Thoughts on my Mind

A Fix on the Mistake August 17, 2009

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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 $(X,d)$ you are given 2 closed disjoint sets $A$ and $B$ . Prove that if one of them is compact then $\inf d(a,b)>0$ with $a\in A$ and $b\in B$ .

proof: Without loss of generality (WLOG) assume that $A$ is compact. Then define $f:A\to{\mathbb{R}}$ as $f(a)=\inf d(a,B)$ . Otherwise, define it to be the infimum of the distance from the point in $A$ to a point in the set $B$ . Now $f$ 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.

Sum Function on Countable Set August 9, 2009

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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 $f:X\to\mathbb{R}$ , and $f(x)\ge{0}$ , such that $f(x_1)+f(x_2)+\cdots+f(x_n)\le{1}$ for any sequence $\{x_n\}\in{X}$ . Prove that the set of points where $f$ is not zero is countable.

Proof:

Suppose for the sake of contradiction that the set is not countable, thus uncountable. Define $E_n=\{x\in{X}:f(x)>\frac{1}{n}\}$ . It should be obvious that $\displaystyle\bigcup_{n=1}^\infty E_n=X$ . So one of the $E_n$ must be uncountable. Thus if we choose points from this set (one of the uncountable sets) we can make $f(x_1)+f(x_2)+\cdots+f(x_n)$ as large as we want. Hence we must have that the set of points wehre $f(x)>0$ is at most countable.

Not that in fact we can ‘weaken’ the assumption on $f$ so that it reads $f(x_1)+f(x_2)+\cdots+f(x_n)\le{M}<\infty$ .

Results March 24, 2009

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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.

Break December 26, 2008

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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”.

Post Putnam December 12, 2008

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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.

Simple Equation December 2, 2008

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Well we all know that the only solutions to $x^2+y^2=(x+y)^2$ are $(0,r)$ and $(r,0)$ where $r\in\mathbb{R}$ . So let us move onto something a little more “difficult”. Find all integer triplets satisfying the following equation.

$a^3+b^3+c^3=(a+b+c)^3$

We begin by doing the obvious thing and expand the right hand side and then cancel terms, thus leaving us with $3ab(a+b)+3ac(a+c)+3bc(b+c)+6abc=0$ . Then divide by 3 to get $ab(a+b)+ac(a+c)+bc(b+c)+2abc=0$ , which after some simple regrouping factors into $(a+b)(a+c)(b+c)=0$ . From here it is obvious that the only solutions are those where one of $a,b,c$ 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.

Solutions to IVT Problems November 26, 2008

Posted by putnam120 in Math Related, Uncategorized.
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(a) Since $f$ is continuous on a compact set ( $\left[a,b\right]$ it attains both its maximum and minimum values, so $m\le f(x)\le M$ for all $x\in\left[a,b\right]$ . Now using these poor estimates we have that $m\int_a^b g(x)dx\le\int_a^b f(x)g(x)dx\le M\int_a^b g(x)dx$ . So if we consider the function $V(x)=f(x)\int_a^b g(x)dx$ we have that $V$ is continuous on the same interval as $f$ . Thus we just apply the intermediate value theorem to $V$ and the desired result follows.

(b) WLOG assume that $f(0)=f(1)=0$ . Let $n$ be given and define new function by $L(x)=n\left(f(x+1/n)-f(x)\right)$ . This is just the equation for the slope of the line connecting the points $f(x)$ and $f(x+1/n)$ and is continuous because $f$ is continuous. Now consider the set of points $x_i=\frac{i}{n}$ for $i=0,1,\dots,n-1$ . Now look at the values $L$ takes at these points. If $L(x_i)=0$ for any $i$ then we have found a solution, so assume that $L(x_i)\neq 0$ for any chose of $i$ . So WLOG assume that $L(x_0)>0$ , since $f(0)=f(1)$ we know that there must be some $i$ such that $L(x_i)<0$ , call this $x_k$ . Then we there is a $x\in\left(x_0,x_k\right)$ such that $L(x)=0$ .

(c) For this one I will prove something a little stronger. All that we need to assume about $f$ is that it is integrable, and has what I shall call the extreme value property on our interval. Basically this property says that on the interval there exists a point $c$ such that $f(c)=\sup f(x)=M$ and a point $d$ such that $f(d)=\inf f(x)=m$ , where $x$ ranges over all the values in our interval. So clearly an increasing function satisfies these conditions. Now define a function by

$V(x)=m\int_a^xg(x)dx+M\int_x^bg(x)dx.$

Now $V$ is continuous and $V(b)\le\int_a^bf(x)g(x)dx\le V(a)$ , thus by the intermediate value theorem we are done.

Some after thoughts: Problem (a) was pretty straight forward and didn’t take too much insight to actually solve. However, problem (b) was a little more challenging until I visualized what I was being asked to prove. This led me to the idea to look at the slope of the connecting line segment. Finally problem (c), I at first wondered why you were given that $f$ was increasing and not continuous, or even just integrable. As can be seen in my proof continuity imposes more conditions than are necessary, while integrability does not provide you with enough. This led me to think about what makes increasing functions special (well one of the things that makes them special).