## Putnam 2014 Problems

**A1.** Prove that every nonzero coefficient of the Taylor series of about is a rational number whose numerator (in lowest terms) is either or a prime number.

**A2.** Let be the matrix whose entry in the -th row and -th column is

for Compute

**A3.** Let and for Compute

in closed form.

**A4.** Suppose is a random variable that takes on only nonnegative integer values, with and (Here denotes the expectation of the random variable ) Determine the smallest possible value of the probability of the event

**A5.** Let Prove that the polynomials and are relatively prime for all positive integers and with

**A6.** Let be a positive integer. What is the largest for which there exist matrices and with real entries such that for all and the matrix product has a zero entry somewhere on its diagonal if and only if

**B1.** A *base over-expansion* of a positive integer is an expression of the form with and for all For instance, the integer has two base 10 over-expansions: and the usual base 10 expansion Which positive integers have a unique base 10 over-expansion?

**B2.** Suppose that is a function on the interval such that for all and How large can be?

**B3.** Let be an matrix with rational entries. Suppose that there are at least distinct prime numbers among the absolute values of the entries of Show that the rank of is at least

**B4.** Show that for each positive integer all the roots of the polynomial

are real numbers.

**B5.** In the 75th Annual Putnam Games, participants compete at mathematical games. Patniss and Keeta play a game in which they take turns choosing an element from the group of invertible matrices with entries in the field of integers modulo where is a fixed positive integer and is a fixed prime number. The rules of the game are:

(1) A player cannot choose an element that has been chosen by either player on any previous turn.

(2) A player can only choose an element that commutes with all previously chosen elements.

(3) A player who cannot choose an element on his/her turn loses the game.

Patniss takes the first turn. Which player has a winning strategy?

**B6.** Let be a function for which there exists a constant such that for all Suppose also that for each rational number there exist integers and such that Prove that there exist finitely many intervals such that is a linear function on each and

## Five points on a circle, centroids and perpendiculars

Suppose you have different points on a circle. For each triangle formed by three of these points, consider the line passing through its centroid, perpendicular to the line determined by the other two.

## Variations on Fatou’s Lemma – Part 2

As we have seen in a previous post, Fatou’s lemma is a result of measure theory, which is strong for the simplicity of its hypotheses. There are cases in which we would like to study the convergence or lower semicontinuity for integrals of the type where converges pointwise to and converges to in some fashion, but not pointwise. For example, we could have that converges to in . In this case we could write the integral as where is the measure defined by . All measures considered in this post will be positive measures.

Certain hypotheses on the measures allow us to find a result similar to Fatou’s Lemma for varying measures. In the following, we define a type of convergence for the measures , named *setwise convergence*, which will allow us to prove the lower semicontinuity result. We say that converges setwise to if for every measurable set . The following proof is taken from Royden, H.L., *Real Analysis*, Chapter 11, Section 4. It is very similar to the proof of Fatou’s lemma given here.

**Theorem A.** Let be a sequence of measures defined on which converges setwise to a measure and a sequence of nonnegative measurable functions which converge pointwise (or almost everywhere in ) to the function . Then

## Variations on Fatou’s Lemma – Part 1

One of the most basic and powerful results of measure theory is Fatou’s Lemma. It provides information on the convergence of the integrals of functions which converge pointwise. The power of this lemma is the simplicity of its hypothesis. One of its interpretations is that the functional which to a positive function associates its integral is lower semicontinuous with respect to the pointwise convergence. Among other things, it allows us to prove the monotone convergence theorem, or Lebesgue’s dominated convergence theorem in just one line.

**Fatou’s Lemma** Suppose is a measure space and is a sequence of integrable functions converging pointwise almost everywhere to . Then is measurable and

*Proof:* One of the classical proofs of this result goes as follows. Using the definition

we note that it is enough to prove that

for every simple function .

Pick and . Suppose . Denote . Note that and . If we denote the support of , we may choose such that for every we have . Thus

Since is bounded above by a constant and , we have

Taking we have

Take and then supremum over to finish the proof. \hfill

**Monotone convergence** Suppose that is a sequence of function defined on such that

for every and . Denote . Then is measurable and

*Proof:* The comes from the monotonicity, while the is just Fatou’s Lemma.

**Dominated convergence** Suppose is a sequence of measurable functions defined on which converges almost everywhere to . Moreover, suppose there exists an integrable function such that for every . Then

*Proof:* Apply Fatou’s lemma to .

It is straightforward to see that the positivity condition can be replaced with a domination from below. If and is integrable, then apply Fatou’s lemma to to deduce that the conclusion still holds for . One may ask what happens with the part. If with integrable, then the reverse Fatou Lemma works. To see this, apply Fatou’s lemma to .

Of course, every one of the few hypotheses made for the Fatou’s lemma are essential. If we allow negative functions, then is a counterexample. converges to zero almost everywhere, while the integrals of are equal to . The inequality can be strict. Consider the similar functions .

Seeing that the hypotheses needed for Fatou lemma to hold are simple enough, we may try and see which generalizations of Fatou lemma hold? One interesting direction is to consider not only variable functions, but also variable measures. This will be treated in a future post.

## Problems of the Miklos Schweitzer Competition 2014

**Problem 1.** Let be a positive integer. Let be a familiy of sets that contains more than half of all subsets of an -element set . Prove that from we can select sets that form a separating family of , i.e., for any two distinct elements of there is a selected set containing exactly one of the two elements.

**Problem 2.** let and let be non-degenerate subintervals of the interval . Prove that

where the summation is over all pairs of indices such that and are not disjoint.

**Problem 3.** We have points in the plane, no three of them collinear. The points are colored with two colors. Prove that from the points we can form empty triangles (they have no colored points in their interiors) with pairwise disjoint interiors, such that all points occuring as vertices of the triangles have the same color.

**Problem 4.** For a positive integer , let be the number of sequences of positive integers such that and for . We make the convention . Let be the unique real number greater than such that . Prove that

- (i) .
- (ii) There exists no number such that .

**Problem 5.** Let be a non-real algebraic integer of degree two, and let be the set of irreducible elements of the ring . Prove that

**Problem 6.** Let be a representation of a finite -group over a field of characteristic . Prove that if the restriction of the linear map to a finite dimensional subspace of is injective, then the subspace spanned by the subspaces () is the direct sum of these subspaces.

**Problem 7.** Lef be a continuous function and let be arbitrary. Suppose that the Minkowski sum of the graph of and the graph of (i.e. the set has Lebesgue measure zero. Does it follow then that the function is of the form , with suitable constants ?

**Problem 8.** Let be a fixed integer. Calculate the distance

where runs over polynomials of degree less than with real coefficients and runs over functions of the form

defined on the closed interval , where and .

**Problem 9.** Let , and let be a convex body, i.e. a compact convex set with nonempty interior. Define the barycenter of the body with respect to the weight function by the usual formula

Prove that the translates of the body have pairwise distinct barycenters with respect to .

**Problem 10.** To each vertex of a given triangulation of the two dimensional sphere, we assign a convex subset of the plane. Assume that the three convex sets corresponding to the three vertices of any two dimensional face of the triangulation have at least one point in common. Show that there exist four vertices such that the corresponding convex sets have at least one point in common.

**Problem 11.** Let be a random variable that is uniformly distributed on the interval , and let

Show that, as , the limit distribution of is the Cauchy distribution with density function .

## Monotonic bijection from naturals to pairs of natural numbers

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 .

## Solving Poisson’s equation on a general shape using finite differences

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:

Note that doing this we do not need to impose the boundary condition on . This is already imposed by , and the fact that is forced to be zero outside .