Oh happiness and felicity! Classes are done!

Ahh, today is a day of great rejoicings! I have finished and turned in the last homework assignment for this semester (true, I have a take-home test for Partial Differential Equations due on Monday, but that is much preferable to an in-class test in that class). I was highly honored today by Dr. Figueroa asking for copies of my last two homework sets to use as the basis for the answer key he was compiling for them, since I typically get most of the problems right (I may have mentioned that the free textbook we're using has no answers, so he has to compile answer keys for the homework he assigns himself).
Speaking of the most recent homework we had for Complex Analysis, I would be remiss if I didn't mention the Residue Theorem as promised. The Residue Theorem is often called "the crown jewel of complex analysis", and now I know why. Problems that previously would have been incredibly difficult, or even impossible, are now a snap. I didn't get a chance to work on the homework until Monday, and expected to be spending upwards of ten hours on it, and was very pleasantly surprised as problem after problem fell to the power of the Residue Theorem. I finished approximately 90% of the problems in less than two hours, and took only about another hour for the remaining ones. The feeling was one of delight, as previously-impossible problems became simple ones. I can hardly wait to actually use it in a physics application now.
The details are a bit deep (it took a whole semester to build up to them!), but basically, it says that when integrating over a region that contains one or more singularities (places where division by zero occurs), the integral is equal to 2πi times the sum of the residues, which are the values of the function being integrated attains at the singularities. Since plugging in the number directly would be undefined, residues have to be found another way, which can be as easy as a single differentiation (though they can certainly be difficult to compute, as well). The beauty of the theorem, is that it completely removes the need to actually integrate anything, which is very nice (differentiation is a science; integration is an art [integration is the inverse of differentiation, if you haven't taken calculus, and while you learn differentiation in first-semester calculus, it takes the rest of that semester plus another three to really get a handle on integration]).
Tomorrow night, I have been invited by Dr. Takamiya to go on an observing run. We won't be going up Mauna Kea; instead, we'll be controlling the telescope from the Institute for Astronomy building near campus (I'm not sure which telescope it is, though. I'll let you know when I find out). I also don't know how long I'll be there, it could be an all-nighter! I'll be sure to let you know all about it Friday.

Excursions into the infinite complex plane.

So much has happened since I last wrote ... I'm still working on making sense out of it all.

Homework has been keeping me busy, as usual. I once again have several things due over the weekend, so I will be busy working on that. Between electromagnetism, partial differential equations, modern physics, and calculus IV, I have plenty to keep me occupied.
As if that weren't enough, I got my computer back from the campus tech guy, who was unable to find anything wrong with it other than the fact that the computer simply refuses to acknowledge the wireless module, and it still locks up, with no pattern that I can seem to figure out, yet in an oddly non-random way. I'm unsure which way to go at this point ... I hardly have time to deal with warranties and such, but I could really use my computer working fully as the end of the semester approaches.

Before I forget, I'd like to say a big 'thank you' to my aunt and uncle who were out here this week for everything -- the dinners were great, and the conversation and catching up were awesome. It was really nice to take a break and focus on something other than school for a while. So, mahalo nui loa! (thank you very much)

 In other news, I managed to solve another problem for my Complex Analysis professor today, which was the question of the convergence or lack thereof of the geometric series in the complex plane on the unit circle. If you don't know what I'm talking about, don't worry, you're not missing much. Infinite series and sequences are some of my least favorite subjects from Calculus II, and from the people I've talked to, many others feel exactly the same way. They inspire...strong feelings, to put it mildly. This is because they are very complicated, and most of the time you work with them is spent trying to figure out if a particular series converges or not. (Convergence means that if you add up an infinite number of terms from the series, you will get a finite number. Divergence simply means that you get infinity. This can be a surprisingly difficult thing to do) The geometric series is one of the few series for which you can actually find what it converges to; its terms look like \(x^n\), for n from zero to infinity. (Its first few terms are \(1 + x + x^2 + x^3 + ...\)) It is well known that for any x less than 1, the series  converges, while for any x greater than 1 it diverges. This is because its radius of convergence is equal to 1, and it turns out that every infinite series has such a radius, inside which it converges and outside which it diverges. On the real number line, the word 'radius' doesn't seem to make much sense, because there are only two points for the boundary, but in the complex plane it makes perfect sense because you are dealing with a plane in which numbers can approach from any direction, instead of merely from the left or right.
Anyway, in the real case, it can be shown that the geometric series diverges at \(x = 1\) and \(x =-1\), for different reasons. But that's only two points. The question in the complex case is, what happens for all the other infinite number of points on the circle of radius 1 around the origin? (written symbolically as \(|z|=1\)) However, there is a rather simple (once you see it) and elegant proof that the geometric series diverges for all \(|z|=1\), and I was able to be the inspiration for my professor that enabled him to discover it today (though as usual, the proof he came up with was much more elegant and simple than the clumsy one I had cobbled together). 

Building a working cloud chamber, part 2.

Today was a rather busy day for me. I and my partners in Modern Physics met again today to try and get our cloud chamber project working. It was a rather mixed session. Initially, we had high hopes, as we had corrected several of the problems that plagued our first generation attempts. However, despite doing everything right that we could think of, we were still unable to see even the faintest wisp of cloud. After repeated attempts, we were about to give up in frustration and think about finding another project altogether, when I suggested for lack of any better idea that we invert the setup to have the heat source on the bottom and the dry ice on top, to create conditions more in line with the earth's hydrological cycle. Lo and behold, almost immediately after flipping everything over and applying a slight vacuum we observed some extremely faint clouds. However, this success was tempered by the fact that the clouds were too faint to actually observe any particle tracks, even after repeated attempts fiddling with various parameters.
We finally stopped again after a while, with several more ideas for improvements and the fact that we had managed to observe something. I'm attaching a humorous picture of our setup below; let me stress that the vapor in the picture is not the cloud we saw, it's from some dry ice I put in the container to create a nice visual effect. We never saw anywhere near that much cloud from the alcohol.

It's the intrepid radiation-carrying Lego man of science!

Yes, that's our radiation source that he's carrying on his head. It fits perfectly into Lego-sized openings, so we used a Lego man to hold it up and keep us from losing it. What you actually see is just the aluminum casing, the americium is only exposed in a tiny portion on the very top.

My classmate whose apartment we were running the experiment at lives right on the beach, and I actually saw a whale breaching while I was there. It was out in the ocean a couple hundred meters offshore, just expelling a huge cloud of water vapor (a lot more cloud than we ever saw in our experiment!).

Doing my latest round of complex analysis homework today, I had some really mixed feelings. Now that I've reached this point in the semester, I know enough theorems and propositions to actually accomplish some impressive things. I'm struck by how often a really complicated looking integral can be evaluated very simply, and turn out to be quite easy in the end. It's taken two months to get to this point, two months of building the scaffolding theorem by theorem, problem by problem, with little (up to this point) to show for it, so it's nice to finally be able to really do stuff.

The Residue Theorem

Today was, all in all, a rather interesting day.

To understand why, I need to back up to Thursday night, when I was finishing up my homework for Complex Analysis. Of the ten problems assigned, nine of them took my twenty minutes or less to complete (because we'd gone over how to work them in class already). The last one was a stumper. In principle, it was a piece of cake after the problems I'd been doing: simply perform an integration over a square region over four different complex functions. Yet over the next three hours I attacked that problem with every trick, technique, or theorem I know of in both the real and complex plane, to no avail. I finally gave up frustrated at 10 o`clock, because I had been up late for the last few nights and really couldn't stay awake any longer, with the intention of working on it in the morning before class.

Before retiring for the night, though, I hit upon a method to compute the integral, in principle, assuming several conditions were met. So this morning I began chipping away at the problem again. After removing all the difficulties to my satisfaction, I proceeding to the tedious (for me) work of actually writing out the problem, made all the more tedious because I anticipated several pages of computations.

And that's when it hit me: after doing a bit of preliminary setup, the closed nature of the path of integration meant that most of the results canceled in pairs, leaving only a simple complex integral, which I did in about half a minute. You can hardly imagine my elation when I realized that I was only dealing with a page or two of writing for all four integrals, instead of the the two pages each I had been envisioning.

Even with the reduced amount of writing, I still wasn't able to finish it before class, but the first thing Dr. Figueroa did was to announce that the homework was postponed until after Spring Break, because he hadn't been able to work out that very same problem with the material we've covered so far, despite staying up till one in the morning working on it (we're using a free textbook from off the Internet, which doesn't come with solutions). He mentioned that the problem could be solved using techniques later on in the book, using something called 'the method of residues' which caught my ear because it sounded like an apt description of what I had found: most of the problem disappeared, leaving only a 'residue', a single, simple, complex integral. When I mentioned I had figured out how to solve the problem, he was intrigued, and had me demonstrate it at the white-board. When he figured out where I was going, he got quite excited, jumped up from his chair, and immediately proceeded to put it all on a firm theoretical footing, then discovered a way to reduce it from the clumsy sum of four parameterizations I had been using to an elegant single parameterization over a circle, which reduced the problem down to a few lines. I think I made his day; he was still pleased as punch when class ended.

I did a little digging, and apparently what I discovered (arithmetic errors aside; I didn't have time to fully develop it) is know as the Residue Theorem, and is a standard and well-known tool in complex analysis.

The other interesting part of my day was the part where I and two classmates attempted to build a working cloud chamber for our project for my Modern Physics class and the fun I had with some of the leftover dry ice, but it's getting late, and that actually has some pictures to go with it, so I will save it for tomorrow when I can really do it justice.

More busyness.

I realize I haven't written anything for a couple days, mostly due to the sheer number of homework assignments that seem to have conspired to be due around now. Right now, I have Modern Physics homework due tomorrow (thankfully completed tonight), Partial Differential Equations homework due on Friday and a 2-page writing assignment in the same class due on Monday (it's a writing intensive course). I also have an Electromagnetism test due on Friday, as well as a currently unknown(!) number of Complex Analysis problems ("at least ten") due the same day. So that is why I haven't written anything recently.

However, before wrapping up tonight, I'd like to mention the results of a test I ran to compare the processor speed of my old and new computer. I found a rather large and detailed scene I'd made in Blender, a 3-D modeling program, and rendered it on both computers. The results? Old computer: ~2:05. New computer: 0:09.75, or a whopping 12.82 times faster. It's interesting that both processors are rated at the same speed, 1.6GHz, the main difference is that my new computer has 4 cores running at that speed compared to one in my old (there are a few other small differences as well). Now, the speed increase is probably not due entirely to the processor, since this computer also has 16 times more RAM and a 1GB video card (my old computer doesn't even have a discrete video card), but the processor is certainly a large part of it. So next time you're considering a new computer, take it from me: quad-core is the way to go. Dual-core, at least.


(I hope to eventually expand this into a sort of series of reviews doing point-by-point comparisons of my new and old computers in various categories. We'll see how far I get.)

On the method of finding roots of unity.

Remember that method of root-finding I worked out last night? Turns out it was an important part of an attempt to prove Fermat's Last Theorem, one of the most famous theorems in the history of mathematics. In other words, probably every mathematician worth their salt already knows about it. (Fermat's Last Theorm says that if you have the equation \(x^n + y^n = z^n\),  there are no integer values for x, y, and z (other than 0) that will solve the equation for \(n\ge3\). It was proposed in 1637, but not actually proven until 1995. Yes, that recently.) I found this out because it got brought up in a talk by our Mathematics department chair Dr. Lee that I attended this afternoon, complete with illustration on the board of the method I was using. I learned a few other useful tidbits of information about it while listening, so I'm going to set down a formal notation of the process here, before I forget it.

Start with your equation, \(z^n=\pm1\) Depending on whether 1 is positive or negative, the formulas for the roots take different forms. Since the roots will share several features in common, it is helpful to denote them by a common symbol. I'm going to use omega, \(\omega\), for variety (and 'cause Dr. Lee was using it).

In both cases the roots are given by \[\omega = e^{(i\pi k/n)}\] however, if \(z^n=+1\), the number k is given by \(k = 0, 2, 4, ... , (2n-2)\), whereas if \(z^n=-1\), k is given by \(k = 1, 3, 5,  ... , (2n-1)\).

I had originally thought that the evenness or oddness of n would effect the formulas, but upon looking over the examples I worked out, I see that it doesn't. The only thing that changes is the range of the number k. Note that one interesting thing happens if n is even and 1 is negative: all the roots will be complex numbers, which makes them very difficult to find without using an explicit method like this.

 Well, once again, it looks like I have managed to independently discover something that was already worked out long before me. Perhaps, someday, I will actually be able to produce something original.

A novel method for finding nth roots.

By dint of hard work, I have tonight come across a previously unknown (to me at least) method by which one can find higher order roots of equations of the form x to the nth power equals a constant. Since in several hours of Internet searching for just such a solution I did not come across anything comparable, I'm going to set it down here so I don't forget it later on.

The problem arose while I was working on my Complex Analysis homework. I needed to find the roots to the equation z to the 6th power equals 1. In other words, I needed to find 6 roots of 1, each of which, when raised to the 6th power, would equal 1. Now, I know that I learned procedures to do just that back in algebra, but I could not for the life of me remember them tonight. After some fruitless searching on the Internet, I came across a Wikipedia article on roots of one. It too did not contain what I was looking for, but it did contain some very useful pieces of information, among them that the n roots of one were always evenly spaced on the unit circle in the complex plane, and that they formed a regular n-sided polygon within the unit circle.

Now, if you remember complex numbers from algebra, you know that they can be written in the form \(a+bi\), where a and b are real numbers. However, if you do some playing around with very complicated math like infinite series, you can derive Euler's formula, which says that e to the \(i\times x\) power equals the cosine of x plus i times the sine of x, or, in symbols, \(e^{ix} = \cos(x) + i \sin(x)\). This allows complex numbers to be written in the equivalent polar form \(re^{ix}\), where r is the absolute distance (positive) from the origin to the point in question, and x is the angle from the real axis in the complex plane.

Since the vertices of the regular n-sided polygon formed by the roots of the equation you're looking at are evenly spaced, it's a trivial matter to quickly write down all the roots of the equation in polar form just by reading the values of the angles for the polar form off a well-drawn picture. You can then undertake the almost-as-simple process of converting them back into the more familiar rectangular form of \(a+bi\).

As an example, take my original problem of \(z^6 = 1\). The picture, hopefully, will make things a little clearer. There, I've marked the polygon that the 6 roots make inscribed on the unit circle. In the complex plane, the x-axis is referred as the "real" axis and the y-axis as the "imaginary" axis. With a number of the form \( a+bi\), a is the real part, b the imaginary part. When working the problem, start off by checking if 1 is a solution. In this case it is, so one vertex of the polygon must be at the point (1,0), and the coordinates of the rest of the vertices can be determined from there. With the polar form, only the angle is necessary, and I've marked the angles of all the vertices for you. Once you have the polar form, you can convert it to a rectangular form using the equations
Real part = \(r \cos(x)\)
Imaginary part = \(r \sin(x)\)
where \(r = 1\) since we're using a unit circle.

Anyway, to cut a long process short, since it's getting late, I will simply give you the 6 roots of the equation given first in their polar forms, derived simply by reading off the diagram, then in their more familiar rectangular forms, arrived at using the definitions just given.
\[z=1,e^{i\pi/3},e^{i2\pi/3},-1,e^{i4\pi/3},e^{i5\pi/3}\\
z=1,\frac{1+\sqrt{3}i}{2},\frac{-1+\sqrt{3}i}{2},-1,\frac{-1-\sqrt{3}i}{2},\frac{1-\sqrt{3}i}{2}\]When I have a bit more time, I'll work out a formal notation for the process, so it can be a bit more rigorous and useful to people encountering it for the first time. Until then, a hui hou!

Final musing: what do you think the chances are that this has probably been worked out hundreds of years ago already by much greater mathematicians than I? I'll ask my professor tomorrow...