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\title[Matematika 4]
{Matematika 4}

\subtitle{3. vaja}

\author[Borut] % 
{B.~Jur\v{c}i\v{c}~Zlobec\inst{1}}

\institute[Univerza v Ljubljani Slovenia] % 
{
  \inst{1}%
    Univerza v Ljubljani,\\Fakulteta za Elektrotehniko\\ 
    1000 Ljubljana, Tr\v{z}a\v{s}ka 25, Slovenija
}


\date[\today] % 
{Matematika FE, Ljubljana, Slovenija \today}

\subject{Matematika}

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\section{Specialne funkcije}
\subsection{Re\v{s}evanje diferencialnih ena\v{c}b z vrsto}

\begin{frame}{Re"si diferencialno ena"cbo z nastavkom 
$y(x)=\sum_{i=0}^\infty a_i x^i$\\
in  zapi"si prvih pet "clenov vrste razli"cnih od ni"c. 
	$y''(x)+x^2y(x)=0,\quad y(0)=1,\quad\text{in}\quad y'(0)=-1/2.$
}
\begin{itemize}
\item $y(x)=\sum_{i=0}^\infty a_i x^i$
in $y''(x)=\sum_{i=2}^\infty i (i-1) a_i x^{i-2}$.
\item $\sum_{i=2}^\infty i (i-1) a_i x^{i-2} + \sum_{i=0}^\infty a_i x^{i+2}=0$,
\item $(i+1)(i+2)a_{i+2} + a_{i-2}=0$, $i=2,3,\dots$ $a_2=a_3=0$,
\item $\displaystyle a_{i+2}=-\frac{a_{i-2}}{(i+1)(i+2)}$, $i=2,3,\dots$, 
\item \formulas{$\displaystyle
y(x)=1-\frac{x}{2}-\frac{x^4}{12}+\frac{x^5}{40}+\frac{x^8}{672}\dots$}. 
\end{itemize}
\end{frame}

\begin{frame}{Re"si diferencialno ena"cbo z nastavkom
$y(x)=x^r\sum_{i=0}^\infty a_i x^i$.\\
        $4 x y''(x)+2y'(x)+y(x)=0$, $y(0)=1$ in
	$\lim_{x\to 0} y'(x)=-\frac{1}{2}.$}
\begin{itemize}
\item $\sum_{i=0}^\infty \left(4 (i+r)(i+r-1)a_i+
	2 a_i (i+r) + a_i x\right)x^{i+r-1}=0.$
\item Prvi pogoj $(4r(r-1)+2r)a_0=0$
\item $(4(i+r)(i+r-1)+ 2 (i+r))a_i + a_{i-1}=0$, $i=1,3,\dots$
\item $a_{i}=-\frac{a_{i-1}}{2(i+r)(2i+2r-1)}$, $i=1,3,\dots$
\item "Ce je $a_0\ne0$, potem je $r(2r-1)=0$, $r=0$ ali $r=\frac{1}{2}$. 
\item Za $r=0$, $a_n=-\frac{a_n-1}{2i(2i-1)}=(-1)^i\frac{a_0}{(2i)!}$,
\item za $r=\frac{1}{2}$, $a_n=(-1)^i\frac{a_0}{(2i+1)!}$,
\item $a_0^{(1)}\left( 1-\frac{x}{2!}+\frac{x^2}{4!}-\dots\right)
	+a_0^{(2)}\left(\sqrt{x}-\frac{\sqrt{x}x}{3!}+
	\frac{\sqrt{x}x^2}{5!}-\dots\right)$.
\item \formulas{$y(x)=\cos\sqrt{x}$}. 
\end{itemize}
\end{frame}

\subsection{Re\v{s}evanje diferencialnih ena\v{c}b z uvebo nove spremenljivke}

\begin{frame}{Uvedi novo odvisno spremenljivko v diferencialno ena"cbo.
\formulas{$(x^2-1)y''(x)+\left(4x-\frac{2}{x}\right)y'(x)-10y(x),\quad y=\frac{z}{x}.$}}
\begin{itemize}
	\item $y'(x)=\frac{z'(x)}{x}-\frac{z}{x^2}$, $y''(x)=\frac{z''(x)}{x}-\frac{2 z'(x)}{x^2}+\frac{2 z(x)}{x^3}$.
	\item $(x^2-1)\left(\frac{z''}{x}-\frac{2 z'}{x^2}+\frac{2 z}{x^3}\right)+\left(4x-\frac{2}{x}\right)
	\left(\frac{z'}{x}-\frac{z}{x^2}\right)-10\frac{z}{x}$.
	\item \formulas{$\displaystyle (1-x^2)z''(x)-2x z'(x)+12z(x)=0$}.
\end{itemize}
\end{frame}

\begin{frame}{Besslova diferencialna ena"cba}
\formulas{
$$
	x^2y''(x)+xy'(x)+(x^2-\nu^2)y(x)=0
$$
}
\begin{itemize}
	\item Parameter $\nu$ ni celo "stevilo, $y(x)=AJ_\nu(x)+BJ_{-\nu}(x)$.
	\item Parameter $\nu=n$ je celo "stevilo, $f(x)=AJ_n(x)+BN_n(x)$.
        \item Parameter $\nu=\frac{1}{2}$, $y(x)=
		\sqrt{\frac{2}{\pi x}}(A\sin(x)+B\cos(x))$
	\item Rekurzivne formule.
		$$J_{\nu-1}+J_{\nu+1}=\frac{2\nu}{x}J_\nu(x)$$
	\item Odvodi Besslovih funkcij.
		$$\frac{dJ_\nu(x)}{dx}=-\frac{\nu}{x}J_{\nu}(x)+J_{\nu-1}(x) $$
\end{itemize}
\end{frame}


\begin{frame}{Poi"s"ci re"sitev diferencialne ena"cbe
\formulas{$x^2y''(x) -xy'(x) +(1+x^2)y(x)=0$, $y(0)=0$, $y'(0)=1$,\\ z uvedbo  $y(x)=x z(x)$.}}
\begin{itemize}
	\item $y'(x)=z(x)+x z'(x)$, $y''(x)=2z'(x)+x z''(x)$. 
	\item $x^2(2z'(x)+x z''(x))-x(z(x)+x z'(x))+(1+x^2)x z(x)=0$
	\item $2x z'(x) + x^2 z''(x)-z(x) -x z'(x) +z(x) +x^2 z(x)=0$
	\item $x^2z''(x)+x z'(x)+x^2 z(x)=0$
	\item $y(x)=x z(x)=A x J_0(x)+ B x N_0(x)$, $z(0)=0$, $z'(0)=1$.
	\item \formulas{$y(x)=x J_0(x)$}.
\end{itemize}
\end{frame}

\begin{frame}{Poi"s"ci re"sitev diferencialne ena"cbe
\formulas{$x^2y''(x) + xy'(x) +(4x^4-1)y(x)=0$,\\ 
$\lim_{x\to0}y(x)=0$, $\lim_{x\to0}y'(x)=1$, z uvedbo  $\xi=x^2$.}}
\begin{itemize}
	\item $2x\,dx=d\xi$, $\frac{d\xi}{dx}=2\sqrt{\xi}$, 
	$\frac{d}{dx}=\frac{d}{d\xi}\frac{d\xi}{dx}=2\sqrt{\xi}\frac{d}{d\xi}$. 
	\item $y'(x)=2\sqrt{\xi}\frac{dy}{d\xi}=2\sqrt{\xi}y_\xi$. 
	\item $y''(x)=2\sqrt{\xi}\left(2\sqrt{\xi}y_\xi\right)_\xi=
	2\sqrt{\xi}\left(\frac{1}{\sqrt{\xi}}y_\xi+2\sqrt{\xi}y_{\xi\xi}\right)=\rightarrow$
	\item $2y_\xi+4\xi y_{\xi\xi}$, $4\xi^2y_{\xi\xi}(\xi)+4\xi y_\xi(\xi)+(4\xi^2-1)y(\xi)=0$.
	\item $\displaystyle \xi^2y_{\xi\xi}(\xi)+\xi y_\xi(\xi)+(\xi^2-\frac{1}{4})y(\xi)=0$
	\item $y(\xi)=A J_{1/2}(\xi)+ B J_{-1/2}(\xi)=
	A'\frac{\sin\xi}{\sqrt{\xi}}+B'\frac{\cos\xi}{\sqrt{\xi}}$.
	\item \formulas{$y(x)=\frac{1}{x}\sin(x^2)$}.
\end{itemize}
\end{frame}


\subsection{Ortogonalni polinomi}
\begin{frame}{Legendrovi polinomi}
	\begin{itemize}
		\item Diferencialna ena"cba: 
		\formulas{$\displaystyle
		(1-x^2)y''(x)-2x y'(x)=-n(n+1)y(x)$}.
		\item Interval: 
		\formulas{$[-1,1]$}. Ute"z: \formulas{$\rho(x)=1$}.
		\item Rodriguesova formula: 
		\formulas{$\displaystyle P_n(x)=\frac{1}{2^n n!}
		\frac{d^n}{dx^n}\left(x^2-1\right)^{n} $}. 
		\item Ortogonalnost: 
		\formulas{$\displaystyle(P_m,P_n)=\int_{-1}^1P_m(x)P_n(x)dx=
		\frac{2}{2n+1}\delta_{mn}$}. 
	\end{itemize}
\end{frame}

\begin{frame}{Poi"s"ci re"sitev diferencialne ena"cbe
\formulas{$y''(x) - \tan(x)y'(x) + 2y(x)=0$,\\ 
$y(x)=0$, $y'(x)=1$, z uvedbo  $\xi=\sin(x)$.}}
\begin{itemize}
	\item $d\xi=\cos(x)dx$,  
	$\frac{d}{dx}=\frac{d}{d\xi}\frac{d\xi}{dx}=\cos(x)\frac{d}{d\xi}$. 
	\item $y'(x)=\cos(x)\frac{dy}{d\xi}$. 
	\item $y''(x)=\cos(x)\frac{d}{d\xi}\left(\cos(x)\frac{dy}{d\xi}\right)$
	\item $y''(x)=\cos^2{x}\frac{d^2y}{d\xi^2}+\cos(x)\frac{dy}{d\xi}\frac{d\cos(x)}{d\xi}$.
	\item $\frac{d\cos(x)}{d\xi}=-\sin(x)\frac{dx}{d\xi}=-\tan(x)$
	\item $y''(x)=\cos^2(x)\frac{d^2y}{d\xi^2}-\sin(x)\frac{dy}{d\xi}.$
	\item $\displaystyle (1-\xi^2)y''(\xi)-2\xi y'(\xi)+2y(\xi)=0.$
	\item \formulas{$y(\xi)=\xi\rightarrow y(x)=\sin(x)$}.
\end{itemize}
\end{frame}

\begin{frame}{Poi"s"ci polinomsko re"sitev diferencialne ena"cbe
\formulas{$(x^2-1)y''(x)+\frac{2}{x}y'(x) -(2+\frac{2}{x^2})y(x)=0$, z uvedbo  $y(x)=x z(x)$.}}
\begin{itemize}
	\item $y'(x)=z(x)+x z'(x)$, $y''(x)=2z'(x)+x z''(x)$. 
	\item $(x^2-1)\left(2z'(x)+xz''(x)\right)+\frac{2}{x}\left(z(x)+xz'(x)\right)-\left(2+\frac{2}{x^2}\right)xz(x)=0$.
	\item $x(x^2-1)z''(x)+(2x^2-2+2)z'(x)+(\frac{2}{x}-2x-\frac{2}{x})z(x)$.
	\item $\displaystyle(1-x^2)z''(x)-2xz'(x)+2z(x)=0$, $2=l(l+1)$, $l=1$ 
	od tod $z(x)=P_1(x)=x$ in $y(x)=x^2$.
	\item \formulas{$\displaystyle y(x)=x^2$}.
\end{itemize}

\end{frame}


\begin{frame}{Hermitovi polinomi}
	\begin{itemize}
		\item Diferencialna ena"cba: \formulas{$\displaystyle y''(x)-x y'(x)=-ny(x)$}.
                \item Interval: 
		\formulas{$(-\infty,\infty)$}. 
		Ute"z: \formulas{$\rho(x)=e^{-x^2/2}$}.
                \item Rodriguesova formula: 
		\formulas{$\displaystyle H_n(x)=(-1)^ne^{x^2/2}
                \frac{d^n}{dx^n}e^{-x^2/2} $}.
                \item Ortogonalnost: \formulas{$\displaystyle(H_m,H_n)=
		\int_{-\infty}^\infty H_m(x)H_n(x)e^{-x^2/2}dx=
                \sqrt{\pi}n!\delta_{mn}$}.
	\end{itemize}
\end{frame}

\begin{frame}{Laguerrovi polinomi}
	\begin{itemize}
		\item Diferencialna ena"cba: 
		\formulas{$\displaystyle xy''(x)+(1-x)y'(x)=-ny(x)$}.
                \item Interval: 
		\formulas{$[0,\infty)$}. Ute"z: \formulas{$\rho(x)=e^{-x}$}.
                \item Rodriguesova formula: 
		\formulas{$\displaystyle L_n(x)=\frac{1}{n!}e^{x}
                \frac{d^n}{dx^n}\left(x^ne^{-x}\right) $}.
                \item Ortogonalnost: \formulas{$\displaystyle(L_m,L_n)=
		\int_{0}^\infty L_m(x)L_n(x)e^{-x}dx=\delta_{mn}$}.
	\end{itemize}
\end{frame}

\begin{frame}{Polinomi "Cebi"seva}
	\begin{itemize}
		\item Diferencialna ena"cba: \formulas{$\displaystyle (1-x^2)y''(x)-xy'(x)=-n^2y(x)$}.
                \item Interval: \formulas{$[-1,1]$}. Ute"z: 
		\formulas{$\rho(x)=\frac{1}{\sqrt{1-x^2}}$}.
                \item Trigonometri"cna formula: 
		\formulas{$\displaystyle T_n(x)=cos(n\arccos(x))$}.
		\item Rekurzivna formula: 
		\formulas{$\displaystyle T_0(x)=1,\quad T_1(x)=x,\quad 
			T_{n+1}=2xT_n(x)-T_{n-1}(x)$}.
                \item Ortogonalnost: 
		\formulas{$\displaystyle (m | n>0), (T_m,T_n)=
		\int_{-1}^1T_m(x)T_n(x)\frac{dx}{\sqrt{1-x^2}}=
		\frac{\pi}{2}\delta_{mn}, (T_0,T_0)=\pi$}.
		\item Lastnosti. 
		\emph{Med vsemi polinomi iste stopnje z istim vodilnim 
		koeficientom ima polinom "Cebi"seva na intervalu $[-1,1]$ 
		najmanj"se absolutne 
		vrednosti ekstremov.}
	\end{itemize}
\end{frame}


\begin{frame}{Dolo"ci konstante $\alpha_{ij}$, tako, da bodo funkcije:}
\formulas{$f_0(x)=\alpha_{00}$, $f_1(x)=\alpha_{10}+\alpha_{11}x$ 
	in $f_2(x)=\alpha_{20}+\alpha_{21}x+\alpha_{22}x^2$ 
	ortonormirane na intervalu $[-1,1]$}.\\[2mm]
\begin{itemize}
	\item $\hat{f}_0(x)=1$, $\|\hat{f}_0(x)\|^2=
	\int_{-1}^1\hat{f}_0^2(x)dx=2$. 
	\item $\hat{f}_1(x)=\alpha\hat{f}_0(x)+x\rightarrow$
	\item $\left(\hat{f}_1(x),\hat{f}_0(x)\right)=
		\alpha\left(\hat{f}_0(x),\hat{f}_0(x)\right)+
		\left(x,\hat{f}_0(x)\right)=0$.
	\item $\alpha=-\frac{\left(x,\hat{f}_0(x)\right)}
		{\left(\hat{f}_0(x),\hat{f}_0(x)\right)}=
	-\frac{\int_{-1}^{1}x\,dx}{\int_{-1}^1dx}=0$,
	$\hat{f}_1(x)=x$, $\|\hat{f}_1(x)\|^2=\frac{2}{3}$.
	\item $\hat{f}_2(x)=\alpha_0\hat{f}_0+
		\alpha_1\hat{f}_1(x)+x^2$, $\alpha_i=
	-\frac{\left(\hat{f}_i(x),x^2\right)}{\left(\hat{f}_i(x),
		\hat{f}_i(x)\right)}$.
	\item $\hat{f}_2(x)=-\frac{1}{3}+x^2$, 
		$\|\hat{f}_2(x)\|^2=\frac{8}{45}$. 
		\formulas{Normirani $\rightarrow$}
	\item \formulas{$f_0(x)=\frac{1}{\sqrt{2}},
		\quad f_1(x)=\sqrt{\frac{3}{2}}x,\quad 
		f_2(x)=\sqrt{\frac{5}{8}}(3x^2-1).$}
\end{itemize}
\end{frame}

\begin{frame}{Dolo"ci konstante $\alpha_{ij}$, tako, da bodo funkcije:}
\formulas{$f_0(x)=\alpha_{00}$, $f_1(x)=\alpha_{10}+\alpha_{11}x$ 
	in $f_2(x)=\alpha_{20}+\alpha_{21}x+\alpha_{22}x^2$ 
	ortonormirane na intervalu $[0,2]$}.\\[2mm]
\begin{itemize}
	\item $\hat{f}_0(x)=1$, $\|\hat{f}_0(x)\|^2=\int_0^2\hat{f}_0^2(x)dx=2$. 
	\item $\hat{f}_1(x)=\alpha\hat{f}_0(x)+x\rightarrow$
	\item $\left(\hat{f}_1(x),\hat{f}_0(x)\right)=
		\alpha\left(\hat{f}_0(x),\hat{f}_0(x)\right)+\left(x,\hat{f}_0(x)\right)=0$.
	\item $\alpha=-\frac{\left(x,\hat{f}_0(x)\right)}{\left(\hat{f}_0(x),\hat{f}_0(x)\right)}=
	-\frac{\int_{0}^{2}x\,dx}{\int_{0}^2dx}=-1$,
	$\hat{f}_1(x)=-1+x$, $\|\hat{f}_1(x)\|^2=\frac{2}{3}$.
	\item $\hat{f}_2(x)=\alpha_0\hat{f}_0+\alpha_1\hat{f}_1(x)+x^2$, $\alpha_i=
	-\frac{\left(\hat{f}_i(x),x^2\right)}{\left(\hat{f}_i(x),\hat{f}_i(x)\right)}$.
	\item $\hat{f}_2(x)=\frac{2}{3}-2x+x^2$, 
	$\|\hat{f}_2(x)\|^2=\frac{8}{45}$. \formulas{Normirani $\rightarrow$}
	\item \formulas{$f_0(x)=\frac{1}{\sqrt{2}},\quad
	f_1(x)=\sqrt{\frac{3}{2}}(x-1),\quad 
	f_2(x)=\sqrt{\frac{5}{8}}(3(x-1)^2-1).$}
\end{itemize}
\end{frame}

\begin{frame}{Dolo"ci konstante $\alpha_{ij}$, tako, da bodo funkcije:}
\formulas{$f_0(x)=\alpha_{00}$, $f_1(x)=\alpha_{10}+\alpha_{11}x$ in
$f_2(x)=\alpha_{20}+\alpha_{21}x+\alpha_{22}x^2$ 
ortonormirane na intervalu $[0,2]$, z ute"zjo $\rho(x)=x$}.\\[2mm]
\begin{itemize}
	\item $\hat{f}_0(x)=1$,
	$\|\hat{f}_0(x)\|^2=\int_0^2x\hat{f}_0^2(x)dx=2$. 
	\item $\hat{f}_1(x)=\alpha\hat{f}_0(x)+x\rightarrow$
	\item $\left(\hat{f}_1(x),\hat{f}_0(x)\right)=
		\alpha\left(\hat{f}_0(x),\hat{f}_0(x)\right)+
		\left(x,\hat{f}_0(x)\right)=0$.
	\item $\alpha=-\frac{\left(x,\hat{f}_0(x)\right)}
		{\left(\hat{f}_0(x),\hat{f}_0(x)\right)}=
	-\frac{\int_{0}^{2}x^2\,dx}{\int_{0}^2x\,dx}=-\frac{4}{3}$,
	$\hat{f}_1(x)=-\frac{4}{3}+x$, $\|\hat{f}_1(x)\|^2=\frac{4}{9}$.
	\item $\hat{f}_2(x)=\alpha_0\hat{f}_0+
		\alpha_1\hat{f}_1(x)+x^2$, $\alpha_i=
	-\frac{\left(\hat{f}_i(x),x^2\right)}
		{\left(\hat{f}_i(x),\hat{f}_i(x)\right)}$.
	\item $\hat{f}_2(x)=\frac{6}{5}-\frac{12}{5}x+x^2$, 
		$\|\hat{f}_2(x)\|^2=\frac{8}{45}$. 
	\formulas{Normirani $\rightarrow$} 
	\item \formulas{$f_0(x)=\frac{1}{\sqrt{2}},\quad
	f_1(x)=-2+\frac{3x}{2},\quad 
	f_2(x)=\sqrt{\frac{5}{8}}(6-12x+5x^2).$}
\end{itemize}

\end{frame}

\begin{frame}{Preizkusi ortogonalnost Lagendrejevih polinomov\\ 
	$P_2(x)$ in $P_3(x)$}
\begin{itemize}
	\item $P_2(x)=\frac{1}{2} \left(3 x^2-1\right)$,
	\item $P_3(x)=\frac{1}{2} \left(5 x^3-3 x\right)$.
	\item $(P_2(x),P_3(x)=\int_{-1}^1P_2(x)P_3(x)dx=\rightarrow$
	\item $\int_{-1}^1\frac{15 x^5}{4}-\frac{7 x^3}{2}+\frac{3 x}{4}=0$.
	\item \formulas{$\left(P_2(x),P_3(x)\right)=0$}.
\end{itemize}
\end{frame}	

\begin{frame}{Poi"s"ci normo Legendrovih polinomov $P_1(x)$ in $P_3(x)$}
\begin{itemize}
        \item $P_1(x)=x\rightarrow$
	\item $\|P_1(x)\|^2=\int_{-1}^1 |P_1(x)|^2dx=\frac{2}{3}$.
        \item $P_3(x)=\frac{1}{2} \left(5 x^3-3 x\right)\rightarrow$
	\item $\|P_3(x)\|^2=\frac{1}{4}\int_{-1}^1\left(5 x^3-3 x\right)^2dx=
		\frac{2}{7}$.
        \item \formulas{$\displaystyle\|P_1(x)\|=\sqrt{\frac{2}{3}},\|P_3(x)\|=
		\sqrt{\frac{2}{7}}$}.
\end{itemize}
\end{frame}

\begin{frame}{Dolo"ci koeficienta $a_1$ in $a_3$ v razvoju funkcije $f(x)$}
	\formulas{po Legendrovih polinomih na intervalu $[-1,1]$. 
	Funkcija $f(x)=\begin{cases} -1,& -1\le x < 0\\ 1, & 0<x\le 1\\ 0,& x=0\end{cases}$}. 
	\\[2mm]
	\begin{itemize}
		\item $\displaystyle f(x)=\sum_{n=0}^\infty a_n P_n(x)$, za $x\in[-1,1]$. 
		\item $a_1=(f(x),P_1(x))/\|P_1(x)\|^2$ in $a_3=(f(x),P_3(x))/\|P_3(x)\|^2$.
		\item $\displaystyle a_1=\frac{3}{2}\int_{-1}^1 x
		f(x)dx=\frac{3}{2}\int_0^1x\,dx=\frac{3}{4}$.
		\item $\displaystyle a_3=\frac{7}{2}\int_{-1}^1 
		\frac{1}{2}\left(5 x^3-3 x\right) f(x)=\frac{7}{8}
		\int_0^1 x^3-3 x\,dx=-\frac{7}{16}$
		\item \formulas{$\displaystyle a_1=\frac{3}{4},
		\quad a_3=-\frac{7}{16}$}
	\end{itemize}
\end{frame}

\begin{frame}{Aproksimacija $f(x)$ z Legendrovimi polynomi}
\begin{center}
	\includegraphics[scale=1]{legendre_poly.pdf}
\end{center}
\end{frame}


\begin{frame}{Ena izmed re"sitev diferencialne ena"cbe je polinom.
Dolo"ci njegovo stopnjo.}
\formulas{$\displaystyle y''(x)-xy'(x)+3y=0$}\\[2mm]
\begin{itemize}
	\item  Hermitova diferencialna ena"cba za $n=3$.
	\item  $y(x)=H_3(x)$.
	\item  Stopnja je 3.  
\end{itemize}
\end{frame}

\begin{frame}{Ena izmed re"sitev diferencialne ena"cbe je polinom.
Dolo"ci njegovo stopnjo.}
\formulas{$\displaystyle y''(x)-2xy'(x)+6y=0$}\\[2mm]
\begin{itemize}
	\item  $a_{n+2}=2\frac{(n-3)a_n}{(n+2)(n+1)}$. 
	\item  Stopnja je 3. 
\end{itemize}
\end{frame}




\subsection{Besslove funkcije}
\begin{frame}{Poi"s"ci Laplaceovo transformacijo Besslove funkcije $J_0(x)$.}
\begin{itemize}
	\item Funkcija $y=J_0(x)$ je re"sitev diferencialne ena"cbe 
	$$xy''(x)+y'(x)+xy(x)=0,\quad y(0)=1\quad \text{in}\quad y'(0)=0.$$
	\item Laplaceova transformacija diferencialne ena"cbe je ena"cba 
	$$-\frac{d}{ds}\left(s^2 Y(s)-s\right)+sX(s)-1-Y'(s)=0,$$
	$$Y'(s)(s^2+1)+sY(s)=0.$$
	\item Rezultat je diferencialna ena"cba prvega reda z lo"cljivimi 
	spremenljivkami. 
	$$\frac{d Y}{Y}=-\frac{s\,ds}{1+s^2}\rightarrow \ln 
		Y(s)=-\frac{1}{2}\ln(1+s^2)\rightarrow.$$ 
	\item \formulas{$\displaystyle Y(s)=\frac{1}{\sqrt{1+s^2}}$.}
\end{itemize}
\end{frame}

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