Archive

Posts Tagged ‘SEEMOUS’

SEEMOUS 2016 Problem 4 – Solution

March 6, 2016 3 comments

Problem 4. Let {n \geq 1} be an integer and set

\displaystyle I_n = \int_0^\infty \frac{\arctan x}{(1+x^2)^n}dx.

Prove that

a) {\displaystyle \sum_{i=1}^\infty \frac{I_n}{n} =\frac{\pi^2}{6}.}

b) {\displaystyle \int_0^\infty \arctan x \cdot \ln \left( 1+\frac{1}{x^2}\right) dx = \frac{\pi^2}{6}}.

Read more…

SEEMOUS 2016 – Problems

March 5, 2016 3 comments

Problem 1. Let {f} be a continuous and decreasing real valued function defined on {[0,\pi/2]}. Prove that

\displaystyle \int_{\pi/2-1}^{\pi/2} f(x)dx \leq \int_0^{\pi/2} f(x)\cos x dx \leq \int_0^1 f(x) dx.

When do we have equality?

Problem 2. a) Prove that for every matrix {X \in \mathcal{M}_2(\Bbb{C})} there exists a matrix {Y \in \mathcal{M}_2(\Bbb{C})} such that {Y^3 = X^2}.

b) Prove that there exists a matrix {A \in \mathcal{M}_3(\Bbb{C})} such that {Z^3 \neq A^2} for all {Z \in \mathcal{M}_3(\Bbb{C})}.

Problem 3. Let {A_1,A_2,...,A_k} be idempotent matrices ({A_i^2 = A_i}) in {\mathcal{M}_n(\Bbb{R})}. Prove that

\displaystyle \sum_{i=1}^k N(A_i) \geq \text{rank} \left(I-\prod_{i=1}^k A_i\right),

where {N(A_i) = n-\text{rank}(A_i)} and {\mathcal{M}_n(\Bbb{R})} is the set of {n \times n} matrices with real entries.

Problem 4. Let {n \geq 1} be an integer and set

\displaystyle I_n = \int_0^\infty \frac{\arctan x}{(1+x^2)^n}dx.

Prove that

a) {\displaystyle \sum_{i=1}^\infty \frac{I_n}{n} =\frac{\pi^2}{6}.}

b) {\displaystyle \int_0^\infty \arctan x \cdot \ln \left( 1+\frac{1}{x^2}\right) dx = \frac{\pi^2}{6}}.

Some hints follow.

Read more…

SEEMOUS 2014

March 18, 2014 Leave a comment

Problem 1. Let {n} be a nonzero natural number and {f:\Bbb{R} \rightarrow \Bbb{R}\setminus \{0\}} be a function such that {f(2014)=1-f(2013)}. Let {x_1,..,x_n} be distinct real numbers. If

\displaystyle \left| \begin{matrix} 1+f(x_1)& f(x_2)&f(x_3) & \cdots & f(x_n) \\ f(x_1) & 1+f(x_2) & f(x_3) & \cdots & f(x_n)\\ f(x_1) & f(x_2) &1+f(x_3) & \cdots & f(x_n) \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ f(x_1)& f(x_2) & f(x_3) & \cdots & 1+f(x_n) \end{matrix} \right|=0

prove that {f} is not continuous.

Problem 2. Consider the sequence {(x_n)} given by

\displaystyle x_1=2,\ \ x_{n+1}= \frac{x_n+1+\sqrt{x_n^2+2x_n+5}}{2},\ n \geq 2.

Prove that the sequence {y_n = \displaystyle \sum_{k=1}^n \frac{1}{x_k^2-1} ,\ n \geq 1} is convergent and find its limit.

Problem 3. Let {A \in \mathcal{M}_n (\Bbb{C})} and {a \in \Bbb{C},\ a \neq 0} such that {A-A^* =2aI_n}, where {A^* = (\overline A)^t} and {\overline A} is the conjugate matrix of {A}.

(a) Show that {|\det(A)| \geq |a|^n}.

(b) Show that if {|\det(A)|=|a|^n} then {A=aI_n}.

Problem 4. a) Prove that {\displaystyle \lim_{n \rightarrow \infty} n \int_0^n \frac{\arctan(x/n)}{x(x^2+1)}dx=\frac{\pi}{2}}.

b) Find the limit {\displaystyle \lim_{n \rightarrow \infty} n\left(n \int_0^n \frac{\arctan(x/n)}{x(x^2+1)}dx-\frac{\pi}{2} \right)}

SEEMOUS 2013 + Solutions

March 23, 2013 9 comments

Here are some of the problems of SEEMOUS 2013. Update: the 4th problem has arrived; it is number 3 below.

1. Let {f:[1,8] \rightarrow \Bbb{R}} be a continuous mapping, such that

\displaystyle \int_1^2 f^2(t^3)dt+2\int_1^2f(t^3)dt=\frac{2}{3}\int_1^8 f(t)dt-\int_1^2 (t^2-1)^2 dt.

Find the form of the map {f}.

Solution: Change the variable from {t} to {t^3} in the RHS integral and DO NOT calculate the last integral in the RHS. Get all the terms in the left and find that it is in fact the integral of a square equal to zero.

2. Let {M,N \in \mathcal{M}_2(\Bbb{C})} be nonzero matrices such that {M^2=N^2=0} and {MN+NM=I_2}. Prove that there is an invertible matrix {A \in \mathcal{M}_2(\Bbb{C})} such that {M=A\begin{pmatrix} 0&1\\ 0&0\end{pmatrix}A^{-1}} and {N=A\begin{pmatrix} 0&0 \\ 1&0\end{pmatrix}A^{-1}}.

Solution: One solution can be given using the fact that {M,N} can be written in that form, but for different matrices {A}.

Another way to do it is to consider applications {f,g: \Bbb{C}^2 \rightarrow \Bbb{C}^2,\ f(x)=Mx,\ g(x)=Nx}. We get at once {f^2=0,g^2=0,fg+gf=Id} and from these we deduce that {(fg)^2=fg} and {(gf)^2=gf}. First note that {fg} is not the zero application. Then there exists {u \in Im(fg) \setminus\{0\}}, i.e. there exists {w (\neq 0)} such that {f(g(w))=v}. We have {fg(u)=(fg)^2(w)=fg(w)=u}. Consider {v=g(u)}.

Then {u,v} are not collinear, {f(u)=0,f(v)=u, g(u)=v,g(v)=0}. Consider now the basis formed by {u,v} and take {A} to be the change of base matrix from the canonical base to {\{u,v\}}.

3. Find the maximum possible value of

\displaystyle \int_0^1 |f'(x)|^2|f(x)|\frac{1}{\sqrt{x}}dx

over all continuously differentiable functions {f:[0,1] \rightarrow \Bbb{R}} with {f(0)=0} and {\int_0^1|f'(x)|^2 dx\leq 1}.

4. Let {A \in \mathcal{M}_2(\Bbb{Q})} such that there is {n \in \Bbb{N},\ n\neq 0}, with {A^n=-I_2}. Prove that either {A^2=-I_2} or {A^3=-I_2}.

Solution: Consider {p \in \Bbb{Q}[X]} the minimal polynomial of {A}, which has degree at most {2}. The eigenvalues of {A} satisfy {\lambda_1^n=\lambda_2^n=-1}. We have two cases: either the eigenvalues are real and therefore they are both equal to {-1} either they are complex and conjugate of modulus one. In both cases the determinant of {A} is equal to {1}. Therefore, by Cayley Hamiltoh theorem {A} satisfies an equation of the type {A^2-qA+I_2=0}.

By hypothesis the minimal polynomial {p} divides {X^n+1}. If {p} has degree one then {A=\lambda I_2} and {\lambda \in \Bbb{Q},\lambda^n=-1} so {A=-I_2}.

If not, then the minimal polynomial is {X^2-qX+1} and we must have

\displaystyle X^2-qX+1 | X^n+1.

It can be proved that {q} is in fact an integer. (using the fact that the product of two primitive polynomials is primitive)

Since {q=2\cos \theta} it follows that {q \in \{0,\pm 1,\pm 2\}}. The rest is just casework.

Categories: Algebra, Analysis, Undergraduate Tags:

SEEMOUS 2012 Problem 3

March 8, 2012 Leave a comment

a) Prove that if k is an even positive integer and A is a real symmetric n \times n matrix such that (\text{Tr}\, (A^k))^{k+1}=(\text{Tr}\, (A^{k+1}))^k then A^n=\text{Tr}\, (A)A^{n-1}.

b) Does this assertion from a) also hold for odd positive integers k?

Read more…

SEEMOUS 2012 Problem 2

March 8, 2012 Leave a comment

Let a_n>0, \ n \geq 1. Consider the right triangles \Delta A_0A_1A_2,\Delta A_0A_2A_3,...,\Delta A_0A_{n-1}A_n,... as in the figure. (More precisely, for every n \geq 2 the hypotenuse A_0A_n of \Delta A_0A_{n-1}A_n is a leg of \Delta A_0A_nA_{n+1} with right angle \angle A_0A_nA_{n+1}, and the vertices A_{n-1} and A_{n+1} lie on the opposite sides of the straight line A_0A_n; also |A_{n-1}A_n|=a_n for every n \geq 1.)

It is possible for the set of points \{A_n | n \geq 0\} to be unbounded, but the series \sum_{n=2}^\infty m(\angle A_{n-1}A_0A_n) to be convergent? Here m(\angle ABC) denotes the measure of \angle ABC.

(A subset of the plane is bounded if, for example is contained in a disk.)

Read more…

SEEMOUS 2012 Problem 1

March 8, 2012 Leave a comment

Let A=(a_{ij}) be the n\times n matrix, where a_{ij} is the remainder of the division of i^j+j^i by 3 for i,j=1,2,...,n. Find the greatest n for which \det(A)\neq 0.

Read more…

Categories: Uncategorized Tags: , ,
%d bloggers like this: