\documentclass{article}
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\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{xcolor}
\usepackage{verbatim}
\usepackage{pgf,tikz,pgfplots}
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\usepackage{mathrsfs}
\usetikzlibrary{arrows}

\usepackage{subfigure}

\newcommand{\rank}{\operatorname{rank}}
\DeclareMathOperator{\Mat}{Mat}
\newcommand{\e}{\varepsilon}

\begin{document}

\title{Tartu Ülikooli üliõpilaste matemaatikaolümpiaad}
\date{31.05.2019}
\maketitle
\thispagestyle{empty}
\begin{enumerate}



\item
Olgu funktsioon $f: (0, \infty) \to \mathbb R$ pidev ja rangelt 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
Tõesta, et determinant
\[
\begin{vmatrix}
51237 & 79922 & 55538 & 39177\\
46152 & 16596 & 37189 & 82561\\
71489 & 23165 & 26563 & 61372\\
44350 & 42391 & 91185 & 64809
\end{vmatrix}
\]
erineb nullist.
\item
Arvuta
\[
\sum_{n=1}^\infty \frac{3n^2 - n}{3^n}.
\]

\item
Ütleme, et funktsioon $f: \mathbb R \to \mathbb R$ on \emph{perioodiline}, kui
leidub $t > 0$ nii, et $f(x+t) = f(x)$ iga $x \in \mathbb R$ korral ning
\emph{peaaegu perioodiline}, kui
iga $\e > 0$ korral leidub $t > 0$ nii, et $\sup_{ x \in \mathbb R} |f(x+t) - f(x)| < \e$.
Tähistagu $P, A, B, C \subset \mathbb R^{\mathbb R}$ vastavalt kõikide perioodiliste, peaaegu perioodiliste, tõkestatud ja
pidevate funktsioonide osahulki. Vaatleme hulka $\mathcal S$, mille elemendid on hulgad $P$, $A$, $B$, $C$
ning ka kõikvõimalikud nende ühisosad. On selge, et hulgas $\mathcal S$
on ülimalt $2^4 - 1 = 15$ elementi. Tõesta, et hulgas $\mathcal S$ on täpselt $10$ elementi.

\item
Hulk $M \subset \mathbb R^2$ on sümmeetriline iga sirge suhtes, mis läbib vähemalt 2 tema punkti.
Tõesta, et $M$ on kas ühe sirge peal või on kõikjal tihe tasandis $\mathbb R^2$.


\item
Olgu $n \in \mathbb N$.
Öeldakse, et maatriksid $A,B \in \Mat_n(\mathbb R)$ on kommuteeruvad,
kui $AB = BA$. Olgu maatriksid $A,B,C \in \Mat_n(\mathbb R)$ paarikaupa kommuteeruvad.
Tõesta, et leiduvad kolm reaalarvu $a,b,c$ nii, et
$|a| + |b| + |c| > 0$ ja
$|aA + bB + cC| = 0$.

\end{enumerate}
\newpage
\thispagestyle{empty}
\begin{enumerate}
\item
Let a function $f: (0, \infty) \to \mathbb R$ be continuous and strictly increasing such that
\[
\lim_{x \to \infty} \frac{f(x)}{x \ln x} = 1.
\]
Prove that the inverse function $f^{-1}$ satisfies
\[
\lim_{x \to \infty} \frac{f^{-1}(x)}{x / \ln x} = 1.
\]

\item
Prove that the determinant
\[
\begin{vmatrix}
51237 & 79922 & 55538 & 39177\\
46152 & 16596 & 37189 & 82561\\
71489 & 23165 & 26563 & 61372\\
44350 & 42391 & 91185 & 64809
\end{vmatrix}
\]
is not $0$.
\item
Calculate
\[
\sum_{n=1}^\infty \frac{3n^2 - n}{3^n}.
\]

\item
Let us call a function $f: \mathbb R \to \mathbb R$ \emph{periodic}, if there exists $t > 0$ such that $f(x+t) = f(x)$ for all $x \in \mathbb R$ and
\emph{almost periodic}, if for every $\e > 0$ there exists $t > 0$ such that $\sup_{ x \in \mathbb R} |f(x+t) - f(x)| < \e$.
Let $P, A, B, C \subset \mathbb R^{\mathbb R}$ denote the subsets of all periodic, almost periodic, bounded, and continuous functions, respectively. Consider the set $\mathcal S$, whose elements are the sets $P$, $A$, $B$, $C$ along with all possible intersections of them. Clearly $\mathcal S$ contains at most $2^4 - 1 = 15$ elements. Prove that $\mathcal S$ contains exactly $10$ elements.

\item
A set $M \subset \mathbb R^2$ is symmetric with respect to every straight line, which passes some 2 of its points. Prove that $M$ either lies on some straight line or is everywhere dense in $\mathbb R^2$.


\item
Let $n \in \mathbb N$.
Matrices $A,B \in \Mat_n(\mathbb R)$ are said to commute if $AB = BA$. Consider matrices
$A,B,C \in \Mat_n(\mathbb R)$ which commute with each other. Prove that there exist
three real numbers $a,b,c$ such that $|a| + |b| + |c| > 0$ and $|aA + bB + cC| = 0$.

\end{enumerate}
\newpage
\textbf{Vihjed}

\begin{enumerate}
\item
Esimeseks sammuks proovi näidata, et $\ln f(x) \sim \ln x$ protsessis $x \to \infty$.
\item
Arvuta determinant modulo mõni algarv. 

\item
Diferentseeri rida $\sum_{n=1}^\infty x^{3n} = \frac 1 {1-x^3}$ kaks korda. 

\item
Siin peab tekitama mitu funktsioonide näidet, mille alusel võib väita, et vastavad funktsioonide klassid on
erinevad.

On kasulik vaadelda ka ühtlaselt pidevaid funktsioone, mis on alati peaaegu perioodilised.

\item
Leia hulgas $M$ romb, tema põhjal leia ristkülik, tekita võrk, leia palju väiksem romb.

\item
Tõesta, et leidub ühine omavektor, 3 vastavat (kompleksset) omaväärtust on $\mathbb R$-lineaarselt sõltuvad.
\end{enumerate}
\newpage
\textbf{Solutions}

\begin{enumerate}
\item
Kuna $f(x) \sim x \ln x$ protsessis $x \to \infty$, siis $\ln f(x) \sim \ln(x\ln x) = \ln x + \ln \ln x \sim \ln x$ (sest üldiselt kui $f(x) \sim g(x)$, siis $\ln( \frac {f(x)}{g(x)})  = \ln(f(x)) - \ln(g(x)) \to 0$, 
kust $\frac{\ln(f(x))}{\ln(g(x))} \to 1$; NB! Siin ei saa asendada $\ln$ suvalise pideva lõpmatuseni kasvava funktsiooniga).
Seega $f(x) \sim x \ln f(x)$ ehk $\frac{f(x)}{\ln f(x)} \sim x$.

Tähistades $x = f^{-1}(y)$, saame et $f^{-1}(y) \sim \frac y {\ln y}$ protsessis $y \to \infty$, sest
$f$ on rangelt kasvav.

\item
Arvutame determinanti modulo $5$. Saame
\[
\begin{vmatrix}
2 & 2 & -2 & 2\\
2 & 1 & -1 & 1\\
-1 & 0 & -2 & 2\\
0 & 1 & 0 & -1
\end{vmatrix}
\equiv
2
\begin{vmatrix}
1 & 1 & -1 & 1\\
0 & -1 & 1 & -1\\
0 & 1 & 2 & -2\\
0 & 1 & 0 & -1
\end{vmatrix}
\equiv
2
\begin{vmatrix}
0 & 1 & -2\\
0 & 2 & -1\\
1 & 0 & -1
\end{vmatrix}
= 2(-1 + 4) \equiv 1 \not \equiv 0.
\]

\item
Diferentseerime rida $\sum_{n=1}^\infty x^{3n} = \frac 1 {1-x^3}$ kaks korda (seda võib teha vahemikus $x \in (-1,1)$). Saame
\[
\sum_{n=1}^\infty 3n(3n-1)x^{3n-2} = \frac{6x(1+2x^3)}{(1-x^3)^3}
\]
ehk
\[
\sum_{n=1}^\infty n(3n-1)x^{3n} = \frac{2x^3(1+2x^3)}{(1-x^3)^3}.
\]
Võttes $x = 3^{-\frac 1 3}$, saame paremal poolel vastuseks $\frac{15}4$.
\item
Näitame, et $|\mathcal S| \leq 10$. Ilmselt $P \subset A$, mistõttu kõik ühisosad, kus esineb $P \cap A$
tegelikult ühtivad vastavate ühisosadega, kus esineb ainult $P$. Neid on täpselt 4 tükki. Seega 
$|\mathcal S| \leq 11$. Iga pidev ja perioodiline funktsioon on pidev igas lõigus, mille pikkus on suurem
kui funktsiooni periood. Seega ta on tõkestatud selles lõigus ning järelikult tõkestatud ka kogu reaalteljel.
Tõestasime, et $P \cap C = P \cap C \cap B$. Seega $|\mathcal S| \leq 10$.

Nüüd näitame, et kõik ülejäänud klassid on erinevad. Mainime kohe, et ühtlaselt pidev funktsioon
on ilmselt peaaegu perioodiline.  


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\newpage
$B\cap P\cap C$ < $P \cap B$:

$B\cap C < B$:

$B\cap C \cap A < B \cap A$:

$C \cap A < A$: Ratsionaalarvude hulga $\mathbb Q$ karateristlik funktsioon $\chi_{\mathbb Q}$.

$P \cap B < P$: $f(x) = \frac{1}{(x - [x])}$, kui $x \neq [x]$ ja $f(x) = 1$ muidu ($[x]$ on täisosa).

$P < A$: 

$B \cap C < C$:

$B \cap A < A$:

$B \cap C \cap A < C \cap A$: $f(x) = x$. 

$P \cap C < B \cap C \cap A$:

$P \cap B < B \cap A$: $f(x) = 1+x$, kui $x \in [-1,0]$, $f(x) = 1 - x$, kui $x \in [0, 1]$, $f(x) = 0$ muidu.

$C \cap A < C$: $f(x) = \ln(x)$, kui $x$ on piisavalt suur, $f(x) = 0$ muidu: $|f(2t) - f(t)| = \ln 2$.

$B \cap A < B$: $f(x) = \operatorname{sign}(x)$.

$B \cap C \cap A < B \cap C$: Arendame eelmist näidet. Kasutame pidevaid sammusid kuitahes väikeste pikkustega.
Olgu funktsioon konstantselt $1$ poollõigul $(-\infty,1]$. 
Seejärel teeme pideva sammu $1$ pealt $-1$ peale pikkusega $1/n$ (alustame hetkel $n=1$). 
Seejärel lõigul pikkusega $n$ olgu funktsioon jälle konstantne. Siis jälle pidev
samm $-1$ pealt $1$ peale pikkusega $1/(n+1)$ ning seejärel funktsioon konstantne lõigul pikkusega $n+1$.
Niimoodi saame jätkata lõpmatuseni, kusjuures $n \to \infty$. 
Suvalise $t > 0$ korral leidub $n$ nii, et $1/n < t < n$. Võttes arvuks $x$ pideva sammu alguspunkti, kus
selle sammu pikkus on $1/n$, saame, et $|f(x+t) - f(x)| = 2$.


\item
Kui $M$ ei ole ühe sirge peal, siis leiduvad kolm punkti, mis ei asu ühel sirgel. Vähemalt kahe peegeldusega
saab siis tekitada rombi. Seejärel rombis $ABCD$ saab peegeldada punkti $A$ sirgete $BC$ ja $CD$ suhtes
ning punkti $C$ sirgete $AB$ ja $AD$ suhtes. Saadud punktid $A_1, B_1, C_1, D_1$ moodustavad ristküliku,
kusjuures selle ristküliku külgede pikkused ei ületa $5d$, kus $d = |AC|$ on diagonaali pikkus.
Tõepoolest, nt $|B_1C_1| \leq |CC_1| + |AC| + |AB_1| \leq 2d + d + 2d = 5d$, sest ilmselt $|CC_1| \leq 2d$.
Saadud ristküliku saab peegeldada tema külgede suhtes. Nii saame katta tasandi ühtlase võrguga.
Võrgus aga saame valida diagonaali ja punkti, mis on kuitahes lähedane selle diagonaalini. Nii saame konstrueerida rombi kuitahes väikese diagonaaliga ning selle rombi alusel jälle võrku kuitahes väikestest ristkülikutest, nagu vaja.

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\end{figure}

\item
Kui üks meie maatriksitest on singulaarne, siis ilmselt vastav kolmik leidub. 
Eeldame, et nad on regulaarsed.

Tõestame, et leidub ühine omavektor. Olgu $\lambda_1$ maatriksi $A$ mingi (kompleksne) omaväärtus
ja $V_1$ vastav omavektorite alamruum (ruumis $\mathbb C^n$). Siis $B(V_1) \subset V_1$. Tõepoolest,
kui $x \in V_1$, siis $A(Bx) = B(Ax) = B(\lambda_1 x) = \lambda_1 Bx$, mistõttu $Bx \in V_1$.
Seega lineaarsel operaatoril $B|_{V_1}$ (mis on regulaarne $B$ regulaarsuse tõttu) omakorda leidub omaväärtus $\lambda_2$ ning vastav omavektorite
alamruum $V_2 \subset V_1$. Analoogiliselt saame, et $C(V_2) \subset V_2$ ning operaatoril $C|_{V_3}$
leidub mingi omaväärtus $\lambda_3$ ja vastav nullist erinev omavektor $x \in V_3 \subset V_2 \subset V_1$.

Vaatleme kompleksseid omaväärtusi $\lambda_i$ kui vektoreid ruumis $\mathbb R^2$. Nad on kindlasti lineearselt sõltuvad ning seega leidub kolmik $(a,b,c) \in \mathbb R^3 \setminus 0$ nii, et
$a \lambda_1 + b \lambda_2 + c\lambda_3 = 0$. Nüüd
\[
(aA + bB + cC)(x) = (a \lambda_1 + b \lambda_2 + c\lambda_3)x = 0,
\]
mis näitabki, et maatriks $aA + bB + cC$ on singulaarne.
\end{enumerate}
\end{document}
