BWM – Talking With An Expert

The day of truth: 5 February. 33 young mathematicians travelled to Schmitten to attend the final round of the Bundeswettbewerb Mathematik (BWM).

While it didn’t turn out as I had hoped, it remains a worthwhile experience. The contest consists of a mathematical discussion with a professor and a teacher. Imagine sitting between two mathematicians with some paper and a pen. The professor starts posing questions and you… well you respond. Or struggle.

The talk started with a typical question.

“Which parts of mathematics have you spent time with?”

“Real Analysis.”

That was my plan! I haven’t had time to intensively study higher theories yet – Real Analysis is what I’m most comfortable with at the moment.

So the professor posed the first question to Real Analysis.

“Does the series $\displaystyle \sum_{n=3}^\infty \frac{1}{n \log{n}}$ converge?”

I should have been happy with this question. I was familiar with this problem. I have solved it. I have even blogged about it. But I struggled. I started by spamming some intuition (growth rate of $\log{n}$).

“Let’s try the integral test”

What was I doing? I know it can be solved with Cauchy’s Condensation Test and yet… I just didn’t want to rely on it. In that situation, though, I wasn’t able to immediately solve the problem using the integral test. “What’s the antiderivative?”  I thought. It’d be easy but I just didn’t try writing it on the paper. It’s $\log \log{n}$ by the way.

“Or… Cauchy’s Condensation Test would also work”

A change of plan. So soon, we’d have proven divergence with the Condensation test.

“Do you see what happens for $\displaystyle \sum_{n=3}^\infty \frac{1}{n \log{n} \log \log n}?$

Yeah. It reduces to the earlier question.

Apparently, those were too easy. (Indeed, these series were very simple.) So my professor decided to step it up a notch.

$\displaystyle \sum_n \frac{1}{n \log{n} \log \log{n} \dots \underbrace{\log \log \dots \log{n}}_{\text{Take }\log \text{ so often that this term is }\leq 1}}.$

Does this one still diverge?”

Uhhm what? I have to solve that? I didn’t even understand how the series looks like at the beginning. If only I had time for myself to think about it… to scribble some attempts… So I tried a few ideas that came to my mind. With a lot of help from my professor, we eventually got near to a solution.

“Do you know the antiderivative of $\arctan?$

“Yes, like $\frac{1}{1+x^2}.$

“Alright, then we can ignore that question.”

Few days after the contest, I noticed that the answer was foolish – I gave him the derivative, the antiderivative actually is

$\displaystyle \int \arctan{x} \ dx = x \arctan{x} - \frac12 \log \left(x^2 + 1 \right) + C.$

But apparently, he didn’t notice that either at that moment.

“What else about Real Analysis did you enjoy?”

“There is a really nice proof for the Fundamental Theorem of Algebra using only Real Analysis”

The proof is beautiful. In fact, I’ve also blogged about that! So I’ve demonstrated this proof. And again, he surprised me with his improvisation skills:

“Does your proof generalize? Does it translate to two-dimensional polynomials?”

Uhm?

“So take two polynomials of two variables. Does there always exist a root that solves both polynomials?”

I was a bit confused by the question statement, so I began writing it down. It turns out, he was interested in two polynomials $P, Q : \mathbb{C} \times \mathbb{C} \to \mathbb{C}$ and wanted to know whether we can always find $(w,z) \in \mathbb{C} \times \mathbb{C}$ giving us $P(w,z) = Q(w,z) = 0.$

You can probably already see the answer. But I did not. My intuition failed me at that moment – I had no time to clearly overthink the question! I just rarely work with two-dimensional polynomials, so I was intimidated by the question in that situation!

It took us a while to jot down why the argument doesn’t generalize – and so it was time to look for a counterexample. I’m sure you can immediately find a counterexample. It’s THAT easy to see that the statement can’t be true.

“Do you know how to approximate complex roots numerically?”

“No.”