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

\definition{
	�adu funkc�
	$$ \sum_{n=0}^\infty a_n(x-x_0)^n\qquad x_0 \in \real,\ a_n \in \real $$
	naz�v�me {\bf mocninnou �adou} se {\bf st�edem} v bod� $x_0$
	a {\bf koeficienty} $a_n$.\footnote*{Zde pou��v�me konvenci $0^0 := 1$}
}

Zaj�m� n�s, zda, kde, jak a kam mocninn� �ada konverguje.

\shortobs{Ka�d� mocninn� �ada konverguje pro $x=x_0$ (tj. ve sv�m st�edu).}

Odpov�� na ot�zku, zda �ada konverguje, p��padn� kde, n�m d�
{\it oblast konvergence} $M$, ve kter� $x_0 \in M$.

Jak mocninn� �ada konverguje? Bodov�, stejnom�rn�, lok�ln� stejnom�rn�
�i absolutn�.

A kam konverguje?
$$ \existss f(x): f(x) = \sum_{n=0}^\infty a_n(x-x_0)^n \textbox{?} $$
V n�kter�ch p��padech je �ada vytvo�ena n�jakou funkc�.
Nap�. Taylor�v polynom:
$$T^{f,n}_{x_0}(x) = \sum{f^k(x_0)\over k!}(x-x_0)^k$$
d�le t�eba \ref{Fourierova �ada} (bude), Martaminova �ada (nebude),
Bersteinovy polynomy (tak� nebudou).


\theorem{1}{o existenci a jednozna�nosti polom�ru konvergence mocninn� �ady}{
	Nech� $\sum_{n=0}^\infty a_n(x-x_0)^n$ je mocninn� �ada.
	Pak $$\onexists R \in [0,\infty]$$ (pozor, $R$ m��e b�t
	i nekone�no) takov�, �e �ada konverguje pro $\forall x$ takov�,
	�e $|x-x_0|<R$, a diverguje pro $\forall x$ takov�, �e
	$|x-x_0|>R$. Co se d�je v krajn�ch hodnot�ch, nev�me --- tam se
	m��e st�t cokoliv.

	\proof{
		$$ R := \sup\{|y-x_0|: \sum_{n=0}^\infty a_n(y-x_0)^n \hbox{\ konverguje}\} $$

		Nech� $|x-x_0|<R$. Potom
		$$ \sum_{n=0}^\infty |a_n||x-x_0|^n $$
		a proto�e $|x-x_0|<R$,
		existuje $y$ takov�, �e
		$$ |y - x_0| > |x - x_0|\quad \et\quad |y - x_0| < R\quad \et\quad
		 \sum_{n=0}^\infty a_n(y-x_0)^n \textbox{konverguje} $$
		$$ \sum_{n=0}^\infty |a_n||x-x_0|^n
			= \sum_{n=0}^\infty |a_n||y-x_0|^n\left(|x-x_0|\over|y-x_0|\right)^n $$

		\penalty-100

		Proto�e �ada $\sum a_n(y-x_0)^n$ konverguje,
		$$\existss K: |a_n||y-x_0|^n \le K $$
		$$ \sum_{n=0}^\infty |a_n||x-x_0|^n \le K\sum_{n=0}^\infty \left(|x-x_0|\over|y-x_0|\right)^n $$
		Proto�e v�ak \M{{|x-x_0|\over|y-x_0|} < 1}, jde o konvergentn� geometrickou �adu.

		\medskip

		Nech� $|x-x_0|>R$. Z definice $R$ plyne, �e $x$ diverguje.

		Jednozna�nost $R$ plyne z jednozna�nosti suprema.
		\qed
	}
}

\theorem{2}{o v�po�tu polom�ru konvergence}{
	Nech� $\sum_{n=0}^\infty a_n(x-x_0)^n$ je mocninn� �ada
	s polom�rem konvergence $R \in [0,\infty]$.
	Potom
	$$ R = {1 \over \limsup\limits_{n\to\infty} \broot{n}\of{|a_n|}} $$

	\proof{
		Ozna�me si
		$$ R^* := {1 \over \limsup\limits_{n\to\infty} \broot{n}\of{|a_n|}} $$

		Budeme dokazovat, �e
		�ada konverguje pro $|x-x_0| < R^*$
		a diverguje pro $|x-x_0| > R^*$,
		tedy �e $R^* = R$.
		Jakmile toto dok�eme,
		tvrzen� vyplyne z jednozna�nosti $R$,
		dok�zan� v \ias{}{1}{}.

		Nech� $R^* \in (0,\infty)$.
		D�le nech� $|x-x_0|<R^*$.
		Potom
		$$ \limsup_{n\to\infty} \broot{n}\of{|a_n||x-x_0|^n}
			= |x-x_0|\limsup_{n\to\infty} \broot{n}\of{|a_n|}
			= {|x-x_0| \over R^*} < 1 $$
		Dle \ref{Cauchy}ova krit�ria �ada (��sel)
		$\sum_{n=0}^\infty a_n(x-x_0)^n$
		konverguje.

		Nech� $|x-x_0| > R^*$.
		Pak
		$$ \limsup_{n\to\infty} \broot{n}\of{|a_n||x-x_0|^n} > 1 $$
		a tedy op�t dle \ref{Cauchy}ho
		$\sum_{n=0}^\infty a_n(x-x_0)^n$
		diverguje.

		P��pady $R^* = 0$ a $R^* = \infty$ odbobn� (\DCV).
		\qed
	}
}


\examples{
\list{
	\listItem{\M{\sum_{n=0}^\infty {x^n \over n!}}

		St�ed je v $x_0 = 0$, polom�r
		$$ R = {1 \over \limsup\limits_{n\to\infty} \broot{n}\of{1/n!}} = 1/0 = \infty $$
		(z konvence {\it (pouze v t�to pas�i)} $1/0 = +\infty$ a $1/\infty = 0$).

		Tedy tato �ada konverguje (bodov�)
		pro $\forall x \in \real$ k funkci $\exp x$.
	}

	\listItem{\M{\sum_{n=0}^\infty n^n(x-1)^n}

		$$ x_0 = 1 $$
		$$ R = {1 \over \limsup\limits_{n\to\infty} \broot{n}\of{n^n}} = 1/\infty = 0 $$

		�ada konverguje jen pro $x = 1$.
	}

	\listItem{\M{\sum_{n=0}^\infty \arctg(2n)(x-3)^n}
		$$ x_0 = 3 $$
		$$ R = 1 $$

		�ada konverguje pro $x \in (2,4)$, diverguje pro $x \in \real \setminus [2,4]$.

		V bodech $x = 2$, $x = 4$ �ada diverguje ($\arctg$ rychle odkonverguje n�kam vysoko nad nulu).

		Tedy �ada konverguje pro $x \in (2,4)$.
	}

	\listItem{\M{\sum_{n=2}^\infty {(x+e)^n \over \log n}}
		$$ x_0 = -e $$
		$$ \broot{n}\of{1/\log n} \in \left(\broot{n}\of{1/n},\broot{n}\of{1}\right) \Rightarrow R = 1 $$

		Konverguje pro $x \in (-e-1, -e+1)$.

		V bod� $x = -e-1$: $\sum_{n=2}^\infty {(-1)^n\over\log n}$ konverguje, Leibniz.

		V bod� $x = -e+1$: $\sum_{n=2}^\infty {1\over\log n}$ diverguje, srovn�n� s harmonickou.

		Tedy �ada konverguje pro $x \in [-e-1,-e+1)$.
	}

	\listItem{\M{\sum_{n=1}^\infty {(-1)^n(2x)^{2n}\over2n}}
		$$ x = 0 $$
		$$ a_m = \cases{\displaystyle {(-1)^n2^{2n}\over2n} & $m=2n$, $n \in \nat$\cr
		                0 & $m=2n+1$, $n \in \nat$} $$

		$$ \limsup_{n\to\infty} \broot{n}\of{|a_n|}
			= \limsup_{n\to\infty} \broot{2n}\of{2^{2n}\over2n}
			= \limsup_{n\to\infty} {2 \over \broot{2n}\of{2n}}
			= 2 $$
		Tedy $R = 1/2$, �ada konverguje pro $x \in (-1/2,1/2)$.

		V bodech $x = \pm1/2$: $\sum {(-1)^n\over2n}$ konverguje
		dle Leibnize, tud�� �ada konverguje pro $x \in [-1/2,1/2]$.
	}
}
}


\shortdef{Ot�zka}{Jak vy�eto�ovat stejnom�rnou (lok�ln� stejnom�rnou) konvergenci mocninn�ch �ad?}

\theorem{3}{o lok�ln� stejnom�rn� konvergenci mocninn� �ady}{
	Nech� $\sum_{n=0}^\infty a_n(x-x_0)^n$ je mocninn� �ada
	s polom�rem konvergence $R \in (0,\infty]$.
	Potom �ada konverguje absolutn� a lok�ln� stejnom�rn�
	na intervalu $(x_0-R,x_0+R)$ je-li $R < \infty$ a na
	intervalu $\real$, je-li $R = +\infty$.

	\proof{
		Sta�� dok�zat, �e $\sum \conS$ na ka�d�m
		intervalu $[x_0-K,x_0+K]$, kde $K \in (0,R)$.

		M�jme tedy $K \in (0,R)$, pak
		$$ \sup_{x\in[x_0-K,x_0+K]} |a_n||(x-x_0)^n| = |a_n|K^n $$
		ale $\sum_{n=0}^\infty |a_n|K^n$ konverguje
		(viz d�kaz \ias{}{1}{}).

		Tedy dle \ref{Weistrass}ova kriteria
		$\sum \conS$ na $[x_0-K,x_0+K]$.
		\qed
	}
}


\penalty-100

\theorem{4}{o derivaci mocninn� �ady}{
	Nech� $\sum_{n=0}^\infty a_n(x-x_0)^n$ je mocninn� �ada
	s polom�rem konvergence $R \in (0,\infty]$.
	Potom �ada
	$$ \sum_{n=1}^\infty na_n(x-x_0)^{n-1} $$
	je tak� mocninn�, se stejn�m st�edem i polom�rem konvergence,
	a nav�c
	$$ \left(\sum_{n=0}^\infty a_n(x-x_0)^n\right)' = \sum_{n=1}^\infty na_n(x-x_0)^{n-1}
		\qquad x \in (x_0 - R, x_0 + R) $$

	\proof{
		Nech� $R'$ je polom�r konvergence �ady
		$\sum_{n=1}^\infty na_n(x-x_0)^{n-1}$,
		pak
		$$ \limsup_{n\to\infty} \broot{n}\of{n|a_n|}
			\becauseof{\ref{VOAL}}{=}
			1 \cdot \limsup_{n\to\infty} \broot{n}\of{|a_n|} $$
		a tedy $R' = R$.
		Z�rove� d�ky tomu
		$$ \sum_{n=1}^\infty na_n(x-x_0)^{n-1}
			\conSLi{}{(x_0-R,x_0+R)} \textnote{viz p�edchoz� v�ta} $$
		a v z�chytn�m bod� $x = x_0$ konverguje �ada $\sum_{n=0}^\infty a_n(x-x_0)^n$.

		Tedy dle v�ty o z�m�n� sumy a derivace
		$$ \left(\sum a_n(x-x_0)^n\right)' = \sum a_nn(x-x_0)^{n-1} $$
		\qed
	}
}

\theorem{5}{o integraci mocninn� �ady}{
	Nech� $\sum_{n=0}^\infty a_n(x-x_0)^n$ je mocninn� �ada
	s polom�rem konvergence $R \in (0,\infty]$.
	Potom
	$$ \int \left(\sum_{n=0}^\infty a_n(x-x_0)^n\right)\,dx
		= c + \sum_{n=0}^\infty {a_n \over n+1}(x-x_0)^{n+1}
		\qquad x \in (x_0-R,x_0+R) $$

	\shortproof{Nechceme. \sqed}
}


\theorem{6}{Abelova}{
	{\it Nejd�le�it�j�� v�ta pas�e.}

	Nech� m�
	mocninn� �ada $\sum_{n=0}^\infty a_nx^n$
	se st�edem $x_0 = 0$
	polom�r konvergence $R \in (0,\infty)$.
	Nech� $\sum_{n=0}^\infty a_nR^n$ konverguje.
	Potom
	$$ \sum_{n=0}^\infty a_nx^n \conSi{}{[0,R]} $$
	$$ !\qquad\sum_{n=0}^\infty a_nR^n = \lim_{x\to R_-} \sum_{n=0}^\infty a_nx^n\qquad! $$

	\proof{
		$$ \sum_{n=0}^\infty a_nx^n
			= \sum_{n=0}^\infty a_nR^n (x/R)^n $$

		�ada $\sum a_nR^n$ konverguje,
		tedy tak� $\sum a_nR^n \conS$.
		D�le $\left|(x/R)^n\right| \le 1$,
		a nav�c je $(x/R)^n$ monot�nn� (v $n$) pro ka�d� $x$.
		Podle \ref{Abel}ova krit�ria tedy $\sum a_nx^n \conS$.

		Vzorec s limitou plyne ihned z \ref{Moore-Osgood}ovy v�ty.
		\qed
	}
}

\penalty-100

\examples{
\list{
\listItem{
	Se�t�te �adu \M{\sum_{n=1}^\infty {(-1)^{n-1} \over n} = 1 - 1/2 + 1/3 - 1/4 + \cdots}

	\smallskip

	Zave�me si um�lou prom�nnou $x$. Definujeme �adu (funkc�)
	$$ \sum_{n=1}^\infty {(-1)^{n-1}\over n}x^n $$

	To je mocninn� �ada s polom�rem konvergence $R = 1$.
	Pro $x = R$ �ada konverguje (\ref{Leibniz}).
	Tedy dle Abelovy v�ty
	$$ f(x) := \sum_{n=1}^\infty {(-1)^n\over n}x^n $$
	a tato funkce spl�uje
	$$ f'(x) = \sum_{n=1}^\infty {(-1)^{n-1}\over n}nx^{n-1}
		= \sum_{n=1}^\infty (-x)^{n-1}
		= \sum_{n=0}^\infty (-x)^n
		= {1 \over 1+x}
		\qquad x \in (-1,1) $$
	a nav�c plat�
	$$ \sum_{n=1}^\infty {(-1)^{n-1}\over n} = \lim_{x\to1_-} f(x) $$

	$$ f(x) = \int f'(x)\,dx = \int {dx\over1+x} = \log(1+x) + c $$
	a zb�v� dopo��tat $c$:
	$$ f(0) = \sum_{n=1}^\infty {(-1)^{n-1}\over n}\cdot0^n = 0
		\Longrightarrow \log (1+0)+c = 0
		\Longrightarrow c = 0 $$

	$$ \Longrightarrow f(x) = \log(1+x) $$

	Tedy
	$$ \sum_{n=1}^\infty {(-1)^{n-1}\over n} = \lim_{x\to1_-} f(x) = \log 2 $$
}

\medskip

\listItem{
	Se�t�te �adu \M{\sum_{n=1}^\infty (-1)^{n} {n^2 \over n!} =: s}

	\smallskip

	$$s = \sum_{n=1}^\infty (-1)^n {n\over(n-1)!}
		= \sum_{n=0}^\infty (-1)^{n+1} {n+1\over n!}$$

	Definujme si:
	$$ f(x) := \sum_{n=0}^\infty (-1)^{n+1} {n+1\over n!}x^n $$
	Pak dle \ias{}{5}{} pro $F(x) = \int f(x)\,dx$ plat�
	$$ F(x) = c + \sum_{n=0}^\infty(-1)^{n+1}{x^{n+1}\over n!}
		= c + \sum_{n=0}^\infty{(-x)^{n+1}\over n!}
		= c - x\sum_{n=0}^\infty{(-x)^n \over n!}
		= c - xe^{-x} $$
	$$ F'(x) = -e^{-x} + xe^{-x} = (x-1)e^{-x} = f(x) $$
	a tedy dle \ref{Abel}ovy v�ty
	$$ s = \lim_{x\to1_-} f(x) = 0 $$

	\shortexerc{\M{\sum_{n=1}^\infty {n^2 \over n!} = 2e}}
}
}
}


\bend
