This is a cute problem I found this evening.
Suppose is a bijection such that if and , then .
Prove that if then .
Proof: The trick is to divide the pairs of positive integers into families with the same product.
Note that the -th column contains as many elements as the number of divisors of . Now we just just use a simple observation. Let be on the -th column (i.e. ). If then cannot be on one of the first columns. Indeed, the monotonicity property implies . The fact that is a bijection assures us that cover the first columns. Moreover, one element from the -th column is surely covered, namely . This means that
where we have denoted by the number of positive divisors of .
One of the questions I received in the comments to my old post on solving Laplace equation (in fact this is Poisson’s equation) using finite differences was how to apply this procedure on arbitrary domains. It is not possible to use this method directly on general domains. The main problem is the fact that, unlike in the case of a square of rectangular domain, when we have a general shape, the boudary can have any orientation, not only the orientation of the coordinate axes. One way to avoid approach this problem would be using the Finite Element Method. Briefly, you discretize the boundary, you consider a triangulation of the domain with respect to this discretization, then you consider functions which are polynomial and have support in a few number of triangles. Thus the problem is reduced to a finite dimensional one, which can be written as a matrix problem. The implementation is not straightforward, since you need to conceive algorithms for doing the discretization and triangulation of your domain.
One other approach is to consider a rectangle which contains the shape and add a penalization on the exterior of your domain . The problem to solve becomes something like:
The following variant of the Lebesgue dominated convergence theorem may be useful in the case we can not dominate a sequence of functions by only one integrable function, but by a convergent sequence of integrable functions.
Let be a sequence of measurable functions defined on a measurable set with real values, which converges pointwise almost everywhere to . Suppose that there exists a sequence of positive, integrable functions such that the two following conditions hold:
- for almost every ;
- There exists an integrable function such that in .
Let be a decreasing sequence of positive numbers such that the series
is uniformly convergent. Then must satisfy .
Note that this result implies that the series is not uniformly convergent on . It is surprisngly similar to the following result:
Suppose that is a decreasing sequence of positive real numbers such that the series is convergence. Then . It is no surprise that the proofs of these two results are similar.
Here’s a nice problem inspired from a post on MathOverflow: link
We call a polygon shrinkable if any scaling of itself with a factor can be translated into itself. Characterize all shrinkable polygons.
It is easy to see that any star-convex polygon is shrinkable. Pick the point in the definition of star-convex, and any contraction of the polygon by a homothety of center lies inside the polygon.
Consider the following Cauchy problem:
Prove that any solution of the above problem cannot be bounded on .
For every positive integer , denote by the number of permutations of such that for every . For , denote by the number of permutations of such that for every and for every . Prove that
IMC 2014 Day 2 Problem 5