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

\subchapter{Konvergence �ad funkc�}{

\definition{
	�ekneme, �e �ada funkc� $\sum_{n=1}^\infty f_n(x)$ definovan�ch
	na $\emptyset \ne M \subset \real$, {\bf konverguje bodov�}
	({\bf stejnom�rn�}, {\bf lok�ln� stejnom�rn�}),
	jestli�e posloupnost funkc�
	$$ \left\{ \sum_{k=1}^n f_k(x) \right\}_{n=1}^\infty $$
	konverguje p��slu�n�m zp�sobem.

	Zna��me $\sum f_n \conP$, $\sum f_n \conS$, $\sum f_n \conSL$.

	M�me-li limitn� funkci $f$, zna��me
	$\sum f_n \conP f$, $\sum f_n \conS f$, $\sum f_n \conSL f$.
}

\theorem{7}{Weistrassovo krit�rium}{

	Nech� $f_n$ jsou definov�ny na $\emptyset \ne M \subset \real$,
	nech�
	$$ s_n := \sup_{x \in m} |f_n(x)|\qquad n \in \nat $$
	$$ \sum_{n=1}^\infty s_n < \infty \Longrightarrow \sum_{n=1}^\infty f_n(x) \conS \textbox{na $M$} $$

	\proof{
		Zvol�me $\eps > 0$, pak
		$$\existss n_0 \in \nat: \sum_{n=n_0}^\infty s_n < \eps$$

		Nech� $m,n \ge n_0$, kde $m,n \in \nat$;
		bez �jmy na obecnosti nech� $m < n$.
		Tedy
		$$ \left| \sum_{k=1}^n f_k(x) - \sum_{k=1}^m f_k(x) \right|
			= \left| \sum_{k=m+1}^n f_k(x) \right|
			\becauseof{$\triangle$}{\le} \sum_{k=m+1}^n |f_k(x)|
			\le \sum_{k=n_0}^\infty |f_k(x)|
			\le \sum_{k=n_0}^\infty s_k
			< \eps $$

		Tud�� posloupnost $\left\{\sum_{k=1}^n f_k\right\}$
		je \ref{stejnom�rn� cauchyovsk�} na $M$, a tedy
		$\sum f_n \conS$ na $M$.

		\qed
	}

}

\theorem{8}{o z�m�n� sumy a derivace}{

	Nech� $f_n$, $n \in \nat$ jsou definov�ny a maj� vlastn� derivace
	$f_n'$ na intervalu $(a,b) \subset \real$. Nech� plat�:

	\list{
	\listItem{\M{\sum_{n=1}^\infty f_n' \conSL} na $(a,b)$}
	\listItem{$\existss x_0 \in (a,b)$ takov�,
		�e \M{\sum_{n=1}^\infty f_n(x_0) < \infty}
		(konverguje)}
	}

	Potom $\sum f_n \conSL$ na $(a,b)$
	a nav�c $\left(\sum f_n\right)' = \sum f_n'$ na $(a,b)$.

	\shortproof{P��m� d�sledek \ias{}{3}{} a definice konvergence �ady. \sqed}

}

\penalty-100

\theorem{9}{o z�m�n� sumy a integr�lu}{

	Nech� $f_n \in N(a,b)$, $n \in \nat$, $(a,b) \subset \real$
	a nech� $\sum f_n \conS$ na $(a,b)$.
	Pak limitn� funkce $f := \sum f_n$ spl�uje $f \in N(a,b)$ a
	$$ \int_a^b f(x)\,dx = \sum_{n=1}^\infty \int_a^b f_n(x)\,dx $$

	\shortproof{P��m� d�sledek \ias{}{4}{}. \sqed}

}

\theorem{10}{Abelovo krit�rium}{

	Nech� $\sum f_n(x) \conS$ na $M$,
	nech� $\{g_n\}$ je {\bf stejnom�rn� omezen�} na $M$:
	$$\existss K > 0,\ \forall n \in \nat\ \forall x \in M: |g_n(x)| \le K$$
	Nech� $\{g_n(x)\}_{n=1}^\infty$ je nav�c pro ka�d� pevn� $x \in M$ monot�nn�
	(bu� nerostouc� nebo neklesaj�c�, dokonce v ka�d�m $x$ jinak).

	Potom $\sum f_ng_n \conS$ na $M$.

	\scpart{\textbfsc{N�znak d�kazu}:}{
		Definujme si pro $N \in \nat$:
		$$ s_N(x) := \sum_{n=1}^N f_n(x) $$
		$$ t_N(x) := \sum_{n=N+1}^\infty f_n(x) $$

		$$ f_ng_n = -t_n(g_n-g_{n+1}) + t_{n-1}g_n - t_ng_{n+1} $$
			(\DCV; \ref{Abelova parci�ln� sumace})
		$$ \sum_{n=M}^N f_ng_n = \sum_{n=M}^N - t_n(g_n-g_{n+1} + t_{M-1}g_M - t_Ng_{N+1} $$
		$$ \left|\sum_{n=M}^N f_ng_n\right| \becauseof{$\triangle$}{\le}
			\sum_{n=M}^N |t_n||g_n-g_{n+1}| + |t_{M-1}||g_M| + |t_N||g_{N+1}| $$

		Zvolme $\eps > 0$. K n�mu $\existss N_0$ tak,
		aby pro $\forall N \ge N_0$ bylo $|t_N| < \eps$.
		Nav�c v�me, �e $|g_N| \le K$ pro $\forall N \in \nat$.
		$$ \becauseof{\strut$M,N\ge N_0$}{\Longrightarrow}
			\left|\sum_{n=M}^N f_ng_n\right|
			\le \sum_{n=M}^N \eps|g_n-g_{n+1}| + \eps K + \eps K $$

		$g_n(x)$ je monot�nn�, tedy (!)
		$$ \sum_{n=M}^N |g_n(x) - g_{n+1}(x)| = |g_{N+1}(x) - g_M(x)| $$
		(Tvrd�m, �e s��t�m-li jednotliv� rozd�ly, je to stejn�, jako kdybych
		ode�etl nejvy��� od nejni���ho. To samoz�ejm� plat�, pouze m�m-li
		monotonii.)
		$$ \Longrightarrow
			\left|\sum_{n=M}^N f_ng_n\right|
			\le \eps\left(|g_{N+1}| + |g_M|\right) + 2\eps K
			\le \eps(K+K) + 2\eps K = 4\hskip1pt\eps K
			\becauseof{($s\to0_+$)}{\longrightarrow} 0 $$
		Tedy $\sum$ je stejnom�rn� \ref{Bolzano-Cauchy}ovsk�,
		tud�� $\sum \conS$.

		\qed
	}
}

\penalty-100

\theorem{11}{Dirichletovo krit�rium}{

	Nech� $\{f_n\}$ m� stejnom�rn� omezen� ��ste�n� sou�ty
	(tj. posloupnost $$\left\{\sum_{k=1}^n f_k(x)\right\}_{n=1}^\infty$$ je stejnom�rn� omezen�)
	na $\emptyset \ne M \subset \real$.
	Nech� $g_n \conS 0$ na $M$.
	Nech� $\{g_n(x)\}_{n=1}^\infty$ je nav�c pro ka�d� pevn� $x \in M$ monot�nn�
	(na typu monotonie op�t nez�le��).

	Potom $\sum f_ng_n \conS$ na $M$.

	\scpart{\textbfsc{N�znak d�kazu}:}{
		Definujme si pro $N \in \nat$
		$$ s_N(x) := \sum_{n=1}^N f_n(x) $$
		(ale $t_N$ nem�me).

		$$ f_ng_n = s_n(g_n-g_{n+1}) - s_{n-1}g_n + s_ng_{n+1} $$
			(\DCV; \ref{Abelova parci�ln� sumace})
		$$ \left|\sum_{n=M}^N f_ng_n\right| \le \sum_{n=M}^N |s_n||g_n-g_{n+1}| + |s_{M-1}||g_M| + |s_N||g_{N+1}| $$
		$$ \sesac{\eqalign{|s_N| &\le K\cr
		                   |g_N| &< \eps}}
		         {\Longrightarrow}
		   \eqmalign{
		   &\le K\sum_{n=M}^N |g_n-g_{n+1}| + K\eps + K\eps\cr
		   &\le K(|g_M| + |g_{N+1}|) + K\eps + K\eps \le 4K\eps} $$
		atd.

		\qed
	}
}

\example{
	Vy�et�ete konvergenci �ady
	\M{\sum_{n=0}^\infty (-1)^n {x^{2n+1} \over 2n+1}}
	na $\real$ a tam, kde konverguje, �ady se�t�te.

	\smallskip

	$$\sum_{n=0}^\infty f_n(x) = \sum_{n=0}^\infty (-1)^n {x^{2n+1} \over 2n+1}$$

	{\bf Pro kter� $x \in \real$ konverguje tato �ada bodov�?}
	$x$ vezmeme pevn�, pak pan \ref{Cauchy} ��k�:
	$$ \lim_{n\to\infty} \broot{n}\of{|a_n|} = \lim_{n\to\infty} \broot{n}\of{|x|^{2n+1}\over2n+1} = x^2 $$
	Tedy �ada konverguje pro $|x| < 1$ a diverguje pro $|x| > 1$. A kdy� $x = \pm1$?
	$$ x = \pm1 \Longrightarrow \sum f_n(x) = \pm \sum {(-1)^n \over 2n+1} $$
	konverguje dle \ref{Leibniz}e.

	\smallskip

	Tud�� \M{\sum_{n=0}^\infty f_n(x)} konverguje {\bf bodov�} pro $x \in [-1,1]$.

	\smallskip

	\penalty-1000

	Spo��t�me derivace:
	$$ f_n'(x) = (-1)^n x^{2n} = \left(-x^2\right)^n\qquad x \in (-1,1) $$
	$$ \sum_{n=0}^\infty f_n'(x) = \sum_{n=0}^\infty \left(-x^2\right)^n
		= {1 \over 1+x^2}\qquad x \in (-1,1) $$

	{\bf Je tato konvergence lok�ln� stejnom�rn�?} Pak by platilo
	$\sum f_n'(x) \conSL$ na $(-1,1)$, tedy
	$\sum f_n'(x) \conS$ na $[-\delta,\delta]$ pro $\forall \delta \in (0,1)$.
	Nech� $\delta \in (0,1)$, pak
	$$ \sup_{x\in[-\delta,\delta]} |f_n'(x)|
		= \sup_{x\in[-\delta,\delta]} \left|\left(-x^2\right)^n\right|
		= \delta^{2n} =: s_n $$
	a jasn� $\sum s_n < \infty$.

	Tak�e dle Weistrassova kriteria $\sum f_n' \conSL$ na $(-1,1)$.
	Vezmeme si n�jak� z�chytn� bod, nap�. $x=0$: $\sum f_n(0) = 0$.
	A dle \ias{}{8}{} pak $\sum f_n \conS$ na $(-1,1)$
	a $\left(\sum f_n\right)' = \sum f_n'$.

	Ozna��me
	$$f(x) := \sum_{n=0}^\infty (-1)^n {x^{2n+1}\over2n+1}\qquad x \in [-1,1]$$
	Pak pro $x \in (-1,1)$:
	$$f'(x) = \left(\sum f_n(x)\right)' \becauseof{\ias{}{8}{}}{=} \sum f_n'(x)
		= \sum_{n=0}^\infty \left(-x^2\right)^n = {1 \over 1+x^2} $$
	$$ f(x) = \int f'(x)\,dx = \arctg x + c \textnote{Taylor} $$
	$$ f(0) = 0 \Longrightarrow 0 = \arctg 0 + c \Longrightarrow c = 0 $$

	\scpart{Z�v�r}{
		$$ \sum_{n=0}^\infty (-1)^n {x^{2n+1}\over2n+1} \conSL \arctg x \textbox{na $(-1,1)$} $$
	}

	{\bf Je tato konvergence stejnom�rn� na $[-1,1]$ nebo nen�?}
	Zvolme $$f_n(x) \equiv (-1)^n,\ |\sum f_n(x)| \le 1$$
	a dejme �anci panu \ref{Dirichlet}ovi:
	$$ g_n(x) = {x^{2n+1}\over2n+1}
		\Longrightarrow |g_n(x)| \le {1 \over 2n+1} \to 0
		\Longrightarrow g_n \conS 0 \textbox{na $[-1,1]$} $$

	Tedy podle Dirichletova krit�ria
	$$ \sum_{k=0}^\infty (-1)^n {x^{2n+1}\over2n+1} \conS \arctg x\qquad x \in [-1,1] $$


	Nyn� se pokusme �adu se��st.

	Pan Euler (1735) vymyslel hezkou formulku
	$$x = 1: \pi/4 = \arctg 1 = 1 - 1/3 + 1/5 - 1/7 + 1/9 - 1/11 \cdots,$$
	ta je ale velmi pomal�,
	teprve 300 �len� d� alespo� aproximaci 22/7,
	ke kter� do�el 2000 let p�ed t�m Archim�des.

	Existuje ale hezk� vzorec:
	$$ \tg (x+y) = {\tg x + \tg y\over 1-\tg x\tg y} $$
	$$ {1/2 + 1/3 \over 1 - 1/2\cdot1/3} = 1 $$
	$$ \arctg 1 = \arctg 1/2 + \arctg 1/3 $$

	Vyjde n�m tedy mnohem rychlej�� formule:
	$$ \pi/4 = (1/2+1/3) - 1/3(1/2^3 + 1/3^3) - 1/5(1/2^5 - 1/3^5) + \cdots $$
}

\example{
	Typick� vyu��t� \ref{Dirichlet}ova krit�ria:
	\list{
	\listItem{$f_n \equiv (-1)^n$}
	\listItem{$f_n(x) = \sin(nx)$}
	}

	Vy�et�ete konvergenci �ady \M{\sum_{n=1}^\infty {\sin(nx) \over n^\alpha}},
	$\alpha \in (0,\infty)$, $x \in (0,2\pi)$.

	Zvolme $f_n(x) = \sin(nx)$, pak element�rn� trigonometri� a indukc�:
	$$ \sum_{k=1}^n f_k(x) = {\sin{nx\over2}\cdot\sin{(n+1)x\over2}\over\sin{x\over2}} $$

	$$ x \in [\delta, 2\pi-\delta], \delta \in (0,\pi): \left|\sum_{k=1}^n f_k(x)\right| \le {1 \over \sin (\delta/2)} $$
	(stejnom�rn� omezenost)

	$g_n(x) \equiv 1/n^\alpha \conS 0$, tedy dle \ref{Dirichlet}a:
	$$ \sum {\sin(nx)\over n^\alpha} \conSL \textbox{na $(0,2\pi)$} $$

	\shortnote{Z Weistrasse lze toto dostat pouze pro $\alpha>1$, tedy je Dirichlet v tomto p��pad� lep��.}
}

}

\bend
