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\input cviceni
\vcorrection{12 mm}
\NoBlackBoxes

{
\centerline{\smc I. Vyj\'ad\v{r}ete primitivn\'{\i} funkce pomoc\'{\i} element\'arn\'{\i}ch
funkc\'{\i}}
\centerline{\smc na maxim\'aln\'{\i}ch intervalech existence}

\pr $\int x^3+2x+\frac{17}x \d x$
\pr $\int 18e^x+16e^{8x}-\frac1x+3\cos x \d x$
\pr $\int\root 3\of {1-3x}\d x$
\pr $\int{(1-x)^3\over x\root 3\of x}\d x$

\pr $\int {x^2\over(1-x)^{100}}\d x$
\pr $\int \sin^7x\cos x\d x$
\pr $\int x e^{-x^2} \d x$
\pr $\int \tg x\d x$
\pr $\int \cotg x\d x$

\pr $\int\sqrt{x^6}\d x$
\pr $\int\vert\cos x\vert\d x$
\pr $\int \arcsin\sin \frac{4x}{x^2+1} \d x$
\pr $\int{2^{x+1}-5^{x-1}\over10^x}\d x$
\pr $\int{e^{3x}+1\over e^x+1}\d x$
\pr $\int \tg^2x\d x$
\pr $\int\cotg^2x\d x$
\pr $\int {\d x\over2+3x^2}$
\pr $\int\sin^2x\d x$
\pr $\int \cos^4x\d x$
\pr $\int{x^2\d x\over\cos^2\left(x^3\right)}$
\pr $\int {x\over 1+4x^2}\d x$
\pr $\int{x\over1+x^4}\d x$
\pr $\int {\d x\over x\log x\log\log x}$
\pr $\int {\d x\over\sin x}$
\pr $\int {\d x\over\cos x}$
\pr $\int x e^x\d x$
\pr $\int \log x\d x$
\pr $\int \arctg x \d x$
\pr $\int{\sin x\over e^x}\d x$
\pr $\int e^{ax}\cos{bx}\d x\ ,\, a, b \in \Bbb R$
\pr $\int x^\alpha\log x\d x$

\pr $\int x^3\log^2x\d x$
\pr $\int e^{\sqrt{x}}\d x$
\pr $\int\log\left(x+\sqrt{x^2+1}\right)\d x$
}

\oddel

@ V\'ysledky jsou uvedeny \clqq a\v{z} na konstantu\crqq.
\pr $\frac14x^4+x^2+16\log|x|$ na $(-\infty,0)$ a na
$(0,\infty)$ \pr $18e^x+2e^{8x}-\log|x|+3\sin x$ na $(-\infty,0)$ a na
$(0,\infty)$
\pr $-\frac14(1-3x)^{\frac43}$, na $\Bbb R$
\pr $-\frac3{\root3\of x}-\frac92\root
3\of{x^2}+\frac95x^{\frac53}-\frac38x^{\frac83}$ na $(-\infty,0)$
a na $(0,\infty)$
\pr ${1\over99(1-x)^{99}}-{1\over 49(1-x)^{98}}+{1\over
97(1-x)^{97}}$, na $(-\infty,1)$ a na $(1,\infty)$
  \pr $\frac18\sin^8x$ na $\Bbb R$ \pr
$-\frac12e^{-x^2}$ na $\Bbb R$ \pr $-\log |\cos x|$ na ka\v{z}d\'em
z interval\accent23u $(-\frac\pi2+k\pi,\frac\pi2+k\pi)$, $k\in\Bbb Z$
\pr $\log|\sin x|$ na ka\v{z}d\'em z interval\accent23u $(k\pi,(k+1)\pi)$,
$k\in\Bbb Z$
\pr $\frac14|x|\cdot x^3$, na $\Bbb R$
\pr $\sin\left(x-\pi\cdot\left[\frac x\pi+\frac12\right]\right)+
2\cdot\left[\frac x\pi+\frac12\right]$, na $\Bbb R$
\pr $F(x)=\cases
2\log(1+x^2)+\pi(x_2-x_1)+ 4\log\frac{x_1}{x_2} &
x\in[-\infty,-x_2] \\
-\pi(x+x_1)-2\log(1+x^2)+4\log(\frac8\pi x_1) & x\in[-x_2,-x_1] \\
2\log(1+x^2) & x\in[-x_1,x_1] \\
\pi(x-x_1)-2\log(1+x^2)+4\log(\frac8\pi x_1) & x\in[x_1,x_2] \\
2\log(1+x^2)+\pi(x_2-x_1)+ 4\log\frac{x_1}{x_2} &
x\in[x_2,\infty] \endcases$\qquad a \quad $\matrix x_1={4-\sqrt{16-\pi^2}\over\pi}
\\ x_2={4+\sqrt{16-\pi^2}\over\pi}\endmatrix$, na $\Bbb R$
\pr ${-10\cdot 5^{-x}\log2+2^{-x}\log5\over 5\log5\log2}$,
na $\Bbb R$
\pr $\frac12e^{2x}-e^x+x$, na $\Bbb R$
\pr $\tg x-x$, na ka\v{z}d\'em z interval\accent23u
$(-\frac\pi2+k\pi,\frac\pi2+k\pi)$, $k\in\Bbb Z$
\pr $-\cotg x-x$, na ka\v{z}d\'em z interval\accent23u $(k\pi,(k+1)\pi)$,
$k\in\Bbb Z$
\pr $\frac1{\sqrt6}\arctg{x\sqrt6\over2}$, na $\Bbb R$
\pr $\frac x2-\frac14\sin 2x$, na $\Bbb R$
\pr $\frac38x+\frac38\sin x\cos x+\frac14\sin x\cos^3x$, na $\Bbb
R$
\pr $\frac13\tg(x^3)$, na ka\v{z}d\'em z interval\accent23u
$(\root 3\of{-\frac\pi2+k\pi},\root3\of{\frac\pi2+k\pi})$,
$k\in\Bbb Z$
\pr $\frac18\log(1+4x^2)$, na $\Bbb R$
\pr $\frac12 \arctg(x^2)$, na $\Bbb R$
\pr $\log|\log\log x|$, na $(1,e)$ a na $(e,\infty)$
\pr $\log\left|\tg\frac x2\right|$, na ka\v{z}d\'em z interval\accent23u $(k\pi,(k+1)\pi)$,
$k\in\Bbb Z$
\pr $-\log\left|\tg(\frac\pi4-\frac x2)\right|$,  na ka\v{z}d\'em z interval\accent23u
$(-\frac\pi2+k\pi,\frac\pi2+k\pi)$, $k\in\Bbb Z$
\pr $(x-1)e^x$ na $\Bbb R$
\pr $x\log x-x$ na $(0,\infty)$
\pr $x\arctg x -\frac12\log(1+x^2)$ na $\Bbb R$
\pr $-\frac12e^{-x}\cdot(\sin x+\cos x)$, na $\Bbb R$
\pr ${e^{ax}\over a^2+b^2}\cdot(a\cos bx+b\sin bx)$ na $\Bbb
R$, pokud $a^2+b^2\ne0$; $x$ na $\Bbb R$, pokud $a=b=0$
\pr ${x^{1+\alpha}\over 1+\alpha}\cdot\left(\log
x-\frac1{1+a}\right)$, na $(0,\infty)$ pro $\alpha\ne1$;
$\frac12\ln^2x$, na $(0,\infty)$ pro $\alpha=-1$
\pr $\frac14x^4\ln^2x-\frac18x^4\ln x+\frac1{32}x^4$, na
$(0,\infty)$
\pr $2e^{\sqrt x}\cdot(\sqrt x-1)$, na $(0,\infty)$
\pr $x\log\left(x+\sqrt{x^2+1}\right)-\sqrt{x^2+1}$, na $\Bbb R$


\newpage

{
\centerline{\smc II. Spo\v{c}t\v{e}te primitivn\'{\i} funkce}

\pr $\int \frac{x^{17}-5}{x-1}\d x$
\pr $\int \frac{x^{17}-5}{x^2-1}\d x$
\pr $\int \frac{x^3+1}{x^3-5x^2+6x}\d x$
\pr $\int \frac{x}{x^3-1}\d x$
\pr $\int {\d x\over x^4+1}$

\pr $\int {\d x\over (x^2+1)^2}$ pomoc\'{\i} $x=\tg y$
\pr $\int{x^8+x-1\over x^6+1}\d x$
\pr $\int{\d x\over\left(x^2+3x+4\right)^2}$

\pr Pro jak\'y vztah mezi parametry $a$, $b$, $c\in \Bbb R$
je primitivn\'{\i} funkce k funkci $f$ racion\'aln\'{\i}, je-li
$f(x)=\frac{ax^2+bx+c}{x^3(x-1)^2}$?
\pr $\int\frac{x\,\d x}{\sqrt{x+1}+\root 3\of{x+1}}$
\pr $\int\sqrt{\frac{1-x}{1+x}}\frac{\d x}{x}$
\pr $\int\root3\of{\frac{1-x}{1+x}}\frac{\d x}{x}$
\pr $\int{\d x\over\left(4+x^2\right)\sqrt{4-x^2}}$
\pr $\int{1-\sqrt{x+1}\over 1-\root 3 \of{x+1}}\d x$
\pr $\int{\sqrt{x+1}-\sqrt{x-1}\over\sqrt{x+1}+\sqrt{x-1}}\d x$
\pr $\int \frac{\exp x}{\exp x + 1}\d x$
\pr $\int{e^{2x}\over 1+e^x}\d x$
\pr $\int\frac{\d x}{1+\sqrt{x^2 + 2x + 2}}$
\pr $\int\,\sqrt{x^2 - 2x - 1}\d x$
\pr $\int\frac{(2x + 3)\,\d x}{(x^2+2x+3)\sqrt{x^2+2x+4}}$
\pr $\int{\d x\over\sqrt{x^2+1}}$
\pr $\int\sqrt{x^2-1}\d x$

\pr $\int\sqrt{1-x^2}\d x$
\pr $\int {\sin x\cos x\over 1+\sin^4x}\d x$
\pr $\int {\sin^2x\over\sin x+2\cos x}$
\pr $\int{\d x\over 2\sin x-\cos x+5}$

\pr $\int{\d x\over\left(\sin^2x+2\cos^2x\right)^2}$
\pr $\int{\d x\over(2+\cos x)\sin x}$
\pr $\int{\sin^2 x\over 1+\sin^2 x}\d x$
\pr $\int{\sin x\cos x\over \sin x+\cos x}\d x$

\pr $\int{\sin x\over\sin^3x+\cos^3x}\d x$
\pr $\int{\d x\over\sin^4x+\cos^4x}$
\pr $\int{\d x\over 1+\varepsilon\cos x}$, $\varepsilon>0$
\pr $\int{\d x \over a^2\sin^2x+b^2\cos^2x}$, $a,b\in\Bbb R$
}

\oddel

@ \pr  $ \left(\sum_1^{17}\frac{x^k}{k}\right) -
4\log|x-1|$,  $x \in (-\infty, 1)$ nebo  $x \in (1,
+\infty)$
\pr $\left(\sum_{1}^{8}\frac{1}{2k}x^{2k}\right)
- 2\log |x - 1| +  3\log |x+1|$,
 $x \in (-\infty, -1)$  nebo $x \in (-1,1)$
nebo  $x \in (1,+\infty)$
\pr  $ x + \frac{1}{6}\log|x| -\frac{9}{2}\log|x-2|+
\frac{28}{3}\log|x-3|$,
$x \in (-\infty, 0)$  nebo  $x \in(0,2)$ nebo  $x \in (2,3)$ nebo
$x \in (3, +\infty)$
\pr $  \frac16\log\frac{(x-1)^2}{x^2+x+1}
+\frac{\sqrt3}{3}\arctg\frac{2x+1}{\sqrt3}$, $x \in (-\infty, 1)$
 nebo  $x \in (1, +\infty)$
\pr $\frac{\sqrt2}8\log{x^2+x\sqrt2+1\over
x^2-x\sqrt2+1}+\frac{\sqrt2}4\arctg(x\sqrt2+1)+\frac{\sqrt2}4
\arctg(x\sqrt2-1)$
\pr ${x\over 2(x^2+1)}+\frac12\arctg x$
\pr $\frac13x^3+\frac16\log(1+x^2)+{\sqrt3-1\over2}\log(x^2-x\sqrt3+1)
-{\sqrt3+1\over2}\log(x^2+x\sqrt3+1)-{3-\sqrt3\over6}\arctg(2x-
\sqrt3)-{3+\sqrt3\over6}\arctg(2x+\sqrt3)$
\pr $\frac17\cdot{2x+3\over
x^2+3x+4}+{4\sqrt7\over49}\arctg{2x+3\over\sqrt7}$
\pr  $ a+2b+3c=0 $
\pr $6\Big(\frac{1}{9}u^{(3/2)}- \frac{1}{8}u^{(4/3)}+
\frac{1}{7}u^{(7/6)}- \frac{1}{6}u+ \frac{1}{5}u^{(5/6)}-
\frac{1}{4}u^{(2/3)}\Big)$,  kde   $u=x+1$,  $x \in
(-1, +\infty) $
\pr $\log\Big|\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+
\sqrt{1-x}}\Big| +2\arctg\sqrt{\frac{1-x}{1+x}}$, $x \in
(-1,1)$
\pr $\log\frac{|u^2-1|}{\sqrt{u^4+u^2+1}} -
\sqrt3\arctg\frac{2u + 1}{\sqrt3}\ + \sqrt3\arctg\frac{2u -
1}{\sqrt3}$, kde $u= \root3\of{\frac{1-x}{1+x}}$,
$x \in (-\infty,-1)$ nebo   $x \in (-1,0)$  nebo $x \in (1,+\infty)$
\pr  $t=\sqrt{2-x\over 2+x}$
\pr $t=\root 6\of{x+1}$
\pr $t=\sqrt{x-1\over x+1}$
\pr $ \log {(e^x + 1)}$, $x \in \Bbb R$
\pr $e^x-\log(1+e^x)$
\pr $ \frac{2}{x - \sqrt{x^2 + 2x +2}}
- \log (\sqrt{x^2 + 2x + 2} -x -1)$, $x \in \Bbb R $
\pr  $ \frac 12 (x - 1)\sqrt{x^2 -2x -1} - \log
|x-1 + \sqrt{x^2 - 2x - 1}|$, $x\in (-\infty, 1 - \sqrt 2)$
 nebo  $x \in (1 + \sqrt 2, +\infty)$
\pr $\log \frac{t^2 - 4t + 6}{t^2 + 2} +
\frac{1}{\sqrt 2} \arctg \frac{t}{\sqrt 2} -
\frac{1}{\sqrt 2} \arctg \frac{t-2}{\sqrt 2}$,  kde
$t = \sqrt{x^2 + 2x + 4} - x$, $x \in \Bbb R$
\pr $\argsinh x=\log(x+\sqrt{x^2+1})$ na $\Bbb R$
\pr $\frac12x\sqrt{x^2-1}-\frac12\log|x+\sqrt{x^2-1}|$ na
$(-\infty,-1)$ a na $(1,+\infty)$
\pr $\frac x2\sqrt{1-x^2}+\frac12\arcsin x$ na $(-1,1)$
\pr $\frac12\arctg\sin^2x$
\pr a \pr substituce $y=\tg\frac x2$
\pr substituce $y=\tg x$
\pr substituce $y=\cos x$
\pr substituce $y=\tg x$
\pr substituce $y=\tg\frac x2$
\pr a \pr substituce $y=\tg x$
\pr substituce $y=\tg\frac x2$
\pr substituce $y=\tg x$\par

\newpage

{\let\e\varepsilon
\centerline{\smc III. Vy\v{s}et\v{r}ete bodovou, stejnom\v{e}rnou a lok\'aln\v{e}
stejnom\v{e}rnou}
\centerline{\smc konvergenci posloupnost\'{\i}}

\pr $x^{n+1}-x^{n-1},\,x\in\langle0,1\rangle,$
\pr $x^n-x^{3n},\,x\in\langle0,1\rangle$
\pr ${1\over x+n}$ a) $x\in(0,+\infty)$, b) $x\in\bold R$

\pr ${nx\over 1+n+x},\,x\in\langle0,1\rangle$
\pr ${x^n\over 1+x^n},\
a)\,x\in\langle0,1-\e\rangle,\ b)\,x\in\langle1-\e,1+\e\rangle,\,c)\,x\in\langle1+\e,+\infty)$,

kde $\e\in(0,1)$
\pr ${2nx\over 1+n^2x^2},\,a)\,x\in\langle0,1\rangle,\,b)\,x\in(1,+\infty)$
\pr $\sqrt{x^2+{1\over n^2}},\,x\in\bold R$
\pr ${\sin nx\over n},\,x\in\bold R$,
\pr $\sin\frac x n,\,x\in\bold R$.
\pr $n\cdot\left(\sqrt{x+\frac1n}-\sqrt x\right)$, $x\in \langle0,\infty)$

\pr Nech\softt{} $f_n\rightrightarrows f$ na $M$ a $g_n\rightrightarrows g$ na
$M$. Plat\'{\i}
(a) $f_n+g_n\rightrightarrows f+g$ na $M$;
(b) $f_n\cdot g_n\rightrightarrows f\cdot g$ na $M$;
(c) ${f_n\over g_n}\rightrightarrows {f\over g}$ na $M$?
}

\oddel

{\let\tost\rightrightarrows
@ \pr $f_n\tost 0$ na $\langle0,1\rangle$ \pr $f_n\to 0$ na $\langle0,1\rangle$,
stejnom\v{e}rn\v{e} na $\langle0,1-\varepsilon\rangle$ pro $\varepsilon\in(0,1)$, na
\v{z}\'adn\'em $\langle1-\varepsilon,1\rangle$ konvergence nen\'{\i} stejnom\v{e}rn\'a.
\pr $f_n\to 0$ na $\bold R$, stejnom\v{e}rn\v{e} na $(k,\infty)$ pro ka\v{z}d\'e
$k\in\bold R$, na \v{z}\'adn\'em $(-\infty,k)$ konvergence nen\'{\i}
stejnom\v{e}rn\'a.
\pr $f_n\tost x$ na $\langle0,1\rangle$ (dokonce na $\langle0,k\rangle$ pro $k\in
(0,\infty)$, ale ne na okol\'{\i}  $+\infty$.
\pr $f_n(x)\to \cases 0 & x\in\langle0,1) \\ \frac12 & x=1 \\ 1 & x>1
\endcases$, a) stejnom\v{e}rn\v{e}, b) nestejnom\v{e}rn\v{e}, c) stejnom\v{e}rn\v{e}.
\pr $f_n\to 0$ na $\bold R$, stejnom\v{e}rn\v{e} na $\langle\varepsilon,\infty\rangle$
pro $\varepsilon>0$, nestejnom\v{e}rn\v{e} na $\langle0,\varepsilon\rangle$.
\pr $f_n\tost|x|$ na $\bold R$
\pr $f_n\tost0$ na $\bold R$
\pr $f_n\to0$ na $\bold R$, stejnom\v{e}rn\v{e} na omezen\'ych intervalech,
nestejnom\v{e}rn\v{e} na okol\'{\i}ch $\pm\infty$.
\pr $f_n(x)\to \cases 0 & x=0 \\ {1\over 2\sqrt x} & x>0\endcases$,
stejnom\v{e}rn\v{e} na $\langle\varepsilon,\infty\rangle$, $\varepsilon>0$, nikoli na
$\langle0,\varepsilon\rangle$.
\pr (a) ANO. (b) NE. Nap\v{r}\'{\i}klad $f_n(x)=g_n(x)=x+\frac1n$ na
$\bold R$.
Tvrzen\'{\i} v\v{s}ak plat\'{\i}, jsou-li $f$ a $g$ omezen\'e na $M$.
(c) NE (ani za p\v{r}edpokladu, \v{z}e v\v{s}echny pod\'{\i}ly maj\'{\i} smysl).
Nap\v{r}\'{\i}klad $f_n(x)=1$, $g_n(x)=\frac1x+\frac1n$ na $(0,+\infty)$.
Tvrzen\'{\i} v\v{s}ak plat\'{\i}, jsou-li funkce $f$ a $\frac1g$ omezen\'e na
$M$.
\par }

\varoddel

{
\centerline{\smc IV. Vy\v{s}et\v{r}ete bodovou, stejnom\v{e}rnou a lok\'aln\v{e}
stejnom\v{e}rnou}
\centerline{\smc konvergenci \v{r}ad}

\pr $\sk1 {kx\over 1+k^5x^2}$
\pr $\sk1 \log\left(1+{x^2\over k^2}\right)$
\pr $\sk1 {(-1)^k\over x+k}$
\pr $\sk1 {(-1)^k\over x+2^k}$
\pr $\sk1 \arctg{2x\over x^2+k^3}$

\pr $\sk2 {(-1)^k\over k+\sin x}$
\pr $\sk1 2^k\sin{1\over 3^k x}$
\pr* $\sk1 {\sin kx\over k}$
\pr* $\sk1 {\cos kx\over k}$
\pr* $\sk1 {\sin x\sin kx\over\sqrt{k+x}}$

\pr* $\sk1 {\cos kx\over k}\arctg(kx+k)$

\pr Spo\v{c}t\v{e}te limity:
(a) $\lx{1-}\sk1{(-1)^k\over k}\cdot{x^k\over x^k+1}$
(b) $\lx{1-}\sk1(x^k-x^{k+1})$
(c) $\lx{0+}\sk1{1\over 2^k k^x}$

\pr Plat\'{\i} $\lp\int_0^1 f_n(x)\d x=\int_0^1 \lp f_n(x)\d x$,
jestli\v{z}e
(a) $f_n(x)=nxe^{-nx^2}$,

(b) $f_n(x)=nx(1-x)^n$, (c)
$f_n(x)={nx\over 1+n^3x^2}$ ?
Jak je to se stejnom\v{e}rnou konvergenc\'{\i} $f_n$ na $(0,1)$?



\pr Doka\v{z}te, \v{z}e funkce $\zeta(x)=\sk1\frac1{k^x}$ m\'a na intervalu $(1,\infty)$
derivace v\v{s}ech \v{r}\'ad\accent23u.

\pr Nech\softt{} $\bold a=\{a_k\}$ je posloupnost nez\'aporn\'ych re\'aln\'ych \v{c}\'{\i}sel.
Polo\v{z}me $\zeta_{\bold a}(x)=\topsmash{\sk1\frac{a_k}{k^x}}$. Co lze \v{r}\'{\i}ci o
defini\v{c}n\'{\i}m oboru, spojitosti a derivac\'{\i}ch funkce $\zeta_{\bold
a}$ ? (*Jak je to v p\v{r}\'{\i}pad\v{e}, \v{z}e $\bold a=\{a_k\}$ je posloupnost
re\'aln\'ych, ne nutn\v{e} nez\'aporn\'ych \v{c}\'{\i}sel?)


\pr Nech\softt{} $f(x)=\sk1{x\over k^p+x^2 k^q}$, kde $p,q\ge 0$.
Doka\v{z}te, \v{z}e

(i) $f$ je spojit\'a na $(0,\infty)$, pokud
$\max(p,q)>1$, (ii) $f$ je spojit\'a na $\Bbb R$, pokud $p+q>2$.

(iii) Vy\v{s}et\v{r}ete defini\v{c}n\'{\i} obor a spojitost $f$ v
z\'avislosti~na~$p,q$.

\pr V kter\'ych bodech m\'a derivaci funkce
(a) $f(x)=\sk1 {(-1)^k\over k+x}$
(b) $f(x)=\sk1 {|x|\over k^2+x^2}$

\pr* Doka\v{z}te, \v{z}e $f(x)=\sk1{\sin kx\over k^3}$ m\'a spojitou
derivaci na $\Bbb R$ a spojitou $f''$ na $(0,2\pi)$.
}

\newpage

@ \pr Konverguje stejnom\v{e}rn\v{e} na $\bold R$ (tot\'e\v{z} plat\'{\i} pro \v{r}adu
absolutn\'{\i}ch hodnot, lze pou\v{z}\'{\i}t Weierstrassovo krit\'erium).
\pr Konverguje bodov\v{e} na $\bold R$ (absolutn\v{e}). Konvergence je
stejnom\v{e}rn\'a na omezen\'ych intervalech (Weierstrassovo krit\'erium),
nen\'{\i} omezen\'a na okol\'{\i}ch $+\infty$ a $-\infty$ (nutn\'a podm\'{\i}nka).
\pr Konverguje bodov\v{e} na mno\v{z}in\v{e} $M=\{x\in\bold R; -x\notin
\bold N\}$ (neabsolutn\v{e}), v ostatn\'{\i}ch bodech nem\'a smysl. Konvergence
je stejnom\v{e}rn\'a na $(K,+\infty)\cap M$ pro ka\v{z}d\'e $K\in\bold R$
(Leibnizovo krit\'erium), nen\'{\i}
stejnom\v{e}rn\'a na $(-\infty,K)\cap M$ (nutn\'a podm\'{\i}nka).
\pr  Konverguje bodov\v{e} na mno\v{z}in\v{e} $M=\{x\in\bold R; x\ne
-2^k\text{ pro }k\in \bold N\}$ (absolutn\v{e}), v ostatn\'{\i}ch bodech nem\'a smysl.
Konvergence
je stejnom\v{e}rn\'a na $(K,+\infty)\cap M$ pro ka\v{z}d\'e $K\in\bold R$
(Weierstrassovo krit\'erium), nen\'{\i}
stejnom\v{e}rn\'a na $(-\infty,K)\cap M$ (nutn\'a podm\'{\i}nka).
\pr Konverguje stejnom\v{e}rn\v{e} na $\bold R$ (tot\'e\v{z} plat\'{\i} pro \v{r}adu
absolutn\'{\i}ch hodnot, lze pou\v{z}\'{\i}t Weierstrassovo krit\'erium).
\pr Konverguje stejnom\v{e}rn\v{e} na $\bold R$ (neabsolutn\v{e}, Leibnizovo
krit\'erium).
\pr Konverguje bodov\v{e} na $\bold R\setminus\{0\}$ (absolutn\v{e}).
Konvergence je stejnom\v{e}rn\'a na $\bold
R\setminus\langle-\varepsilon,\varepsilon\rangle$ pro
$\varepsilon>0$ (Weierstrassovo krit\'erium), nen\'{\i} stejnom\v{e}rn\'a na
$(-\varepsilon,0)$ ani na $(0,\varepsilon)$ (nutn\'a podm\'{\i}nka).
\qquad
{\bf Pozn\'amka:} V dal\v{s}\'{\i}ch p\v{r}\'{\i}kladech se pou\v{z}\'{\i}v\'a i Dirichletovo a
Abelovo krit\'erium a odhady $\left|\sum_{k=1}^n \sin kx\right|\le
{1\over 2|\sin \frac x2|}$ a $\left|\sum_{k=1}^n \cos kx\right|\le
{1\over 2|\sin \frac x2|}$ pro $x\ne 2k\pi$ ($k\in\bold Z$).
\pr Konverguje bodov\v{e} na $\bold R$ (pro $x=k\pi$ je to z\v{r}ejm\'e,
pro ostatn\'{\i} to plyne z Dirichletova krit\'eria). Konvergence je
stejnom\v{e}rn\'a na intervalech
$(2k\pi+\varepsilon,(2k+2)\pi-\varepsilon)$ pro ka\v{z}d\'e
$\varepsilon\in (0,\pi)$ (Dirichletovo krit\'erium), nen\'{\i}
stejnom\v{e}rn\'a na $(2k\pi,2k\pi+\varepsilon)$ ani na
$(2k\pi-\varepsilon,2k\pi)$ pro $\varepsilon>0$
(Bolzano-Cauchyova podm\'{\i}nka).
\pr  Konverguje bodov\v{e} na $\bold R\setminus \{2k\pi; k\in\bold
Z\}$ (Dirichletovo krit\'erium, divergence v bodech $2k\pi$ je
d\accent23usledkem divergence \v{r}ady $\sk1 \frac1k$). Konvergence je
stejnom\v{e}rn\'a na intervalech
$(2k\pi+\varepsilon,(2k+2)\pi-\varepsilon)$ pro ka\v{z}d\'e
$\varepsilon\in (0,\pi)$ (Dirichletovo krit\'erium), nen\'{\i}
stejnom\v{e}rn\'a na $(2k\pi,2k\pi+\varepsilon)$ ani na
$(2k\pi-\varepsilon,2k\pi)$ pro $\varepsilon>0$
(v\v{e}ta o z\'am\v{e}n\v{e} limit).
\pr Konverguje stejnom\v{e}rn\v{e} na $(-1,+\infty)$ (Dirichletovo
krit\'erium), pro $x\le-1$ nem\'a smysl.
\pr  Konverguje bodov\v{e} na $\bold R\setminus \{2k\pi; k\in\bold
Z\}$ (Dirichletovo a Abelovo krit\'erium, divergence v bodech $2k\pi$ je
d\accent23usledkem divergence \v{r}ady $\sk1 \frac1k$). Konvergence je
stejnom\v{e}rn\'a na intervalech
$(2k\pi+\varepsilon,(2k+2)\pi-\varepsilon)$ pro ka\v{z}d\'e
$\varepsilon\in (0,\pi)$ (Dirichletovo a Abelovo krit\'erium), nen\'{\i}
stejnom\v{e}rn\'a na $(2k\pi,2k\pi+\varepsilon)$ ani na
$(2k\pi-\varepsilon,2k\pi)$ pro $\varepsilon>0$
(v\v{e}ta o z\'am\v{e}n\v{e} limit).
\pr (a) Limita je $\frac12\sk1 {(-1)^k\over k}=-\frac12\log2$
(\v{r}ada konverguje stejnom\v{e}rn\v{e} na $\langle0,+\infty)$ podle Abelova
krit\'eria). (b) Limita je $1$ (lze spo\v{c}\'{\i}tat element\'arn\v{e}, nelze
p\v{r}ehodit limitu a sumu, \v{r}ada nekonverguje stejnom\v{e}rn\v{e} na
$(1-\varepsilon,1)$). (c) Limita je $1$ (\v{r}ada konverguje
stejnom\v{e}rn\v{e} nap\v{r}\'{\i}klad na $(-1,+\infty)$).
\pr (a) $f_n\to0$ na $\bold R$, konvergence nen\'{\i} stejnom\v{e}rn\'a na
$(0,1)$, $\lim_{n\to\infty}\int_0^1 f_n =\frac12\ne0=\int_0^1
\lim_{n\to\infty} f_n$. (b)  $f_n\to0$ na $\langle0,1\rangle$,
konvergence nen\'{\i} stejnom\v{e}rn\'a na $(0,1)$,
$\lim_{n\to\infty}\int_0^1 f_n =0=\int_0^1 \lim_{n\to\infty} f_n$.
(c) $f_n\rightrightarrows 0$ na $(0,1)$, a tedy
$\lim_{n\to\infty}\int_0^1 f_n =0=\int_0^1 \lim_{n\to\infty} f_n$.
\pr Doka\v{z}te pomoc\'{\i} Weierstrassova krit\'eria, \v{z}e \v{r}ada $n$-t\'ych
derivac\'{\i} konverguje lok\'aln\v{e} stejnom\v{e}rn\v{e} na $(1,+\infty)$ pro
v\v{s}echna $n$.
\pr Defini\v{c}n\'{\i} obor je bu\softd{} $\emptyset$, nebo interval
$(\alpha,+\infty)$ pro n\v{e}jak\'e $\alpha\in\bold R$, nebo interval
$\langle \alpha,+\infty)$ pro n\v{e}jak\'e $\alpha\in\bold R$, nebo
$\bold R$. Funkce je spojit\'a na sv\'em defini\v{c}n\'{\i}m oboru. Pro
$x\in(\alpha,+\infty)$ (je-li defini\v{c}n\'{\i} obor $(\alpha,+\infty)$
nebo $\langle\alpha,+\infty)$) nebo pro $x\in\bold R$ (je-li
defini\v{c}n\'{\i} obor $\bold R$) m\'a funkce $\zeta_{\bold a}$ pro v\v{s}echna
$n\in\bold N$ vlastn\'{\i} $n$-tou derivaci ${(\zeta_{\bold
a})}^{(n)}(x)=\sk1 {(-1)^n a_k\log^n k\over k^x}$. Je-li defini\v{c}n\'{\i} obor
$\langle\alpha,+\infty)$, pak $n$-t\'a derivace zprava v bod\v{e}
$\alpha$ je d\'ana t\'ym\v{z} vzorcem, pokud ov\v{s}em \v{r}ada ve vzorci
konverguje (co\v{z} m\accent23u\v{z}e a nemus\'{\i} b\'yt pravda). Pou\v{z}ije se
Weierstrassovo krit\'erium (a v\v{e}ty o z\'am\v{e}n\v{e} limit a derivac\'{\i} atp.).
Pro obecn\'y p\v{r}\'{\i}pad plat\'{\i} stejn\'e z\'av\v{e}ry, jen se pou\v{z}ije Abelovo
krit\'erium m\'{\i}sto Weierstrassova.
\pr (i) Uka\v{z}te, \v{z}e \v{r}ada konverguje lok\'aln\v{e} stejnom\v{e}rn\v{e} na
$(0,+\infty)$.
(ii) Uka\v{z}te, \v{z}e \v{r}ada konverguje stejnom\v{e}rn\v{e} na $\bold R$.
(iii) Je-li $\max(p,q)\le1$, je $D_f=\{0\}$, je-li $\max(p,q)>1$,
je $D_f=\bold R$. Pro $\max(p,q)>1$ je $f$ spojit\'a na $\bold R$.
Je-li nav\'{\i}c $p+q>2$ nebo $p>q$, je $f$ spojit\'a na $\bold R$;
je-li $p+q\le 2$ a $p<q$, nen\'{\i} $f$ spojit\'a v $0$ (ani zleva ani
zprava). (P\v{r}\'{\i}pad $p+q\le2$ a $p=q$ nem\accent23u\v{z}e nastat, je-li
$\max(p,q)>1$.)
\pr (a) Ve v\v{s}ech bodech defini\v{c}n\'{\i}ho oboru, kter\'ym je mno\v{z}ina
$\{x\in\bold R; -x\notin \bold N\}$. (b) $D_f=\bold R$, derivace
existuje (vlastn\'{\i}) v bodech $\bold R\setminus\{0\}$. V $0$
derivace neexistuje, proto\v{z}e $f'_+(0)=\sk1{1\over k^2}$ a
$f'_-(0)=-f'_+(0)$.
\pr Pro prvn\'{\i} derivaci pou\v{z}ijte Weierstrassovo krit\'erium, pro
druhou Dirichletovo krit\'erium (viz p\v{r}\'{\i}klad 8).

\varoddel

{
\let\s\sk
\centerline{\smc V. N\'asleduj\'{\i}c\'{\i} funkce vyj\'ad\v{r}ete jako sou\v{c}et
mocninn\'e \v{r}ady o st\v{r}edu $0$}
\centerline{\smc na maxim\'aln\'{\i}m otev\v{r}en\'em intervalu}

\pr $e^{-x^2}$
\pr $\cos^2 x$
\pr $\sin^3 x$
\pr $x^{10}\over 1-x$
\pr $1\over (1-x)^2$
\pr $x\over \sqrt{1-2x}$
\pr $\log\sqrt{1+x\over 1-x}$
\pr $x\over 1+x-2x^2$
\pr $1\over x^2+x+1$
\pr $\log(1+x+x^2+x^3)$

\centerline{\smc Na kruhu konvergence se\v{c}t\v{e}te mocninn\'e \v{r}ady}

\pr $\s0{x^{2k+1}\over 2k+1}$ \pr $\s0{(-1)^k
x^{2k+1}\over 2k+1}$ \pr $\s1 k x^k$  \pr $\s1 (-1)^k k^2 x^k$
\pr $\s0{x^{2k}\over (2k)!}$

\pr $\s1 {x^k\over k(k+1)}$ \pr $\s1 k(k+1)x^k$\quad
\pr $1+\s1{(2k-1)!!\over(2k)!!}x^k$ \pr $\s0{x^{4k}\over (4k)!}$
}
\oddel

@ \pr $\sk0  {(-1)^k x^{2k}\over k!}$ na $\bold R$.
\pr $1+\sk1 { (-1)^k 2^{2k-1}\over (2k)! } x^{2k}$ na $\bold R$.
\pr $\sk0 \frac 34 \cdot { (-1)^k \over (2k+1)!} (1-3^{2k})
x^{2k+1}$ na $\bold R$.
\pr $\sk{10} x^k$ na $(-1,1)$.
\pr $\sk0 (k+1)x^k$ na $(-1,1)$.
\pr $\sk1 \binom{-\frac12}{k-1} (-1)^{k-1} \cdot 2^{k-1} x^k=\sk1
{(2k-3)!! \over (k-1)!} x^k$ na $(-\frac12,\frac12)$.
\pr $\sk0 {x^{2k+1}\over 2k+1}$ na $(-1,1)$.
\pr $\sk0 {1+(-1)^k 2^{k+1}\over 3} x^k$ na $(-\frac12,\frac12)$.
\pr $\sk0 (-1)^{k+1}{2\over \sqrt3}\sin(k+1)\frac\pi3 \cdot x^k$
na $(-1,1)$. \newline
\pr $\sk1 a_k x^k$, kde $a_k=\cases {1\over 2l-1} & k=2l-1 \\
-{1\over 2l}+{(-1)^{l+1}\over l} & k=2l \endcases$ ($l\in\bold
N$) na $(-1,1)$.
\pr $\frac12\log{1+x\over 1-x}$ na $(-1,1)$.
\pr $\arctg x$ na $(-1,1)$.
\pr ${x\over (1-x)^2}$ na $(-1,1)$.
\pr ${x(1-x)\over (1+x)^3}$ na $(-1,1)$.
\pr ${e^x+e^{-x}\over 2}$ na $\bold R$.
\pr Sou\v{c}et je $f(x)=\cases 0 & x=0 \\ {\log (1+x)\over
x}+\log(1+x)-1 & x\in (-1,0)\cup(0,1).\endcases$
\pr $2x\over(1-x)^3$ na $(-1,1)$.
\pr $1\over \sqrt{1-x}$ na $(-1,1)$.
\pr $\frac12({e^x+e^{-x}\over 2}+\cos x)$ na $\bold R$.
\par

\varoddel

{
\centerline{\smc VI. 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) $\sin^2 x$, \quad
(b) $\cos^3 x$,\quad
(c) $\cos^{2m} x$, \quad
(d) ${\sin mx\over \sin x}$, \quad
(e) $\sum\limits_{m=1}^\infty \alpha^m{\sin m x\over \sin x}$,
kde $|\alpha|<1$.
(f) ${1\over \sin x}$.

\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 $(0,2\pi)$, $x=0$.
(e) $x^2-x$ na $(-\pi,\pi)$.

(f) $x^2+\sin^4 x$ na $(0,2\pi)$.

\pr (a) Napi\v{s}te funkci $f(x)=x$ na $(0,\pi)$ jako sou\v{c}et cosinov\'e
\v{r}ady. Jak\'y je sou\v{c}et \v{r}ady v ostatn\'{\i}ch bodech?
(b) Napi\v{s}te funkci $f(x)=x^2$ na $(-\pi,0)$ jako sou\v{c}et sinov\'e
\v{r}ady. Jak\'y je sou\v{c}et \v{r}ady v ostatn\'{\i}ch bodech?
(c)  Napi\v{s}te funkci $f(x)=\sin x$ na $(0,\pi)$ jako sou\v{c}et cosinov\'e
\v{r}ady. Jak\'y je sou\v{c}et \v{r}ady v ostatn\'{\i}ch bodech?

*\pr Funkci $f(x)=x$ na $(0,\frac\pi2)$ vyj\'ad\v{r}ete jako sou\v{c}et
trigonometrick\'e \v{r}ady obsahuj\'{\i}c\'{\i} jen \v{c}leny tvaru

(a) $c\sin kx$, $k\in\Bbb N$ lich\'e,
(b) $c\cos kx$, $k\in\Bbb N$ lich\'e,
(c) $c\cos kx$, $k\in\Bbb N$ sud\'e,
(d) $c\sin kx$, $k\in\Bbb N$ sud\'e.

Jak vypadaj\'{\i} sou\v{c}ty t\v{e}chto \v{r}ad?
}

\newpage

@ \pr a) $\frac12-\frac12\cos 2x$, konverguje v\v{s}ude
k $f$, je to trigonometrick\'y polynom.
b) $\frac34\sin x-\frac14\sin 3x$, konverguje v\v{s}ude
k $f$, je to trigonometrick\'y polynom.
c) ${1\over 2^{2m}}\cdot\binom{2m}{m}+\sum\limits_{n=1}^m
{1\over 2^{2m-1}}\cdot\binom{2m}{m-n}\cos 2nx$, konverguje v\v{s}ude
k $f$, je to trigonometrick\'y polynom.
d) Pro $m=2k$: $\sum\limits_{n=1}^k 2\cos (2n-1)x$; pro $m=2k+1$:
$1+\sum\limits_{n=1}^k 2\cos 2nx$; konverguje v\v{s}ude
k $f$, je to trigonometrick\'y polynom.
e) ${\alpha\over 1-\alpha^2}+\sn1{2\alpha^{n+1}\over
1-\alpha^2}\sin nx$, konverguje v\v{s}ude k $f$.
f) Toto nen\'{\i} po \v{c}\'astech spojit\'a funkce, nem\'a F\v{R}.
\pr a) ${\pi^2\over 3}+\sn1 {4\cdot(-1)^n\over n^2}\cdot\cos nx$;
$\sn1 {(-1)^{n+1}\over n^2}={\pi^2\over 12}$.
b) $\sn1{2\cdot(-1)^{n+1}\over n}\cdot\sin nx$;
$\sn0{(-1)^n\over 2n+1}=\frac\pi4$.
c) $\sn1{2n\cdot (-1)^{n+1}\over \pi(n^2-a^2)}\cdot \sin\pi
a\cdot \sin nx$;
${\pi\over \cos \pi a}=\sn0 (-1)^n\cdot\left({1\over
n+\frac12-a}+{1\over n+\frac12+a}\right)$.
\hfill\linebreak
d) ${4\pi^2\over 3}+\sn1\left(\frac4{n^2}\cos nx -\frac{4\pi}n\sin
nx\right)$;
$\sn1{1\over n^2}={\pi^2\over 6}$.
\pr a) $\frac\pi2-\sn0{4\over \pi (2n+1)^2}\cos (2n+1)x$;
b) $\sn1\left(2\cdot(-1)^n\cdot\frac\pi n+{4\over \pi
n^3}(1-(-1)^n)\right)\sin nx$.
\pr a) $\sk0{4\cdot (-1)^k\over\pi\cdot(2k+1)^2}\sin(2k+1)x$, sou\v{c}et je
$f^*(x)=\cases -\pi-x & x\in[-\pi,\frac\pi2] \\ x &
x\in[-\frac\pi2,\frac\pi2] \\ \pi-x & x\in[\frac\pi2,\pi]\endcases$, a
$2\pi$-periodicky.
b) $\sk0\left({2\cdot (-1)^k\over
2k+1}-{4\over\pi\cdot(2k+1)^2}\right)\cdot\cos (2k+1)x$, sou\v{c}et je
$f^*(x)=\cases -\pi-x & x\in[-\pi,\frac\pi2) \\ |x| &
x\in(-\frac\pi2,\frac\pi2) \\ x+\pi & x\in(\frac\pi2,\pi]
\\ 0 & x=\pm\frac\pi2\endcases$, a
$2\pi$-periodicky.
c) $\frac\pi4+\sk1{(-1)^k-1\over\pi\cdot k^2}\cos 2kx$, sou\v{c}et je
$f^*(x)=\cases \pi+x & x\in[-\pi,\frac\pi2] \\ |x| &
x\in[-\frac\pi2,\frac\pi2] \\ \pi-x & x\in[\frac\pi2,\pi]\endcases$, a
$2\pi$-periodicky.
d)$\sk1{2\cdot (-1)^k\over k}\sin 2kx$, sou\v{c}et je
$f^*(x)=\cases \pi+x & x\in[-\pi,\frac\pi2) \\ x &
x\in(-\frac\pi2,\frac\pi2) \\ x-\pi & x\in(\frac\pi2,\pi]
\\ 0 & x=\pm\frac\pi2\endcases$, a
$2\pi$-periodicky.
\par

\varoddel

{%
\centerline{\smc VII. Otev\v{r}en\'e a uzav\v{r}en\'e mno\v{z}iny, vnit\v{r}ek,
uz\'av\v{e}r \dots}

\pr Rozhodn\v{e}te, zda n\'asleduj\'{\i}c\'{\i} mno\v{z}iny jsou otev\v{r}en\'e ev.
uzav\v{r}en\'e, zjist\v{e}te jejich vnit\v{r}ek, uz\'av\v{e}r, hranici.

(a) $\Bbb Q$,\qquad (b) $\Bbb N$,\qquad (c) $\{\frac1n\mid
n\in\Bbb N\}$,\qquad (d) $(-\infty,0)\cup\{x\in\Bbb Q\mid x>0\}$

V\v{s}echny tyto mno\v{z}iny uva\v{z}ujte (i) v $\Bbb R$ s obvyklou metrikou
(ii) v rovin\v{e} ($\Bbb R^2$) s euklidovskou metrikou.

\pr Rozhodn\v{e}te, zda n\'asleduj\'{\i}c\'{\i} mno\v{z}iny jsou otev\v{r}en\'e ev.
uzav\v{r}en\'e, zjist\v{e}te jejich vnit\v{r}ek, uz\'av\v{e}r, hranici.

$\matrix\format\l&\quad\l\\
\text{a) $A=\{(x,y)\in\Bbb R^2\mid x>0,\,y\le 0\}$;} &
\text{b) $B=\{(x,y)\in\Bbb R^2\mid x^2+y^2+2xy=5\}$;} \\
\text{c) $C=\{(x,y,z)\in\Bbb R^3\mid x\ge0,y>0,x+y=2,z\le0\}$;} &
\text{d) $D=\{f\in\Cal C[0,1]\mid f(\frac12)=2\}$;} \\
\text{e) $E=\{f\in\Cal C[0,1]\mid f(\frac12)\in(0,2)\}$;} &
\text{f) $F=\{f\in\Cal C[0,1]\mid \int\limits_0^1f(x)\roman
dx=0\}$} \\
\text{g) $G=\{(x,y)\in\Bbb R^2\mid |x+y|>x+y\}$} &
\text{g) $H=\{(x,y)\in\Bbb R^2\mid |x-y|=x-y\}$}
\endmatrix$

\pr Definujme $\Cal N:\Cal C[0,1]\to\Bbb R$ p\v{r}edpisem $\Cal
N(f)=\int\limits^1_0f(x)\roman dx$. Je $\Cal N$ spojit\'e
zobrazen\'{\i}?

\pr Plat\'{\i} rovnosti $\overline{B(a,\delta)}=\{x\mid\rho(a,x)\le
\delta\}$ a $\vn\{x\mid\rho(a,x)\le\delta\}=B(a,\delta)$ pro
ka\v{z}d\'e $\delta>0$ a) v ka\v{z}d\'em metrick\'em prostoru  b) v normovan\'em
line\'arn\'{\i}m prostoru ?
}

\newpage

{\def\partial{\operatorname{bd}}
@ \pr (i) V $\Bbb R$: (a) Nen\'{\i} otev\v{r}en\'a ani uzav\v{r}en\'a, $\vn
\Bbb Q=\emptyset$, $\partial\Bbb Q=\overline{\Bbb Q}=\Bbb R$.
(b) Uzav\v{r}en\'a, $\vn \Bbb N=\emptyset$,
$\overline{\Bbb N}=\partial\Bbb N=\Bbb N$.
(c) Nen\'{\i} otev\v{r}en\'a ani uzav\v{r}en\'a, vnit\v{r}ek je $\emptyset$, uz\'av\v{e}r i
hranice je $\{\frac1n\mid n\in\Bbb N\}\cup\{0\}$.
(d)  Nen\'{\i} otev\v{r}en\'a ani uzav\v{r}en\'a, vnit\v{r}ek je $(-\infty,0)$, uz\'av\v{e}r
$\Bbb R$, hranice $[0,\infty)$.
(ii) V rovin\v{e} (jako\v{z}to podmno\v{z}iny osy $x$, kterou zna\v{c}\'{\i}me $\Bbb
R$):
(a) Nen\'{\i} otev\v{r}en\'a ani uzav\v{r}en\'a, $\vn
\Bbb Q=\emptyset$,
$\partial\Bbb Q=\overline{\Bbb Q}=\Bbb R$.
(b) Uzav\v{r}en\'a, $\vn \Bbb N=\emptyset$,
 $\overline{\Bbb N}=\partial\Bbb N=\Bbb N$.
(c) Nen\'{\i} otev\v{r}en\'a ani uzav\v{r}en\'a, vnit\v{r}ek je $\emptyset$, uz\'av\v{e}r i
hranice je $\{\frac1n\mid n\in\Bbb N\}\cup\{0\}$.
(d)  Nen\'{\i} otev\v{r}en\'a ani uzav\v{r}en\'a, vnit\v{r}ek je $\emptyset$, uz\'av\v{e}r
$\Bbb R$, hranice $\Bbb R$.
\pr
(a) Nen\'{\i} otev\v{r}en\'a ani uzav\v{r}en\'a, $\vn A=\{(x,y)\mid x>0,y<0\}$,
$\overline
A=\{(x,y)\mid x\ge0,y\ge0\}$, $\partial A=\{(x,y)\mid (x\ge
0,y=0) \text{ nebo } (x=0,y\le0)\}$.
(b) Uzav\v{r}en\'a, $\vn B=\emptyset$, $\overline B=\partial B= B$.
(c) Nen\'{\i} otev\v{r}en\'a ani uzav\v{r}en\'a, $\vn C=\emptyset$, $\overline
C=\partial C=\{(x,y,z)\mid x\ge0,y\ge0,x+y=2,z\le0\}$.
(d) Uzav\v{r}en\'a, $\vn D=\emptyset$, $\overline D=\partial D=D$.
(e) Otev\v{r}en\'a, $\vn E=E$, $\overline E=\{f\mid
f(\frac12)\in[0,2]\}$, $\partial E=\{f\mid
f(\frac12)\in\{0,2\}\}$.
(f) Uzav\v{r}en\'a, $\vn F=\emptyset$, $\overline F=\partial F=F$.
(g) Otev\v{r}en\'a, $\vn G=G$, $\overline G=\{(x,y)\mid x+y\le 0\}$,
$\partial G=\{(x,y)\mid x+y= 0\}$.
(h) Uzav\v{r}en\'a, $\vn H=\{(x,y)\mid x>y\}$, $\partial H=\{(x,y)\mid
x=y\}$.
\pr Ano, vyu\v{z}ijte v\v{e}tu o z\'am\v{e}n\v{e} limity a integr\'alu.
%linearitu a monotonii Newtonova integr\'alu.
\pr V metrick\'em prostoru platit nemus\'{\i}
(nap\v{r}. diskr\'etn\'{\i} prostor a $\delta=1$), ale v NLP plat\'{\i}.
\par
}

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