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 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.Share:
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
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.16 August 2015, 3:02 pm
Any chance of sharing the other problem lists mentioned above? I stumbled across these and like them very much, I’ve given the last one as an exercise several times to my calculus students.18 July 2020, 9:00 pm
Sorry, William. Not in the near future.23 July 2020, 4:57 pm