Calculating the limit of a sum using the Riemann-Zeta function

Remember when I said that the next article would be the psychology article or an inequality article? Well...

I took the following problem off of Brilliant and I decided to make it a little bit tougher. I have to admit, the idea of using the Riemann-Zeta function didn't come to mind when solving the problem, but I decided to post it nonetheless, because it's rather interesting. As a reminder, the Riemann-Zeta function is defined as \(ζ(s)=\sum_{i=1}^{\infty} i^{-s}\). The Riemann Hypothesis, an open problem in Mathematics, is about a property of this function, namely that all 0-s of this function lie on the line \(Re(x)=\frac{1}{2}\). 

Back to the problem at hand: what is the value of $$\sum_{n∊ℕ}\frac{2^{w(n)}}{n^2}$$ where \(w(n)\) is the number of prime divisors of n.

Solution:

First off, even though an approximation of \(w(n)\) sounds plausible, believe me, it's not. We shall write this sum in a more comprehensible manner. Let us observe that, for a certain number n, the number of ways of writing n as a product of two coprime numbers is equal to \(2^{w(n)}\), so \(\sum_{n∊ℕ}\frac{2^{w(n)}}{n^2}=\sum_{(a,b)=1}\frac{1}{(ab)^2}\). Now, we know that \(\sum_{a,b∊ℕ}\frac{1}{(ab)^2}=(\sum_{a∊ℕ}\frac{1}{a^2})*(\sum_{b∊ℕ}\frac{1}{b^2})=ζ(2)^2\). In order to find this sum for coprime numbers, let us consider \(a=cm,b=cn\), with \(c\) a natural number and \((m,n)=1\). Thus, $$\sum_{a,b∊ℕ}\frac{1}{(ab)^2}=\sum_{m.n,c∊ℕ}\frac{1}{c^4 m^2 n^2}=$$ $$=(\sum_{c∊ℕ} \frac{1}{c^4})(\sum_{(m,n)=1} \frac{1}{(mn^2)})$$

Thus, the required sum is \(\frac{ζ(2)^2}{ζ(4)}=2.5\), which concludes the proof. Of course, there is a generalization to this problem for \(n^m\) rather than \(n^2\), which is solvable through the same method and yields the result \(\frac{ζ(n)^2}{ζ(2n)}\). This was the first time when I actually used the Riemann-Zeta function in a problem, and it gave me a satisfying feeling, even though, as I said, the proof isn't completely mine. Brilliant has a great stash of great problems like this, so you should check them out.

Apologies for the fact that I haven't posted in two weeks, but I am very stressed out due to this whole application process and, since this is my terminal year of high school, I must prepare for the exams at the end of the year. There's also the fact that I am still thinking about how to solve the inequality I wanted to write about, and the fact that writing anything non-mathematical in nature, such as the psychology article, is a bit tough and requires some background checks... so I suppose that you should expect something completely random in the next article. 

Cris.

Intersection of 3 Proper Subgroups

Welcome back to the world of magic Math tricks! Today we shall be covering an interesting problem that I found a few days ago on AoPS regarding the intersection of 3 proper subgroups (duh...). Apparently the problem was first proposed in the 9th number of 'Gazeta Matematica' from 1985. I surely do love problems that are older than me.

Let \(G\) be a group and \(H_1, H_2, H_3\) be proper subgroups of \(G\) such that \(G= H_1 \cup H_2 \cup H_3\)

Prove that \(x^2 \in H_1 \cap H_2 \cap H_3\) for all \(x \in G\).

Solution:

First of all, let us prove that none of this subgroups includes another. Let's presume by absurd that \(H_2\) is contained in \(H_1\). This would imply that \(H_1 \cup H_3 = G\), which, in turn, would imply that either \(H_1\) or \(H_3\) is \(G\), which contradicts the fact that the subgroups are proper. This argument works if one group is included in the reunion of the others, too.

If \(x \in H_1 \cap H_2 \cap H_3\), then it is obvious that \(x^2 \in H_1 \cap H_2 \cap H_3\).

If \(x \in H_1 \cap H_2 - H_3\), then we shall do a magic trick and write \(x^2=xaa^{-1}x\), where \(a \in H_3 - H_1 \cup H_2\). \(xa\) obviously doesn't belong to \(H_1 \cup H_2\), so it must be in \(H_3\). This argument works for \(a^{-1}x\), too, so \(x^2 \in H_1 \cap H_2 \cap H_3\). The other possible intersections of these groups can be treated the same.

If \(x \in H_1 - H_2 \cup H_3\), then we write \(x^2=xaa^{-1}x\) again, this time with \(a \in H_2 - H_1 \cup H_3\). Obviously, \(xa\) can't be in either \(H_2\) or \(H_1\), so it is in \(H_3\), so \(x^2 \in H_3\). By choosing a suitable \(b \in H_3 - H_1 \cup H_2\), we get that \(x^2 \in H_2\), so \(x^2 \in H_1 \cap H_2 \cap H_3\). All similar cases are treated as above.

This little exercise can work as a Mathematical 'Creativity Test' if you think about it, because it's about discerning a certain pattern and coming up with an idea that solves the problem instantly. More posts might be coming this week, as I am working on 2 more articles atm: the psychology article (finally) and another one with partial derivatives / Lagrange multipliers. Tune in next time for one of these articles (probably the psychology article will come first)!

Cris.