Andrei Zelevinsky’s Problems

Andrei ZelevinskyI was afraid of my advisor Israel Gelfand. He used to place unrealistic demands on me. After each seminar he would ask his students to prove by the next week any open problems mentioned by the speaker. So I got used to ignoring his requests.

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 an such that 1 + 1/2 + 1/3 + … + 1/k = ln k + γ + a1/k + … + an/kn + …

Problem 4. Let x1 not equal to zero, and xk = sin xk-1. Find the asymptotic behavior of xk.

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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One Comment

  1. tanyakh:

    Here is a solution to problem 2 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
    one 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.

    As all polygons have the same center of gravity, the limiting point is the centroid.