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\begin{document}
\title{Treeningvõistlus}
\date{27.10.2017}
\maketitle
\thispagestyle{empty}

\begin{enumerate}
\item
Leia $n \times n$ maatriksi
\[
\left(
\begin{array}{rrrrr}
1 & 1 & \dots & 1 & 0\\
1 & 1 & \dots & 0 & 1\\
& \dots & \dots & \dots &\\
1 & 0 & \dots & 1 & 1\\
0 & 1 & \dots & 1 & 1
\end{array}
\right)
\]
pöördmaatriks.
\item
Polünoomi
\[
P(x) = a_{2018} x^{2018} + a_{2017} x^{2017} + \dots + a_1 x + a_0
\]
kordajad rahuldavad võrdust
\[
\frac{a_{2018}}{2019} + \frac{a_{2016}}{2017} + \dots + \frac{a_{2}}{3} + a_0 = 0.
\]
Tõesta, et leidub $y \in \mathbb R$ nii, et $|y| < 2$ ja $P(y) = 0$.
\item
Ruutmaatriks $A$ on \emph{nilpotentne} kui leidub $k \in \mathbb N$ nii, et
$A^k = 0$. Olgu $B$ ja $C$ nilpotentsed maatriksid, kusjuures $BC = CB$ (st nad kommuteeruvad).
\begin{enumerate}
\item
Tõesta, et $B+C$ on nilpotentne.
\item
Kas väide kehtib alati isegi siis, kui $B$ ja $C$ ei kommuteeru?
\end{enumerate}
\item
Pideva funktsiooni $f : (0,1) \to \mathbb R$ korral
\[
x\frac{f'(x)}{f(x)} \leq \frac{1+x}{1-x}
\]
iga $x \in (0,1)$ jaoks ja
\[
\lim_{x \to 0} \frac{f(x)}{x} = 1.
\]

Tõesta, et
\[
f(x) \leq x + 2x^2 + 3x^3 + \dots
\]
iga $x \in (0,1)$ jaoks.

\item
Leia kõik pidevad funktsioonid $f:(-1,1) \to \mathbb R$ mille puhul
\[
f(x+y) = \frac{f(x) + f(y)}{1-f(x)f(y)}
\]
kõikide selliste $x,y \in (-1, 1)$ korral, et $x+y \in (-1, 1)$.
\end{enumerate}
\newpage

\title{Training competition}
\date{27.10.2017}
\maketitle
\thispagestyle{empty}

\begin{enumerate}
\item
Find the inverse of the $n \times n$ matrix
\[
\left(
\begin{array}{rrrrr}
1 & 1 & \dots & 1 & 0\\
1 & 1 & \dots & 0 & 1\\
& \dots & \dots & \dots &\\
1 & 0 & \dots & 1 & 1\\
0 & 1 & \dots & 1 & 1
\end{array}
\right).
\]
\item
The coefficients of the polynomial
\[
P(x) = a_{2018} x^{2018} + a_{2017} x^{2017} + \dots + a_1 x + a_0
\]
satisfy
\[
\frac{a_{2018}}{2019} + \frac{a_{2016}}{2017} + \dots + \frac{a_{2}}{3} + a_0 = 0.
\]
Prove that $P(y) = 0$ for some $y \in \mathbb R$ with $|y| < 2$.
\item
An $n \times n$ matrix $A$ is called \emph{nilpotent} if there exists $k \in \mathbb N$ such that
$A^k = 0$. Let $B$ and $C$ be nilpotent matrices such that $BC = CB$ (i.e., they commute).
\begin{enumerate}
\item
Prove that $B+C$ is nilpotent.
\item
Does the claim always hold even if $B$ and $C$ do not commute?
\end{enumerate}
\item
Let a continuous function $f : (0,1) \to \mathbb R$ satisfy
\[
x\frac{f'(x)}{f(x)} \leq \frac{1+x}{1-x}
\]
for all $x \in (0,1)$ and
\[
\lim_{x \to 0} \frac{f(x)}{x} = 1.
\]

 Prove that 
\[
f(x) \leq x + 2x^2 + 3x^3 + \dots
\]
for all $x \in (0,1)$.
\item
Find all continuous functions $f:(-1,1) \to \mathbb R$ such that
\[
f(x+y) = \frac{f(x) + f(y)}{1-f(x)f(y)}
\]
for all $x,y \in (-1, 1)$ with $x+y \in (-1, 1)$.
\end{enumerate}
\newpage
\begin{enumerate}
\item
Note that $S = A - J$, where $S$ is the given matrix, $A$ is the matrix of 1's, and $J$ is the matrix of 0's with 1's on the perpendicular
diagonal. 
Clearly, $A^2 = nA$, $J^2 = I$, and $JA = AJ = A$.
If $M = (m_{ij})$ is the inverse of $S$, then $MA - MJ = I$. Let $m_i$ denote $\sum_{j} m_{ij}$. For the first row, this gives
\[
m_1 - m_{1k} = 0, \quad k \neq n,
\]
and
\[
m_1 - m_{1n} = 1,
\]
so
\[
m_1 = \sum_{j} m_{1j} = (n-1)m_1 + m_1 - 1
\]
and $m_1 = 1 / (n-1)$. Hence,
\[
M = \frac{1}{n-1}A - J.
\]
Indeed,
\[
(\frac{1}{n-1}A - J)(A-J) = \frac{1}{n-1}A^2 - JA - \frac{1}{n-1}AJ + J^2 = (\frac{n}{n-1} - 1 - \frac{1}{n-1})A + I = I.
\]
\item
Note that
\[
\int_{-1}^1 P(x) dx = 2\left(\frac{a_{2018}}{2019} + \frac{a_{2016}}{2017} + \dots + \frac{a_{2}}{3} + a_0 \right) = 0.
\]
So either $P(x) \equiv 0$ or $P(x)$ changes its sign in the interval $(-1, 1)$ and by continuity of $P$ there is a point $y$ in $(-1, 1)$
such that $P(y) = 0$.
\item
\begin{enumerate}
\item
Let the nilpotency indeces of $B$ and $C$ be $b$ and $c$, respectively. Since our matrices commute, we get
\[
(B+C)^{m} = \sum_{i = 0}^{m} \left(\begin{array}{r}i\\m\end{array}\right) B^i C^{m-i}.
\]
Choosing $m$ so that in any summand $i \geq b$ or $m-i \geq c$ (e.g., $m = b+c$ is enough), we get that $(B+C)^m = 0$.
\item
Take 
\[
B = \left(
\begin{array}{rr}
0 & 1\\
0 & 0
\end{array}
\right),
\quad
C = \left(
\begin{array}{rr}
0 & 0\\
1 & 0
\end{array}
\right).
\]
\end{enumerate}
\item
We have 
\[
\frac{f'(x)}{f(x)} - \frac 1 x \leq \frac 2 {1-x}
\]
so that
\[
\int_0^x \left(\frac{f'(x)}{f(x)} - \frac 1 x \right) dx \leq \int_0^x \frac 2 {1-x} dx,
\]
i.e.,
\[
\ln \frac{f(x)}{x} \leq \ln \frac 1 {(1-x)^2}.
\]
Hence,
\[
\frac{f(x)}{x} \leq \frac 1 {(1-x)^2} = \sum_{n=1}^\infty n x^{n-1},
\]
as needed.
\item
Clearly, $f(0) = 0$. 
Let $u(x) = \arctan(f(x))$.
Then $u(x)$ is continuous and the equation becomes
\[
\tan(u(x+y)) = \tan(u(x) + u(y)).
\]
So $u(x+y) = u(x) + u(y) + k(x,y)\pi$ with $k(x,y) \in \mathbb Z$.
But the continuity of $u$ implies that $k$ is continuous and hence constant. So $k \equiv 0$ because $u(0) = 0$. 

So we get that $u(x) + u(y) = u(x+y)$ for all $x,y,x+y \in (-1,1)$.
This is the Cauchy equation with additional constraints $x,y,x+y \in (-1,1)$. The solution stays the same though, $u(x) = ax$ for suitable $a \in \mathbb R$.
Hence the answer is $f(x) = \tan(ax)$, where $|a| \leq \pi / 2$ to keep $f$ continuous.

How to solve Cauchy equation: first note that $u(nx) = nu(x)$ for all $n \in \mathbb N$, $x, nx \in (-1,1)$. Now $u(x/n) = u(x)/n$ for all
$n \in \mathbb N$, $x \in (-1,1)$. Then $u((m/n)x) = m u(x/n) = (m/n) u(x)$ forall $m, n \in \mathbb N$, $x \in (-1,1)$. Also, clearly 
$u(-x) = -u(x)$ for all $x \in (-1,1)$. Hence $u(qx) = qu(x)$ for all $q \in \mathbb Q$, $x, qx \in (-1,1)$.
Take $r \in \mathbb R$ such that $x, rx \in (-1,1)$. There exists a sequence $(q_n) \subset \mathbb Q$ such that $q_n \to r$ and $|q_n| \leq r$. By continuity of $u$, one has $q_n u(x) = u(q_n x) \to u(rx)$. On the other hand, $q_n u(x) \to r u(x)$. So $u(rx) = ru(x)$ for all $r \in \mathbb R$, $x, rx \in (-1, 1)$. Denoting $a := u(x_0)/x_0$ for some $x_0 \in (-1,1)$, we get that 
$u(x) = u(x_0 \cdot x/x_0) = (x/x_0)u(x_0) = ax$ for all $x \in (-1, 1)$.

(from https://artofproblemsolving.com/community/c6h386060)
\end{enumerate}
\end{document}
