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

\scpart{Fyzik�ln� motivace}{
	Rovnice veden� tepla n�m ��k�, �e vezmeme-li si t�eba
	m�d�nou ty�, kter� m� na jednom konci teplotu $0\dg C$
	a na druh�m konci ji zah��v�me kahanem na $100\dg C$,
	a n�s zaj�m�, jak vypad� funkce $T(x,t)$.

	To vede na hled�n� $f$ takov�, aby $f'' = -f$
	(nap�. $\sin x$, $\cos x$).

	Je v�hodn� �e�en� hledat ve tvaru �ady, ale ta �ada
	nen� mocninn�. M�sto mocnin $x^n$ pracuje s funkcemi
	$\sin (nx)$ a $(\cos (nx)$.

	\optnote{

		Koup�m si pizzu, hod�m ji do trouby. V $t = 8^{00}$
		bude m�t $T = 110\dg C$. Kdy bude $T = 60\dg C$?

		V $t = 18^{00}$ nastane smrt hlady.
		Je t�eba se sp�lit?

		M�jme �as $t$, teplotu $T$.
		Teplota m�stnosti je $T_0 = 20$.
		$k$ je materi�lov� konstanta.

		$$\eqalign{
		t = 0:\ & T(t) = T(0) = 110\dg\cr
		t = 5:\ & T(t) = T(5) = 90\dg
		}$$

		Newton:
		$$ -{dT\over dt} = k(T(t)-T_0) $$
		$$ \int {dT\over T-T_0} = \int -k\,dt) $$
		$$ \log (T-T_0) = -kt + c $$
		$$ T - T_0 = e^ce^{-kt} =: Ae^{-kt} $$

		Nev�m, co je $k$ a $A$:
		$$ T(t) = T_0 + Ae^{-kt} $$

		Dosad�me:
		$$\eqmalign{
		t = 0:\ & 110 = 20 + Ae^0 &\Rightarrow A = 90 & T = 20 + 90e^{-kt}\cr
		t = 5:\ & 80 = 20 + 90e^{-k5} &\Rightarrow 2/3 = e^{-5k} & e^{-k} = \root{5}\of{2/3}
		}$$
		$$ \Longrightarrow T(t) = 20 + 90(2/3)^{t/5} $$

		$$\eqalign{t = 10:\ & T(10) \le 60 ??\cr
		                   & 20 + 90(2/3)^2 = 20 + 40 = 60} $$

		To bylo o fous.

	}

	Pot�� s ty�� je ta, �e mus�me derivovat parci�ln�, nebo� m�me
	dv� prom�nn�. V�e se pov�liv� komplikuje a na pomoc sp�ch�
	pr�v� pan Fourier.
}

Jean--Baptiste Joseph Fourier (1768---1830)
v roce 1822 vymyslel Fourierovy �ady.
Euler toti� pracoval se syst�mem funkc� $\{1,x,x^2,x^3,\ldots\}$,
ale Fourier si �ekl, �e by se mu hodil sp��e
trigonometrick� syst�m $\{1,\cos x,\sin x,\cos(2x),\sin(2x),\ldots\}$.

\scpart{Ot�zka:}{
	Je-li d�no $f$, chceme naj�t ��sla $a_n$, $b_n$ tak, aby �ada
	$$ a_0/2 + \sum_{n=1}^\infty (a_n\cos(nx) + b_n\sin(nx)) $$
	konvergovala (nejl�pe stejnom�rn�) k funkci $f$.
}


\penalty-1000


\definition{
	Funkce $f$ se naz�v� {\bf $2\pi$-periodick�}, jestli�e
	$$ \forall x \in \real: f(x+2\pi) = f(x) $$

	Mno�inu $2\pi$-periodick�ch funkc� zna��me $P_{2\pi}$.

	�adu funkc� tvaru
	$$ a_0/2 + a_1\cos x + b_1\sin x + a_2\cos(2x) + b_2\sin(2x) + \cdots
		= a_0/2 + \sum_{k=1}^\infty (a_n\cos(nx) + b_n\sin(nx)) $$
	naz�v�me {\bf trigonometrickou �adou}.
	��sla $\{a_n\}_{n=0}^\infty$, $\{b_n\}_{n=1}^\infty$ pak naz�v�me
	{\bf koeficienty trigonometrick� �ady}.
}

\note{
	Jestli�e trigonometrick� �ada bodov� konverguje k funkci $f$, tj.
	$$f(x) = a_0/2 + \sum_{k=1}^\infty (a_k\cos(kx)+b_k\sin(kx))$$
	potom nezbytn� $f \in P_{2\pi}$.
}

\scpart{V�po�et koeficient�}{

Budeme porovn�vat syst�my pro mocninnou �adu $\{1,x,x_2,\ldots\}$
a pro trigonometrickou �adu $\{1,\cos x,\sin x, \ldots\}$.

Jak hledat koeficienty? P�edpokl�dejme, �e
$$f(x) = a_0/2 + \sum_{n=1}^\infty (a_n\cos(nx)+b_n\sin(nx))\qquad x \in \real$$
a nech� �ada konverguje stejnom�rn� v $\real$.

\penalty-1000

\observation{
	Takov� trigonometrick� syst�m je {\it ortogon�ln�} (funkce jsou na sebe kolm�).
	$$ \forall m,n\in\nat \lor n = 0: \int_{-\pi}^\pi \sin(mx)\cos(nx)\,dx = 0 $$
	$$ \int_{-\pi}^\pi \sin(kx)\sin(nx)\,dx = \cases{\pi & $k = n$ \cr 0 & $k \ne n$} $$
	$$ \int_{-\pi}^\pi \cos(kx)\cos(nx)\,dx = \cases{\pi & $k = n$ \cr 0 & $k \ne n$} $$
}

\list{
\namedListItem{$a_0$}{Integrujeme $f(x)$ od $-\pi$ do $\pi$:
	$$ \int_{-\pi}^\pi f(x)\,dx = {a_0 \over 2}\cdot2\pi
		+ \sum_{n=1}^\infty\bigg(
			a_n\undernote{\int_{-\pi}^\pi \cos(nx)\,dx}{0} +
			b_n\undernote{\int_{-\pi}^\pi \sin(nx)\,dx}{0}
		\bigg) = a_0\pi $$
	$$ \Longrightarrow a_0 = 1/\pi \int_{-\pi}^\pi f(x)\,dx $$
}
\namedListItem{$a_m$, $m \ge 1$}{Vyn�sob�me $f(x)$ $\cos(mx)$ a integrujeme od $-\pi$ do $\pi$:
	$$ \int_{-\pi}^\pi f(x)\cos(mx)\,dx
		= a_0/2 \undernote{\int_{-\pi}^\pi\cos(mx)\,dx}{0} + $$
	$$ + \sum_{n=1}^\infty\bigg(
			a_n\undernote{\int_{-\pi}^\pi\cos(nx)\cos(mx)\,dx}{a_m\pi} +
			b_n\undernote{\int_{-\pi}^\pi\sin(nx)\cos(mx)\,dx}{0}
		\bigg) $$
	$$ \Longrightarrow a_m = 1/\pi \int_{-\pi}^\pi f(x)\cos(mx)\,dx $$
}
\namedListItem{$b_m$, $m \ge 1$}{Obdobn� jako $a_n$:
	$$ b_m = 1/\pi \int_{-\pi}^\pi f(x)\sin(mx)\,dx $$
}
}

}


\definition{
	Nech� $f$ je $2\pi$-periodick� a Riemannovsky integrovateln�:
	$$ f \in P_{2\pi} \cap R([-\pi,\pi]) $$
	Pak ��sla $\{a_n\}_{n=0}^\infty$, $\{b_n\}_{n=0}^\infty$, kde
	$$ a_0 = 1/\pi \int_{-\pi}^\pi f(x)\,dx $$
	$$ a_n = 1/\pi \int_{-\pi}^\pi f(x)\cos(nx)\,dx $$
	$$ b_n = 1/\pi \int_{-\pi}^\pi f(x)\sin(nx)\,dx $$
	naz�v�me {\bf Fourierov�mi koeficienty funkce $f$}.
	�adu
	$$ a_0/2 + \sum_{n=1}^\infty \left(a_n\cos(nx) + b_n\sin(nx)\right) $$
	pak naz�v�me {\bf Fourierovou �adou $f$}.

	\medskip

	�ekneme, �e $f \in P_{2\pi}$ je {\bf po ��stech hladk�} ($f \in c_{2\pi}$),
	jestli�e existuje kone�n� mnoho bod� $$ -\pi \le x_1 < x_2 < \cdots < x_m < \pi $$
	takov�ch, �e na intervalech $(x_i,x_{i+1})$ jsou $f$ i $f'$ spojit� a v bodech
	$x_1,\ldots,x_m$ existuj� jednostrann� limity funkc� $f$ i $f'$.

	$f$ je {\bf po ��stech spojit�},
	jestli�e existuj� body $x_0,\ldots,x_n \in [-\pi,\pi)$,
	tak, �e $f$ je spojit� na $(x_j,x_{j+1})$
	a m� jednostrann� limity $\lim_{y\to {x_j}_\pm} f(y)$.

	\shortnote{Pokud $f \in c_{2\pi}$, pak $f'$ je po ��stech spojit�.}
}

\penalty-1000


\theorem{1}{Besselova nerovnost}{
	Nech� $f \in c_{2\pi}$, $a_n,b_n$ jej� Fourierovy koeficienty.
	Potom plat�
	$$ a_0^2/2 + \sum_{n=1}^\infty \left(a_n^2 + b_n^2\right) \le 1/\pi \int_{-\pi}^\pi f(x)^2\,dx $$

	\proof{
		Ozna��me
		$$ s_n(x) := a_0/2 + \sum_{k=1}^n\left(a_k\cos(kx) + b_k\sin(kx)\right) $$

		Z�ejm� plat�:
		$$ \int_{-\pi}^\pi (f(x)-s_n(x))^2\,dx \ge 0 $$
		$$ 2\int_{-\pi}^\pi f(x)s_n(x)\,dx - \int_{-\pi}^\pi s_n(x)^2\,dx \le \int_{-\pi}^\pi f(x)^2\,dx $$

		P�itom plat�:
		$$ \int_{-\pi}^\pi f(x)s_n(x)\,dx
			= a_0/2\int_{-\pi}^\pi \undernote{f(x)}{\pi a_0}\,dx + $$
		$$ + \sum_{k=1}^n \bigg(
				a_k\int_{-\pi}^\pi\undernote{f(x)\cos(kx)}{\pi a_k}\,dx +
				b_k\int_{-\pi}^\pi\undernote{f(x)\sin(kx)}{\pi b_k}\,dx +
			\bigg) = $$
		$$ = \pi\bigg(a_0^2/2 + \sum_{k=1}^n (a_k+b_k)^2\bigg) $$
		Analogicky plat� (\DCV) tak�:
		$$ \int_{-\pi}^\pi s_n(x)^2\,dx = \pi\bigg(a_0^2/2 + \sum_{k=1}^n(a_k^2 + b_k^2)\bigg) $$

		Tedy nutn�:
		$$ \pi\bigg(a_0^2/2 + \sum_{k=1}^n \left(a_k^2 + b_k^2\right) \le \int_{-\pi}^\pi f(x)^2\,dx\bigg) $$
		\qed
	}
}

\theorem{2}{Riemannovo--Lebesgueovo lemma}{
	Nech� $f \in c_{2\pi}$, $a_n,b_n$ jej� Fourierovy koeficienty. Pak
	$$ \lim_{n\to\infty} a_n = \lim_{n\to\infty} b_n = 0 $$
	a speci�ln�
	$$ \lim_{n\to\infty} 1/\pi\int_{-\pi}^\pi f(x)\cos(nx)\,dx
		= \lim_{n\to\infty} 1/\pi \int_{-\pi}^\pi fx)\sin(nx)\,dx
		= 0 \qquad \forall f \in c_{2\pi} $$

	\shortproof{P��m� d�sledek \ias{}{1}{}. \sqed}
}

\penalty-1000

\note{
	Pro v�ty 1 a 2 je p�edpoklad $f \in c_{2\pi}$ zbyte�n� siln�,
	sta�� p�edpokl�dat, �e $f^2 \in R[-\pi,\pi]$.
	Tedy m�me:

	\theorem{1}{}{
	$$ f \in P_{2\pi} \et f^2 \in R[-\pi,\pi] \Rightarrow $$
	$$ a_0^2/2 + \sum_{n=1}^\infty \left(a_n^2 + b_n^2\right) \le 1/\pi\int_{-\pi}^{\pi} f(x)^2\,dx $$
	}

	\theorem{2}{}{
	$$ f \in P_{2\pi} \et f^2 \in R[-\pi,\pi] \Rightarrow $$
	$$ \lim_{n\to\infty} a_n = \lim_{n\to\infty} b_n = 0 $$
	}
}


\note{
	Riemannovo--Lebesgueovo lemma ve skute�nosti tvrd�, �e
	$$ f \in R[-\pi,\pi] \Rightarrow \lim_{n\to\infty} a_n = \lim_{n\to\infty} b_n = 0 $$
}

\lemma{}{
	$$ 1/2 + \cos x + \cos(2x) + \cdots + \cos(nx) = {\sin((n+1/2)x)\over 2 \sin x/2} \qquad x \ne 2k\pi $$

	\shortproof{Jednodu�e indukc�.}

	\impl{
		$$ 1/\pi \int_0^\pi {\sin((n+1/2)t)\over 2\sin t/2}\,dt
			= 1/\pi \int_{-\pi}^0 {\sin((n+1/2)t)\over2\sin t/2}\,dt = 1/2 $$
	}
}

\scpart{Zna�en�:}{
	M�me $f$, pak Fourierovu �adu zna��me
	$$S(f,x) := a_0/2 + \sum_{n=1}^\infty (a_n\cos(nx)+b_n\sin(nx))$$
	$$S_n(f,x) := a_0/2 + \sum_{k=1}^n (a_k\cos(kx)+b_k\sin(kx))$$
	P�ipome�me si, �e
	$$ a_n = 1/\pi \int_{-\pi}^\pi f(t)\cos(nt)\,dx $$
	$$ b_n = 1/\pi \int_{-\pi}^\pi f(t)\sin(nt)\,dx $$

	\penalty-100

	$$ f(x+0) := \lim_{y\to0_+} f(x+y) $$
	$$ f(x-0) := \lim_{y\to0_+} f(x-y) $$

	Pro $f \in P_{2\pi}$ definuji \M{f^*(x) := {f(x+0) + f(x-0) \over 2}}.
}

\notes{
\list{
\listItem{$f$ spojit� v $x$ $\Rightarrow$ $f(x) = f^*(x)$}
\listItem{$f$ m� ``skok'' v $x$ $\Rightarrow$ $f^*(x)$ je aritmetick�m pr�m�rem
	limit zleva a zprava (bez ohledu na hodnotu).}
}
}

\scpart{Nesmysln� vzorec:}{
	$$ {f(x+0)\over2} = 1/\pi \int_0^\pi {f(x+0) \over 2\sin t/2} \sin ((n+1/2)t)\,dt $$
	$$ {f(x-0)\over2} = 1/\pi \int_{-\pi}^0 {f(x-0) \over 2\sin t/2} \sin ((n+1/2)t)\,dt $$
}


\theorem{3}{o bodov� konvergenci Fourierovy �ady}{
	$$ f \in c_{2\pi} \Rightarrow S(f,x) = f^*(x)\qquad x \in \real $$
	Tedy Fourierova �ada po ��stech hladk� funkce
	konverguje {\it bodov�} na cel�m $\real$
	k v�razu $$f^*(x) = {f(x+0) + f(x-0) \over 2}$$

	\proof{
		$$
		\eqalign{
		S_n(f,x) &= 1/\pi \int_{-\pi}^\pi f(t)
				\left\{1/2 + \sum_{k=1}^n(\cos(kt)\cos(kx) + \sin(kt)\sin(kx))\right\}\,dt\cr
		&= 1/\pi \int_{-\pi}^\pi f(t)
				\left\{1/2 + \sum_{k=1}^n\cos(k(t-x))\right\}\,dt \textnote{trigonometrie}\cr
		&= 1/\pi \int_{-\pi}^\pi f(x+t)
				\left\{1/2 + \sum_{k=1}^n\cos(kt)\right\}\,dt \textnote{substituce a $2\pi$-peridodicita}\cr
		&= 1/\pi \int_{-\pi}^\pi f(x+t) {\sin((n+1/2)t)\over2\sin t/2}\,dt \textnote{lemma}} $$

		$$ S_n(f,x) - f^*(x) \becauseof{NV}{=} 1/\pi \int_{-\pi}^0 {f(x+t)-f(x-0)\over2\sin t/2}\sin((n+1/2)t)\,dt + $$
		                                  $$ + 1/\pi \int_0^\pi {f(x+t)-f(x+0)\over2\sin t/2}\sin((n+1/2)t)\,dt $$

		Ozna��me
		$$ G(t) := \cases{\displaystyle{f(x+t) - f(x-0) \over 2\sin t/2} & $t \in [-\pi,0)$\cr
		                  0 & $t = 0$\cr
		                  \displaystyle{f(x+t) - f(x+0) \over 2\sin t/2} & $t \in (0,\pi]$}$$

		$$ \Rightarrow S_n(f,x) - f^*(x) = 1/\pi \int_{-\pi}^\pi G(t) \sin((n+1/2)t)\,dt $$
		$$ = 1/\pi \int_{-\pi}^\pi
			\undernote{G(t) \sin(t/2)}{\textbox{po ��stech}\atop\textbox{spojit�?}}
			\cos(nt)\,dt
		   + 1/\pi \int_{-\pi}^\pi
			\undernote{G(t)\cos(t/2)}{\textbox{po ��stech}\atop\textbox{spojit�?}}
			\sin(nt)\,dt $$

		Chci dok�zat, �e $G$ je po ��stech spojit�.
		K tomu sta�� dok�zat existenci $G(0+0)$, $G(0-0)$.
		$$ G(0+0) = \lim_{t\to0_+} G(t) \becauseof{L'H ``$0\over0$''}{=}
			{\lim_{t\to0_+} f'(x+t) \over \lim_{t\to0_+} 2\cos(t/2)\cdot1/2}
			= f'(x+0) \textnote{v�me, �e existuje} $$
		$$ G(0-0) = f'(x-0) $$

		Tedy $G$ je po ��stech spojit�,
		tud�� tak� $G(t)\sin(t/2)$, $G(t)\cos(t/2)$
		jsou po ��stech spojit�.
		Tedy jejich druh� mocnina je $\in R[-\pi,\pi]$
		a podle \ias{}{2}{} jejich Fourierovy koeficienty jdou k nule.
		$$ \lim_{n\to\infty} (S_n(f,x) - f^*(x)) = 0 $$
		\qed
	}
}


\theorem{4}{o stejnom�rn� konvergenci Fourierovy �ady}{
	$$ f \in c_{2\pi} \et \textbox{spojit� na $\real$} \Longrightarrow
		S(f,x) \conS f(x)\qquad x \in \real $$

	\proof{
		Proto�e
		$$ \sup_{x \in \real} |a_n\cos(nx) + b_n\sin(nx)|
			\le |a_n| + |b_n| $$
		sta�ilo by dok�zat, �e
		$$ \sum_{n=1}^\infty (|a_n|+|b_n|) \textbox{konverguje} $$
		nebo� pak by tvrzen� plynulo z \ref{Weistrass}ova krit�ria.

		V�me, �e $f'$ je definovan� (a spojit�) v�ude
		krom� kone�n� mnoha bod� v $[-\pi,\pi)$.
		Je tak po ��stech spojit� na $\real$, a tedy
		$$ (f')^2 \in R[-\pi,\pi] $$

		Nech� $\alpha_n, \beta_n$ jsou
		Fourierovy koeficienty funkce $f'$,
		tedy
		$$ \alpha_n = 1/\pi \int_{-\pi}^\pi f'(x)\cos(nx)\,dx
			\becauseof{per partes}{=}
			\undernote{1/\pi [f(x)\cos(nx)]_{-\pi}^\pi}{0}
			- 1/\pi \int_{-\pi}^\pi f(x)(-n\sin(nx))\,dx $$
		$$ = n/\pi \int_{-\pi}^\pi f(x)\sin(nx)\,dx = nb_n $$
		a obdobn�
		$$ \beta_n = -na_n $$

		Z \ref{Bessel}ovy nerovnosti pro $f'$ m�me
		$$ \sum_{n=1}^\infty (\alpha_n^2 + \beta_n^2) \le 1/\pi \int_{-\pi}^\pi f'(x)^2\,dx < \infty $$

		V�me, �e
		$$ \sum (\alpha n^2 + \beta n^2) < \infty\qquad \alpha_n = nb_n, \beta_n = -na_n $$
		a chceme
		$$ \sum(|a_n| + |b_n|) < \infty $$

		Tedy jest:
		$$ \sum_{n=1}^\infty (|a_n| + |b_n|)
			= \sum_{n=1}^\infty (|\alpha_n|/n + |\beta_n|/n)
			\becauseof{(na dluh)}{=}
			\sum_{n=1}^\infty |\alpha_n|/n + \sum_{n=1}^\infty |\beta_n|/n $$
		Do hry kone�n� po mnoha m�s�c�ch p�ich�z� d�vn� \ref{Cauchyova nerovnost}:
		$$ \le \left(\sum_{n=1}^\infty \alpha_n^2\right)^{1/2}
		       \left(\sum_{n=1}^\infty 1/n^2\right)^{1/2}
		    +  \left(\sum_{n=1}^\infty \beta_n^2\right)^{1/2}
		       \left(\sum_{n=1}^\infty 1/n^2\right)^{1/2}
		   < \infty $$
		\qed
	}
}


\theorem{5}{o lok�ln� stejnom�rn� konvergenci Fourierovy �ady}{
	Nech� $f \in c_{2\pi}$, pak jej� Fourierova �ada konverguje bodov� k $f^*$.
	Tato konvergence je lok�ln� stejnom�rn� na ka�d�m intervalu $(\alpha,\beta)$,
	na kter�m je $f$ spojit�.

	\shortproof{Bez. \sqed}
}


\example{
	$$ f(x) = \pi - x\qquad x \in [-\pi,\pi) $$

	$2\pi$-periodick� roz���en�: \M{f(x) = \pi - (x+2k\pi)\qquad x \in [(2k-1)\pi,(2k+1)\pi)}

	$$f \in c_{2\pi} \Rightarrow \undernote{s(f,x)}{0} \to f^*(x) = \cases{f(x) & $x \ne 2k\pi$ \cr \pi & $x = 2k\pi$}$$

	\exercise{
		$$ a_0 = 2\pi,\ a_n = 0,\ n \ge 1,\ b_n = {2(-1)^n\over n} $$
		$$ S(f,x) = \pi + 2\sum_{n=1}^\infty {(-1)^n\over n} \sin (nx) $$
		$$ = \pi - 2\sin x + \sin(2x) - 2/3\sin(3x) + \cdots $$
	}
}

\penalty-1000

\theorem{6}{o derivov�n� Fourierovy �ady}{
	Nech� $f \in P_{2\pi}$ a m� Fourierovy koeficienty
	$\{a_n\}_{n=0}^\infty$, $\{b_n\}_{n=1}^\infty$.
	D�le nech� $$ \sum_{n=1}^\infty n(|a_n|+|b_n|) $$
	konverguje. Potom $f \in c^1$ (tj. m� spojitou derivaci)
	a �ada $$ \sum_{n=1}^\infty n(\cos(nx)b_n - \sin(nx)a_n) \conSi{f'}{\real} $$

	\proof{
		$$ f_n(x) := a_n\cos(nx) + b_n\sin(nx) $$
		$$ f_n'(x) = n(b_n\cos(nx)-a_n\sin(nx)) $$
		$$ \sup_{x \in \real} |f_n'(x)| \le n(|a_n| + |b_n|) =: s_n $$
		p�i�em� $\sum s_n < \infty$ (z p�edpokladu).
		Tedy dle \ref{Weistrass}ova krit�ria
		$\sum_{n=1}^\infty f_n' \conSi{}{\real}$.

		$$ g:=\sum_{n=1}^\infty f_n'(x) $$
		Nav�c pro pevn� bod $x=0$ plat�
		$$ \left|\sum_{n=1}^\infty f_n(0)\right| = \left|\sum a_n\right| \le \sum|a_n| < \infty $$
		Tedy dle v�ty o z�m�n� sumy a derivace $\sum_{n=1}^\infty f_n \conSL f$,
		proto�e ale $f \in P_{2\pi}$, tak $\sum f_n \conSi{f}{\real}$ a nav�c $f'=g$.
		\qed
	}
}


\note{
	Je-li $f$ sud�, $b_n=0$ a $a_n = 2/\pi\int_0^\pi f(x)\cos(nx)\,dx$.

	Je-li $f$ lich�, $a_n=0$ a $b_n = 2/\pi\int_0^\pi f(x)\sin(nx)\,dx$.

	Fourierovy koeficienty se tedy kr�sn� po��taj�, je-li funkce symetrick�.
}



\bend
