The first problem of the day is a limit asked at a contest. This problem was given to me at a party organized by the halls I live in. While the party continued in the background, for about 15 minutes, my universe consisted of me and the limit. I guess that's what happens when you ask STEM people to have fun. The limit is more of an icebreaker for the harder pure Analysis thingy that follows. The limit goes something like this:
Find the limit of \((\frac{a_1^{1/n}+a_2^{1/n}+a_3^{1/n}}{3})^n\) as n approaches infinity, where \(a_1, a_2, a_3\) are real positive numbers.
Solution:
I'm not really a fan of calculating limits or integrals, I'm more of an Algebra person, but here it goes. Write the fraction as \((\frac{a_1^{1/n}+a_2^{1/n}+a_3^{1/n}}{3})^n=(1+\frac{a_1^{1/n}+a_2^{1/n}+a_3^{1/n}-3}{3})^n\). Then, we shall use the exponential limits. Thus, after a bit of calculations, we end up finding that the limit of the initial sequence is the limit of \(e^{\frac{n}{3}(a_1^{1/n}+a_2^{1/n}+a_3^{1/n}-3)}\). Thus, what is left for us to do is calculate the upper limit. However, it is easy to prove that, as \(x\) tends towards 0, the limit of \(\frac{a^x-1}{x}\) is \(ln(a)\), using L'Hospital's rule. Thus, after applying this to our limit, we get that the limit is \((a_1*a_2*a_3)^{\frac{1}{3}}\). Of course, we can generalize the problem by adding m variables instead of 3, and we would get the geometric mean of all the m variables.
The problem appeared on a multi-choice contest, as far as I've been told. The guy who told me the problem gave me the result, and asked me to prove it. I think I solved it in about 20 minutes, after a few failed attempts at trying some inequalities. I'm certain that someone more passionate about Calculus could have solved it way quicker than I did, but the limit was fun to think about nonetheless.
The next problem, and the actual purpose of this article, is a problem about rotating bodies that appeared on this year's Traian Lalescu contest, at the B section; the section for Engineering departments. It caused some troubles among students, as it used a trick that is not that well known. Here it goes:
Let \(f:[a,\infty]→[0,\infty]\) a monotone and continuous function, such that \(\int_{a}^{\infty} f(x) dx\) is convergent. The graph of the function, alongside the coordinate axes, contain a surface of area A.
1. Prove that, for all natural numbers n, there exist the points \(x_1, x_2,...x_n∊[a,\infty]\) such that the lines \(x=x_k\) split the domain in n parts of equal area.
2. Let \(a_n=\frac{\sum_{k=1}^{n} f(x_k)}{n}\) Prove that V, the volume obtained by rotating the curve determined by f around the \(OX\) axis, is finite, and that \(\lim_{n\to\infty} a_n=\frac{V}{πA}\).
Solution:
Part 1 is fairly straightforward to prove, so it is left as an exercise to the reader (mainly because I am too lazy to write it down lmao). The second part is what the problem is all about.
The volume of the rotated solid is \(V=π*\int_{a}^{\infty}f^2(x) dx\), so we want to prove that the limit is equal to \(\frac{\int_{a}^{\infty} f^2(x) dx}{\int_{a}^{\infty} f(x) dx}\).
Now, here comes a dank meme. Kids, don't let your parents try this at home.Now, here comes a dank meme. Kids, don't let your parents try this at home.
Let \(F(x)\) be the integral from a to x of the function. Thus, \(F(x_k)=\frac{k}{n}A\). Also, F is clearly bijective, so we can write \(x_k=F^{-1}(\frac{k}{n}A)\), so \(f(x_k)=f(F^{-1}(\frac{k}{n}A)\). Thus, after a few calculations, we can write \(a_n=A^{-1}\frac{\sum A*f(F^{-1}(\frac{kA}{n})}{n}\). As n tends towards infinity, the sum becomes an integral, so the limit of \(a_n\) will become \(A^{-1}\int_{0}^{A}f(F^{-1}(x))dx\).
We shall prove that \(\int_{0}{A}f(F^{-1}(x))dx=\int_{a}^{\infty}f^2(x)dx\). In order to do this, we apply the change of variable \(t=F^{-1}(x)\), and, after calculations, get the required result.
The big trick with this problem is that it required that \(F^{-1}\) part. That is a well-known trick in Romania, as a problem similar to this one was asked in 2009 at the Olympiad (I think it was in 2009). A similar problem was also asked at the entrance exam for a programme at the Ecole Polytechnique in France. As always, the problem was pretty fun to solve, yet it required a trick fit for the Olympiads.
Before ending the article, I want to make an announcement: I'm finally launching a theory page! I wish to share my knowledge of stuff among students who wish to partake at College competitions such as the IMC. The first theory page will be for Linear Algebra and it will cover everything from basic vector spaces to the Jordan decomposition. I will update that theory page from time to time, whenever I can; don't expect updates to be constant though, since it is pretty tough to cope with courses, out of class learning (I'm studying Complex Analysis atm; expect the next problem to be about Complex Analysis) and social life all at once. I will, however, do my best. So, as always, tune in next time, for something completely random!
Cris.
