## Andrei Zelevinsky’s Problems

He also had an idea that it is good to learn mathematics through problem solving. So he asked different mathematicians to compile a list of math problems that are important for undergraduate students to think through and solve by themselves. I still have several lists of these problems.

Here I would like to post the list by Andrei Zelevinsky. This is my favorite list, partially because it is the shortest one. Andrei was a combinatorialist, and it is surprising that the problems he chose are not combinatorics problems at all. This list was compiled many years ago, but I think it is still useful, just keep in mind that by calculating, he meant calculating by hand.

**Problem 1.** Let G be a finite group of order |G|. Let H be its subgroup, such that the index (G:H) is the smallest prime factor of |G|. Prove that H is a normal subgroup.

**Problem 2.** Consider a procedure: Given a polygon in a plane, the next polygon is formed by the centers of its edges. Prove that if we start with a polygon and perform the procedure infinitely many times, the resulting polygon will converge to a point. In the next variation, instead of using the centers of edges to construct the next polygon, use the centers of gravity of k consecutive vertices.

**Problem 3.** Find numbers a_{n} such that 1 + 1/2 + 1/3 + … + 1/k = ln k + γ + a_{1}/k + … + a_{n}/k^{n} + …

**Problem 4.** Let x_{1} not equal to zero, and x_{k} = sin x_{k-1}. Find the asymptotic behavior of x_{k}.

**Problem 5.** Calculate the integral from 0 to 1 of x^{−x} over x with the precision 0.001.

I regret that I ignored Gelfand’s request and didn’t even try to solve these problems back then.

I didn’t have any photo of Andrei, so his widow, Galina, sent me one. This is how I remember him.

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## tanyakh:

Here is a solution to problem 3 received from Dan Klain:

First, to show a limit exists: Since the sequence P_0, P_1, P_2, …

is bounded (all contained in P_0), compactness tells us there is a

convergent subsequence. But since the sequence is also monotone P_0 >

P_1 > … the original sequence must have the same limit as the

subsequence. Call this limiting set Q.

If the original polygon has k vertices, then so do all its successors

in the sequence. In the limit the number of vertices will stay the

same or decrease, but cannot increase. So Q is a polygon with at most

k vertices.

Applying the operation to the whole sequence just shifts it forward

16 August 2015, 3:02 pmone step, so the tail is the same, and Q must be invariant under the

operation. This is impossible unless Q is a single point. Moreover,

the operation preserves the mean of the (original) vertices, so Q must

be that point.