Annoyance

Well I decided that this year I would like to keep better track of what I listen to, so that when I make the "Best of 2009" post the numbers will be more accurate. Only problem is that I can't for the love of me enable the Last.fm scrobbling program while using Ubuntu. Well that's not entirely true. I've gotten it so that it recognizes the song, the only problem is that when I get it to do this I am unable to actually play the song with my media player. I've looked around and have a general idea of what is causing this problem. Only thing is that I haven't been able to find a solution, well I'm sure I could but right now it's not at the top of my priority list. In addition to this I use pandora quite a bit, and I'm not sure if it keeps track of your listening trends the say way Last.fm does. Oh well.

So I'm using my Windows XP computer for the first time in a few months. One thing I will say is that I miss having a gcc compiler. I have Dev-C++ which usually pretty close to being the same but it's just not the same. I looked around and found a way to get the compiler I want for XP but don't really feel like doing all the necessary work right now, I'll probably end up doing it around the time of finals. The only reason this is a bother is because when I submit solutions to SPOJ problems there is a small chance that there might be a compilation error. It would be really cool if there was an online "compiler" so that I could check before submitting (this is when I'm not writing the program under Ubuntu of course).

One last thing, for those of you who continually ask "What are you going to do with a math degree?" or "Why math?" Here is an answer I'm sure you'll like. Notice how the top rated jobs are mostly math/science related.

Efficient Finals

Well I was reading an algorithms book and here is something that was pointed out that I never considered before.

Assigning times for Final Exams is equivalent to coloring a map.

How so? Well assume that we only have the following restriction, two finals can not be at the same time if they share at least one student. Now let the classes be the vertices of the graph, and place an edge between any two vertices if they share a student. So basically you assign colors (or times) in a way such that no edge has vertices of the same color.

Obviously the four color theorem does not apply when coloring this graph. Just consider the case where a student is taking 5 classes. (go and look up to see how this breaks the necessary assumptions for the theorem). None the less it is still an interesting connection, though I am sure Universities do not use this method to schedule their finals (for some obvious reasons).

IMPORTANT NEWS

Well for unknown reasons (well not totally unknown) I can no longer use LaTex on this blog. I did look up ways to get this back but ran into a few problems. Most of the solutions only worked for Unix like systems, and I would like to be able to use LaTex on the blog even if I was on a Windows machine. Also most of these solutions would have required me to add another version of Tex to my linux system and I didn't really want to do that. There was however, one solution that would work with all systems (supposedly). All I had to do was edit the source code for a GreaseMonkey script, but after doing that I was still running into the same issues as before so I quickly gave up on that dead end.

I will still be using this blog but whenever I want to post anything mathematically related I shall post a link to my WordPress blog in the post. The reason I am using WordPress is becasuse it has LaTex built in and thus less work on my part.

I Have Wireless

So I have finally managed to get my wireless card working with Ubuntu 8.04. You have no idea how happy I am about this. I have been wanting to really give Ubuntu a try but have always been reluctant to because of the fact that it wouldn't work with my wireless card and I would need an ethernet cable in order to connect to the internet. Now I know this happiness will not be too long lived because once I update to the next version of Ubuntu (which in my case will probably be 9.04 and not 8.10) I will most likely have to go through the same process again and hope that it still works on the new version.

On a more "depressing" note; I managed to finish only half of one my my 5 analysis homework problems. They all seem to be about connected sets. One of the funny things is that in Rundin for a set to be connected it must not be able to be written as the union of two separated sets. However, on Wikipedia, Mathworl, and one of my other analysis text they also say that the separated sets must be open sets. Now I know that if two open sets are disjoint they are separated so I don't see the real need for this extra assumption. In any case the problem I managed to solve was as follows.

Let A and B be two connected subsets of a metric space X. Show that is A and B have nonempty intersection then their union is also connected.

The second half of the problem asked us to state and prove a generalized version of this for the union of arbitrary connected sets. I came up with:

Let {A} be collection of connected sets. If for each F in {A} there exist a G in {A} such that the intersection of F and G is nonempty then the union of all the members of {A} is a connected set.

solution

This is the solution for problem #1 in my previous post.

Let , since is a Cauchy sequence there exist an such that if if Similarly since a subsequence converges to we have that when . Now let be the larger of and .

Now .

Here is the solution for #2 in my previous post (this one took a little longer for me to think of).

Let be a sequence in such that . Now from properties of we have that for all Now let , and since we have that for large enough. So since is complete. Additionally, is a limit point of each , so .

From the Integers to Infinite Series

In the upcoming fall semester I will be taking the first year graduate analysis class. Hopefully, this class will be more "rewarding" than the undergraduate analysis sequence I completed last year. The undergraduate sequence was not a total waste of time since I actually did learn some new things, but the class moved so slowly and the test were not all that difficult. I never really had to exert myself in order to obtain my A. As a result in the second semester I tried to get a 100% on every assignment and test. I almost achieved this goal, however, I did not make a small clarification on one of my proof thus resulting in me losing 1 or 2 point for the problem.

The graduate class will be using the same book as the undergraduate sequence. However, we will hopefully cover all of the sections. Last year the professor would pick and choose which topics he though were necessary and possible for the majority of the class to understand. As a result some of our proofs were longer than needed but I suppose knowing multiple ways to prove something isn't all that bad.

Right now I am reading through as many chapters as I can and working all the exercises. My only requirement is that I can only be working on problems from at most two different chapters and that I can only be reading one chapter ahead of the problems I am solving. So basically right now I am working problems from chapter 2 and 3, and reading chapter 4. I am almost done with the problems from chapter 3, but the ones from chapter 2 are going to take a little more time. I suppose that it has to do with the fact that dealing with topology is still a little "new" to me.

Tomorrow I will hopefully be able to do the following problems:

1) Suppose is a Cauchy sequence in a metric space , and some subsequence converges to a point . Prove that the full sequence converges to .
2) Let be a sequence of closed, nonempty, bounded sets in a complete metric space . Also and . Prove that contains exactly one point.

I will eventually get around to making a post about my REU in Michigan. Sorry to disappoint, but I just need some more time to better collect my thought on this topic so that I will have a post worth reading.

REU 2008 - Week 1

Don't expect an entry for every week of the REU but I figured that the first week was important enough to have its own.

When I first moved into my apartment I was pleasantly surprised by the size. My first impression was that it was way too small for four people to live comfortably. However, after spending some time here I quickly realized that I was mistaken. The roommates are pretty cool, well actually I have only met 2 of them, the last one will be arriving on Monday. The first one I met had just recently returned from Africa where he was helping install water filtration systems in a village. I don't remember the name of the country but it was a French speaking one. The other roommate is a year younger than me and it kind of shows in ways, mainly when we are working in groups (he is a member of my group in addition to being my roommate). The three of us that live here so far are all math majors and the one remaining roommate is an engineering major.

The research is going extremely well. At the beginning we would "play around" with the problems and find a solution. Then our mentors (I can't think of a better term for them, I guess advisers works but that makes it seem like we are PhD students) would present us with a paper which contained the result we had just discovered. You would think that this was disheartening but it actually was not. Having actually come to the conclusion ourselves made reading some of the proofs much easier since we had a better understanding of what they were doing and could relate some of their ideas to our own. However, we have so far have come across a result which we have not been able to find stated in any paper. I guess you could say we have our first theorem, and it's quite a wonderful feeling. We were surprised by the theorem actually. This is because based on the question we were answering one would expect the result to be complicated. However, it is quite the opposite. When we first saw it we were all skeptical and furiously searched for a counterexample but could not find one. After going through it again we were certain that it was correct. All that is left to do for this finding is to actually write up the proof of the theorem and the necessary lemmas.

Well I'll keep this one relatively short and end by saying that it seems like I will be going to the beach quite often (even though we have to climb over a "mountain" to get there). So far this week I have gone three different times.

Sum of Two Squares

I know that I haven't had a post about math or anything math related for a while, so I think it is time to change this.

Well the problem discussed here really isn't that difficult but it is interesting enough to mention. "Given an integer determine if can be written as the sum of two squares." For example if , however, if instead we had it can be shown that it is not possible to have where .

Sure you could try using "brute force" to solve this but that is boring and definitely not worthy of its own post. The necessary key insight is a theorem by Fermat. The theorem states that a prime number can be expressed as the sum of two squares if and only if . The proof of one direction is not that difficult but a proof of the other direction (that if then is the sum of two squares) requires a little more work. I don't feel like posting them here but feel free to look on Wikipedia, where you will find numerous proofs. The proof by Euler is quite straight forward but is quite long and split into many sections (your call on whether this is a good thing or not). My personal favorite is the first proof by Dedekind using Gaussian Integers.

Now another important fact to know is the following. If two integers, and , can each be written as the sum of two squares, then their product, can be written as the sum of two squares. The proof is just a basic exercise in equation manipulation, if you have not proven this fact for yourself I suggest you try to in your spare time.

Using these two facts the problem basically boils down to "factor ." Actually our task is even easier than this, we only need to find the prime factors, , of such that . Thus if factors as and for some , we have that and is odd, then can not be expressed as the sum of two squares.

Pretty simple, now writing a program to actually do all of these things is a little harder but not by that much.