Home > Uncategorized > IMC 2017 – Day 2 – Solutions

IMC 2017 – Day 2 – Solutions


See the previous post for the statements of the problems.

Problem 6. We have that

\displaystyle \int_0^1 f(nx) dx = \frac{1}{n} \int_0^n f(x)dx

Suppose {f(x) \geq L} for {x > M}. Then for {n \gg M} we have

\displaystyle \frac{1}{n} \int_0^n f(x) = \frac{1}{n} \int_0^M f(x) dx + \frac{1}{n}\int_M^n f(x) dx \geq \frac{1}{n} \int_0^M f(x)dx + \frac{n-M}{n}L.

This shows that {\displaystyle \liminf\limits_{n \rightarrow \infty} \frac{1}{n} \int_0^n f(x) dx \geq L}. In the same way, having {f(x) \leq L} for {x>M} leads to {\displaystyle \limsup_{n \rightarrow \infty} \frac{1}{n} \int_0^n f(x)dx \leq L}.

Now all we need to do is to apply the above procedure to inequalities of the form {f(x) \leq L+\delta} for {x>M} and {f(x) \geq L-\delta} for {x>M}, which come from the definition of the limit when {x \rightarrow \infty}. The case {L} infinite is similar.

Problem 7. If all roots of {q_n} are real then the sum of their square is non-negative. Suppose {p} is a polynomial of degree {d>0}. If we denote {w_1,...,w_{n+d}} are the roots of {q_n} then

\displaystyle 0 \leq \sum w_i^2 = \left(\sum w_i\right)^2-2\sum_{i<j} w_i w_j.

If {q_n(x) = ax^{n+d}+bx^{n+d-1}+cx^{n+d-2}+...} then the previous inequality translates to

\displaystyle 0 \leq \sum w_i^2 = \left( \frac{b}{a}\right)^2 - 2 \frac{c}{a},

which is equivalent to {b^2 - 2ac \geq 0}. Now if we compute the coefficients {a,b,c} for {q_n} we find something which for {n \rightarrow \infty} contradicts the previous inequality.

Problem 8. We start by observing that the eigenvalues of {A_1} are {1} and {-1}. Next, using the properties of block matrix determinants, we have the following equalities

\displaystyle \det(A_{n+1}-\lambda I) = \det\begin{pmatrix} A_n-\lambda I & I \\ I & A_n-\lambda I \end{pmatrix} = \det((A_n-\lambda I)^2-I)

\displaystyle = \det(A_n-(\lambda+1)I)\det(A_n-(\lambda-1)I)

The block determinant formula holds if {(A_n-\lambda I)} is invertible. This happens for all but finitely many values of {\lambda} (the eigenvalues of {A_n)}). Moreover, since the above equalities are polynomial equalities, this implies equality for every {\lambda}. Therefore, for each eigenvalue {\lambda} of {A_n} (multiplicity accounted) we have {\lambda+1} and {\lambda-1} as eigenvalues for {A_{n+1}}.

Therefore the eigenvalues of {A_n} are

  • {A_2: \ \{-2,0,0,2\}}
  • {A_3: \ \{-3,-1,-1,-1,1,1,1,3 \}}
  • {A_4: \ \{-4,-2,-2,-2,-2,0,0,0,0,0,0,2,2,2,2,4\}}

Note that a phenomenon similar to the Pascal triangle appears, which makes that adjacent eigenvalues of {A_n} (which have a difference equal to {2}) with multiplicities {{n\choose k}} and {{n\choose k+1}} generate an eigenvalue of {A_{n+1}} with multiplicity {{n\choose k}+{n\choose k+1}={n+1 \choose k+1}}

Problem 9. Elementary techniques in ODE show that {\displaystyle f_{n+1}(x) = \exp(\int_0^x f_n(t)dt)}. We note that {f_1 = 1} and {f_2(x) = e^x}. Therefore {f_1\leq f_2} on {[0,1]}. This implies, using a recurrence argument, that {f_n(x) \leq f_{n+1}(x)}. This shows that {\lim_{n \rightarrow \infty} f_n(x)} exists for {x\in [0,1]}, 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 {F} of {f_n} as {n \rightarrow \infty}. We arrive at the ODE {F' = F^2,\ F(0)=1}. It is immediate to see that {F(x) = \frac{1}{1-x}}. Now we would like to prove that {f_n \leq F} on {[0,1)}, and for this we define {g_n = f_n-F}. Using {f_n \leq f_{n+1}} and the hypothesis we get {f_{n+1}' \leq f_{n+1}^2}. Using this we have

\displaystyle g_n' = f_n'-F' \leq f_n^2 - F^2=g_n(f_n+F)

and {g_n(0)=0}. This implies that

\displaystyle \left( g_n \exp(-\int_0^x (f_n+F)) \right)'\leq 0

This implies easily that {g_n \leq 0} on {[0,1)}, which means that {f_n \leq F} on {[0,1)}. Thus the pointwise limit of {f_n} exists. Let’s denote it by {G}. Since {f_n \leq f_{n+1}} using the monotone convergence theorem we have the convergence of integrals {\int_0^x f_n(t)dt = \int_0^x G(t)dt}. Thus the limit satisfies

\displaystyle G(x) = \exp\left(\int_0^x G(t)dt\right).

Therefore {G} satisfies {G'=G^2} and {G(0)=1} which means that the limit is indeed {G(x) = 1/(1-x)} for {x \in [0,1)}.

Advertisements
Categories: Uncategorized
  1. No comments yet.
  1. No trackbacks yet.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: