\input /home/pasky/school/fastex/lib.tex

\newcount\oldlistlevel

%\chapter{Zaveden� element�rn� funkce}{

Chceme definovat transcendentn� funkce, zejm�na:
\list{
	\olistItem{$\exp$, $\log$, $a^x$}
	\olistItem{goniometrick� funkce ($\sin$, $\cos$, $\tg$, $\cotg$)}
	\olistItem{cyklometrick� funkce ($\arcsin$, $\arccos$, $\arctg$, $\arccotg$)}
	\olistItem{hyperbolick� funkce ($\sinh$, $\cosh$, $\tgh$, $\cotgh$)}
}


\theorem{13}{zaveden� exponenci�ln� funkce}{

	$\onexists$ re�ln� funkce ``$\exp$'' spl�uj�c� axiomy:
	\list{
		\listItem{$\exp(x+y) = \exp x \cdot \exp y\qquad \forall x,y \in \real$}
		\listItem{$\exp x \ge 1 + x\qquad \forall x \in \real$}
	}

	\proof{
	\list{
	\namedListItem{A}{Jednozna�nost (p�edpokl�d�me, �e existuje):

		\list{
			\oldlistlevel=\listlevel
			\listlevel=2

			\listItem{%1
				$\exp(mx) = (\exp x)^m\quad \forall n \in \nat, x \in \real$: Trivi�ln� indukc� z (i).
			}

			\listItem{%2
				$\exp(0) = 1$:
				$$ \exp(0) = \exp(0+0) \becauseof{(i)}{=} \exp(0)^2
					\Rightarrow \exp(0) = \cases{1&\cr
					                             0&nikdy --- (ii)} $$
			}

			\listItem{%3
				\M{\exp(-x) = {1\over\exp x}}:

				$$ 1 = \exp(0) = \exp(x-x) = \exp(x) \cdot \exp(-x) $$
			}

			\listItem{%4
				$\exp x \ne 0\quad \forall x \in \real$: Ihned plyne z (3).
			}

			\listItem{%5
				\M{\lim_{x\to \infty} \exp x = +\infty}: Ihned plyne z (ii) --- doln� policajt $(1+x)$.
			}

			\listItem{%6
				\M{\lim_{x\to -\infty} \exp x = 0}: Kombinace (3) a (5).
			}

			\listItem{%7
				$\exp x > 1\quad \forall x > 0$: Ihned z (ii).
			}

			\listItem{%8
				$\exp \nearrow$ na $\real$:

				$$ x < y \Rightarrow 1 \becauseof{(7)}{<} \exp(y-x) = {\exp y\over \exp x}
				         \Rightarrow \exp y > \exp x $$
			}

			\listItem{%9
				\M{\lim_{x\to 0} {\exp x - 1\over x} = 1} (!)

				$$ {1\over\exp x} \becauseof{(3)}{=} \exp(-x) \becauseof{(ii)}{\ge} 1-x $$

				To znamen�:

				$$ 1+x \becauseof{(ii)}{\le} \exp x \le {1\over 1-x} $$

				$$ \Rightarrow x \le \exp x - 1 \le {1\over1-x}-1 = {x \over 1-x} $$
				$$ \Rightarrow \undernote{1}{\to1} \le \undernote{\exp x - 1\over x}{\to1}
					\le \undernote{1\over1-x}{\to1} $$
			}

			\listItem{%10
				$\exp$ je spojit� na $\real$:

				Pro $\forall a \in \real$ mus� platit:

				$$ \lim_{x\to a} (\exp x - \exp a)
					= \lim_{x\to a} \exp a \cdot \left({\exp x \over \exp a} - 1\right) = $$
				$$
					= \lim_{x\to a} \undernote{\exp a}{\to \exp a}
					                \undernote{\left(\exp(x-a)-1\over x-a\right)}{\becauseof{(9)}{\to} 1}
							\undernote{(x-a)}{\to 0}
					\becauseof{\ref{VOAL}}{=} 0 $$
			}

			\listItem{%11
				\M{\exp x = \lim_{x\to \infty} \left(1+{x\over n}\right)^n\quad \forall x \in \real} (!!)

				Takto p�i definici limitou tak� dok�eme jednozna�nost --- limity jsou jednozna�n�.

				$$ \exp\left(-{x\over n}\right) \becauseof{(ii)}{\ge} 1-{x\over n} $$

				To znamen�:

				$$ \forall x \in \real, n \in \nat:
					\left(1+{x \over n}\right)^n \le \exp x \le \left(1-{x\over n}\right)^{-n} $$

				$$ \forall x > 0:\ 
					\undernote{1}{\to1}
					  \le {\exp x \over \left(1+{x \over n}\right)^n}
					  \le \undernote{\left(1-{x^2\over n^2}\right)^{-n}}{\to1}
					  \becauseof{\ref{Bernoulli}}{\le} \undernote{\left(1-{x^2\over n}\right)^{-1}}{\to1} $$
				(pro dost velk� $n$)
			}

			\listlevel=\oldlistlevel
		}

	}


	\namedListItem{B}{Existence:

	Dokazujeme existenci \M{\lim_{n\to \infty} \left(1+{x\over n}\right)^n} pro $\forall x \in \real$.


	\medskip

	\scpart{1. krok}{
		\M{\{\left(1+{x\over n}\right)^n\}} je rostouc� pro $x>0$.

		Pou�ijeme \ref{AG-nerovnost} ve tvaru
		$$\root{n+1}\of{\strut a_1\ldots a_{n+1}} \le {a_1 + a_2 + \cdots + a_{n+1}\over n+1}$$
		pro $a_1 = a_2 = \cdots = a_n = 1+ {x \over n}$, $a_{n+1} = 1$:

		$$ \root{n+1}\of{\left(1+{x\over n}\right)^n} \becauseof{AG}{\le} {\left(1+{x\over n}\right)n + 1\over n + 1} = 1+ {x\over n+1} $$

		$$ \left(1+{x\over n}\right)^n \le \left(1+{x\over n+1}\right)^{n+1} $$
		Tedy je posloupnost \M{\{\left(1+{x\over n}\right)^n\}} rostouc�.

	}


	\medskip

	\scpart{2. krok}{
		Posloupnost \M{\{\left(1+{x\over n}\right)^n\}} je omezen�.

		V�me, �e je monot�nn�, tedy sta�� dok�zat,
		�e n�jak� podposloupnost je omezen�.

		\shortdef{Tvrzen�}{$\{a_n\}$ monot�nn� a $\{a_{n_k}\}$ omezen� $\Rightarrow$ $\{a_n\}$ omezen�. D�kaz: \DCV.}

		Dok�eme, �e
		$$\forall x \in \real^+\ \exists k \in \nat,\ \forall n \in \nat:
			\left(1+{x\over nk}\right)^{nk} \le \left(1-{x\over k}\right)^k$$

		Pot�ebuji $k$ tak velk�, aby \M{{x\over k + x} < 2}, tj. aby platil \ref{Bernoulli}.

		$$\left(1+{x\over nk}\right)^{-n} = \left(nk+x\over nk\right)^{-n} = \left(nk\over nk+x\right)^{n}
		  = \left(x-{x\over nk+x}\right)^{n} $$

		$$\becauseof{\ref{Bernoulli}}{\ge} 1 - {nx\over nk+x} \ge 1 - {x\over k}, \textbox{umocnit na $k$} $$

		Posledn� nerovnost plat�, nebo� $nkx \le nkx + x^2$.
	}

	}


	\medskip

	\qed

	}
	}

	\notes{
	\list{
		\listItem{$\exp(0) = 1$ }
		\listItem{$\exp(1) = \lim_{n\to \infty} \left(1+{1\over n}\right)^n = e$ }
		\listItem{$\exp$ je rostouc�, $\exp\colon \real \to (0,\infty)$
			$\Rightarrow$ $\exp$ m� inverzn� funkci na $(0,\infty)$.
			Tuto funkci nazveme p�irozen�m logaritmem, zna��me ``$\log$''.
			}
		\listItem{$a > 0$, def. $\log_a x = {\log x \over \log a}$ }
	}
	}
}



\theorem{14}{z�kladn� vlastnosti logaritmu}{
	Funkce $\log$, definovan� p�edpisem
	$$\log = \exp^{-1}$$
	m� n�sleduj�c� vlastnosti:

	\list{
		\listItem{$D(\log) = (0,\infty)$, $\log\colon (0,\infty) \to \real$ }
		\listItem{$\log(xy) = \log x + \log y\quad \forall x,y\in(0,\infty)$, $\log(x^n) = n\log x$ }
		\listItem{$\log$ je spojit�, rostouc� na $(0,\infty)$, $\log 1 = 0$, $\log e = 1$, $\log {1/x} = -\log x$ }
		\listItem{Z�kladn� limity:
			$$\lim_{x\to 0_+} \log x = -\infty$$
			$$\lim_{x\to \infty} \log x = \infty$$
			$$\lim_{x\to 1} {\log x\over x - 1} = \lim_{x\to 1} {\log (x+1)\over x} = 1$$
			}
	}

	\proof{
		Plyne z odpov�daj�c�ch vlastnost� funkce $\exp$. \DCV
		\qed
	}
}



\scpart{Obecn� mocnina}{
	$a > 0$ (!), $b \in \real$, pak definuji
	$$ a^b \defined \exp(b\log a) $$

	Speci�ln�, definuji funkci $a^x \defined \exp(x\log a)$. Pro $a = e:\ e^x = \exp x$.

	\penalty-100
	\note{
		$$x^2 = e^{2\log x} = e^{\log x + \log x} = e^{\log x}\cdot e^{\log x} = x\cdot x = x^2$$
	}
}



\scpart{Sinus}{
	\scpart{Co od n�j chceme?}{
		\list{
			\olistItem{definovan� na $\real$ }
			\olistItem{spojit� }
			\olistItem{$2\pi$-periodick� }
			\olistItem{omezen� }
			\olistItem{$\lim_{x\to0}{\sin x\over x} = 1$ }
			\olistItem{sou�tov� vzorce }
			\olistItem{pevn� body:
				$$ \eqalign{\sin(0) = 0,&\ \sin({\pi/2})=1,\ \sin(\pi) = 0\cr
				            \sin(0) = 1,&\ \sin({\pi/2})=0,\ \sin(\pi) = -1} $$
				}
		}
	}

	\scpart{St�edo�kolsk� definice}{
		$$\sin \a = {a\over c}$$

		Ale v �em m���m $\a$? Ve stupn�ch vznikne n�co ``rozh��can�ho'', ale s radi�ny
		se dost�v�me do kruhu, nebo� nem�me definov�nu obloukovou vzd�lenost. Tu
		bych musel matematicky definovat zase p�es $\sin$.
	}

	\theorem{15}{goniometrick� funkce}{
		$\onexists$ re�ln� funkce $s$ a $\onexists$ re�ln� funkce $c$ takov�, �e:

		\list{
			\listItem{$$\eqalign{s(x+y) &= s(x)c(y) + c(x)s(y)\cr
					     c(x+y) &= c(x)c(y) - s(x)s(y)}$$}
			\listItem{$s$ lich�, $c$ sud� (na $\real$)}
			\listItem{$s > 0$ na $(0,1)$, $s(1) = 0$}
		}

		\proof{
			Nech� takov� funkce existuj�. Pak:
			\list{
			\oldlistlevel=\listlevel
			\listlevel=2

			\listItem{%1
				$s(0) = 0$ (z lichosti)

				$c(0) = 1$:
				$$ c(x + (-x)) = c(x)c(-x) - s(x)s(-x) \becauseof{sudost}{=} c^2(x) + s^2(x) $$
				$$ c(0) = c(0+0) = c^2(0) + s^2(0) = c^2(0) \Rightarrow c(0) = 1 $$

				\TODO{Ale to plat� i pro $c(0) = 0$ $\ldots$?!}
			}

			\listItem{%2
				$c^2(x) + s^2(x) = 1\quad \forall x \in \real$

				Plyne z p�edchoz�ho --- ekvivalentn� k $c(x + (-x)) = c(0) = 1$.
			}

			\listItem{%3
				\M{\eqalign{\forall x \in \real:\ s(2x) &= 2s(x)c(x)\cr
				                                c(2x) &= c^2(x)-s^2(x)}}

				Ihned plyne z (i).
			}

			\listItem{%4
				$s\left(1/2\right) = 1,\ c\left(1/2\right) = 0$

				$$ \eqalign{0 \becauseof{(iii)}{=} s(1)
				              \becauseof{(3)}{=} 2 \* \undernote{s\left(1/2\right)}{> 0 \textbox{(iii)}}
					                           \* c\left(1/2\right)
				            &\Rightarrow c\left(1/2\right) = 0\cr
					    &\becauseof{(2)}{\Rightarrow} s\left(1/2\right) = 1} $$
			}

			\listItem{%5
				$c(1) = -1$

				$$ c(1) = c\left(2\cdot{1/2}\right)
				     \becauseof{(3)}{=} c^2\left(1/2\right) - s^2\left(1/2\right)
				     = 0 - 1 $$
			}

			\listItem{%6
				$s(x+1)=-s(x)$ (antiperiodicita):

				$$ s(x+1) = s(x)\undernote{c(1)}{-1} + c(x)\undernote{s(1)}{0} = -s(x) $$

				$s(x+2)=s(x)$ (periodicita):

				$$ \eqalign{s(x+2) &= s(x)c(2\cdot1) + c(x)s(2\cdot1)\cr
				                   &= s(x)\undernote{(c^2(1) - s^2(1))}{1} + c(x)\undernote{(2s(1)c(1))}{0}\cr
						   &= s(x) } $$
			}

			\listItem{%7
				$s\left({1/2}-x\right) = c(x)$

				$$ s\left({1/2}-x\right) = \undernote{s\left(1/2\right)}{1}c(x)
				                  - \undernote{c\left(1/2\right)}{0}s(x) = c(x) $$
			}

			\listItem{%8
				$c>0$ na $\left(0,{1/2}\right)$ --- z (7) a (iii)
			}

			\listItem{%9
				$s$ je rostouc� na $\left(0,{1/2}\right)$,
				$c$ je klesaj�c� na $\left(0,{1/2}\right)$.

				Vol�me $x$, $y$ tak, aby $0 < x - y < x < {1/2}$ (tedy tak� $0 < y < x$),
				$$ \eqalign{s(x-y) &= s(x)\undernote{c(y)}{\le1}
				                      - \undernote{c(x)s(y)}{>0 \textbox{(8), (iii)}}\cr
						   &\le s(x)} $$
			}

			\namedListItem{10-}{
				$$\eqalign{c\left(1/2^n\right) &= \sqrt{1+c{1/2^{n-1}}\over2}\cr
				           s\left(1/2^n\right) &= \sqrt{1-c{1/2^{n-1}}\over2}} $$

				\DCV
			}

			\listItem{%10
				$s$, $c$ spojit� v bod� 0.

				V bodech $k2^{-n}$ definujeme funkce $s$, $c$ dle (10--).
				V ostatn�ch bodech ($x \ne k2^{-n}$) pak limitn� supremem:
				$$ x \ne k2^{-n}:\ s(x) = \sup \{s(y), y = k2^{-n}, y < x\},\ x \in \left(0,{1/2}\right)$$

				Sta�� definovat na $\left(0,{1/2}\right)$, na zbytku
				nadefinujeme z periodicity a antiperiodicity.

				$a_n := c\left(1/2^n\right)$, pak $a_{n+1} = \sqrt{1+a_n\over2}$.
				Podle (9) je $a_n$ rostouc� a omezen�
				$\Rightarrow$ $\lim a_n = A$, $A = \sqrt{1+A\over2}$
				$\Rightarrow$ $A = 1$
				$\Rightarrow$ $\lim_{x\to0} s(0) = 0$
				$\Rightarrow$ $c$ je spojit�.
			}

			\listItem{%11
				$s(a) - s(b) = 2s\left(a-b\over2\right)c\left(a+b\over2\right)$

				\DCV: Z (i).
			}

			\listItem{%12
				$s$, $c$ spojit� na $\real$:

				$$ \lim_{x\to a} (s(x)-s(a)) \becauseof{(ii)}{=}
					2\lim_{x\to a} \undernote{s\left(x-a\over2\right)}{\to0}
					               \undernote{c\left(x+a\over2\right)}{\le1} $$
			}

			\listItem{%13
				Existuje vlastn� \M{\lim_{x\to 0} {s(x)\over x} = \pi}. {\bf (D�le�it�!)}

				$$ t(x) \defined {s(x) \over c(x)},\ x \in \real \setminus \{{\hbox{$1/2$}}+k\} $$

				Potom $t$ je lich�, rostouc� na $\left(-{1/2},{1/2}\right)$ a
				$$ t(x + y) = {t(x)+t(y) \over 1 - t(x)t(y)} $$

				$$ t\left(1/4\right) = 1 $$

				$$ \eqalign{x,y \in \left(0,{1/4}\right):\ t(x+y) &\ge t(x)+t(y)\cr
				                                               s(x+y) &\le s(x)+s(y)} $$
				$$ \Rightarrow t(x) \ge s(x) $$
				$$ \Rightarrow\ \forall k,m,n \in \nat,\ k2^{-n} \in \left(0,{1/4}\right): $$
					$$ t(k2^{-n}) \ge kt(2^{-n}) \ge ks(2^{-n}) \ge {k\over m}s(2^{-n}m) $$
				$$ \Rightarrow {t(k2^{-n})\over k2^{-n}} \ge {s(2^{-n}m) \over m2^{-n}} $$
				$$ \Rightarrow {t(y)\over y} \ge {s(y)\over y}\qquad \forall x,y \in \left(0, {1/2}\right) $$

				Definujeme \M{\pi := \inf_{y\in\left(0,{1/2}\right)} {t(y)\over y}}.

				$$ \undernote{\pi}{\to\pi} \ge \undernote{{s(x)\over x} = {t(x)\over x}c(x)}{\to\pi (policie)}
				     \ge \undernote{\pi c(x)}{\to\pi1} $$

				\qed
			}

			\listlevel=\oldlistlevel
			}
		}

		Definujeme:

		$$ \eqalign{\sin(x) &= s\left(x\over\pi\right)\cr
		            \cos(x) &= c\left(x\over\pi\right)\cr
		            \tg(x)  &= {\sin x\over\cos x}\cr
		            \cotg(x)&= {\cos x\over\sin x}} $$

		Pak bude platit:

		\list{
			\listItem{$\sin, \cos$ jsou $2\pi$--periodick�, $\pi$--antiperiodick� a spojit� na $\real$}
			\listItem{$\sin$ je rostouc� na $\left(-{\pi/2},{\pi/2}\right)$}
			\listItem{$\cos$ je klesaj�c� na $\left(0,\pi\right)$}
			\listItem{Z�kladn� limity:
				$$\lim_{x\to 0} {\sin x\over x} = 1$$
				$$\lim_{x\to 0} {\tg x\over x} = 1$$
				$$\lim_{x\to 0} {1-\cos x\over x^2} = \lim_{x\to 0} {1-\cos^2 x\over x^2(1+\sin x)} = {1/2}$$
			}
		}
	}
}



\notes{
\list{
	\listItem{Alternativn� zp�sob:

	$$ \exp(x) = \sum_{n=0}^\infty {x^n\over n!} = 1 + x + {x^2\over2} + {x^3\over6} + \cdots $$
	$$ E(x) := \sum_{n=0}^\infty {x^n\over n!} \hypothetically{=} \lim_{n\to \infty} \left(1+{x\over n}\right)^n $$
	Bu� m��eme dok�zat pomoc� odhad�, �e
	$$\lim_{n\to \infty} \left(1+{x\over n}\right)^n = \lim_{n\to \infty} \sum_{k=0}^n {x^k\over k!},$$
	nebo dok�zat, �e:
	$$ E(x+y) = E(x) \cdot E(y) \textnote{sou�in �ad} $$
	$$ E(x) \ge 1 + x \textnote{pro $y>0$ trivi�ln�} $$

	$$ \sin(x) = \sum_{n=0}^\infty (-1)^n {x^{2n+1}\over(2n+1)!} = x - {x^3\over6} + {x^5\over5!} - \cdots $$
	$$ \cos(x) = \sum_{n=0}^\infty (-1)^n {x^2n\over(2n)!} = 1 - {x^2\over2} + {x^4\over4!} - \cdots $$

	}

	\listItem{Z�kladn� limity:

	$$ \lim_{x\to0} {e^x-1\over x} = 1 $$
	$$ \lim_{x\to0} {\sin x\over x} = 1 $$
	$$ a>0:\ \lim_{x\to0} {a^x-1\over x} = \lim_{x\to 0} {e^{x\log a}-1\over \undernote{x\log a}{\to 1}}\log a = \log a $$
	$$ \lim_{x\to0} {1 - \cos x\over x^2} = \lim_{x\to 0} {1 - \cos^2 x\over x^2(1+\cos x)} =
	   \lim_{x\to0} \undernote{\sin^2x\over x^2}{\to 1} \cdot \undernote{1\over1+\cos x}{\to 1/2} = {1/2} $$

	Tedy v nule se $e^x-1$ a $\sin x$ chovaj� v okol� nuly stejn� jako line�rn�, $1-\cos x$ jako kvadratick�.

	$$ \lim_{x\to0} {1-\cos x\over x} = \lim_{x\to0} {1-\cos x\over x^2}x = 0 $$

	}

}


}


\bend
