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\begin{document}
\title{Sonsuzk\"u\c c\"uk Analiz}
\subtitle{\textit{Infinitesimal Analysis}}
\author{David Pierce}
\date{30 Aral\i k 2018}
\publishers{Matematik B\"ol\"um\"u\\
Mimar Sinan G\"uzel Sanatlar \"Universitesi\\
\url{mat.msgsu.edu.tr/~dpierce/}\\
\url{david.pierce@msgsu.edu.tr}}

\maketitle

\section*{\"Ons\"oz}

Bu notlar, kalk\"ul\"us \"o\u grencileri ve \"o\u gretmenleri i\c cin
yaz\i lm\i\c st\i r.
``Epsilon--delta'' tan\i m\i n\i\ kullanmadan
limitleri tan\i mlayaca\u g\i z ve
onlar\i n temel teoremlerini kan\i tlayaca\u g\i z.

Limitlerin ``standart,'' ``epsilon-delta'' tan\i m\i\ karma\c s\i kt\i r.
Verece\u gimiz ``standart olmayan,''
``sonsuzk\"u\c c\"uk'' tan\i m daha basittir,
ama arkas\i ndaki cebir karma\c s\i kt\i r.
Kalk\"ul\"us\"un kendisi karma\c s\i kt\i r,
ve farkl\i\ \"o\u grenciler ve \"o\u gretmenler,
farkl\i\ y\"ontemleri tercih edebilirler.

A\c sa\u g\i da baz\i\ teoremlerin kan\i t\i, okuyana b\i rak\i l\i yor.
Teknik terimler, \emph{italik} olabildi\u gi halde
tan\i mlan\i nca \textbf{siyaht\i r.}

Abraham Robinson \cite{MR1373196}
sonsuzk\"u\c c\"uk y\"ontemi
sert bir \c sekilde meydana koydu.
Keisler \cite{Keisler-calculus} taraf\i ndan
ve Henle ile Kleinberg \cite{MR1999278} taraf\i ndan yaz\i lm\i\c s
sonsuzk\"u\c c\"uk y\"ontemi kullanan ders kitaplar\i\ vard\i r.


\tableofcontents
\listoffigures

\addpart{Ger\c cel\"ust\"u say\i lar}

\section{Ger\c cel say\i lar ve sayma say\i lar\i}

\emph{Tam s\i ralanm\i\c s} bir \emph{cismin} var oldu\u gunu varsay\i yoruz,
ve o cismi
\begin{equation*}
  \R
\end{equation*}
ile yaz\i yoruz; $\R$'nin elemanlar\i, \textbf{ger\c cel say\i lard\i r.}
Bu say\i lar, s\i n\i rs\i z bir do\u gru olu\c sturur,
ve bu y\"ontemle
say\i lar\i n i\c slemlerine geometrik tan\i mlar verilebilir.
\"Orne\u gin \c Sekiller \ref{fig:add} ve \numarada{fig:mul}
\begin{figure}
  \centering
  \begin{pspicture}(-5,-1.5)(2,0)
    \psset{PosAngle=-90}
    \pstGeonode(0,0)O(2,0)A(-5,0)B
    \pstLineAB[nodesep=-1]BA
    \pstTranslation OAB[C]
\end{pspicture}

    \begin{pspicture}(-2.2,-1.5)(3.6,0)
    \psset{PosAngle=-90}
    \pstGeonode(0,0)O(3.6,0)A(-2.2,0)B
    \pstTranslation OAB[C]
    \pstLineAB[nodesep=-1]BA
\end{pspicture}

    \begin{pspicture}(0,-1.5)(5.9,0)
    \psset{PosAngle=-90}
    \pstGeonode(0,0)O(4.1,0)A(1.8,0)B
    \pstTranslation OAB[C]
    \pstLineAB[nodesep=-1]OC
\end{pspicture}

    \begin{pspicture}(0,-0.5)(6.8,0)
    \psset{PosAngle=-90}
    \pstGeonode(0,0)O(2.5,0)A(4.3,0)B
    \pstTranslation OAB[C]
    \pstLineAB[nodesep=-1]OC
\end{pspicture}
\caption{Ger\c cel say\i lar\i n toplanmas\i}
  \label{fig:add}
\end{figure}
$O$, $A$, $B$, ve $C$ noktalar\i,
$0$, $a$, $b$, ve $c$ ger\c cel say\i lar\i\ olarak d\"u\c s\"un\"uls\"un,
ve \Sekilde{fig:mul} $I$ noktas\i,
$1$ ger\c cel say\i s\i\ olarak d\"u\c s\"un\"uls\"un.
\Sekilde{fig:add}
e\u ger
\begin{equation*}
  \overrightarrow{BC}=\overrightarrow{OA}
\end{equation*}
ise,
o zaman
\begin{equation*}
  a+b=c.
\end{equation*}
Toplaman\i n birle\c stirme ve de\u gi\c stirme \"ozellikleri,
ve her ger\c cel say\i n\i n negatif\/inin var oldu\u gu,
apa\c c\i kt\i r.
\Sekilde{fig:mul}
\begin{figure}
  \centering
  \psset{unit=3cm,PointSymbol=none}
  \begin{pspicture}(0,-0.25)(2,1.25)
    \pstGeonode[PosAngle={-90,-90,180}](2,0)B(0,0)O(0,1)J
    \pstInterLC[PosAngle=-90,PointSymbolA=none,PointNameA={},
    PointSymbolB=*] OBOJKI
    \pstHomO[HomCoef=0.33,PosAngle={180}]O{J}[D]
    \pstInterLC[PosAngle=-90,PointSymbolA=none,PointNameA={},
    PointSymbolB=*] OBODLA
    \pstTranslation[PointSymbol=none,PointName=none] JBD[E]
    \pstInterLL[PosAngle=-90] DEOBC
    \pstLineAB[nodesep=-0.5]OB
    \pstLineAB JB
    \pstLineAB DC
    \pstLineAB OJ
  \end{pspicture}

    \psset{unit=2cm}
    \begin{pspicture}(-0.88,-0.5)(2.2,1.25)
    \pstGeonode[PosAngle={-90,45,180}](2.2,0)B(0,0)O(0,1)J
    \pstInterLC[PosAngle=-90,PointSymbolA=none,PointNameA={},
    PointSymbolB=*] OBOJKI
    \pstHomO[HomCoef=-0.4,PosAngle={0}]O{J}[D]
    \pstInterLC[PosAngle=90,PointSymbolB=none,PointNameB={},
    PointSymbolB=*] OBODAL
    \pstTranslation[PointSymbol=none,PointName=none] JBD[E]
    \pstInterLL[PosAngle=90] DEOBC
    \pstLineAB[nodesep=-0.5]CB
    \pstLineAB JB
    \pstLineAB DC
    \pstLineAB DJ
  \end{pspicture}
\caption{Ger\c cel say\i lar\i n \c carp\i lmas\i}
  \label{fig:mul}
\end{figure}
e\u ger
\begin{align*}
  OD&=OA,&OJ&=OI,&DC&\parallel JB
\end{align*}
ise, o zaman
\begin{equation*}
  ab=c.
\end{equation*}
Ger\c cel say\i lar\i n \c carp\i lmas\i n\i n geometrik tan\i m\i,
Descartes \cite{Descartes-Geometrie} taraf\i ndan verildi.
\c Carpman\i n birle\c stirme, de\u gi\c stirme,
ve toplama \"uzerinde da\u g\i lma \"ozellikleri,
ve s\i f\i r olmayan her ger\c cel say\i n\i n tersinin var oldu\u gu,
apa\c c\i k de\u gildir.
\"Oklid'de bulunan orant\i\ kuram\i n\i\ kullanmadan
Hilbert \cite{Hilbert-10}, \c carpman\i n \"ozelliklerini
sert bir \c sekilde kurur.
Bu i\c s i\c cin,
``Thales and the Nine-point Conic'' \cite{Pierce-Thales-9}
ba\c sl\i kl\i\ makalemdeki gibi
\"Oklid'in birinci kitab\i\ \cite{Oklid-2014-T}
yeter.


Ger\c cel say\i lar\i n \textbf{taml\i\u g\i na} g\"ore,
e\u ger
\Sekilde{fig:completeness}ki gibi
\begin{figure}
  \centering
  \begin{pspicture}(-5,0)(5,1)
    \psset{dotsize=4pt 2}
    \psline{o->}(-1,1)(-5,1)
    \psline{o->}(1,0)(5,0)
    \psline{*->}(1,1)(5,1)
    \psline{*->}(-1,0)(-5,0)
  \end{pspicture}
  \caption{Ger\c cel say\i lar\i n taml\i\u g\i}
  \label{fig:completeness}
\end{figure}
ger\c cel say\i lar do\u grusu iki par\c caya k\i r\i l\i rsa,
o zaman par\c calar\i n birinin u\c c noktas\i\ vard\i r.
Taml\i\u g\i\
\Teoremde{thm:stp} kullanaca\u g\i z.

\textbf{Sayma say\i lar\i} k\"umesini $\N$ ile yazal\i m.
O zaman
\begin{equation*}
  \N=\{1,2,3,\dots\}.
\end{equation*}
Bu k\"umesi,
$\R$'nin bir altk\"umesi olarak d\"u\c s\"un\"ulebilir.
Bu varsay\i m\i, \Teoremde{thm:ivt} kullanaca\u g\i z.

\section{K\"u\c c\"uk ve b\"uy\"uk k\"umeler}

$\N$'nin baz\i\ altk\"umeleri \textbf{k\"u\c c\"ukt\"ur.}
Tan\i ma g\"ore
\begin{compactenum}[i)]
\item
  her sonlu k\"ume k\"u\c c\"ukt\"ur;
\item
  k\"u\c c\"uk bir k\"ume taraf\i ndan kapsanan her k\"ume de k\"u\c c\"ukt\"ur;
\item
  iki k\"u\c c\"uk k\"umenin birle\c simi de k\"u\c c\"ukt\"ur.
\end{compactenum}
Ayr\i ca
$\N$'nin her sonlu altk\"umesinin t\"umleyeni
\textbf{k\"u\c c\"ukt\"ur,}
dolay\i s\i yla
\begin{compactenum}[i)]
  \setcounter{enumi}3
\item
  sonlu t\"umleyeni olan her k\"ume b\"uy\"ukt\"ur;
\item
  b\"uy\"uk bir k\"ume kapsayan her k\"ume de b\"uy\"ukt\"ur;
\item
  iki b\"uy\"uk k\"umenin kesi\c simi de b\"uy\"ukt\"ur.
\end{compactenum}
$\N$'nin her altk\"umesinin
 ya k\"u\c c\"uk ya da b\"uy\"uk oldu\u gunu
ama ikisi olmad\i\u g\i n\i,
B\"ol\"um \numarada{ch:algebra} bahsedilen
\emph{Se\c cim Aksiyomunu}
kullanarak varsay\i yoruz.

\section{Diziler ve denklikleri}

Bir \textbf{dizi,}
$\N$'den $\R$'ye giden bir fonksiyondur.
Dizilerin olu\c sturdu\u gu k\"ume
\begin{equation*}
  \R^{\N}
\end{equation*}
olarak yazabiliriz.
\begin{itemize}
  \item
Ya $(a_k\colon k\in\N)$,
\item
  ya da $(a_1,a_2,a_3,\dots)$,
  \item
    ya da k\i saca $\seq a$
\end{itemize}
olarak ayn\i\ diziyi
yazabiliriz.
Girdilerine g\"ore,
\begin{align*}
  \seq a+\seq b&=(a_k+b_k\colon k\in\N),&
    \seq a\seq b&=(a_kb_k\colon k\in\N)
\end{align*}
kurallar\i n\i\ kullanarak
iki dizinin toplam\i n\i\ ve \c carp\i m\i n\i\
tan\i mlar\i z.
\c Simdi tan\i ma g\"ore
\begin{equation*}
  \seq a\sim\seq b\iff\{k\in\N\colon a_k=b_k\}\text{ b\"uy\"ukt\"ur}
\end{equation*}
olsun.

\begin{theorem}
  $\sim$ ba\u g\i nt\i s\i\ bir denklik ba\u g\i nt\i s\i d\i r.
\end{theorem}

Verilen $\sim$ ba\u g\i nt\i s\i na g\"ore
$[\seq a]$ olarak
bir $\seq a$ dizisinin denklik s\i n\i f\i n\i\ yazal\i m.
B\"oylece
\begin{equation*}
  [\seq a]=\{\seq x\in\R^{\N}\colon\seq a\sim\seq x\}.
\end{equation*}
B\"ut\"un dizilerin denklik s\i n\i flar\i,
$\stR$ k\"umesini olu\c stursun.  B\"oylece
\begin{equation}\label{eqn:stR}
  \stR=\{[\seq x]\colon\seq x\in\R^{\N}\}.
\end{equation}

\begin{theorem}
  $\sim$ ba\u g\i nt\i s\i,
  $\R^{\N}$ halkas\i n\i n i\c slemlerine sayg\i\ g\"osterir,
  yani
  \begin{equation*}
    \seq a\sim\seq c\And\seq b\sim\seq d
    \implies\seq a+\seq b\sim\seq c+\seq d\And\seq a\seq b\sim\seq c\seq d.
  \end{equation*}
  Bu \c sekilde $\stR$, iyitan\i mlanm\i\c s bir cisim olur.
  Ayr\i ca
  \begin{equation*}
    [\seq a]<[\seq b]\iff\{k\in\N\colon a_k<b_k\}\text{ b\"uy\"ukt\"ur}
  \end{equation*}
  kural\i na g\"ore $\stR$,
   iyitan\i mlanm\i\c s \emph{s\i ralanm\i\c s} bir cisim olur.
   Bu s\i ralanm\i\c s cisme
   \begin{equation*}
     x\mapsto[x,x,x,\dots]
   \end{equation*}
\emph{k\"o\c segen} g\"ommesi ile
  $\R$ g\"om\"ul\"ur.
\end{theorem}

$\stR$ cisminin elemanlar\i na \textbf{ger\c cel\"ust\"u}
(\emph{hyperreal}) denebilir.
Son teoremi kullanarak
$\R$'yi $\stR$'nin bir altk\"umesi olarak d\"u\c s\"unece\u giz;
b\"oylece her ger\c cel say\i, ger\c cel\"ust\"ud\"ur.

\section{Sonsuz ve sonsuzk\"u\c c\"uk say\i lar}

Ger\c cel bir say\i\ gibi  her $[\seq a]$ ger\c cel\"ust\"u say\i n\i n,
\begin{equation*}
  \left.
  \begin{array}{rr}
    [\seq a]\geq0\text{ ise}&[\seq a]\\{}
    [\seq a]<0\text{ ise}&-[\seq a]
  \end{array}
  \right\}=
  \bigl|[\seq a]\bigr|
\end{equation*}
kural\i n\i\ sa\u glayan
$\bigl|[\seq a]\bigr|$ \textbf{mutlak de\u geri} vard\i r.
E\u ger ger\c cel\"ust\"u bir say\i n\i n mutlak de\u geri
\begin{compactitem}
\item
  \emph{bir} pozitif ger\c cel say\i dan
k\"u\c c\"uk ise,
ger\c cel\"ust\"u say\i ya \textbf{sonlu} densin;
\item
  \emph{her} pozitif ger\c cel say\i dan
k\"u\c c\"uk ise,
ger\c cel\"ust\"u say\i ya \textbf{sonsuzk\"u\c c\"uk}
(\emph{infinitesimal}) densin.
\end{compactitem}

\begin{example}
  $0$ sonsuzk\"u\c c\"ukt\"ur.
\end{example}

\begin{theorem}
  $0$'dan farkl\i\ olan
  hi\c c ger\c cel say\i\ sonsuzk\"u\c c\"uk de\u gildir.  
\end{theorem}

\begin{example}\label{ex:k-inv}
  $[k\inv\colon k\in\N]$,
      sonsuzk\"u\c c\"uk ger\c cel\"ust\"u bir say\i d\i r.
\end{example}

Tabii ki sonlu olmayan ger\c cel\"ust\"u bir say\i, \textbf{sonsuzdur.}

\begin{example}
  $[k\colon k\in\N]$,
  sonsuz ger\c cel\"ust\"u bir say\i d\i r.
\end{example}

\begin{theorem}
  S\i f\i r olmayan bir $[\seq a]$ ger\c cel\"ust\"u say\i s\i\ i\c cin
  \begin{equation*}
    \text{$[\seq a]$ sonsuzk\"u\c c\"ukt\"ur}
    \iff
    \text{$[\seq a]\inv$ sonsuzdur.}
  \end{equation*}
\end{theorem}

\section{Sonsuzyak\i nl\i k}

Tan\i ma g\"ore
\begin{equation*}
  [\seq a]\approx[\seq b]\iff
  [\seq a]-[\seq b]\text{ sonsuzk\"u\c c\"ukt\"ur}
\end{equation*}
olsun.
E\u ger $[\seq a]\approx[\seq b]$ ise
$[\seq b]$, $[\seq a]$'ya \textbf{sonsuzyak\i nd\i r}
(\emph{infinitely close}).

\begin{theorem}\mbox{}%\label{thm:0}
  \begin{enumerate}
  \item 
  $\stR$ cisminde $\approx$ ba\u g\i nt\i s\i\
    bir denklik ba\u g\i nt\i s\i d\i r.
  \item
    $\text{$[\seq a]$ sonsuzk\"u\c c\"ukt\"ur}\iff[\seq a]\approx0$.
\item
  $[\seq a]\approx[\seq b]\iff[\seq a]-[\seq b]\approx0$.
  \end{enumerate}
\end{theorem}

\begin{theorem}
    \.Iki farkl\i\ ger\c cel say\i\ sonsuzyak\i n olamaz.
\end{theorem}

\begin{theorem}\label{thm:finite}
  Bir ger\c cel say\i ya sonsuzyak\i n olan
  her ger\c cel\"ust\"u say\i,
  sonludur.
\end{theorem}

\begin{theorem}\label{thm:stp}
  Her sonlu ger\c cel\"ust\"u say\i,
  bir ger\c cel say\i ya sonsuzyak\i nd\i r.
\end{theorem}

\begin{proof}
  E\u ger $[\seq a]$ sonlu ise, o zaman
  $\{x\in\R\colon x<[\seq a]\}$ k\"umesinin $b$ 
  \"ust s\i n\i r\i\ vard\i r.
  Bu durumda verilen k\"ume $-b$ eleman\i n\i\ i\c cerdi\u ginden
  bo\c s de\u gildir.
  O halde verilen k\"umenin supremumu vard\i r,
  ve bu supremum, $[\seq a]$'ya sonsuzyak\i nd\i r.
\end{proof}

Kan\i tta bulunan ger\c cel say\i,
$[\seq a]$'n\i n \textbf{standart par\c cas\i d\i r}
(\emph{standard part}).

\begin{theorem}\label{thm:inf-close}
  E\u ger
  $[\seq a]\approx[\seq c]$, $[\seq b]\approx[\seq d]$, ve $[\seq e]$ sonlu ise,
  o zaman
  \begin{align*}
    [\seq a]+[\seq b]&\approx[\seq c]+[\seq d],&[\seq a][\seq e]&\approx[\seq c][\seq e].
  \end{align*}
  Sonlu ger\c cel\"ust\"u bir say\i ya sonsuzyak\i n say\i lar da
  sonludur.
\end{theorem}

\addpart{Analiz}

\section{Limitler}

\begin{theorem}
  E\u ger $f\colon\R\to\R$ ise, o zaman
  \begin{equation*}
    [\seq x]\mapsto[f(x_k)\colon k\in\N],
  \end{equation*}
  $\stR$'den kendisine giden
  iyitan\i mlanm\i\c s bir fonksiyondur.
\end{theorem}

\c Simdi
\begin{equation*}
  \stf
\end{equation*}
olarak
teoremde bulunan fonksiyonu
yazal\i m;
ayr\i ca $a$ ve $L$, ger\c cel say\i\ olsun.
E\u ger her $[\seq x]$ ger\c cel\"ust\"u say\i s\i\ i\c cin
\begin{equation*}
  a\approx[\seq x]\And a\neq[\seq x]\implies\stf[\seq x]\approx L
\end{equation*}
gerektirmesi sa\u glan\i rsa, o zaman tan\i m\i m\i za g\"ore $L$,
$f$'nin $a$'daki \textbf{limitidir.}
Bu limiti
\begin{equation*}
  \lim_{x\to a}f(x)\quad\text{ veya }\quad\lim_af
\end{equation*}
ile
yazar\i z.
Limit tan\i m\i m\i z\i n
``standart'' tan\i ma denk oldu\u gunu,
B\"ol\"um \numarada{ch:ed}
Teorem \numarada{thm:ed} kan\i tlayaca\u g\i z.

\begin{theorem}\label{thm:+.}
  $\lim_af=L$ ve $\lim_ag=M$ ise
  \begin{equation*}
    \lim_a(f+g)=L+M\And\lim_a(fg)=LM.
  \end{equation*}
\end{theorem}

\begin{proof}
  E\u ger $[\seq x]\approx a$ ve $[\seq x]\neq a$ ise,
  o zaman varsay\i ma g\"ore
  \begin{align*}
    \stf[\seq x]&\approx L,&\stg[\seq x]&\approx M.
  \end{align*}
  \"Ozellikle \Teorem{thm:finite} sayesinde
  $\stf[\seq x]$ sonludur.  \c Simdi
  \Teorem{thm:inf-close} sayesinde
  \begin{align*}
    \stf[\seq x]+\stg[\seq x]&\approx L+M,&
    \stf[\seq x]\stg[\seq x]&\approx\stf[\seq x]M\approx LM.\qedhere
  \end{align*}
\end{proof}

\section{S\"ureklilik}

E\u ger her $a$ ger\c cel say\i s\i\ i\c cin $f(a)=\lim_af$ ise,
o zaman $f$ \textbf{s\"ureklidir.}

\begin{theorem}\label{thm:cont}
  $f$'nin s\"urekli olmas\i\ i\c cin gerek ve yeter bir ko\c sul,
  t\"um sonlu $[\seq x]$ ve $[\seq y]$
  ger\c cel\"ust\"u say\i lar\i\ i\c cin
  \begin{equation*}
    [\seq x]\approx[\seq y]\implies\stf[\seq x]\approx\stf[\seq y].
  \end{equation*}
\end{theorem}

\begin{example}
  E\u ger $f(x)=x^2$ ise,
  o zaman $f$ s\"ureklidir,
  ama \"Ornek \numarada{ex:k-inv}ki gibi
  $[k\inv\colon k\in\N]$ ger\c cel\"ust\"u say\i s\i n\i n
  sonsuzk\"u\c c\"uk oldu\u gu halde
  \begin{equation*}
    \stf[k+k\inv\colon k\in\N]=\stf[k\colon k\in\N]+2.
  \end{equation*}
\end{example}

\begin{theorem}[Arade\u ger]\label{thm:ivt}
  $f$ s\"urekli, $a<b$, ve $f(a)<c<f(b)$ %veya $f(a)>c>f(b)$
  olsun.
  O zaman bir $d$ i\c cin $a<d<b$ ve $f(d)=c$.
\end{theorem}

\begin{proof}
\.Istedi\u gimiz $d$ ger\c cel say\i s\i,
bir $[d_n\colon n\in\N]$ ger\c cel\"ust\"u say\i s\i nin
standart par\c cas\i\ olacak.
\emph{\"Ozyineleme} ile
$(d_n\colon n\in\N)$ dizisini tan\i mlayaca\u g\i z.
\"Once
\begin{equation*}
  \delta_1=b-a
\end{equation*}
olsun,
ve bir $m$ i\c cin $\delta_m$ tan\i mlanm\i\c s ise
\begin{equation*}
  \delta_{m+1}=\frac{\delta_m}2
\end{equation*}
olsun.
K\i saca her $n$ sayma say\i s\i\ i\c cin
  \begin{equation*}%\label{eqn:delta-def}
    \delta_n=\frac{b-a}{2^{n-1}}
  \end{equation*}
  olsun.
  \c Simdi
  \begin{equation*}
    d_1=a
  \end{equation*}
olsun.
  O zaman $n=1$ durumunda
  \begin{align}\label{eqn:delta}
    a&\leq d_n<d_n+\delta_n\leq b,&
    f(d_n)\leq c<f(d_n+\delta_n)
  \end{align}
  e\c sitsizlikleri sa\u glan\i r.
  \c Simdi bir $m$ sayma say\i s\i\ i\c cin $n=m$ durumunda
  \eqref{eqn:delta} e\c sitsizliklerinin sa\u gland\i\u g\i n\i\
  varsayal\i m.
  O zaman
  \begin{equation*}
    \left.
    \begin{array}{ll}
      \text{$f(d_m+\delta_{m+1})\leq c$ ise}&d_m+\delta_{m+1}\\
      \text{$c<f(d_m+\delta_{m+1})$ ise}&d_m
    \end{array}
    \right\}=d_{m+1}
  \end{equation*}
  olsun.
  Bu durumda $n=m+1$ durumunda
  \eqref{eqn:delta} e\c sitsizlikleri sa\u glan\i r.

  \"Ozyineleme ile her $n$ i\c cin $d_n$ tan\i mlanm\i\c st\i.
  T\"umevar\i mdan her $n$ i\c cin
  \eqref{eqn:delta} e\c sitsizlikleri sa\u glan\i r.
  O zaman
  \begin{equation*}
\stf[d_n\colon n\in\N]\leq c<\stf[d_n+\delta_n\colon n\in\N],    
  \end{equation*}
  ama $[\delta_n\colon n\in\N]$ sonsuzk\"u\c c\"ukt\"ur.
  \Teorem{thm:cont} sayesinde
  \begin{equation*}
    \stf[d_n\colon n\in\N]\approx\stf[d_n+\delta_n\colon n\in\N].
  \end{equation*}
  Sonu\c c olarak
  \begin{equation*}
    \stf[d_n\colon n\in\N]\approx c.
  \end{equation*}
  \c Simdi $d$,
  $[d_n\colon n\in\N]$ ger\c cel\"ust\"u say\i s\i nin
  standart par\c cas\i\ olsun.
  O zaman
  \begin{equation*}
    f(d)\approx\stf[d_n\colon n\in\N],
  \end{equation*}
dolay\i s\i yla $f(d)=c$.
\end{proof}

\begin{theorem}[S\i n\i rl\i l\i k]
  $f$ s\"urekli ve $a<b$ olsun.
  O zaman  $\R$'nin $\{f(x)\colon a\leq x\leq b\}$
  altk\"umesi s\i n\i rl\i d\i r.
\end{theorem}

\begin{proof}
  S\i n\i rl\i\ olmas\i n.
  O zaman her $n$ i\c cin, bir $c_n$ i\c cin,
  \begin{align*}
    a&\leq c_n\leq b,&
    \abs{f(c_n)}&>n.
  \end{align*}
  O zaman $\stf[\seq c]$ sonsuzdur.
  E\u ger $[\seq c]$ ger\c cel\"ust\"u say\i s\i n\i n standart par\c cas\i\
  $c$ ise, o zaman $f(c)$ sonlu oldu\u gundan $f$ s\"urekli olamaz.
\end{proof}

\begin{theorem}[Maksimum-minimum]\label{thm:max-min}
  $f$ s\"urekli ve $a<b$ olsun.
  O zaman baz\i\ $c$ ve $d$ i\c cin,
  \begin{align*}
    a\leq c&\leq b,&a\leq d&\leq b,
  \end{align*}
  ve ayr\i ca her $x$ i\c cin,
  \begin{equation*}
    a\leq x\leq b\implies f(c)\leq f(x)\leq f(d).
  \end{equation*}
\end{theorem}

\section{T\"urevler}

Tan\i ma g\"ore
\begin{equation*}
  \lim_{x\to a}\frac{f(a)-f(x)}{a-x}=f'(a).
\end{equation*}
Burada $f'$ fonksiyonu, $f$'nin \textbf{t\"urevidir.}
O halde
\begin{equation*}
  [\seq h]\approx0\And[\seq h]\neq0\implies
  \frac{\stf(x+[\seq h])-f(x)}{[\seq h]}\approx f'(x).
\end{equation*}
Bu nedenle $f'$ fonksiyonu
\begin{equation*}
  \frac{\dee f(x)}{\dee x}
\end{equation*}
olarak yaz\i labilir.
Tabii ki $f$'nin t\"urevi olmayabilir.

\begin{theorem}\label{thm:x'}
  $\dee1/\dee x=0$ ve
  $\dee x/\dee x=1$.
\end{theorem}

\begin{theorem}\label{thm:+'}
  $(f+g)'=f'+g'$.
\end{theorem}

\begin{theorem}\label{thm:.'}
  $(fg)'=f'g+fg'$.
\end{theorem}

\begin{proof}
  \Teoremi{thm:+.} kullanarak
  \begin{equation*}
    \frac{f(a)g(a)-f(x)g(x)}{a-x}
    =f(a)\frac{g(a)-g(x)}{a-x}+\frac{f(a)-f(x)}{a-x}g(x)
  \end{equation*}
  e\c sitli\u ginden istedi\u gimiz sonucu elde ederiz.
\end{proof}

\begin{example}
  Teoremler \ref{thm:x'}, \ref{thm:+'}, ve \ref{thm:.'} sayesinde
  \begin{equation*}
    \frac{\dee(a_0+a_1x+\dots+a_nx^n)}{\dee x}
    =a_1+a_2x+\dots+na_nx^{n-1}.
  \end{equation*}
  Burada $a_1+a_2x+\dots+na_nx^{n-1}$,
  \begin{equation*}
    f(x)=a_1+a_2x+\dots+na_nx^{n-1}
  \end{equation*}
  denkli\u ginin tan\i mlad\i\u g\i\ teorem olarak
  anla\c s\i l\i r.
\end{example}

\begin{theorem}
  E\u ger $f$'nin t\"urevi varsa,
  o zaman $f$ s\"ureklidir.
\end{theorem}

\section{Integraller}

\textbf{Do\u gal say\i lar} k\"umesi $\upomega$
(\emph{omega}) ile yazal\i m.
O zaman
\begin{equation*}
  \upomega=\{0,1,2,\dots\}=\{0\}\cup\N.
\end{equation*}
$\stR$'nin \eqref{eqn:stR} tan\i m\i nda $\R$'nin yerine
s\i ras\i yla $\N$ ve $\upomega$ konularak
$\stN$ ve $\sto$ elde edilir.
O zaman
\begin{equation*}
  \sto=\{0\}\cup\stN.
\end{equation*}
Burada $\sto$'n\i n elemanlar\i na
\textbf{do\u gal\"ust\"u say\i} densin;
Her do\u gal say\i,
do\u gal\"ust\"u olarak d\"u\c s\"unebilir.

Tan\i m k\"umesi
bir $[a,b]$ kapal\i\ s\i n\i rl\i\ aral\i\u g\i n\i\ kapsayan
bir $f$ fonksiyonu i\c cin,
e\u ger bir $n$ sayma say\i s\i\ i\c cin
baz\i\ $x_i$ ger\c cel say\i lar\i
\begin{equation*}
  a=x_0<x_1<\dots<x_n=b
\end{equation*}
ko\c sulunu sa\u glarsa,
o zaman $\{x_0,x_1,\dots,x_n\}$ k\"umesi,
$[a,b]$ aral\i\u ginin bir \textbf{par\c calanmas\i d\i r.}
E\u ger ayr\i ca
$t_i$ ger\c cel say\i lar\i
\begin{equation*}
  x_{i-1}\leq t_i\leq x_i
\end{equation*}
ko\c sulunu sa\u glarsa, o zaman
\begin{equation}\label{eqn:Rs}
  \sum_{i=1}^nf(t_i)(x_i-x_{i-1})
\end{equation}
toplam\i,
$[a,b]$ aral\i\u g\i nda $f$'nin bir
\textbf{Riemann toplam\i d\i r.}

\c Simdi
girdileri sayma say\i s\i\ olan bir $\seq n$ dizisi i\c cin,
girdileri $[a,b]$ aral\i\u g\i nda
$f$'nin Riemann toplamlar\i\ olan bir
\begin{equation*}
  \left(\sum_{i=1}^{n_k}
  f(t_{k,i})(x_{k,i}-x_{k,i-1})\colon k\in\N\right)
\end{equation*}
dizisi verilsin.
O zaman
\begin{align*}
  \seq t_{\seq i}&=(t_{k,i_k}\colon k\in\N),&
  \seq x_{\seq i}&=(x_{k,i_k}\colon k\in\N),
\end{align*}
ve
\begin{equation*}
  \seq x_{\seq i-1}=(x_{k,i_k-1}\colon k\in\N)
\end{equation*}
tan\i mlanabilir.
Burada bir $k$ i\c cin girdiler tan\i ms\i z olabilir,
ama $1\leq i_k\leq n_k$ ise tan\i ml\i d\i r,
ve $1\leq[\seq i]\leq[\seq n]$ ise
\begin{equation*}
  \{k\in\N\colon i_k\leq n_k\}
\end{equation*}
k\"umesi b\"uy\"ukt\"ur.
Bu durumda $[\seq t_{\seq i}]$,
$[\seq x_{\seq i}]$, ve
$[\seq x_{\seq i-1}]$ iyitan\i ml\i d\i r, ve
\begin{equation*}
  [\seq x_{\seq i-1}]\leq[\seq t_{\seq i}]\leq[\seq x_{\seq i}].
\end{equation*}
Ayr\i ca
\begin{equation*}
  [\seq i]=[\seq j]\implies[\seq t_{\seq i}]=[\seq t_{\seq j}]
  \And[\seq x_{\seq i}]=[\seq x_{\seq j}]
  \And[\seq x_{\seq i-1}]=[\seq x_{\seq j-1}].
\end{equation*}
\c Simdi
\begin{equation*}
  \left[\sum_{i=1}^{n_k}f(t_{k,i})(x_{k,i}-x_{k,i-1})\colon
    k\in\N\right]
\end{equation*}
ger\c cel\"ust\"u say\i s\i
\begin{equation*}%\label{eqn:inf-Rs}
  \sum_{[\seq i]=1}^{[\seq n]}
  \stf[\seq t_{\seq i}]([\seq x_{\seq i}]-[\seq x_{\seq i-1}])
\end{equation*}
toplam\i\ olarak
ve k\i saca
\begin{equation*}
  \RS TX
\end{equation*}
yaz\i labilir.
E\u ger her durumda
\begin{equation*}
  [\seq x_{\seq i}]
  \approx
  [\seq x_{\seq i-1}]
\end{equation*}
ise, o zaman toplam,
$[a,b]$ aral\i\u g\i nda $f$'nin \textbf{sonsuz
  bir Riemann toplam\i} densin.
O zaman
\begin{equation*}
  \{[\seq x_0],[\seq x_1],\dots,[\seq x_{\seq n}]\}
\end{equation*}
k\"umesi, yani
\begin{equation*}
\{[\seq x_{\seq i}]\colon[\seq i]\in\sto\And
    [\seq i]\leq[\seq n]\}
\end{equation*}
  k\"umesi,
$[a,b]$ aral\i\u g\i n\i n
\textbf{sonsuz bir par\c calanmas\i d\i r.}


E\u ger
$[a,b]$ aral\i\u g\i nda $f$'nin her sonsuz Riemann toplam\i\
ayn\i\ $I$ ger\c cel say\i s\i na sonsuzyak\i n ise,
o zaman $I$,
$[a,b]$ aral\i\u g\i nda $f$'nin \textbf{integralidir,}
ve
\begin{equation*}
  I=\int_a^bf=\int_a^bf(x)\dee x
\end{equation*}
yaz\i l\i r.

\begin{theorem}
  E\u ger bir fonksiyon bir aral\i kta s\"urekli ise
  o aral\i kta fonksiyonun integrali vard\i r.
\end{theorem}

\begin{proof}
  Bir
  $\RS TX$ sonsuz Riemann toplam\i\ verilsin.
  \Teorem{thm:max-min} sayes\i nda
  her $[x_{k,i-1},x_{k,i}]$ aral\i\u ginda
  $f$'nin en k\"u\c c\"uk de\u geri
  bir $c_{k,i}$ noktas\i nda
  ve en b\"uyuk de\u geri bir $d_{k,i}$ noktas\i nda al\i n\i r.
  Bu durumda
  \begin{align*}
    \RS CX&=\RSmin X,&\RS DX&=\RSmax X
  \end{align*}
  olsun.
  O zaman
  \begin{equation*}
    \RSmin X\leq\RS TX\leq\RSmax X.
  \end{equation*}
    Ayr\i ca her $k$ i\c cin, bir $j_k$ i\c cin,
  \begin{equation*}
    \maks_{1\leq i\leq n_k}(f(d_{k,i})-f(c_{k,i}))
    =f(d_{k,i_k})-f(c_{k,i_k}),
  \end{equation*}
dolay\i s\i yla
  \begin{equation*}
    \RSmax X-\RSmin X\leq
    (\stf[\seq d_{\seq j}]-\stf[\seq c_{\seq j}])(b-a).
  \end{equation*}
  Ayr\i ca $[\seq d_{\seq j}]$ ve $[\seq c_{\seq j}]$
  sonsuzyak\i nd\i r,
  \c c\"unk\"u
  $[\seq x_{\seq j-1}]$ ve $[\seq x_{\seq j}]$'nin
  aras\i ndad\i r.
  Bundan dolay\i\
  $f$ s\"urekli oldu\u gundan
  \begin{equation*}
    \stf[\seq d_{\seq j}]-\stf[\seq c_{\seq j}]\approx0,
  \end{equation*}
  ve sonu\c c olarak
  \begin{equation*}
    \RSmin X\approx\RS TX\approx\RSmax X.
  \end{equation*}
  E\u ger farkl\i\ bir $Y$ sonsuz par\c calanmas\i\ verilirse,
  o zaman
  \begin{equation*}
    \RSmin X\leq\RSmin{X\cup Y}\leq\RSmax{X\cup Y}\leq\RSmax X
  \end{equation*}
  Bundan dolay\i\ $[a,b]$ aral\i\u g\i nda
  $f$'nin b\"ut\"un sonsuz Riemann toplam\i\
  sonsuzyak\i nd\i r.
\end{proof}

Tan\i ma g\"ore
\begin{equation*}
  \int_b^af=-\int_a^bf.
\end{equation*}

\begin{theorem}
  \begin{equation*}
    \int_a^bf=\int_a^cf+\int_c^bf.
  \end{equation*}
\end{theorem}

\begin{theorem}
  E\u ger $f$ s\"urekli ise, o zaman
  \begin{equation*}
    \frac{\dee}{\dee x}\int_a^xf=f.
  \end{equation*}
\end{theorem}

\begin{proof}
  $F(x)=\int_a^xf$ olsun, ve $[\seq h]$,
  s\i f\i r olmas\i n ama sonsuzk\"u\c c\"uk olsun.
  E\u ger $h_k\neq0$ ise,
  o zaman $f$ s\"urekli oldu\u gundan
  $[x,x+h_k]$ veya $[x+h_k,x]$ aral\i\u g\i nda,
  $f$ en k\"u\c c\"uk de\u gerini bir $c_k$ noktas\i nda,
  ve en b\"uy\"uk de\u gerini bir $d_k$ noktasinda, al\i r.
  O zaman
  \begin{equation*}
    f(c_k)\leq\frac{F(x+h_k)}{h_k}\leq f(d_k),
  \end{equation*}
  dolay\i s\i yla
  \begin{equation*}
    \stf[\seq c]
    \leq\frac{\stF(x+[\seq h])}{[\seq h]}\leq\stf[\seq d].
  \end{equation*}
  \.Istedi\u gimiz sonu\c c \c c\i kar
  \c c\"unk\"u $[\seq c]\approx x\approx[\seq d]$.
\end{proof}

\addpart{Ek}

%\appendix

\section{Cebir}\label{ch:algebra}

Cebir a\c c\i s\i ndan
$\N$'nin altk\"umeleri,
bir \emph{Boole halkas\i} olu\c sturur,
ve $\N$'nin k\"u\c c\"uk altk\"umeleri,
o halkan\i n asal bir ideali (\emph{prime ideal}) olu\c sturur.
Her halkan\i n asal bir ideali var oldu\u gu,
Se\c cim Aksiyomu ile kan\i tlanabilir,
ama bu aksiyomdan az g\"u\c c\"ud\"ur
\cite{MR0284328}.

Ayr\i ca $\{x\in\R^{\N}\colon x\sim 0\}$ k\"umesi,
$\R^{\N}$ halkas\i n\i n as\i l olmayan (\emph{non-principal})
asal bir idealidir, ve
\begin{equation*}
  \stR=\R^{\N}/\{x\in\R^{\N}\colon x\sim 0\}.
\end{equation*}
E\u ger tam tersine $P$,
$\R^{\N}$ halkas\i n\i n as\i l olmayan asal bir ideali ise,
o zaman $\N$'nin k\"u\c c\"uk altk\"umeleri,
$P$'nin $a$ elemanlar\i\ i\c cin
\begin{equation*}
  \{k\in\N\colon a_k\neq0\}
\end{equation*}
olarak tan\i mlanabilir.

$\stR$ cisminin sonlu elemanlar\i,
bir $\mathfrak O$ de\u gerlendirme halkas\i\
(\emph{valuation ring}) olu\c sturur,
ve bunun $\mathfrak M$ maksimal idealinin elemanlar\i,
$\stR$'nin sonsuzk\"u\c c\"uk elemanlar\i d\i r.
Bu durumda
tan\i m k\"umesi $\R$ olan
\begin{equation*}
  x\mapsto[x,x,x,\dots]+\mathfrak M
\end{equation*}
g\"ondermesi,
$\mathfrak O/\mathfrak M$ \"ust\"une bir $\phi$ izomorf\/izmas\i d\i r,
ve $[\seq a]$ sonlu ise $\phi^{-1}([\seq a]+\mathfrak M)$,
$[\seq a]$'n\i n standart par\c cas\i d\i r.

\section{Epsilon-delta tan\i m\i}\label{ch:ed}

\begin{theorem}\label{thm:ed}
  $f$'nin $a$'daki limitinin $L$ olmas\i\ i\c cin
  $\lim_af=L$ gerek ve yeter bir ko\c sul,
  $\R$'de her pozitif $\epsilon$ i\c cin,
  bir pozitif $\delta$ i\c cin,
  her $x$ i\c cin
  \begin{equation*}
    0<\abs{x-a}<\delta\implies\abs{f(x)-L}<\epsilon.
  \end{equation*}
\end{theorem}

\begin{proof}
  \.Ilk olarak
  verilen ko\c sulun do\u gru olmas\i n\i\ varsayal\i m.
  Bir $\seq x$ \emph{dizisi} i\c cin
  $[\seq x]\approx a$ ama $[\seq x]\neq a$ olsun.
  O halde $\stf[\seq x]\approx L$ g\"osterece\u giz.
  \c Simdi $\epsilon$, pozitif ger\c cel bir say\i\ olsun;
  $\{k\in\N\colon\abs{f(x_k)-L}<\epsilon\}$
  k\"umesinin b\"uy\"uk oldu\u gunu g\"ostermek yeter.
  Varsay\i ma g\"ore pozitif ger\c cel bir $\delta$ i\c cin
  o k\"ume $\{k\in\N\colon0<\abs{x_k-a}<\delta\}$ k\"umesini kapsar;
  ayr\i ca bu k\"ume b\"uy\"ukt\"ur.

  Tam tersine $\epsilon$,
  pozitif ger\c cel bir say\i\ olsun.
  M\"umk\"unse her pozitif ger\c cel $\delta$ i\c cin,
  ger\c cel bir $x$ i\c cin,
  \begin{equation*}
    0<\abs{x-a}<\delta\And\abs{f(x)-L}\geq\epsilon
  \end{equation*}
  olsun.
  O zaman bir $\seq x$ \emph{dizisi} i\c cin
  her $k$ sayma say\i s\i\ i\c cin
  \begin{equation*}
    0<\abs{x_k-a}<\frac1k\And\abs{f(x_k)-L}\geq\epsilon.
  \end{equation*}
  Bu durumda $[\seq x]\approx a$ ve $[\seq x]\neq a$,
  ama $\stf[\seq x]\not\approx L$.
\end{proof}

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