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\begin{center}
  {\Large Übungsblatt 2} \\
  Jan Hendrik Dithmar \\
  Übungsgruppe 1 - Mo, 14-16 Uhr \\
  Janko Böhm
\end{center}

\newpage

\subsection* {Aufgabe 1}
$ \mathbb{R} $-Vektorraum $ U = \mathbb{R}[x]_{\le 3} $ mit der Basis $ \{ 1,x,x^2,x^3 \} $, \\
also $ dim \, U = 4 $. \\
(a) \\
Skalarprodukt $ \displaystyle <p,q> = \int_{-1}^1 p(x)q(x)x^2 dx, \; \; \; x^2 \ge 0 $ und stetig in $ [-1,1] $. \\
Bestimmung einer Orthonormalbasis von $ U $: \\
$ \{ 1=x^0, x^1, x^2, x^3 \} $ Basis von $ U = \mathbb{R}[x]_{\le 3} $. \\
$ \{ b_0(x), b_1(x), b_2(x), b_3(x) \} $ Orthogonalbasis von $U$. \\
$ \{ e_0(x), e_1(x), e_2(x), e_3(x) \} $ Orthonormalbasis von $U$. \\
Beginn: $\displaystyle 1=b_0(x) \Rightarrow e_0(x) = \frac{1}{||b_0(x)||} \cdot b_0(x)$. \\
Fortsetzung:
\begin{eqnarray*}
b_{n+1}(x) & = & x^{n+1} - \sum_{\nu=0}^n <x^{n+1}, e_{\nu}> e_{\nu}(x) \; n=0,1,2 \\
e_{n+1}(x) & = & \frac{1}{||b_{n+1}(x)||} \cdot b_{n+1}(x)
\end{eqnarray*}
wobei $ \displaystyle ||b_{n+1}(x)|| = \sqrt{<b_{n+1}(x), b_{n+1}(x)} $. \\ \\
0. Bestimmung von $e_0(x)$:
\begin{eqnarray*}
b_0(x) & = & 1 \\
<b_0,b_0> = <1,1> & = & \int_{-1}^1 x^2 dx = \left[ \frac{1}{3}x^3 \right]_{-1}^1 = \frac{1}{3} + \frac{1}{3} = \frac{2}{3} \\
\Rightarrow ||b_0|| & = & \sqrt{<b_0,b_0>} = \sqrt{\frac{2}{3}} \\
\Rightarrow e_0(x) & = & \frac{1}{||b_0||}\cdot b_0(x) = \frac{1}{\sqrt{\frac{2}{3}}} \cdot 1 = \sqrt{\frac{3}{2}} = \frac{1}{2} \sqrt{6} \\
e_0(x) & = & \frac{1}{2} \sqrt{6}
\end{eqnarray*}
1. Bestimmung von $e_1(x)$:
\begin{eqnarray*}
b_1(x) & = & x-<x,e_0>e_0(x) \\
<x,e_0> & = & <x, \frac{1}{2} \sqrt{6}> = \int_{-1}^1 x \frac{1}{2} \sqrt{6}x^2 dx = \frac{1}{2} \sqrt{6}\cdot \int_{-1}^1 x^3 dx = 0 \; \; \text{ da } \int_{-1}^1 x^3 dx = 0 \\
\Rightarrow b_1(x) & = & x \\
<b_1,b_1> & = & <x,x> = \int_{-1}^1 x\cdot x\cdot x^2 dx = \int_{-1}^1 x^4 dx = \left[ \frac{1}{5}x^5 \right]_{-1}^1 = \frac{2}{5} \\
\Rightarrow ||b_1|| & = & \sqrt{<b_1,b_1>} = \sqrt{\frac{2}{5}} \\
\Rightarrow e_1(x) & = & \frac{1}{||b_1||} \cdot b_1(x) = \frac{1}{\sqrt{\frac{2}{5}}} \cdot x = \sqrt{\frac{5}{2}} \cdot x = \frac{1}{2}\sqrt{10}\cdot x \\
e_1(x) & = & \frac{1}{2}\sqrt{10} \cdot x
\end{eqnarray*}
2. Bestimmung von $e_2(x)$:
\begin{eqnarray*}
b_2(x) & = & x^2 - \sum_{\nu = 0}^1 <x^2, e_{\nu}>e_{\nu}(x) \\
b_2(x) & = & x^2 - <x^2,e_0>e_0(x) - <x^2,e_1>e_1(x) \\
<x^2,e_0> & = & <x^2, \frac{1}{2}\sqrt{6}> = \int_{-1}^1 x^2 \frac{1}{2}\sqrt{6}x^2 dx = \frac{1}{2}\sqrt{6} \int_{-1}^1 x^4 dx = \frac{1}{2}\sqrt{6} \left[ \frac{1}{5}x^5 \right]_{-1}^1 = \frac{1}{2}\sqrt{6}\cdot \frac{2}{5} \\
& = & \frac{1}{5}\sqrt{6} \\
<x^2,e_1> & = & <x^2,\frac{1}{2}\sqrt{10}x> = \int_{-1}^1 x^2 \frac{1}{2}\sqrt{10} \cdot x \cdot x^2 dx = \frac{1}{2}\sqrt{10} \cdot \int_{-1}^1 x^5 dx = \frac{1}{2}\sqrt{10} \cdot 0 = 0 \\
\Rightarrow b_2(x) & = & x^2 - \frac{1}{5}\sqrt{6} \cdot \frac{1}{2}\sqrt{6} = x^2 - \frac{3}{5} \\
<b_2,b_2> & = & <x^2 - \frac{3}{5}, x^2 - \frac{3}{5}> = \int_{-1}^1 \left( x^2-\frac{3}{5} \right) ^2 x^2 dx \\
& = & \int_{-1}^1 \left( x^6-\frac{6}{5}x^4 + \frac{9}{25}x^2 \right) dx = \left[ \frac{1}{7}x^7 - \frac{6}{25} x^5 + \frac{3}{25}x^3 \right] _{-1}^1 = \frac{2}{7}-\frac{6}{25} = \frac{50-42}{7\cdot 25} \\
& = & \frac{8}{175} \\
\Rightarrow ||b_2|| & = & \sqrt{<b_2,b_2>} = \sqrt{\frac{8}{175}} \Rightarrow e_2(x) = \frac{1}{||b_2||} \cdot b_2(x) \\
\Rightarrow e_2(x) & = & \frac{1}{\sqrt{\frac{8}{175}}} \cdot \left( x^2 -\frac{3}{5} \right) = \sqrt{\frac{175}{8}} \cdot \left( x^2 - \frac{3}{5} \right) = \frac{5}{4}\sqrt{14} \cdot \left( x^2 - \frac{3}{5} \right) \\
e_2(x) & = & \frac{5}{4}\sqrt{14} \cdot \left( x^2 - \frac{3}{5} \right)
\end{eqnarray*}
3. Bestimmung von $e_3(x)$: \\
\begin{eqnarray*}
b_3(x) & = & x^3 - \sum_{\nu = 0}^2 <x^3,e_\nu> e_\nu(x) \\
b_3(x) & = & x^3 - <x^3,e_0>e_0(x)-<x^3,e_1>e_1(x)-<x^3,e_2>e_2(x) \\
<x^3,e_0> & = & <x^3, \frac{1}{2}\sqrt{6}> = \int_{-1}^1 x^3 \frac{1}{2}\sqrt{6}x^2 dx = \frac{1}{2}\sqrt{6} \int_{-1}^1 x^5 dx \\
<x^3,e_0> & = & 0 \\
<x^3,e_1> & = & <x^3, \frac{1}{2}\sqrt{10}\cdot x> = \int_{-1}^1 x^3 \frac{1}{2}\sqrt{10}\cdot x \cdot x^2 dx = \frac{1}{2}\sqrt{10} \cdot \int_{-1}^1 x^6 dx \\
& = & \frac{1}{2}\sqrt{10} \cdot \left[ \frac{1}{7}x^7 \right]_{-1}^1 = \frac{1}{2}\sqrt{10} \cdot \frac{2}{7} = \frac{1}{7} \sqrt{10} \\
<x^3,e_2> & = & <x^3, \frac{5}{4}\sqrt{14} \cdot \left(x^2 - \frac{3}{5} \right)> = \int_{-1}^1 x^3 \cdot \frac{5}{4} \sqrt{14} \cdot \left( x^2 - \frac{3}{5} \right) \cdot x^2 dx \\
& = & \frac{5}{4} \sqrt{14} \cdot \int_{-1}^1 x^5 \cdot \left( x^2 - \frac{3}{5} \right) dx \\
& = & \frac{5}{4} \sqrt{14} \cdot \int_{-1}^1 \left( x^7 - \frac{3}{5} x^5 \right) dx \\
<x^3, e_2> & = & 0 \\
\Rightarrow b_3(x) & = & x^3 - \frac{1}{7}\sqrt{10} \cdot \frac{1}{2}\sqrt{10} \cdot x = x^3 - \frac{5}{7}x \\
<b_3,b_3> & = & <x^3 - \frac{5}{7}x, x^3, \frac{5}{7} x> = \int_{-1}^1 \left( x^3 - \frac{5}{7} x \right) ^2 \cdot x^2 dx \\
& = & \int_{-1}^1 \left( x^8 -\frac{10}{7}x^6 + \frac{25}{49}x^4 \right) dx = \left[ \frac{1}{9}x^9 - \frac{10}{49}x^7 + \frac{5}{49}x^5 \right]_{-1}^1 = \left( \frac{1}{9}-\frac{5}{49} \right) \cdot 2 \\
& = & \frac{49-45}{9 \cdot 49} \cdot 2 = \frac{8}{441} \\
\Rightarrow ||b_3|| & = & \sqrt{<b_3,b_3>} = \sqrt{\frac{8}{441}} \\
\Rightarrow e_3(x) & = & \frac{1}{||b_3||} \cdot b_3(x) = \sqrt{\frac{441}{8}} \cdot b_3(x) = \frac{21}{2 \cdot \sqrt{2}}b_3(x) \\
e_3(x) & = & \frac{21}{4}\sqrt{2} \cdot \left( x^3 - \frac{5}{7}x \right)
\end{eqnarray*}
Ergebnis: \\
$ \{ e_0(x), e_1(x), e_2(x), e_3(x) \} $ ist eine Orthonormalbasis des Vektorraums $ U = \mathbb{R}[x]_{\le 3} $ bezüglich des Skalarprodukts $ \displaystyle <p,q> = \int_{-1}^1 p(x)q(x)x^2dx $. \\
(b) \\
0. Bestimmung von $e_0(x)$:
\begin{eqnarray*}
b_0(x) & = & 1 \\
<b_0,b_0> & = & <1,1> = \int_{-1}^1 (1-x^2) dx = \left[ x-\frac{1}{3}x^3 \right]_{-1}^1 = \left( 1- \frac{1}{3} \right) \cdot 2 = \frac{4}{3} \\
\Rightarrow ||b_0|| & = & \sqrt{<b_0,b_0>} = \sqrt{\frac{4}{3}} = \frac{2}{\sqrt{3}} \\
\Rightarrow e_0(x) & = & \frac{1}{2}\sqrt{3} \cdot b_0(x) \\
\Leftrightarrow e_0(x) & = & \frac{1}{2}\sqrt{3}
\end{eqnarray*}
1. Bestimmung von $e_1(x)$:
\begin{eqnarray*}
b_1(x) & = & x - <x,e_0>e_0(x) \\
<x,e_0> & = & <x, \frac{1}{2}\sqrt{3}> = \int_{-1}^1 x \frac{1}{2}\sqrt{3} \cdot (1-x^2)dx = \frac{1}{2}\sqrt{3} \cdot 0 = 0 \\
\Rightarrow b_1(x9 & = & x \\
<b_1,b_1> & = & <x,x>  =  \int_{-1}^1 x \cdot x \cdot (1-x^2)dx = \int_{-1}^1 (x^2-x^4)dx \\
& = & \left[ \frac{1}{3}x^3 - \frac{1}{5}x^5 \right]_{-1}^1 = \left( \frac{1}{3}-\frac{1}{5} \right) \cdot 2 = \frac{4}{15} \\
\Rightarrow ||b_1|| & = & \sqrt{<b_1,b_1>} = \sqrt{\frac{4}{15}} = \frac{2}{\sqrt{15}} \\
\Rightarrow e_1(x) & = & \frac{1}{2}\sqrt{15} \cdot b_1(x) \\
\Leftrightarrow e_1(x) & = & \frac{1}{2}\sqrt{15} \cdot x
\end{eqnarray*}
2. Bestimmung von $e_2(x)$:
\begin{eqnarray*}
b_2(x) & = & x^2 - <x^2,e_0>e_0(x) - <x^2,e_1>e_1(x) \\
<x^2,e_0> & = & <x^2, \frac{1}{2}\sqrt{3}> = \int_{-1}^1 x^2 \frac{1}{2}\sqrt{3} \cdot (1-x^2) dx \\
& = & \frac{1}{2}\sqrt{3} \cdot \int_{-1}^1 (x^2-x^4) dx = \frac{1}{2}\sqrt{3} \cdot \frac{4}{15} = \frac{2}{15}\sqrt{3} \\
<x^2,e_1> & = & <x^2, \frac{1}{2}\sqrt{15} \cdot x> = \int_{-1}^1 x^2 \frac{1}{2}\sqrt{15} \cdot x \cdot (1-x^2) dx \\
& = & \frac{1}{2}\sqrt{15} \cdot \int_{-1}^1 (x^3-x^5) dx = \frac{1}{2}\sqrt{15} \cdot 0 = 0 \\
\Rightarrow b_2(x) & = & x^2 - \frac{2}{15}\sqrt{3}\cdot\frac{1}{2}\sqrt{3} = x^2 - \frac{1}{5} \\
<b_2,b_2> & = & <x^2 - \frac{1}{5}, x^2 - \frac{1}{5}> = \int_{-1}^1 \left( x^2 - \frac{1}{5} \right)^2 \cdot (1-x^2) dx \\
& = & \int_{-1}^1 \left( x^4 - \frac{2}{5}x^2 + \frac{1}{25} \right) \cdot (1-x^2) dx \\
& = & \int_{-1}^1 \left( -x^6 + \frac{7}{5} x^4 - \frac{11}{25} x^2 + \frac{1}{25} \right) dx \\
& = & \left[ -\frac{1}{7}x^7 + \frac{7}{25}x^5 - \frac{11}{75} x^3 + \frac{1}{25}x \right]_{-1}^1 \\
& = & \frac{-75+147-77+21}{7\cdot 75} \cdot 2 = \frac{32}{525} \\
\Rightarrow ||b_2|| & = & \sqrt{\frac{32}{525}} \Rightarrow e_2(x) = \sqrt{\frac{525}{32}} \cdot \left( x^2 - \frac{1}{5} \right) \\
e_2(x) & = & \frac{5}{8}\sqrt{42} \cdot \left( x^2 - \frac{1}{5} \right)
\end{eqnarray*}
3. Bestimmung von $e_3(x)$:
\begin{eqnarray*}
b_3(x) & = & x^3 - <x^3,e_0>e_0(x) - <x^3,e_1>e_1(x) - <x^3,e_2>e_2(x) \\
<x^3,e_0> & = & <x^3, \frac{1}{2}\sqrt{3}> = \int_{-1}^1 x^3\frac{1}{2}\sqrt{3} \cdot (1-x^2) dx \\
& = & \frac{1}{2}\sqrt{3} \int_{-1}^1 (x^3-x^5) dx = 0 \\
<x^3,e_1> & = & <x^3, \frac{1}{2}\sqrt{15}\cdot x> = \int_{-1}^1 x^3\frac{1}{2}\sqrt{15}\cdot x\cdot (1-x^2) dx \\
& = & \frac{1}{2}\sqrt{15} \cdot \int_{-1}^1 (x^4-x^6) dx = \frac{1}{2}\sqrt{15} \cdot \left[ \frac{1}{5}x^5 - \frac{1}{7} x^7 \right]_{-1}^1 \\
& = & \frac{1}{2}\sqrt{15}\cdot \left[ \frac{1}{5}-\frac{1}{7} \right] \cdot 2 = \sqrt{15} \frac{2}{35} = \frac{2}{35} \sqrt{15} \\
<x^3,e_2> & = & <x^3, \frac{5}{8}\sqrt{42}\cdot \left( x^2 - \frac{1}{5} \right)> = \int_{-1}^1 x^3\frac{5}{8}\sqrt{42}\cdot \left( x^2 - \frac{1}{5} \right) \cdot (1-x^2) dx \\
& = & \frac{5}{8}\sqrt{42} \cdot \int_{-1}^1 \left( -x^7+\frac{6}{5}x^5 - \frac{1}{5}x^3 \right) dx = \frac{5}{8}\sqrt{42} \cdot 0 = 0 \\
\Rightarrow b_3(x) & = & x^3 - \frac{2}{35}\sqrt{15} \cdot \frac{1}{2}\sqrt{15} \cdot x = x^3 - \frac{3}{7} x \\
<b_3,b_3> & = & <x^3 - \frac{3}{7}x, x^3 - \frac{3}{7}x> = \int_{-1}^1 \left( x^3 - \frac{3}{7}x \right)^2 \cdot (1-x^2) dx \\
& = & \int_{-1}^1 \left( x^6 - \frac{6}{7} x^4 + \frac{9}{49}x^2 \right) \cdot (1-x^2) dx \\
& = & \int_{-1}^1 \left( -x^8 + \frac{13}{7}x^6 - \frac{51}{49}x^4 + \frac{9}{49}x^2 \right) dx \\
& = & \left[ -\frac{1}{9}x^9+\frac{13}{49} x^7 - \frac{51}{245}x^5 + \frac{3}{49} x^3 \right]_{-1}^1 \\
& = & \left( -\frac{1}{9} + \frac{16}{49} - \frac{51}{245} \right) \cdot 2 \\
& = & \frac{-245+720-459}{9\cdot 49 \cdot 5}\cdot 2 = \frac{32}{2205} \\
\Rightarrow ||b_3|| & = & \sqrt{\frac{32}{2205}} \\
\Rightarrow e_3(x) & = & \frac{21}{8}\sqrt{10} \cdot \left( x^3- \frac{3}{7}x \right)
\end{eqnarray*}
Ergebnis: \\
$ \{ e_0(x), e_1(x), e_2(x), e_3(x) \} $ ist eine Orthonormalbasis des Vektorraums $ U = \mathbb{R}[x]_{\le 3} $ bezüglich des Skalarprodukts $ \displaystyle <p,q> = \int_{-1}^1 p(x)q(x)(1-x^2)dx $.

\subsection*{Aufgabe 2}
(a) \\
$ V = \SetR_[x]_{\le 4}; \; \; U = \SetR[x]_{\le 3} $ UR von $ V $. \\
Auf Aufgabe 1 (a): \\
$ \{ e_0(x), e_1(x), e_2(x), e_3(x) \} $ ist eine Orthonormalbasis von $ U $ mit
\begin{eqnarray*}
e_0(x) & = & \frac{1}{2}\sqrt{6} \\
e_1(x) & = & \frac{1}{2}\sqrt{10}\cdot x \\
e_2(x) & = & \frac{5}{4}\sqrt{14} \cdot \left( x^2 - \frac{3}{5} \right) \\
e_3(x) & = & \frac{21}{4}\sqrt{2}\cdot \left( x^3- \frac{5}{7}x \right)
\end{eqnarray*}
bezügl. des Skalarprodukts $ \displaystyle <p,q> = \int_{-1}^1 p(x)q(x)x^2 dx $ \hfill $(1)$ \\
Sei $f=x^2\cdot(x^2-1) \in V$ \\
Für die orthogonale Projektion $\pi_1(f)\in U$ von $f$ auf $U$ bezügl. des Skalarproduktes $(1)$ gilt:
$$ \pi_1(f) = \sum_{\nu=0}^3 <f,e_\nu>e_\nu $$
Für $f=x^2\cdot(x^2-1) = x^4-x^2 $ folgt:
\begin{eqnarray*}
<f,e_\nu> & = & <x^4-x^2,e_\nu> = <x^4,e_\nu> - <x^2,e_\nu> \\
\Rightarrow \pi_1(f) & = & \sum_{\nu=0}^3 <x^4,e_\nu>e_\nu - \sum_{\nu=0}^3 <x^2,e_\nu>e_\nu
\end{eqnarray*}
Aus Aufgabe 1 (a): \\
$ <x^2,e_0> = \frac{1}{5}\sqrt{6} \; \; \AND \; \; <x^2,e_1> = 0 $.
\begin{eqnarray*}
<x^2,e_2> & = & \int_{-1}^1 x^2\frac{5}{4}\sqrt{14} \cdot \left( x^2 - \frac{3}{5} \right) \cdot x^2 dx \\
& = & \frac{5}{4}\sqrt{14}\cdot \int_{-1}^1 \left( x^6 - \frac{3}{5} x^4 \right) dx = \frac{5}{4} \sqrt{14} \cdot \left[ \frac{1}{7} x^7 - \frac{3}{25}x^5 \right]_{-1}^1 \\
& = & \frac{5}{4}\sqrt{14} \left( \frac{1}{7} - \frac{3}{25} \right) \cdot 2 = \frac{5\cdot4}{2\cdot7\cdot25}\cdot \sqrt{14} \\
& = & \frac{2}{35}\sqrt{14} \\
<x^2,e_3> & = & \int_{-1}^1 x^2 \cdot \frac{21}{4}\sqrt{2} \cdot \left( x^3 - \frac{5}{7}x \right) \cdot x^2 dx = \frac{21}{4}\sqrt{2} \cdot 0 \\
& = & 0 \\
\Rightarrow \sum_{\nu=0}^3 <x^2,e_\nu>e_\nu & = & \frac{1}{5}\sqrt{6}\frac{1}{2}\sqrt{6} + \frac{2}{35}\sqrt{14}\frac{5}{4}\sqrt{14}\cdot \left( x^2 - \frac{3}{5} \right) \\
& = & \frac{3}{5} + \left( x^2 - \frac{3}{5} \right) = x^2 \; \; \; (I) \\
<x^4,e_0> & = & \int_{-1}^1 x^4\frac{1}{2}\sqrt{6}x^2dx = \frac{1}{2}\sqrt{6} \cdot \int_{-1}^1x^6dx = \frac{1}{2}\sqrt{6} \cdot \left( \frac{1}{7}x^7 \right)_{-1}^1 \\
& = & \frac{1}{7}\sqrt{6} \\
<x^4,e_1> & = & \int_{-1}^1 x^4 \frac{1}{2}\sqrt{10}\cdot x \cdot x^2dx = \frac{1}{2}\sqrt{10} \cdot 0 \\
& = & 0 \\
<x^4,e_2> & = & \int_{-1}^1 x^4 \frac{5}{4}\sqrt{14} \cdot \left( x^2 - \frac{3}{5} \right) x^2dx = \frac{5}{4}\sqrt{14} \cot \int_{-1}^1 \left(x^8-\frac{3}{5}x^6\right)dx \\
& = & \frac{5}{4}\sqrt{14} \cdot \left[ \frac{1}{9}x^9 - \frac{3}{35}x^7 \right]_{-1}^1 = \frac{5}{2}\sqrt{14} \cdot \frac{35-27}{9\cdot35} = \frac{5\cdot8}{2\cdot9\cdot35}\cdot\sqrt{14} \\
& = & \frac{4}{63}\sqrt{14} \\
<x^4,e_3> & = & \int_{-1}^1x^4\frac{21}{4}\sqrt{2} \cdot \left( x^3 - \frac{5}{7}x \right) \cdot x^2 dx \\
& = & \frac{21}{4}\sqrt{2} \cdot \int_{-1}^1 \left(x^9-\frac{5}{7}x^7\right) dx \\
& = & 0
\end{eqnarray*}
\begin{eqnarray*}
\Rightarrow \sum_{\nu=0}^3 <x^4,e_\nu>e_\nu & = & \frac{1}{7} \sqrt{6} \frac{1}{2}\sqrt{6} + \frac{4}{63}\sqrt{14}\frac{5}{4}\sqrt{14}\cdot \left( x^2 - \frac{3}{5} \right) \\
& = & \frac{3}{7}+\frac{5\cdot14}{63}\cdot \left(x^2-\frac{3}{5}\right) \\
& = & \frac{3}{7}+\frac{10}{9}\cdot\left(x^2-\frac{3}{5}\right) \\
& = & \frac{10}{9}x^2 - \frac{5}{21} \; \; \; (II) \\
\stackrel{(I),(II)}{\Longrightarrow} \pi_1(f) & = & \frac{10}{9}x^2 - \frac{5}{21}-x^2 = \frac{1}{9}x^2-\frac{5}{21} \\
\Rightarrow f- \pi_1(f) & = & x^4-x^2 - \left(\frac{1}{9}x^2-\frac{5}{21}\right) \\
\Leftrightarrow f-\pi_1(f) = x^4 - \frac{10}{9}x^2 + \frac{5}{21}
\end{eqnarray*}
\begin{center}
\begin{figure}
\includegraphics[width=10cm]{2a}
\caption{Zeichnung zu Aufgabe 2a}
\end{figure}
\end{center}
(b) \\
Aus Aufgabe 1 (b): \\
$ \{ e_0, e_1, e_2, e_3 $ ist eine Orthonormalbasis von $ U = \SetR[x]_{\le 3} $ mit
\begin{eqnarray*}
e_0(x) & = & \frac{1}{2}\sqrt{3} \\
e_1(x) & = & \frac{1}{2}\sqrt{15}x \\
e_2(x) & = & \frac{5}{8}\sqrt{42} \left( x^2 - \frac{1}{5} \right) \\
e_3(x) & = & \frac{21}{8}\sqrt{10} \left( x^3 - \frac{3}{7}x \right)
\end{eqnarray*}
bezügl. des Skalarprodukts $ \displaystyle <p,q> = \int_{-1}^1 p(x)q(x)(1-x^2)dx $ \hfill $ (2) $. \\
Sei $ f = x^2(x^2-1) \in V = \SetR[x]_{\le 4} $. \\
Für die orthogonale Projektion $ \pi_2(f) \in U $ von $ f $ auf $U$ bezügl. des Skalarproduktes (2) gilt: \\
\begin{eqnarray*}
\pi_2(f) & = & \sum_{\nu=0}^3 <f,e_\nu>e_\nu \\
<f,e_0> & = & \int_{-1}^1 x^2(x^2-1)\frac{1}{2}\sqrt{3}(1-x^2)dx \\
& = & \frac{1}{2}\sqrt{3} \cdot \int_{-1}^1 (-x^6+2x^4-x^2)dx = \frac{1}{2}\sqrt{3} \left[-\frac{1}{7}x^7 + \frac{2}{5}x^5 - \frac{1}{3}x^3 \right]_{-1}^1 \\
& = & \sqrt{3}\frac{-15+42-35}{7\cdot5\cdot3} \\
& = & -\frac{8}{105}\sqrt{3} \\
<f,e_1> & = & \int_{-1}^1 x^2(x^2-1)\frac{1}{2}\sqrt{15}x(1-x^2)dx = 0 \\
<f e_2> & = & \int_{-1}^1 x^2(x^2-1)\frac{5}{8}\sqrt{42}\left(x^2-\frac{1}{5}\right)(1-x^2)dx \\
& = & \frac{5}{8}\sqrt{42} \cdot \int_{-1}^1 \left(x^4-\frac{1}{5}x^2\right)\cdot(-x^4+2x^2-1)dx \\
& = & \frac{5}{8}\sqrt{42}\cdot \int_{-1}^1 \left( -x^8+\frac{11}{5}x^6-\frac{7}{5}x^4 +\frac{1}{5}x^2 \right) dx \\
& = & \frac{5}{8}\sqrt{42} \cdot \left[ -\frac{1}{9}x^9 + \frac{11}{35}x^7 - \frac{7}{25}x^5 + \frac{1}{15}x^3 \right]_{-1}^1 \\
& = & \frac{5}{4}\sqrt{42}\cdot \frac{-175+495-441+105}{3^2\cdot5^2\cdot7} \\
& = & -\frac{5\cdot16}{4\cdot3^2\cdot5^2\cdot7}\cdot\sqrt{42} = -\frac{4}{315}\sqrt{42} \\
<f,e_3> & = & \int_{-1}^1 x^2(x^2-1) \cdot \frac{21}{8}\sqrt{10}\left(x^3-\frac{3}{7}x\right)(1-x^2)dx = 0 \\
\Rightarrow \pi_2(f) & = & -\frac{8}{105}\sqrt{3}\frac{1}{2}\sqrt{3} - \frac{4}{315}\sqrt{42}\frac{5}{8}\sqrt{42}\left(x^2-\frac{1}{5}\right) \\
\pi_2(f) & = & -\frac{4}{35}-\frac{4\cdot42\cdot5}{315\cdot8}\cdot \left(x^2-\frac{1}{5}\right) \\
\pi_2(f) & = & -\frac{4}{35}-\frac{1}{3}\cdot\left(x^2-\frac{1}{5}\right) = -\frac{1}{3}x^2-\frac{1}{21} \\
\Rightarrow f - \pi_2(f) & = & x^4 - \frac{2}{3}x^2+\frac{1}{21}
\end{eqnarray*}
\begin{center}
\begin{figure}
\includegraphics[width=10cm]{2b}
\caption{Zeichnung zu Aufgabe 2b}
\end{figure}
\end{center}

\subsection*{Aufgabe 4}
Gegeben:
$ A_n = \left(
\begin{array}{rrrr}
2 & -1 & & 0 \\
-1 & \ddots & \ddots \\
& \ddots & \ddots & -1 \\
0 & & -1 & 2
\end{array}
\right) \in M(n\times n, \SetR), n \in \mathbb{N}_{\ge 1} $ \\
Alle Matrizen $ A_n $ sind symmetrisch. \\
Bez.: $ A_k $ sei die linke obere k-reihige Teilmatrix von $A_n$. \\
Nach dem Hauptminoren-Kriterium für Definitheit gilt dann: \\
$A_n$ positiv definit $ \Leftrightarrow $ $ det A_k > 0 $ für $ k=1,...,n$. (Fischer, S. 327) \hfill $(*)$ \\
Behauptung: $ det \, A_k = k+1 \; \forall k\in \{1,...,n\} $, $n\in \SetN_{\ge 1}$ beliebig. \\
Beweis:
\begin{enumerate}
\item Induktionsanfang: $ k=1 $: $ A_1 = (2) \Rightarrow det \, A_1 = 2 = 1+1 $

\item
\begin{enumerate}
\item Induktionsannahme:
Für $ k \ge 1 $ gelte: $det \, A_k = k+1 \Leftrightarrow$ für $k\ge0$ gelte: $det\,A_{k-1}=k+2$.
\item Induktionsbehauptung:
Dann gilt für $k+1$: $det\,A_{k+1}=k+2$
\item Induktionsbegründung:
$$ A_{k+1} = \left(
\begin{array}{rrrrr}
2 & -1 & & & 0 \\
-1 & \ddots & \ddots \\
& \ddots & 2 & -1 & 0 \\
& & -1 & 2 & -1 \\
0 & & 0 & -1 & 2
\end{array}
\right) \in M((k+1)\times(k+1),\SetR)
$$
Berechnung von $ det \, A_{k+1} $ durch Entwicklung nach der $(k+1)$-ten Zeile:
$$
det \, A_{k+1} = -(-1) \cdot \left|
\begin{array}{rrrrr}
& & & & 0 \\
& & & & \vdots \\
& & A_{k-1} & & \vdots \\
& & & & \vdots \\
& & & & 0 \\
0 & \cdots & 0 & -1 & -1
\end{array}
\right| + 2\cdot det\,A_k \\
\text{nach der k-ten Spalte entwickeln}
$$
\begin{eqnarray*}
\Leftrightarrow A_{k+1} & = & (+1) \cdot (-1) \cdot det \, A_{k-1} + 2 \cdot det \, A_k \\
\Leftrightarrow A_{k+1} & = & 2 \cdot det \, A_k - det \, A_{k-1} \\
& \stackrel{(2a)}{=} & 2 \cdot (k+1) - k \\
& = & 2 \cdot k + 2 - k \\
& = & k+2
\end{eqnarray*}
Wegen $ det \, A_k = k+1 $ ist $ det \, A_k > 0 $ für $ k=1,...,n $ \\
$ \stackrel{*}{\Rightarrow} $ Alle Matrizen $ A_n $ sind positiv definit. \hfill $ \Box $
\end{enumerate}
\end{enumerate}

\end{document}