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\newcommand{\Mat}{\operatorname{Mat}}
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\begin{document}
\title{Treeningvõistlus}
\date{16.03.2018}
\maketitle
\thispagestyle{empty}

\begin{enumerate}
\item
Olgu funktsioon $f : \mathbb R \times \mathbb R \to \mathbb R$ ja tema osatuletised $f_x$ ja $f_y$
pidevad, kusjuures $f(0,0) = 0$ ning $|f_x(t,s)| \leq 2|t-s|$ ja $|f_y(t,s)| \leq 2|t-s|$ kõikide $t,s \in \mathbb R$ korral.
Tõesta, et $|f(5,4)| \leq 1$.
\item
Leia kõik sellised $n \in \mathbb N$, mille korral
\[
\sum_{k=1}^n (k+2)^n = (n+3)^n.
\]
\item
Olgu $n,k \in \mathbb N$.
Leia determinant
\[
\begingroup
\renewcommand*{\arraystretch}{2}
\begin{vmatrix}
\binom n0 & \binom n1 & \dots & \binom nk\\
\binom {n+1}0 & \binom {n+1}1 & \dots & \binom {n+1}k\\
\dots & \dots & \dots & \dots\\
\binom {n+k}0 & \binom {n+k}1 & \dots & \binom {n+k}k\\
\end{vmatrix}.
\endgroup
\]
%\item
%Olgu funktsioon $f: (0, \infty) \to \mathbb R$ pidev ja kasvav, kusjuures
%\[
%\lim_{x \to \infty} \frac{f(x)}{x \ln x} = 1.
%\]
%Tõesta, et pöördfunktsiooni $f^{-1}$ korral
%\[
%\lim_{x \to \infty} \frac{f^{-1}(x)}{x / \ln x} = 1.
%\]
\item
Olgu $\gamma = e^{\frac{2\pi i}{9}} = \cos\frac{2\pi}{9}+i\sin\frac{2\pi}9$ (s.o. 9. astme algjuur). Leia neljanda astme täisarvuliste kordajatega polünoom, mille juureks on $\gamma + \gamma^{-1}$.
\item
Olgu funktsioon $f : (a,b) \to \mathbb R$ \emph{kumer}, s.t.  
$f(\lambda x + (1-\lambda)y) \leq \lambda f(x) + (1-\lambda)f(y)$ kõikide $x,y \in (a,b)$ ja $\lambda \in [0,1]$ korral.
\begin{enumerate}
\item
Tõesta, et iga $x \in (a,b)$ korral funktsioon $t \mapsto \frac{f(x+t) - f(x)}{t}$ on mittekahanev hulgas $(0, b-x)$.
\item
Tõesta, et $f$ on diferentseeruv kõikides intervalli $(a,b)$ punktides, välja arvatud mingis ülimalt loenduvas hulgas.
\end{enumerate}

\end{enumerate}
\newpage

\title{Training competition}
\date{16.03.2018}
\maketitle
\thispagestyle{empty}

\begin{enumerate}
\item
Let a function $f : \mathbb R \times \mathbb R \to \mathbb R$ and its partial derivatives $f_x$ and $f_y$
be continuous. Assume that $f(0,0) = 0$ and $|f_x(t,s)| \leq 2|t-s|$ and $|f_y(t,s)| \leq 2|t-s|$ for all $t,s \in \mathbb R$.
Prove that $|f(5,4)| \leq 1$.
\item
Find all $n \in \mathbb N$ such that
\[
\sum_{k=1}^n (k+2)^n = (n+3)^n.
\]
\item
Let $n,k \in \mathbb N$.
Find the determinant
\[
\begingroup
\renewcommand*{\arraystretch}{2}
\begin{vmatrix}
\binom n0 & \binom n1 & \dots & \binom nk\\
\binom {n+1}0 & \binom {n+1}1 & \dots & \binom {n+1}k\\
\dots & \dots & \dots & \dots\\
\binom {n+k}0 & \binom {n+k}1 & \dots & \binom {n+k}k\\
\end{vmatrix}.
\endgroup
\]
%\item
%Olgu funktsioon $f: (0, \infty) \to \mathbb R$ pidev ja kasvav, kusjuures
%\[
%\lim_{x \to \infty} \frac{f(x)}{x \ln x} = 1.
%\]
%Tõesta, et pöördfunktsiooni $f^{-1}$ korral
%\[
%\lim_{x \to \infty} \frac{f^{-1}(x)}{x / \ln x} = 1.
%\]
\item
Let $\gamma = e^{\frac{2\pi i}{9}} = \cos\frac{2\pi}{9}+i\sin\frac{2\pi}9$ (this is a primite ninth root of unity). Find a polynomial with integer coefficients of fourth degree having $\gamma + \gamma^{-1}$ as a root.
\item
Let a function $f : (a,b) \to \mathbb R$ be \emph{convex}, i.e.  
$f(\lambda x + (1-\lambda)y) \leq \lambda f(x) + (1-\lambda)f(y)$ for all $x,y \in (a,b)$ and $\lambda \in [0,1]$.
\begin{enumerate}
\item
Prove that for all $x \in (a,b)$ the function $t \mapsto \frac{f(x+t) - f(x)}{t}$ is non-decreasing on $(0, b-x)$.
\item
Prove that $f$ is differentiable in all points of $(a,b)$ except for some at most countable set of points.
\end{enumerate}

\end{enumerate}
\newpage
\textbf{Solutions.}

\begin{enumerate}
\item
Note that $f_x(x,y)=f_y(x,y) = 0$ whenever $x=y$. So differentiating the function $g(x) := f(x,x)$ we get
\[
g'(x) = f_x(x,x)x' + f_y(x,x)x' = 0.
\]
Hence, $f(4,4) = g(4) = 0$. Now
\[
|f(5,4)| = \left|\int_4^5 f(x,4)dx + f(4,4)\right|
 \leq \int_4^5 |f(x,4)|dx \leq \int_4^5 2(x-4) dx = 1.
\]
\item
\emph{Answer:} $n=2,3$.

The cases $n=2,3$ satisfy the equation. The cases $n=1,4,5$ do not satisfy the equation because of different parity of sides. Let us show that for $n \geq 6$ one has
\[
3^n + 4^n + \dots + (n+2)^n < \left(\frac 1{2^n} + \frac 1{2^{n-1}} + \dots + \frac 1 2\right) (n+3)^n < (n+3)^n.
\]
So let us first show that
\[
(n+2)^n \leq \frac 1 2 (n+3)^n.
\]
This is true for $n=6$ because $2\cdot 8^6 = 524288 < 531441 = 9^6$. Since 
\[
\frac{(n+3)^n}{(n+2)^n} = \left(1 + \frac 1 {n+2}\right)^n = \left(1 + \frac 1 {n+2}\right)^{n+2} \left(1 + \frac 1 {n+2}\right)^{-2}
\]
is increasing, it is also true for all $n \geq 6$.

Let us prove now by induction that
\[
(n+3-k)^n < \frac 1 {2^k} (n+3)^n
\]
for $k \geq 2$. Assume
\[
(n+3-(k-1))^n < \frac 1 {2^{k-1}} (n+3)^n.
\]
Since
\[
\frac{n+3-k}{n+3 - (k-1)} \leq \frac{n+2}{n+3},
\]
one has
\[
\left(\frac{n+3-k}{n+3 - (k-1)}\right)^n \leq \frac{(n+2)^n}{(n+3)^n} < \frac 1 2.
\]
The assumption of the induction now yields the claim.
\item
\emph{Answer: 1}.

Denote the corresponding matrix by $A(n,k)$. Note that
for $n =0$ we get a lower triangular matrix, so
\[
|A_{0,k}| = 
\begingroup
\renewcommand*{\arraystretch}{2}
\begin{vmatrix}
1 & 0 & 0 & \dots &0\\
1 & 1 & 0 & \dots &0 \\
1 & 2 & 1 & \dots &0\\
\dots & \dots & \dots & \dots & \dots\\
1 & n & \dots & \dots &1\\
\end{vmatrix}
\endgroup
= 1.
\]
Note that 
\[
A(n+1,k) = A(n,k) \cdot 
\begin{pmatrix}
1 & 1 & 0 & 0 &\dots &0 & 0\\
0 & 1 & 1 & 0 &\dots &0 & 0\\
0 & 0 & 1 & 1 & \dots &0 & 0\\
\dots & \dots & \dots & \dots & \dots & \dots & \dots \\
0 & 0 & 0 & 0 & \dots &1 & 1\\
0 & 0 & 0 & 0 & \dots &0 & 1\\
\end{pmatrix},
\]
because $\binom mi + \binom m {i+1} = \binom {m+1} {i+1}$. 
So $|A(n+1,k)| = |A(n,k)| \cdot 1 = 1$.
\item
Denote $\alpha_n = \gamma^n + \gamma^{-n}$ and $\alpha = \alpha_1$.

We have $\gamma^9 - 1 = 0$ and $\gamma - 1 \neq 0$, so
\[
\gamma^8 + \gamma^7 + \dots + \gamma + 1 = 0.
\]
Divide it by $\gamma^4$ to get
\[
\alpha_4 + \alpha_3 + \alpha_2 + \alpha + 1 = 0.
\]
Note that $\alpha^2 = \alpha_2 + 2$,
$\alpha^3 = \alpha_3 + 3 \alpha$, and
$\alpha^4 = \alpha_4 + 4 \alpha_2 + 6$. From this it is easy to get
\[
\alpha^4 + \alpha^3 - 3\alpha^2 - 2\alpha - 11 = 0.
\]
\item
\begin{enumerate}
\item
Let $0 < t < s$. Then
\[
x+t = \lambda x + (1-\lambda)(x+s)
\]
for $\lambda = 1 - \frac t s$. It remains to apply convexity.
\item
Note that for $s < u < t$ one has
\[
\frac{f(u) - f(s)}{u-s} \leq \frac{f(t) - f(u)}{t-u}.
\]
Hence also for $s <u \leq v < t$ one has
\[
\frac{f(u) - f(s)}{u-s} \leq \frac{f(t) - f(v)}{t-v}.
\]
In other words, denoting 
\[
g(x,t) = \frac{f(x+t) - f(x)}{t}
\]
we get that $g(x,s) \geq g(y,t)$ whenever $x > y$. 
This means that $f$ has one-sided derivatives at every point and 
\[
f'_-(x) \leq f'_+(x) \leq f'_-(y) \leq f'_+(y)
\]
for all $y > x$. If $f$ is not differentiable at $x$, then there is $q_x \in \mathbb Q$ such that
\[
f'_-(x) < q_x < f'_+(x).
\]
Note that $q_x$ are distinct for distinct values of $x$.
\end{enumerate}
\end{enumerate}
\end{document}
