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Infinity Norm June 18, 2009

Posted by putnam120 in Math Related.
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Well here is a question that I used to wonder about for a while.

If f is integrable over U, then

\displaystyle\lim_{n\to\infty}\left\{\int_U|f|^nd\mu\right\}^{\frac 1n}=\sup_U|f|.

Well here is a proof of a simplified version.  I only assume that M=\sup|f|<\infty, and \mu(U)<\infty

proof: By “the world’s most obvious integral inequality” \int_U|f|^nd\mu\le M^n\mu(U). So obviously ||f||_\infty\le M.

For the reverse direction let \epsilon>0 then |f|>M-\epsilon on some set K\subset U. Therefore \int_U|f|^n\ge (M-\epsilon)^n\mu(K). To get the result take n\to\infty then \epsilon\to 0. \mathbb{Q.E.D.}

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

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

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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).

Some Problems Involving I.V.T. November 25, 2008

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Well I am in a big Analysis mood. I guess that it has something to do with the fact that at this point in my life Analysis is the topic in math that spikes my interest the most. This can be accredited to my Calculus BC teacher from high school (Mrs. Johnson) and my undergraduate Advanced Calculus professor (Dr. Shen). They both did a wonderful job of presenting material and showing applications and fascinating consequences.

Anyway onto the post, I was reading through the intermediate real analysis section of Problem Solving Through Problems. Here are some of the problems I found interesting.

(a) Suppose that f:\left[a,b\right]\to\mathbb{R} is continuous and g:\left[a,b\right]\to\mathbb{R} is integrable and such that g(x)\ge{0} for all x\in\left[a,b\right]. Prove that there is a number c in \left[a,b\right] such that

\int_a^bf(x)g(x)dx=f(c)\int_a^bg(x)dx.

(b) Let f:\left[0,1\right]\to\mathbb{R} be continuous and suppose that f(0)=f(1). Prove that for each positive integer n there is an x in \left[0,1-\frac{1}{n}\right] such that f(x)=f(x+1/n).

(c) Assume the same conditions on f,g as in part (a) except that instead of being continuous f is now assumed to be increasing. Prove that there is a c\in\left[a,b\right] such that

\int_a^bf(x)g(x)dx=f(a)\int_a^cg(x)dx+f(b)\int_c^bg(x)dx.

My thoughts: The solution to (a) is a pretty straight forward application of the intermediate value theorem. For problem (b) I am considering looking at the slope of the line connecting f(x) and f(x+1/n). Finally for (c) I am going to try and generalize it in an appropriate way. My solutions should be posted in the near future.

Hölder’s Inequality November 23, 2008

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I am only going to do the case where f,g are real functions. The result however, still holds if they are complex.

Statement: Suppose that f,g are integrable functions with respect to \alpha on the interval \left[a,b\right]. Additionally p,q\in\mathbb{R^+} such that \frac 1p+\frac 1q=1. Then we have the following inequality:

\displaystyle\left|\int_a^bfgd\alpha\right|\le\left\{\int_a^b|f|^pd\alpha\right\}^{\frac 1p}\left\{\int_a^b|g|^qd\alpha\right\}^{\frac 1q}.

This can also be stated as ||fg||_1\le ||f||_p||g||_q.

Lemma: If u,v\ge{0} then uv\le\frac{u^p}{p}+\frac{v^q}{q}, where we have the same conditions on p,q as before.

Proof of lemma: Just apply Jensen’s Inequality to e^x and the fact that e^{\ln x}=x. \mathbb{Q.E.D.}

Proof: Without loss of generality we can assume that ||f||_p=||g||_q=1, if not we can just divide f,g by the appropriate constants and make it so. Now from the lemma we have that \forall x\in\left[a,b\right] \displaystyle |f(x)g(x)|\le \frac{|f(x)|^p}{p}+\frac{|g(x)|^q}{q} we then integrate both sides of the inequality and the result follows. \mathbb{Q.E.D.}

Integral Test November 22, 2008

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Well it’s been a while since I have posted a math related post. So I am going to do this one on one of the problems from our analysis homework.  Basically we were asked to prove the integral test, not too difficult but definitely something that should be done.

Statement: Assume that f(x)\ge 0 and that f decreases monotonically on \left[1,\infty\right). Then \displaystyle\int_1^\infty f(x)dx converges if and only if \displaystyle\sum_{n=1}^\infty f(n) converges.

Aside: When I submitted this to my professor for grading in proved the theorem in both direction. Here I am going to try and combine them, thus saving time on my part.

Proof: Consider the interval \left[m,n\right] where m,n are integers with m<n. Additionally let P be the partition \left\{m,m+1,\dots ,n\right\}. Now because f is monotonically decreasing and f\ge 0 we have

(1)   \displaystyle 0\le\sum_{k=m+1}^nf(k)=L(P,f)\le\sum_{k=m}^{n-1}f(k)=U(P,f)\le\sum_{k=m}^nf(k)

Let \epsilon>0. If \displaystyle\sum_{n=1}^\infty f(n) converges then there exists an $N$ such that \displaystyle\sum_{k=m}^n f(k)<\epsilon whenever m,n>N. Similarly if \int_1^\infty f(x)dx converges we have that there exists W such that \int_m^nf(x)dx<\epsilon whenever m,n>W. The theorem follows from combining these facts with (1). \mathbb{Q.E.D.}

I would like to mention that I have left out some of the small details, such as proving that the integral actually does exist if the sum converges.

IMPORTANT NEWS November 22, 2008

Posted by putnam120 in Life Events, Math Related.
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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.