Well I’ve taken the Putnam exam for the last time.  I think that I did better than last year (20) but if not then I did just as well.  The highest score I could possibly get is a 40, but I think that a 30 is more realistic.

In the next post I hope to post all the problems from the exam and then solve some of then in the proceeding post.  Though I’ll have solutions, or hints for most of the problems, I actually didn’t think of most of them during the examination.  I guess at this point I’ll tell you which ones I solved, well submitted solutions for at least.  A1, A2, B1, and B5.

Some sad news, I knew the answers to; B2 but didn’t know how to show that my lower bound was in fact correct, B4 but forgot everything I learned in Fourier Series so got stuck on an integral, and A4 and once again I was headed in the right direction but gave up too early.

Regardless I feel pretty good about my performance.  The other team members seemed to all have done at least as good, so we should hopefully place just as high as last year (13 I believe).

Well I am almost done with all my graduate school applications.  All that remains is to finalize my personal statements for each school, but that shouldn’t take too long.  The list of of schools is as follows (in no particular order):

1. NYU
2. Cornell
3. Columbia
4. Carnegie Mellon
5. Duke
6. Wisconsin-Madison

This boils down to 5 math programs and 1 in operations research.  Part of me wants to also apply to UF but as things are right now I don’t think that I will.  After doing a little preliminary number crunching I realized that I will easily exceed $600 for the whole process.

So here is one proof for the question from the previous post.  In case you forgot I shall now restate it:

Assume that in a metric space you are given 2 closed disjoint sets and .  Prove that if one of them is compact then with and .

proof: Without loss of generality (WLOG) assume that is compact.  Then define as .  Otherwise, define it to be the infimum of the distance from the point in to a point in the set .  Now is continuous and thus attains its infimum.  But this must be positive, because if it were not we would have contradicted the fact that the sets were disjoint and closed.

This proof works, but I would also like to find one that relies on a set being compact if any open cover has a finite subcover.  Sure it might be more work than necessary, but I feel that it should be good exercise.

I was on a math help forum and was trying to help someone with the following problem.

Given two disjoint closed sets and in a metric space , prove that there exists disjoint open sets and such that and .

In my proof I made the following mistake: Let for , . Then .

I should have stopped at this time, since I remember talking about something very similar to this in my analysis class.  It turns out that to gaurantee you need that at least one of to be compact.

Here is an interesting problem that I saw a few days ago.  I don’t really know why I found this more interesting that other problems/solutions I’ve seen but it just is.

Here is the problem statement (generalized).  Consider a function , and ,  such that for any sequence .  Prove that the set of points where is not zero is countable.

Proof:

Suppose for the sake of contradiction that the set is not countable, thus uncountable.  Define .  It should be obvious that . So one of the must be uncountable.  Thus if we choose points from this set (one of the uncountable sets) we can make as large as we want. Hence we must have that the set of points wehre is at most countable.

Not that in fact we can ‘weaken’ the assumption on so that it reads .

Well here is a question that I used to wonder about for a while.

If is integrable over , then

Well here is a proof of a simplified version.  I only assume that and

proof: By “the world’s most obvious integral inequality” So obviously

For the reverse direction let then on some set Therefore To get the result take then .

The only part that bothers me is making sure that is measurable, but that shoudn’t be too difficult to fix.

Well I got my Putnam results back today. I managed to score a 20 (which is double my score from last year). I was hoping to make the top 15% but just missed and am instead in the top 17%. In addition to this I missed being in the top 500 by 119 places (which is probably the equivalent of 1 or 2 points).

Hopefully next year I’ll get top 500.

Well I haven’t posted anything here in a while. The main reason is that it is currently winter break and I am using the time to do other things. For instance I am finally sitting down to solve some SPOJ problems I have marked as interesting or “useful”.

Well it has been a while since I have metioned my math adventures.  A few days ago I took the Putnam exam. Personally I found the morning seesion to be kind of “easy” ( for the Putnam that is). I was able to answer 2 questions completely and half of a third. The afternoon session was more of what I expected. I knew the answer to 2 of the questions but couldn’t think of a proof for one of them, and then the other I was making a computational mistake (how depressing).

I’m not going to post the problems since it shouldn’e be too difficult to find them with a simple Google search or looking on the AoPS fourms.

After talking to my school mates I felt that I could have answered 4 of the morning session questions, if I didn’t have a thing against using the integral test. My favorite problem on the whole test was as follows: What is the maximum number of rational points that can be on a circle with a center who’s is irrational.

Oh well I am actually pretty happy with my 20, an improvement from last year’s 10, who knows maybe I’ll get a 21 but that’s unlikely.

Well we all know that the only solutions to are and where . So let us move onto something a little more “difficult”. Find all integer triplets satisfying the following equation.

We begin by doing the obvious thing and expand the right hand side and then cancel terms, thus leaving us with . Then divide by 3 to get , which after some simple regrouping factors into . From here it is obvious that the only solutions are those where one of is euqal to the negation of one of the others. See pretty simple, the only real work is being able to expand the first expression. The factoring might be a little tricky but just from looking at the equation you should have a general idea of what the solution should be and that should help you get pretty far with the factoring.