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\title{Cvi\v{c}en\'i  Matematick\'a anal\'yza 2, MFF}   % tyto dvě položky jsou zde v podstatě formálně, ve skutečnosti nejsou nikde 
\author{Marek C\'uth}

\begin{document}
 \centerline{\textbf{V\'YSLEDKY}}\ \\	
 \centerline{\MakeUppercase{i.  taylor\r{u}v polynom}}\ \\
 \textbf{1.}\resetpr \pr $x+\frac{1}{3}x^3$
	\pr $1-\frac{1}{2}x^2+\frac{5}{24}x^4$
	\pr $x-\frac{1}{3}x^3+\frac{1}{10}x^5$
	\pr $\frac{x^2}{2}$
	\pr $1+2x+2x^2-2x^4$
	\pr $1+2x+x^2-\frac{2}{3}x^3-\frac{5}{6}x^4-\frac{1}{15}x^5$
	\pr $1-\frac{1}{2}x+\frac{1}{12}x^2-\frac{1}{720}x^4$
	\pr $-\frac{x^2}{2}-\frac{1}{12}x^4-\frac{1}{45}x^6$\\[3pt]
 %\textbf{3.} a) $\frac{1}{3840}$, b) $2^{-4}$\\
 %\textbf{4.} 2,236076354981\\
 \textbf{2.}\resetpr \pr $-\tfrac{1}{12}$ \pr $\tfrac{1}{3}$ \pr $\tfrac{1}{3}$ \pr -$\tfrac{1}{4}$ \pr $\log^2 a$ \pr 0 \pr $\tfrac{1}{2}$
	\pr $\tfrac{1}{3}$ \pr $\tfrac{1}{6}$ \pr $\tfrac{1}{2}$ \pr -6\\[3pt]
\textbf{3.}\resetpr \pr $\frac{7}{12}$ \pr $-\tfrac{3}{2}$\\[3pt]
 \textbf{4.}\resetpr \pr AK (absolutn\v{e} konverguje) \pr D (diverguje) \pr D \pr AK \pr D \pr AK \pr AK\\[20pt]
 \centerline{\MakeUppercase{ii. mocninn\'e \v{r}ady}}\ \\
\textbf{1.}\resetpr \pr R=1; pokud $p > 1$ pak AK pro $|x|\leq 1$ a D jinak; pokud $p\in(0,1]$ pak AK pro $|x| < 1$, K pro $x = -1$ a D pro $x=1$; pokud $p\leq 0$, pak AK pro $|x| < 1$ a D jinak\\
\pr R=$1/3$, AK pro $|x+1| < R$, K pro $x = -4/3$, D pro $x = -2/3$\\
\pr R=$\frac{1}{e}$, AK pro $|x| < R$, jinak D
\pr R=1, pokud $a > 1$ pak AK pro $|x| \leq R$ a jinak D, pokud $a\leq 1$ pak AK pro $|x| < R$ a jinak D
\pr R = 1/4, AK pro $|x| < R$, jinak D
\pr R=$\max\{a,b\}$, AK pro $|x|<R$, jinak D
\pr R=$\frac{1}{\max\{a,b\}}$, AK pro $|x|<R$, D pro $|x| > R$; pokud $b > a$ pak AK pro $|x| = R$; pokud $a \geq b$ pak K pro $x = -1/a$ a D pro $x = 1/a$
\pr R=$1/3$, AK pro $|x| < R$, jinak D
\pr R=$1$, AK pro $|x| \leq R$, jinak D\\[5pt]
\textbf{2.}\resetpr \pr $f(x) = \sumo x^n$, $g(x) = \sumo (n+1)x^n$, $h(x) = \sumo nx^n$, polom\v{e}r konvergence $1$. \pr $f(x) = \sumo \frac{1}{2^{n+1}}(x+1)^n$, $g(x) = \sumo \frac{n+1}{2^{n+2}}(x+1)^n$, $h(x) = -\frac{1}{4} + \sum_{n=1}^\infty \left(\frac{n}{2^{n+1}} - \frac{n+1}{2^{n+2}}\right)(x+1)^n$, polom\v{e}r konvergence $2$. \pr $f(x) = \sumo \frac{(-1)^{n+1}}{6^{n+1}}(x-7)^n$, $g(x) = \sumo \frac{(n+1)(-1)^{n}}{6^{n+2}}(x-7)^n$, $h(x) = \frac{7}{36} + \sum_{n=1}^\infty \left(\frac{7(n+1)(-1)^{n}}{6^{n+2}} + \frac{n(-1)^{n+1}}{6^{n+1}}\right)(x-7)^n$, polom\v{e}r konvergence $6$.\\[5pt]
\textbf{3.}\resetpr Polom\v{e}r konvergence je v\v{z}dy $+\infty$. \pr $f(x) = \sumo \frac{x^n}{n!}$, $g(x) = \sumo \frac{x^{n+1}}{n!}$, $h(x) = \sumo \frac{x^{n+2}}{n!}$
	\pr $f(x) = \sumo \frac{e}{n!}(x-1)^n$, $g(x) = (x-1)f(x) + f(x) = e + \suma e\left(\frac{1}{(n-1)!} + \frac{1}{n!}\right)(x-1)^n$, $h(x) = (x-1)^2f(x) + 2(x-1)f(x) + f(x) = e + 3e(x-1) + \sum_{n=2}^\infty e\left(\frac{1}{(n-2)!} + \frac{2}{(n-1)!} + \frac{1}{n!}\right)(x-1)^n$
	\pr $f(x) = \sumo \frac{(x+8)^n}{e^8n!}$, $g(x) = (x+8)f(x) - 8f(x) = -\frac{8}{e^8} + \suma \frac{1}{e^8}\left(\frac{1}{(n-1)!} - \frac{8}{n!}\right)(x+8)^n$, $h(x) = (x+8)g(x) - 8g(x) = \frac{64}{e^8} + \frac{48}{e^8}(x+8) + \sum_{n=2}^\infty \frac{1}{e^8}\left(\frac{1}{(n-2)!} - \frac{16}{(n-1)!} + \frac{64}{n!}\right)(x+8)^n$\\[5pt]
	\textbf{4.}\resetpr Polom\v{e}r konvergence je v\v{z}dy $+\infty$. \pr $f(x) = \sumo (-1)^n\frac{x^{2n+1}}{(2n+1)!}$, $g(x) = \sumo (-1)^n\frac{x^{2n+2}}{(2n+1)!}$, $h(x) = \sumo (-1)^n\frac{x^{2n+3}}{(2n+1)!}$\\
	\pr $\sin x = \sin (x-\tfrac{\pi}{6} + \tfrac{\pi}{6}) = \sin(x-\tfrac{\pi}{6})\cos(\tfrac{\pi}{6}) + \cos(x-\tfrac{\pi}{6})\sin(\tfrac{\pi}{6}) = \frac{\sqrt{3}}{2}\sumo (-1)^n\frac{(x-\tfrac{\pi}{6})^{2n+1}}{(2n+1)!} + \frac{1}{2}\sumo (-1)^n\frac{(x-\tfrac{\pi}{6})^{2n}}{(2n)!}$, $g(x) = (x-\tfrac{\pi}{6})f(x) + \tfrac{\pi}{6}f(x) = \ldots$, $h(x) = (x-\tfrac{\pi}{6})g(x) + \tfrac{\pi}{6}g(x) = \ldots$ (nebo take $h(x) = (x-\tfrac{\pi}{6})^2f(x) + \tfrac{\pi}{3}(x-\tfrac{\pi}{6})f(x) + \tfrac{\pi^2}{36}f(x)$)\\
	\pr $\sin x = \sin(x-2)\cos(2) + \cos(x-2)\sin(2) = \cos 2\sumo (-1)^n\frac{(x-2)^{2n+1}}{(2n+1)!} + \sin 2\sumo (-1)^n\frac{(x-2)^{2n}}{(2n)!}$, $g(x) = (x-2)f(x) + 2f(x) = \ldots$, $h(x) = (x-2)g(x) + 2g(x) = \ldots$ (nebo take $h(x) = (x-2)^2f(x) + 4(x-2)f(x) + 4f(x)$)\\[5pt]
	\textbf{5.}\resetpr
	\pr $\sum_{n=0}^\infty (-1)^n\frac{x^{2n}}{n!},\;x\in\er$
	\pr $\sum_{n=0}^\infty (-1)^n\frac{x^{2n+1}}{2n+1},\;x\in(-1,1)$\\
	\pr $\frac{1 - \cos(2x)}{2} = \sum_{n=1}^\infty \frac{(-1)^{n+1}2^{2n-1}}{(2n)!}x^{2n},\;x\in\er$
	\pr $x + \suma (-1)^{n+1}\frac{x^{n+1}}{n(n+1)}$
	%\pr $\frac{1}{3}\suma (1 - (-2)^n)x^n,\;|x| < 1/2$\\
	\pr $\arctg 2 + \suma \frac{(-1)^n 2^{2n-1}}{2n-1}x^{2n-1},\; |x| < 1/2$
	%\pr $\suma nz^{n-1},\;|z|<1$
	\pr $\sum_{n=0}^\infty \frac{x^{2n+1}}{2n+1},\;|x| < 1$\\[20pt]
	 \centerline{\MakeUppercase{iii. hled\'an\'i primitivn\'i funkce - \'uvod}}\ \\ 
V\'ysledky jsou uvedeny v\v{z}dy "a\v{z} na konstantu":\\
  \textbf{1.}
  	a) $\frac{x^{10}}{10} + \log{|x|} - 5e^x - \frac{1}{2x^2} - \sin{x}$ na $(-\infty,0)$ a $(0,\infty)$\quad
  	b) $\frac 23 e^{3x} + \frac{5(5-x)^{\frac 65}}{6},\; x\in\er$\\
  	c) $-\frac{1}{x} - \frac{3}{2x^2} - \frac{2}{x^3}$ na $(-\infty,0)$ a $(0,\infty)$\quad
		d) $-\frac{(1-x)^{11}}{11} + \frac{(1-x)^{12}}{12}$\\[10pt]
  \textbf{2.}
  	a) $\frac 12 |x|x,\; x\in\er$\\  	
  	b) $F(x) = \left\{
  		\begin{array}{ll} \sin{x} + 4k & x\in[-\frac \pi2 + 2k\pi, \frac \pi2 + 2k\pi], k\in\zet\\
  		
			-\sin{x}+4k+2 & x\in(\frac \pi2 + 2k\pi, 3\frac \pi2 + 2k\pi), k\in\zet\\ \end{array} \right.$\qquad
		c) $\frac 14 |x|x^3,\; x\in\er$\\
  	d) $F(x) = \left\{ 
  		\begin{array}{ll} -\frac 12\cos(2x-1) & x\geq \frac 12\\
			\frac 12\cos(2x-1) - 1 & x < \frac 12\\ \end{array} \right.$\\
		e) $F(x) = (-1)^k(-\cos x + \sin x) + k2\sqrt{2},\;x\in[-\frac \pi4 + k\pi,-\frac \pi4 + (k+1)\pi],\;k\in\zet$\\
		f) $F(x) = \left\{ 
  		\begin{array}{ll} \frac {x^3}{3} & x\in(-\infty,0)\\
			 \frac {x^2}{2}& x\in[0,1]\\
			 \frac {x^3}{3} + \frac 16& x\in(1,\infty) \end{array} \right.$\qquad\quad
		g) $F(x) = \left\{ 
  		\begin{array}{ll} e^x - 2 & x < 0\\
			-e^{-x} & x \geq 0\\ \end{array} \right.$\qquad
		h) $F(x) = \left\{ 
  		\begin{array}{ll} -(x^2+x) & x < -\frac 12\\
			x^2+x+\frac 12 & x \geq -\frac 12\\ \end{array} \right.$\\
  \textbf{3.}
  	a) $-\log|\cos{x}|$ na ka\v{z}d\'em z interval\r{u} $(-\frac \pi2 + k\pi, \frac \pi2 + k\pi), k\in\zet$\\
  	b) $\log|\sin{x}|$ na ka\v{z}d\'em z interval\r{u} $(k\pi, (k+1)\pi), k\in\zet$\\
  	c) $\frac 13 \tg{x^3}$ na ka\v{z}d\'em z interval\r{u} $(\sqrt[3]{- \frac \pi2 + k\pi}, \sqrt[3]{\frac \pi2 + k\pi}), k\in\zet$\quad
  	d) $\frac 12\arctg{x^2},\; x\in\er$\\
  	e) $\log|\log{x}|$ na $(0,1)$ a $(1,\infty)$\quad
  	f) $\sqrt{x^2 + 5},\; x\in\er$\quad  	
  	g) $\log|\log(\log{x})|$ na $(1,e)$ a $(e,\infty)$\\[10pt]
	\textbf{4.}
		a) $\frac{4(x^2 + 7)}{7\sqrt[4]{x}},\; D_f = (0,\infty)$\quad
		b) $2x - \frac{12}{5}\sqrt[6]{72x^5} + \frac 32\sqrt[3]{9x^2},\; D_f = \er\setminus\{0\}$\quad		
		c) $\frac{4^x}{\log{4}} + 2\frac{6^x}{\log{6}} + \frac{9^x}{\log{9}},\; D_f = \er$\\
		d) $x - \arctg{x},\; D_f = \er$\quad
		e) $ -\frac{2}{5}\sqrt{2 - 5x},\; D_f = (-\infty,\frac 25)$\quad
		f) $\frac 14\arctg(x^4),\; D_f = \er$\\
		g) $\cos(\frac 1x),\; D_f = \er\setminus\{0\}$\quad
		h) $\frac x2 + \frac{\sin(2x)}{4},\; D_f = \er$\quad
		i) $\arctg(x^2+1),\; D_f = \er$\quad
		j) $\frac 12\arctg(\sin^2x),\; D_f = \er$\\
		k) $F(x)= \left\{ 
			\begin{array}{ll}	2\log(1+x^2)+\pi(x_2-x_1)+ 4\log\frac{x_1}{x_2} & x\in(-\infty,-x_2] \\
											-\pi(x+x_1)-2\log(1+x^2)+4\log(x^2_1+1) & x\in[-x_2,-x_1] \\
											2\log(1+x^2) & x\in[-x_1,x_1] \\
											\pi(x-x_1)-2\log(1+x^2)+4\log(x^2_1+1) & x\in[x_1,x_2] \\
											2\log(1+x^2)+\pi(x_2-x_1)+ 4\log\frac{x_1}{x_2} &	x\in[x_2,\infty)\\ \end{array} \right.$\qquad a \quad $\begin{array}{l} x_1=\frac{4-\sqrt{16-\pi^2}}{\pi}\\ x_2=\frac{4+\sqrt{16-\pi^2}}{\pi}\end{array}$, na $\er$\\[10pt]
	\textbf{5.}
  	a) $-x^3\cos{x} + 3x^2\sin{x} + 6x\cos{x} - 6\sin{x},\; x\in\er$\quad
  	b) $\frac 12(e^x\sin{x} + e^x\cos{x}),\; x\in\er$\quad
		c) $x\log{x} - x,\; x\in(0,\infty)$\quad
  	d) $I_n:=\int x^n e^x \dx = x^n e^x - nI_{n-1};\,I_{1} := xe^x - e^x,\; x\in\er$\\		
  	e) $\frac 14(2x^2\log{x} - x^2),\; x\in(0,\infty)$\quad
  	f) $\frac 12e^x(x\sin{x} + x\cos{x} - \sin{x}),\; x\in\er$\\[10pt]
  \textbf{6.}
  	a) $\frac {2}{\sqrt{\cos{x}}},\; D_f = \bigcup_{k\in\zet}(-\frac \pi2 + 2k\pi,\frac \pi2 + 2k\pi)$\quad
		b) $\frac 18\sqrt[3]{8x^3 + 27},\; D_f = \er\setminus\{-\frac 32\}$\quad
		c) $\frac 12\arctg^2{x},\; D_f = \er$\\
		d) $-\frac{1}{2\log{\frac 23}}\log|1 - (\frac 23)^{2x}|,\; D_f = \er\setminus\{0\}$\quad
		e) $x\arctg{x} - \frac 12\log(1 + x^2),\; D_f = \er$\\
		f) $-\frac{2x^2 - 1}{4}\cos(2x) + \frac x2\sin(2x),\; D_f = \er$\quad
		g) $\frac 23 x^{3/2}\left(\log^2x - \frac 43\log{x} + \frac 89\right),\; D_f = (0,\infty)$\\
		h) $-\frac{e^{-2x}}{2}(x^2 + x + \frac 12),\; D_f = \er$\quad
		i) $-\frac 1x(\log^2x + 2\log{x} + 2),\; D_f = (0,\infty)$\quad
		j) $\frac 13(x^3 - 1)e^{x^3},\; D_f = \er$\\
		k) $2(\sqrt{x}-1)e^{\sqrt{x}},\; D_f = (0,\infty)$\quad
		l) $2(6-x)\sqrt{x}\cos{\sqrt{x}} - 6(2-x)\sin{\sqrt{x}},\; D_f = (0,\infty)$\\[10pt]
  %\textbf{7.}
  %	a) $2\arcsin \frac x2 + \sin(2\arcsin \frac x2) = 2\arcsin \frac x2 + \frac x2\sqrt{4-x^2},\;D_f = (-2,2)$\\
  %	b) $\tg(\arcsin x) = \frac{x}{\sqrt{1-x^2}},\;D_f = (-1,1)$\quad
  %	c) $\frac{1}{a^2}\sin(\arctg \frac xa) = \frac{x}{a^2\sqrt{a^2+x^2}},\;D_f = \er$\\
  %	d) $a(\arcsin \frac xa - \cos(\arcsin \frac xa)) = a\arcsin \frac xa -\sqrt{a^2 - x^2},\;x\in(-a,a)$\\
  %	e) $2a^2(\frac{3\arcsin\sqrt{\frac{x}{2a}}}{2} - \sin(2\arcsin\sqrt{\frac{x}{2a}}) + \frac{1}{8}\sin(4\arcsin\sqrt{\frac{x}{2a}})),\;D_f = (0,2a)$\\[20pt]
	\centerline{\MakeUppercase{iv. hled\'an\'i primitivn\'i funkce - pokra\v{c}ov\'an\'i}}\ \\
V\'ysledky jsou uvedeny v\v{z}dy "a\v{z} na konstantu":\\
  \textbf{1.}\resetpr 
		\pr $2\arcsin \frac x2 + \sin(2\arcsin \frac x2) = 2\arcsin \frac x2 + \frac x2\sqrt{4-x^2},\;D_f = (-2,2)$
		\pr $\frac{1}{a^2}\sin(\arctg \frac xa) = \frac{x}{a^2\sqrt{a^2+x^2}},\;D_f = \er$
		\pr $\frac{x}{2}\sqrt{x^2+a^2} + \frac{a^2}{2}\log(x+\sqrt{x^2+a^2}),\;D_f = \er$
		\pr $5\frac{x^2}{2} - 7x + 8\log{|x + 1| + 2\frac{1}{x+1}},\; D_f = \er\setminus\{-1\}$
		\pr $\frac{1}{6}\left(\frac {12}{5}\log{|x + \frac 32|} + \frac {18}{5}\log{|x + \frac 23|} - 6\log{|x + 1|}\right),\; D_f = \er\setminus\{-\frac{3}{2}, -1, -\frac{2}{3}\}$
		\pr $\frac{x^3}{3} + 2x^2 + 10x + 20\log{|x - 1|} - 15\frac{1}{x - 1} - 3\frac{1}{(x - 1)^2},\; D_f = \er\setminus\{1\}$
		\pr $\frac{x}{2}\sqrt{x^2-a^2} - \sgn(x) \frac{a^2}{2}\log(|x|+\sqrt{x^2-a^2}),\;D_f = (-\infty,a]\cup [a,\infty)$
		\pr $\frac{x}{2}\sqrt{x^2-2} + \sgn(x) \log(|x|+\sqrt{x^2-2}),\;D_f = (-\infty,-\sqrt{2}]\cup [\sqrt{2},\infty)$\\[5pt]
		%\pr $2a^2(\frac{3\arcsin\sqrt{\frac{x}{2a}}}{2} - \sin(2\arcsin\sqrt{\frac{x}{2a}}) + \frac{1}{8}\sin(4\arcsin\sqrt{\frac{x}{2a}})),\;D_f = (0,2a)$
	\textbf{2.}\resetpr
		\pr $\frac{5x+2}{3(x^2 + x + 1)} + \frac{2}{9}\log{|x-1|} - \frac{1}{9}\log{(x^2 + x + 1)} + \frac{8}{3\sqrt{3}}\arctg{(\frac{2x+1}{\sqrt{3}})},\; D_f = \er\setminus\{1\}$
		\pr $\frac{\sqrt{2}}{8}\log(x^2+\sqrt{2}x+1) + \frac{1}{2\sqrt{2}}\arctg(\sqrt{2}x+1) - \frac{\sqrt{2}}{8}\log(x^2-\sqrt{2}x+1) +  \frac{1}{2\sqrt{2}}\arctg(\sqrt{2}x-1),\;D_f = \er$
		\pr $-\frac x2 + \frac 13\log{|e^x-1|} + \frac 16\log(e^x+2),\;D_f = \er\setminus\{0\}$
		\pr $x-3\log(e^{x/6}+1)-3\log(\sqrt{e^{x/3}+1}) - 3\arctg(e^{x/6}),\;D_f = \er$
		\pr $\frac{1}{5\sqrt{2}}\arctg(\log(x)/\sqrt{2}) + \frac{9}{10\sqrt{3}}\log\left|\frac{\log(x) - \sqrt{3}}{\log(x) + \sqrt{3}}\right|,\;D_f = (0,\infty)\setminus\{e^{\sqrt{3}},e^{-\sqrt{3}}\}$\\[5pt]
	\textbf{3.}\resetpr
		\pr $\frac{1}{4}\log{\left|\frac{1-\cos{x}}{1+\cos{x}}\right|} - \frac{1}{2(\cos{x} + 1)},\; D_f = \er\setminus\bigcup_{k\in\zet}\{k\pi,\frac \pi2 + k\pi\}$ \texttt{[d\'a se \v{r}e\v{s}it substituc\'i $t = \cos{x}$]}\\
		\pr $\tg{x} + \log{\left|\frac{\tg{x}}{(\tg{x} + 1)^2}\right|},\; D_f = \er\setminus\bigcup_{k\in\zet}\{k\pi,\frac \pi2 + k\pi,3\frac \pi4 + k\pi\}$ \texttt{[d\'a se \v{r}e\v{s}it substituc\'i $t = \tg{x}$]}\\
		\pr $\frac 32\log(\cos^2x + 1) - \log(\cos^2{x}),\; D_f = \er\setminus\bigcup_{k\in\zet}\{\frac \pi2 + k\pi\}$\\
		\pr $F(x) = \left\{ 
  		\begin{array}{ll} \arctg(\tg x) - \frac{1}{\sqrt{2}}\arctg(\sqrt{2}\tg x) + k\pi(1-1/\sqrt{2}) & x\in(-\frac \pi2 + k\pi,\frac \pi2 + k\pi),\quad k\in\zet\\
			\frac \pi2 - \frac{\pi}{2\sqrt{2}} + k\pi(1-1/\sqrt{2}) & x = \frac \pi2 + k\pi,\quad k\in\zet\\ \end{array} \right.$\\	
		\pr  $F(x) = \left\{ 
  		\begin{array}{ll} \frac 16\log\left(\frac{\tg^2x-\tg x+1}{(\tg x+1)^2}\right) + \frac{\sqrt{3}}{3}\arctg\left(\frac{2\tg x - 1}{\sqrt{3}}\right) & x\in(-\frac \pi4+k\pi, \frac \pi2+k\pi)\\
  		\frac{\sqrt{3}\pi}{6} & x = \frac \pi2+k\pi\\
			\frac 16\log\left(\frac{\tg^2x-\tg x+1}{(\tg x+1)^2}\right) + \frac{\sqrt{3}}{3}\arctg\left(\frac{2\tg x - 1}{\sqrt{3}}\right) + \frac{\sqrt{3}\pi}{3} & x\in(\frac \pi2+k\pi, 3\frac \pi4+k\pi)\\ \end{array} \right.\quad\text{ pro }k\in\zet$\\
		\pr $F(x) = \left\{
  		\begin{array}{ll} \frac {1}{\sqrt{6}} \arctg\left(\sqrt{\frac 23}\tg{(\frac x2)}\right)  + k\frac{\pi\sqrt{6}}{6} & \text{ pro }x\in(-\pi + 2k\pi, \pi + 2k\pi), k\in\zet\\
			\frac {\pi}{2\sqrt{6}} + k\frac{\pi\sqrt{6}}{6} & \text{ pro }x = \pi + 2k\pi, k\in\zet\\ \end{array} \right.$\\
		\pr $F(x) = \left\{
  		\begin{array}{ll} \frac{\sqrt{5}}{5}\arctg\left(\frac{3\tg \frac{x}{2} + 1}{\sqrt{5}}\right) + k\pi\frac{\sqrt{5}}{5} & \text{ pro }x\in(-\pi + 2k\pi, \pi + 2k\pi), k\in\zet\\
			\frac{\pi}{2}\frac{\sqrt{5}}{5} + k\pi\frac{\sqrt{5}}{5} & \text{ pro }x = \pi + 2k\pi, k\in\zet\\ \end{array} \right.$\\[5pt]
			\textbf{4.}\resetpr
		\pr $2\sqrt{x} - 2\log(1 + \sqrt{x}),\; D_f = (0,\infty)$
		\pr $\frac{2}{x-\sqrt{x^2+2x+2}} - \log(\sqrt{x^2+2x+2} - x - 1),\;D_f = \er$\\
		\pr $2\sgn(x-1)\sqrt{\frac{x-2}{x+1}},\;D_f = (-\infty,1) \cup (2,\infty)$\\[5pt]
			\textbf{5.}\resetpr \pr $\frac{-2(t^2+3t+1)^2}{(2t+3)^3}$ \pr $x_1 = \frac{3+\sqrt{5}}{2}$, $x_2 = \frac{3-\sqrt{5}}{2}$. Pak na intervalu $(-\infty,x_2)$ vede na integr\'al z $\frac{2t^2(x_1-x_2)^2}{(t^2-1)^3}$; na intervalu $(x_1,\infty)$ vede na integr\'al z $-\frac{2t^2(x_1-x_2)^2}{(t^2-1)^3}$.\\[5pt]
			\textbf{6.}\resetpr
			\pr $F(x) = \left\{
				\begin{array}{ll} \log\left(\sqrt{\frac{\tg^2\frac{x}{2}+1}{2\tg^2\frac{x}{2} - \tg\frac{x}{2} + 1}}\right) + \arctg(\tg\frac{x}{2}) - \frac{6}{\sqrt{28}}\arctg\left(\frac{8\tg\frac{x}{2}-2}{\sqrt{28}}\right) - & k\pi\left(\frac{6}{\sqrt{28}}-1\right), \\ & x\in(-\pi + 2k\pi, \pi + 2k\pi), k\in\zet\\
				-\frac{1}{2}\log(2) + (-k\pi-\pi/2)\left(\frac{6}{\sqrt{28}}-1\right), & x = \pi + 2k\pi, k\in\zet\\ \end{array} \right.$\\
		\pr $\frac{1}{2}\log(|2\sqrt{x^2+x} + 2x + 1|) - \frac{1}{2(2\sqrt{x^2+x} + 2x + 1)},\;D_f = (-\infty,-1)\cup (0,\infty)$\\[20pt]
		\centerline{\MakeUppercase{v. newton\r{u}v integr\'al}}\ \\
		\noindent \textbf{1.}\resetpr \pr $1$ \pr $\frac{1-\log 2}{2}$ \pr $4\pi$ \pr $2-\frac2e$ \pr $200\sqrt{2}$ \pr $2 -
\frac{\pi}{2}$ \pr $\frac{a^4}{16}\pi$ \pr $\frac{2\pi}{\sqrt{1-\varepsilon^2}}$ \pr $\frac{2}{ab}\arctg(\frac{a\tg(1)}{b})$ \pr
$\frac{\log 3}{2}-\frac{\sqrt{3}}{6}\pi$ \pr $2\sqrt{2}\pi$\\[5pt]
\textbf{2.}\resetpr\pr $\tfrac{\pi}{\sqrt{3}}$ \pr $\pi$ \pr $\frac{5}{2}\sqrt{2}\pi$\\ \pr $57(\pi - 2\sqrt{3}\frac \pi3) + \frac{\sqrt{3}\pi}{9} + \arctg(\tg(18\pi^2)) - \frac{2}{\sqrt{3}}\arctg\left(\frac{2\tg(18\pi^2) + 1}{\sqrt{3}}\right)$\\
\pr $\frac{\pi^2}{4}$ (substituce $x = y + \frac{\pi}{2}$, pot\'e rozd\v{e}lit na sou\v{c}et dvou integr\'al\r{u}, z nich\v{z} jeden je nulov\'y jako\v{z}to integr\'al z lich\'e funkce a druh\'y se spo\v{c}te standardn\v{e})
\pr $\frac{(e^{-\pi} - e^{-(2n+1)\pi})(e^{\pi/2} - \tfrac{1}{2} + \tfrac{1}{2}e^{\pi})}{1 - e^{-\pi}}$ (substituce $x = e^y$, pak rozd\v{e}lit na sou\v{c}et $2n$ integr\'al\r{u}, ka\v{z}d\'y z nich p\v{r}ev\'est substituc\'i [posunut\'i] na integr\'al p\v{r}es $(0,\pi)$, pou\v{z}\'i vlastnosti exponenci\'aly a dopo\v{c}\'itat)
\pr $-\frac{1}{2n} + \frac{1}{2n-2} + \ldots + \frac{(-1)^n}{2} + \frac{(-1)^{n+1}}{2}\log 2$ (substituc\'i $y = \tg x$ lze p\v{r}ev\'est na integr\'al z racion\'aln\'i funkce, pak aplikovat substituci $z = \frac{y-1}{y+1}$ a dopo\v{c}\'itat)
\pr $-1$ (rozd\v{e}l\'ime na integr\'al p\v{r}es intervaly $(\tfrac{\pi}{6},\tfrac{\pi}{4})$ a $(\tfrac{\pi}{4},\tfrac{5\pi}{6})$)\\[20pt]
		\centerline{\MakeUppercase{vi. mocninn\'e \v{r}ady - s\v{c}\'it\'an\'i \v{c}\'iseln\'ych \v{r}ad}}\ \\
		\textbf{1.}\resetpr \pr $x^5e^{x^4},\;x\in\er$ \pr $\frac{x}{(x-1)^2},\;x\in(-1,1)$ \pr $\log(\frac{1-x}{x}) - \log(1-x) + 1,\;x\in[-1,1]\setminus\{0\}$, sou\v{c}et je $0$ pro $x = 0$ \pr $\frac{x(1+x)}{(1-x)^3},\;x\in(-1,1)$ \pr $\frac{x(1-x)}{(1+x)^3},\;x\in(-1,1)$ \pr $\frac{x(3-x)}{(1-x)^3},\;x\in(-1,1)$
		%\pr $\cosh x,\;x\in\er \texttt{[hint: $\sum_{n=0}^\infty \frac{x^{2n}}{(2n)!} = \frac 12\sum_{n=0}^\infty\frac{1 + (-1)^n}{n!}\;x^n$]}$ \pr $\frac{1}{2}(\cos x + \cosh x),\;x\in\er$ \pr $\frac{x^4(5-3x^2)}{(1-x^2)^3},\;x\in(-1,1)$\\ 
		\pr $2x\arctg x - \log(1+x^2),\;x\in[-1,1]$ 
		%\pr $\frac{e^{x/3}}{9}(x^2 + 3x - 9) + 1,\;x\in\er$\\
		\\[5pt]
		\textbf{2.}\resetpr \pr $-\log 2$ \pr $2$ \pr $\frac \pi4 - 1$ \pr $8$ \pr $\frac{3}{128}$ \pr $2\arctg(1/2) - 1$ \pr $\frac \pi4 - \frac 12$ \pr $-2\log(\frac 23) - \frac 53$ \pr $-\frac{1}{4\sqrt{e}}$\\[20pt]
		%\textbf{3.}\resetpr \pr $\sum_{n=0}^\infty (-1)^n\frac{x^{2n}}{n!},\;x\in\er$ \pr $\sum_{n=0}^\infty (-1)^n\frac{x^{2n+1}}{2n+1},\;x\in(-1,1)$ \pr $\sum_{n=0}^\infty \frac{(-1)^{n+1}2^{2n-1}}{(2n)!}x^{2n},\;x\in\er$ \pr $\sum_{n=10}^\infty x^n$ \pr $x + \suma (-1)^{n+1}\frac{x^{n+1}}{n(n+1)}$\\[20pt]
\centerline{\MakeUppercase{vii. konvergence newtonova integr\'alu}}\ \\
 \textbf{1.}\resetpr \pr konverguje \pr diverguje \pr konverguje \pr konverguje \pr diverguje \pr konverguje \pr konverguje, pokud $p,q < 1$ \pr konverguje, pokud $q < 1$ a $p < \frac 12$ \pr konverguje \pr konverguje, pokud $p,q > 0$ \pr konverguje, pokud $p > -1$\\
 \textbf{2.}\resetpr \pr konverguje, pokud $m < 3$ \pr konverguje \pr konverguje, pokud $k < -1$ \pr konverguje, pokud $\alpha < -1 < \alpha + \beta$ \pr konverguje, pokud $\alpha\in(-1,1)$ \pr konverguje, pokud $\alpha + \gamma > -1$ a $\beta - \gamma > -1$ \pr konverguje, pokud $p > 1$ nebo $p = 1$ a $q > 1$\\
 \textbf{4.} oba integr\'aly jsou konverguj\'i\\
 \textbf{5.}\resetpr \pr konverguje (neabsolutn\v{e}) \pr konverguje (neabsolutn\v{e}) \pr konverguje, pokud $\alpha\in(1,3)$ \pr neabsolutn\v{e} konverguje \pr konverguje (neabsolutn\v{e})\\
 \textbf{6.}\resetpr \pr konverguje (neabsolutn\v{e}) \pr konverguje, pokud $\alpha\in(-2,-1)$ \pr konverguje pro $\alpha < 0$ a (neabsolutn\v{e}) pro $\alpha > 1$\\[20pt]
\centerline{\MakeUppercase{viii. aplikace ur\v{c}it\'eho integr\'alu}}\ \\
\textbf{1.}\resetpr \pr $2\arctg 1 - \frac{1}{3}$ \pr $4.5$ \pr $\frac{1}{3}$ \pr $\frac{16}{3}$ \pr $\pi ab$\\
\textbf{2.}\resetpr \pr $\frac{8}{27}(10^{2/3} - 1)$ \pr $-\frac{1}{2}\log(\sqrt{2} - 1) + \frac{\sqrt{2}}{2}$\\
\textbf{3.}\resetpr \pr $1 + \frac{1}{2}\log(3/2)$ \pr $\frac{e^2+1}{4}$ \pr $2\pi$ \pr $6a$\\
\textbf{4.} Objem $= 2a^2\pi^2b$, povrch $= 4\pi^2ab$\\
\textbf{5.}\resetpr \pr diverguje \pr konverguje\\[20pt]
\centerline{\MakeUppercase{viii. metrick\'e prostory}}\ \\
\noindent \textbf{1.}\resetpr \pr ano \pr ne\\
\textbf{2.}\resetpr \pr ano \pr ne \pr ano\\
\textbf{3.} \pr $\frac 14$ \pr $f_n$ ano, $g_n$ ne\\
\textbf{4.}\resetpr \pr $\frac 16$ \pr $a = \frac 1{\sqrt{2}}$\\
\textbf{5.} ob\v{e} tvrzen\'i plat\'i pokud je metrika generovan\'a normou, jinak ne (jako p\v{r}\'iklad lze vz\'it prostor s diskr\'etn\'i metrikou)\\
\textbf{6.}\resetpr \pr je uzav\v{r}en\'a, m\'a pr\'azdn\'y vnit\v{r}ek, $\partial \en = \ov{\en} = \en$
	\pr $\Int \qe=\emptyset,\,\partial \qe =\overline{\qe}=\er$, $\qe$ nen\'{\i} otev\v{r}en\'a ani uzav\v{r}en\'a.
\pr Mno\v{z}ina nen\'{\i} uzav\v{r}en\'a ani otev\v{r}en\'a, vnit\v{r}ek je pr\'azdn\'y,
hranice i uz\'av\v{e}r jsou $\{\frac1n\setsep n\in\en\}\cup\{0\}$.
\pr Mno\v{z}ina nen\'{\i} uzav\v{r}en\'a ani otev\v{r}en\'a. Vnit\v{r}ek je
$\{[x,y]\in\er^2\setsep x>0,\,y< 0\}$, uz\'av\v{e}r $\{[x,y]\in\er^2\setsep
x\ge0,\,y\le 0\}$,  hranice$\{[x,y]\in\er^2\setsep
x\ge0\ \&\ y\le 0\ \&\ (x=0\vee y=0)\}$.
\pr Otev\v{r}en\'a, uz\'av\v{e}r $\{[x,y]\setsep x+y\le0\}$, hranice
$\{[x,y]\setsep x+y=0\}$.
\pr Uzav\v{r}en\'a, vnit\v{r}ek $\{[x,y]\setsep x > y\}$, hranice $\{[x,y]\setsep
x = y\}$.
\pr Uzav\v{r}en\'a, pr\'azdn\'y vnit\v{r}ek.
\pr Ani uzav\v{r}en\'a ani otev\v{r}en\'a, vnit\v{r}ek pr\'azdn\'y, hranice i uz\'av\v{e}r
$\{[x,y,z]\in\er^3\setsep x\ge0,y\ge0,x+y=2,z\le0\}$.
\pr Uzav\v{r}en\'a, pr\'azdn\'y vnit\v{r}ek.
\pr Otev\v{r}en\'a, uz\'av\v{e}r $\{f\in\C[0,1]:\; f(\frac12)\in [0,2]\}$, hranice $\{f\in\C[0,1]:\; f(\frac12)\in\{0,2\}\}$
\pr Uzav\v{r}en\'a, pr\'azdn\'y vnit\v{r}ek.\\[10pt]
\newpage
\centerline{\MakeUppercase{ix. funkce v\'ice prom\v{e}nn\'ych - limity, spojitost, derivace}}\ \\
\textbf{1.} a)-v), b)-i), c)-vi), d)-ii), e)-iii), f)-iv)\\[10pt]
\textbf{2.}\resetpr \pr 0 \pr Limita neexistuje \pr Limita neexistuje \pr 0 \pr 0 \pr 2 \pr Limita neexistuje \pr Limita neexistuje \pr 1 \pr 1 \pr Limita neexistuje, proto\v{z}e funkce nen\'i definov\'ana na prstencov\'em okol\'i bodu $(0,3)$, limita p\v{r}es mno\v{z}inu $\{(x,y)\in\er^2:\; x\neq 0\}$ je 3. \pr Limita neexistuje, proto\v{z}e funkce nen\'i definov\'ana na prstencov\'em okol\'i bodu $(0,0)$, limita p\v{r}es mno\v{z}inu $\{(x,y)\in\er^2:\; x\neq -y\}$ tak\'e neexistuje. \pr 0 \pr Limita neexistuje, proto\v{z}e funkce nen\'i definov\'ana na prstencov\'em okol\'i bodu $(0,0)$, limita p\v{r}es mno\v{z}inu $\{(x,y)\in\er^2:\; xy\neq 0\}$ je $+\infty$. \pr 0\\[5pt]
\textbf{3.}\resetpr \pr ano, $f(x,x) = 3x^2$ \pr ne \pr ano, $f(0,y)=y$ \pr ne\\[5pt]
\textbf{4.}\resetpr
	\pr $\pdx=m x^{m-1} y^n$, $\pdy=n x^m y^{n-1}$ pro $(x,y)\in\er^2$.
	\pr $\pdx=y e^{xy}$, $\pdy=x e^{xy}$ pro $(x,y)\in\er^2$.
	\pr $\pdx=y+z$, $\pdy=x+y$, $\pdz=x+y$ pro $(x,y,z)\in\er^3$.
	\pr $\pdx(x,y)= |y|\sgn x$ pro $x\ne 0$. $\pdy(x,y)=|x|\sgn y$ pro $y\ne 0$. $\pdx(0,0)=\pdy(0,0)=0$. $\pdx(0,y)$ pro $y\ne 0$ a $\pdy(x,0)$ pro $x\ne0$ neexistuj\'{\i}.
	\pr $\pdx(x,y)=-\sgn(y+\cos x)\cdot\sin x$, $\pdy(x,y)=\sgn(y+\cos x)$, pokud $y\neq -\cos x$. $\pdy(x,-\cos x)$ neexistuje pro $x\in\er$. $\pdx(k\pi,(-1)^{k+1})=0$ pro $k\in\zet$. $\pdx(x,-\cos x)$ neexistuje pro $x\neq k\pi$.
	\pr $\pdx(x,y)=\cos x\sgn(\sin x-\sin y)$, $\pdy(x,y)=-\cos y\sgn(\sin x-\sin y)$, pokud $\sin x\ne\sin y$. $\pdx(\frac\pi2+k\pi,\frac\pi2+l\pi)=\pdy(\frac\pi2+k\pi,\frac\pi2+l\pi)=0$. V ostatn\'{\i}ch bodech parci\'aln\'{\i} derivace neexistuj\'{\i}.
	\pr $\pdx(x,y)=-\cos x\sgn(\cos y-\sin x)$, $\pdy(x,y)=-\sin y\sgn(\cos y-\sin x)$, pokud $\sin x\neq\cos x$. 
$\pdx(\frac\pi2+k\pi,(k+2l)\pi)=\pdy(\frac\pi2+k\pi+2l\pi,k\pi)=0$. V ostatn\'{\i}ch bodech parci\'aln\'{\i} derivace neexistuj\'{\i}.
\pr Pokud $x,y>0$ nebo $x,y<0$, pak $\pdx= \frac zy\cdot \left(\frac xy\right)^{z-1}$; $\pdy=-\frac{zx}{y^2}\cdot \left(\frac xy\right)^{z-1}$; $\pdz=\left(\frac xy\right)^z\cdot\log\frac xy$.
\pr Pokud $x>0$ a $z\ne 0$, pak $\pdx=\frac yz \cdot x^{\frac yz-1}$; $\pdy=x^{\frac yz}\cdot\log x\cdot \frac1z$;
$\pdz=-x^{\frac yz}\cdot\log x\cdot \frac y{z^2}$.
\pr Pokud $x>0$, pak $\pdx=\cos(x^y)\cdot x^{y-1}\cdot y$; $\pdy=\cos(x^y)\cdot x^y\cdot \log x$.
\pr $\pdx=e^{-\frac{\pi}{x^2+3xy+3y^2}} \cdot \frac{\pi(2x+3y)}{(x^2+3xy+3y^2)^2}$, $\pdy=e^{-\frac{\pi}{x^2+3xy+3y^2}} \cdot \frac{\pi(3x+6y)}{(x^2+3xy+3y^2)^2}$ pro $(x,y)\ne(0,0)$; v bod\v{e} $(0,0)$ jsou ob\v{e} parci\'aln\'{\i} derivace nulov\'e.
\pr Pokud $x>-y^2$, pak $\pdx=\frac{1}{2\sqrt{x+y^2}}$; $\pdy=\frac{y}{\sqrt{x+y^2}}$. Jinak parci\'aln\'i derivace nemaj\'i smysl.\\[10pt]
\textbf{5.} viz. v\'ysledky zkou\v{s}kov\'ych p\'isemek zde:\\ \url{http://www.karlin.mff.cuni.cz/~kalenda/edu.php?edutype=archpis}\\[10pt]
\textbf{6.}\resetpr
  \pr ne \pr ne \pr ano \pr ne \pr ano \pr ano \\[10pt]
\textbf{7.}\resetpr
  \pr $\nabla f (x,y) = (\tfrac{5x^4(x^4 + y^4) - (x^5 - y^5)4x^3}{(x^4 + y^4)^2}, \tfrac{-5y^4(x^4 + y^4) - (x^5 - y^5)4y^3}{(x^4 + y^4)^2})$ pro $(x,y)\neq (0,0)$, v bod\v{e} $(0,0)$ tot\'aln\'i diferenci\'al neexistuje
  \pr $\nabla f(x,y) = (3x^2,0)$ pro $x > y$, $\nabla f(x,y) = (0,3y^2)$ pro $x < y$, $\nabla f(0,0) = (0,0)$, v bodech $(a,a)$ kde $a\neq 0$ tot\'aln\'i diferenci\'al neexistuje.\\[10pt]
\textbf{8.} a) $F'(\pi,1,0) = \left( \begin{array}{ccc}
                            0 & 0 & \pi\sin1\\
                            2\pi\sin\frac{1}{1+\pi^2} - \frac{2\pi^3}{(1+\pi^2)^2}\cos\frac{1}{1+\pi^2} & - \frac{2\pi^2}{(1+\pi^2)^2}\cos\frac{1}{1+\pi^2} & 0
                           \end{array}\right)
	      $\\[5pt]
	     b) $\frac{\partial F_2}{\partial x}(0,0,0) = 0$.\\[5pt]
\textbf{9.} a) $(G\circ F)'(0,0) = \left( \begin{array}{ccc}
                            3 & 4 & 6\\
                            0 & -2 & -3\\
                            1 & 0 & 0\\
                           \end{array}\right)
	      $\\[5pt]
	    b) $\partial_{(2,0,1)}F_1(0,0,0) = 5$.\\[5pt]
\textbf{10.} a) $F'(1,-1) = \left( \begin{array}{cc}
                            1/2 & 1 \\
                            1 & 1\\
                            2 & -\frac{1}{2}\\
                            -e & e\\
                           \end{array}\right)
	      $\\[5pt]
	      b) $\frac{\partial F_3}{\partial x}(0,0) = 0$.\\[5pt]
\textbf{11.}\resetpr
\pr $\frac{\partial F}{\partial r} = 8r + 4s$, $\frac{\partial F}{\partial s} = 4r + 2s$, $\frac{\partial F}{\partial t} = 2t$
\pr $\frac{\partial F}{\partial s} = 4s^3 + 8st + 3t$, $\frac{\partial F}{\partial t} = 4s^2 + 3s + 8t + 2$\\[10pt]
\end{document}
