## IMC 2017 – Day 2 – Solutions

See the previous post for the statements of the problems.

**Problem 6.** We have that

Suppose for . Then for we have

This shows that . In the same way, having for leads to .

Now all we need to do is to apply the above procedure to inequalities of the form for and for , which come from the definition of the limit when . The case infinite is similar.

**Problem 7.** If all roots of are real then the sum of their square is non-negative. Suppose is a polynomial of degree . If we denote are the roots of then

If then the previous inequality translates to

which is equivalent to . Now if we compute the coefficients for we find something which for contradicts the previous inequality.

**Problem 8.** We start by observing that the eigenvalues of are and . Next, using the properties of block matrix determinants, we have the following equalities

The block determinant formula holds if is invertible. This happens for all but finitely many values of (the eigenvalues of ). Moreover, since the above equalities are polynomial equalities, this implies equality for every . Therefore, for each eigenvalue of (multiplicity accounted) we have and as eigenvalues for .

Therefore the eigenvalues of are

Note that a phenomenon similar to the Pascal triangle appears, which makes that adjacent eigenvalues of (which have a difference equal to ) with multiplicities and generate an eigenvalue of with multiplicity

**Problem 9.** Elementary techniques in ODE show that . We note that and . Therefore on . This implies, using a recurrence argument, that . This shows that exists for , but, for now, might be infinite.

In order to find an upper bound we may look what is the equation satisfied by an eventual limit of as . We arrive at the ODE . It is immediate to see that . Now we would like to prove that on , and for this we define . Using and the hypothesis we get . Using this we have

and . This implies that

This implies easily that on , which means that on . Thus the pointwise limit of exists. Let’s denote it by . Since using the monotone convergence theorem we have the convergence of integrals . Thus the limit satisfies

Therefore satisfies and which means that the limit is indeed for .