...Prove Their Worth...

"Problems worthy of attack
prove their worth
by hitting back." - Piet Hein

A kind of running diary and rambling pieces on my struggles with assorted books, classes, and other things, as they happen. You must be pretty bored to be reading this...

Tuesday, May 21, 2002

To quote Mr. Cartman, "Fuck fuckety fuck fuck fuck." This is really terribly frustrating. What, you may ask yourself, has reduced me to quoting South Park? It's not still not knowing how to prove that there's an M(z) taking two non-concentric non-intersecting circles to concentric circles. (Though I still haven't solved that.) It's not having a Tom Petty song stuck in my head. (That's actually the case, but I don't mind.) And it's not even the (relatively stale by now) news that Steven Jay Gould died. (Though that's definetly a downer.)

No, indeedy. See, as talked about in the last entry, I was struggling with a particular problem on Mobius transforms. I've made no headway. So I thought I'd go on ahead. And there are some neat problems that crop up toward the end of that chapter, let me tell you. Problem is, I've found that I'm temporarily sick of M(z)s. So I decided to just read the next chapter. This was a good decision, as it turns out, because just as I remember, this is a fun chapter - basically, 'differentiation of complex functions' (I'd read it before).

So, you ask, what's neat about it? Well, for instance, there's the concept of the derivative as an amplitwist. What this means is that if you take the derivative of a complex function at a point (f'(z)), you get a complex number. Now, what does that number mean? Well, multiplication of something by a complex number (r*ei*theta) means you're expanding that something (that's the 'r' part), and rotating it (that's the ei*theta). Another way of saying that is that you're 'amplifying and twisting', or amplitwisting, in short. So, that complex number that you get from taking the derivative tells you how you need to amplitwist an infinitesmal vector poking out from z to get the 'image vector' poking out from f(z). So, if instead of thinking about points, let's think of tiny little figures, triangles, say. Now the 'amplitwist' tells us how a teeny weeny little triangle around z is going to twisted and amplified by the transformation we've just differentiated. As expected, it tells us how an infinitesemal chunk of a the transform behaves.

Along the way in figuring all this out, there's a brilliantly simple derivation of the Jacobian, and an extremely illuminating derivation of the Cauchy-Riemann equations. (Basically, take the concept of the amplitwist, combine it with the concept of the Jacobian, and the Cauchy-Riemann equations pop right out!). There is other good stuff, too.

But what got me peeved is the following. The derivative of a constant better be zero, right? Right. It's the same in complex analysis. So, back when I was reading the chapter, it had this statement, and warned that I must make sure to see it, and that it isn't obvious. So I sat, and huffed, and puffed, and concluded that I did see it, and while not entirely obvious, it was really quite simple, and even obvious-like in a crafty way. Then I moved on. Now I'm doing the associated exercises. And lo and behold, I'm asked to show why the derivative of a constant is zero. And I've forgotten how to do this! What's worse is that I KNEW how to do it. I just forgot. And I can't seem to remember.

"Fuck fuckety fuck fuck fuck"

Saturday, May 18, 2002

Hrm. Well, I'm still stuck on that particular problem, and it's gotten frustrating enough that I spend about twenty minutes on it in the evenings, then move on to something else, lest I go more nuts with frustration, and assault any rabbits and/or cats I run into at night while jogging with warbling [ala Morissette] romantic serenades to cactuses.

The something else in this case means rereading Chapter 3. Currently, I'm reading about the visualization of a generalized M(z). It turns out to be quite an interesting matter, with lots of nice pictures involved - various whorls and curves and so on. I saw that it was interesting the first time through, but now I find that I'm following it a lot better (perhaps it helps having read the chapter on complex differentiation and analytic functions and so on, which comes later).

In true form for myself, I've of course run into problems with the material. Amusingly, it's not with the core of the material as such. As part of the setup-attack, Needham juggles various constructions and transforms and compositions of transforms, and uses 'and clearly' several times. Somehow, I always get bogged down when people start farting around with rapid combinations and discombinations of transforms. If I go slow, furrow my brow, and carefully and deliberately plow through, not forgetting to alliterate compulsively, I follow things, but if i just 'read', I don't. :-( Anyways, onto things I do follow.

It makes sense to me that you can pull a 'multiplier' out of Mobius transform (it's not unlike pulling a tooth, but it's not like pulling a tooth, either, as it's not fundamentally and irrevoccably unpleasant, and the weaponry of basic linear algebra* [i.e., squeezing an eigenvalue out of the matrix representation of the transform] acts as an anaestetic.]) I'd even puff my chin out enough to say that I grok how the nature of that multiplier determines the 'spirit' of the transform - if it's a dilation, the transform is elliptic (curves concentrating around the fixed points of the transform), if it's a rotation the transform is hyperbolic (curves from one fixed point to another), and if the multiplier is a combo of a rotation and a dilation, the transform is loxodromic (looks like cool whorling curves from one fixed point to the other). The first two types of transform, it's neat to note, look a whole hell of a lot like EM field lines, and the loxodromic one might look kind of like the field of one EM 'pole' moving past another (?). Who wants to bet there's a major application here, and they actually are EM field lines?

Here's the picture I've got in my head right now. It's probably completely wrong and insane, but so it goes. A Mobius transform can be looked at a linear transform on C2 acting on homogenous coordinates in C2 of a point z, which lives in C (note: homogenous coordinates just means you label a point z by a ratio of two complex numbers, which are then your 'homogenous coordinates'. Sounds pedantic, confused, and fancy, but it's all made clear on p. 158) That reminds me of the obscene volumes of delicious Japanese food I consumed today at the Hinode during the 10 buck/person lunch-time buffet. Mmm. But moving on...

Now, if you want to avoid headaches (a good thing in general if one isn't Zeus - not everyone can get someone as cool as Athena out of a headache), you'll deal with Maxwell's equations in an explicitly relativistic mindset - with four-vectors and Fuv and all that shit. I think. I.e., you're now playing on Minkowski's playground, and need to deal with four dimensions.

Now, I also remember reading a snippet about spinors somewhere - they're column vectors of two complex numbers. Now, I know Penrose came up with them, and I know for a FACT Mobius transforms have a lot to do with relativity (the book says so, and refers me to Penrose's work on the subject - I looked at it, but it's too hard-core for me. I'm still on the soft-core diet, and I'm not ready for the 'German scheisse video' (in South Park lingo) of math books. So, apparently, you can do EM with spinors, and there are advantages to this. Now, what I'm ponderously siddling up to is that you can pick up this 'homogenous coordinate' shit by its collar, give it a good shake and poke it with a sharp stick in its tender spots, until it admits it's actually a spinor, and then do EM from this perspective. Which is why pictures of stuff like Mobius transforms ends up looking suspiciously like EM field lines.

Note to the gullible (i.e., self): This is quite likely to be amusingly wrong, so don't actually bet on it! But it'd be funny if it was even remotely close to the truth.

In other news, I have a new watch, one that I can actually read, due to the face not being so scratched up as to raise questions of whether I tame almost wild grizzly bears in a circus for fun and profit. And it's pretty, so I'm actually at some risk of getting mugged for it - the previous watch was more likely to shake a would-be mugger into giving me his watch out of pity. Also, I got a book (as a present) called "The Glass Bead Game", by Hermann Hesse. It's quite thick. It won the Nobel Prize in Literature in 1946, and people I trust say it's really, really good. So it's probably really good. Look forward to reading it.

My fingers are tired.

* -- Note to self: a decent online text for linear algebra exists. Read it sometime. You really, really need a refresher. [cough]

Wednesday, May 15, 2002

OK, so I'm stuck on a problem tonight (again), not unlike a moth that wandered into an attractive shoebox, and can't get out (moths are pretty stupid, if you'll recall). Now I'm buzzing around inside the shoebox, hitting its walls at random times with surprisingly solid thumps. That's figurative, of course, for all i readers out there - I don't normally buzz, as such, and I rarely hit walls (that only tends to happen in the dark, or when I'm distracted by, say, attractive human females or some particularly interesting representative of the local fauna ... for quite different reasons, I assure you!) Though I'd have to admit on cross that I actually might be said to 'buzz' after a solid dose of some hypergolic substance, such as Mexican food. Said buzz means it's generally time for any bystanders to head for the shelters, by the way. But that's getting rather off-topic (good thing I caught myself before I ended up talking about the hot and heavy relationship between a certain comely Floridian manatee and Donald Rumsfeld in a third-degree parenthetical). Which is amusing, because I haven't actually defined any topic yet. My rambling talents are impressive, n'est pas?

So, you i readers are probably wondering what problem it is that has me thinking about being a moth. It's a simple-sounding one, really. For future reference, the problem in question is Exercise #10 in the Chapter of Pleasure Spiked With a Considerable Amount of Pain<, a.k.a. Chapter 3: Mobius Transformations And Inversion. So. I'm being asked to prove the following: Any two non-concentric non-intersecting circles can be mapped to concentric circles by means of a suitable Mobius transformation. (From here on out, I'll refer to a Mobius Transformation as M(z), to save typing and make things more cryptic, and also to save me the headache of trying to figure out how to get a double-dot above the 'o' in Mobius, which ought to be there.)

Now, as the first step of the proof, it's suggested I consider the following: Grab two circles, A and B, as specified above, by the lapels, and show that there exist two points that are symmetrical with respect to both A and B. Now, the first time I sketched this, while sitting on the Throne Of Power, this seemed as helpful as single-ply TP to a guy with hemmorhoids and a spelling problem. Further reflection (read: taking care of nature's business, which I promise not to discuss further here) allowed me to come up with something interesting. Draw a line through the centers of A and B. If these mystical symmetrical points exist, they damn well BETTER be on this line, right? I mean, it just makes sense, if you squint at it enough!

Now, squint some more, and don't forget to blink thoughtfully while furrowing your brow. It'll start to make a hazy kind of sense that one of the points ought to be 'close' to the center of the inner circle, which means that the symmetric point ought to be 'far away' from the small circle's circumference. But if you get the position of the point right, the symmetrical point ought to be so 'far' away that it'll be beyond the circumference of the outer circle. Fiddle with the position of the point on the line a bit, and you should be able to get it so that the image point is symmetrical with respect to both the inner and outer circles. Or so the voices in my head tell me.

The above, for the uninitiated, is known as a handwaving argument, due to the intellectual similarity of the user of said argument to a lizard that just found out from a book that it shares an ancestor with birds. It kinda-sorta makes sense, but it's quite far from an actual proof, or anything concrete and analytical and mathematical. How to actually get from the "I think I'm a turkey, I can fly if just flap hard enough!" stage to the "Eureka! Here's the solution" stage is too complex a process to fit inside this margin sometimes referred to as teh Intarweb.

Ok, enough with the testing. I can't get mozblog to work, which is a shame (and given the fact that it seems to work for others, it's a poor reflection on my l33t hax0r-ness.) Let's try a more substantive update.

Since I last blabbed, I did get the decomposition of the Mobius transform composed. As it turned out, I was simply being a huge tit in my attempts to reverse the given decomposition, managing to completely fuck up applying the very basic concept of what the fuck a transformation is. Through the highly technical techniques* of vigorous squinting, mumbling, and the consumption of three (3) strawberries, I saw the errors of my calculations, carefully applied the definition of a transformation four times in succession, and got the Mobius transformation. I could then backtrack and actually do what I set out to do, which is figure out exactly where the decomposition comes from. Surprisingly, it does not come from Mozambique.

* - Somewhere, somewhen, an English teacher just got a stronger than usual urge to violate a pillow with an axe, and doesn't know why.


testing, again.

Monday, May 06, 2002

So, I'm kind of stuck on figuring out how to decompose a Mobius Transformation. I know how it's supposed to end up: three dilation rotations, and one inversion. And I even know the exact formulas for all of them - they're given. And I can even say that it makes sense that there ought to be an inversion and some rotations making up the Mobius Transformation. But I can't seem to derive the decomposition. I've tried going backwards, from the given decomposition to the transformation, but even that failed: I ended up in a bizarre thicket of thorny and foul algebra, which didn't look right. Grr. I suspect I fucked up in applying the transformations successively, though - wasn't paying enough attention. So I'll need to try again. Sigh. So it goes.

Testing, Testing, 1 2 3.