\def\softd{{\leavevmode\setbox1=\hbox{d}%
\hbox to 1.05\wd1{d\kern-0.4ex{\char039}\hss}}}%cstocs
\def\softt{{\leavevmode\setbox1=\hbox{t}%
\hbox to \wd1{t\kern-0.6ex{\char039}\hss}}}%cstocs
\def\softl{l\kern-0.45ex\raise0.1ex\hbox{'}\kern-0.10ex}%cstocs
\def\softL{L\kern-0.8ex\raise0.1ex\hbox{'}\kern0.1ex}%cstocs
% endofdefsfromcstocs %

\input cviceni
\vcorrection{12 mm}
\NoBlackBoxes
{
\centerline{\smc XI. Ur\v cete $y'$ a $y''$ v~bodech,
v~jejich\v z~okol\'i je uvedenou rovnic\'i}
\centerline{\smc zad\'ana diferencovateln\'a
 funkce $y=y(x)$}

\pr $x^2+2xy-y^2=a^2\quad a\in\Bbb R$
\pr $\log\sqrt{x^2+y^2} = \arctg \frac y x$
\pr $y-\varepsilon\sin y =x\quad 0<\varepsilon<1$

\pr $x^y=y^x$
\pr $y=2x\arctg\frac y x$

\centerline{\smc Zjist\v ete, v~okol\'i kter\'ych bod\accent23u je
n\'asleduj\'ic\'im p\v redpisem zad\'ana diferencovateln\'a}
\centerline{\smc  funkce
$z=z(x,y)$ a najd\v ete jej\'i parci\'aln\'i derivace prvn\'iho a
druh\'eho \v r\'adu}

\pr $\frac x z~=\log\frac z~y$
\pr $xyz=x+y+z$
\pr $F(x+y+z,x^2+y^2+z^2)=0 \quad({\partial^2z\over\partial
x\partial y})$ ($F$ je t\v r\'\i dy $C^2$)
\pr $F(xz,yz)=0\quad({\partial^2z\over\partial
x^2})$  ($F$ je t\v r\'\i dy $C^2$)

\pr Doka\v{z}te, \v{z}e existuj\'{\i} funkce $u=u(x,y)$ a $v=v(x,y)$, pro
kter\'e plat\'{\i} $u(1,2)=v(1,2)=0$, a kter\'e na n\v{e}jak\'em okol\'{\i} bodu
$(1,2)$ spl\v{n}uj\'{\i} rovnice $x e^{u+v}+2uv=1$, $y e^{u-v}-{u\over
1+v}=2x$ a jsou tam t\v{r}\'{\i}dy $\Cal C^1$. Spo\v{c}t\v{e}te jejich diferenci\'al
v~bod\v{e} $(1,2)$.

\pr Nech\softt{} $M=\{(x,y,z)\mid x^2+y^2+z^2-1=0, x^2+y^2-x=0\}$.
Najd\v{e}te $a\in M$, aby $P=M\setminus\{a\}$ byla jednorozm\v{e}rn\'a
plocha t\v{r}\'{\i}dy $\Cal C^1$.

\pr Vy\v{s}et\v{r}ete extr\'emy funkce $z$ zadan\'e implicitn\v{e}.
(a) $x^2+y^2+z^2-2x+2y-4z-10=0$

(b) $x^2+y^2+z^2-xz-yz+2x+2y+2z-2=0$\quad
(c) $(x^2+y^2+z^2)^2=a^2(x^2+y^2-z^2)$
}

@ \pr $y'={y+x\over y-x}$,
$(x,y)\ne({a\over\sqrt2},{a\over\sqrt2}),
({-a\over\sqrt2},{-a\over\sqrt2})$
\pr $y'={x+y\over x-y}$, $(x,y)\ne (\sqrt2 e^{\frac\pi4},\sqrt2
e^{\frac\pi4}),(-\sqrt2 e^{\frac\pi4},-\sqrt2 e^{\frac\pi4})$
\pr $y'={1\over 1-\varepsilon\cos y}$
\pr $y'=\frac yx\cdot{x\log y+y\over y\log x+x}$, $(x,y)\ne(e,e)$
\pr $y'=\frac yx$
\pr $\pdx z={z\over z+x},\,\pdy z={z^2\over y(x+z)}$ s~v\'yjimkou
bod\accent23u $(-\frac ye,y,\frac ye)$
\pr $\pdx z={1-yz\over xy-1},\,\pdy z={1-xz\over xy-1}$
\pr {\def\J{\left(\partial_1F+2z\partial_2F\right)}
  $\pd{^2z}{x\partial y}={(z-x)(z-y)\left(\partial_{11}F\cdot
(\partial_2 F)^2-2\partial_{12}F\cdot\partial_1F\cdot\partial_2F
+\partial_{22}F\cdot(\partial_1F)^2\right)+2\partial_2F\cdot
(\partial_1F+2x\partial_2F)\cdot(\partial_1F+2y\partial_2F) \over
\J^3}$, $\J\ne0$ }
\pr {\def\J{\left(x\partial_1F+y\partial_2F\right)}
$\pd{^2z}{x\partial y}={zy\left(\partial_{11}F\cdot
(\partial_2 F)^2-2\partial_{12}F\cdot\partial_1F\cdot\partial_2F
+\partial_{22}F\cdot(\partial_1F)^2\right)+2z^2(\partial_1F)^2
 \over \J^3}$,\hfill \linebreak $\J\ne0$}
\pr $\pdx u=0,\,\pdy u=-1,\,\pdx v=-1,\,\pdy v=1$
\pr  $(1,0,0)$

\oddel


{
\centerline{\smc XII. Implicitn\'{\i} funkce a geometrie}

\pr Nech\softt{} $x_0$ je jednon\'asobn\'y ko\v{r}en rovnice
$x^n+a_1 x^{n-1}+\cdots+a_n=0$. Doka\v{z}te, \v{z}e existuj\'{\i}
$\varepsilon>0$ a $\delta>0$ takov\'e, \v{z}e kdykoli
$|b_i-a_i|<\delta$ pro $i=1,\dots,n$, pak rovnice
$x^n+b_1 x^{n-1}+\cdots+b_n=0$  m\'a v~intervalu
$(x_0-\varepsilon,x_0+\varepsilon)$ pr\'av\v{e} jeden ko\v{r}en. Spo\v{c}t\v{e}te
$\pd x{a_i}(a_1,\dots,a_n)$.

\pr Najd\v{e}te rovnice te\v{c}ny a norm\'alov\'e roviny k~parametricky
zadan\'ym k\v{r}ivk\'am

(a) $x=a\cos\alpha\cos t$, $y=a\sin\alpha\cos t$, $z=a\sin t$
v~bod\v{e} $t=t_0$,

(b) $x=a\sin^2t$, $y=b\sin t\cos t$, $z=c\cos^2 t$ v~bod\v{e}
$t=\frac\pi4$.

\pr Najd\v{e}te rovnice te\v{c}ny a norm\'alov\'e roviny k~implicitn\v{e}
zadan\'ym k\v{r}ivk\'am

(a) $x^2+z^2=10$, $y^2+z^2=10$ v~bod\v{e} $(1,1,3)$,
(b) $x^2+y^2+z^2=6$, $x+y+z=0$ v~bod\v{e} $(1,-2,1)$.

\pr Najd\v{e}te rovnice te\v{c}n\'e roviny a norm\'aly k~ploch\'am

(a) $z=x^2+y^2$ v~bod\v{e} $(1,2,5)$,
(b) $z=y+\log\frac xz$ v~bod\v{e} $(1,1,1)$,
(c) $2^{\frac xz}+2^{\frac yz}=8$ v~bod\v{e} $(2,2,1)$,
(d) $x=u\cos v$, $y=u\sin v$, $z=av$ v~bod\v{e} $(u_0,v_0)$.

\pr Najd\v{e}te rovnici te\v{c}n\'e a norm\'alov\'e roviny  k~plo\v{s}e $e^{\frac
ux}\cos\frac vy=\frac x{\sqrt2}$, $e^{\frac
ux}\sin\frac vy=\frac y{\sqrt2}$ v~bod\v{e}  $(1,1,0,\frac\pi4)$.
}


@ \pr $\pd x{a_i}=-{x_0^{n-i}\over n
x_0^{n-1}+(n-1)a_1x_0^{n-2}+\cdots+a_{n-1}}$
\pr (a) te\v{c}na: $x\sin\alpha-y\cos\alpha=0$, $y\cos t_0+z\sin\alpha\sin
t_0=a\sin\alpha$, norm\'alov\'a rovina: $x\cos\alpha\sin t_0+y\sin\alpha\sin
t_0-z\cos t_0=0$, pokud $a\ne 0$.
(b) te\v{c}na: $y=\frac b2$, $xc+za=ac$, norm\'alov\'a rovina:
$ax-cz={a^2-c^2\over 2}$, pokud $ac\ne0$.
\pr (a) te\v{c}na: $x+y=2$, $y+z=4$, norm\'alov\'a rovina: $x-y+z=3$.
(b) te\v{c}na: $x-2y+z=6$, $x+y+z=0$, norm\'alov\'a rovina: $x-z=0$.
\pr (a) te\v{c}n\'a rovina: $z=2x+4y-5$, norm\'ala: $2x+y=4$, $4x-8y+5z=13$.
(b) te\v{c}n\'a rovina: $x-y+2z=2$, norm\'ala: $x+y=2$, $-x+y+z=1$.
(c) te\v{c}n\'a rovina: $x+y-4z=0$, norm\'ala: $x-y=0$, $2x+2y+z=9$.
(d) te\v{c}n\'a rovina: $ax\sin v_0-ay\cos v_0+u_0z=au_0v_0$, norm\'ala: $x\cos
v_0+y\sin v_0=u_0$, $-x u_0\sin v_0+y u_0\cos v_0+az=av_0$, pokud
$a\ne0$ nebo $u_0\ne0$.
\pr te\v{c}n\'a rovina: $-x+\frac\pi4\cdot y+u-v=-1$,
$-\left(\frac\pi4+1\right)y+u+v=-1$, norm\'alov\'a rovina:
$2y+u+\left(1+\frac\pi2\right)v=2+\frac\pi4+\frac{\pi^2}8$,
$-\left(\frac{\pi^2}4+\pi+6\right)x+\frac\pi2y+\left(\frac{\pi^2}8+
\frac34\pi+3\right)u-\left(\frac\pi4+3\right)v=-\frac5{16}\pi^2+\frac34\pi+6$.

\newpage

{
\centerline{\smc XIII. Je\v{s}t\v{e} implicitn\'{\i} funkce a geometrie}

\pr Spo\v{c}t\v{e}te $\pdxy z$ \clqq v~bod\v{e} $u=2$, $v=1$\crqq, jestli\v{z}e
$x=u+v^2$, $y=u^2-v^3$, $z=2uv$.

\pr Na plo\v{s}e $x^2+2y^2+3z^2+2xy+2xz+4yz=8$ najd\v{e}te body, v~nich\v{z}
je te\v{c}n\'a rovina rovnob\v{e}\v{z}n\'a s~n\v{e}kterou ze sou\v{r}adnicov\'ych rovin.

\pr Najd\v{e}te te\v{c}n\'e roviny k~plo\v{s}e $x^2+2y^2+3z^2=1$, kter\'e jsou
rovnob\v{e}\v{z}n\'e s~rovinou $x+4y+6z=0$.
\pr Doka\v{z}te, \v{z}e te\v{c}n\'e roviny k~plo\v{s}e $\sqrt x+\sqrt y+\sqrt
z=\sqrt a$ $(a>0)$ ur\v{c}uj\'{\i} na os\'ach \'useky s~konstantn\'{\i}m sou\v{c}tem
d\'elek.

\smc
\centerline{Najd\v{e}te k\v{r}ivku, kter\'a je spole\v{c}nou te\v{c}nou syst\'emu
k\v{r}ivek}
%\vskip-5pt
\centerline{\eightpoint\smc(t.j.
ka\v{z}d\'y jej\'{\i}\v{z} bod le\v{z}\'{\i} na n\v{e}kter\'e k\v{r}ivce ze syst\'emu,}
%\vskip-5pt
\centerline{\eightpoint\smc a jej\'{\i}\v{z} te\v{c}na
v~tomto bod\v{e} spl\'yv\'a s~te\v{c}nou on\'e k\v{r}ivky ze syst\'emu)}
\rm

\pr $x\cos\alpha+y\sin\alpha=p$, $p=\roman{const}$, $\alpha\in\Bbb R$
\pr $(x-a)^2+y^2=\frac12a^2$, $a\in\Bbb R$
\pr $y=kx+\frac ak$, $a=\roman{const}$, $k\ne0$
\pr $y^2=2px+p^2$, $p\in\Bbb R$
\pr Pro \'use\v{c}ky d\'elky $l$, jejich\v{z} konce jsou na os\'ach.

\pr Pro elipsy ${x^2\over a^2}+{y^2\over b^2}=1$ s~konstantn\'{\i}m
obsahem $S$.

\centerline{\smc
Najd\v{e}te plochu, kter\'a je spole\v{c}nou te\v{c}nou k~syst\'emu ploch. \rm}

\pr Pro syst\'em sf\'er o~polom\v{e}ru $r$ se st\v{r}edy na kru\v{z}nici $x=R\cos
t$, $y=R\sin t$, $z=0$ ($R>r$).

\pr  Pro elipsoidy ${x^2\over a^2}+{y^2\over b^2}+{z^2\over c^2}=1$
s~konstantn\'{\i}m objemem $V$.

\pr Pro roviny $z-z_0=p(x-x_0)+q(y-y_0)$, kde $x_0,y_0,z_0$ jsou
konstanty a $p^2+q^2=1$.}

@ \pr $\frac{26}{121}$
\pr $(0,\sqrt8,-\sqrt8)$, $(0,-\sqrt8,\sqrt8)$, $(2,-4,2)$, $(-2,4,-2)$,
$(-4,2,0)$, $(4,-2,0)$
\pr $x+4y+6z=\sqrt{21}$, $x+4y+6z=-\sqrt{21}$
\pr Sou\v{c}et je $a$.
\quad{\smc N\'avod k~dal\v{s}\'{\i}m p\v{r}\'{\i}klad\accent23um:} Nech\softt{} $F$ je hladk\'a funkce a pro ka\v{z}d\'e
$\alpha\in(a,b)$ je rovnic\'{\i}
$F(x,y,\alpha)=0$ ur\v{c}ena hladk\'a k\v{r}ivka. Pak spole\v{c}n\'a te\v{c}na
$x=x(\alpha)$, $y=y(\alpha)$ mus\'{\i} spl\v{n}ovat $F(x,y,\alpha)=0$ a $\pd
F\alpha(x,y,\alpha)=0$. Analogicky pro plochy.
\pr $x^2+y^2=p^2$
\pr $y=x$, $y=-x$
\pr $y^2=4ax$ pro $a\ne0$
\pr Neexistuje
\pr $x^{\frac23}+y^{\frac23}=l^{\frac23}$
\pr $|xy|={S\over 2\pi}$
\pr $\left(\sqrt{x^2+y^2}-R\right)^2 +z^2=r^2$
\pr $|xyz|={V\over4\sqrt3\pi}$
\pr $(x-x_0)^2+(y-y_0)^2=(z-z_0)^2$

\oddel

{
\smc \centerline{XIV. V~n\'asleduj\'{\i}c\'{\i}ch \'uloh\'ach zjist\v{e}te $\sup$
$\inf$ funkce $f$ na mno\v{z}in\v{e} $M$}

\centerline{a vy\v{s}et\v{r}ete, zda t\v{e}chto hodnot $f$ na $M$ nab\'yv\'a.}
\rm


\pr $f\left(x,y,z\right) = x - 2y + 2z$,

\quad a) $M = \{ \left(x,y,z\right), x^2 + y^2 + z^2 = 1 \}$,
 b) $M = \{ \left(x,y,z\right), x^2 + y^2 + z^2 = 1, x+y+z = 0 \}$

\pr $f\left(x,y,z\right) = xyz$,

\quad a) $M = \{ \left(x,y,z\right), x^2 + y^2 +z^2 = 1 \}$,
 b) $M = \{\left(x,y,z\right), x^2 + y^2 + z^2 =1, \, x + y + z~= 0\}$

\pr $f\left(x,y,z\right) = \sin {x} \sin {y} \sin {z}$,
    $ M = \{\left(x,y,z\right); x + y +z = {\pi\over 2}, x > 0, y > 0, z~>0 \}$

\pr $f\left(x,y,z\right) = xy^2z^3$,
   $M = \{\left(x,y,z\right); x + 2y + 3z = a, x >0, y >0 \}$, kde  $a>0$.

\pr $f\left(x_1, \dots, x_n\right) = x_1^p +  \dots +~x_n^p$;
   $M = \{ \left(x_1,  \dots, x_n\right); x_1 + \dots + x_n =
   a$, $x_1>0$, $\dots, x_n > 0 \};$ kde $a>0, p > 0.$
\pr $f\left(x,y\right) = x + y$,
   $M=\{\left(x,y\right); x^3 + y^3 - 2xy = 0, x \ge 0, y \ge 0 \}$

\pr $f(x,y,z)=10z+x-y$,
$M=\{(x,y,z), x^2+y^2+z^2\le1, y+x\ge0\}$
\pr $f(x,y)=\frac1x+\frac1y$, $M=\{(x,y),\frac1{x^2}+\frac4{y^2}=1\}$
\pr $f(x,y)=y$, $M=\{(x,y), \left(x^2+y^2\right)^2=Kxy\}$, kde $K>0$

\pr Nech\softt{} $M=\{(x,y,z), x^2+y^2+z^2=1, x^2+y^2=x\}$. Doka\v{z}te, \v{z}e funkce
$f(x,y,z)=x+y+z$ neb\'yv\'a maxima na $M$ v~bod\v{e} $(x_0,y_0,z_0)$, kde $z_0$
je algebraick\'e \v{c}\'{\i}slo.

\pr Nech\softt{} $A=(a_{ij})_{i,j=1}^n$ je re\'aln\'a symetrick\'a matice. Vy\v{s}et\v{r}ete
extr\'emy kvadratick\'e formy $Q(x)=(Ax,x)$ na jednotkov\'e sf\'e\v{r}e, a odvo\softd{}te
odsud, \v{z}e $A$ m\'a alespo\v{n} jedno vlastn\'{\i} \v{c}\'{\i}slo.

\pr Pomoc\'{\i} v\v{e}ty o~multiplik\'atorech doka\v{z}te:

\quad  a) nerovnost ${x^n + y^n \over 2} \ge \left({x+y \over 2}\right)^n$,
  kde  $n \ge 1, x \ge 0, y \ge 0$;

\quad b) nerovnost mezi aritmetick\'ym a geometrick\'ym pr\accent23um\v{e}rem;

\quad c) H\"olderovu nerovnost: $\sum\limits_{i=1}^n a_ib_i\le\left(\sum
\limits_{i=1}^n a_i^p\right)^{\frac1p}\cdot\left(\sum
\limits_{i=1}^n b_i^q\right)^{\frac1q}$, kde $a_i,b_i\ge0$, $p>1$,
$\frac1p+\frac1q=1$;

\quad d) Hadamardovu nerovnost: $(\det(a_{ij}))^2\le\prod\limits_{i=1}^n
\left(\sum\limits_{j=1}^n a_{ij}^2\right)$
}

\newpage

@ \pr a) max $3$ v~$(\frac13,-\frac23,\frac23)$, min $-3$
v~$(-\frac13,\frac23,-\frac23)$;
b) max $\sqrt{26\over3}$ v $(\frac2{\sqrt{78}},-\frac7{\sqrt{78}},
\frac5{\sqrt{78}})$,
min  $-\sqrt{26\over3}$ v $(-\frac2{\sqrt{78}},\frac7{\sqrt{78}},
-\frac5{\sqrt{78}})$;
\pr a) max $\frac1{3\sqrt3}$ v~bodech
$(\frac1{\sqrt3},\frac1{\sqrt3},\frac1{\sqrt3})$,
$(-\frac1{\sqrt3},-\frac1{\sqrt3},\frac1{\sqrt3})$,
$(\frac1{\sqrt3},-\frac1{\sqrt3},-\frac1{\sqrt3})$,
$(-\frac1{\sqrt3},\frac1{\sqrt3},-\frac1{\sqrt3})$;
 min $-\frac1{3\sqrt3}$ v~bodech
$(-\frac1{\sqrt3},-\frac1{\sqrt3},-\frac1{\sqrt3})$,
$(-\frac1{\sqrt3},\frac1{\sqrt3},\frac1{\sqrt3})$,
$(\frac1{\sqrt3},-\frac1{\sqrt3},\frac1{\sqrt3})$,
$(\frac1{\sqrt3},\frac1{\sqrt3},-\frac1{\sqrt3})$;
b) max $\frac1{3\sqrt6}$ v~bodech
$(\frac2{\sqrt6},-\frac1{\sqrt6},-\frac1{\sqrt6})$,
$(-\frac1{\sqrt6},\frac2{\sqrt6},-\frac1{\sqrt6})$,
$(-\frac1{\sqrt6},-\frac1{\sqrt6},\frac2{\sqrt6})$;
 min $-\frac1{3\sqrt6}$ v~bodech
$(-\frac2{\sqrt6},\frac1{\sqrt6},\frac1{\sqrt6})$,
$(\frac1{\sqrt6},-\frac2{\sqrt6},\frac1{\sqrt6})$,
$(\frac1{\sqrt6},\frac1{\sqrt6},-\frac2{\sqrt6})$;
\pr max $\frac18$ v~$(\frac\pi6,\frac\pi6,\frac\pi6)$,
inf $0$, nenab\'yv\'a se
\pr max ${a^6\over 6^6}$ v~$(\frac a6,\frac a6,\frac a6)$, inf $-\infty$
\pr pro $p=1$ je $f$ na $M$ konstantn\'{\i}; pro $p>1$ je sup $a^p$, nenab\'yv\'a
se, min ${a^p\over n^{p-1}}$ v~$(\frac an,\dots,\frac an)$; pro
$p\in(0,1)$ je inf $a^p$, nenab\'yv\'a se, max ${a^p\over n^{p-1}}$
v~$(\frac an,\dots,\frac an)$;
\pr max $2$ v~$(1,1)$, min $0$ v~(0,0)
\pr max $\sqrt{102}$
v~$(\frac1{\sqrt{102}},-\frac1{\sqrt{102}},\frac{10}{\sqrt{102}}$;
 min $-\sqrt{102}$
v~$(-\frac1{\sqrt{102}},\frac1{\sqrt{102}},-\frac{10}{\sqrt{102}}$;
\pr max $\frac{\sqrt5}2$ v~$(\frac{\sqrt5}2,2\sqrt5)$;
 min $-\frac{\sqrt5}2$ v~$(-\frac{\sqrt5}2,-2\sqrt5)$;
\pr max $\frac14\sqrt k\cdot15^{\frac34}$ v~$(\frac14\sqrt k\root
4\of{15},\frac14\sqrt k\cdot15^{\frac34})$;
 min $-\frac14\sqrt k\cdot15^{\frac34}$ v~$(-\frac14\sqrt k\root
4\of{15},-\frac14\sqrt k\cdot15^{\frac34})$;
\pr Doka\v{z}te, \v{z}e maximum se nab\'yv\'a, a \v{z}e t\v{r}et\'{\i} sou\v{r}adnice p\v{r}\'{\i}slu\v{s}n\'eho
bodu je ko\v{r}enem vhodn\'eho polynomu.
\pr max je $\lambda_{\text{max}}$, t.j. nejv\v{e}t\v{s}\'{\i} vlastn\'{\i} \v{c}\'{\i}slo matice
$A$, a nab\'yv\'a se ve v\v{s}ech vlastn\'{\i}ch vektorech jemu p\v{r}\'{\i}slu\v{s}n\'ych;
 min je $\lambda_{\text{min}}$, t.j. nejmen\v{s}\'{\i} vlastn\'{\i} \v{c}\'{\i}slo matice
$A$, a nab\'yv\'a se ve v\v{s}ech vlastn\'{\i}ch vektorech jemu p\v{r}\'{\i}slu\v{s}n\'ych;
\pr a) pou\v{z}ijte p\v{r}\'{\i}klad 5, ev. vy\v{s}et\v{r}ete minimum funkce ${x^n+y^n\over
2}$ na mno\v{z}in\v{e} $\{(x,y),x+y=c,x\ge0,y\ge0\}$ pro $c>0$;
b) zkoumejte extr\'emy funkce $\root n\of{x_1\cdot\dots\cdot x_n}$ na
mno\v{z}in\v{e} $\{(x_1,\dots,x_n), x_1+\dots+x_n=c,x_1\ge0,\dots,x_n\ge0\}$ pro
$c>0$;
c) zkoumejte extr\'emy funkce
$f(a_1,\dots,a_n,b_1,\dots,b_n)=\sum\limits_{i=1}^n a_ib_i$ na mno\v{z}in\v{e}
$\{(a_1,\dots,a_n,b_1,\dots,b_n), \sum\limits_{i=1}^n a_i^p=A,
\sum\limits_{i=1}^n b_i^p=B,a_i\ge0,b_i\ge0\}$ pro $A,B>0$, a
doka\v{z}te, \v{z}e maximum je v bodech, kde $b_i^q=ca_i^p$ pro vhodn\'e
$c$ a v\v{s}echna $i$;
d) zkoumejte extr\'emy funkce $(\det(a_{ij}))^2$ ($n^2$ prom\v{e}nn\'ych) na mno\v{z}in\v{e}
$\{(a_{ij}), \sum\limits_{j=1}^n a_{ij}^2=S_i, i=1,\dots,n\}$, kde
$S_i>0$, p\v{r}eve\softd{}te na p\v{r}\'{\i}pad $S_i=1$, a pak doka\v{z}te, \v{z}e maximum se nab\'yv\'a
pro ortogon\'aln\'{\i} matice.

\oddel

{\centerline{\smc XV. \v{R}e\v{s}te soustavy diferenci\'aln\'{\i}ch rovnic}

\pr $z'+y=0$, $z'-y'=3z+y$
\pr $z'+z-y=e^x$, $y'-z+y=e^x$
\pr $5z'-2y'+4z-y=e^{-x}$, $z'+8z-3y=5e^{-x}$
\pr $z'+3z+y=0$, $y'-z+y=0$, $y(0)=z(0)=1$
\pr $z'=y-7z$, $y'+2z+5y=0$
\pr $z'=2y-5z+e^x$, $y'=z-6y+e^{-2x}$
\pr $z''+y'+z=e^x$, $z'+y''=1$
\pr $u'=w+v-u$, $v'=w+u-v$, $w'=u+v+w$, ($u(0)=1$, $v(0)=w(0)=0$)
\pr $u'=v+w$, $v'=u+w$, $w'=u+v$, ($u(0)=-1$, $v(0)=1$, $z(0)=0$)
}

@ \pr $z={3B-A\over4}e^x+{A+B\over 4}e^{-3x}$,
$y={A-3B\over4}e^x+{3A+3B\over 4}e^{-3x}$
\pr $z=e^x+de^{-4x}+c$, $y=e^x-de^{-4x}+c$
\pr $z=-2e^{-x}+c e^x+d e^{-2x}$, $y=3e^{-x}+3c e^x+2d e^{-2x}$
\pr $z=-(c+d)xe^{-2x}+d e^{-2x}$, $y=(c+d)xe^{-2x}+c e^{-2x}$, s
po\v{c}. podm. : $z=-2xe^{-2x}+e^{-2x}$, $y=2xe^{-2x}+e^{-2x}$
\pr $z=(c-d)e^{-6x}\sin x +d e^{-6x}\cos x$, $y=c e^{-6x}\cos x
+(c-2d)e^{-6x}\sin x$
\pr $z=\frac7{40}e^x+\frac15 e^{-2x}+\frac23(c+d)e^{-4x}
-\frac13(2c-d)e^{-7x}$,
$y=\frac1{40}e^x+\frac3{10}e^{-2x}+\frac13(c+d)e^{-4x}+\frac13
(2c-d)e^{-7x}$
\pr $z(x)=e^x-\frac16x^3-\frac {b+c}2 x^2+c+dx$,
$y=-e^x+\frac1{24}x^4+a+bx+\frac{b+c}6x^3+\frac d2 x^2$
\pr
$u={a+b-c\over3}e^{-x}+{a+b+2c\over6}e^{2x}+{a-b\over2}e^{-2x}$,
$v={a+b-c\over3}e^{-x}+{a+b+2c\over6}e^{2x}+{b-a\over2}e^{-2x}$,
$w={c-a-b\over3}e^{-x}+{a+b+2c\over 3}e^{2x}$, s po\v{c}.podm.
$u=\frac13e^{-x}+\frac16e^{2x}+\frac12e^{-2x}$,
$v=\frac13e^{-x}+\frac16e^{2x}-\frac12e^{-2x}$,
$w=-\frac13e^{-x}+\frac13e^{2x}$
\pr $u={a+b+c\over 3}e^{2x}-{b+c-2a\over3}e^{-x}$,
$v={a+b+c\over 3}e^{2x}-{a+c-2b\over3}e^{-x}$,
$w={a+b+c\over 3}e^{2x}-{a+b-2c\over3}e^{-x}$
s po\v{c}. podm. $u=-e^{-x}$, $v=e^{-x}$, $w=0$

\newpage

{
\centerline{\smc XVI. Diferenci\'aln\'{\i} rovnice}

\pr Uka\v{z}te (na p\v{r}\'{\i}kladu), \v{z}e v\v{e}ta o spojit\'e z\'avislosti \v{r}e\v{s}en\'{\i}
(line\'arn\'{\i} diferenci\'aln\'{\i} rovnice s konstantn\'{\i}mi koeficienty)
neplat\'{\i} na neomezen\'em intervalu (nap\v{r}. $[0,\infty)$).

\pr Nech\softt{} $f:[0,\infty)\times\Bbb R^n\to\Bbb R^n$ je t\v{r}\'{\i}dy $\Cal
C^1$ a existuj\'{\i} konstanty $M,A>0$ takov\'e, \v{z}e $\|f(t,u)\|\le
A+M\|u\|$ plat\'{\i} pro v\v{s}echna $t\ge0$ a $u\in\Bbb R^n$. Doka\v{z}te, \v{z}e
potom pro libovolnou po\v{c}\'ate\v{c}n\'{\i} podm\'{\i}nku $y(0)=y_0$ existuje
maxim\'aln\'{\i} \v{r}e\v{s}en\'{\i} rovnice $y'=f(y,t)$ spl\v{n}uj\'{\i}c\'{\i} po\v{c}\'ate\v{c}n\'{\i} podm\'{\i}nku
a definovan\'e na cel\'em $[0,\infty)$.

\pr Uva\v{z}ujme rovnici $y'=Ay$, kde $A$ je \v{c}tvercov\'a matice.
Ur\v{c}ete, za jak\'ych podm\'{\i}nek na vlastn\'{\i} \v{c}\'{\i}sla matice $A$ plat\'{\i}:

(a) Ka\v{z}d\'e \v{r}e\v{s}en\'{\i} m\'a v $+\infty$ limitu $0$ (t.j. $0$ je stabiln\'{\i}
\v{r}e\v{s}en\'{\i}).

(b) Existuje nenulov\'e periodick\'e \v{r}e\v{s}en\'{\i}.

(c) Ka\v{z}d\'e \v{r}e\v{s}en\'{\i} je periodick\'e.

*\pr Uva\v{z}ujme rovnici $y'=Ay+g(y)$, kde $A$ je \v{c}tvercov\'a matice a
$g:\Bbb R^n\to\Bbb R^n$ je funkce t\v{r}\'{\i}dy $\Cal C^1$, kter\'a je u
nuly \clqq velmi mal\'a\crqq ve smyslu, \v{z}e $g(0)=0$ a $g'(0)=0$.
Co lze \v{r}\'{\i}ci o stabilit\v{e} nulov\'eho \v{r}e\v{s}en\'{\i} (pou\v{z}ijte i p\v{r}edchoz\'{\i}
p\v{r}\'{\i}klad)?

\pr Uva\v{z}ujme rovnici $y''=f(y)$. Nalezn\v{e}te funkci $F:\Bbb
R^2\to\Bbb R$ takovou, \v{z}e pro ka\v{z}d\'e \v{r}e\v{s}en\'{\i} $y$ na\v{s}\'{\i} rovnice je
$F(y(t),y'(t))$ konstantn\'{\i}. ($y(t)$ m\accent23u\v{z}e znamenat v\'ychylku
hmotn\'eho bodu v \v{c}ase $t$, $f(y(t))$ pak s\'{\i}lu p\accent23usob\'{\i}c\'{\i} na n\v{e}j v
\v{c}ase $t$, a rovnice $F(y(t),y'(t))=c$ pak znamen\'a z\'akon zachov\'an\'{\i}
energie.)

\pr Aplikujte p\v{r}edchoz\'{\i} p\v{r}\'{\i}klad na rovnici kyvadla $y''=\sin y$.

\pr Uva\v{z}ujme rovnici $y''+\alpha y'+y=0$. Vy\v{s}et\v{r}ete rozd\'{\i}l v
chov\'an\'{\i} jej\'{\i}ch \v{r}e\v{s}en\'{\i} pro $\alpha=0$ a $\alpha>0$. Uka\v{z}te, \v{z}e pro
$\alpha>0$ energie syst\'emu (ve smyslu p\v{r}edminul\'eho p\v{r}\'{\i}kladu) s
\v{c}asem kles\'a (a tedy $\alpha$ m\'a v\'yznam t\v{r}en\'{\i}).
}

@ \pr Nap\v{r}. rovnice $y'=y$. Pak pro dv\v{e} r\accent23uzn\'a \v{r}e\v{s}en\'{\i} $y_1$, $y_2$
je $\sup\limits_{t\ge0}|y_1(t)-y_2(t)|=+\infty$
\pr Pou\v{z}ijte v\v{e}tu o existenci maxim\'aln\'{\i}ho \v{r}e\v{s}en\'{\i}, Gronwallovo
lemma a \clqq v\v{e}tu o opou\v{s}t\v{e}n\'{\i} kompaktu\crqq.
\pr (a) $\Re\lambda<0$ pro ka\v{z}d\'e vlastn\'{\i} \v{c}\'{\i}slo $\lambda$.
(b) $\Re\lambda=0$ pro n\v{e}jak\'e vlastn\'{\i} \v{c}\'{\i}slo $\lambda$, pro kter\'e
existuje vlastn\'{\i} vektor $\bold{v}$ takov\'y, \v{z}e $(\lambda
I-A)\bold{u}=\bold{v}$ nem\'a \v{r}e\v{s}en\'{\i}, t.j. jedna z Jordanov\'ych
bun\v{e}k p\v{r}\'{\i}slu\v{s}n\'ych k $\lambda$ je $1\times1$ (chceme-li
nekonstantn\'{\i}, pak nav\'{\i}c $\lambda\ne0$).
(c) $\Re\lambda=0$ pro ka\v{z}d\'e vlastn\'{\i} \v{c}\'{\i}slo a nav\'{\i}c existuje b\'aze
z vlastn\'{\i}ch vektor\accent23u.
\pr
\pr $F(u,v)=\frac12v^2-g(u)$, kde $g$ je primitivn\'{\i} funkce k $f$.
\pr $\frac12(y')^2+\cos y=\const$
\pr Pro $\alpha=0$ jsou \v{r}e\v{s}en\'{\i} periodick\'a, pro $\alpha>0$ maj\'{\i} v
$+\infty$ limitu $0$.

\oddel

{\centerline{\smc Test \v{c}\'{\i}slo II/3 -- prvn\'{\i} verze}

\pr Polo\v{z}me $F(x,y)=x^x-y^y$ pro $x>0$ a $y>0$.

(a) Doka\v{z}te, \v{z}e existuje pr\'av\v{e} jedno $a\in(0,\frac12)$, pro kter\'e
plat\'{\i} $F(a,\frac12)=0$. \hfill(5 bod\accent23u)

(b) Doka\v{z}te, \v{z}e mno\v{z}ina $\{(x,y)\mid F(x,y)=0\}$ je v okol\'{\i} bodu
$(a,\frac12)$ grafem n\v{e}jak\'e hladk\'e funkce $y=f(x)$ spl\v{n}uj\'{\i}c\'{\i}
$f(a)=\frac12$.\hfill(7 bod\accent23u)

(c) Rozhodn\v{e}te, zda je funkce $f$ na n\v{e}jak\'em okol\'{\i} bodu $a$
rostouc\'{\i} ev. klesaj\'{\i}c\'{\i}.\hfill(6 bod\accent23u)

\bf \v{R}e\v{s}en\'{\i}. \rm (a) Polo\v{z}me
$g(x)=F(x,\frac12)=x^x-\left(\frac12\right)^{\frac12}$. Pak $g$
je funkce spojit\'a na $(0,\infty)$ a p\v{r}itom $g(\frac12)=0$ a
$\lx{0+}g(x)=\lx{0+}\left(e^{x\log
x}-\left(\frac12\right)^{\frac12}\right)=1-\left(\frac12\right)^{\frac12}
>0$. D\'ale $g'(x)=x^x(\log x+1)$, tedy $g$ je klesaj\'{\i}c\'{\i} na
$(0,\frac1e)$ a rostouc\'{\i} na $(\frac1e,\infty)$. Proto\v{z}e
$\frac1e<\frac12$, dost\'av\'ame $g(x)<0$ pro
$x\in[\frac1e,\frac12)$, a tedy existuje jedin\'e $a\in(0,\frac1e)$
spl\v{n}uj\'{\i}c\'{\i} $g(a)=0$, co\v{z} je tot\'e\v{z}, co $F(a,\frac12)=0$.

(b) Pou\v{z}ijeme v\v{e}tu o implicitn\'{\i}ch funkc\'{\i}ch. Ov\v{e}\v{r}me jej\'{\i}
p\v{r}edpoklady. Funkce $F$ je z\v{r}ejm\v{e} t\v{r}\'{\i}dy $\Cal C^\infty$ na
$(0,,\infty)\times(0,\infty)$, $F(a,\frac12)=0$ z volby $a$, a
$$\pdy
F(a,\frac12)=\left(-y^y(\log y+1)\right)_{\smallmatrix
x=a\\y=\frac12\endsmallmatrix}=-\left(\frac12\right)^{\frac12}
(\log\frac12+1)\ne0.$$
Nyn\'{\i} u\v{z} tvrzen\'{\i} plyne okam\v{z}it\v{e} z v\v{e}ty o implicitn\'{\i}ch funkc\'{\i}ch.

\newpage

(c) Funkce $f$ z (b) spl\v{n}uje na okol\'{\i} bodu $a$ rovnici
$x^x-f(x)^{f(x)}=0$, zderivujeme-li tuto rovnost, dostaneme
$$x^x(\log x+1)-f(x)^{f(x)}(\log f(x)\cdot f'(x)+f'(x))=0,$$
neboli
$$f'(x)={x^x(\log x+1)\over f(x)^{f(x)}(\log f(x)+1)}=
{\log x+1\over \log f(x)+1}.$$
Pro posledn\'{\i} rovnost jsme
pou\v{z}ili, \v{z}e $x^x=f(x)^{f(x)}$ d\'{\i}ky volb\v{e} $f$. Tedy $f'(a)=
{\log a+1\over \log\frac12 +1}$. P\v{r}itom $\log\frac12+1>0$, nebo\softt{}
$\frac12>\frac1e$, zat\'{\i}mco $\log a+1<0$, nebo\softt{} $a<\frac1e$ (viz
(a)). Tedy $f'(a)<0$, a proto\v{z}e $f$ je $\Cal C^1$, je $f'$
spojit\'a, a tedy $f'<0$ na n\v{e}jak\'em okol\'{\i} bodu $a$. Na tomto okol\'{\i}
je pak $f$ klesaj\'{\i}c\'{\i}.

\pr Ur\v{c}ete supremum a infimum funkce $f(x,y)=x+y^2$ na mno\v{z}in\v{e}

$M=\{(x,y)\mid x^2+y<\frac34, y\ge-\frac14\}$ a zjist\v{e}te, zda $f$
t\v{e}chto hodnot na $M$ nab\'yv\'a.\hfill (16 bod\accent23u)

\bf \v{R}e\v{s}en\'{\i}. \rm Funkce $f$ je spojit\'a, dokonce $\Cal C^\infty$ na
cel\'em $\Bbb R^2$.
Mno\v{z}ina $M$ je omezen\'a, nebo\softt{} pro $(x,y)\in M$ nutn\v{e} plat\'{\i}
$-\frac14\le y<\frac34$ a $x^2<1$. Tedy $\overline M$ je
kompaktn\'{\i}, a $f$ nab\'yv\'a extr\'em\accent23u na $\overline M$. Hledejme
podez\v{r}el\'e body.

Proto\v{z}e $\pdx f=1\ne0$, ve vnit\v{r}ku \v{z}\'adn\'e podez\v{r}el\'e body nejsou.
Zb\'yv\'a vy\v{s}et\v{r}it hranici. Ta sest\'av\'a ze dvou \v{c}\'ast\'{\i}, z \'use\v{c}ky
$y=-\frac14, x\in[-1,1]$ a \v{c}\'asti paraboly $x^2+y=\frac34,
y\ge-\frac14$. Na zm\'{\i}n\v{e}n\'e \'use\v{c}ce m\'ame
$f(x,-\frac14)=x+\frac1{16}$, co\v{z} je nejv\v{e}t\v{s}\'{\i} pro $x=1$ a
nejmen\v{s}\'{\i} pro $x=-1$, p\v{r}itom $f(1,-\frac14)=\frac{17}{16}$ a
$f(-1,-\frac14)=-\frac{15}{16}$.

Nyn\'{\i} vy\v{s}et\v{r}eme chov\'an\'{\i} $f$ na parabole $x^2+y=\frac34$. Proto\v{z}e
gradient funkce $x^2+y-\frac34$ je $(2x,1)$, co\v{z} nikdy nen\'{\i}
nulov\'y vektor, podle v\v{e}ty o multiplik\'atorech dost\'av\'ame, \v{z}e, je-li
v $(x,y)$ lok\'aln\'{\i} extr\'em vzhledem k na\v{s}\'{\i} parabole, pak existuje
$\lambda\in\Bbb R$, \v{z}e $1+\lambda\cdot 2x=0$ a $2y+\lambda=0$,
tedy $1-4xy=0$, neboli $y={1\over 4x}$. Pokud takov\'y bod le\v{z}\'{\i} na
na\v{s}\'{\i} parabole, plat\'{\i} $x^2+{1\over 4x}=\frac34$, neboli
$4x^3-3x+1=0$. Tato rovnice m\'a ko\v{r}en $x=-1$ a dvojn\'asobn\'y ko\v{r}en
$x=\frac12$. Pro $x=-1$ dost\'av\'ame $y=-\frac14$, kter\'y\v{z}to bod u\v{z}
mezi podez\v{r}el\'ymi body m\'ame. Pro $x=\frac12$ dost\'av\'ame $y=\frac12$
a $f(\frac12,\frac12)=\frac34$. Je\v{s}t\v{e} bychom m\v{e}li za\v{r}adit druh\'y
krajn\'{\i} bod $(1,-\frac14)$, ale ten ji\v{z} mezi podez\v{r}el\'ymi body je.

Tedy z\v{r}ejm\v{e} $f$ nab\'yv\'a na $\overline M$ maxima $\frac{17}{16}$ v
bod\v{e} $(1,-\frac14)$ a minima $-\frac{15}{16}$ v bod\v{e}
$(-1,-\frac14)$. Tato \v{c}\'{\i}sla jsou z\'arove\v{n} supremem a infimem
funkce $f$ na $M$, ale $f$ na $M$ t\v{e}chto hodnot nenab\'yv\'a.

\pr Uva\v{z}ujme soustavu diferenci\'aln\'{\i}ch rovnic $y'+z=2y$,
$z'+2z=3y$. Ur\v{c}ete mno\v{z}iny

$S=\{(a,b)\in\Bbb R^2\mid y(0)=a,z(0)=b, (y,z) \text{ je \v{r}e\v{s}en\'{\i}m
 }\Rightarrow
\lx{+\infty}y(x)=\lx{+\infty}z(x)=0\}$ a

$U=\{(a,b)\in\Bbb R^2\mid y(0)=a,z(0)=b, (y,z) \text{ je \v{r}e\v{s}en\'{\i}m
 }\Rightarrow
\lx{-\infty}y(x)=\lx{-\infty}z(x)=0\}$.
\vskip-10pt
\hfill (16~bod\accent23u)

\bf \v{R}e\v{s}en\'{\i}. \rm  Tato soustava m\'a matici $\pmatrix 2 & -1 \\ 3 &
-2 \endpmatrix$. Spo\v{c}t\v{e}me jej\'{\i} vlastn\'{\i} \v{c}\'{\i}sla:
$$\det\pmatrix
\lambda-2 & 1 \\ -3 &
\lambda+2\endpmatrix=\lambda^2-4+3=\lambda^2-1=(\lambda-1)
(\lambda+1),$$ tedy vlastn\'{\i} \v{c}\'{\i}sla jsou $1$ a $-1$. Vlastn\'{\i} vektor
k \v{c}\'{\i}slu $1$ je nap\v{r}\'{\i}klad $\pmatrix 1 \\ 1 \endpmatrix$ a k \v{c}\'{\i}slu
$-1$ nap\v{r}\'{\i}klad $\pmatrix 1\\ 3\endpmatrix$. Tedy substituc\'{\i}
$y=u+v$, $z=u+3v$ dostaneme soustavu $u'=u$, $v'=-v$, jej\'{\i}\v{z}
\v{r}e\v{s}en\'{\i} m\'a tvar $u=c
e^x$, $v=d e^{-x}$, a tak p\accent23uvodn\'{\i} soustava m\'a \v{r}e\v{s}en\'{\i}
$y=c e^x+d e^{-x}$, $z= c e^x+ 3d e^{-x}$, kde $c,d\in\Bbb R$.
P\v{r}itom z\v{r}ejm\v{e} $\lx{+\infty}y(x)=\lx{+\infty}z(x)=0$, pr\'av\v{e} kdy\v{z}
$c=0$; a $\lx{-\infty}y(x)=\lx{-\infty}z(x)=0$, pr\'av\v{e} kdy\v{z} $d=0$.
Nav\'{\i}c $c={3y(0)-z(0)\over2}$ a $d={z(0)-y(0)\over 2}$, tedy
$S=\{(a,b)\mid b=3a\}$ a $U=\{(a,b)\mid a=b\}$.
}

\newpage
%\oddel

{\centerline{\smc Test \v{c}\'{\i}slo II/3 -- druh\'a verze}

\pr Polo\v{z}me $F(x,y)=x^x-y^y$ pro $x>0$ a $y>0$.

(a) Doka\v{z}te, \v{z}e existuje pr\'av\v{e} jedno $b\in(\frac13,1)$, pro kter\'e
plat\'{\i} $F(\frac13,b)=0$. \hfill(5 bod\accent23u)

(b) Doka\v{z}te, \v{z}e mno\v{z}ina $\{(x,y)\mid F(x,y)=0\}$ je v okol\'{\i} bodu
$(\frac13,b)$ grafem n\v{e}jak\'e hladk\'e funkce $x=f(y)$ spl\v{n}uj\'{\i}c\'{\i}
$f(b)=\frac13$.\hfill(7 bod\accent23u)

(c) Rozhodn\v{e}te, zda je funkce $f$ na n\v{e}jak\'em okol\'{\i} bodu $b$
rostouc\'{\i} ev. klesaj\'{\i}c\'{\i}.\hfill(6 bod\accent23u)

\bf \v{R}e\v{s}en\'{\i}. \rm (a) Polo\v{z}me $h(y)=F(\frac13,y)$ a postupujme
analogicky jako v prvn\'{\i} verzi. $h$ je spojit\'a, $h(\frac13)=0$,
$h(1)<0$, h roste na $(0,\frac1e)$ a kles\'a na $(\frac1e,\infty)$,
p\v{r}i\v{c}em\v{z} $\frac13<\frac1e<1$, tedy existuje jedin\'e $b$ dle zad\'an\'{\i},
nav\'{\i}c $b\in(\frac1e,1)$.

(b) Jako v prvn\'{\i} verzi: $F$ je $\Cal C^\infty$, $F(\frac13,b)=0$,
$\pdx
F(\frac13,b)=\left(\frac13\right)^{\frac13}(\log\frac13+1)\ne0$.

(c) Stejn\v{e} jako v prvn\'{\i} verzi spo\v{c}teme, \v{z}e
$f'(a)= {\log b+1\over \log\frac13 +1}$, tentokr\'at $\log b+1>0$ a
$\log\frac13>0$, tedy $f$ je klesaj\'{\i}c\'{\i} na okol\'{\i} $b$.

\pr Ur\v{c}ete supremum a infimum funkce $f(x,y)=x^2+y$ na mno\v{z}in\v{e}

$M=\{(x,y)\mid x+y^2<\frac34, x\ge0\}$ a zjist\v{e}te, zda $f$
t\v{e}chto hodnot na $M$ nab\'yv\'a.\hfill (16 bod\accent23u)

\bf \v{R}e\v{s}en\'{\i}. \rm Podobn\v{e} jako v prvn\'{\i} verzi $M$ je omezen\'a, a tedy
$\overline M$ je kompaktn\'{\i}. Proto\v{z}e $\pdy f=1$, nejsou extr\'emy
uvnit\v{r}, zb\'yv\'a hranice. Hranice je tvo\v{r}ena op\v{e}t \'use\v{c}kou a \v{c}\'ast\'{\i}
paraboly. Na parabole najdeme op\v{e}t podez\v{r}el\'e body $(-\frac14,1)$
a $(\frac12,\frac12)$, ten prvn\'{\i} nele\v{z}\'{\i} v $\overline M$, pro
druh\'y plat\'{\i} $f(\frac12,\frac12)=\frac34$. \'Use\v{c}ka m\'a tvar $x=0$,
$y\in[-\frac{\sqrt3}2,\frac{\sqrt3}2]$, $f(0,y)=y$, tedy nejv\v{e}t\v{s}\'{\i}
hodnota je v $(0,\frac{\sqrt3}2)$ a nejmen\v{s}\'{\i} v
$(0,-\frac{\sqrt3}2)$. Tedy supremum je $\frac{\sqrt3}2$ a
infimum $-\frac{\sqrt3}2$, p\v{r}i\v{c}em\v{z} \v{z}\'adn\'e z t\v{e}chto hodnot $f$ na
$M$ nenab\'yv\'a.

\pr Uva\v{z}ujme soustavu diferenci\'aln\'{\i}ch rovnic $y'+y=4z$,
$z'-2y=z$. Ur\v{c}ete mno\v{z}iny

$S=\{(a,b)\in\Bbb R^2\mid y(0)=a,z(0)=b, (y,z) \text{ je \v{r}e\v{s}en\'{\i}m
 }\Rightarrow
\lx{+\infty}y(x)=\lx{+\infty}z(x)=0\}$ a

$U=\{(a,b)\in\Bbb R^2\mid y(0)=a,z(0)=b, (y,z) \text{ je \v{r}e\v{s}en\'{\i}m
 }\Rightarrow
\lx{-\infty}y(x)=\lx{-\infty}z(x)=0\}$.
\vskip-10pt
\hfill (16~bod\accent23u)

\bf \v{R}e\v{s}en\'{\i}. \rm  Tato soustava m\'a matici $\pmatrix -1 & 4 \\ 2 &
1 \endpmatrix$. Spo\v{c}t\v{e}me jej\'{\i} vlastn\'{\i} \v{c}\'{\i}sla:
$$\det\pmatrix
\lambda+1 & -4 \\ -2 &
\lambda-1\endpmatrix=\lambda^2-1-8=\lambda^2-9=(\lambda-3)
(\lambda+3),$$
tedy vlastn\'{\i} \v{c}\'{\i}sla jsou $3$ a $-3$. Vlastn\'{\i} vektor
k \v{c}\'{\i}slu $3$ je nap\v{r}\'{\i}klad $\pmatrix 1 \\ 1 \endpmatrix$ a k \v{c}\'{\i}slu
$-3$ nap\v{r}\'{\i}klad $\pmatrix 2\\ -1\endpmatrix$. Tedy substituc\'{\i}
$y=u+2v$, $z=u-v$ dostaneme soustavu $u'=3u$, $v'=-3v$, jej\'{\i}\v{z}
\v{r}e\v{s}en\'{\i} m\'a tvar $u=c
e^{3x}$, $v=d e^{-3x}$, a tak p\accent23uvodn\'{\i} soustava m\'a \v{r}e\v{s}en\'{\i}
$y=c e^{3x}+2d e^{-3x}$, $z= c e^{3x}-d e^{-3x}$, kde $c,d\in\Bbb R$.
P\v{r}itom z\v{r}ejm\v{e} $\lx{+\infty}y(x)=\lx{+\infty}z(x)=0$, pr\'av\v{e} kdy\v{z}
$c=0$; a $\lx{-\infty}y(x)=\lx{-\infty}z(x)=0$, pr\'av\v{e} kdy\v{z} $d=0$.
Nav\'{\i}c $c={y(0)+2z(0)\over3}$ a $d={y(0)-z(0)\over 3}$, tedy
$S=\{(a,b)\mid a=-2b\}$ a $U=\{(a,b)\mid a=b\}$.
}

\oddel

{
\centerline{\smc XVII. Najd\v{e}te v\v{s}echna maxim\'aln\'{\i} \v{r}e\v{s}en\'{\i}
diferenci\'aln\'{\i}ch rovnic}

\pr $y'' + 4y' + 4y = 0$
\pr $y'' - 3y' + 2y = 0$
\pr $y'' - 6y' +13 = 0$
\pr $y''-2y'-3y = e^{4x}$

\pr $y'' - 2y' + 2y = \cos x$
\pr $y''' - y'' - 2y' = e^{2x} + x^3 + 3x^2 +1$
\pr $y''  + 3y' + 2y = \sin x  + \sin 2x$
\pr $4y'''  + y' = 3e^x + 2 \sin \frac x2$
\pr $y'' - 3y' + 2y = \frac{e^{2x}}{\sqrt{1 - e^{2x}}}$
\pr $y'' - y = \frac{e^x - e^{-x}}{e^x + e^{-x}}$

\pr $y'' - 2y' + y = \frac{e^x}{x^2+x+1}$
\pr $y''  + y = \tg x$
\pr $y''  - 2y' + y  = \frac{e^x}{\sqrt{4-x^2}}$}

\newpage

{
\centerline{\smc XVIII. Trigonometrick\'e a Fourierovy \v{r}ady}

\pr (a) Se\v{c}t\v{e}te trigonometrickou \v{r}adu $\sn0 {\cos nx \over n!}$.
(b) Spo\v{c}t\v{e}te $\int_{\frac\pi2}^{\frac52\pi} e^{\cos x}\cos(\sin
x)\cos nx\d x$ ($n=0,1,2,\dots$)

\pr Se\v{c}t\v{e}te trigonometrickou \v{r}adu $\sn1 q^n \sin nx$, kde
$0<q<1$. Je tato \v{r}ada Fourierovou \v{r}adou sv\'eho sou\v{c}tu?

\pr Pro n\'asleduj\'{\i}c\'{\i} funkce najd\v{e}te trigonometrickou \v{r}adu, jej\'{\i}m\v{z}
jsou sou\v{c}tem, a v\'ysledek aplikujte v uveden\'ych bodech.
\qquad (a) $x^2$ na $(-\pi,\pi)$, $x=0$; \qquad
(b) $x$ na $(-\pi,\pi)$, $x=\frac\pi2$;

(c) $\sin ax$ na $(-\pi,\pi)$ ($a\in\Bbb R\setminus\Bbb Z$),
$x=\frac\pi2$; \qquad
(d) $x^2$ na $(0,2\pi)$, $x=0$.

\pr (a) Napi\v{s}te funkci $f(x)=x$ na $(0,\pi)$ jako sou\v{c}et cosinov\'e
\v{r}ady.

(b) Napi\v{s}te funkci $f(x)=x^2$ na $(-\pi,0)$ jako sou\v{c}et sinov\'e
\v{r}ady.

\pr Doka\v{z}te, \v{z}e pro $|x|<1$ plat\'{\i} $\sin\pi x=\pi
x\cdot\prod\limits_{n=1}^\infty\left(1-\frac{x^2}{n^2}\right)$.
(Rozvi\v{n}te ve Fourierovu \v{r}adu funkci $\cos\alpha x$ na
$(-\pi,\pi)$, aplikujte pro $x=\pi$, vyj\'ad\v{r}ete $\cotg \alpha\pi
-{1\over\alpha\pi}$ a v\'yslednou rovnost integrujte.)
}

\oddel

{\centerline{\smc Test \v{c}\'{\i}slo II/4 -- prvn\'{\i} verze}

\pr  Najd\v{e}te v\v{s}echna maxim\'aln\'{\i} \v{r}e\v{s}en\'{\i} diferenci\'aln\'{\i} rovnice

\qquad\qquad $y'''-y''+4y'-4y=\sin x+\sin 2x$ \hfill (18 bod\accent23u)

\bf \v{R}e\v{s}en\'{\i}. \rm Charakteristick\'y polynom t\'eto rovnice je
$$\lambda^3-\lambda^2 + 4\lambda -4 =(\lambda^2+4)(\lambda-1),$$
tedy m\'a t\v{r}i r\accent23uzn\'e ko\v{r}eny $1$, $2i$, $-2i$, tud\'{\i}\v{z} fundament\'aln\'{\i}
syst\'em homogenn\'{\i} rovnice tvo\v{r}\'{\i} funkce $e^x$, $\cos 2x$, $\sin
2x$. Prav\'a strana je sou\v{c}tem dvou kvazipolynom\accent23u ($\sin x$ a $\sin
2x$), a tedy hledejme partikul\'arn\'{\i} \v{r}e\v{s}en\'{\i} $y_1$ rovnice
$y'''-y''+4y'-4y=\sin x$
a partikul\'arn\'{\i} \v{r}e\v{s}en\'{\i} $y_2$ rovnice $y'''-y''+4y'-4y=\sin 2x$.
Proto\v{z}e $i$ nen\'{\i} ko\v{r}enem charakteristick\'eho polynomu, m\'a $y_1$
tvar $y_1=a\cos x+b\sin x$ pro n\v{e}jak\'e konstanty $a$, $b$.
Pak plat\'{\i}
$$ \matrix
y_1=a\cos x+b\sin x, & y_1^\prime=-a\sin x+b\cos x, \\
y_1^{\prime\prime}=-a\cos x-b\sin x, &
y_1^{\prime\prime\prime}=a\sin x-b\cos x,
\endmatrix$$
a tedy
$$y_1^{\prime\prime\prime}-y_1^{\prime\prime}+4y_1^{\prime}-4y_1
=(-b+a+4b-4a)\cos x + (a+b-4a-4b)\sin x=3(b-a)\cos x-3(a+b)\sin
x,$$
odkud dost\'av\'ame
$$3(b-a)=0,\quad -3(a+b)=1, \text{ neboli }\quad a=b=-\frac16.$$
Proto\v{z}e $2i$ je ko\v{r}enem charakteristick\'eho polynomu n\'asobnosti
$1$, m\'a $y_2$ tvar $y_2=x\cdot(c\cos 2x+d\sin 2x)$ pro n\v{e}jak\'e
konstanty $c$, $d$. Pak plat\'{\i}:
{\eightpoint
$$ \matrix \format \l &\quad \l \\
y_2=x(c\cos 2x+d\sin 2x), & y_2^\prime=x(-2c\sin 2x+2d\cos 2x)+
(c\cos 2x+d\sin 2x), \\
y_2^{\prime\prime}=x(-4c\cos 2x-4d\sin 2x)+2(-2c\sin 2x+2d\cos 2x), &
y_2^{\prime\prime\prime}=x(8c\sin 2x-8d\cos 2x)+3(-4c\cos 2x-4d\sin 2x),
\endmatrix$$}
a tedy
$$y_2^{\prime\prime\prime}-y_2^{\prime\prime}+4y_2^{\prime}-4y_2
=(-12c-4d+4c)\cos 2x + (-12d+4c+4d)\sin 2x=-4(2c+d)\cos
2x+4(c-2d)\sin 2x,$$
odkud dost\'av\'ame
$$-4(2c+d)=0,\quad 4(c-2d)=1, \text{ neboli } d=-\frac{1}{10},\
c=\frac1{20}.$$

\newpage

Obecn\'e \v{r}e\v{s}en\'{\i} na\v{s}\'{\i} rovnice m\'a tedy tvar
$$y=\frac x{20}(\cos 2x-2 \sin 2x) -\frac16(\sin x+\cos x) + A e^x + B \cos 2x +
C \sin 2x,\quad A,B,C\in\Bbb R.$$

\pr  Se\v{c}t\v{e}te \v{r}adu $\sn0 {(-1)^n\cos 3(n+\frac12)x\over (2n+1)!}$.
\hfill (14 bod\accent23u)

\bf \v{R}e\v{s}en\'{\i}. \rm Plat\'{\i} $\cos 3(n+\frac12)x= \Re e^{3(n+\frac12)ix}$,
a tedy m\accent23u\v{z}eme ps\'at

{%\eightpoint
$$\align
\sn0 &{(-1)^n\cos 3(n+\frac12)x\over (2n+1)!}=\sn0 {(-1)^n\Re
e^{3(n+\frac12)ix}\over (2n+1)!}=
\Re \sn0 {(-1)^n e^{3(n+\frac12)ix}\over (2n+1)!}\\& = \Re \sn0 {(-1)^n
\left(e^{\frac32ix}\right)^{2n+1}\over (2n+1)!}
=\Re \sin\left(e^{\frac32ix}\right)
= \Re \sin\left(\cos\frac32x+i\sin\frac32x\right)
\\&=\Re \left(\sin\left(\cos\frac32x\right)\cos\left(i\sin\frac32x\right)+
\cos\left(\cos\frac32x\right)\sin\left(i\sin\frac32x\right)\right)
\\ &=\Re \left(\sin\left(\cos\frac32x\right)\cdot{e^{-\sin\frac32x}
+e^{\sin\frac32x} \over2}+\cos\left(\cos\frac32x\right)
\cdot{e^{-\sin\frac32x}-e^{\sin\frac32} \over2i}\right)
\\ &=\sin\left(\cos\frac32x\right)\cdot{e^{-\sin\frac32x}
+e^{\sin\frac32x} \over2}
\endalign$$}

Pou\v{z}ili jsme rozvoj funkce $\sin$ v mocninnou \v{r}adu a vzore\v{c}ky
$\sin z={e^{iz}-e^{-iz}\over 2i}$, $\cos z={e^{iz}+e^{-iz}\over
2}$, kter\'e plat\'{\i} pro v\v{s}echna komplexn\'{\i} $z$.

\pr Polo\v{z}me $f(x)=\cos^3 x+\sgn(x-1)$.

(a) Rozvi\v{n}te $f$ ve Fourierovu \v{r}adu na intervalu $(-\pi,\pi)$.
\hfill (10 bod\accent23u)

(b) Ur\v{c}ete sou\v{c}et \v{r}ady z (a) v\v{s}ude, kde \v{r}ada konverguje.
\hfill (8 bod\accent23u)

\bf \v{R}e\v{s}en\'{\i}. \rm (a) Z definice Fourierov\'ych koeficient\accent23u je
z\v{r}ejm\'e, \v{z}e Fourierovy koeficienty funkce $f$ jsou sou\v{c}tem
Fourierov\'ych koeficient\accent23u funkc\'{\i} $\cos^3 x$ a $\sgn(x-1)$. Nejprve
se v\v{e}nujme prvn\'{\i} funkci. Podle goniometrick\'ych vzore\v{c}k\accent23u plat\'{\i}
$$\align
\cos^3 x &= \cos x \cdot {\cos 2x-1\over 2}=-\frac12 \cos x
+\frac12 \cos x\cdot\cos 2x\\ & =-\frac12 \cos x +\frac14(\cos 3x
+\cos x)=-\frac14\cos x+\frac14\cos 3x,\endalign$$
co\v{z} je z\'arove\v{n} rozvoj funkce $\cos^3 x$ ve Fourierovu \v{r}adu (v\'yraz
vpravo je kone\v{c}n\'a (a tedy nap\v{r}. stejnom\v{e}rn\v{e} konvergentn\'{\i})
trigonometrick\'a \v{r}ada, a proto je Fourierovou \v{r}adou sv\'eho sou\v{c}tu).

D\'ale po\v{c}\'{\i}tejme koeficienty $a_n'$, $b_n'$ funkce $\sgn(x-1)$.
$$\align
a_0'&=\frac1\pi\int_{-\pi}^\pi \sgn(x-1)\d
x=\frac1\pi\cdot\left(\int_{-\pi}^1 -1 \d x+\int_1^\pi 1 \d
x\right)=
\frac1\pi(-(1+\pi)+(\pi-1))=-\frac2\pi \\
a_n'&=\frac1\pi\int_{-\pi}^\pi \sgn(x-1)\cos nx\d
x=\frac1\pi\cdot\left(\int_{-\pi}^1 -\cos nx \d x+\int_1^\pi \cos
nx \d x\right) \\ &=  \frac1\pi\left(\left[-{\sin nx\over
n}\right]_{-\pi}^1+\left[{\sin nx\over n}\right]_1^{\pi}\right) =
\frac1\pi\left(-{\sin n\over n}-{\sin n\over n}\right)= -{2\sin
n\over n\pi}\text{\quad pro\quad} n\ge1 \\
b_n'&=\frac1\pi\int_{-\pi}^\pi \sgn(x-1)\sin nx\d
x=\frac1\pi\cdot\left(\int_{-\pi}^1 -\sin nx \d x+\int_1^\pi \sin
nx \d x\right) \\ &=  \frac1\pi\left(\left[{\cos nx\over
n}\right]_{-\pi}^1+\left[-{\cos nx\over n}\right]_1^{\pi}\right) =
\frac1\pi\left({\cos n\over n}-{(-1)^n\over n}-{(-1)^n\over n}
+{\cos n\over n}\right) \\ &= {2(\cos n-(-1)^n)\over n\pi}
\text{\quad pro\quad} n\ge1
\endalign$$

Tedy Fourierova \v{r}ada funkce $f$ m\'a tvar:

$$\multline
-\frac1\pi - \left(\frac2\pi\sin1+\frac14\right)\cos x
+\frac2\pi(\cos 1+1)\sin x -\frac1\pi\sin 2\cos 2x +\frac1\pi(\cos
2-1)\sin 2x\\ \ -\left(\frac2{3\pi}\sin 3-\frac14\right)\cos 3x
+\frac2{3\pi}(\cos 3+1)\sin 3x +\sn4 \left(-{2\sin n \cos nx\over
n\pi}+{2(\cos n-(-1)^n)\sin nx\over n\pi}\right)
\endmultline$$

(b) Sou\v{c}et Fourierovy \v{r}ady funkce $\cos^3 x$ je zase $\cos^3 x$,
proto\v{z}e tato \v{r}ada je kone\v{c}n\'a (tedy je to vlastn\v{e} trigonometrick\'y
polynom).  D\'ale si v\v{s}imn\v{e}me funkce $\sgn(x-1)$. Pokud
$x\in(-\pi,1)\cup(1,\pi)$, je na okol\'{\i} bodu $x$ funkce
$\sgn(x-1)$ konstantn\'{\i}, a tedy spojit\'a a nav\'{\i}c m\'a kone\v{c}nou
variaci (je dokonce monot\'onn\'{\i}), a proto podle Jordan-Dirichletova
krit\'eria Fourierova \v{r}ada konverguje k funk\v{c}n\'{\i} hodnot\v{e}.
V bodech $1$ a $\pi$ lze rovn\v{e}\v{z} pou\v{z}\'{\i}t Jordan-Dirichletovo
krit\'erium, funkce je na okol\'{\i} monot\'onn\'{\i} (a tedy m\'a kone\v{c}nou
variaci), a proto sou\v{c}tem Fourierovy \v{r}ady je aritmetick\'y pr\accent23um\v{e}r
limity zleva a limity zprava, co\v{z} je v obou p\v{r}\'{\i}padech $0$ (v bod\v{e}
$1$ je to z\v{r}ejm\'e, v bod\v{e} $\pi$ se po\v{c}\'{\i}taj\'{\i} limity funkce
periodicky dodefinovan\'e). Proto\v{z}e sou\v{c}et Fourierovy \v{r}ady je
funkce $2\pi$-periodick\'a, plyne z \v{r}e\v{c}en\'eho, \v{z}e Fourierova \v{r}ada m\'a
sou\v{c}et v ka\v{z}d\'em bod\v{e}, a tento se rovn\'a
$$\matrix \format \r\qquad &\l \\
\cos^3 x+\sgn(x-2k\pi-1), & x\in (-\pi+2k\pi,\pi+2k\pi), k\in\Bbb
Z \\ -1, & x=(2k+1)\pi, k\in\Bbb Z\endmatrix$$

}
\oddel
{\centerline{\smc Test \v{c}\'{\i}slo II/4 -- druh\'a verze}

\pr Najd\v{e}te v\v{s}echna maxim\'aln\'{\i} \v{r}e\v{s}en\'{\i} diferenci\'aln\'{\i} rovnice

\qquad\qquad $y'''+y''+4y'+4y=\cos x+\cos 2x$ \hfill (18 bod\accent23u)

\bf \v{R}e\v{s}en\'{\i}. \rm Charakteristick\'y polynom t\'eto rovnice je
$$\lambda^3+\lambda^2 + 4\lambda +4 =(\lambda^2+4)(\lambda+1),$$
tedy m\'a t\v{r}i r\accent23uzn\'e ko\v{r}eny $-1$, $2i$, $-2i$, tud\'{\i}\v{z} fundament\'aln\'{\i}
syst\'em homogenn\'{\i} rovnice tvo\v{r}\'{\i} funkce $e^{-x}$, $\cos 2x$, $\sin
2x$. Prav\'a strana je sou\v{c}tem dvou kvazipolynom\accent23u ($\cos x$ a $\cos
2x$), a tedy hledejme partikul\'arn\'{\i} \v{r}e\v{s}en\'{\i} $y_1$ rovnice
$y'''+y''+4y'+4y=\cos x$
a partikul\'arn\'{\i} \v{r}e\v{s}en\'{\i} $y_2$ rovnice $y'''+y''+4y'+4y=\cos 2x$.
Proto\v{z}e $i$ nen\'{\i} ko\v{r}enem charakteristick\'eho polynomu, m\'a $y_1$
tvar $y_1=a\cos x+b\sin x$ pro n\v{e}jak\'e konstanty $a$, $b$.
Pak plat\'{\i}
$$ \matrix
y_1=a\cos x+b\sin x, & y_1^\prime=-a\sin x+b\cos x, \\
y_1^{\prime\prime}=-a\cos x-b\sin x, &
y_1^{\prime\prime\prime}=a\sin x-b\cos x,
\endmatrix$$
a tedy
$$y_1^{\prime\prime\prime}+y_1^{\prime\prime}+4y_1^{\prime}+4y_1
=(-b-a+4b+4a)\cos x + (a-b-4a+4b)\sin x=3(b+a)\cos x+3(b-a)\sin
x,$$
odkud dost\'av\'ame
$$3(b+a)=1,\quad 3(b-a)=0, \text{ neboli }\quad
a=b=\frac16.$$
Proto\v{z}e $2i$ je ko\v{r}enem charakteristick\'eho polynomu n\'asobnosti
$1$, m\'a $y_2$ tvar $y_2=x\cdot(c\cos 2x+d\sin 2x)$ pro n\v{e}jak\'e
konstanty $c$, $d$. Pak plat\'{\i}:
{\eightpoint
$$ \matrix \format \l &\quad \l \\
y_2=x(c\cos 2x+d\sin 2x), & y_2^\prime=x(-2c\sin 2x+2d\cos 2x)+
(c\cos 2x+d\sin 2x), \\
y_2^{\prime\prime}=x(-4c\cos 2x-4d\sin 2x)+2(-2c\sin 2x+2d\cos 2x), &
y_2^{\prime\prime\prime}=x(8c\sin 2x-8d\cos 2x)+3(-4c\cos 2x-4d\sin 2x),
\endmatrix$$}
a tedy
$$y_2^{\prime\prime\prime}-y_2^{\prime\prime}+4y_2^{\prime}-4y_2
=(-12c+4d+4c)\cos 2x + (-12d-4c+4d)\sin 2x=-4(2c-d)\cos
2x-4(c+2d)\sin 2x,$$
odkud dost\'av\'ame
$$-4(2c-d)=1,\quad -4(c+2d)=0, \text{ neboli } d=\frac{1}{20},\
c=-\frac1{10}.$$
Obecn\'e \v{r}e\v{s}en\'{\i} na\v{s}\'{\i} rovnice m\'a tedy tvar
$$y=\frac x{20}(\sin 2x-2 \cos 2x) +\frac16(\sin x+\cos x) + A
e^{-x} + B \cos 2x + C \sin 2x,\quad A,B,C\in\Bbb R.$$

\pr Se\v{c}t\v{e}te \v{r}adu $\sn0 {(-1)^n\sin 3nx\over (2n)!}$.
\hfill (14 bod\accent23u)

\bf \v{R}e\v{s}en\'{\i}. \rm Plat\'{\i} $\sin 3nx= \Im e^{3nix}$,
a tedy m\accent23u\v{z}eme ps\'at

$$\align
\sn0 {(-1)^n\sin 3nx\over (2n)!}&=\sn0 {(-1)^n\Im e^{3nix}\over (2n)!}=
\Im \sn0 {(-1)^n e^{3nix}\over (2n)!}= \Im \sn0 {(-1)^n
\left(e^{\frac32ix}\right)^{2n}\over (2n)!}\\ &=
\Im \cos\left(e^{\frac32ix}\right)
= \Im\cos\left(\cos\frac32x+i\sin\frac32x\right)\\ &=
\Im\left(\cos\left(\cos\frac32x\right)\cos\left(i\sin\frac32x
\right)-\sin\left(\cos\frac32x\right)\sin\left(i\sin\frac32x
\right)\right) \\ &
=\Im\left(\cos\left(\cos\frac32x\right)\cdot{e^{-\sin\frac32x}+
e^{\sin\frac32x}\over2}-\sin\left(\cos\frac32x\right)
\cdot{e^{-\sin\frac32x}-e^{\sin\frac32x}\over2i}\right)\\ &=
\sin\left(\cos\frac32x\right)
\cdot{e^{-\sin\frac32x}-e^{\sin\frac32x}\over2}
\endalign$$

Pou\v{z}ili jsme rozvoj funkce $\cos$ v mocninnou \v{r}adu a vzore\v{c}ky
$\sin z={e^{iz}-e^{-iz}\over 2i}$, $\cos z={e^{iz}+e^{-iz}\over
2}$, kter\'e plat\'{\i} pro v\v{s}echna
komplexn\'{\i} $z$.


\pr Polo\v{z}me $f(x)=\sin^3 x-\sgn(x+1)$.

(a) Rozvi\v{n}te $f$ ve Fourierovu \v{r}adu na intervalu $(-\pi,\pi)$.
\hfill (10 bod\accent23u)

(b) Ur\v{c}ete sou\v{c}et \v{r}ady z (a) v\v{s}ude, kde \v{r}ada konverguje.
\hfill (8 bod\accent23u)

\bf \v{R}e\v{s}en\'{\i}. \rm (a) Z definice Fourierov\'ych koeficient\accent23u je
z\v{r}ejm\'e, \v{z}e Fourierovy koeficienty funkce $f$ jsou
rozd\'{\i}lem
Fourierov\'ych koeficient\accent23u funkc\'{\i} $\sin^3 x$ a $\sgn(x+1)$. Nejprve
se v\v{e}nujme prvn\'{\i} funkci. Podle goniometrick\'ych vzore\v{c}k\accent23u plat\'{\i}
$$\align
\sin^3 x &= \sin x \cdot {1-\cos 2x\over 2}=\frac12 \sin x
-\frac12 \sin x\cdot\cos 2x\\ & =\frac12 \sin x -\frac14(\sin 3x
-\sin x)=\frac34\sin x-\frac14\sin 3x,\endalign$$
co\v{z} je z\'arove\v{n} rozvoj funkce $\sin^3 x$ ve Fourierovu \v{r}adu (v\'yraz
vpravo je kone\v{c}n\'a (a tedy nap\v{r}. stejnom\v{e}rn\v{e} konvergentn\'{\i})
trigonometrick\'a \v{r}ada, a proto je Fourierovou \v{r}adou sv\'eho sou\v{c}tu).

D\'ale po\v{c}\'{\i}tejme koeficienty $a_n'$, $b_n'$ funkce $\sgn(x+1)$.
$$\align
a_0'&=\frac1\pi\int_{-\pi}^\pi \sgn(x+1)\d
x=\frac1\pi\cdot\left(\int_{-\pi}^{-1} -1 \d x+\int_{-1}^\pi 1 \d
x\right)=
\frac1\pi(-(-1+\pi)+(\pi+1))=\frac2\pi \\
a_n'&=\frac1\pi\int_{-\pi}^\pi \sgn(x+1)\cos nx\d
x=\frac1\pi\cdot\left(\int_{-\pi}^{-1} -\cos nx \d x+\int_{-1}^\pi \cos
nx \d x\right) \\ &=  \frac1\pi\left(\left[-{\sin nx\over
n}\right]_{-\pi}^{-1}+\left[{\sin nx\over n}\right]_{-1}^{\pi}\right) =
\frac1\pi\left(-{\sin(-n)\over n}-{\sin(-n)\over n}\right)= {2\sin
n\over n\pi}\text{\quad pro\quad} n\ge1 \\
b_n'&=\frac1\pi\int_{-\pi}^\pi \sgn(x+1)\sin nx\d
x=\frac1\pi\cdot\left(\int_{-\pi}^{-1} -\sin nx \d x+\int_{-1}^\pi \sin
nx \d x\right) \\ &=  \frac1\pi\left(\left[{\cos nx\over
n}\right]_{-\pi}^{-1}+\left[-{\cos nx\over n}\right]_{-1}^{\pi}\right) =
\frac1\pi\left({\cos(-n)\over n}-{(-1)^n\over n}-{(-1)^n\over n}
+{\cos(-n)\over n}\right) \\ &= {2(\cos n-(-1)^n)\over n\pi}
\text{\quad pro\quad} n\ge1
\endalign$$

Tedy Fourierova \v{r}ada funkce $f$ m\'a tvar:

$$\multline
-\frac1\pi - \frac2\pi\sin1\cos x
+\left(\frac34-\frac2\pi(\cos 1+1)\right)\sin x
-\frac1\pi\sin 2\cos 2x - \frac1\pi(\cos
2-1)\sin 2x\\  - \frac2{3\pi}\sin 3\cos 3x
+\left(\frac14-\frac2{3\pi}(\cos 3+1)\right)\sin 3x
-\sn4 \left({2\sin n \cos nx\over
n\pi}+{2(\cos n-(-1)^n)\sin nx\over n\pi}\right)
\endmultline$$

(b) Ze stejn\'ych d\accent23uvod\accent23u jako v prvn\'{\i} verzi dostaneme, \v{z}e
Fourierova \v{r}ada konverguje ve v\v{s}ech bodech a m\'a sou\v{c}et
$$\matrix \format \r\qquad &\l \\
\sin^3 x-\sgn(x-2k\pi+1), & x\in (-\pi+2k\pi,\pi+2k\pi), k\in\Bbb
Z \\ 0, & x=(2k+1)\pi, k\in\Bbb Z\endmatrix$$
}

\oddel

{\def\dist{\operatorname{dist}}
 \centerline{\smc XIX. Je\v{s}t\v{e} trigonometrick\'e a Fourierovy \v{r}ady}

\pr Rozvi\v{n}te n\'asleduj\'{\i}c\'{\i} funkce ve Fourierovu \v{r}adu na intervalu
$(-\pi,\pi)$ a ur\v{c}ete sou\v{c}et, v\v{s}ude, kde \v{r}ada konverguje.
(a) $\cos^{2m} x$, \quad
(b) ${\sin mx\over \sin x}$, \quad
(c) $\sum\limits_{m=1}^\infty \alpha^m{\sin m x\over \sin x}$,
kde $|\alpha|<1$, \quad
(d) ${1\over\sin x}$.

\pr Je trigonometrick\'a \v{r}ada $\sn2{\sin nx\over\log n}$
Fourierovou \v{r}adou n\v{e}jak\'e funkce (nap\v{r}. sv\'eho sou\v{c}tu)?

\pr Pro n\'asleduj\'{\i}c\'{\i} funkce ur\v{c}ete, v kter\'ych bodech konverguje
jejich Fourierova \v{r}ada, a jak\'y m\'a sou\v{c}et.

(a) $\sgn\sin 100x$,\, (b) $\sin\frac1x$, $x\in(-\pi,\pi)$,\,
(c) $\dist(x, C)$, $x\in(0,2\pi)$, kde $C$ je Cantorovo
diskontinuum.

\pr Se\v{c}t\v{e}te \v{r}ady $\sn1{\sin^2 n\alpha\over n^2}$,
$\sn1{\cos^2 n\alpha\over n^2}$. (Pou\v{z}ijte Parsevalovu rovnost
pro $\chi_{[-\alpha,\alpha]}$, $|\alpha|\le\pi$.)}

\end