Dan Shved

My mathematics blog

Month: April, 2013

An exercise in transfinite magic

In my hometown there is a university called SUSU. And in this university regular math contests (link in Russian) are held, organized by A. Yu. Evnin. The following problem appeared in one of these contests:

Problem 1. A function f: \mathbb {R} \to \mathbb {R} has the following property: every straight line on the plane \mathbb {R}^2 intersects the graph of f and the parabola y = x^2 at the same number of points. Prove that f(x) = x^2 for every x \in \mathbb {R}.

The solution is quite straightforward, I am not going to write it down. It is probably enough to say that it heavily exploits the convexity of y = x^2.

This post is not about problem 1. It is about the next thing that comes to mind:

Problem 2. Two functions f, g: \mathbb {R} \to \mathbb {R} have graphs F and G such that for every straight line l \subset \mathbb {R}^2: |l \cap F| = |l \cap G| < \infty. Does it follow that f and g are the same function?

If you think about this one for a minute, I’m sure you’ll find it much more challenging than problem 1. This whole post is about solving problem 2. If you want to give it a try yourself, this is the last point in the text where you can still do this without knowing the answer.

Read the rest of this entry »


A problem with polynomial functions

Here is a problem that popped up in my head about a year ago.

Problem 1. A function f: \mathbb {Z} \times \mathbb {Z} \to \mathbb {Z} is polynomial of each argument, i.e. for every x_0 \in \mathbb {Z} the function y \to f(x_0, y) can be represented by a polynomial with integer coefficients, and the same goes for x \to f(x, y_0) for every y_0. Is it true then that f must be polynomial of both arguments simultaneously, i.e. representable by an element of \mathbb {Z}[x,y]?

If you’ve ever been a freshman in an analysis course, you are bound to be aware of a common trap with the notion of differentiability. The trap is like this: you are asked to prove that a function g: \mathbb {R} \times \mathbb {R} \to \mathbb {R} is differentiable everywhere. You have several months of single-variable analysis behind your back. So your first instinct may be to prove that \partial g / \partial x and \partial g / \partial y exist at every point and be done with it. But this is not enough, as demonstrated by

\displaystyle g(x,y) = \left\{ \begin{array}{ll}0 &  \quad \mathrm{at}\, \mathrm{point}\, (0,0) \\ \frac{xy}{\sqrt {x^2+y^2}} &  \quad \mathrm{everywhere}\, \mathrm{else.}\end{array}\right.

Problem 1 asks if there is the same trap with polynomiality (not a word?) as there is with differentiability. When I came up with problem 1 I was busy with more “serious” math, so the question didn’t get any real thought. I always suspected that it would’t take long to solve, but somehow I was too lazy to do the job. That is, I was until this morning, when I finally put in the required several minutes and got this thing sorted out. If you’re curious, you might want to do the same before moving on to the next paragraph.

Read the rest of this entry »