master > master: notes - A3-2
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notes/notes.tex
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@ -31,6 +31,8 @@
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%% |____ body/woche3/A4.tex
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%% |____ body/woche4/index.tex
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%% |____ body/woche4/A1.tex
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%% |____ body/woche4/A2.tex
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%% |____ body/woche4/A3.tex
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%% |____ body/woche4/A4.tex
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%% |____ back/index.tex
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%% |____ back/sources.bib
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@ -117,6 +119,7 @@
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dvipsnames,
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}{xcolor}
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\usepackage{makecmds} % need for \provideenvironment
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\usepackage{inputenc}
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\usepackage{babel}
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\usepackage{geometry}
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@ -306,6 +309,9 @@
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\providecommand{\powerSet}{}
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\renewcommand{\powerSet}[1]{\mathop{\powerset}(#1)}
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\providecommand{\blockCases}{}
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\renewenvironment{blockCases}[2]{\left\{\begin{array}[#1]{#2}}{\end{array}\right.}
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\newcommand{\fieldK}[0]{\mathbb{K}}
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\newcommand{\reals}[0]{\mathbb{R}}
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\newcommand{\complex}[0]{\mathbb{C}}
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@ -1069,7 +1075,7 @@ Per Definition gilt $\norm{f}_{\infty} \in \reals$ gdw. $f$ beschränkt ist.
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\begin{schattierteboxdunn}
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\begin{claim}
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\setblocklabel{claim:1:ex:2.1:raj-analysis-ii-notes}
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\setblocklabel{claim:1:ex:3.1:raj-analysis-ii-notes}
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Angenommen,
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so dass $(f_{n})_{n\in\naturals}$
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genüge folgender Eigenschaft
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@ -1238,6 +1244,463 @@ Per Definition gilt $\norm{f}_{\infty} \in \reals$ gdw. $f$ beschränkt ist.
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%% ********** END OF FILE: body/woche4/A1.tex **********
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\setcounterafter{section}{2}
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%% ********************************************************************************
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%% FILE: body/woche4/A2.tex
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%% ********************************************************************************
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\let\oldsectionname\sectionname
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\def\sectionname{Aufgabe}
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\section[]{}
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\let\sectionname\oldsectionname
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\label{sec:6}
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\begin{schattierteboxdunn}
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\begin{claim}
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\setblocklabel{claim:1:ex:3.2:raj-analysis-ii-notes}
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Seien $\alpha \in (0, \infty)$.
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Dann
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$\displaystyle\sum_{k=1}^{\infty}k^{-\alpha} < \infty$
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gdw. $\alpha > 1$.
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\end{claim}
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\end{schattierteboxdunn}
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\begin{beweis}
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Sei $N\in\naturals$ mit $N > 1$.
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Dann
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\begin{maths}[mc]{rcl}
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\eqtag[eq:1:\beweislabel]
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\displaystyle\sum_{k=1}^{N}k^{-\alpha}
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&= &k^{-N} + \displaystyle\sum_{k=1}^{N - 1}
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\displaystyle\int_{t=k}^{k + 1} k^{-\alpha} \dee t\\
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&\geq &0 + \displaystyle\sum_{k=1}^{N - 1}
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\displaystyle\int_{t=k}^{k + 1} t^{-\alpha} \dee t
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= \displaystyle\int_{t=1}^{N} t^{-\alpha} \dee t\\
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\end{maths}
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und
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\begin{maths}[mc]{rcl}
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\eqtag[eq:2:\beweislabel]
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\displaystyle\sum_{k=1}^{N}k^{-\alpha}
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&= &1 + \displaystyle\sum_{k=2}^{N} \displaystyle\int_{t=k-1}^{k} k^{-\alpha} \dee t\\
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&\leq &1 + \displaystyle\sum_{k=2}^{N}
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\displaystyle\int_{t=k-1}^{k} t^{-\alpha} \dee t\\
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&= &1 + \displaystyle\int_{t=1}^{N} t^{-\alpha} \dee t.\\
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\end{maths}
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Laut Skript (siehe \cite[\S{}11.3,~Bsp.~(c)]{pogorzelskiVLSkript})
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wissen wir nun, dass
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\begin{maths}[mc]{rcl}
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\eqtag[eq:3:\beweislabel]
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\displaystyle\int_{t=1}^{\infty} t^{-\alpha} \dee t
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&= &\begin{blockCases}{mc}{rcl}
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+\infty &: &\alpha \leq 1\\
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\frac{1}{1-\alpha} &: &\alpha > 1\\
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\end{blockCases}.\\
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\end{maths}
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Folglich gelten
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\begin{maths}[mc]{rcl}
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\displaystyle\sum_{k=1}^{\infty}k^{-\alpha}
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&= &\displaystyle\limsup_{N \to \infty}\displaystyle\sum_{k=1}^{N}k^{-\alpha}\\
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&\eqcrefoverset{eq:2:\beweislabel}{\leq}
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&\displaystyle\limsup_{N \to \infty}\displaystyle\int_{t=1}^{N} t^{-\alpha} \dee t + 1\\
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&= &\displaystyle\displaystyle\int_{t=1}^{\infty} t^{-\alpha} \dee t + 1\\
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&\eqcrefoverset{eq:3:\beweislabel}{=}
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&\frac{1}{1-\alpha} + 1
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< \infty,\\
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\end{maths}
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falls $\alpha > 1$, und
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\begin{maths}[mc]{rcl}
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\displaystyle\sum_{k=1}^{\infty}k^{-\alpha}
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&= &\displaystyle\limsup_{N \to \infty}\displaystyle\sum_{k=1}^{N}k^{-\alpha}\\
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&\eqcrefoverset{eq:1:\beweislabel}{\geq}
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&\displaystyle\limsup_{N \to \infty}\displaystyle\int_{t=1}^{N} t^{-\alpha} \dee t\\
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&= &\displaystyle\int_{t=1}^{\infty} t^{-\alpha} \dee t\\
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&\eqcrefoverset{eq:3:\beweislabel}{=}
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&+\infty,\\
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\end{maths}
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falls $0 < \alpha \leq 1$.
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Darum gilt die Behauptung.
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\end{beweis}
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%% ********** END OF FILE: body/woche4/A2.tex **********
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\setcounterafter{section}{3}
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%% ********************************************************************************
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%% FILE: body/woche4/A3.tex
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%% ********************************************************************************
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\let\oldsectionname\sectionname
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\def\sectionname{Aufgabe}
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\section[]{}
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\let\sectionname\oldsectionname
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\label{sec:7}
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Für diese Aufgaben brauchen wir zunächst einmal Lemma, um unsere Arbeit zu erleichtern.
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\begin{lemm}
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\setblocklabel{lemm:uneigentlich:ex:3.3:raj-analysis-ii-notes}
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Sei $a\in\reals$ und sei ${h:[a,\infty)\to\reals}$ eine Funktion
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mit ${h(t) \longrightarrow 0}$ für ${t \longrightarrow +\infty}$.
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Sei außerdem
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$(T_{n})_{n} \subseteq [a,\infty)$
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eine monoton wachsende Folge mit ${T_{n} \longrightarrow \infty}$
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und so,
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dass $(T_{n+1}-T_{n})_{n}$ beschränkt ist.
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Dann ist
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${\displaystyle\int_{a}^{\infty} h \dee t}$
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konvergent
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gdw.
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$(\displaystyle\int_{a}^{T_{n}} h \dee t)_{n}$
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konvergent ist.
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\end{lemm}
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\begin{einzug}[\mytab][\mytab]
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\begin{beweis}
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Offensichtlich gilt die \enclosedquote{nur dann wenn}-Richtung.
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Für die \enclosedquote{wenn}-Richtung,
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sei angenommen, $(\displaystyle\int_{a}^{T_{n}} h \dee t)_{n}$ konvergiert
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mit Grenzwert $I \in \reals$.
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Setze $C \in (0,\infty)$ mit $\sup_{n}T_{n+1}-T_{n} \leq C$.
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Sei $\eps > 0$.
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Sei $N$ ein genügend großer Index
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mit $\abs{\int_{a}^{T_{n}} h \dee t - I} < \frac{\eps}{2}$
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für alle $n \geq N$.
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Da $f$ gegen $+\infty$ verschwindend ist,
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existiert ein $\tilde{T}\in[a,\infty)$,
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so dass $\abs{f(\cdot)} \leq \frac{\eps}{2C}$
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auf $[\tilde{T},\infty]$.
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Sei nun $T \in [\max\{T_{N},\tilde{T}\},\infty)$ beliebig.
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Da $T \geq T_{N}$ und ${(T_{n})_{n}\longrightarrow +\infty}$ monoton,
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existiert ein $n\geq N$ so, dass $T \in [T_{n},T_{n+1}]$.
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Da $n \geq N$ und $[T_{n},T]\subseteq[\tilde{T},\infty)$
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erhält man die Abschätzung
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\begin{maths}[mc]{rcl}
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\absLong{
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\displaystyle\int_{a}^{T} h(t) \dee t
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-
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I
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}
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&\leq
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&\absLong{
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\displaystyle\int_{a}^{T} h(t) \dee t
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-
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\displaystyle\int_{a}^{T_{n}} h(t) \dee t
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}
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+ \absLong{
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\displaystyle\int_{a}^{T_{n}} h(t) \dee t
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-
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I
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}\\
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&< &\absLong{
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\displaystyle\int_{T_{n}}^{T} h(t) \dee t
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}
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+ \frac{\eps}{2}\\
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&< &\displaystyle\int_{T_{n}}^{T}
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\underbrace{
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\abs{h(t)}
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}_{\leq \frac{\eps}{2C}}
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\dee t
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+ \frac{\eps}{2}\\
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&\leq &(T-T_{n})\frac{\eps}{2C}
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+ \frac{\eps}{2}\\
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&\leq &(T_{n+1}-T_{n})\frac{\eps}{2C}
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+ \frac{\eps}{2}\\
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&\leq &C\frac{\eps}{2C}
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+ \frac{\eps}{2}
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=\eps.\\
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\end{maths}
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Darum gilt für genügend großes
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$T \in [a,\infty)$
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(in Abhängigkeit von $\eps$),
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dass $\abs{
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\displaystyle\int_{a}^{T} h(t) \dee t
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-
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I
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} < \eps$.
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Da dies für alle $\eps > 0$ gilt
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ist
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$(\displaystyle\int_{a}^{T} h \dee t)_{T\in[a,\infty)}$
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konvergent
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(mit Grenzwert $I$).
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Also konvergiert $\displaystyle\int_{a}^{\infty} h \dee t$.
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\end{beweis}
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\end{einzug}
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\begin{enumerate}{\bfseries (a)}
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%% A3(a)
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\item
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\begin{schattierteboxdunn}
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\begin{claim}
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\setblocklabel{claim:1:ex:3.3a:raj-analysis-ii-notes}
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Sei $f:\reals\ohne\{0\}\to\reals$
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definiert durch $f(x)=\frac{\sin(x)}{x}$.
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Dann konvergiert
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$\displaystyle\int_{x=1}^{\infty} f \dee x$.
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\end{claim}
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\end{schattierteboxdunn}
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\begin{beweis}
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Für $c \in (0, \infty)$ beobachte, dass
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\begin{maths}[mc]{rcl}
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\eqtag[eq:1:\beweislabel]
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\absLong{
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\displaystyle\int_{c}^{c + 2\pi}
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\frac{\sin(x)}{x}
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\dee x
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}
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&= &\absLong{
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\displaystyle\int_{c}^{c + 2\pi}
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\frac{\sin(x)}{x}
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\dee x
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-
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\displaystyle\int_{c}^{c + 2\pi}
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\frac{\sin(x)}{c}
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\dee x
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}\\
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&&\text{da $\int_{c}^{c + 2\pi}\sin(x)\dee x = [-\cos(x)]_{c}^{c+2\pi}=0$}\\
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&= &\absLong{
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\displaystyle\int_{c}^{c + 2\pi}
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\sin(x)\cdot(\frac{1}{x} - \frac{1}{c})
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\dee x
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}\\
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&\leq &\displaystyle\int_{c}^{c + 2\pi}
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\underbrace{
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\abs{\sin(x)}
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}_{\leq 1}
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\underbrace{
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\abs{\frac{1}{x} - \frac{1}{c}}
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}_{\substack{
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\leq \frac{1}{c} - \frac{1}{c + 2\pi}\\
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= \frac{2\pi}{c(c + 2\pi)}
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}}
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\dee x\\
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&\leq &2\pi \cdot \frac{2\pi}{c(c + 2\pi)}
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\leq (\frac{2\pi}{c})^{2}.\\
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\end{maths}
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Da $f$ offensichtlich gegen $+\infty$ verschwindend ist,
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laut \Cref{lemm:uneigentlich:ex:3.3:raj-analysis-ii-notes}
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reicht es aus zu zeigen,
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dass
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$(\int_{1}^{2\pi n}f\dee x)_{n\in\naturals}$ konvergiert.
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Sei hiefür $\eps > 0$ beliebig.
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Da $\sum_{k=1}^{\infty}\frac{1}{k^{2}}<\infty$, wissen wir,
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dass ein $N\in\naturals$ existiert
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so dass
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$\abs{
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\sum_{k=1}^{n_{1}} \frac{1}{k^{2}}
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-
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\sum_{k=1}^{n_{2}} \frac{1}{k^{2}}
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} < \frac{\eps}{3}$
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für alle $n_{1},n_{2} \geq N$.
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\OE wähle $N > \frac{3}{\eps}$.
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Für $n_{1},n_{2} \geq N$ berechnen wir daher:
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\begin{maths}[mc]{rcl}
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\absLong{
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\displaystyle\int_{1}^{2\pi n_{1}} f \dee x
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-
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\displaystyle\int_{1}^{2\pi n_{2}} f \dee x
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}
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&=
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&\absLong{
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\displaystyle\int_{2\pi \min\{n_{1},n_{2}\}}^{2\pi \max\{n_{1},n_{2}\}} f \dee x
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}\\
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&=
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&\absLong{
|
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\displaystyle\sum_{k=\min\{n_{1},n_{2}\}}^{\max\{n_{1},n_{2}\}-1}
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\displaystyle\int_{2\pi k}^{2\pi (k+1)} f \dee x
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}\\
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&=
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&\displaystyle\sum_{k=\min\{n_{1},n_{2}\}}^{\max\{n_{1},n_{2}\}-1}
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\absLong{
|
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\displaystyle\int_{2\pi k}^{2\pi (k+1)} f \dee x
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}\\
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&\eqcrefoverset{eq:1:\beweislabel}{\leq}
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&\displaystyle\sum_{k=\min\{n_{1},n_{2}\}}^{\max\{n_{1},n_{2}\}-1}
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(\frac{2\pi}{2\pi k})^{2}\\
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&=
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&\absLong{
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\displaystyle\sum_{1}^{n_{1}} \frac{1}{k^{2}}
|
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-
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\displaystyle\sum_{1}^{n_{2}} \frac{1}{k^{2}}
|
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}
|
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< \eps.\\
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\end{maths}
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Es folgt, dass
|
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$(\int_{1}^{2\pi n}f\dee x)_{n\in\naturals}\subseteq\reals$
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eine Cauchy-Folge ist und wegen Vollständigkeit von $\reals$ konvergent.
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Laut \Cref{lemm:uneigentlich:ex:3.3:raj-analysis-ii-notes}
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existiert damit $\int_{1}^{\infty} f \dee x$.
|
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\end{beweis}
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|
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%% A3(b)
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\item
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\begin{schattierteboxdunn}
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\begin{claim}
|
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\setblocklabel{claim:1:ex:3.3b:raj-analysis-ii-notes}
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$\int_{x=1}^{\infty} \abs{\frac{\sin(x)}{x}} \dee x$
|
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divergiert gegen $+\infty$.
|
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\end{claim}
|
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\end{schattierteboxdunn}
|
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|
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\begin{beweis}
|
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\end{beweis}
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|
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%% A3(c)
|
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\item
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\begin{schattierteboxdunn}
|
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\begin{claim}
|
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\setblocklabel{claim:1:ex:3.3c:raj-analysis-ii-notes}
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Sei $g:\reals\ohne\{0\}\to\reals$
|
||||
definiert durch $g(x)=\frac{\sin(x)}{x}$.
|
||||
Dann divergiert
|
||||
$\displaystyle\int_{x=1}^{\infty} g \dee x$
|
||||
gegen $+\infty$.
|
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\end{claim}
|
||||
\end{schattierteboxdunn}
|
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|
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Um dies zu beweisen brauchen wir ein kleines Zwischenresultat.
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|
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\begin{lemm}
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\setblocklabel{lemm:1:ex:3.3c:raj-analysis-ii-notes}
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Für $k, c \in [0, \infty)$
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mit $k \geq 1$
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||||
ist
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||||
${I_{k,c} \coloneq \displaystyle\int_{x=1}^{\infty} \frac{\sin(kx - c)}{x} \dee x}$
|
||||
konvergent.
|
||||
\end{lemm}
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||||
|
||||
\begin{einzug}[\mytab][\mytab]
|
||||
\begin{beweis}
|
||||
Schreibe
|
||||
${I_{k,c}(T) \coloneq \displaystyle\int_{x=1}^{T} \frac{\sin(kx - c)}{x} \dee x}$
|
||||
für $T \in [1, \infty)$.
|
||||
Wegen Aufgabe 3(a) wissen wir bereits,
|
||||
dass
|
||||
${I_{1,0}(T) \longrightarrow I_{1,0} \in \reals}$
|
||||
für ${T \longrightarrow +\infty}$.
|
||||
Für $T \in [1, \infty)$ gilt
|
||||
|
||||
\begin{maths}[mc]{rcl}
|
||||
I_{k,c}(T)
|
||||
&=
|
||||
&\displaystyle\int_{1}^{T}
|
||||
\frac{\sin(kx - c)}{kx-c + c}
|
||||
\cdot u^{\prime}
|
||||
\dee u\\
|
||||
&&\text{mit $u(x) \coloneq kx - c$}\\
|
||||
&=
|
||||
&\displaystyle\int_{k + c}^{kT + c}
|
||||
\frac{\sin(u)}{u + c}
|
||||
\dee u\\
|
||||
&=
|
||||
&\displaystyle\int_{1}^{kT + c}
|
||||
\frac{\sin(u)}{u}
|
||||
\dee u
|
||||
-\underbrace{
|
||||
\displaystyle\int_{1}^{kT + c}
|
||||
\sin(u)\cdot(\frac{1}{u}-\frac{1}{u+c})
|
||||
\dee u
|
||||
}_{\eqcolon D_{c}(kT + c)}\\
|
||||
&&-\underbrace{
|
||||
\displaystyle\int_{1}^{k + c}
|
||||
\frac{\sin(u)}{u + c}
|
||||
\dee u
|
||||
}_{\eqcolon E_{k,c}}\\
|
||||
&= &I_{1,0}(kT + c) + D_{c}(kT + c) + E_{k,c}.\\
|
||||
\end{maths}
|
||||
|
||||
Wegen Stetigkeit von ${[1,k+c] \ni x \mapsto \frac{\sin(u)}{u + c}}$,
|
||||
ist diese Funktion Riemann-integrierbar.
|
||||
Also ist $E_{k,c} \in \reals$ wohldefiniert.
|
||||
Wie oben wissen wir, dass
|
||||
${T \longrightarrow +\infty}$
|
||||
$\Rightarrow$
|
||||
${kT + c \longrightarrow +\infty}$
|
||||
$\Rightarrow$
|
||||
${I_{1,0}(kT+c) \longrightarrow I_{1,0} \in \reals}$.
|
||||
Darum, um die Konvergenz von $I_{k,c}$ zu zeigen,
|
||||
müssen wir lediglich die Konvergenz von
|
||||
$(D_{c}(kT + c))_{T\in[1,\infty)}$
|
||||
zeigen.
|
||||
|
||||
Nebenrechnung:
|
||||
$\abs{\sin(u)\cdot(\frac{1}{u}-\frac{1}{u+c})}
|
||||
=\abs{\sin(u)}\cdot\frac{c}{u(u+c)}
|
||||
\leq \frac{c}{u^{2}}$
|
||||
für $u\in[1,\infty)$.
|
||||
Da $\int_{u=1}^{\infty} \frac{c}{u^{2}} \dee u$ existiert,
|
||||
ist somit der Bertrag der stetigen Funktion
|
||||
${[1,\infty)\ni u \mapsto \sin(u)\cdot(\frac{1}{u}-\frac{1}{u+c})}$
|
||||
uneigentlich Riemann-integrierbar.
|
||||
Folglich ist diese Funktion (ohne Betrag)
|
||||
uneigentlich Riemann-integrierbar.
|
||||
Insbesondere ist $(D_{c}(T))_{T\in[1,\infty)}$
|
||||
und somit auch $(D_{c}(kT + c))_{T\in[1,\infty)}$
|
||||
konvergent.
|
||||
|
||||
Darum konvergiert $(I_{k,c}(T))_{T\in[1,\infty)}$.
|
||||
\Dh $\displaystyle\int_{x=1}^{\infty} \frac{\sin(kx - c)}{x} \dee x$ existiert.
|
||||
\end{beweis}
|
||||
\end{einzug}
|
||||
|
||||
\def\beweislabel{claim:1:ex:3.3c:raj-analysis-ii-notes}
|
||||
Jetzt können wir mit dem Beweis von \Cref{\beweislabel} fortsetzen.
|
||||
|
||||
\begin{beweis}[von \Cref{\beweislabel}]
|
||||
Sei $T \in [1,\infty)$.
|
||||
Schreibe ${J(T) \coloneq \int_{x=1}^{T} g \dee x}$.
|
||||
Für $x\in[1,\infty)$ gilt
|
||||
$%
|
||||
g(x) = \frac{\sin^{2}(x)}{x}
|
||||
= \frac{1}{2}(\frac{1}{x} - \frac{\cos(2x)}{x})
|
||||
= \frac{1}{2}(\frac{1}{x} - \frac{\sin(\frac{\pi}{2} - 2x)}{x})
|
||||
= \frac{1}{2}(\frac{1}{x} + \frac{\sin(2x - \frac{\pi}{2})}{x})
|
||||
= \frac{1}{2}(\frac{1}{x} + h_{2,\frac{\pi}{2}}(x))
|
||||
$.
|
||||
Darum gilt
|
||||
|
||||
\begin{maths}[mc]{rcl}
|
||||
J(T)
|
||||
&=
|
||||
&\frac{1}{2}
|
||||
\displaystyle\int_{x=1}^{T}\frac{1}{x}\dee x
|
||||
+
|
||||
\frac{1}{2}
|
||||
\displaystyle\int_{x=1}^{T}h_{2,\frac{\pi}{2}}(x) \dee x\\
|
||||
&= &\frac{1}{2}\log(T) - \frac{1}{2}I_{2,\frac{\pi}{2}}(T).\\
|
||||
\end{maths}
|
||||
|
||||
Da
|
||||
${(I_{2,\frac{\pi}{2}}(T))_{T\in[1,\infty)} \longrightarrow I_{2,\frac{\pi}{2}} \in \reals}$
|
||||
(siehe \Cref{lemm:1:ex:3.3c:raj-analysis-ii-notes})
|
||||
und
|
||||
${(\log(T))_{T\in[1,\infty)} \longrightarrow +\infty}$,
|
||||
folgt ${(J(T))_{T\in[1,\infty)} \longrightarrow +\infty}$.
|
||||
\Dh $\displaystyle\int_{x=1}^{\infty} g \dee x = +\infty$.
|
||||
\end{beweis}
|
||||
\end{enumerate}
|
||||
|
||||
%% ********** END OF FILE: body/woche4/A3.tex **********
|
||||
|
||||
\setcounterafter{section}{4}
|
||||
|
||||
%% ********************************************************************************
|
||||
|
|
@ -1248,7 +1711,7 @@ Per Definition gilt $\norm{f}_{\infty} \in \reals$ gdw. $f$ beschränkt ist.
|
|||
\def\sectionname{Aufgabe}
|
||||
\section[]{}
|
||||
\let\sectionname\oldsectionname
|
||||
\label{sec:6}
|
||||
\label{sec:8}
|
||||
|
||||
\begin{enumerate}{\bfseries (a)}
|
||||
\setcounterafter{enumi}{2}
|
||||
|
|
@ -1484,7 +1947,7 @@ Otto Forster.
|
|||
\newblock {Grundkurs Mathematik}. Springer Spektrum, Wiesbaden, 12 edition,
|
||||
2016.
|
||||
|
||||
\bibitem[Pog 2]{pogorzelski}
|
||||
\bibitem[Pog 2]{pogorzelskiVLSkript}
|
||||
Felix Pogorzelski.
|
||||
\newblock {Vorlesungsskript: Analysis I--II}, 2021--2.
|
||||
\newblock {basierend auf dem Skript von Daniel Lenz 2013--14 + 2020--21}.
|
||||
|
|
|
|||
Loading…
Reference in New Issue