow this blog's descent into nothingness, I thought I'd post an announcement that I finally finished my responsibilities for this quarter with a few hours of grading exams. It's very frustrating having to deal with people who lie to your face about turning things in that they didn't and people who cheat just cleverly enough not to get caught, and even worse, people who do both. Ah, well, good luck getting into med school with a C in what's basically a high school math course.

The weather was fantastic today, so after spending hours in a gray windowless office, I took the opportunity to read the internet and promptly fall asleep. Maybe I'll come up with something to put up to match Dan's bizarre short story-like post. Recombobulation indeed.

Update: Since apparently Sarah is in Milwaukee, she should be careful:

No Posts

I know I haven't posted anything in forever. I am working on that, but it is nearing the end of the quarter so it is unlikely I will have adequate time to type a bunch of stuff up because I am busy typing up other stuff when I do finally manage to figure it out. Also, grading. I just finished a marathon session of it and that was only half of the problems on half the exams. Fortunately I am only responsible for half the problems. Peace!!!!!!!

New Post

I'd had this update sitting on my hard drive for a few weeks, incomplete, before today, but I just got around to finishing it because of being sick and finals and whatnot keeping me out of the blogosphere. Now I'm on break, so "enjoy."

I've talked about topologies before, so if you are interested in reading my ramblings but don't know what it is, you can find it in some old post. Or you can just use Wikipedia like a normal person.

Anyway, many useful topologies are defined by a function called a metric, which just measures distance between two points. A metric d is defined as having three properties:

1) d(x,y) >= 0 and d(x,y) = 0 iff x = y (if we relax this last condition, it is a pseudo-metric

2) d(x,y) = d(y,x)

3) d(x,z) <= d(x,y) + d(y,z)

You can see that distance in the normal sense meets all of these conditions. In fact, property 3) is called the triangle inequality and you have to use it all the time in analysis.

The way that a metric induces a topology is pretty straightforward. You just define an "open ball" as the set of all points less than some distance from a point, which you can call the center of the ball. Picturing this in the Euclidean 3-space, known to non math nerds as just 3 dimensions in the usual sense, means spheres of some radius, not including any points on the surface. From here, you just say that a set is open if every point in the set has an open ball containing it which is entirely contained in the set. In Euclidean space, this again translates into anything that is missing its boundary. I'm not going to define boundary for you, although it is a rigorously defined thing in general topology, but it should be clear in Euclidean space what that means. For example, in Euclidean 1-space, heretofore known as the real line, the boundary of the interval [a,b) is {a,b}. So, this set is not open, since it contains some of its boundary. More rigorously, open balls in 1-space just mean open intervals, so if you try to put an open ball around a, it will contain things to the left of a, which can't be in the original interval. Then that set can't be open. It's not closed, either, but I won't get into that.

You'll note that the way we defined this topology does indeed give us a topology:

1) The empty set has no points, so it vacuously meets the condition to be open. Obviously, any open ball of a point will be contained in the entire space, so the entire space is also open.

2) If a point is in a union of open sets, it's in one of them, so it's got one of these open balls, which as a subset of one of the sets, is a subset of the union, so a union of open sets is open.

3) If you intersect two open sets and choose a point in their intersection, then there is an open ball centered at that point corresponding to each of the two sets. Just choose the minimum of those two radii and you've got yourself the open ball you wanted.

Anyway, I was just dragging my feet on undergrad stuff until now. It's time to step it up to measure theory, but not really. I'm just going to glide over all the annoying parts of trying to set up integration the Lebesgue way because it doesn't matter for what I want to get to.

Intuitively, how "big" is the interval [0,1]? It should have length 1-0 = 1, right? How about the interval (a,b)? If you said b-a, you know how to generalize, but aren't Stieltjes (which just means you are normal). Anyway, how about the interval (a,b]? It should still be b-a, right? All we added was one point, and that point should be infinitely small in a certain geometric sense.

Now, how about a union of intervals, like, say, [0,1] U [2,3]? It should just be 2, to my mind since it's just two (disjoint) intervals of length 1. And how about [a,b]U[c,d], assuming c>b; that is those intervals are disjoint? If you said b-a+d-c, congrats. So I think we're clear on how to "measure" intervals, and I'll let you work out for yourself how to do it if you want to union a countable number of intervals.

But, how do you measure something that's weird looking? For example, how big is the set of integers? Well, it should work out to be 0, since if you think about it, they don't really take up any space on the line. They're just like inch marks on an infinitely long ruler. To cut to the chase, the way that Lebesgue thought to do this was to put intervals (which we know how to measure) around sets, and call the measure of the set the infimum of the intervals we can put around it. The infimum of an ordered set is just its greatest lower bound for my intro analysis students out there. So, revisiting the integers question above with this new definition of measure makes it obvious because we can certainly cover the integers with a bunch of very small intervals, arbitrarily small, in fact. That's a sort of baby analysis problem for you, so I won't bother working out all the details without TeX handy.

Alright, so I jumped around a bit, going from metrics to measures, which seem from the names like they should be the same thing, but aren't. Now I'm going to tell you how to induce a metric from a measure, which as you recall, we used to induce a topology. Stick with me through all this terminology.

Some of the properties of a measure make it very attractive as a candidate to induce a topology not on say, the real line itself, but rather on the set of its subsets, called its power set, since a measure measures sets, not points. How would we do that? Intuitively, sets are close together if they overlap quite a bit and far apart if they don't. Less intuitively but still pretty clear, what we are really concerned with there is the parts of sets that DON'T overlap. For example, [0,2] and [1,3] have as their intersection [1,2], but so do [-100,2] and [1, 100] but this second set of measures seems more far apart than the first one. So what we want to measure is the symmetric difference of two sets, (A/B)U(B/A).

This seems to work out nicely, as (A/A)U(A/A) is empty, so it has measure 0. Furthermore, using the symmetric difference makes our would-be metric symmetric. You can check the triangle inequality for yourself, but you'll note that there is a little problem with our definition.

What is the "distance" between [0,1]U{2} and [0,1]? These are not the same sets (not the same points in a metric way of thinking), but their symmetric difference is {2}, which has measure zero, so according to our "metric" these points (sets to a measure way of thinking) are the same. That's a peculiarity.

So what do we do? What mathematicians always do in this kind of situation. Mod out.

What I mean is, these sets aren't equal, but the "metric" tells us that they are, so let's just say that they are and work from there. More precisely, let's define an equivalence relation R by saying that two sets are equivalent if their symmetric difference is 0. Now we have a new space, the set of equivalence classes mod R of [measurable]* sets of real numbers. Using the our "metric" based on the symmetric difference of two sets now gives an actual metric. So the question is, what are open sets in our induced topology? What are closed sets? Compact? Etc., Etc.

Have fun with that for a while, I'm on break.

*I say measurable because it turns out that not all sets are measurable (in the Lebesgue sense). What kind of sets aren't measurable? I don't know, but I can tell you what kind of sets are: Borel sets. These are the kind of sets that you get by performing countable set operations (unions, intersections, complements) on intervals. So, you are going to have to work hard to find a set that isn't measurable, but they are out there. In fact, because there are non-measurable sets in Euclidean 3-space, you can take apart the surface of a sphere and rotate the parts around without stretching them or anything and reassemble them into two spheres of the same size as the original sphere. I know this makes no sense, but it's called the Banach-Tarski paradox and it's one of the coolest results out there.

Converge Slow, Homie

Since a string of digits recently asked me to mention something about math on here, I'll bring up a problem I am working on that should be easier than it is.

Hopefully, we are all familiar with convergence in the numerical sense, but if not, I'll try to hand-wave at it so that even an eighth grader can understand it. I've been told a good teacher can explain anything so that an eighth grader can understand it. It seems like an arbitrary line to me, but maybe a good enough one. Perhaps that is when people start displaying abstract thinking ability.

So, we can start with a sequence. A sequence is just a special kind of function, and for our purposes, we'll stick to sequences of real numbers. As to what a real number is, it's just about any kind of number you can think of that doesn't involve i somewhere. So whole numbers, 0, fractions, even irrational stuff, like 2^(1/2) or pi.

That said, a sequence is just a function of natural numbers, so something like

1, 2, 4, 9, 16, ... you can see how this sequence "goes to infinity," in that it just keeps getting bigger (I am purposefully being vague about this concept). On the other hand, the sequence s(n) = 1/n, that is

1, 1/2, 1/3, 1/4, ... doesn't keep getting bigger; it keeps getting smaller. However, it doesn't "go to negative infinity." In fact it demonstrates the central idea of calculus, which is convergence. In particular, it is said to converge to 0, or that the limit as n approaches infinity of s(n) is 0. What do I mean by that? I mean that we can think of this sequence as approximating 0, as if I didn't know what 0 was, but I was guessing at it, and each time I guessed, my guess got closer. The sequence is said to converge to a number if it approximates that number to any error. More formally,

A sequence s(n) is said to converge to a number L if and only if for all E > 0, there exists an index N such that if n > N, |s(n) -L| < E.

If you think about it, it just says that the sequence gets as close to L as we would like and stays at least that close; that we can approximate L infinitely well with a big enough term of s(n).

A proof of convergence usually goes like this: Given E > 0 (we actually use epsilon, usually)

|s(n) - L |

(bunch of algebra with inequalities)

< E for n such and such

That is, we usually set E and find an expression for N in terms of E that suffices. In the simple case above, you just set N = 1/E and you're good.

Sometimes it isn't so easy, and that is what I'm dealing with at the moment. The sequence I'm looking at is

S(n) = 1, 1/2, (1/2)(3/4), (1/2)(3/4)(5/6), ...

And so on. The denominators are just the product of the even numbers and the numerators are the product of the odds, always smaller. Each factor is less than 1, so each term is smaller than the last term, but that's not enough to show that it converges to 0. One idea might be a comparison test.

That is, if I can show that for any positive integer k, there's a positive integer N such that S(N) < 1/k, I can just compare it with the previous limit problem and say the limits must be the same. [It is easy enough to show that a sequence of positive numbers cannot converge to a negative number, so the new sequence must be "squeezed" between the old 1/n sequence and 0.]

Just working out some terms of the sequence explicitly, I've found that the limit must be less than .15, and I'm convinced that it is actually 0; that I can somehow show it is squeezed down by 1/n if we look far enough along the sequence. The problem is that this convergence is very slow. You'll note that each subsequent factor is bigger than the last, in fact, the last factor converges to 1. However, they still are less than one, so they make each term decrease, just by less and less. It is rather annoying and making it hard for me to find the right expression or technique.

Anyway, it is just part of a somewhat bigger problem related to a theorem of Tauber.