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\begin{document}
\title{Matemaatika võistlus}
\date{Tartu, 11.10.2019}
\maketitle
\thispagestyle{empty}

\begin{enumerate}
\item
Olgu $A$, $B$ ja $C$ reaalväärtustega maatriksid suurusega $5\times 5$, kusjuures
\[
\left(B + \left(A - \left(A^{-1} + \left(B^{-1}-A\right)^{-1}\right)^{-1}\right)^{-1}\right)A = C + C^{-1}
\]
ning kõik ülalolevas avaldises esinevad pöördmaatriksid eksisteerivad. 
Leia vähemalt üks sobiv maatriks $C$ ja avalda see maatriksite $A$ ja $B$ kaudu.

\item
Tõesta, et kõikide positiivsete täisarvude $n$ korral
\[
\left(\frac{7+\sqrt{61}}2\right)^n + \left(\frac{7-\sqrt{61}}2\right)^n
\]
on täisarv. Milliste $n$ korral see jagub $7$-ga?

\item
Olgu funktsioon $f: \mathbb R \to \mathbb R$ diferentseeruv.
Kas alati leidub $x \in [0,1]$ nii, et
\[
4\left(f(1) - f(0)\right)^2 + (\pi xf'(x))^2 = \pi^2 \cdot f'(x)^2?
\]
\item
Olgu $A$ ja $B$ sama järku reaalväärtustega ruutmaatriksid, kusjuures 
\[
AB - BA = 2A.
\]
\begin{enumerate}
\item
Tõesta, et $A^k B - BA^k = 2kA^k$ iga $k \in \mathbb N$ korral.
\item
Tõesta, et leidub $k \in \mathbb N$ nii, et $A^k = 0$.
\end{enumerate}
\item
\begin{enumerate}
\item
Tõesta, et 
\[
\frac{\sin n}n \left(\sum_{k=1}^n \frac 1 {\sqrt k}\right) \xrightarrow[n \to \infty]{} 0.
\]
\item
Kas 
\[
\sum_{n=1}^\infty \frac{\sin n}n \left(\sum_{k=1}^n \frac 1 {\sqrt k}\right)
\]
koondub?
\end{enumerate}

\end{enumerate}
\newpage
\title{Math Competition}
\date{Tartu, 11.10.2019}
\maketitle
\thispagestyle{empty}
\begin{enumerate}
\item
Let $A$, $B$, and $C$ be $5\times 5$ realvalued matrices such that
\[
\left(B + \left(A - \left(A^{-1} + \left(B^{-1}-A\right)^{-1}\right)^{-1}\right)^{-1}\right)A = C + C^{-1}
\]
and such that all inverses appearing above exist. 
Find at least one such matrix $C$ and represent it as formula depending on $A$ and $B$.

\item
Prove that 
\[
\left(\frac{7+\sqrt{61}}2\right)^n + \left(\frac{7-\sqrt{61}}2\right)^n
\]
is an integer for all positive integers $n$. For which $n$ is this number divisible by $7$?

\item
Let $f: \mathbb R \to \mathbb R$ be differentiable.
Does there always exist $x \in [0,1]$ such that
\[
4\left(f(1) - f(0)\right)^2 + (\pi xf'(x))^2 = \pi^2 \cdot f'(x)^2?
\]
\item
Let $A$ and $B$ be square realvalued matrices of the same size such that 
\[
AB - BA = 2A.
\]
\begin{enumerate}
\item
Prove that $A^k B - BA^k = 2kA^k$ for all $k \in \mathbb N$.
\item
Prove that there exists $k \in \mathbb N$ such that $A^k = 0$.
\end{enumerate}
\item
\begin{enumerate}
\item
Prove that
\[
\frac{\sin n}n \left(\sum_{k=1}^n \frac 1 {\sqrt k}\right) \xrightarrow[n \to \infty]{} 0.
\]
\item
Does the series
\[
\sum_{n=1}^\infty \frac{\sin n}n \left(\sum_{k=1}^n \frac 1 {\sqrt k}\right)
\]
converge?
\end{enumerate}

\end{enumerate}
\newpage
\paragraph{Vihjed}
\begin{enumerate}
\item
Vt millega võrdub nende kolme suure pöördmaatriksi (ehk vastavate maatriksite) korrutis.
\item
Need liidetavad on mingi polünoomi juured! Kas tuleb rekurrentne seos?
\item
Lagrange'i kvt rakendatud kahe sobiva funktsiooni vahele.
\item
\begin{enumerate}
\item Lihtne.
\item Uuri operaatori $L: A \mapsto AB - BA$ omaväärtused.
\end{enumerate}
\item
\begin{enumerate}
\item
Paremat kordajat saab hinnata integraaltestiga.
\item
$\sum \sin (n)$ on tõkestatud (kui mõelda $e$ astmetega). Ülejäänud kordaja monotoonselt läheb nulli.
\end{enumerate}
\end{enumerate}
\newpage
\begin{enumerate}
\item
Denote $G_1 = B^{-1}-A$, $G_2 = A^{-1} + G_1^{-1}$, and $G_3 = A - G_2^{-1}$. 
Note that
\[
G_3 \cdot G_2 \cdot G_1 = \left(I + AG_1^{-1} - I\right) \cdot G_1 = A.
\]
So the left hand side of the original equation is equal to
\[
(B + G_2 \cdot G_1 \cdot A^{-1})A = BA + G_2 \cdot G_1,
\]
where
\[
G_2 \cdot G_1 = A^{-1}G_1 + I = A^{-1}B^{-1} - I + I = (BA)^{-1}.
\]
So we can take $C = BA$.
\item
Note that $x_i = \frac{7\pm\sqrt{61}}2$ are the roots of $x^2 - 7x - 3 = 0$. So $x_i^{n+2} - 7x_i^{n+1} - 3x_i^{n} = 0$ for all $n = 0, 1, 2, \dots$. Hence also
\[
x_1^{n+2} + x_2^{n+2} = 7(x_1^{n+1} + x_2^{n+1}) +  3(x_1^{n} + x_2^n)
\]
for all $n = 0, 1, 2, \dots$. Since $x_1 + x_2 = 7$, it is clear that the claim follows. Moreover, $7$ divides $x_1^n + x_2^n$ if and only if $n$ is odd. 
\item
If $f(1) = f(0)$ then take $x=1$. Otherwise assume $x \neq 1$ and write
\[
\frac 4{\pi^2 (1-x^2)} = \frac{f'(x)^2}{(f(1) - f(0))^2}.
\]
So it is enough to find $x$ such that
\[
\frac 2{\pi \sqrt{1-x^2}} = \frac{f'(x)}{f(1) - f(0)},
\]
which is the same as
\[
\left(\frac {2 \arcsin x}{\pi} - \frac{f(x)}{f(1) - f(0)}\right)' = 0.
\]
Denote the left hand side by $F'(x)$ and note that the claim follows from Rolle's (or mean value) theorem,
because $F(1) = F(0)$. Indeed, $F$ is continuous on $[0,1]$, differentiable on $(0,1)$, and
\[
F(1) - F(0) = \frac{2}{\pi} (\arcsin(1)-\arcsin(0)) - \frac{f(1) - f(0)}{f(1)-f(0)} = 1 - 1 = 0.
\]
\item
\begin{enumerate}
\item
\begin{align*}
A^{n+1}B - BA^{n+1} = A(A^nB) - A(BA^n) + (AB)A^n - (BA)A^n = A(2nA^n) + (2A)A^n = 2(n+1)A^{n+1}. 
\end{align*}
\item
\emph{Solution 1.}
Consider a linear operator $L:\operatorname{Mat}_{n} \to \operatorname{Mat}_{n}$ defined by $X \mapsto XB - BX$.
By the above, $2k$ is its eigenvalue whenever $A^k \neq 0$. But it has at most $n^2$ eigenvalues, so $A^k$ must be $0$ for some $k$.

\emph{Solution 2 (found by Urve).} Consider matrix norms in (a) (recall that the matrix norm satisfies the triangle inequality and a similar inequality for products of matrices):
\[
 2k \| A^k \| = \|2k A^k\|= \| A^k B-B A^k \| \leq \|A^k B\| + \|BA^k\| \leq 2 \| A^k \| \| B\| 
\]
for all $k \in \mathbb N$.
So either claim holds or $\|B\| \geq k$ for all $k \in \mathbb N$, which is impossible.
\end{enumerate}
\item
\begin{enumerate}
\item
\emph{Solution 1.}
The proof of the integral test gives
\[
\sum_{k=1}^n \frac 1 {\sqrt k} \leq 1 + \int_1^n \frac 1 {\sqrt x} dx = 1 + \left(2\sqrt x \Big|_1^n\right)
= 2\sqrt n - 1,
\]
so that
\[
\frac{1}n \left(\sum_{k=1}^n \frac 1 {\sqrt k}\right) \leq \frac 2 {\sqrt n} \xrightarrow[n \to \infty]{} 0,
\]
which implies the claim, because $\sin n$ is bounded.

\emph{Solution 2 (found by Natalia).}
Let us estimate using the over-harmonic series:
\[
\left|\frac{\sin n}{n}\sum_{k=1}^n \frac 1 {\sqrt k}\right| \leq \frac{1}{n^{1/4}}\sum_{k=1}^n \frac 1 {k^{5/4}} \leq \frac{1}{n^{1/4}} \zeta\left(\frac{5}{4}\right) \xrightarrow[n \to \infty]{} 0,
\]
where $\zeta\left(\frac{5}{4}\right) = \sum_{k=1}^\infty \frac 1 {k^{5/4}} < \infty$.
\item
The above convergence is actually monotone. Indeed,
\[
\frac{1}{n+1} \left(\sum_{k=1}^{n+1} \frac 1 {\sqrt k}\right) < \frac{1}n \left(\sum_{k=1}^n \frac 1 {\sqrt k}\right) \iff
n \left(\sum_{k=1}^{n+1} \frac 1 {\sqrt k}\right) < (n+1) \left(\sum_{k=1}^n \frac 1 {\sqrt k}\right) 
\iff \frac{n}{\sqrt{n+1}} < \sum_{k=1}^n \frac 1 {\sqrt k},
\]
which is clear.
On the other hand,
\begin{align*}
\left|\sum_{n=1}^N \sin n\right| &= \left|\sum_{n=1}^N \frac{e^{in} - e^{-in}}2\right|
= \frac 12 \left|\frac{e^{i(N+1)}-1}{e^i - 1} - \frac{e^{-i(N+1)}-1}{e^{-i} - 1}\right|
\leq
\frac 12 \left(\frac{|e^{i(N+1)}| + 1}{|e^i - 1|} + \frac{|e^{-i(N+1)}| + 1}{|e^{-i} - 1|}\right)\\
& \leq \frac 12 \left(\frac{2}{m} + \frac{2}{m}\right) = \frac 2 m,
\end{align*}
where $m = |e^i - 1| = |e^{-i} - 1| > 0$ (in fact, $ m = \sqrt{(\cos(1) - 1)^2 + \sin^2(1)} = \sqrt{2 - 2\cos(1)}$).

By the way, there is another way to see that these partial sums are bounded, using trigonometry instead of the complex representation (\emph{found by Natalia}):
\begin{align*}
\sum_{n=1}^N \sin n &= \frac 1{2\sin(\frac 1 2)}\sum_{n=1}^N 2\sin(\frac 1 2)\sin n
= \frac 1{2\sin(\frac 1 2)}\sum_{n=1}^N \left[\cos\left(\frac{2n-1}2\right) - \cos\left(\frac{2n+1}2\right)\right]\\
&=\frac{\cos(\frac 1 2) - \cos(\frac{1+2N}{2})}{2\sin(\frac 1 2)}.
\end{align*}

The convergence of the original series now follows from Dirichlet's test: if $(a_n)$ and $(b_n)$ are sequences such that $(a_n)$ monotonely goes to $0$ and $|\sum_{n=1}^N b_n|$ is bounded (by, e.g., $M > 0$), then
\[
\sum_{n=1}^\infty b_n \cdot a_n 
\] 
converges. 

Dirichlet's test can be proven relatively easily using summation by parts:
\[
\sum_{n=1}^N b_n \cdot a_n = \sum_{n=1}^N \left(\sum_{k=1}^n b_k\right) \cdot a_n - \sum_{n=2}^{N} \left(\sum_{k=1}^{n-1} b_k\right) a_{n} = a_{N+1}\cdot \left(\sum_{k=1}^N b_k\right) + \sum_{n=1}^N \left(\sum_{k=1}^n b_k\right) \cdot (a_n - a_{n+1}).
\]
Here the first summand goes to $0$ and the second converges absolutely:
\[
%\left|\sum_{n=1}^N \left(\sum_{k=1}^n b_k\right) \cdot (a_n - a_{n+1})\right| \leq 
\sum_{n=1}^N \left|\sum_{k=1}^n b_k\right| \cdot |a_n - a_{n+1}|
\leq M \sum_{n=1}^N (a_n - a_{n+1}) = M(a_1 - a_{N+1}) \to Ma_1.
\]
\end{enumerate}

\end{enumerate}
\end{document}
