\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
\loadbold
\vcorrection{12 mm}
\NoBlackBoxes
\def\uv#1{\clqq #1\crqq{}}


{%
\centerline{\smc I. Zjist\v{e}te, zda $(V,\oplus,\odot)$ je vektorov\'y
prostor nad $\bold K$, jestli\v{z}e}

\pr $\bold K=\bold R$,
$V$ je mno\v{z}ina v\v{s}ech re\'aln\'ych funkc\'{\i} definovan\'ych na $\bold
R$,

a) $(f\oplus g)(x)=f(x)+g(x)$, $(\lambda\odot g)(x)=\lambda g(x)$
pro $f,g\in V$, $\lambda\in\bold R$, $x\in\bold R$.


b) $(f\oplus g)(x)=f(x)g(x)$, $(\lambda\odot g)(x)=\lambda g(x)$
pro $f,g\in V$, $\lambda\in\bold R$, $x\in\bold R$.

\pr $\bold K=\bold R$, $V=(0,+\infty)$, $u\oplus v=uv$,
$\lambda\odot v=\lambda v$.
\pr $\bold K=\bold R$, $V=\bold C$, $u\oplus v=u+v$,
$\lambda\odot v=\lambda v$.

\pr $\bold K=\bold C$, $V=\bold R$, $u\oplus v=u+v$,
$\lambda\odot v=|\lambda| v$.

\pr $\bold K=\bold R$, $V=\bold R^2$, $u\oplus v=u+v$,
$\lambda\odot [u_1,u_2]=[\lambda^3 u_1,\lambda u_2]$.

\pr $\bold K=\bold C$, $V=\bold C^2$, $[u_1,u_2]\oplus
[v_1,v_2]=[u_1+v_1,u_2+v_2]$, a) $\lambda\odot[u_1,u_2]=[\bar\lambda
u_1,\bar\lambda u_2]$; b)
$\lambda\odot[u_1,u_2]=[\lambda
u_1,\bar\lambda u_2]$.
\pr $\bold K=\bold C$, $V=\bold R^2$, $[u_1,u_2]\oplus
[v_1,v_2]=[u_1+v_1,u_2+v_2]$, $\lambda\odot[u_1,u_2]=[u_1\Re\lambda
-u_2\Im\lambda,u_2\Re\lambda +u_1\Im\lambda]$.

\centerline{\smc Zjist\v{e}te, zda je $U$ vektorov\'ym podprostorem $V$, jestli\v{z}e}

\pr $V=\bold R^3$;
a) $U=\{\boldsymbol x\in\bold R^3\: x_1-x_2=0\}$,
b) $U=\{\boldsymbol x\in\bold R^3\: x_1=x_2^2\}$,
c) $U=\{\boldsymbol x\in\bold R^3\: x_1+x_3=1\}$.
\pr $V=\bold R^2$, $U=\{\boldsymbol x\in\bold R^2\: x_1\ge 0\}$.
\pr $V=\Cal C(\langle0,1\rangle)$;
a) $U=\{f\in V\: f(0)+f(1)=f(1/2)\}$,
b) $U=\{f\in V\: f(0)=(f(1))^2\}$,
c) $U=\{f\in V\: \forall x\in[0,1]\:|f(x)|\le 1\}$,
d) $U=\{f\in V\: \int_0^1 f(x)\d x=0\}$,
e) $U=\{f\in V\: \int_0^{1/2} f(x)\d x=17\}$,
f) $U=\{f\in V\: \int_0^1 (f(x))^3\d x=0\}$,
g) $U=\{f\in V\: \int_{1/2}^1 (f(x))^2\d x=0\}$.
}

\oddel

@ \pr a) ANO, b) NE (neplat\'{\i} nap\v{r}\. (iv)).
\pr NE ($\odot$ nen\'{\i} operace). \pr ANO \pr NE (neplat\'{\i} (vi)).
\pr NE (neplat\'{\i} (vi)). \pr a) ANO, b) ANO. \pr ANO.
\pr a) ANO, b) NE ($[1,1,0]\in U$, $2\cdot[1,1,0]\notin U$),
c) NE ($[0,0,0]\notin U$).
\pr NE ($[1,0]\in U$, $-[1,0]\notin U$).
\pr a) ANO, b) NE ($1\in U$, $2\cdot 1\notin U$),
c) NE ($1\in U$, $2\cdot 1\notin U$),
d) ANO, e) NE ($0\notin U$), f) NE (nech\softt{} $f(x)=\frac12-x$ a
$g(x)=|2x-1|-\frac12$; pak $f,g\in U$, $f+g\notin U$),
g) ANO ($U=\{f\in V\:\forall x\in\langle\frac12,1\rangle\:
f(x)=0\}$).


\varoddel

{%
\def\R{{\bold R}}
\def\C{{\bold C}}
\def\lin{\operatorname{lin}}

\centerline{\smc II. Ur\v{c}ete dimenzi a nalezn\v{e}te n\v{e}jakou b\'azi
vektorov\'eho prostoru $V$, kde}

\pr $V=\lin_{\R} \{[1,1,1],[1,2,2],[10,11,11]\}\subset \R^3$.

\pr $V=\lin_{\R} \{[1,1,1,2,2],[-1,-2,-2,-2,2],[19,1,1,1,1],
[-1,-3,-3,-2,6]\}\subset\R^5$.

\pr $V=\lin_{\R} \{ x+7,x^2-x-1,x^2+3,x-5 \}\subset C(\langle
0,1\rangle)$.

\pr $V=\{\sin x,\cos x,\sin 2x, \cos 2x\} \subset C(\langle
0,\pi\rangle)$.
\pr $V=\C^2$ jako vektorov\'y prostor nad~$\R$.

\pr $V=M(2\times 3)$.
\pr $V=\{ {\Bbb A}\in M(3\times 3)\: {\Bbb A}^T=\Bbb A\}$.


\pr $V=\{[x_1,x_2,x_3]\in\R^3\: x_1+x_2-x_3=0\}$.

\pr $V=\{[x_1,x_2,x_3,x_4]\in\R^4\: x_1+x_3-x_4=0,x_2+x_4=0,
x_1+x_2+x_4=0\}$.

\pr $V=\{[x_1,x_2,x_3,x_4]\in\R^4\: \pmatrix 1 & -2 & 2 & 3 \\ 2
& 8 & -1 & 0 \\ 1 & 2 & 2 & 2 \\ 1 & 1 & 1 & 1\endpmatrix
\pmatrix x_1\\x_2\\x_3\\x_4\endpmatrix=\bold o\}$.

\pr $V=\{f\in C(\R)\: f\text{ m\'a v\v{s}ude p\'atou derivaci a
$f^{(5)}=0$ na }\R\}$.}

\oddel

@ \pr $\dim V=2$, b\'aze je nap\v{r}\'{\i}klad $\{[1,1,1],[0,1,1]\}$
(nebo t\'e\v{z} $\{[1,1,1],[1,2,2]\}$).
\pr $\dim V=3$, b\'aze je nap\v{r}\'{\i}klad
$\{[1,1,1,2,2],[0,-1,-1,0,4],[0,0,0,37,109]\}$.
\pr $\dim V=3$, b\'aze je nap\v{r}\'{\i}klad $\{1,x,x^2\}$.
\pr $\dim V=4$, mno\v{z}ina ze zad\'an\'{\i} je b\'aze.
\pr $\dim V=4$, b\'aze je nap\v{r}\'{\i}klad $\{[1,0],[i,0],[0,1],[0,i]\}$.
\pr $\dim V=6$, b\'azi tvo\v{r}\'{\i} nap\v{r}\'{\i}klad \v{s}estice matic, kter\'e maj\'{\i} na
jednom m\'{\i}st\v{e} $1$ a na ostatn\'{\i}ch $0$.
\pr $\dim V=6$, b\'azi tvo\v{r}\'{\i} nap\v{r}\'{\i}klad \newline $\left\{
\left(\smallmatrix 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0
\endsmallmatrix\right),
\left(\smallmatrix 0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0
\endsmallmatrix\right),
\left(\smallmatrix 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1
\endsmallmatrix\right),
\left(\smallmatrix 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0
\endsmallmatrix\right),
\left(\smallmatrix 0 & 0 & 1 \\ 0 & 0 & 0 \\ 1 & 0 & 0
\endsmallmatrix\right),
\left(\smallmatrix 0 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0
\endsmallmatrix\right)\right\}$.
\pr $\dim V=2$, b\'aze je nap\v{r}\'{\i}klad $\{[1,0,1],[0,1,1]\}$.
\pr $\dim V=1$, b\'aze je nap\v{r}\'{\i}klad $\{[0,-1,1,1]\}$.
\pr $\dim V=0$, b\'aze je $\emptyset$ ($V=\{[0,0,0,0]\}$).
\pr $\dim V=5$, b\'aze je nap\v{r}\'{\i}klad $\{1,x,x^2,x^3,x^4\}$.

\newpage

{
\def\Ker{\operatorname{Ker}}
\centerline{\smc III. Je $L\:U\to V$ line\'arn\'{\i} zobrazen\'{\i}? Pokud
ano, jak vypad\'a $\Ker(L)$ a $\Im(L)$?}

\pr $U=\bold R^3$, $V=\bold R^4$;
a) $L(u,v,w)=[u+w,w-v+7u,u,u]$,
b) $L(u,v,w)=[u^2,v,0,0]$,

c) $L(u,v,w)=[0,0,0,0]$,
d) $L(u,v,w)=[u,v-1,4w,u+w]$,
e) $L(u,v,w)=[u,v,w,w]$,

f) $L(u,v,w)=[u+v,u-v,w,10u]$.

\pr $U=C(\bold R)$, $V=C(\bold R)$;
a) $L(f)(x)=f(x+1)-f(x)$,
b) $L(f)(x)=\int_0^x f(t)\d t$,

c) $L(f)(x)=f(x^2)$,
d) $L(f)(x)=\int_{x-1}^{x+1} f(t) \d t$.

\pr $U=C^1(\bold R)$, $V=C(\bold R)$;
a) $L(f)(x)=f'(x)$,
b) $L(f)(x)=f(x)-f'(x)$,
c) $L(f)(x)=\int_0^x f'(t)\d t$.
}

\oddel

{\def\Ker{\operatorname{Ker}}
\def\lin{\operatorname{lin}}
@ \pr a) ANO, $\Ker(L)=\{[0,0,0]\}$,

$\Im(L)=\lin_{\bold R}\{[1,7,1,1],[0,-1,0,0],[1,1,0,0]\}$;
b) NE;
c) ANO, $\Ker(L)={\bold R}^3$, $\Im(L)=\{[0,0,0,0]\}$;
d) NE;
e) ANO, $\Ker(L)=\{[0,0,0]\}$, $\Im(L)=\lin_{\bold
R}\{[1,0,0,0],[0,1,0,0],[0,0,1,1]\}$;
f) ANO, $\Ker(L)=\{[0,0,0]\}$, $\Im(L)=\lin_{\bold
R}\{[1,1,0,10],[1,-1,0,0],[0,0,1,0]\}$.
\pr a) ANO, $\Ker(L)$ tvo\v{r}\'{\i} $1$-periodick\'e funkce,
$\Im(L)=C(\bold R)$ (m\'{\i}rn\v{e} obt\'{\i}\v{z}n\v{e}j\v{s}\'{\i});
b) ANO, $\Ker(L)=\{0\}$, $\Im(L)=\{f\in C^1(\bold R)\: f(0)=0\}$;
c) ANO, $\Ker(L)=\{f\in C(\bold R)\: f\restriction\langle
0,+\infty)=0\}$, $\Im(L)$ tvo\v{r}\'{\i} sud\'e funkce;
d) ANO, $\Ker(L)$ tvo\v{r}\'{\i} $2$-periodick\'e funkce spl\v{n}uj\'{\i}c\'{\i} nav\'{\i}c
$\int_{-1}^1 f(x)\d x=0$, $\Im(L)=C^1(\bold R)$ (oboj\'{\i} m\'{\i}rn\v{e}
obt\'{\i}\v{z}n\v{e}j\v{s}\'{\i}).
\pr a) ANO, $\Ker(L)$ tvo\v{r}\'{\i} konstantn\'{\i} funkce, $\Im(L)=C(\bold
R)$;
b) ANO, $\Ker(L)=\lin_{\bold R}\{e^x\}=\{a e^x\:a\in\bold R\}$,
$\Im(L)=C(\bold R)$;
c) ANO, $\Ker(L)$ tvo\v{r}\'{\i} konstantn\'{\i} funkce,
$\Im(L)=\{f\in C^1(\bold R)\: f(0)=0\}$.}

\varoddel

{%
\centerline{\smc IV. Biline\'arn\'{\i} formy}

\pr Zjist\v{e}te, zda $B$ je biline\'arn\'{\i} forma, pokud ano, napi\v{s}te
reprezentuj\'{\i}c\'{\i} matici a rozhodn\v{e}te, zda $B$ je symetrick\'a.
a) $B([x_1,x_2],[y_1,y_2])=x_1+y_2$;
b) $B([x_1,x_2],[y_1,y_2])=(x_1+y_2)(x_2+y_1)$;

c) $B([x_1,x_2],[y_1,y_2])=x_1 y_2+x_2 y_1$;
d) $B([x_1,x_2],[y_1,y_2])=x_1^2+x_2^2 $;

e) $B([x_1,x_2],[y_1,y_2])=x_1 x_2 + y_1 y_2 $;
f) $B([x_1,x_2],[y_1,y_2])=(x_1 - x_2)(y_1+9 y_2) $.

\pr Zjist\v{e}te, zda n\'asleduj\'{\i}c\'{\i} matice jsou pozitivn\v{e} definitn\'{\i},
negativn\v{e} definitn\'{\i} atd.

a) $\left(\smallmatrix 1 & 2 \\ 2 & 1 \endsmallmatrix\right)$;
b) $\left(\smallmatrix 2 & 1 \\ 1 & 2 \endsmallmatrix\right)$;
c) $\left(\smallmatrix 1 & 2 \\ 0 & 1 \endsmallmatrix\right)$;
d) $\left(\smallmatrix 1 & 1 \\ 1 & 1 \endsmallmatrix\right)$;
e) $\left(\smallmatrix 1 & 1 & 1 \\ 1 & 3 & 5 \\ 1 & 5 & 1 \endsmallmatrix\right)$;
f) $\left(\smallmatrix -10 & 1 & 1 \\ 1 & -10 & 2 \\ 1 & 2 & -1 \endsmallmatrix\right)$;

g) $\left(\smallmatrix 10 & 1 & 3 \\ 1 & 10 & 1 \\ 3 & 1 & 1 \endsmallmatrix\right)$;
h) $\left(\smallmatrix 3 & 1 & 3 \\ 1 & 10 & 1 \\ 3 & 1 & 3 \endsmallmatrix\right)$;
i) $\left(\smallmatrix -3 & -1 & 3 \\ -1 & 3 & 1 \\ 3 & 1 & -3 \endsmallmatrix\right)$;
j) $\left(\smallmatrix 1& 1 & 3 & -1 \\ 1 & 2 & 1 & 2 \\ 3 & 1 & 3
& 1 \\ -1 & 2 & 1 & 4 \endsmallmatrix\right)$.
}

\oddel

@ \pr a), b), d), e) NE; c) ANO, $\pmatrix 0 & 1 \\ 1&
0\endpmatrix$, je symetrick\'a;
f) ANO, $\pmatrix 1 & 9 \\ -1 & -9 \endpmatrix$, nen\'{\i} symetrick\'a.
\pr a) ID, b) PD, c),d) PSD, ne PD; e) ID, f) ND, g) PD, h) PSD,
ne PD, i),j) ID.


\varoddel

{%
\centerline{\smc V. Najd\v{e}te vlastn\'{\i} \v{c}\'{\i}sla a v\v{s}echny jim p\v{r}\'{\i}slu\v{s}n\'e
vlastn\'{\i} vektory pro matice}

\pr $\pmatrix 3 & -1 \\ 2 & 0 \endpmatrix$
\pr $\pmatrix 3 & 6 \\ 4 & -2 \endpmatrix$
\pr $\pmatrix 1 & 1 \\ -1 & 1 \endpmatrix$
\pr $\pmatrix 1 & 2 \\ -2 & 5 \endpmatrix$
\pr $\pmatrix -2& 1 & 2 \\ -4 & 3 & 2 \\ -5 & 1 & 5 \endpmatrix$

\pr $\pmatrix 5 & 0 & -2 \\ 2 & 3 & -2 \\ 4 & 0 & -1 \endpmatrix$
\pr $\pmatrix 8 & -1 & -3 \\ 8 & 1 & -4 \\ 7 & -1 & -2 \endpmatrix$
\pr $\pmatrix 3& -1 & -1 \\ 6 & -2 & -2 \\ 3 & -1 & -1 \endpmatrix$
\pr $\pmatrix 2& 0 & -1 \\ 5 & -1 & -2 \\ 1 & 1 & -1 \endpmatrix$

\pr $\pmatrix -1 & 1 & 1 \\ -3 & 2 & 2 \\ -2 & 1 & 2 \endpmatrix$
\pr $\pmatrix -23 & 21 & 3 & -17 \\ -40 & 35 & 4 & -31 \\
58 & -50 & -5 & 47 \\ -8 & 6 & 0 & -7 \endpmatrix$}

\newpage

@ (ve form\'atu: (vlastn\'{\i} \v{c}\'{\i}slo, n\'asobnost, mno\v{z}ina vlastn\'{\i}ch
vektor\accent23u))
\pr $(1,1,\{t\cdot[1,2]\:t\in\bold C\setminus\{0\}\})$,
$(2,1,\{t\cdot[1,1]\:t\in\bold C\setminus\{0\}\})$.
\pr $(6,1,\{t\cdot[2,1]\:t\in\bold C\setminus\{0\}\})$,
$(-5,1,\{t\cdot[3,-4]\:t\in\bold C\setminus\{0\}\})$.
\pr $(1+i,1,\{t\cdot[1,i]\:t\in\bold C\setminus\{0\}\})$,
$(1-i,1,\{t\cdot[1,-i]\:t\in\bold C\setminus\{0\}\})$.
\pr $(3,2,\{t\cdot[1,1]\:t\in\bold C\setminus\{0\}\})$.
\pr $(1,1,\{t\cdot[1,1,1]\:t\in\bold C\setminus\{0\}\})$,
$(2,1,\{t\cdot[1,2,1]\:t\in\bold C\setminus\{0\}\})$,
$(3,1,\{t\cdot[1,1,2]\:t\in\bold C\setminus\{0\}\})$.
\pr $(1,1,\{t\cdot[1,1,2]\:t\in\bold C\setminus\{0\}\})$,
$(3,2,\{t\cdot[1,0,1]+s\cdot[0,1,0]\:[s,t]\in\bold
C^2\setminus\{[0,0]\}\})$.
\pr $(1,1,\{t\cdot[1,1,2]\:t\in\bold C\setminus\{0\}\})$,
$(3,2,\{t\cdot[1,2,1]\:t\in\bold C\setminus\{0\}\})$.
\pr $(0,3,\{t\cdot[1,3,0]+s\cdot[0,-1,1]\:[s,t]\in\bold
C^2\setminus\{[0,0]\}\})$.
\pr $(0,3,\{t\cdot[1,1,2]\:t\in\bold C\setminus\{0\}\})$.
\pr $(1,3,\{t\cdot[1,1,1]\:t\in\bold C\setminus\{0\}\})$.
\pr $(i,2,\{t\cdot[3,4,-5+i,0]+s\cdot[-7-i,0,-8-10i,8]\:[s,t]\in\bold
C^2\setminus\{[0,0]\}\})$,
$(-i,2,\{t\cdot[3,4,-5-i,0]+s\cdot[-7+i,0,-8+10i,8]\:[s,t]\in\bold
C^2\setminus\{[0,0]\}\})$.

\varoddel

{%
\centerline{\smc VI. Rozvoj funkce v bod\v{e}}

\pr Najd\v{e}te hlavn\'{\i} \v{c}\'ast rozvoje $k$-t\'eho \v{r}\'adu v bod\v{e} $a$ pro
funkce $f+g$, $f-g$, $fg$, $f/g$, $g/f$ (pokud existuj\'{\i}), jsou li
zad\'any hlavn\'{\i} \v{c}\'asti rozvoje funkc\'{\i} $f$ a $g$:

a) $k=1$, $a=0$, $f$: $x$, $g$: $1+x$;
b) $k=1$, $a=1$, $f$: $1-2(x-1)$, $g$: $1+(x-1)$;
c) $k=2$, $a=3$, $f$: $7+(x-3)+\frac12(x-3)^2$, $g$: $6(x-3)$;
d) $k=3$, $a=-2$, $f$: $1+(x+2)^3$, $g$: $(x+2)-\frac12(x+2)^2$;

\pr Najd\v{e}te hlavn\'{\i} \v{c}\'ast rozvoje $k$-t\'eho \v{r}\'adu funkce $f\circ g$ v
bod\v{e} $a$, je-li zad\'ana hlavn\'{\i} \v{c}\'ast rozvoje funkce $g$ v bod\v{e} $a$
a funkce $f$ v bod\v{e} $g(a)$.
a) $k=1$, $f$, $g$: $1+(x-1)$;
b) $k=1$, $f$: $2+(x-7)$, $g$: $7+(x-1)$;
c) $k=2$, $f$: $(x-6)+3(x-6)^2$, $g$: $6+x+x^2$;
d) $k=3$, $f$: $(x-2)$, $g$: $2+(x-3)^2+(x-3)^3$;
}

\oddel

@ \pr a) $f+g$: $1+2x$; $f-g$: $-1$; $fg$: $x$; $f/g$: $x$.
b) $f+g$: $2-(x-1)$; $f-g$: $-3(x-1)$; $fg$: $1-(x-1)$; $f/g$:
$1-3(x-1)$; $g/f$: $1+3(x-1)$.
c) $f+g$: $7+7(x-3)^2+\frac12(x-3)^2$; $f-g$: $7-5(x-3)^2+\frac12(x-3)^2$;
$fg$: $42(x-3)+6(x-3)^2$; $g/f$:
$\frac67(x-3)-\frac6{49}(x-3)^2$.
d) $f+g$: $1+(x+2)-\frac12(x+2)^2+(x+2)^3$;
$f-g$: $1-(x+2)+\frac12(x+2)^2+(x+2)^3$;
$fg$: $(x+2)-\frac12(x+2)^2$;
$g/f$: $(x+2)-\frac12(x+2)^2$.
\pr a) $1+(x-1)$; b) $2+(x-1)$; c) $4x^2$; d) $(x-3)^2+(x-3)^3$.

\varoddel

{\let\l\lx
\centerline{\smc VII. Najd\v{e}te Taylor\accent23uv polynom $k$-t\'eho \v{r}\'adu v
bod\v{e} $0$ pro funkce}

\pr $\tg x$, $k=4$.
\pr $\cos(\sin x)$, $k=5$.
\pr $\sin(\sin x)$, $k=6$.
\pr $\sin(1-\cos x)$, $k=3$.
\pr $\arctg x$, $k\in\bold N$.
\pr $\arcsin x$, $k\in\bold N$.

\centerline{\smc Spo\v{c}t\v{e}te limity}

\pr $\l0 {\cos x- e^{-x^2\over 2}\over x^4}$
\pr $\l0{e^x \sin x - x(1+x)\over x^3}$
\pr $\l{+\infty} \left(\root 6 \of{x^6+x^5} - \root 6 \of {x^6-x^5}\right)$

\pr $\l{+\infty} x^{\frac32}\left(\sqrt{x+1}+\sqrt{x-1}-2\sqrt
x\right)$
\pr $\l0 {a^x+a^{-x}-2\over x^2} \ (a>0)$
\pr $\l0\left(\frac1x-{1\over\sin x}\right)$

\pr $\l{+\infty}\left(x-x^2\ln\left(1+\frac1x\right)\right)$
\pr $\l0 \frac1x\left(\frac1x -{\cos x\over \sin x}\right)$
\pr $\l{+\infty}\left(\left(x^3-x^2+\frac x2\right)
e^{\frac1x}-\sqrt{x^6+1}\right)$

\pr Najd\v{e}te $n\in\Bbb N$, aby limita
a) $\l0{\tg(\sin x)-\sin(\tg x)\over x^n}$, \quad
b) $\l0{(1+x)^x-1\over x^n}$
c) $\l0{\cos x-\cos(\tg x)\over x^n}$, \quad

d) $\l0{e-(1+x)^{\frac1x}\over x^n}$
byla kone\v{c}n\'a a r\accent23uzn\'a od $0$, a spo\v{c}t\v{e}te tuto limitu.

\pr Najd\v{e}te $a,b\in\Bbb R$, aby $\l0 {x-(a+b\cos x)\sin x\over x^4}=0$.

\pr Najd\v{e}te $a,b\in\Bbb R$, aby $\l0 {x-a\sin
x-b\tg x\over x^4}=0$ a spo\v{c}t\v{e}te $\l0 {x-a\sin
x-b\tg x\over x^5}$.
}

\varoddel

{%
\centerline{\smc VIII. Taylor�v polynom -- aplikace Lagrangeova tvaru
zbytku}

\pr Spo�t�te $e^2$ s chybou men� ne� $10^{-4}$.
\pr Spo�t�te $\log 1.1$ s chybou men� ne� $10^{-4}$.
\pr Spo�t�te $\sqrt 5$ s chybou men� ne� $10^{-3}$.
\pr Spo�t�te $\root 3\of{1.01}$ s chybou men� ne� $10^{-8}$.
\pr Kolik �len� Taylorova polynomu funkce $\cos$ v $0$ sta��
vz�t, aby pro ka�d� $x\in(-0.5,0.5)$ byla chyba men� ne�
$10^{-3}$?
}

\newpage

{
\let\prik\pr
\def\pr#1\right){\prik}
\centerline{\smc IX. Nalezn�te lok�ln� extr�my n�sleduj�c�ch funkc� v uveden�ch
mno�in�ch: \rm}
\pr1\right) $f\left(x,y\right) = x^4 + y^4 - x^2 -2xy - y^2; M = {\Bbb R}^2$

\pr2\right) $f\left(x,y\right) = xy + \frac{50}{x} + \frac{20}{y}; M = \{\left(x,y\right); x > 0; y
> 0 \}$

\pr3\right) $z\left(x,y\right) = x^2y^3\left(6 - x - y\right); M = \{\left(x,y\right); x > 0; y > 0\}$

*\pr4\right) $z\left(x,y\right) = x^2y^3\left(6 - x - y\right); M = {\Bbb R}^2$

\pr5\right) $f\left(x,y,z\right) = x + {y^2 \over 4x} + {z^2 \over y} + {2 \over z}$,
 $M = \{\left(x,y,z\right); x > 0, y > 0, z > 0 \}.$

\pr6\right) $f\left(x,y,z\right)  = x^4 - y^4 - x^2 - 2xy -
y^2,\quad M=\Bbb R^2$

\pr8\right) Doka�te, �e funkce
$
f\left(x,y\right) = \left(1 + e^y\right)\cos x - y e^y
$
nab�v� lok�ln�ho maxima v nekone�n� mnoha bodech, av�ak
lok�ln�ho minima ani v~jednom bod�.}

\varoddel

{
\centerline{\smc X. Najd�te v�echna �e�en� diferen�n�ch rovnic (s
po��te�n�mi podm�nkami)}

\pr $y(n+2) + 4y(n+1) + 4y(n) = 0$
\pr $y(n+2) - 3y(n+1) + 2y(n) = 0$

\pr $y(n+2) - 6y(n+1) +13y(n) = 0$, $y(1)=0$
\pr $y(n+2)-2y(n+1)-3y(n) = 0$, $y(1)=2$, $y(2)=1$
\pr $y(n+2)-y(n+1)-y(n)=0$, $y(1)=y(2)=1$
\pr $y(n+4) + 6y(n+2) + 9y(n) = 0$


\pr $y(n+2) - 2y(n+1) + 2y(n) = \cos n$
\pr $y(n+3) - y(n+2) - 2y(n+1)+2y(n) = n + 2^n$

\pr $y(n+2)  + 3y(n+1) + 2y(n) = \sin n  + \sin 2n$
\pr $8y(n+3)  + y(n) = 3n + 1/2^n$

\pr $y(n+2) - 3y(n+1) + 2y(n) = n^2$, $y(1)=3$, $y(2)=2$.
\pr $y(n+2) - y(n) = 17$, $y(1)=y(2)=0$.
}

\end