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\begin_body
\begin_layout Title
Teorija Analize 1 —
IŠRM 2023/24
\end_layout
\begin_layout Author
\noun on
Anton Luka Šijanec
\end_layout
\begin_layout Date
\begin_inset ERT
status open
\begin_layout Plain Layout
\backslash
today
\end_layout
\end_inset
\end_layout
\begin_layout Abstract
Povzeto po zapiskih s predavanj profesorja Oliverja Dragičevića.
\end_layout
\begin_layout Standard
\begin_inset CommandInset toc
LatexCommand tableofcontents
\end_inset
\end_layout
\begin_layout Section
Števila
\end_layout
\begin_layout Definition*
Množica je matematični objekt,
ki predstavlja skupino elementov.
Če element
\begin_inset Formula $a$
\end_inset
pripada množici
\begin_inset Formula $A$
\end_inset
,
pišemo
\begin_inset Formula $a\in A$
\end_inset
,
sicer pa
\begin_inset Formula $a\not\in A$
\end_inset
.
Množica
\begin_inset Formula $B$
\end_inset
je podmnožica množice
\begin_inset Formula $A$
\end_inset
,
pišemo
\begin_inset Formula $B\subset A$
\end_inset
,
če
\begin_inset Formula $\forall b\in B:b\in A$
\end_inset
.
Presek
\begin_inset Formula $B$
\end_inset
in
\begin_inset Formula $C$
\end_inset
označimo
\begin_inset Formula $B\cap C\coloneqq\left\{ x;x\in B\wedge x\in C\right\} $
\end_inset
.
Unijo
\begin_inset Formula $B$
\end_inset
in
\begin_inset Formula $C$
\end_inset
označimo
\begin_inset Formula $B\cup C\coloneqq\left\{ x;x\in B\vee x\in C\right\} $
\end_inset
.
Razliko/komplement
\begin_inset Quotes gld
\end_inset
\begin_inset Formula $B$
\end_inset
manj/brez
\begin_inset Formula $C$
\end_inset
\begin_inset Quotes grd
\end_inset
označimo
\begin_inset Formula $B\setminus C\coloneqq\left\{ x;x\in B\wedge x\not\in C\right\} $
\end_inset
.
\end_layout
\begin_layout Subsection
Realna števila
\end_layout
\begin_layout Standard
Množico realnih števil označimo
\begin_inset Formula $\mathbb{R}$
\end_inset
.
V njej obstajata binarni operaciji seštevanje
\begin_inset Formula $a+b$
\end_inset
in množenje
\begin_inset Formula $a\cdot b$
\end_inset
.
\end_layout
\begin_layout Subsubsection
Lastnosti seštevanja
\end_layout
\begin_layout Axiom
Komutativnost:
\begin_inset Formula $\forall a,b\in\mathbb{R}:a+b=b+a$
\end_inset
.
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Axiom
Asociativnost:
\begin_inset Formula $\forall a,b,c\in\mathbb{R}:a+\left(b+c\right)=\left(a+b\right)+c$
\end_inset
,
torej je
\begin_inset Formula $a+\cdots+z$
\end_inset
dobro definiran izraz.
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Axiom
Obstoj enote:
\begin_inset Formula $\exists0\in\mathbb{R}\forall a\in\mathbb{R}:a+0=a$
\end_inset
.
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Axiom
Obstoj inverzov:
\begin_inset Formula $\forall a\in\mathbb{R}\exists b\in\mathbb{R}\ni a+b=0$
\end_inset
.
\end_layout
\begin_deeper
\begin_layout Claim*
Inverz je enoličen.
\end_layout
\begin_layout Proof
Naj bo
\begin_inset Formula $a,b,c\in\mathbb{R}$
\end_inset
in
\begin_inset Formula $a+b=0$
\end_inset
in
\begin_inset Formula $a+c=0$
\end_inset
.
Tedaj
\begin_inset Formula $b=b+0=b+a+c=0+c=c$
\end_inset
.
\end_layout
\begin_layout Corollary*
Inverz je funkcija in aditivni inverz
\begin_inset Formula $a$
\end_inset
označimo z
\begin_inset Formula $-a$
\end_inset
.
Pri zapisu
\begin_inset Formula $a+\left(-b\right)$
\end_inset
običajno
\begin_inset Formula $+$
\end_inset
izpustimo in pišemo
\begin_inset Formula $a-b$
\end_inset
,
čemur pravimo odštevanje
\begin_inset Formula $b$
\end_inset
od
\begin_inset Formula $a$
\end_inset
.
\end_layout
\begin_layout Claim*
\begin_inset Formula $\forall a\in\mathbb{R}:a=-\left(-a\right)$
\end_inset
.
\end_layout
\begin_layout Proof
Naj bo
\begin_inset Formula $b=-a$
\end_inset
in
\begin_inset Formula $c=-b=-\left(-a\right)$
\end_inset
.
Tedaj velja
\begin_inset Formula $c-a=c+b=-\left(-a\right)-a=0$
\end_inset
in
\begin_inset Formula $a=0+a=c-a+a=c+\left(-a\right)+a=c+0=c=-\left(-a\right)$
\end_inset
.
\end_layout
\begin_layout Claim*
\begin_inset Formula $-\left(b+c\right)=-b-c$
\end_inset
\end_layout
\begin_layout Proof
Velja
\begin_inset Formula $b+c+\left(-b-c\right)=b+c+\left(\left(-b\right)+\left(-c\right)\right)=b+\left(-b\right)+c+\left(-c\right)=0$
\end_inset
,
torej je
\begin_inset Formula $b+c$
\end_inset
inverz od
\begin_inset Formula $\left(-b-c\right)$
\end_inset
,
torej je
\begin_inset Formula $-\left(b+c\right)=-b-c$
\end_inset
.
\end_layout
\end_deeper
\begin_layout Subsubsection
Lastnosti množenja
\end_layout
\begin_layout Axiom
Komutativnost:
\begin_inset Formula $\forall a,b\in\mathbb{R}:ab=ba$
\end_inset
.
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Axiom
Asociativnost:
\begin_inset Formula $\forall a,b,c\in\mathbb{R}:a\left(bc\right)=\left(ab\right)c$
\end_inset
,
torej je
\begin_inset Formula $a\cdots z$
\end_inset
dobro definiran izraz.
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Axiom
Obstoj enote:
\begin_inset Formula $\exists1\in\mathbb{R}\forall a\in\mathbb{R}:a1=a$
\end_inset
.
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Axiom
Obstoj inverzov:
\begin_inset Formula $\forall a\in\mathbb{R}\setminus\left\{ 0\right\} \exists b\in\mathbb{R}\setminus\left\{ 0\right\} \ni:ab=1$
\end_inset
\end_layout
\begin_deeper
\begin_layout Claim*
Inverz je enoličen.
\end_layout
\begin_layout Proof
Naj bo
\begin_inset Formula $a,b,c\in\mathbb{R}\setminus\left\{ 0\right\} $
\end_inset
in
\begin_inset Formula $ab=1$
\end_inset
in
\begin_inset Formula $ac=1$
\end_inset
.
Tedaj
\begin_inset Formula $b=b1=bac=1c=c$
\end_inset
.
\end_layout
\begin_layout Corollary*
Inverz je funkcija in multiplikativni inverz
\begin_inset Formula $a$
\end_inset
označimo z
\begin_inset Formula $a^{-1}$
\end_inset
.
Pri zapisu
\begin_inset Formula $a\cdot b^{-1}$
\end_inset
lahko
\begin_inset Formula $\cdot$
\end_inset
izpustimo in pišemo
\begin_inset Formula $a/b$
\end_inset
,
čemur pravimo deljenje
\begin_inset Formula $a$
\end_inset
z
\begin_inset Formula $b$
\end_inset
za neničeln
\begin_inset Formula $b$
\end_inset
.
\end_layout
\end_deeper
\begin_layout Subsubsection
Skupne lastnosti v
\begin_inset Formula $\mathbb{R}$
\end_inset
\end_layout
\begin_layout Axiom
\begin_inset Formula $1\not=0$
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Axiom
Distributivnost:
\begin_inset Formula $\forall a,b,c\in\mathbb{R}:\left(a+b\right)c=ac+bc$
\end_inset
\end_layout
\begin_layout Paragraph
Urejenost
\begin_inset Formula $\mathbb{R}$
\end_inset
\end_layout
\begin_layout Standard
Realna števila delimo na pozitivna
\begin_inset Formula $\mathbb{R}_{+}\coloneqq\left\{ x\in\mathbb{R};x>0\right\} $
\end_inset
,
negativna
\begin_inset Formula $\mathbb{R}_{-}\coloneqq\left\{ x\in\mathbb{R};x<0\right\} $
\end_inset
in ničlo
\begin_inset Formula $0$
\end_inset
.
Če je
\begin_inset Formula $x\in\mathbb{\mathbb{R}}_{+}\cup\left\{ 0\right\} $
\end_inset
,
pišemo
\begin_inset Formula $x\geq0$
\end_inset
,
če je
\begin_inset Formula $x\in\mathbb{R}_{-}\cup\left\{ 0\right\} $
\end_inset
,
pišemo
\begin_inset Formula $x\leq0$
\end_inset
.
\end_layout
\begin_layout Axiom
Če je
\begin_inset Formula $a\not=0$
\end_inset
,
je natanko eno izmed
\begin_inset Formula $\left\{ a,-a\right\} $
\end_inset
pozitivno,
imenujemo ga absolutna vrednost
\begin_inset Formula $a$
\end_inset
(pišemo
\begin_inset Formula $\left|a\right|$
\end_inset
),
in natanko eno negativno,
pišemo
\begin_inset Formula $-\left|a\right|$
\end_inset
.
\end_layout
\begin_layout Definition*
\begin_inset Formula $\left|0\right|=0$
\end_inset
.
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Definition*
Za
\begin_inset Formula $a,b\in\mathbb{R}$
\end_inset
se
\begin_inset Formula $\left|a-b\right|$
\end_inset
imenuje razdalja.
\end_layout
\begin_layout Axiom
\begin_inset Formula $\forall a,b\in\mathbb{R}:a,b>0\Rightarrow\left(a+b>0\right)\wedge\left(ab>0\right)$
\end_inset
.
\end_layout
\begin_layout Definition*
Za
\begin_inset Formula $a,b\in\mathbb{R}$
\end_inset
:
\begin_inset Formula $a$
\end_inset
je večje od
\begin_inset Formula $b$
\end_inset
,
oznaka
\begin_inset Formula $a>b\Leftrightarrow a-b>0$
\end_inset
.
\begin_inset Formula $a$
\end_inset
je manjše od
\begin_inset Formula $b$
\end_inset
,
oznaka
\begin_inset Formula $a<b\Leftrightarrow a-b<0$
\end_inset
.
Podobno
\begin_inset Formula $\leq$
\end_inset
in
\begin_inset Formula $\geq$
\end_inset
.
\end_layout
\begin_layout Claim*
Trikotniška neenakost.
\begin_inset Formula $\forall a,b\in\mathbb{R}$
\end_inset
:
\begin_inset Formula $\left|\left|a\right|-\left|b\right|\right|\leq\left|a+b\right|\leq\left|a\right|+\left|b\right|$
\end_inset
.
\end_layout
\begin_layout Proof
Dokažimo desni neenačaj.
Vemo
\begin_inset Formula $ab\leq\left|ab\right|$
\end_inset
in
\begin_inset Formula $\left|a\right|=\sqrt{a^{2}}$
\end_inset
.
Naj bo
\begin_inset Formula $a^{2}+2ab+b^{2}\leq\left|a\right|^{2}+2\left|a\right|\left|b\right|+\left|b\right|^{2}$
\end_inset
,
torej
\begin_inset Formula $\left(a+b\right)^{2}\leq\left(\left|a\right|+\left|b\right|\right)^{2}$
\end_inset
,
korenimo:
\begin_inset Formula $\left|a+b\right|\leq\left|a\right|+\left|b\right|$
\end_inset
.
\end_layout
\begin_layout Subsubsection
Intervali
\end_layout
\begin_layout Definition*
Naj bo
\begin_inset Formula $a<b$
\end_inset
.
Označimo odprti interval
\begin_inset Formula $\left(a,b\right)\coloneqq\left\{ x\in\mathbb{R};a<x<b\right\} $
\end_inset
,
zaprti
\begin_inset Formula $\left[a,b\right]\coloneqq\left\{ x\in\mathbb{R};a\leq x\leq b\right\} $
\end_inset
,
polodprti
\begin_inset Formula $(a,b]\coloneqq\left\{ x\in\mathbb{R};a<x\leq b\right\} $
\end_inset
in podobno
\begin_inset Formula $[a,b)$
\end_inset
.
\begin_inset Formula $\left(a,\infty\right)\coloneqq\left\{ x\in\mathbb{R};x>a\right\} $
\end_inset
in podobno
\begin_inset Formula $[a,\infty)$
\end_inset
.
\end_layout
\begin_layout Subsection
Temeljne številske podmnožice
\end_layout
\begin_layout Subsubsection
Naravna števila
\begin_inset Formula $\mathbb{N}$
\end_inset
\end_layout
\begin_layout Definition*
\begin_inset Formula $\mathbb{N}\coloneqq\left\{ 1,1+1,1+1+1,1+1+1+1,\dots\right\} $
\end_inset
\end_layout
\begin_layout Paragraph
Matematična indukcija
\end_layout
\begin_layout Standard
Če je
\begin_inset Formula $A\subseteq\mathbb{N}$
\end_inset
in velja
\begin_inset Formula $1\in A$
\end_inset
(baza) in
\begin_inset Formula $a\in A\Rightarrow a+1\in A$
\end_inset
(korak),
tedaj
\begin_inset Formula $A=\mathbb{N}$
\end_inset
.
\end_layout
\begin_layout Claim*
\begin_inset Formula $1+2+3+\cdots+n=\frac{n\left(n+1\right)}{2}$
\end_inset
.
\end_layout
\begin_layout Proof
\begin_inset Formula $A\coloneqq\left\{ n\in\mathbb{N};\text{velja trditev za }n\right\} $
\end_inset
.
Dokažimo
\begin_inset Formula $A=\mathbb{N}$
\end_inset
.
\end_layout
\begin_deeper
\begin_layout Itemize
Baza:
\begin_inset Formula $1=\frac{1\cdot2}{2}=1$
\end_inset
.
\end_layout
\begin_layout Itemize
Korak:
Predpostavimo
\begin_inset Formula $1+2+3+\cdots+n=\frac{n\left(n+1\right)}{2}$
\end_inset
.
Prištejmo
\begin_inset Formula $n+1$
\end_inset
:
\begin_inset Formula
\[
1+2+3+\cdots+n+\left(n+1\right)=\frac{n\left(n+1\right)}{2}+\left(n+1\right)=\frac{n\left(n+1\right)}{2}+\frac{2\left(n+1\right)}{2}=\frac{\left(n+2\right)\left(n+1\right)}{2}
\]
\end_inset
\end_layout
\end_deeper
\begin_layout Subsubsection
Cela števila
\begin_inset Formula $\mathbb{Z}$
\end_inset
\end_layout
\begin_layout Standard
Množica
\begin_inset Formula $\mathbb{N}$
\end_inset
je zaprta za seštevanje in množenje,
torej
\begin_inset Formula $\forall a,b\in\mathbb{N}:a+b\in\mathbb{N}\wedge ab\in\mathbb{N}$
\end_inset
,
ni pa zaprta za odštevanje,
ker recimo
\begin_inset Formula $5-3\not\in\mathbb{N}$
\end_inset
.
Zapremo jo za odštevanje in dobimo množico
\begin_inset Formula $\mathbb{Z}$
\end_inset
.
\end_layout
\begin_layout Definition*
\begin_inset Formula $\mathbb{Z}\coloneqq\left\{ a-b;b,a\in\mathbb{N}\right\} $
\end_inset
\end_layout
\begin_layout Subsubsection
Racionalna števila
\begin_inset Formula $\mathbb{Q}$
\end_inset
\end_layout
\begin_layout Standard
Najmanjša podmnožica
\begin_inset Formula $\mathbb{R}$
\end_inset
,
ki vsebuje
\begin_inset Formula $\mathbb{Z}$
\end_inset
in je zaprta za deljenje,
je
\begin_inset Formula $\mathbb{Q}$
\end_inset
.
\end_layout
\begin_layout Definition*
\begin_inset Formula $\mathbb{Q}\coloneqq\left\{ a/b;a\in\mathbb{Z},b\in\mathbb{Z}\setminus\left\{ 0\right\} \right\} $
\end_inset
.
\end_layout
\begin_layout Standard
Velja
\begin_inset Formula $\mathbb{N}\subset\mathbb{Z}\subset\mathbb{Q}\subset\mathbb{R}$
\end_inset
.
\end_layout
\begin_layout Claim*
Za
\begin_inset Formula $a\in\mathbb{Q}$
\end_inset
,
\begin_inset Formula $b\not\in\mathbb{Q}$
\end_inset
velja
\begin_inset Formula $a+b\not\in\mathbb{Q}$
\end_inset
in
\begin_inset Formula $a\not=0\Rightarrow ab\not\in\mathbb{Q}$
\end_inset
.
\end_layout
\begin_layout Proof
PDDRAA
\begin_inset Formula $a+b\in\mathbb{Q}$
\end_inset
.
Tedaj
\begin_inset Formula $a+b-a\in\mathbb{Q}$
\end_inset
,
tedaj
\begin_inset Formula $b\in\mathbb{Q}$
\end_inset
,
kar je
\begin_inset Formula $\rightarrow\!\leftarrow$
\end_inset
.
PDDRAA
\begin_inset Formula $ab\in\mathbb{Q}$
\end_inset
.
Tedaj
\begin_inset Formula $\frac{ab}{a}\in\mathbb{Q}$
\end_inset
,
tedaj
\begin_inset Formula $b\in\mathbb{Q}$
\end_inset
,
kar je
\begin_inset Formula $\rightarrow\!\leftarrow$
\end_inset
.
\end_layout
\begin_layout Subsection
\begin_inset CommandInset label
LatexCommand label
name "subsec:Omejenost-množic"
\end_inset
Omejenost množic
\end_layout
\begin_layout Definition*
Naj bo
\begin_inset Formula $A\subset\mathbb{R}$
\end_inset
.
\begin_inset Formula $A$
\end_inset
je navzgor omejena
\begin_inset Formula $\Leftrightarrow\exists m\in\mathbb{R}\forall a\in A:a\leq m$
\end_inset
.
Takemu
\begin_inset Formula $m$
\end_inset
pravimo zgornja meja.
Najmanjši zgornji meji
\begin_inset Formula $A$
\end_inset
pravimo supremum ali natančna zgornja meja množice
\begin_inset Formula $A$
\end_inset
,
označimo
\begin_inset Formula $\sup A$
\end_inset
.
Če je zgornja meja
\begin_inset Formula $A$
\end_inset
(
\begin_inset Formula $m$
\end_inset
) element
\begin_inset Formula $A$
\end_inset
,
je maksimum množice
\begin_inset Formula $A$
\end_inset
,
označimo
\begin_inset Formula $m=\max A$
\end_inset
.
Če množica ni navzgor omejena,
pišemo
\begin_inset Formula $\sup A=\infty$
\end_inset
.
\end_layout
\begin_layout Standard
Če
\begin_inset Formula $s=\sup A\in\mathbb{R}$
\end_inset
,
mora veljati
\begin_inset Formula $\forall a\in A:a\leq s$
\end_inset
in
\begin_inset Formula $\forall\varepsilon>0\exists b\in A\ni:b>s-\varepsilon$
\end_inset
,
torej za vsak neničeln
\begin_inset Formula $\varepsilon$
\end_inset
\begin_inset Formula $s-\varepsilon$
\end_inset
ni več natančna zgornja meja za
\begin_inset Formula $A$
\end_inset
.
\end_layout
\begin_layout Definition*
Naj bo
\begin_inset Formula $A\subset\mathbb{R}$
\end_inset
.
\begin_inset Formula $A$
\end_inset
je navzdol omejena
\begin_inset Formula $\Leftrightarrow\exists m\in\mathbb{R}\forall a\in A:a\geq m$
\end_inset
.
Takemu
\begin_inset Formula $m$
\end_inset
pravimo spodnja meja.
Največji spodnji meji
\begin_inset Formula $A$
\end_inset
pravimo infimum ali natančna spodnja meja množice
\begin_inset Formula $A$
\end_inset
,
označimo
\begin_inset Formula $\inf A$
\end_inset
.
Če je spodnja meja
\begin_inset Formula $A$
\end_inset
(
\begin_inset Formula $m$
\end_inset
) element
\begin_inset Formula $A$
\end_inset
,
je minimum množice
\begin_inset Formula $A$
\end_inset
,
označimo
\begin_inset Formula $m=\min A$
\end_inset
.
Če množica ni navzdol omejena,
pišemo
\begin_inset Formula $\inf A=-\infty$
\end_inset
.
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Definition*
Množica
\begin_inset Formula $A\subset\mathbb{R}$
\end_inset
je omejena,
če je hkrati navzgor in navzdol omejena.
\end_layout
\begin_layout Axiom
\begin_inset CommandInset label
LatexCommand label
name "axm:Dedekind.-Vsaka-navzgor"
\end_inset
Dedekind.
Vsaka navzgor omejena množica v
\begin_inset Formula $\mathbb{R}$
\end_inset
ima natančno zgornjo mejo v
\begin_inset Formula $\mathbb{R}$
\end_inset
.
\end_layout
\begin_layout Remark*
Za
\begin_inset Formula $\mathbb{Q}$
\end_inset
aksiom
\begin_inset CommandInset ref
LatexCommand ref
reference "axm:Dedekind.-Vsaka-navzgor"
plural "false"
caps "false"
noprefix "false"
nolink "false"
\end_inset
ne velja.
Če
\begin_inset Formula $B\subset\mathbb{Q}$
\end_inset
,
se lahko zgodi,
da
\begin_inset Formula $\sup B\not\in\mathbb{Q}$
\end_inset
.
Primer:
\begin_inset Formula $B\coloneqq\left\{ q\in\mathbb{Q};q^{2}\leq2\right\} $
\end_inset
.
\begin_inset Formula $\sup B=\sqrt{2}\not\in\mathbb{Q}$
\end_inset
.
\end_layout
\begin_layout Example*
\end_layout
\begin_layout Subsection
Decimalni zapis
\end_layout
\begin_layout Definition*
\begin_inset Formula $\forall x\in\mathbb{R}^{+}\exists!m\in\mathbb{N}\cup\left\{ 0\right\} ,d_{1},d_{2},\dots\in\left\{ 0..9\right\} $
\end_inset
,
ki število natančno določajo.
Pišemo
\begin_inset Formula $x=m,d_{1}d_{2}\dots$
\end_inset
.
Natančno določitev mislimo v smislu:
\end_layout
\begin_deeper
\begin_layout Itemize
\begin_inset Formula $m\leq x<m+1$
\end_inset
—
s tem se izognemo dvojnemu zapisu
\begin_inset Formula $1=0,\overline{9}$
\end_inset
in
\begin_inset Formula $1=1,\overline{0}$
\end_inset
.
\end_layout
\begin_layout Itemize
\begin_inset Formula $[m,m+1)$
\end_inset
razdelimo na 10 enako dolgih polodprtih intervalov
\begin_inset Formula $I_{0},\dots,I_{9}$
\end_inset
.
\begin_inset Formula $x$
\end_inset
leži na natanko enem izmed njih,
indeks njega je
\begin_inset Formula $d_{1}$
\end_inset
.
Nadaljujemo tako,
da
\begin_inset Formula $I_{d_{1}}$
\end_inset
razdelimo zopet na 10 delov itd.
\end_layout
\end_deeper
\begin_layout Definition*
Števila
\begin_inset Formula $x\in\mathbb{R^{-}}$
\end_inset
pišemo tako,
da zapišemo decimalni zapis števila
\begin_inset Formula $-x$
\end_inset
in predenj zapišemo
\begin_inset Formula $-$
\end_inset
.
\end_layout
\begin_layout Definition*
Če se decimalke v zaporedju
\begin_inset Formula $\left(d_{n}\right)_{n\in\mathbb{N}}$
\end_inset
ponavljajo,
uporabimo periodični zapis,
denimo
\begin_inset Formula $5,01\overline{763}\in\mathbb{Q}$
\end_inset
.
\end_layout
\begin_layout Subsection
Kompleksna števila
\end_layout
\begin_layout Definition*
Vpeljimo število
\begin_inset Formula $i$
\end_inset
z lastnostjo
\begin_inset Formula $i^{2}=-1$
\end_inset
,
da je
\begin_inset Formula $i$
\end_inset
rešitev enačbe
\begin_inset Formula $x^{2}+1=0$
\end_inset
.
\end_layout
\begin_layout Claim*
\begin_inset Formula $i\not\in\mathbb{R}$
\end_inset
\end_layout
\begin_layout Proof
Sicer bi veljajo
\begin_inset Formula $i^{2}\geq0$
\end_inset
,
kar po definiciji ne velja.
\end_layout
\begin_layout Definition*
Kompleksna števila so
\begin_inset Formula $\mathbb{C}\coloneqq\left\{ a+bi;a,b\in\mathbb{R}\right\} $
\end_inset
.
\begin_inset Formula $bi$
\end_inset
je še nedefinirano,
zato za kompleksna števila definirano seštevanje in množenje za
\begin_inset Formula $z=a+bi$
\end_inset
in
\begin_inset Formula $w=c+di$
\end_inset
:
\end_layout
\begin_deeper
\begin_layout Itemize
\begin_inset Formula $z+w\coloneqq\left(a+c\right)+\left(b+d\right)i$
\end_inset
\end_layout
\begin_layout Itemize
\begin_inset Formula $zw\coloneqq\left(a+bi\right)\left(c+di\right)=ac+adi+bic+bidi=ac+adi+bic-bd=\left(ac-bd\right)+\left(ad+bc\right)i$
\end_inset
\end_layout
\end_deeper
\begin_layout Definition*
Definiramo še konjugirano vrednost
\begin_inset Formula $z\in\mathbb{C}$
\end_inset
:
\begin_inset Formula $\overline{z}\coloneqq a-bi$
\end_inset
in označimo
\begin_inset Formula $\left|z\right|^{2}=z\overline{z}$
\end_inset
.
\end_layout
\begin_layout Claim*
\begin_inset Formula $z\overline{z}=a^{2}+b^{2}\geq0$
\end_inset
za
\begin_inset Formula $z=a+bi$
\end_inset
.
\end_layout
\begin_layout Proof
\begin_inset Formula $\left(a+bi\right)\left(a-bi\right)=a^{2}+abi-bia-bibi=a^{2}+b^{2}$
\end_inset
.
\end_layout
\begin_layout Standard
Velja,
da je
\begin_inset Formula $\mathbb{R}\subset\mathbb{C}$
\end_inset
v smislu identifikacije
\begin_inset Formula $\mathbb{R}$
\end_inset
z množico
\begin_inset Formula $\mathbb{C}$
\end_inset
:
\begin_inset Formula $\mathbb{R}=\left\{ a+0i;a\in\mathbb{R}\right\} $
\end_inset
,
torej smo
\begin_inset Formula $\mathbb{R}$
\end_inset
razširili v
\begin_inset Formula $\mathbb{C}$
\end_inset
,
kjer ima vsak polinom vedno rešitev.
\end_layout
\begin_layout Subsubsection
Deljenje v
\begin_inset Formula $\mathbb{C}$
\end_inset
\end_layout
\begin_layout Standard
Za
\begin_inset Formula $w,z\in\mathbb{C},w\not=0$
\end_inset
iščemo
\begin_inset Formula $x\in\mathbb{C}\ni:wx=z$
\end_inset
.
Ločimo dva primera:
\end_layout
\begin_layout Itemize
\begin_inset Formula $w\in\mathbb{R}\setminus\left\{ 0\right\} $
\end_inset
:
definiramo
\begin_inset Formula $x=\frac{z}{w}\coloneqq\frac{a}{w}+\frac{b}{w}i$
\end_inset
\end_layout
\begin_layout Itemize
\begin_inset Formula $w\in\mathbb{C}\setminus\left\{ 0\right\} $
\end_inset
(splošno):
\begin_inset Formula $wx=z\overset{/\cdot\overline{w}}{\Longrightarrow}w\overline{w}x=z\overline{w}\Rightarrow\left|w\right|^{2}x=z\overline{w}\Rightarrow x=\frac{z\overline{w}}{\left|w\right|^{2}}$
\end_inset
,
z
\begin_inset Formula $\left|w\right|^{2}$
\end_inset
pa znamo deliti,
ker je realen.
\end_layout
\begin_layout Subsubsection
Lastnosti v
\begin_inset Formula $\mathbb{C}$
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Formula $+$
\end_inset
in
\begin_inset Formula $\cdot$
\end_inset
sta komutativni,
asociativni,
distributivni,
\begin_inset Formula $0$
\end_inset
je aditivna enota,
\begin_inset Formula $1$
\end_inset
je multiplikativna.
\end_layout
\begin_layout Definition*
Za
\begin_inset Formula $z=a+bi$
\end_inset
vpeljemo
\begin_inset Formula $\Re z=a$
\end_inset
in
\begin_inset Formula $\Im z=b$
\end_inset
.
\end_layout
\begin_layout Remark*
Opazimo
\begin_inset Formula $\Re z=\frac{z+\overline{z}}{2}$
\end_inset
,
\begin_inset Formula $\Im z=\frac{z-\overline{z}}{2}$
\end_inset
.
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Remark*
\begin_inset Formula $\mathbb{C}$
\end_inset
si lahko predstavljamo kot urejene pare;
\begin_inset Formula $a+bi$
\end_inset
ustreza paru
\begin_inset Formula $\left(a,b\right)$
\end_inset
.
Tako
\begin_inset Formula $\mathbb{C}$
\end_inset
enačimo/identificiramo z
\begin_inset Formula $\mathbb{R}^{2}\coloneqq\left\{ \left(a,b\right);a,b\in\mathbb{R}\right\} $
\end_inset
,
s čimer dobimo geometrično predstavitev
\begin_inset Formula $\mathbb{C}$
\end_inset
kot vektorje v
\begin_inset Formula $\mathbb{R}^{2}$
\end_inset
.
\end_layout
\begin_layout Definition*
Za
\begin_inset Formula $z=a+bi$
\end_inset
,
predstavljen z vektorjem s komponentami
\begin_inset Formula $\left(a,b\right)$
\end_inset
,
velja
\begin_inset Formula $a=\left|z\right|\cos\varphi$
\end_inset
in
\begin_inset Formula $v=\left|z\right|\sin\varphi$
\end_inset
.
Kotu
\begin_inset Formula $\varphi$
\end_inset
pravimo argument kompleksnega števila
\begin_inset Formula $z$
\end_inset
,
oznaka
\begin_inset Formula $\arg z$
\end_inset
.
\end_layout
\begin_layout Corollary*
\begin_inset Formula $z=$
\end_inset
\begin_inset Formula $\left|z\right|\left(\cos\varphi+i\sin\varphi\right)$
\end_inset
.
Velja
\begin_inset Foot
status open
\begin_layout Plain Layout
TODO DOPISATI ZAKAJ (v bistvu še jaz ne vem).
ne razumem.
\end_layout
\end_inset
\begin_inset Formula $\left(\cos\varphi+i\sin\varphi\right)\left(\cos\psi+i\sin\psi\right)=\cos\left(\varphi+\psi\right)+i\sin\left(\varphi+\psi\right)$
\end_inset
,
zato lahko pišemo
\begin_inset Formula $e^{i\varphi}=\cos\varphi+i\sin\varphi$
\end_inset
.
Množenje kompleksnh števil
\begin_inset Formula $z=\left|z\right|e^{i\varphi}$
\end_inset
in
\begin_inset Formula $w=\left|w\right|e^{i\psi}$
\end_inset
vrne število
\begin_inset Formula $zw$
\end_inset
,
za katero velja:
\end_layout
\begin_deeper
\begin_layout Itemize
\begin_inset Formula $\left|zw\right|=\left|z\right|\left|w\right|$
\end_inset
\end_layout
\begin_layout Itemize
\begin_inset Formula $\arg zw=\arg z+\arg w$
\end_inset
(do periode
\begin_inset Formula $2\pi$
\end_inset
natančno)
\end_layout
\end_deeper
\begin_layout Section
Zaporedja
\end_layout
\begin_layout Definition*
Funkcija
\begin_inset Formula $a:\mathbb{N}\to\mathbb{R}$
\end_inset
se imenuje realno zaporedje,
oznaka
\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
\end_inset
.
\begin_inset Formula $a_{n}$
\end_inset
je funkcijska vrednost pri
\begin_inset Formula $n$
\end_inset
.
\end_layout
\begin_layout Example*
\begin_inset Formula $a_{n}=n$
\end_inset
:
\begin_inset Formula $1,2,3,\dots$
\end_inset
;
\begin_inset Formula $a_{n}=\left(-1\right)^{n}n^{2}$
\end_inset
:
\begin_inset Formula $-1,4,-9,16,-25,\dots$
\end_inset
;
\begin_inset Formula $a_{n}=\cos\left(\frac{\pi}{2}n\right)=0,-1,0,1,0,-1,\dots$
\end_inset
\end_layout
\begin_layout Standard
Zaporedje lahko podamo rekurzivno.
Podamo prvi člen ali nekaj prvih členov in pravilo,
kako iz prejšnjih členov dobiti naslednje.
\end_layout
\begin_layout Example*
\begin_inset Formula $a_{1}=0,a_{n+1}=a_{n}+n$
\end_inset
da zaporedje
\begin_inset Formula $a_{n}=\frac{n\left(n+1\right)}{2}$
\end_inset
.
\begin_inset Formula $a_{0}=0,a_{n+1}=\sqrt{b+a_{n}}$
\end_inset
da zaporedje
\begin_inset Formula $0,\sqrt{b},\sqrt{b+\sqrt{b}},\sqrt{b+\sqrt{b+\sqrt{b}}},\dots$
\end_inset
.
Fibbonacijevo zaporedje:
\begin_inset Formula $a_{1}=a_{2}=1,a_{n+1}=a_{n}+a_{n-1}$
\end_inset
da zaporedje
\begin_inset Formula $1,1,2,3,5,8,\dots$
\end_inset
\end_layout
\begin_layout Subsection
Posebni tipi zaporedij
\end_layout
\begin_layout Definition*
Zaporedje
\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
\end_inset
je aritmetično,
če
\begin_inset Formula $\exists k\in\mathbb{R}\forall n\in\mathbb{N}:a_{n+1}-a_{n}=k$
\end_inset
.
Tedaj
\begin_inset Formula $a_{n+1}=a_{n}+k=a_{1}+nd$
\end_inset
.
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Definition*
Zaporedje
\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
\end_inset
je geometrično,
če
\begin_inset Formula $\exists\lambda\in\mathbb{R}\forall n\in\mathbb{N}:a_{n+1}=a_{n}\lambda$
\end_inset
.
Tedaj
\begin_inset Formula $a_{n}=\lambda^{n-1}a_{1}$
\end_inset
.
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Definition*
Zaporedje
\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
\end_inset
je navzdol oz.
navzgor omejeno,
če je množica vseh členov tega taporedja navzgor oz.
navzdol omejena (glej
\begin_inset CommandInset ref
LatexCommand ref
reference "subsec:Omejenost-množic"
plural "false"
caps "false"
noprefix "false"
nolink "false"
\end_inset
).
Podobno z množico členov definiramo supremum,
infimum,
maksimum in infimum zaporedja.
\end_layout
\begin_layout Definition*
Zaporedje je naraščajoče,
če
\begin_inset Formula $\forall n\in\mathbb{N}:a_{n+1}\geq a_{n}$
\end_inset
,
padajoče,
če
\begin_inset Formula $\forall n\in\mathbb{N}:a_{n+1}\leq a_{n}$
\end_inset
,
strogo naraščajoče,
če
\begin_inset Formula $\forall n\in\mathbb{N}:a_{n+1}>a_{n}$
\end_inset
,
strogo padajoče podobno,
monotono,
če je naraščajoče ali padajoče in strogo monotono,
če je strogo naraščajoče ali strogo padajoče.
\end_layout
\begin_layout Subsection
Limita zaporedja
\end_layout
\begin_layout Definition*
Množica
\begin_inset Formula $U\subseteq\mathbb{R}$
\end_inset
je odprta,
če
\begin_inset Formula $\forall u\in U\exists r>0\ni:\left(u-r,u+r\right)\subseteq U$
\end_inset
.
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Definition*
Množica
\begin_inset Formula $U\subseteq\mathbb{R}$
\end_inset
je zaprta,
če je
\begin_inset Formula $U^{\mathcal{C}}\coloneqq\mathbb{R}\setminus U$
\end_inset
odprta.
\end_layout
\begin_layout Claim*
Odprt interval je odprta množica.
\end_layout
\begin_layout Proof
Za poljubna
\begin_inset Formula $a,b\in\mathbb{R}$
\end_inset
,
\begin_inset Formula $b>a$
\end_inset
,
naj bo
\begin_inset Formula $u\in\left(a,b\right)$
\end_inset
poljuben.
Ustrezen
\begin_inset Formula $r$
\end_inset
je
\begin_inset Formula $\min\left\{ \left|r-a\right|,\left|r-b\right|\right\} $
\end_inset
,
da je
\begin_inset Formula $\left(u-r,u+r\right)\subseteq U$
\end_inset
.
\end_layout
\begin_layout Claim*
Zaprt interval je zaprt.
\end_layout
\begin_layout Proof
Naj bosta
\begin_inset Formula $a,b\in\mathbb{R}$
\end_inset
poljubna in
\begin_inset Formula $b>a$
\end_inset
.
Dokazujemo,
da je
\begin_inset Formula $\left[a,b\right]$
\end_inset
zaprt,
torej da je
\begin_inset Formula $\left[a,b\right]^{\mathcal{C}}=\left(-\infty,a\right)\cup\left(b,\infty\right)$
\end_inset
odprta množica.
Za poljuben
\begin_inset Formula $u\in\left[a,b\right]^{\mathcal{C}}$
\end_inset
velja,
da je bodisi
\begin_inset Formula $\in\left(-\infty,a\right)$
\end_inset
bodisi
\begin_inset Formula $\left(b,\infty\right)$
\end_inset
,
kajti
\begin_inset Formula $\left(-\infty,a\right)\cap\left(b,\infty\right)=\emptyset$
\end_inset
.
Po prejšnji trditvi v obeh primerih velja
\begin_inset Formula $\exists r>0\ni:\left(u-r,u+r\right)\subseteq U$
\end_inset
,
torej je
\begin_inset Formula $\left[a,b\right]^{\mathcal{C}}$
\end_inset
res odprta,
torej je
\begin_inset Formula $\left[a,b\right]$
\end_inset
res zaprta.
\end_layout
\begin_layout Definition*
Množica
\begin_inset Formula $B$
\end_inset
je okolica točke
\begin_inset Formula $t\in\mathbb{R}$
\end_inset
,
če vsebuje kakšno odprto množico
\begin_inset Formula $U$
\end_inset
,
ki vsebuje
\begin_inset Formula $t$
\end_inset
,
torej
\begin_inset Formula $t\in U^{\text{odp.}}\subseteq B\subseteq\mathbb{R}$
\end_inset
.
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Definition*
\begin_inset Formula $L\in\mathbb{R}$
\end_inset
je limita zaporedja
\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{R}}$
\end_inset
\begin_inset Formula $\Leftrightarrow\forall\varepsilon>0\exists n_{0}\in\mathbb{N}\forall n\in\mathbb{N}:n\geq n_{0}\Rightarrow\left|a_{n}-L\right|<\varepsilon$
\end_inset
.
ZDB
\begin_inset Formula $\forall V$
\end_inset
okolica
\begin_inset Formula $L\in\mathbb{R}\exists n_{0}\in\mathbb{N}\forall n\in\mathbb{N}:n\geq n_{0}\Rightarrow a_{n}\in V$
\end_inset
,
pravimo,
da
\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
\end_inset
konvergira k
\begin_inset Formula $L$
\end_inset
in pišemo
\begin_inset Formula $L\coloneqq\lim_{n\to\infty}a_{n}$
\end_inset
ali drugače
\begin_inset Formula $a_{n}\underset{n\to\infty}{\longrightarrow}L$
\end_inset
.
Če zaporedje ima limito,
pravimo,
da je konvergentno,
sicer je divergentno.
\end_layout
\begin_layout Claim*
Konvergentno zaporedje v
\begin_inset Formula $\mathbb{R}$
\end_inset
ima natanko eno limito.
\end_layout
\begin_layout Proof
Naj bosta
\begin_inset Formula $J$
\end_inset
in
\begin_inset Formula $L$
\end_inset
limiti zaporedja
\begin_inset Formula $\left(a_{n}\right)_{n\to\infty}$
\end_inset
.
Torej
\begin_inset Foot
status open
\begin_layout Plain Layout
Ko trdimo,
da obstaja
\begin_inset Formula $n_{0}$
\end_inset
,
še ne vemo,
ali sta za
\begin_inset Formula $L$
\end_inset
in
\begin_inset Formula $J$
\end_inset
ta
\begin_inset Formula $n_{0}$
\end_inset
ista.
Ampak trditev še vedno velja,
ker lahko vzamemo večjega izmed njiju,
ako bi bila drugačna.
\end_layout
\end_inset
po definiciji
\begin_inset Formula $\forall\varepsilon>0\exists n_{0}\in\mathbb{N}\forall n\in\mathbb{N}:n\geq n_{0}\Rightarrow\left|a_{n}-L\right|<\varepsilon\wedge\left|a_{n}-J\right|<\varepsilon$
\end_inset
.
Velja torej
\begin_inset Formula $\forall\varepsilon>0:\left|J-L\right|<\varepsilon$
\end_inset
.
PDDRAA
\begin_inset Formula $J\not=L$
\end_inset
.
Tedaj
\begin_inset Formula $\left|J-L\right|\not=0$
\end_inset
,
naj bo
\begin_inset Formula $\left|J-L\right|=k$
\end_inset
.
Tedaj
\begin_inset Formula $\exists\varepsilon>0:\left|J-L\right|\not<\varepsilon$
\end_inset
,
ustrezen
\begin_inset Formula $\varepsilon$
\end_inset
je na primer
\begin_inset Formula $\frac{\left|J-L\right|}{2}$
\end_inset
.
\end_layout
\begin_layout Claim*
\begin_inset CommandInset label
LatexCommand label
name "Konvergentno-zaporedje-v-R-je-omejeno"
\end_inset
Konvergentno zaporedje v
\begin_inset Formula $\mathbb{R}$
\end_inset
je omejeno.
\end_layout
\begin_layout Proof
Naj bo
\begin_inset Formula $L=\lim_{n\to\infty}a_{n}$
\end_inset
.
Znotraj intervala
\begin_inset Formula $\left(L-1,L+1\right)$
\end_inset
so vsi členi zaporedja razen končno mnogo (
\begin_inset Formula $\left\{ a_{1},\dots,a_{n_{0}}\right\} $
\end_inset
).
\begin_inset Formula $\left\{ a_{n}\right\} _{n\in\mathbb{N}}$
\end_inset
je unija dveh omejenih množic;
\begin_inset Formula $\left(L-1,L+1\right)$
\end_inset
in
\begin_inset Formula $\left\{ a_{1},\dots,a_{n_{0}}\right\} $
\end_inset
,
zato je tudi sama omejena.
\end_layout
\begin_layout Theorem*
\begin_inset ERT
status open
\begin_layout Plain Layout
\backslash
hypertarget{pmkdlim}{Naj bosta}
\end_layout
\end_inset
\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
\end_inset
in
\begin_inset Formula $\left(b_{n}\right)_{n\in\mathbb{N}}$
\end_inset
konvergentni zaporedji v
\begin_inset Formula $\mathbb{R}$
\end_inset
.
Tedaj so tudi
\begin_inset Formula $\left(a_{n}*b_{n}\right)_{n\in\mathbb{N}}$
\end_inset
konvergentna in velja
\begin_inset Formula $\lim_{n\to\infty}a_{n}*b_{n}=\lim_{n\to\infty}a_{n}*\lim_{n\to\infty}b_{n}$
\end_inset
za
\begin_inset Formula $*\in\left\{ +,-,\cdot\right\} $
\end_inset
.
Če je
\begin_inset Formula $\lim_{n\to\infty}b_{n}\not=0$
\end_inset
,
isto velja tudi za deljenje.
\end_layout
\begin_layout Proof
Naj bo
\begin_inset Formula $a_{n}\to A$
\end_inset
in
\begin_inset Formula $b_{n}\to B$
\end_inset
oziroma
\begin_inset Formula $\forall\varepsilon>0\exists n_{1},n_{2}\ni:\left(n>n_{1}\Rightarrow\left|a_{n}-A\right|<\varepsilon\right)\wedge\left(n>n_{2}\Rightarrow\left|b_{n}-B\right|<\varepsilon\right)$
\end_inset
,
torej
\begin_inset Formula $\forall\varepsilon>0\exists n_{0}=\max\left\{ n_{1},n_{2}\right\} \ni:n>n_{0}\Rightarrow\left|a_{n}-A\right|<\varepsilon\wedge\left|b_{n}-B\right|<\varepsilon$
\end_inset
.
Dokažimo za vse operacije:
\end_layout
\begin_deeper
\begin_layout Labeling
\labelwidthstring 00.00.0000
\begin_inset Formula $+$
\end_inset
Po predpostavki velja
\begin_inset Formula $\forall\varepsilon>0\exists n_{0}\ni:n>n_{0}\Rightarrow\left|a_{n}-A\right|+\left|a_{n}-B\right|<2\varepsilon$
\end_inset
.
Oglejmo si sedaj
\begin_inset Formula
\[
\left|\left(a_{n}+b_{n}\right)-\left(A+B\right)\right|=\left|\left(a_{n}-A\right)+\left(b_{n}-B\right)\right|\leq\left|a_{n}-A\right|+\left|b_{n}-B\right|
\]
\end_inset
in uporabimo še prejšnjo trditev,
torej
\begin_inset Formula $\forall2\varepsilon\exists n_{0}\ni:\left|\left(a_{n}+b_{n}\right)-\left(A+B\right)\right|\leq2\varepsilon$
\end_inset
,
s čimer dokažemo
\begin_inset Formula $\left(a_{n}+b_{n}\right)\to A+B$
\end_inset
.
\end_layout
\begin_layout Labeling
\labelwidthstring 00.00.0000
\begin_inset Formula $-$
\end_inset
Oglejmo si
\begin_inset Formula
\[
\left|\left(a_{n}-b_{n}\right)-\left(A-B\right)\right|=\left|a_{n}-b_{n}-A+B\right|=\left|\left(a_{n}-A\right)+\left(-\left(b_{n}-B\right)\right)\right|\leq\left|a_{n}-A\right|+\left|b_{n}-B\right|
\]
\end_inset
in nato kot zgoraj.
\end_layout
\begin_layout Labeling
\labelwidthstring 00.00.0000
\begin_inset Formula $\cdot$
\end_inset
Oglejmo si
\begin_inset Formula
\[
\left|a_{n}b_{n}-AB\right|=\left|a_{n}b_{n}-Ab_{n}+Ab_{n}-AB\right|=\left|\left(a_{n}-A\right)b_{n}+A\left(b_{n}-B\right)\right|\leq\left|a_{n}-A\right|\left|b_{n}\right|+\left|A\right|\left|b_{n}-B\right|.
\]
\end_inset
Od prej vemo,
da sta zaporedji omejeni,
ker sta konvergentni,
zato
\begin_inset Formula $\exists M>0\forall n\in\mathbb{N}:\left|b_{n}\right|\leq M$
\end_inset
.
Naj bo
\begin_inset Formula $\varepsilon>0$
\end_inset
poljuben
\begin_inset Formula $n_{1},n_{2}\in\mathbb{N}$
\end_inset
taka,
da
\begin_inset Formula $n\geq n_{1}\Rightarrow\left|a_{n}-A\right|<\frac{\varepsilon}{2M}$
\end_inset
in
\begin_inset Formula $n\geq n_{2}\Rightarrow\left|b_{n}-B\right|<\frac{\varepsilon}{2\left|A\right|}$
\end_inset
.
Tedaj za
\begin_inset Formula $n_{0}\coloneqq\max\left\{ n_{1},n_{2}\right\} $
\end_inset
velja
\begin_inset Formula $\left|a_{n}b_{n}-AB\right|\leq\left|a_{n}-A\right|\left|b_{n}\right|+\left|A\right|\left|b_{n}-B\right|<\frac{\varepsilon}{2M}M+\left|A\right|\frac{\varepsilon}{2\left|A\right|}=\varepsilon$
\end_inset
,
skratka
\begin_inset Formula $\left|a_{n}b_{n}-AB\right|<\varepsilon$
\end_inset
,
s čimer dokažemo
\begin_inset Formula $\left(a_{n}+b_{n}\right)\to A+B$
\end_inset
.
\end_layout
\begin_layout Labeling
\labelwidthstring 00.00.0000
\begin_inset Formula $/$
\end_inset
Ker je
\begin_inset Formula $B\not=0$
\end_inset
,
\begin_inset Formula $\exists n_{0}\in\mathbb{N}\forall n\geq n_{0}:\left|b_{n}\right|\geq\frac{\left|B\right|}{2}>0$
\end_inset
.
ZDB vsi členi zaporedja razen končno mnogo so v poljubno majhni okolici
\begin_inset Formula $\left|B\right|$
\end_inset
.
Če torej vzamemo točko na polovici med 0 in
\begin_inset Formula $\left|B\right|$
\end_inset
,
to je
\begin_inset Formula $\frac{\left|B\right|}{2}$
\end_inset
,
bo neskončno mnogo absolutnih vrednosti členov večjih od
\begin_inset Formula $\frac{\left|B\right|}{2}$
\end_inset
.
Pri razumevanju pomaga številska premica.
Nadalje uporabimo predpostavko z
\begin_inset Formula $\varepsilon=\frac{\left|B\right|}{2}$
\end_inset
,
torej je za
\begin_inset Formula $n>n_{0}:$
\end_inset
\begin_inset Formula $\left|B-b_{n}\right|<\frac{\left|B\right|}{2}$
\end_inset
in velja
\begin_inset Formula
\[
\left|b_{n}\right|=\left|B-\left(B-b_{n}\right)\right|=\left|B+\left(-\left(B-b_{n}\right)\right)\right|\overset{\text{trik. neen.}}{\geq}\left|\left|B\right|-\left|B-b_{n}\right|\right|=\left|B\right|-\left|B-b_{n}\right|\overset{\text{predp.}}{>}\left|B\right|-\frac{\left|B\right|}{2}=\frac{\left|B\right|}{2},
\]
\end_inset
skratka
\begin_inset Formula $\left|b_{n}\right|>\frac{\left|B\right|}{2}$
\end_inset
.
Če spet izpustimo končno začetnih členov,
velja
\begin_inset Formula
\[
\frac{a_{n}}{b_{n}}-\frac{A}{B}=\frac{a_{n}B-Ab_{n}}{b_{n}B}\overset{\text{prištejemo in odštejemo člen}}{=}\frac{\left(a_{n}-A\right)B+A\left(B-b_{n}\right)}{Bb_{n}}=\frac{1}{b_{n}}\left(a_{n}-A\right)+\frac{A/B}{b_{n}}\left(B-b_{n}\right)
\]
\end_inset
sedaj uporabimo na obeh straneh absolutno vrednost:
\begin_inset Formula
\[
\left|\frac{a_{n}}{b_{n}}-\frac{A}{B}\right|=\left|\frac{1}{b_{n}}\left(a_{n}-A\right)+\frac{A}{Bb_{n}}\left(B-b_{n}\right)\right|\leq\frac{1}{\left|b_{n}\right|}\left|a_{n}-A\right|+\frac{\left|A\right|}{\left|B\right|\left|b_{n}\right|}\left|B-b_{n}\right|<\frac{2}{\left|B\right|}\left|a_{n}-A\right|+\frac{2\left|A\right|}{\left|B\right|^{2}}\left|B-B_{n}\right|
\]
\end_inset
skratka
\begin_inset Formula $\left|\frac{a_{n}}{b_{n}}-\frac{A}{B}\right|<\frac{2}{\left|B\right|}\left|a_{n}-A\right|+\frac{2\left|A\right|}{\left|B\right|^{2}}\left|B-B_{n}\right|$
\end_inset
.
Opazimo,
da
\begin_inset Formula $\frac{2}{\left|B\right|}$
\end_inset
in
\begin_inset Formula $\frac{2\left|A\right|}{\left|B\right|^{2}}$
\end_inset
nista odvisna od
\begin_inset Formula $n$
\end_inset
.
Sedaj vzemimo poljuben
\begin_inset Formula $\varepsilon>0$
\end_inset
in
\begin_inset Formula $n_{1},n_{2}\in\mathbb{N}$
\end_inset
takšna,
da velja:
\end_layout
\begin_deeper
\begin_layout Itemize
\begin_inset Formula $n\geq n_{1}\Rightarrow\left|a_{n}-A\right|<\frac{\varepsilon\left|B\right|}{4}$
\end_inset
\end_layout
\begin_layout Itemize
\begin_inset Formula $n\geq n_{2}\Rightarrow\left|b_{n}-B\right|<\frac{\varepsilon\left|B\right|^{2}}{4\left|A\right|}$
\end_inset
\end_layout
\begin_layout Standard
Tedaj iz zgornje ocene sledi za
\begin_inset Formula $n\geq\max\left\{ n_{0},n_{1},n_{2}\right\} $
\end_inset
:
\begin_inset Formula
\[
\left|\frac{a_{n}}{b_{n}}-\frac{A}{B}\right|<\frac{\cancel{2}}{\cancel{\left|B\right|}}\cdot\frac{\varepsilon\cancel{\left|B\right|}}{\cancelto{2}{4}}+\frac{\cancel{2}\cancel{\left|A\right|}}{\cancel{\left|B\right|^{2}}}\cdot\frac{\varepsilon\cancel{\left|B\right|^{2}}}{\cancelto{2}{4}\cancel{\left|A\right|}}=\frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon,
\]
\end_inset
s čimer dokažemo
\begin_inset Formula $a_{n}/b_{n}\to A/B$
\end_inset
.
\end_layout
\end_deeper
\end_deeper
\begin_layout Example*
Naj bo
\begin_inset Formula $a>0$
\end_inset
.
Izračunajmo
\begin_inset Formula
\[
\sqrt{a+\sqrt{a+\sqrt{a+\sqrt{\cdots}}}}\eqqcolon\alpha.
\]
\end_inset
\end_layout
\begin_layout Example*
\begin_inset Formula $\alpha$
\end_inset
je torej
\begin_inset Formula $\lim_{n\to\infty}x_{n}$
\end_inset
,
kjer je
\begin_inset Formula $x_{0}=0,x_{1}=\sqrt{a},x_{2}=\sqrt{a+\sqrt{a}},x_{3}=\sqrt{a+\sqrt{a+\sqrt{a}}},\dots,x_{n+1}=\sqrt{a+x_{n}}$
\end_inset
.
Iz zadnjega sledi
\begin_inset Formula $x_{n+1}^{2}=a+x_{n}$
\end_inset
.
Če torej limita
\begin_inset Formula $\alpha\coloneqq\lim x_{n}$
\end_inset
obstaja,
mora veljati
\begin_inset Formula $\alpha^{2}=a+\alpha$
\end_inset
oziroma
\begin_inset Formula $\alpha_{1,2}=\frac{1\pm\sqrt{1+4a}}{2}$
\end_inset
.
Opcija z minusom ni mogoča,
ker je zaporedje
\begin_inset Formula $\left(x_{n}\right)_{n\in\mathbb{N}}$
\end_inset
očitno pozitivno.
Če torej limita obstaja (
\series bold
česar še nismo dokazali
\series default
),
je enaka
\begin_inset Formula $\frac{1+\sqrt{1+4a}}{2}$
\end_inset
,
za primer
\begin_inset Formula $a=2$
\end_inset
je torej
\begin_inset Formula $\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{\cdots}}}}=2$
\end_inset
.
\end_layout
\begin_layout Remark*
Lahko se zgodi,
da limita rekurzivno podanega zaporedja ne obstaja,
čeprav jo znamo izračunati,
če bi obstajala.
Na primer
\begin_inset Formula $y_{1}\coloneqq1$
\end_inset
,
\begin_inset Formula $y_{n+1}=1-2y_{n}$
\end_inset
nam da zaporedje
\begin_inset Formula $1,-1,3,-5,11,\dots$
\end_inset
,
kar očitno nima limite.
Če bi limita obstajala,
bi zanjo veljalo
\begin_inset Formula $\beta=1-2\beta$
\end_inset
oz.
\begin_inset Formula $3\beta=1$
\end_inset
,
\begin_inset Formula $\beta=\frac{1}{3}$
\end_inset
.
Navedimo torej nekaj zadostnih in potrebnih pogojev za konvergenco zaporedij.
\end_layout
\begin_layout Theorem*
\begin_inset ERT
status open
\begin_layout Plain Layout
\backslash
hypertarget{kmoz}{Konvergenca monotonega in omejenega zaporedja}
\end_layout
\end_inset
.
Naj bo
\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
\end_inset
monotono realno zaporedje.
Če narašča,
ima limito
\begin_inset Formula $\lim_{n\to\infty}a_{n}=\sup\left\{ a_{n},n\in\mathbb{N}\right\} $
\end_inset
.
Če pada,
ima limito
\begin_inset Formula $\lim_{n\to\infty}a_{n}=\inf\left\{ a_{n},n\in\mathbb{N}\right\} $
\end_inset
.
(
\begin_inset Formula $\sup$
\end_inset
in
\begin_inset Formula $\inf$
\end_inset
imata lahko tudi vrednost
\begin_inset Formula $\infty$
\end_inset
in
\begin_inset Formula $-\infty$
\end_inset
—
zaporedje s tako limito ni konvergentno v
\begin_inset Formula $\mathbb{R}$
\end_inset
).
\end_layout
\begin_layout Proof
Denimo,
da
\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
\end_inset
narašča.
Pišimo
\begin_inset Formula $s\coloneqq\sup_{n\in\mathbb{N}}a_{n}$
\end_inset
.
Vzemimo poljuben
\begin_inset Formula $\varepsilon>0$
\end_inset
.
Tedaj
\begin_inset Formula $s-\varepsilon$
\end_inset
ni zgornja meja za
\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
\end_inset
,
zato
\begin_inset Formula $\exists n_{0}\in\mathbb{N}\ni:s-\varepsilon<a_{n_{0}}$
\end_inset
.
Ker pa je
\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
\end_inset
naraščajoče,
sledi
\begin_inset Formula $\forall n\geq n_{0}:a_{n}\geq a_{n_{0}}>s-\varepsilon$
\end_inset
.
Hkrati je
\begin_inset Formula $a_{n}\leq s$
\end_inset
,
saj je
\begin_inset Formula $s$
\end_inset
zgornja meja.
Torej
\begin_inset Formula $\forall n\geq n_{0}:a_{n}\in(s-\varepsilon,s]\subset\left(s-\varepsilon,s+\varepsilon\right)$
\end_inset
,
s čimer dokažemo konvergenco.
\end_layout
\begin_layout Proof
Denimo sedaj,
da
\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
\end_inset
pada.
Dokaz je povsem analogen.
Pišimo
\begin_inset Formula $m\coloneqq\inf_{n\in\mathbb{N}}a_{n}$
\end_inset
.
Vzemimo poljuben
\begin_inset Formula $\varepsilon>0$
\end_inset
.
Tedaj
\begin_inset Formula $m+\varepsilon$
\end_inset
ni spodnja meja za
\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
\end_inset
,
zato
\begin_inset Formula $\exists n_{0}\in\mathbb{N}\ni:m+\varepsilon>a_{n_{0}}$
\end_inset
.
Ker pa je
\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
\end_inset
padajoče,
sledi
\begin_inset Formula $\forall n\geq n_{0}:a_{n}\leq a_{n_{0}}<m+\varepsilon$
\end_inset
.
Hkrati je
\begin_inset Formula $a_{n}\geq m$
\end_inset
,
saj je
\begin_inset Formula $m$
\end_inset
spodnja meja.
Torej
\begin_inset Formula $\forall n\geq n_{0}:a_{n}\in[m,m+\varepsilon)\subset\left(m-\varepsilon,m+\varepsilon\right)$
\end_inset
.
\end_layout
\begin_layout Corollary*
Za monotono zaporedje velja,
da je v
\begin_inset Formula $\mathbb{R}$
\end_inset
konvergentno natanko tedaj,
ko je omejeno.
\end_layout
\begin_layout Example*
Naj bo,
kot prej,
\begin_inset Formula $a>0$
\end_inset
in
\begin_inset Formula $x_{0}=0,x_{n+1}=\sqrt{a+x_{n}}$
\end_inset
.
Dokažimo,
da je
\begin_inset Formula $\left(x_{n}\right)_{n}$
\end_inset
konvergentno.
Dovolj je pokazati,
da je naraščajoče in navzgor omejeno.
\end_layout
\begin_deeper
\begin_layout Itemize
Naraščanje z indukcijo:
Baza:
\begin_inset Formula $0=x_{0}>x_{1}=\sqrt{a}$
\end_inset
.
Dokažimo
\begin_inset Formula $x_{n+1}-x_{n}>0$
\end_inset
.
\begin_inset Formula
\[
\left(x_{n+1}-x_{n}\right)\left(x_{n+1}+x_{n}\right)=x_{n+1}^{2}-x_{n}^{2}=\left(a+x_{n}\right)-\left(a+x_{n-1}\right)=x_{n}-x_{n-1}
\]
\end_inset
Ker je zaporedje pozitivno,
je
\begin_inset Formula $x_{n+1}+x_{n}>0$
\end_inset
.
Desna stran je po I.
P.
pozitivna,
torej tudi
\begin_inset Formula $x_{n+1}-x_{n}>0$
\end_inset
.
\end_layout
\begin_layout Itemize
Omejenost:
Če je zaporedje res omejeno,
je po zgornjem tudi konvergentno in je
\begin_inset Formula $\sup_{n\in\mathbb{N}}x_{n}=\lim_{n\to\infty}x_{n}=\frac{1+\sqrt{1+4a}}{2}\leq\frac{1+\sqrt{1+4a+4a^{2}}}{2}=\frac{1+\sqrt{\left(2a+1\right)^{2}}}{}=1+a$
\end_inset
.
Uganili smo neko zgornjo mejo.
Domnevamo,
da
\begin_inset Formula $\forall n\in\mathbb{N}:x_{n}\leq1+a$
\end_inset
.
Dokažimo to z indukcijo:
Baza:
\begin_inset Formula $0=x_{0}<1+a$
\end_inset
.
Po I.
P.
\begin_inset Formula $x_{n}>1+a$
\end_inset
.
Korak:
\begin_inset Formula
\[
x_{n+1}=\sqrt{x_{n}+a}\leq\sqrt{1+a+a}=\sqrt{1+2a}<\sqrt{1+2a+2a^{2}}=1+a
\]
\end_inset
\end_layout
\end_deeper
\begin_layout Example*
S tem smo dokazali,
da
\begin_inset Formula $\lim_{n\to\infty}x_{n}=\frac{1+\sqrt{1+4a}}{2}$
\end_inset
.
\end_layout
\begin_layout Example*
To lahko dokažemo tudi na alternativen način.
Vidimo,
da je edini kandidat za limito,
če obstaja
\begin_inset Formula $L=\frac{1+\sqrt{1+4a}}{2}$
\end_inset
in da torej velja
\begin_inset Formula $L^{2}=a+L$
\end_inset
.
Preverimo,
da je
\begin_inset Formula $L$
\end_inset
res limita:
\begin_inset Formula
\[
x_{n+1}-L=\sqrt{a+x_{n}}-L=\frac{\left(\sqrt{a+x_{n}}-L\right)\left(\sqrt{a+x_{n}}+L\right)}{\sqrt{a+x_{n}}+L}=\frac{\left(a+x_{n}\right)-L^{2}}{\sqrt{a+x_{n}}+L}=\frac{\left(a+x_{n}\right)-\left(a+L\right)}{\sqrt{a+x_{n}}+L}=\frac{x_{n}-L}{\sqrt{a+x_{n}}+L}.
\]
\end_inset
Vpeljimo sedaj
\begin_inset Formula $y_{n}\coloneqq x_{n}-L$
\end_inset
.
Sledi
\begin_inset Formula $\left|y_{n+1}\right|\leq\frac{\left|y_{n}\right|}{\sqrt{a+x_{n}}+L}\leq\frac{\left|y_{n}\right|}{L}$
\end_inset
.
Ker je
\begin_inset Formula $\left|y_{0}\right|=L$
\end_inset
,
dobimo
\begin_inset Foot
status open
\begin_layout Plain Layout
Za razumevanje si oglej nekaj členov rekurzivnega zaporedje
\begin_inset Formula $y_{0}=L,y_{n}=\frac{\left|y_{n+1}\right|}{L}$
\end_inset
.
Začnemo z 1 in nato vsakič delimo z
\begin_inset Formula $L$
\end_inset
.
\end_layout
\end_inset
oceno
\begin_inset Formula $\left|y_{n}\right|\leq\frac{1}{L^{n-1}}$
\end_inset
oziroma
\begin_inset Formula $\left|x_{n}-L\right|\leq\frac{1}{L^{n-1}}$
\end_inset
.
Ker iz definicije
\begin_inset Formula $L$
\end_inset
sledi
\begin_inset Formula $L>1$
\end_inset
,
je
\begin_inset Formula $L^{n}\to\infty$
\end_inset
za
\begin_inset Formula $n\to\infty$
\end_inset
,
torej smo dokazali,
da
\begin_inset Formula $\left|x_{n}-L\right|$
\end_inset
eksponentno pada proti 0 za
\begin_inset Formula $n\to\infty$
\end_inset
.
Eksponentno padanje
\begin_inset Formula $\left|x_{n}-L\right|$
\end_inset
proti 0 je dovolj,
da rečemo,
da zaporedje konvergira k
\begin_inset Formula $L$
\end_inset
\begin_inset Foot
status open
\begin_layout Plain Layout
a res,
vprašaj koga.
ne razumem.
zakaj.
TODO.
\end_layout
\end_inset
.
\end_layout
\begin_layout Claim*
\begin_inset Formula $\lim_{n\to\infty}\sin n$
\end_inset
in
\begin_inset Formula $\lim_{n\to\infty}\cos n$
\end_inset
ne obstajata.
\end_layout
\begin_layout Proof
Pišimo
\begin_inset Formula $a_{n}=\sin n$
\end_inset
in
\begin_inset Formula $b_{n}=\cos n$
\end_inset
.
Iz adicijskih izrekov dobimo
\begin_inset Formula $a_{n+1}=\sin\left(n+1\right)=\sin n\cos1+\cos n\sin1=a_{n}\cos1+b_{n}\sin1$
\end_inset
.
Torej
\begin_inset Formula $b_{n}=\frac{a_{n+1}-a_{n}\cos1}{\sin1}$
\end_inset
.
Torej če
\begin_inset Formula $\exists a\coloneqq\lim_{n\to\infty}a_{n},a\in\mathbb{R}$
\end_inset
,
potem tudi
\begin_inset Formula $\exists b\coloneqq\lim_{n\to\infty}b_{n},b\in\mathbb{R}$
\end_inset
.
Podobno iz adicijske formule za
\begin_inset Formula $\cos\left(n+1\right)$
\end_inset
sledi
\begin_inset Formula $a_{n}=\frac{b_{n}\cos1-b_{n+1}}{\sin1}$
\end_inset
,
torej če
\begin_inset Formula $\exists b$
\end_inset
,
potem tudi
\begin_inset Formula $\exists a$
\end_inset
.
Iz obojega sledi,
da
\begin_inset Formula $\exists a\Leftrightarrow\exists b$
\end_inset
.
Posledično,
če
\begin_inset Formula $a$
\end_inset
in
\begin_inset Formula $b$
\end_inset
obstajata,
iz zgornjih obrazcev za
\begin_inset Formula $a_{n}$
\end_inset
in
\begin_inset Formula $b_{n}$
\end_inset
sledi,
da za
\begin_inset Formula
\[
\lambda=\frac{1-\cos1}{\sin1}\in\left(0,1\right)
\]
\end_inset
velja
\begin_inset Formula $b=\lambda a$
\end_inset
in
\begin_inset Formula $a=-\lambda b$
\end_inset
in zato
\begin_inset Formula $b=\lambda\left(-\lambda b\right)$
\end_inset
oziroma
\begin_inset Formula $1=-\lambda^{2}$
\end_inset
,
torej
\begin_inset Formula $-1=\lambda^{2}$
\end_inset
,
torej
\begin_inset Formula $\lambda=i$
\end_inset
,
kar je v protislovju z
\begin_inset Formula $\lambda\in\left(0,1\right)$
\end_inset
.
Podobno za
\begin_inset Formula $a=-\lambda\left(\lambda a\right)$
\end_inset
oziroma
\begin_inset Formula $1=-\lambda^{2}$
\end_inset
,
kar je zopet
\begin_inset Formula $\rightarrow\!\leftarrow$
\end_inset
.
Edina druga opcija je,
da je
\begin_inset Formula $a=b=0$
\end_inset
.
Hkrati pa vemo,
da
\begin_inset Formula $a_{n}^{2}+b_{n}^{2}=1$
\end_inset
,
zato
\begin_inset Formula $a+b=1$
\end_inset
,
kar ni mogoče za ničelna
\begin_inset Formula $a$
\end_inset
in
\begin_inset Formula $b$
\end_inset
.
Torej
\begin_inset Formula $a$
\end_inset
in
\begin_inset Formula $b$
\end_inset
ne obstajata.
\end_layout
\begin_layout Subsection
Eulerjevo število
\end_layout
\begin_layout Theorem*
Bernoullijeva neenakost.
\begin_inset Formula $\forall\alpha\leq1,n\in\mathbb{N}$
\end_inset
velja
\begin_inset Formula $\left(1-\alpha\right)^{n}\geq1-n\alpha$
\end_inset
.
\end_layout
\begin_layout Proof
Z indukcijo na
\begin_inset Formula $n$
\end_inset
ob fiksnem
\begin_inset Formula $\alpha$
\end_inset
.
\end_layout
\begin_deeper
\begin_layout Itemize
Baza:
\begin_inset Formula $n=1$
\end_inset
:
\begin_inset Formula $\left(1-\alpha\right)^{1}=1-1\alpha$
\end_inset
.
Velja celo enakost.
\end_layout
\begin_layout Itemize
I.
P.:
Velja
\begin_inset Formula $\left(1-\alpha\right)^{n}\geq1-n\alpha$
\end_inset
\end_layout
\begin_layout Itemize
Korak:
\begin_inset Formula $\left(1-\alpha\right)^{n+1}=\left(1-\alpha\right)\left(1-\alpha\right)^{n}\geq\left(1-\alpha\right)$
\end_inset
\begin_inset Formula $\left(1-n\alpha\right)=1-n\alpha-\alpha-n\alpha^{2}=1-\left(n+1\right)\alpha-n\alpha^{2}\geq1-\left(n+1\right)\alpha$
\end_inset
,
torej ocena velja tudi za
\begin_inset Formula $n+1$
\end_inset
.
\end_layout
\end_deeper
\begin_layout Definition*
Vpeljimo oznaki:
\end_layout
\begin_deeper
\begin_layout Itemize
Za
\begin_inset Formula $n\in\mathbb{N}$
\end_inset
označimo
\begin_inset Formula $n!=1\cdot2\cdot3\cdot\cdots\cdot n$
\end_inset
(pravimo
\begin_inset Formula $n-$
\end_inset
faktoriala oziroma
\begin_inset Formula $n-$
\end_inset
fakulteta).
Ker velja
\begin_inset Formula $n!=n\cdot\left(n-1\right)!$
\end_inset
za
\begin_inset Formula $n\geq2$
\end_inset
,
je smiselno definirati še
\begin_inset Formula $0!=1$
\end_inset
.
\end_layout
\begin_layout Itemize
Za
\begin_inset Formula $n,k\in\mathbb{N}$
\end_inset
označimo še binomski simbol:
\begin_inset Formula $\binom{n}{k}\coloneqq\frac{n!}{k!\left(n-k\right)!}$
\end_inset
(pravimo
\begin_inset Formula $n$
\end_inset
nad
\begin_inset Formula $k$
\end_inset
).
\end_layout
\begin_layout Itemize
Če je
\begin_inset Formula $\left(a_{k}\right)_{k}$
\end_inset
neko zaporedje (lahko tudi končno),
lahko pišemo
\begin_inset Formula $\sum_{k=1}^{n}a_{k}\coloneqq a_{1}+a_{2}+\cdots+a_{n}$
\end_inset
(pravimo summa) in
\begin_inset Formula $\prod_{k=1}^{n}a_{k}\coloneqq a_{1}\cdot\cdots\cdot a_{n}$
\end_inset
(pravimo produkt).
\end_layout
\end_deeper
\begin_layout Example*
\begin_inset Formula $\sum_{k=1}^{n}\frac{1}{k}=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}$
\end_inset
in
\begin_inset Formula $\prod_{k=1}^{n}k=1\cdot2\cdot3\cdot\cdots\cdot n=n!$
\end_inset
.
\end_layout
\begin_layout Theorem*
Binomska formula.
\begin_inset Formula $\forall a,b\in\mathbb{R},n\in\mathbb{N}:\left(a+b\right)^{n}=\sum_{k=0}^{n}\binom{n}{k}a^{k}b^{n-k}$
\end_inset
.
\end_layout
\begin_layout Proof
Indukcija po
\begin_inset Formula $n$
\end_inset
.
\end_layout
\begin_deeper
\begin_layout Itemize
Baza
\begin_inset Formula $n=1$
\end_inset
:
\begin_inset Formula $\sum_{k=0}^{1}\binom{n}{k}a^{k}b^{n-k}=\binom{1}{0}a^{0}b^{1-0}+\binom{1}{1}a^{1}b^{1-1}=a+b=\left(a+b\right)^{1}$
\end_inset
\end_layout
\begin_layout Itemize
I.
P.
\begin_inset Formula $\sum_{k=0}^{n}\binom{n}{k}a^{k}b^{n-k}=\left(a+b\right)^{n}$
\end_inset
\end_layout
\begin_layout Itemize
Korak:
\begin_inset Formula
\[
\left(a+b\right)^{n+1}=\left(a+b\right)\left(a+b\right)^{n}=\left(a+b\right)\sum_{k=0}^{n}\binom{n}{k}a^{k}b^{n-k}=\sum_{k=0}^{n}\binom{n}{k}a^{k+1}b^{n-k}+\sum_{k=0}^{n}\binom{n}{k}a^{k}b^{n-k+1}=
\]
\end_inset
sedaj naj bo
\begin_inset Formula $m=k+1$
\end_inset
v levem členu:
\begin_inset Formula
\[
=\sum_{m=1}^{n+1}\binom{n}{m-1}a^{m}b^{n-\left(m-1\right)}+\sum_{k=0}^{n}\binom{n}{k}a^{k}b^{n-k+1}=a^{n+1}+\sum_{k=1}^{n}\left[\binom{n}{k-1}+\binom{n}{k}\right]a^{k}b^{n-k+1}+b^{n+1}=
\]
\end_inset
Sedaj obravnavajmo le izraz v oglatih oklepajih:
\begin_inset Formula
\[
\binom{n}{k-1}+\binom{n}{k}=\frac{n!}{\left(k-1\right)!\left(n-k+1\right)!}+\frac{n!}{k!\left(n-k\right)!}=\frac{kn!}{k!\left(n-k+1\right)!}+\frac{n!\left(n-k+1\right)}{k!\left(n-k+1\right)!}=\frac{kn!+n!\left(n-k+1\right)}{k!\left(n-k+1\right)!}=
\]
\end_inset
\begin_inset Formula
\[
=\frac{n!\left(\cancel{k+}n\cancel{-k}+1\right)}{k!\left(n+1-k\right)!}=\frac{n!\left(n+1\right)}{k!\left(n+1-k\right)!}=\frac{\left(n+1\right)!}{k!\left(n+1-k\right)}=\binom{n+1}{k}
\]
\end_inset
in skratka dobimo
\begin_inset Formula $\binom{n}{k-1}+\binom{n}{k}=\binom{n+1}{k}$
\end_inset
.
Vstavimo to zopet v naš zgornji račun:
\begin_inset Formula
\[
\cdots=a^{n+1}+\sum_{k=1}^{n}\left[\binom{n}{k-1}+\binom{n}{k}\right]a^{k}b^{n-k+1}+b^{n+1}=a^{n+1}+\sum_{k=1}^{n}\binom{n+1}{k}a^{k}b^{n-k+1}+b^{n+1}=
\]
\end_inset
\begin_inset Formula
\[
=a^{n+1}+\sum_{k=1}^{n}\binom{n+1}{k}a^{k}b^{n-k+1}+b^{n+1}=a^{n+1}+\sum_{k=0}^{n}\binom{n+1}{k}a^{k}b^{n-k+1}=\sum_{k=0}^{n+1}\binom{n+1}{k}a^{k}b^{n-k+1}
\]
\end_inset
\end_layout
\end_deeper
\begin_layout Theorem*
Bernoulli.
Zaporedje
\begin_inset Formula $a_{n}\coloneqq\left(1+\frac{1}{n}\right)^{n}$
\end_inset
je konvergentno.
\end_layout
\begin_layout Proof
Dokazali bomo,
da je naraščajoče in omejeno.
\end_layout
\begin_deeper
\begin_layout Itemize
Naraščanje:
Dokazujemo,
da za
\begin_inset Formula $n\geq2$
\end_inset
velja
\begin_inset Formula $a_{n}\geq a_{n-1}$
\end_inset
oziroma
\begin_inset Formula
\[
\left(1+\frac{1}{n}\right)^{n}\overset{?}{\geq}\left(1+\frac{1}{n-1}\right)^{n-1}=\left(\frac{n-1}{n-1}+\frac{1}{n-1}\right)^{n-1}=\left(\frac{n}{n-1}\right)^{n-1}=\left(\frac{n-1}{n}\right)^{1-n}=\left(1-\frac{1}{n}\right)^{1-n}
\]
\end_inset
\begin_inset Formula
\[
\left(1+\frac{1}{n}\right)^{n}\overset{?}{\geq}\left(1-\frac{1}{n}\right)^{1-n}
\]
\end_inset
\begin_inset Formula
\[
\left(1+\frac{1}{n^{2}}\right)^{n}=\left(\left(1+\frac{1}{n}\right)\left(1-\frac{1}{n}\right)\right)^{n}=\left(1+\frac{1}{n}\right)^{n}\left(1-\frac{1}{n}\right)^{n}\overset{?}{\geq}1-\frac{1}{n}
\]
\end_inset
\begin_inset Formula
\[
\left(1+\frac{1}{n^{2}}\right)^{n}\overset{?}{\geq}1-\frac{1}{n},
\]
\end_inset
kar je poseben primer Bernoullijeve neenakosti za
\begin_inset Formula $\alpha=\frac{1}{n^{2}}$
\end_inset
.
\end_layout
\begin_layout Itemize
Omejenost:
Po binomski formuli je
\begin_inset Formula
\[
a_{n}=\left(1+\frac{1}{n}\right)^{n}=\sum_{k=0}^{n}\binom{n}{k}\left(\frac{1}{n}\right)^{k}=\sum_{k=0}^{n}\frac{n!}{k!\left(n-k\right)!n^{k}}=1+1+\sum_{k=2}^{n}\frac{n!}{k!\left(n-k\right)!n^{k}}=2+\sum_{k=2}^{n}\frac{1}{k!}\cdot\frac{n\left(n-1\right)\left(n-2\right)\cdots\left(n-k+1\right)}{n^{k}}=
\]
\end_inset
\begin_inset Formula
\[
=2+\sum_{k=2}^{n}\frac{1}{k!}\cdot\frac{n}{n}\cdot\frac{n-1}{n}\cdot\frac{n-2}{n}\cdot\cdots\cdot\frac{n-k+1}{n}=2+\sum_{k=2}^{n}\frac{1}{k!}\cdot\cancel{\left(1-0\right)}\cdot\left(1-\frac{1}{n}\right)\cdot\left(1-\frac{2}{n}\right)\cdot\cdots\cdot\left(1-\frac{k-1}{n}\right)=
\]
\end_inset
\begin_inset Formula
\[
=2+\sum_{k=2}^{n}\frac{1}{k!}\prod_{j=1}^{k-1}\left(1-\frac{j}{n}\right)<2+\sum_{k=2}^{n}\frac{1}{k!}<2+\sum_{k=2}^{n}\frac{1}{2^{k-1}}=\cdots
\]
\end_inset
Opomnimo,
da je
\begin_inset Formula $1-\frac{j}{n}<1$
\end_inset
,
zato
\begin_inset Formula $\prod_{j=1}^{k-1}\left(1-\frac{j}{n}\right)<1$
\end_inset
(prvi neenačaj) ter
\begin_inset Formula $k!=1\cdot2\cdot3\cdot\cdots\cdot k\geq1\cdot2\cdot2\cdot\cdots\cdot2=2^{k-1}$
\end_inset
(drugi).
Sedaj si z indukcijo dokažimo
\begin_inset Formula $\sum_{k=2}^{n}\frac{1}{2^{k-1}}=1-\frac{1}{2^{n-1}}$
\end_inset
:
\end_layout
\begin_deeper
\begin_layout Itemize
Baza:
\begin_inset Formula $n=2$
\end_inset
:
\begin_inset Formula $\frac{1}{2^{2-1}}=1-\frac{1}{2^{2-1}}=1-\frac{1}{2}=\frac{1}{2}$
\end_inset
.
Velja!
\end_layout
\begin_layout Itemize
I.
P.:
\begin_inset Formula $\sum_{k=2}^{n}\frac{1}{2^{k-1}}=1-\frac{1}{2^{n-1}}$
\end_inset
\end_layout
\begin_layout Itemize
Korak:
\begin_inset Formula $\sum_{k=2}^{n+1}\frac{1}{2^{k-1}}=1-\frac{1}{2^{n-1}}+\frac{1}{2^{n+1-1}}=1-2\cdot2^{-n}+2^{-n}=1+2^{-n}\left(1-2\right)=1+2^{-n}$
\end_inset
\end_layout
\begin_layout Standard
In nadaljujmo z računanjem:
\begin_inset Formula
\[
\cdots=2+1-\frac{1}{2^{n-1}}=3-\frac{1}{2^{n-1}},
\]
\end_inset
s čimer dobimo zgornjo mejo
\begin_inset Formula $\forall n\in\mathbb{N}:a_{n}<3$
\end_inset
.
Ker je očitno
\begin_inset Formula $\forall n\in\mathbb{N}:a_{n}>0$
\end_inset
,
je torej zaporedje omejeno in ker je tudi monotono po
\begin_inset ERT
status open
\begin_layout Plain Layout
\backslash
hyperlink{kmoz}{prejšnjem izreku}
\end_layout
\end_inset
konvergira.
\end_layout
\end_deeper
\end_deeper
\begin_layout Definition*
Označimo število
\begin_inset Formula $e\coloneqq\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^{n}$
\end_inset
in ga imenujemo Eulerjevo število.
Velja
\begin_inset Formula $e\approx2,71828$
\end_inset
.
\end_layout
\begin_layout Remark*
V dokazu vidimo moč izreka
\begin_inset Quotes gld
\end_inset
omejenost in monotonost
\begin_inset Formula $\Rightarrow$
\end_inset
konvergenca
\begin_inset Quotes grd
\end_inset
,
saj nam omogoča dokazati konvergentnost zaporedja brez kandidata za limito.
Jasno je,
da ne bi mogli vnaprej uganiti,
da je limita ravno
\shape italic
transcendentno število
\shape default
\begin_inset Formula $e$
\end_inset
.
\end_layout
\begin_layout Definition*
Podzaporedje zaporedja
\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
\end_inset
je poljubno zaporedje oblike
\begin_inset Formula $\left(a_{\varphi\left(n\right)}\right)_{n\in\mathbb{N}}$
\end_inset
,
kjer je
\begin_inset Formula $\varphi:\mathbb{N}\to\mathbb{N}$
\end_inset
strogo naraščajoča funkcija.
\end_layout
\begin_layout Theorem*
Če je
\begin_inset Formula $L=\lim_{n\to\infty}a_{n}$
\end_inset
,
tedaj je
\begin_inset Formula $L$
\end_inset
tudi limita vsakega podzaporedja.
\end_layout
\begin_layout Proof
Po predpostavki velja
\begin_inset Formula $\forall\varepsilon>0\exists n_{0}\in\mathbb{N}\forall n\in\mathbb{N}:n\geq n_{0}:\left|a_{n}-L\right|<\varepsilon$
\end_inset
.
Vzemimo poljuben
\begin_inset Formula $\varepsilon>0$
\end_inset
.
Po predpostavki obstaja
\begin_inset Formula $n_{0}\in\mathbb{N}$
\end_inset
,
da bodo vsi členi zaporedja po
\begin_inset Formula $n_{0}-$
\end_inset
tem v
\begin_inset Formula $\left(L-\varepsilon,L+\varepsilon\right)$
\end_inset
.
Iz definicijskega območja
\begin_inset Formula $\varphi$
\end_inset
vzemimo poljuben element
\begin_inset Formula $n_{1}$
\end_inset
,
da velja
\begin_inset Formula $n_{1}\geq n_{0}$
\end_inset
.
Gotovo obstaja,
ker je definicijsko območje števno neskončne moči in s pogojem
\begin_inset Formula $n_{1}\geq n_{0}$
\end_inset
onemogočimo izbiro le končno mnogo elementov.
\begin_inset Note Note
status open
\begin_layout Plain Layout
Če slednji ne obstaja,
je v
\begin_inset Formula $D_{\varphi}$
\end_inset
končno mnogo elementov,
tedaj vzamemo
\begin_inset Formula $n_{1}\coloneqq\max D_{\varphi}+1$
\end_inset
in je pogoj za limito izpolnjen na prazno.
Sicer pa v
\end_layout
\end_inset
Velja
\begin_inset Formula $\forall n\in\mathbb{N}:n>n_{1}\Rightarrow\left|a_{\varphi n}-L\right|<\varepsilon$
\end_inset
,
ker je
\begin_inset Formula $\varphi$
\end_inset
strogo naraščajoča in izbiramo le elemente podzaporedja,
ki so v izvornem zaporedju za
\begin_inset Formula $n_{0}-$
\end_inset
tim členom in zato v
\begin_inset Formula $\left(L-\varepsilon,L+\varepsilon\right)$
\end_inset
.
\end_layout
\begin_layout Example*
\begin_inset Formula $\lim_{n\to\infty}\frac{1}{2n+3}=\lim_{n\to\infty}\frac{1}{n}=0$
\end_inset
za zaporedje
\begin_inset Formula $a_{n}=\frac{1}{n}$
\end_inset
in podzaporedje
\begin_inset Formula $a_{\varphi n}$
\end_inset
,
kjer je
\begin_inset Formula $\varphi\left(n\right)=2n+3$
\end_inset
.
\end_layout
\begin_layout Theorem*
Karakterizacija limite s podzaporedji.
Naj bo
\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
\end_inset
realno zaporedje in
\begin_inset Formula $L\in\mathbb{R}$
\end_inset
.
Tedaj
\begin_inset Formula $L=\lim_{n\to\infty}a_{n}\Leftrightarrow$
\end_inset
za vsako podzaporedje
\begin_inset Formula $\left(a_{n_{k}}\right)_{k\in\mathbb{N}}$
\end_inset
zaporedja
\begin_inset Formula $\left(a_{n}\right)_{n}$
\end_inset
obstaja njegovo podzaporedje
\begin_inset Formula $\left(a_{n_{k_{l}}}\right)_{l\in\mathbb{N}}$
\end_inset
,
ki konvergira k
\begin_inset Formula $L$
\end_inset
.
\end_layout
\begin_layout Proof
Dokazujemo ekvivalenco:
\end_layout
\begin_deeper
\begin_layout Labeling
\labelwidthstring 00.00.0000
\begin_inset Formula $\left(\Rightarrow\right)$
\end_inset
Dokazano poprej.
Limita se pri prehodu na podzaporedje ohranja.
\end_layout
\begin_layout Labeling
\labelwidthstring 00.00.0000
\begin_inset Formula $\left(\Leftarrow\right)$
\end_inset
PDDRAA
\begin_inset Formula $a_{n}\not\to L$
\end_inset
.
Tedaj
\begin_inset Formula $\exists\varepsilon>0$
\end_inset
in podzaporedje
\begin_inset Formula $\left(a_{n_{k}}\right)_{k\in\mathbb{N}}\ni:\forall k\in\mathbb{N}:\left|a_{n_{k}}-K\right|>\varepsilon$
\end_inset
(*)
\begin_inset Note Note
status open
\begin_layout Plain Layout
tu je na
\begin_inset Quotes gld
\end_inset
Zaporedja 2
\begin_inset Quotes grd
\end_inset
napaka,
neenačaj obrne v drugo smer
\end_layout
\end_inset
.
Po predpostavki sedaj
\begin_inset Formula $\exists\left(a_{n_{k_{l}}}\right)_{l\in\mathbb{N}}\ni:\lim_{l\to\infty}a_{n_{k_{l}}}=L$
\end_inset
.
To pa je v protislovju z (*),
torej je začetna predpostavka
\begin_inset Formula $a_{n}\not\to L$
\end_inset
napačna,
torej
\begin_inset Formula $a_{n}\to L$
\end_inset
.
\end_layout
\end_deeper
\begin_layout Subsection
Stekališča
\end_layout
\begin_layout Definition*
Točka
\begin_inset Formula $s\in\mathbb{R}$
\end_inset
je stekališče zaporedje
\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}\subset\mathbb{R}$
\end_inset
,
če v vsaki okolici te točke leži neskončno členov zaporedja.
\end_layout
\begin_layout Remark*
Pri limiti zahtevamo več;
da izven vsake okolice limite leži le končno mnogo členov.
\end_layout
\begin_layout Example*
Primeri stekališč.
\end_layout
\begin_deeper
\begin_layout Enumerate
\begin_inset Formula $L=\lim_{n\to\infty}a_{n}\Rightarrow L$
\end_inset
je stekališče za
\begin_inset Formula $a_{n}$
\end_inset
\end_layout
\begin_layout Enumerate
\begin_inset Formula $0,1,0,1,\dots$
\end_inset
stekališči sta
\begin_inset Formula $\left\{ 0,1\right\} $
\end_inset
in zaporedje nima limite (ni konvergentno)
\end_layout
\begin_layout Enumerate
\begin_inset Formula $1,1,2,1,2,3,1,2,3,4,\dots$
\end_inset
ima neskončno stekališč,
\begin_inset Formula $\mathbb{N}$
\end_inset
.
\end_layout
\begin_layout Enumerate
\begin_inset Formula $b_{n}=n-$
\end_inset
to racionalno število
\begin_inset Foot
status open
\begin_layout Plain Layout
Racionalnih števil je števno mnogo,
zato jih lahko linearno uredimo in oštevilčimo.
\end_layout
\end_inset
ima neskončno stekališč,
\begin_inset Formula $\mathbb{R}$
\end_inset
.
\end_layout
\end_deeper
\begin_layout Remark*
Limita je stakališče,
stekališče pa ni nujno limita.
Poleg tega,
če se spomnimo,
velja,
da vsota konvergentnih zaporedij konvergira k vsoti njunih limit,
ni pa nujno res,
da so stekališča vsote dveh zaporedij paroma vsote stekališč teh dveh zaporedij.
Primer:
\begin_inset Formula $a_{n}=\left(-1\right)^{n}$
\end_inset
in
\begin_inset Formula $b_{n}=-\left(-1\right)^{n}$
\end_inset
.
Njuni stekališči sta
\begin_inset Formula $\left\{ -1,1\right\} $
\end_inset
,
toda
\begin_inset Formula $a_{n}+b_{n}=0$
\end_inset
ima le stekališče
\begin_inset Formula $\left\{ 0\right\} $
\end_inset
,
ne pa tudi
\begin_inset Formula $\left\{ 1,-1,2,-2\right\} $
\end_inset
.
\end_layout
\begin_layout Theorem*
\begin_inset Formula $S$
\end_inset
je stekališče
\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}\Leftrightarrow S$
\end_inset
je limita nekega podzaporedja
\begin_inset Formula $a_{n}$
\end_inset
.
\end_layout
\begin_layout Proof
Dokazujemo ekvivalenco.
\end_layout
\begin_deeper
\begin_layout Labeling
\labelwidthstring 00.00.0000
\begin_inset Formula $\left(\Leftarrow\right)$
\end_inset
Očitno.
\end_layout
\begin_layout Labeling
\labelwidthstring 00.00.0000
\begin_inset Formula $\left(\Rightarrow\right)$
\end_inset
Definirajmo
\begin_inset Formula $\forall m\in\mathbb{N}:U_{m}\coloneqq\left(S-\frac{1}{m},S+\frac{1}{m}\right)$
\end_inset
.
Ker je
\begin_inset Formula $S$
\end_inset
stekališče,
\begin_inset Formula $\forall m\in\mathbb{N}\exists a_{k_{m}}\in U_{m}$
\end_inset
.
Podzaporedje
\begin_inset Formula $\left(a_{k_{m}}\right)_{m\in\mathbb{N}}$
\end_inset
konvergira k
\begin_inset Formula $S$
\end_inset
,
kajti
\begin_inset Formula $\forall\varepsilon>0\exists n_{0}\in\mathbb{N}\forall n\in\mathbb{N}:n\geq n_{0}\Rightarrow\left|a_{k_{n}}-S\right|<\frac{1}{n}<\varepsilon$
\end_inset
.
\end_layout
\end_deeper
\begin_layout Corollary*
Če je
\begin_inset Formula $L$
\end_inset
limita nekega zaporedja,
je
\begin_inset Formula $L$
\end_inset
edino njegovo stekališče.
\end_layout
\begin_layout Proof
Naj bo
\begin_inset Formula $a_{n}\to L$
\end_inset
.
Naj bo
\begin_inset Formula $S$
\end_inset
stekališče za
\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
\end_inset
.
Po izreku zgoraj je
\begin_inset Formula $S$
\end_inset
limita nekega podzaporedja
\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
\end_inset
.
Toda limita vsakega podzaporedja je enaka limiti zaporedja,
iz katerega to podzaporedje izhaja,
če ta limita obstaja.
Potemtakem je
\begin_inset Formula $S=L$
\end_inset
.
\end_layout
\begin_layout Theorem*
\begin_inset ERT
status open
\begin_layout Plain Layout
\backslash
hypertarget{bw}{Bolzano-Weierstraß}
\end_layout
\end_inset
.
Eksistenčni izrek.
Vsako omejeno zaporedje v realnih številih ima kakšno stekališče v
\begin_inset Formula $\mathbb{R}$
\end_inset
.
\end_layout
\begin_layout Proof
Označimo
\begin_inset Formula $m_{0}\coloneqq\inf_{n\in\mathbb{N}}a_{n},M_{0}\coloneqq\sup_{n\in\mathbb{N}}a_{n},I_{0}\coloneqq\left[m_{0},M_{0}\right]$
\end_inset
.
Očitno je
\begin_inset Formula $\forall n\in\mathbb{N}:a_{n}\in I_{0}$
\end_inset
.
Interval
\begin_inset Formula $I_{0}$
\end_inset
razdelimo na dve polovici:
\begin_inset Formula $I_{0}=\left[m_{0},\frac{m_{0}+M_{0}}{2}\right]\cup\left[\frac{m_{0}+M_{0}}{2},M_{0}\right]$
\end_inset
.
Izberemo polovico (vsaj ena obstaja),
v kateri leži neskončno mnogo členov,
in jo označimo z
\begin_inset Formula $I_{1}$
\end_inset
.
Spet jo razdelimo na pol in z
\begin_inset Formula $I_{2}$
\end_inset
označimo tisto polovico,
v kateri leži neskončno mnogo členov.
Postopek ponavljamo in dobimo zaporedje zaprtih intervalov
\begin_inset Formula $\left(I_{n}\right)_{n\in\mathbb{N}}$
\end_inset
in velja
\begin_inset Formula $I_{0}\supset I_{1}\supset I_{2}\supset\cdots$
\end_inset
ter
\begin_inset Formula $\left|I_{n}\right|=2^{-n}\left|I_{0}\right|$
\end_inset
.
\end_layout
\begin_layout Proof
Označimo sedaj
\begin_inset Formula $I_{n}\eqqcolon\left[l_{n},d_{n}\right]$
\end_inset
.
Iz konstrukcije je očitno,
da
\begin_inset Formula $\left(l_{n}\right)_{n\in\mathbb{N}}$
\end_inset
narašča in
\begin_inset Formula $\left(d_{n}\right)_{n\in\mathbb{N}}$
\end_inset
pada ter da sta obe zaporedji omejeni.
Posledično
\begin_inset Formula $\exists l\coloneqq\lim_{n\to\infty}l_{n},d\coloneqq\lim_{n\to\infty}d_{n}$
\end_inset
.
Iz
\begin_inset Formula $l_{n}\leq l\leq d\leq d_{n}$
\end_inset
sledi ocena
\begin_inset Formula $d-l\leq l_{n}-d_{n}=\left|I_{n}\right|=2^{-n}\left|I_{0}\right|$
\end_inset
,
kar konvergira k 0.
Posledično
\begin_inset Formula $d=l$
\end_inset
.
\end_layout
\begin_layout Proof
Treba je pokazati še,
da je
\begin_inset Formula $d=l$
\end_inset
stekališče za
\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
\end_inset
.
Vzemimo poljuben
\begin_inset Formula $\varepsilon>0$
\end_inset
.
Ker je
\begin_inset Formula $l=\lim_{n\to\infty}l_{n}\Rightarrow\exists n_{1}\in\mathbb{N}\ni:l_{n_{1}}>l-\varepsilon$
\end_inset
in ker je
\begin_inset Formula $d=\lim_{n\to\infty}d_{n}\Rightarrow\exists n_{2}\in\mathbb{N}\ni:d_{n_{2}}<d-\varepsilon$
\end_inset
.
Torej
\begin_inset Formula $\left[l_{n_{1}},d_{n_{2}}\right]\subset\left(l-\varepsilon,d+\varepsilon\right)$
\end_inset
.
Torej za
\begin_inset Formula $n_{0}\coloneqq\max\left\{ n_{1},n_{2}\right\} $
\end_inset
velja
\begin_inset Formula $I_{n_{0}}=\left[l_{n_{0}},d_{n_{n}}\right]\subset\left(l-\varepsilon,d+\varepsilon\right)$
\end_inset
.
Ker
\begin_inset Formula $I_{n_{0}}$
\end_inset
po konstrukciji vsebuje neskončno mnogo elementov,
jih torej tudi
\begin_inset Formula $\left(l-\varepsilon,d+\varepsilon\right)$
\end_inset
oziroma poljubno majhna okolica
\begin_inset Formula $d=l$
\end_inset
,
torej je
\begin_inset Formula $d=l$
\end_inset
stekališče za
\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
\end_inset
.
\end_layout
\begin_layout Corollary*
Če je
\begin_inset Formula $s\in\mathbb{R}$
\end_inset
edino stekališče omejenega zaporedja
\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
\end_inset
,
tedaj je
\begin_inset Formula $s=\lim_{n\to\infty}a_{n}$
\end_inset
.
\end_layout
\begin_layout Proof
Naj bo
\begin_inset Formula $s$
\end_inset
stekališče
\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
\end_inset
.
PDDRAA
\begin_inset Formula $a_{n}\not\to s$
\end_inset
.
Tedaj
\begin_inset Formula $\exists\varepsilon>0\ni:$
\end_inset
izven
\begin_inset Formula $\left(s-\varepsilon,s+\varepsilon\right)$
\end_inset
se nahaja neskončno mnogo členov zaporedja.
Ti členi sami zase tvorijo omejeno zaporedje,
ki ima po
\begin_inset ERT
status open
\begin_layout Plain Layout
\backslash
hyperlink{bw}{B.-W.}
\end_layout
\end_inset
izreku stekališče.
Slednje gotovo ne more biti enako
\begin_inset Formula $s$
\end_inset
,
torej imamo vsaj dve stekališči,
kar je v je v
\begin_inset Formula $\rightarrow\!\leftarrow$
\end_inset
s predpostavko.
\end_layout
\begin_layout Definition*
Pravimo,
da ima realno zaporedje:
\end_layout
\begin_deeper
\begin_layout Itemize
stekališče v
\begin_inset Formula $\infty$
\end_inset
,
če
\begin_inset Formula $\forall M>0:\left(M,\infty\right)$
\end_inset
vsebuje neskončno mnogo členov zapopredja
\end_layout
\begin_layout Itemize
limito v
\begin_inset Formula $\infty$
\end_inset
,
če
\begin_inset Formula $\forall M>0:\left(M,\infty\right)$
\end_inset
vsebuje vse člene zaporedja od nekega indeksa dalje
\end_layout
\begin_layout Standard
in podobno za
\begin_inset Formula $-\infty$
\end_inset
.
\end_layout
\end_deeper
\begin_layout Remark*
Povezava s pojmom realnega stekališča/limite:
okolice
\begin_inset Quotes gld
\end_inset
točke
\begin_inset Quotes grd
\end_inset
\begin_inset Formula $\infty$
\end_inset
so intervali oblike
\begin_inset Formula $\left(M,\infty\right)$
\end_inset
.
To je smiselno,
saj biti
\begin_inset Quotes gld
\end_inset
blizu
\begin_inset Formula $\infty$
\end_inset
\begin_inset Quotes grd
\end_inset
pomeni bizi zelo velik,
kar je ravno biti v
\begin_inset Formula $\left(M,\infty\right)$
\end_inset
za poljubno velik
\begin_inset Formula $M$
\end_inset
.
\begin_inset Quotes gld
\end_inset
Okolica točke
\begin_inset Formula $\infty$
\end_inset
\begin_inset Quotes grd
\end_inset
so torej vsi intervali oblike
\begin_inset Formula $\left(M,\infty\right)$
\end_inset
.
\end_layout
\begin_layout Subsection
Limes superior in limes inferior
\end_layout
\begin_layout Definition*
Naj bo
\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
\end_inset
realno zaporedje.
Tvorimo novo zaporedje
\begin_inset Formula $s_{n}\coloneqq\sup\left\{ a_{k};k\geq n\right\} $
\end_inset
.
Očitno je padajoče (
\begin_inset Formula $s_{1}\geq s_{2}\geq s_{3}\geq\cdots$
\end_inset
),
ker je supremum množice vsaj supremum njene stroge podmnožice.
Zaporedje
\begin_inset Formula $\left(s_{n}\right)_{n\in\mathbb{N}}$
\end_inset
ima limito,
ki ji rečemo limes superior oziroma zgornja limita zaporedja
\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
\end_inset
in označimo
\begin_inset Formula $\limsup_{n\to\infty}a_{n}=\overline{\lim_{n\to\infty}}a_{n}\coloneqq\lim_{n\to\infty}s_{n}$
\end_inset
in velja,
da leži v
\begin_inset Formula $\mathbb{R}\cup\left\{ -\infty,\infty\right\} $
\end_inset
.
Podobno definiramo tudi limes inferior oz.
spodnjo limito zaporedja:
\begin_inset Formula $\liminf_{n\to\infty}a_{n}=\underline{\lim_{n\to\infty}}a_{n}\coloneqq\lim_{n\to\infty}\left(\inf_{k\geq n}a_{k}\right)=\sup_{n\in\mathbb{N}}\left(\inf_{k\geq n}a_{k}\right)$
\end_inset
.
\end_layout
\begin_layout Remark*
Za razliko od običajne limite,
ki lahko ne obstaja,
\begin_inset Formula $\limsup$
\end_inset
in
\begin_inset Formula $\liminf$
\end_inset
vedno obstajata.
\end_layout
\begin_layout Claim*
\begin_inset Formula $\limsup_{n\to\infty}a_{n}$
\end_inset
je največje stekališče zaporedja
\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
\end_inset
in
\begin_inset Formula $\liminf_{n\to\infty}$
\end_inset
najmanjše.
\end_layout
\begin_layout Proof
Označimo
\begin_inset Formula $s\coloneqq\limsup_{n\to\infty}a_{n}$
\end_inset
.
Za
\begin_inset Formula $\liminf$
\end_inset
je dokaz analogen in ga ne bomo pisali.
Dokazujemo,
da je
\begin_inset Formula $s$
\end_inset
stekališče in
\begin_inset Formula $\forall t>s:t$
\end_inset
ni stekališče.
Ločimo primere:
\end_layout
\begin_deeper
\begin_layout Labeling
\labelwidthstring 00.00.0000
\begin_inset Formula $s\in\mathbb{R}$
\end_inset
Naj bo
\begin_inset Formula $\varepsilon>0$
\end_inset
poljuben.
Ker
\begin_inset Foot
status open
\begin_layout Plain Layout
Infimum padajočega konvergentnega zaporedja je očitno njegova limita.
\end_layout
\end_inset
je
\begin_inset Formula $s=\inf s_{n}$
\end_inset
,
\begin_inset Formula $\exists n_{0}\in\mathbb{N}\ni:s_{n_{0}}\in[s,s+\varepsilon)$
\end_inset
.
Ker
\begin_inset Formula $\left(s_{n}\right)_{n\in\mathbb{N}}$
\end_inset
pada proti
\begin_inset Formula $s$
\end_inset
,
sledi
\begin_inset Formula $\forall n\in\mathbb{N}:n\geq n_{0}\Rightarrow s_{n}\in[s,s+\varepsilon)$
\end_inset
.
Po definiciji
\begin_inset Formula $s_{n}$
\end_inset
velja
\begin_inset Formula $\forall n\in\mathbb{N}\exists N\left(n\right)\geq n\ni:s_{n}-\varepsilon<a_{N\left(n\right)}$
\end_inset
.
Torej imamo
\begin_inset Formula $s-\varepsilon\leq s_{n}-\varepsilon<a_{N\left(n\right)}\leq s_{n}<s+\varepsilon$
\end_inset
(zadnji neenačaj za
\begin_inset Formula $n\geq n_{0}$
\end_inset
),
skratka
\begin_inset Formula $a_{N\left(n\right)}-s<\varepsilon$
\end_inset
oziroma
\begin_inset Formula $\forall n\geq n_{0}:\left|a_{N\left(n\right)}-s\right|<\varepsilon$
\end_inset
.
Ker je
\begin_inset Formula $N\left(n\right)\geq n$
\end_inset
,
je
\begin_inset Formula $\left\{ N\left(n\right);n\in\mathbb{N}\right\} $
\end_inset
neskončna množica,
torej je neskončno mnogo členov v poljubni okolici
\begin_inset Formula $s$
\end_inset
.
\end_layout
\begin_deeper
\begin_layout Standard
Treba je dokazati še,
da
\begin_inset Formula $\forall t>s:t$
\end_inset
ni stekališče.
Naj bo
\begin_inset Formula $t>s$
\end_inset
.
Označimo
\begin_inset Formula $\delta\coloneqq t-s>0$
\end_inset
.
Po definiciji
\begin_inset Foot
status open
\begin_layout Plain Layout
\begin_inset Formula $s$
\end_inset
je limita zaporedja
\begin_inset Formula $\left(s_{n}\right)_{n\in\mathbb{N}}$
\end_inset
,
zato v poljubno majhni okolici obstaja tak
\begin_inset Formula $s_{n_{1}}$
\end_inset
.
\begin_inset Formula $s_{n_{1}}$
\end_inset
torej tu najdemo v
\begin_inset Formula $[s,s+\frac{\delta}{2})$
\end_inset
.
\end_layout
\end_inset
\begin_inset Formula $s$
\end_inset
\begin_inset Formula $\exists n_{1}\in\mathbb{N}\ni:s\leq s_{n_{1}}<s+\frac{\delta}{2}<s+t$
\end_inset
.
Ker
\begin_inset Formula $\left(s_{n}\right)_{n\in\mathbb{N}}$
\end_inset
pada proti
\begin_inset Formula $s$
\end_inset
,
sledi
\begin_inset Formula $\forall n\geq n_{1}:s\leq s_{n}<s+\frac{\delta}{2}$
\end_inset
.
Po definiciji
\begin_inset Foot
status open
\begin_layout Plain Layout
\begin_inset Formula $s_{n}$
\end_inset
je supremum členov od vključno
\begin_inset Formula $n$
\end_inset
dalje
\end_layout
\end_inset
\begin_inset Formula $s_{n}$
\end_inset
sledi
\begin_inset Formula $\forall n\geq n_{1}:a_{n}\leq s+\frac{\delta}{2}$
\end_inset
.
Za takšne
\begin_inset Formula $n$
\end_inset
je
\begin_inset Formula $\left|t-a_{n}\right|=t-a_{n}\geq t-\left(s+\frac{\delta}{2}\right)=\frac{\delta}{2}$
\end_inset
.
Torej v
\begin_inset Formula $\frac{\delta}{2}-$
\end_inset
okolici točke
\begin_inset Formula $t$
\end_inset
leži kvečjemu končno mnogo členov zaporedja oziroma členi
\begin_inset Formula $\left(a_{1},a_{2},\dots,a_{n_{1}-1}\right)$
\end_inset
.
Torej
\begin_inset Formula $t$
\end_inset
ni stekališče za
\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
\end_inset
.
\end_layout
\end_deeper
\begin_layout Labeling
\labelwidthstring 00.00.0000
\begin_inset Formula $s=\infty$
\end_inset
Naj bo
\begin_inset Formula $M>0$
\end_inset
poljuben.
Ker je
\begin_inset Formula $s=\inf s_{n}$
\end_inset
,
velja
\begin_inset Formula $\forall n\in\mathbb{N}:s_{n}=\infty$
\end_inset
.
Po definiciji
\begin_inset Formula $s_{n}=\infty$
\end_inset
velja
\begin_inset Formula $\forall n\in\mathbb{N}\exists N\left(n\right):a_{N\left(n\right)}>M$
\end_inset
.
Ker je
\begin_inset Formula $N\left(n\right)\geq n$
\end_inset
,
je
\begin_inset Formula $\left\{ N\left(n\right);n\in\mathbb{N}\right\} $
\end_inset
neskončna množica,
torej je neskončno mnogo členov v
\begin_inset Formula $\left(M,\infty\right)$
\end_inset
za poljuben
\begin_inset Formula $M$
\end_inset
,
torej je
\begin_inset Formula $s=\infty$
\end_inset
res stekališče.
\end_layout
\begin_deeper
\begin_layout Standard
Večjih stekališč od
\begin_inset Formula $\infty$
\end_inset
očitno ni.
\end_layout
\end_deeper
\begin_layout Labeling
\labelwidthstring 00.00.0000
\begin_inset Formula $s=-\infty$
\end_inset
Naj bo
\begin_inset Formula $m<0$
\end_inset
poljuben.
Ker je
\begin_inset Formula $s=\inf s_{n}$
\end_inset
,
\begin_inset Formula $\exists n_{0}\in\mathbb{N}\ni:s_{n_{0}}\in\left(-\infty,m\right)$
\end_inset
Ker
\begin_inset Formula $\left(s_{n}\right)_{n\in\mathbb{N}}$
\end_inset
pada proti
\begin_inset Formula $s=-\infty$
\end_inset
,
sledi
\begin_inset Formula $\forall n\in\mathbb{N}:n\geq n_{0}:s_{n}\in\left(-\infty,m\right)$
\end_inset
.
Po definiciji
\begin_inset Formula $s_{n}$
\end_inset
velja
\begin_inset Formula $\forall n\in\mathbb{N}:a_{n}\in\left(-\infty,m\right)$
\end_inset
.
Ker je za poljuben
\begin_inset Formula $m$
\end_inset
neskončno mnogo členov v
\begin_inset Formula $\left(-\infty,m\right)$
\end_inset
,
je
\begin_inset Formula $s=-\infty$
\end_inset
res stekališče.
\end_layout
\end_deeper
\begin_layout Subsection
Cauchyjev pogoj
\end_layout
\begin_layout Definition*
Zaporedje
\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
\end_inset
ustreza Cauchyjevemu pogoju (oz.
je Cauchyjevo),
če
\begin_inset Formula $\forall\varepsilon>0\exists n_{0}\in\mathbb{N}\ni:\forall m,n\in\mathbb{N}:m,n\geq n_{0}\Rightarrow\left|a_{m}-a_{n}\right|<\varepsilon$
\end_inset
.
ZDB Dovolj pozni členi so si poljubno blizu.
\end_layout
\begin_layout Claim*
Zaporedje v
\begin_inset Formula $\mathbb{R}$
\end_inset
je konvergentno
\begin_inset Formula $\Leftrightarrow$
\end_inset
je Cauchyjevo.
\end_layout
\begin_layout Proof
Dokazujemo ekvivalenco.
\end_layout
\begin_deeper
\begin_layout Labeling
\labelwidthstring 00.00.0000
\begin_inset Formula $\left(\Rightarrow\right)$
\end_inset
Če
\begin_inset Formula $a_{n}\to L$
\end_inset
,
tedaj
\begin_inset Formula $\left|a_{m}-a_{n}\right|=\left|\left(a_{m}-L\right)+\left(L-a_{n}\right)\right|\leq\left|a_{m}-\varepsilon\right|+\left|a_{n}-\varepsilon\right|$
\end_inset
.
Cauchyjev pogoj sledi iz definicije limite za
\begin_inset Formula $\frac{\varepsilon}{2}$
\end_inset
.
\end_layout
\begin_layout Labeling
\labelwidthstring 00.00.0000
\begin_inset Formula $\left(\Leftarrow\right)$
\end_inset
Če je zaporedje Cauchyjevo,
je omejeno:
\begin_inset Formula $\exists n_{0}\in\mathbb{N}\ni:\forall m,n\in\mathbb{N}:m,n\geq n_{0}\Rightarrow\left|a_{m}-a_{n}\right|\leq1$
\end_inset
.
V posebnem,
\begin_inset Formula $m=n_{0}$
\end_inset
,
\begin_inset Formula $\left|a_{n_{0}}-a_{n}\right|\leq1$
\end_inset
oziroma
\begin_inset Formula $\forall n\in\mathbb{N}:n\geq n_{0}\Rightarrow a_{n}\in\left[a_{n_{0}}-1,a_{n_{0}}+1\right]$
\end_inset
.
Preostali členi tvorijo končno veliko množico,
ki ima
\begin_inset Formula $\min$
\end_inset
in
\begin_inset Formula $\max$
\end_inset
,
torej je
\begin_inset Formula $\left\{ a_{k};k\in\mathbb{N}\right\} =\left\{ a_{1},a_{2},\dots,a_{n_{0}-1}\right\} \cup\left\{ a_{k};k\in\mathbb{N},k\geq n_{0}\right\} $
\end_inset
tudi omejena.
Po
\begin_inset ERT
status open
\begin_layout Plain Layout
\backslash
hyperlink{bw}{izreku od prej}
\end_layout
\end_inset
sledi,
da ima zaporedje stekališče
\begin_inset Formula $s$
\end_inset
.
Dokažimo,
da je
\begin_inset Formula $s=\lim_{n\to\infty}a_{n}$
\end_inset
.
Vzemimo poljuben
\begin_inset Formula $\varepsilon>0$
\end_inset
.
Ker je
\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
\end_inset
Cauchyjevo,
\begin_inset Formula $\exists n_{0}\in\mathbb{N}\ni:\forall m,n\in\mathbb{N}:m,n\geq n_{0}\Rightarrow\left|a_{m}-a_{n}\right|<\frac{\varepsilon}{2}$
\end_inset
.
Po definiciji
\begin_inset Formula $s$
\end_inset
\begin_inset Formula $\exists n_{1}\geq n_{0}\ni:\left|a_{n_{1}}-s\right|<\frac{\varepsilon}{2}$
\end_inset
.
Sledi
\begin_inset Formula $\forall n\geq n_{0}:\left|a_{n}-s\right|=\left|a_{n}-s+s-a_{n_{1}}\right|\leq\left|a_{n}-s\right|+\left|s-a_{n_{1}}\right|<\frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon$
\end_inset
.
\end_layout
\end_deeper
\begin_layout Remark*
Moč izreka je v tem,
da lahko konvergenco preverjamo tudi tedaj,
ko nimamo kandidatov za limito.
\end_layout
\begin_layout Section
Številske vrste
\end_layout
\begin_layout Standard
Kako sešteti neskončno mnogo števil?
Nadgradimo pristop končnih vsot na neskončne vsote!
\end_layout
\begin_layout Definition*
Imejmo zaporedje
\begin_inset Formula $\left(a_{k}\right)_{k\in\mathbb{N}},a_{k}\in\mathbb{R}$
\end_inset
.
Izraz
\begin_inset Formula $\sum_{j=1}^{\infty}a_{j}$
\end_inset
se imenuje vrsta s členi
\begin_inset Formula $a_{j}$
\end_inset
.
Pomen izraza opredelimo na naslednjo način:
\end_layout
\begin_layout Definition*
Tvorimo novo zaporedje,
pravimo mu zaporedje delnih vsot vrste:
\begin_inset Formula $s_{1}=a_{1}$
\end_inset
,
\begin_inset Formula $s_{2}=a_{1}+a_{2}$
\end_inset
,
\begin_inset Formula $s_{3}=a_{1}+a_{2}+a_{3}$
\end_inset
,
...,
\begin_inset Formula $s_{n}=a_{1}+a_{2}+\cdots+a_{n}=\sum_{j=1}^{n}a_{j}$
\end_inset
—
številu
\begin_inset Formula $s_{n}$
\end_inset
pravimo
\begin_inset Formula $n-$
\end_inset
ta delna vsota.
\end_layout
\begin_layout Definition*
Vrsta je konvergentna,
če je v
\begin_inset Formula $\mathbb{R}$
\end_inset
konvergentno zaporedje
\begin_inset Formula $\left(s_{n}\right)_{n\in\mathbb{N}}$
\end_inset
.
Številu
\begin_inset Formula $s=\lim_{n\to\infty}s_{n}$
\end_inset
tedaj pravimo vsota vrste in pišemo
\begin_inset Formula $s\eqqcolon\sum_{j=1}^{\infty}a_{j}$
\end_inset
.
Pojem neskončne vsote torej prevedemo na pojem limite pridruženega zaporedja delnih vsot.
Včasih vrsto (kot operacijo) enačimo z njeno vsoto (izidom operacije).
\end_layout
\begin_layout Definition*
Če vrsta ni konvergentna,
rečemo,
da je divergentna.
Enako,
če je
\begin_inset Formula $s\in\left\{ \pm\infty\right\} $
\end_inset
.
\end_layout
\begin_layout Example*
Primeri vrst.
\end_layout
\begin_deeper
\begin_layout Enumerate
\begin_inset Formula $a_{n}=\frac{1}{2^{n}}$
\end_inset
,
torej zaporedje
\begin_inset Formula $\frac{1}{2},\frac{1}{4},\frac{1}{8},\dots$
\end_inset
.
Ali se sešteje v 1?
Velja
\begin_inset Formula $s=\lim_{n\to\infty}\sum_{j=1}^{n}a_{j}$
\end_inset
.
Pišimo
\begin_inset Formula $q=\frac{1}{2}$
\end_inset
,
tedaj
\begin_inset Formula $a_{n}=q^{n}$
\end_inset
in
\begin_inset Formula
\[
s_{n}=q+q^{2}+q^{3}+\cdots+q^{n}=q\left(1+q+q^{2}+\cdots+q^{n-1}\right)=q\frac{\left(1+q+q^{2}+\cdots+q^{n-1}\right)\left(1-q\right)}{1-q}=
\]
\end_inset
\begin_inset Formula
\[
=q\frac{\left(1+q+q^{2}+\cdots+q^{n-1}\right)-\left(q+q^{2}+q^{3}+\cdots+q^{n}\right)}{1-q}=q\frac{1-q^{n}}{1-q}=\frac{q}{1-q}\left(1-q^{n}\right)
\]
\end_inset
Izračunajmo
\begin_inset Formula $\lim_{n\to\infty}s_{n}=\lim_{n\to\infty}\frac{q}{1-q}\left(1-\cancelto{0}{q^{n}}\right)=\frac{q}{1-q}$
\end_inset
(velja,
ker
\begin_inset Formula $q\in\left(-1,1\right)$
\end_inset
),
torej je
\begin_inset Formula $s=\sum_{n=1}^{\infty}q^{n}=\frac{q}{1-q}$
\end_inset
.
\end_layout
\begin_layout Enumerate
Geometrijska vrsta (splošno).
Naj bo
\begin_inset Formula $q\in\mathbb{R}$
\end_inset
.
Vrsta
\begin_inset Formula $\sum_{j=0}^{\infty}q^{j}$
\end_inset
se imenuje geometrijska vrsta.
Velja
\begin_inset Formula $s=\lim_{n\to\infty}\sum_{j=0}^{n}q^{j}$
\end_inset
in
\begin_inset Formula $s_{n}=1+q+q^{2}+q^{3}+\cdots+q^{n}$
\end_inset
.
Če je
\begin_inset Formula $q=1$
\end_inset
,
je
\begin_inset Formula $s_{n}=n+1$
\end_inset
,
sicer množimo izraz z
\begin_inset Formula $\left(1-q\right)$
\end_inset
:
\begin_inset Formula
\[
\left(1+q+q^{2}+\cdots+q^{n}\right)\left(1-q\right)=\left(1+q+q^{2}+\cdots+q^{n}\right)-\left(q+q^{2}+q^{3}+\cdots+q^{n+1}\right)=1-q^{n+1}
\]
\end_inset
torej
\begin_inset Formula $s_{n}=\frac{1-q^{n+1}}{1-q}$
\end_inset
in vrsta konvergira
\begin_inset Formula $\Leftrightarrow q\not=1$
\end_inset
in
\begin_inset Formula $\lim_{n\to\infty}\frac{1-q^{n+1}}{1-q}\exists$
\end_inset
v
\begin_inset Formula $\mathbb{R}$
\end_inset
.
To pa se zgodi natanko za
\begin_inset Formula $q\in\left(-1,1\right)$
\end_inset
,
takrat je
\begin_inset Formula $\lim_{n\to\infty}\frac{1-\cancelto{0}{q^{n+1}}}{1-q}=\frac{1}{1-q}$
\end_inset
.
\end_layout
\begin_layout Enumerate
Harmonična vrsta.
Je vrsta
\begin_inset Formula $\sum_{j=1}^{\infty}\frac{1}{j}$
\end_inset
.
Velja
\begin_inset Formula $\frac{1}{j}\underset{j\to\infty}{\longrightarrow}0$
\end_inset
,
toda vrsta divergira.
Dokaz sledi kmalu malce spodaj.
\end_layout
\end_deeper
\begin_layout Question*
Kako lahko enostavno določimo,
ali dana vrsta konvergira?
\end_layout
\begin_layout Subsection
Konvergenčni kriteriji
\end_layout
\begin_layout Theorem*
\begin_inset ERT
status open
\begin_layout Plain Layout
\backslash
hypertarget{cauchyvrste}{Cauchyjev pogoj}
\end_layout
\end_inset
.
Vrsta
\begin_inset Formula $\sum_{j=1}^{\infty}a_{j}$
\end_inset
je konvergentna
\begin_inset Formula $\Leftrightarrow$
\end_inset
delne vrste ustrezajo Cauchyjevemu pogoju;
\begin_inset Formula $\forall\varepsilon>0\exists n_{0}\in\mathbb{N}\forall n,m\in\mathbb{N}:n,m\geq n_{0}\Rightarrow\left|s_{m}-s_{n}\right|=\left|\sum_{j=n+1}^{m}a_{j}\right|<\varepsilon$
\end_inset
.
\end_layout
\begin_layout Corollary*
\begin_inset Formula $\sum_{j=1}^{\infty}a_{j}$
\end_inset
konvergira
\begin_inset Formula $\Rightarrow\lim_{j\to\infty}a_{j}=0$
\end_inset
.
\end_layout
\begin_layout Proof
Uporabimo izrek zgoraj za
\begin_inset Formula $n=m-1$
\end_inset
:
\begin_inset Formula $\forall\varepsilon>0\exists n_{0}\in\mathbb{N}\forall n\in\mathbb{N}:n\geq n_{0}\Rightarrow\left|s_{n}-s_{n+1}\right|=\left|a_{n}\right|<\varepsilon$
\end_inset
.
\end_layout
\begin_layout Example*
Vrsti
\begin_inset Formula $\sum_{j=1}^{\infty}\cos n$
\end_inset
in
\begin_inset Formula $\sum_{j=1}^{\infty}\sin n$
\end_inset
divergirata,
saj smo videli,
da členi ne ene ne druge ne konvergirajo nikamor,
torej tudi ne proti 0,
kar je potreben pogoj za konvergenco vrste.
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Example*
Harmonična vrsta divergira.
Protiprimer Cauchyjevega pogoja:
Naj bo
\begin_inset Formula $\varepsilon=\frac{1}{4}$
\end_inset
.
Tedaj ne glede na izbiro
\begin_inset Formula $n_{0}$
\end_inset
najdemo:
\begin_inset Formula
\[
s_{2n}-s_{n}=\sum_{j=n+1}^{2n}\frac{1}{j}=\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2n}>\frac{1}{2n}+\frac{1}{2n}+\cdots+\frac{1}{2n}=\frac{1}{2}
\]
\end_inset
Dokaz divergence brez Cauchyjevega pogoja:
\begin_inset Formula $s_{2^{n}}=a_{1}+\sum_{j=1}^{n}\left(s_{2^{j}}-s_{s^{j-1}}\right)>1+\frac{n}{2}$
\end_inset
in
\begin_inset Formula $\lim_{n\to\infty}1+\frac{n}{2}=\infty$
\end_inset
.
\begin_inset Note Note
status open
\begin_layout Plain Layout
Geometrični argument za divergenco:
TODO XXX FIXME DODAJ
\end_layout
\end_inset
\end_layout
\begin_layout Theorem*
\begin_inset ERT
status open
\begin_layout Plain Layout
\backslash
hypertarget{pk}{Primerjalni kriterij}
\end_layout
\end_inset
.
Naj bosta
\begin_inset Formula $\sum_{n=1}^{\infty}a_{n}$
\end_inset
in
\begin_inset Formula $\sum_{n=1}^{\infty}b_{n}$
\end_inset
vrsti z nenegativnimi členi.
Naj bo
\begin_inset Formula $\forall k\geq k_{0}:a_{k}\leq b_{k}$
\end_inset
(od nekod naprej) —
pravimo,
da je
\begin_inset Formula $\sum_{n=1}^{\infty}b_{n}$
\end_inset
majoranta za
\begin_inset Formula $\sum_{n=1}^{\infty}a_{n}$
\end_inset
od nekod naprej.
\end_layout
\begin_deeper
\begin_layout Itemize
Če
\begin_inset Formula $\sum_{n=1}^{\infty}b_{n}$
\end_inset
konvergira,
tedaj tudi
\begin_inset Formula $\sum_{n=1}^{\infty}a_{n}$
\end_inset
konvergira.
\end_layout
\begin_layout Itemize
Če
\begin_inset Formula $\text{\ensuremath{\sum_{n=1}^{\infty}a_{n}=\infty}}$
\end_inset
,
tedaj tudi
\begin_inset Formula $\sum_{n=1}^{\infty}b_{n}=\infty$
\end_inset
.
\end_layout
\end_deeper
\begin_layout Example*
Videli smo,
da
\begin_inset Formula $\sum_{k=1}^{\infty}\frac{1}{k}$
\end_inset
divergira.
Kaj pa
\begin_inset Formula $\sum_{k=1}^{\infty}\frac{1}{k^{2}}$
\end_inset
?
Preverimo naslednje in uporabimo primerjalni kriterij:
\end_layout
\begin_deeper
\begin_layout Enumerate
\begin_inset Formula $\forall k\in\mathbb{N}:\frac{1}{k^{2}}\leq\frac{2}{k\left(k+1\right)}$
\end_inset
?
Računajmo
\begin_inset Formula $k^{2}\geq\frac{k\left(k+1\right)}{2}\sim k\geq\frac{k+1}{2}\sim\frac{k}{2}\geq\frac{1}{2}$
\end_inset
.
Velja,
ker
\begin_inset Formula $k\in\mathbb{N}$
\end_inset
.
\end_layout
\begin_layout Enumerate
Vrsta
\begin_inset Formula $\sum_{k=1}^{\infty}\frac{2}{k\left(k+1\right)}$
\end_inset
konvergira?
Opazimo
\begin_inset Formula $\frac{1}{k}-\frac{1}{k+1}=\frac{k+1}{k\left(k+1\right)}-\frac{k}{k\left(k+1\right)}=\frac{1}{k\left(k+1\right)}$
\end_inset
.
Za delne vsote vrste
\begin_inset Formula $\sum_{k=1}^{\infty}\frac{1}{k\left(k+1\right)}$
\end_inset
velja:
\begin_inset Formula
\[
\sum_{k=1}^{n}\frac{1}{k\left(k+1\right)}=\sum_{k=1}^{n}\left(\frac{1}{k}-\frac{1}{k+1}\right)=1-\frac{1}{n+1}\underset{n\to\infty}{\longrightarrow}1,
\]
\end_inset
torej
\begin_inset Formula $\sum_{k=1}^{\infty}\frac{2}{k\left(k+1\right)}=2$
\end_inset
.
Posledično po primerjalnem kriteriju tudi
\begin_inset Formula $\sum_{k=1}^{\infty}\frac{1}{k^{2}}$
\end_inset
konvergira.
Izkaže se
\begin_inset Formula $\sum_{k=1}^{\infty}\frac{1}{k^{2}}=\frac{\pi^{2}}{6}\approx1,645$
\end_inset
.
\end_layout
\end_deeper
\begin_layout Theorem*
Kvocientni oz.
d'Alembertov kriterij.
Za vrsto s pozitivnimi členi
\begin_inset Formula $\sum_{n=1}^{\infty}a_{n}$
\end_inset
definirajmo
\begin_inset Formula $D_{n}\coloneqq\frac{a_{n+1}}{a_{n}}$
\end_inset
.
\end_layout
\begin_deeper
\begin_layout Enumerate
\begin_inset Formula $\exists n_{0}\in\mathbb{N},q\in\left(0,1\right)\forall n\geq n_{0}:D_{n}\leq q\Longrightarrow\sum_{n=1}^{\infty}a_{n}<\infty$
\end_inset
(vrsta konvergira)
\end_layout
\begin_layout Enumerate
\begin_inset Formula $\exists n_{0}\in\mathbb{N}\forall n\geq n_{0}:D_{n}\geq1\Longrightarrow\sum_{n=1}^{\infty}a_{n}=\infty$
\end_inset
(vrsta divergira)
\end_layout
\begin_layout Enumerate
\begin_inset Formula $\exists D=\lim_{n\to\infty}D_{n}\in\mathbb{R}\Longrightarrow$
\end_inset
\end_layout
\begin_deeper
\begin_layout Enumerate
\begin_inset CommandInset label
LatexCommand label
name "enu:kvocientni3a"
\end_inset
\begin_inset Formula $D<1\Longrightarrow\sum_{n=1}^{\infty}a_{n}<\infty$
\end_inset
(vrsta konvergira)
\end_layout
\begin_layout Enumerate
\begin_inset Formula $D>1\Longrightarrow\sum_{n=1}^{\infty}a_{n}=\infty$
\end_inset
(vrsta divergira)
\end_layout
\begin_layout Enumerate
\begin_inset Formula $D=1\Longrightarrow$
\end_inset
s tem kriterijem ne moremo določiti konvergence.
\end_layout
\end_deeper
\end_deeper
\begin_layout Proof
Razlaga.
\end_layout
\begin_deeper
\begin_layout Enumerate
\begin_inset Formula $\forall n>n_{0}:D_{n}\leq q$
\end_inset
,
torej
\begin_inset Formula $\frac{a_{n+1}}{a_{n}}\leq q\sim a_{n+1}\leq qa_{n}$
\end_inset
in hkrati
\begin_inset Formula $\text{\ensuremath{\frac{a_{n+2}}{a_{n+1}}\leq q\sim a_{n+2}\leq qa_{n+1}}}$
\end_inset
,
torej skupaj
\begin_inset Formula $a_{n+2}\leq qa_{n+1}\leq qqa_{n}=q^{2}a_{n}$
\end_inset
,
sledi
\begin_inset Formula $q_{n+2}\leq q^{2}a_{n}$
\end_inset
in
\begin_inset Formula $\forall k\in\mathbb{N}:q_{n+k}\leq q^{k}a_{n}$
\end_inset
.
Vrsto smo majorizirali z geometrijsko vrsto,
ki ob
\begin_inset Formula $q\in\left(0,1\right)$
\end_inset
konvergira po primerjalnem kriteriju,
zato tudi naša vrsta konvergira.
\end_layout
\begin_layout Enumerate
\begin_inset Formula $\forall n>n_{0}:\frac{a_{n+1}}{a_{n}}\geq D\geq1$
\end_inset
,
torej
\begin_inset Formula $a_{n+1}\geq a_{n}$
\end_inset
in hkrati
\begin_inset Formula $a_{n+2}\geq a_{n+1}$
\end_inset
,
torej skupaj
\begin_inset Formula $a_{n+2}\geq a_{n}$
\end_inset
,
sledi
\begin_inset Formula $\forall k\in\mathbb{N}:a_{n+k}\geq a_{n}$
\end_inset
.
Naša vrsta torej majorizira konstantno vrsto,
ki očitno divergira;
\begin_inset Formula $\sum_{k=n_{0}}^{\infty}a_{k}\geq\sum_{k=n_{0}}^{\infty}a_{n}=0$
\end_inset
.
Potemtakem tudi naša vrsta divergira.
Poleg tega niti ne velja
\begin_inset Formula $a_{k}\underset{k\to\infty}{\longrightarrow}0$
\end_inset
,
torej vrsta gotovo divergira.
\end_layout
\begin_layout Enumerate
Enako kot 1 in 2.
\end_layout
\end_deeper
\begin_layout Example*
Za
\begin_inset Formula $x>0$
\end_inset
definiramo
\begin_inset Formula $e^{x}=\sum_{k=0}^{\infty}\frac{x^{k}}{k!}$
\end_inset
.
Vrsta res konvergira po točki
\begin_inset CommandInset ref
LatexCommand ref
reference "enu:kvocientni3a"
plural "false"
caps "false"
noprefix "false"
nolink "false"
\end_inset
.
\begin_inset Formula
\[
D_{n}=\frac{\frac{x^{n+1}}{\left(n+1\right)!}}{\frac{x^{n}}{n!}}=\frac{x^{n+1}n!}{x^{n}\left(n+1\right)!}=\frac{x}{n+1}\underset{n\to\infty}{\longrightarrow}0
\]
\end_inset
\end_layout
\begin_layout Theorem*
Korenski oz.
Cauchyjev kriterij.
Naj bo
\begin_inset Formula $\sum_{k=1}^{\infty}a_{k}$
\end_inset
vrsta z nenegativnimi členi.
Naj bo
\begin_inset Formula $c_{n}\coloneqq\sqrt[n]{a_{n}}$
\end_inset
.ž
\end_layout
\begin_deeper
\begin_layout Enumerate
\begin_inset Formula $\exists n_{0}\in\mathbb{N},q\in\left(0,1\right)\forall n>n_{0}:c_{n}\leq q\Longrightarrow\sum_{k=1}^{\infty}a_{k}<\infty$
\end_inset
(vrsta konvergira)
\end_layout
\begin_layout Enumerate
\begin_inset Formula $\exists n_{0}\in\mathbb{N}\forall n>n_{0}:c_{n}\geq1\Longrightarrow\sum_{k=1}^{\infty}a_{k}=\infty$
\end_inset
(vrsta divergira)
\end_layout
\begin_layout Enumerate
\begin_inset Formula $\exists c=\lim_{n\to\infty}c_{n}\in\mathbb{R}\Longrightarrow$
\end_inset
\end_layout
\begin_deeper
\begin_layout Enumerate
\begin_inset Formula $c<1\Longrightarrow\sum_{k=1}^{\infty}a_{k}<\infty$
\end_inset
(vrsta konvergira)
\end_layout
\begin_layout Enumerate
\begin_inset Formula $c>1\Longrightarrow\sum_{k=1}^{\infty}a_{k}=\infty$
\end_inset
(vrsta divergira)
\end_layout
\begin_layout Enumerate
\begin_inset Formula $c=1\Longrightarrow$
\end_inset
s tem kriterijem ne moremo določiti konvergence.
\end_layout
\end_deeper
\end_deeper
\begin_layout Proof
Skica dokazov.
\end_layout
\begin_deeper
\begin_layout Enumerate
Velja
\begin_inset Formula $\forall n>n_{0}:c_{n}\leq q$
\end_inset
.
To pomeni
\begin_inset Formula $\sqrt[n]{a_{n}}\leq q$
\end_inset
,
torej
\begin_inset Formula $a_{n}\leq q^{n}$
\end_inset
in
\begin_inset Formula $a_{n+1}\leq q^{n+1}$
\end_inset
,
torej je vrsta majorizirana z geometrijsko vrsto
\begin_inset Formula $\sum_{n=1}^{\infty}q^{n}$
\end_inset
.
\end_layout
\begin_layout Enumerate
Velja
\begin_inset Formula $\forall n>n_{0}:c_{n}\geq1$
\end_inset
.
To pomeni
\begin_inset Formula $\sqrt[n]{a_{n}}\geq1$
\end_inset
,
torej
\begin_inset Formula $a_{n}\geq1$
\end_inset
,
torej je vrsta majorizirana s konstantno in zato divergentno vrsto
\begin_inset Formula $\sum_{n=1}^{\infty}1$
\end_inset
.
\end_layout
\begin_layout Enumerate
Enako kot 1 in 2.
\end_layout
\end_deeper
\begin_layout Subsection
Alternirajoče vrste
\end_layout
\begin_layout Definition*
Vrsta je alternirajoča,
če je predznak naslednjega člena nasproten predznaku tega člena.
ZDB
\begin_inset Formula $\forall n\in\mathbb{N}:\sgn a_{n+1}=-\sgn a_{n}$
\end_inset
,
kjer je
\begin_inset Formula $\sgn:\mathbb{R}\to\left\{ -1,0,1\right\} $
\end_inset
s predpisom
\begin_inset Formula $\sgn a=\begin{cases}
-1 & ;a<0\\
1 & ;a>0\\
0 & ;a=0
\end{cases}$
\end_inset
.
ZDB
\begin_inset Formula $\forall n\in\mathbb{N}:a_{n+1}a_{n}\leq0$
\end_inset
.
\end_layout
\begin_layout Theorem*
Leibnizov konvergenčni kriterij.
Naj bo
\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
\end_inset
padajoče zaporedje in
\begin_inset Formula $\lim_{n\to\infty}a_{n}=0$
\end_inset
.
Tedaj vrsta
\begin_inset Formula $\sum_{k=1}^{\infty}\left(-1\right)^{k}a_{k}$
\end_inset
konvergira.
Če je
\begin_inset Formula $s\coloneqq\sum_{k=1}^{\infty}\left(-1\right)^{k}a_{k}$
\end_inset
in
\begin_inset Formula $s_{n}\coloneqq\sum_{k=1}^{\infty}\left(-1\right)^{k}a_{k}$
\end_inset
,
tedaj
\begin_inset Formula $\left|s-s_{k}\right|\leq a_{n+1}$
\end_inset
.
\end_layout
\begin_layout Proof
Skica dokaza.
Vidimo,
da delne vsote
\begin_inset Formula $s_{2n}$
\end_inset
padajo k
\begin_inset Formula $s''$
\end_inset
in delne vsote
\begin_inset Formula $s_{2n-1}$
\end_inset
naraščajo k
\begin_inset Formula $s'$
\end_inset
.
Toda ker
\begin_inset Formula $s_{2n}-s_{2n-1}=a_{2n}$
\end_inset
,
velja
\begin_inset Formula $s'=s''$
\end_inset
.
Limita razlike dveh zaporedij je razlika limit teh dveh zaporedij,
torej
\begin_inset Formula $s'=s''=s$
\end_inset
.
\begin_inset Formula $s$
\end_inset
je supremum lihih in infimum sodih vsot.
\begin_inset Formula $\left|s-s_{n}\right|\leq\left|s_{n+1}-s_{n}\right|=a_{n+1}$
\end_inset
.
\end_layout
\begin_layout Example*
Harmonična vrsta
\begin_inset Formula $1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots\to\infty$
\end_inset
,
toda alternirajoča harmonična vrsta
\begin_inset Formula $1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots\to\log2$
\end_inset
.
\end_layout
\begin_layout Subsection
Absolutno konvergentne vrste
\end_layout
\begin_layout Definition*
Vrsta
\begin_inset Formula $\sum_{n=1}^{\infty}a_{n}$
\end_inset
je absolutno konvergentna,
če je
\begin_inset Formula $\sum_{n=1}^{\infty}\left|a_{n}\right|$
\end_inset
konvergentna.
\end_layout
\begin_layout Theorem*
Absolutna konvergenca
\begin_inset Formula $\Rightarrow$
\end_inset
konvergenca.
\end_layout
\begin_layout Proof
Uporabimo
\begin_inset ERT
status open
\begin_layout Plain Layout
\backslash
hyperlink{cauchyvrste}{Cauchyjev pogoj za konvergenco vrst}
\end_layout
\end_inset
in trikotniško neenakost.
\begin_inset Formula
\[
\left|s_{m}-s_{n}\right|=\left|\sum_{j=n+1}^{m}a_{j}\right|\leq\sum_{j=n+1}^{m}\left|a_{j}\right|<\varepsilon
\]
\end_inset
za
\begin_inset Formula $m,n\geq n_{0}$
\end_inset
za nek
\begin_inset Formula $n_{0}$
\end_inset
.
\end_layout
\begin_layout Remark*
Obrat ne velja,
protiprimer je alternirajoča harmonična vrsta.
\end_layout
\begin_layout Subsection
Pogojno konvergentne vrste
\end_layout
\begin_layout Standard
\begin_inset Formula $\sum_{k=0}^{\infty}2-\sum_{k=0}^{\infty}1\not=\sum_{k=0}^{\infty}\left(2-1\right)$
\end_inset
,
temveč
\begin_inset Formula $\infty-\infty=$
\end_inset
nedefinirano.
\end_layout
\begin_layout Question*
Ross-Littlewoodov paradoks.
Ali smemo zamenjati vrstni red seštevanja,
če imamo neskončno mnogo sumandov?
\end_layout
\begin_layout Standard
Najprej vprašanje natančneje opredelimo in vpeljimo orodja za njegovo obravnavo.
\end_layout
\begin_layout Definition*
Naj bo
\begin_inset Formula $\mathcal{M}\subset\mathbb{N}$
\end_inset
.
Permutacija
\begin_inset Formula $\mathcal{M}$
\end_inset
je vsaka bijektivna preslikava
\begin_inset Formula $\pi:\mathcal{M}\to\mathcal{M}$
\end_inset
.
Če je
\begin_inset Formula $\mathcal{M}=\left\{ a_{1},\dots,a_{n}\right\} $
\end_inset
končna množica,
tedaj
\begin_inset Formula $\pi$
\end_inset
označimo s tabelo:
\begin_inset Formula
\[
\left(\begin{array}{ccc}
a_{1} & \cdots & a_{n}\\
\pi\left(a_{1}\right) & \cdots & \pi\left(a_{n}\right)
\end{array}\right)
\]
\end_inset
\end_layout
\begin_layout Example*
\begin_inset Formula
\[
\pi=\left(\begin{array}{ccccc}
1 & 2 & 3 & 4 & 5\\
5 & 3 & 1 & 4 & 2
\end{array}\right)
\]
\end_inset
\end_layout
\begin_layout Definition*
Vrsta
\begin_inset Formula $\sum_{n=1}^{\infty}a_{n}$
\end_inset
je brezpogojno konvergentna,
če za vsako permutacijo
\begin_inset Formula $\pi:\mathbb{N}\to\mathbb{N}$
\end_inset
vrsta
\begin_inset Formula $\sum_{n=1}^{\infty}\pi\left(a_{n}\right)$
\end_inset
konvergira in vsota ni odvisna od
\begin_inset Formula $\pi$
\end_inset
.
Vrsta je pogojno konvergentna,
če je konvergentna,
toda ne brezpogojno.
\end_layout
\begin_layout Example*
\begin_inset Formula $1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\cdots$
\end_inset
je pogojno konvergentna,
ker pri seštevanju z vrstnim redom,
pri katerem tisočim pozitivnim členom sledi en negativen in njemu zopet tisoč pozitivnih itd.,
vrsta ne konvergira.
\end_layout
\begin_layout Theorem*
Absolutna konvergenca
\begin_inset Formula $\Leftrightarrow$
\end_inset
Brezpogojna konvergenca
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Theorem*
Riemannov sumacijski izrek.
Če je vrsta pogojno konvergentna,
tedaj
\begin_inset Formula $\forall x\in\mathbb{R}\cup\left\{ \pm\infty\right\} \exists$
\end_inset
permutacija
\begin_inset Formula $\pi:\mathbb{N}\to\mathbb{N}\ni:\sum_{n=1}^{\infty}a_{\pi\left(n\right)}=x$
\end_inset
.
ZDB Končna vsota je lahko karkoli,
če lahko poljubno spremenimo vrstni red seštevanja.
Prav tako obstaja taka permutacija
\begin_inset Formula $\pi$
\end_inset
,
pri kateri
\begin_inset Formula $\sum_{n=1}^{\infty}a_{\pi\left(n\right)}$
\end_inset
nima vsote ZDB delne vsotee ne konvergirajo.
\end_layout
\begin_layout Example*
\begin_inset Formula $\sum_{n=1}^{\infty}\frac{\left(-1\right)^{n}}{n}$
\end_inset
.
\end_layout
\begin_layout Section
Funkcijske vrste
\end_layout
\begin_layout Standard
Tokrat poskušamo seštevati funkcije.
V prejšnjem razdelku seštevamo le realna števila.
Funkcijska vrsta,
če je
\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
\end_inset
zaporedije funkcij
\begin_inset Formula $X\to\mathbb{R}$
\end_inset
in
\begin_inset Formula $x$
\end_inset
zunanja konstanta,
izgleda takole:
\end_layout
\begin_layout Standard
\begin_inset Formula
\[
\sum_{n=1}^{\infty}a_{n}\left(x\right)
\]
\end_inset
\end_layout
\begin_layout Definition*
Naj bo
\begin_inset Formula $X$
\end_inset
neka množica in
\begin_inset Formula $\Phi=\left\{ \varphi_{n}:X\to\mathbb{R},n\in\mathbb{N}\right\} $
\end_inset
družina funkcij.
\end_layout
\begin_layout Definition*
Pravimo,
da funkcije
\begin_inset Formula $\varphi_{n}$
\end_inset
konvergirajo po točkah na
\begin_inset Formula $X$
\end_inset
,
če je
\begin_inset Formula $\forall x\in X$
\end_inset
zaporedje
\begin_inset Formula $\left(\varphi_{n}\left(x\right)\right)_{n\in\mathbb{N}}$
\end_inset
konvergentno.
\end_layout
\begin_layout Definition*
Označimo limito s
\begin_inset Formula $\varphi\left(x\right)$
\end_inset
.
ZDB to pomeni,
da
\begin_inset Formula
\[
\forall\varepsilon>0,x\in X:\exists n_{0}=n_{0}\left(\varepsilon,x\right)\in\mathbb{N}\forall n\in\mathbb{N}:n\geq n_{0}\Rightarrow\left|\varphi_{n}\left(x\right)-\varphi\left(x\right)\right|<\varepsilon.
\]
\end_inset
\end_layout
\begin_layout Definition*
Pravimo,
da funkcije
\begin_inset Formula $\varphi_{n}$
\end_inset
konvergirajo enakomerno na
\begin_inset Formula $X$
\end_inset
,
če
\begin_inset Formula
\[
\forall\varepsilon>0\exists n_{0}=n_{0}\left(\varepsilon\right)\in\mathbb{N}\forall x\in X,n\in\mathbb{N}:n\geq n_{0}\Rightarrow\left|\varphi_{n}\left(x\right)-\varphi\left(x\right)\right|\leq\varepsilon
\]
\end_inset
oziroma ZDB
\begin_inset Formula
\[
\forall\varepsilon>0\exists n_{0}=n_{0}\left(\varepsilon\right)\in\mathbb{N}\forall n\in\mathbb{N}:n\geq n_{0}\Rightarrow\sup_{x\in X}\left|\varphi_{n}\left(x\right)-\varphi\left(x\right)\right|\leq\varepsilon.
\]
\end_inset
\end_layout
\begin_layout Definition*
Poudariti je treba,
da je pri konvergenci po točkah
\begin_inset Formula $n_{0}$
\end_inset
lahko odvisen od
\begin_inset Formula $\varepsilon$
\end_inset
in
\begin_inset Formula $x$
\end_inset
,
pri enakomerni konvergenci pa le od
\begin_inset Formula $\varepsilon$
\end_inset
.
\end_layout
\begin_layout Note*
Očitno enakomerna konvergenca implicira konvergenco po točkah,
obratno pa ne velja.
\end_layout
\begin_layout Example*
Za
\begin_inset Formula $n\in\mathbb{N}$
\end_inset
definiramo
\begin_inset Formula $\varphi_{n}:\left[0,1\right]\to\left[0,1\right]$
\end_inset
s predpisom
\begin_inset Formula $\varphi_{n}\left(x\right)=x^{n}$
\end_inset
.
Tedaj obstaja
\begin_inset Formula $\varphi\left(x\right)\coloneqq\lim_{n\to\infty}\varphi_{n}\left(x\right)=\begin{cases}
0 & ;x\in[0,1)\\
1 & ;x=1
\end{cases}$
\end_inset
.
Torej po definiciji velja
\begin_inset Formula $\varphi_{n}\to\varphi$
\end_inset
po točkah,
toda ne velja
\begin_inset Formula $\varphi_{n}\to\varphi$
\end_inset
enakomerno.
Za poljubno velik pas okoli
\begin_inset Formula $\varphi\left(x\right)$
\end_inset
bodo še tako pozne funkcijske vrednosti
\begin_inset Formula $\varphi_{n}\left(x\right)$
\end_inset
od nekega
\begin_inset Formula $x$
\end_inset
dalje izven tega pasu.
Če bi
\begin_inset Formula $\varphi_{n}\to\varphi$
\end_inset
enakomerno,
tedaj bi za poljuben
\begin_inset Formula $\varepsilon\in\left(0,1\right)$
\end_inset
in dovolj pozne
\begin_inset Formula $n$
\end_inset
(večje od nekega
\begin_inset Formula $n_{0}\in\mathbb{N}$
\end_inset
) veljalo
\begin_inset Formula $\forall x\in\left[0,1\right]:\left|\varphi_{n}\left(x\right)-\varphi\left(x\right)\right|<\varepsilon$
\end_inset
.
To je ekvivalentno
\begin_inset Formula $\forall x\in\left(0,1\right):\left|x^{n}\right|<\varepsilon\Leftrightarrow n\log x<\log\varepsilon\Leftrightarrow n>\frac{\log\varepsilon}{\log x}$
\end_inset
.
Toda
\begin_inset Formula $\lim_{x\nearrow1}\frac{\log\varepsilon}{\log x}=\infty$
\end_inset
,
zato tak
\begin_inset Formula $n$
\end_inset
ne obstaja.
\end_layout
\begin_layout Definition*
Naj bo
\begin_inset Formula $X$
\end_inset
neka množica in
\begin_inset Formula $\left(f_{j}:X\to\mathbb{R}\right)_{j\in\mathbb{N}}$
\end_inset
dano zaporedje funkcij.
Pravimo,
da funkcijska vrsta
\begin_inset Formula $\sum_{j=1}^{\infty}f_{j}$
\end_inset
konvergira po točkah na
\begin_inset Formula $X$
\end_inset
,
če
\begin_inset Formula $\forall x\in X:\sum_{j=1}^{\infty}f_{j}\left(x\right)<0$
\end_inset
(številska vrsta je konvergentna).
ZDB to pomeni,
da funkcijsko zaporedje delnih vsot
\begin_inset Formula $s_{n}\coloneqq\sum_{j=1}^{n}f_{j}$
\end_inset
konvergira po točkah na
\begin_inset Formula $X$
\end_inset
.
\end_layout
\begin_layout Definition*
Funkcijska vrsta
\begin_inset Formula $s=\sum_{j=1}^{\infty}$
\end_inset
konvergira enakomerno na
\begin_inset Formula $X$
\end_inset
,
če funkcijsko zaporedje delnih vsot
\begin_inset Formula $s_{n}\coloneqq\sum_{j=1}^{n}f_{j}$
\end_inset
konvergira enakomerno na
\begin_inset Formula $X$
\end_inset
.
\end_layout
\begin_layout Definition*
Funkcija oblike
\begin_inset Formula $x\mapsto\sum_{j=1}^{\infty}f_{j}\left(x\right)$
\end_inset
se imenuje funkcijska vrsta.
\end_layout
\begin_layout Exercise*
Dokaži,
da
\begin_inset Formula $\sum_{n=1}^{\infty}x^{n}$
\end_inset
ne konvergira enakomerno!
Vrsta konvergira po točkah le na intervalu
\begin_inset Formula $x\in\left(0,1\right)$
\end_inset
,
za druge
\begin_inset Formula $x$
\end_inset
divergira.
Ko fiksiramo zunanjo konstanto,
gre za geometrijsko vrsto.
Delna vsota
\begin_inset Formula $\sum_{j=1}^{n}x^{j}=\frac{x\left(1-x^{n}\right)}{1-x}$
\end_inset
.
Velja
\begin_inset Formula $\lim_{n\to\infty}\frac{x\left(1-x^{n}\right)}{1-x}=x\lim_{n\to\infty}\frac{1-\cancelto{0}{x^{n}}}{1-x}=\frac{x}{1-x}$
\end_inset
.
Sedaj prevedimo,
ali
\begin_inset Formula $\forall\varepsilon>0\exists n_{0}\in\mathbb{N}\forall x\in\left(-1,1\right),n\geq n_{0}:\left|\frac{x\left(1-x^{n}\right)}{1-x}-\frac{x}{1-x}\right|<\varepsilon$
\end_inset
.
Za začetekk si oglejmo le
\begin_inset Formula $x>0$
\end_inset
.
Ker je tedaj
\begin_inset Formula $\frac{x\left(1-x^{n}\right)}{1-x}<\frac{x}{1-x}$
\end_inset
,
je
\begin_inset Formula $\left|\frac{x\left(1-x^{n}\right)}{1-x}-\frac{x}{1-x}\right|=\frac{x}{1-x}-\frac{x\left(1-x^{n}\right)}{1-x}=\frac{\cancel{x-x+}x^{n+1}}{1-x}$
\end_inset
.
Računajmo sedaj
\begin_inset Formula $\frac{x^{n+1}}{1-x}<\varepsilon\sim x^{n+1}<\varepsilon\left(1-x\right)\sim\left(n+1\right)\log x<\log\left(\varepsilon\left(1-x\right)\right)\sim n+1>\frac{\log\left(\varepsilon\left(1-x\right)\right)}{\log x}\sim n>\frac{\log\left(\varepsilon\left(1-x\right)\right)}{\log x}-1$
\end_inset
.
Ker je
\begin_inset Formula $n$
\end_inset
odvisen od
\begin_inset Formula $x$
\end_inset
,
vsota ni enakomerno konvergentna.
\end_layout
\begin_layout Standard
Poseben primer funkcijskih vrst so funkcijske vrste funkcij oblike
\begin_inset Formula $f_{j}=b_{j}\cdot x^{j}$
\end_inset
,
torej potence (monomi).
\end_layout
\begin_layout Definition*
Potenčna vrsta je funkcijska vrsta oblike
\begin_inset Formula $\sum_{j=1}^{\infty}b_{j}\cdot x^{j}$
\end_inset
,
kjer so a
\begin_inset Formula $\left(b_{j}\right)_{j\in\mathbb{N}}$
\end_inset
dana realna števila.
\end_layout
\begin_layout Theorem*
Cauchy-Hadamard.
Za vsako potenčno vrsto obstaja konvergenčni radij
\begin_inset Formula $R\in\left[0,\infty\right]\ni:$
\end_inset
\end_layout
\begin_deeper
\begin_layout Itemize
vrsta absolutno konvergira za
\begin_inset Formula $\left|x\right|<R$
\end_inset
,
\end_layout
\begin_layout Itemize
vrsta divergira za
\begin_inset Formula $\left|x\right|>R$
\end_inset
.
\end_layout
\end_deeper
\begin_layout Theorem*
Velja
\begin_inset Formula $\text{\ensuremath{\frac{1}{R}=\limsup_{k\to\infty}\sqrt[k]{\left|b_{k}\right|}}}$
\end_inset
,
kjer vzamemo
\begin_inset Formula $\frac{1}{0}\coloneqq\infty$
\end_inset
in
\begin_inset Formula $\frac{1}{\infty}\coloneqq0$
\end_inset
.
\end_layout
\begin_layout Proof
Rezultat že poznamo za zelo poseben primer
\begin_inset Formula $\forall j\in\mathbb{N}:b_{j}=1$
\end_inset
(geometrijska vrsta).
Ideja dokaza je,
da konvergenco vsake potenčne vrste opišemo s pomočjo geometrijske vrste.
\end_layout
\begin_deeper
\begin_layout Itemize
Konvergenca:
Za
\begin_inset Formula $x=0$
\end_inset
vrsta očitno konvergira,
zato privzamemo
\begin_inset Formula $x\not=0$
\end_inset
.
Definirajmo
\begin_inset Formula $R$
\end_inset
s formulo iz definicije (
\begin_inset Formula $R=\frac{1}{\limsup_{k\to\infty}\sqrt[k]{\left|b_{k}\right|}}$
\end_inset
).
Naj bo
\begin_inset Formula $x$
\end_inset
tak,
da
\begin_inset Formula $\left|x\right|<R\leq\infty$
\end_inset
(sledi
\begin_inset Formula $R>0$
\end_inset
).
Naj bo
\begin_inset Formula $\varepsilon>0$
\end_inset
.
Tedaj po definiciji
\begin_inset Formula $R$
\end_inset
velja
\begin_inset Formula $\sqrt[k]{\left|b_{k}\right|}\leq\frac{1}{R}+\varepsilon$
\end_inset
za vse dovolj velike
\begin_inset Formula $k$
\end_inset
.
Za take
\begin_inset Formula $k$
\end_inset
sledi
\begin_inset Formula
\[
\left|b_{k}\right|\left|x\right|^{k}\leq\left(\left(\frac{1}{R}+\varepsilon\right)\left|x\right|\right)^{k}.
\]
\end_inset
Opazimo,
da je desna stran neenačbe člen geometrijske vrste,
s katero majoriziramo vrsto iz absolutnih vrednosti členov naše vrste.
Preverimo,
da desna stran konvergira.
Konvergira,
kadar
\begin_inset Formula $\left(\frac{1}{R}+\varepsilon\right)\left|x\right|<1$
\end_inset
oziroma
\begin_inset Formula $\varepsilon<\frac{1}{\left|x\right|}-\frac{1}{R}$
\end_inset
.
Po
\begin_inset ERT
status open
\begin_layout Plain Layout
\backslash
hyperlink{pk}{primerjalnem kriteriju}
\end_layout
\end_inset
torej naša vrsta absolutno konvergira.
\end_layout
\begin_layout Itemize
Divergenca:
Vzemimo poljuben
\begin_inset Formula $\varepsilon>0$
\end_inset
.
Po definciji
\begin_inset Formula $R$
\end_inset
sledi,
da je
\begin_inset Formula $\sqrt[k]{\left|b_{k}\right|}\geq\frac{1}{R}-\varepsilon$
\end_inset
za vse dovolj velike
\begin_inset Formula $k$
\end_inset
.
Za take
\begin_inset Formula $k$
\end_inset
sledi
\begin_inset Formula
\[
\left|b_{k}\right|\left|x\right|^{k}\geq\left(\left(\frac{1}{R}-\varepsilon\right)\left|x\right|\right)^{k}.
\]
\end_inset
Opazimo,
da je desna stran neenačbe člen geometrijske vrste,
ki je majorizirana z vrsto iz absolutnih vrednosti členov naše vrste.
Desna stran divergira,
ko
\begin_inset Formula $\left(\frac{1}{R}-\varepsilon\right)\left|x\right|=1$
\end_inset
oziroma
\begin_inset Formula $\varepsilon=\frac{1}{R}-\frac{1}{\left|x\right|}$
\end_inset
,
zato tudi naša vrsta divergira.
\end_layout
\end_deeper
\begin_layout Example*
Primer konvergenčnega radija potenčne vrste od prej:
\begin_inset Formula $\sum_{j=1}^{\infty}x^{j}$
\end_inset
.
Velja
\begin_inset Formula $\forall j\in\mathbb{N}:b_{j}=1$
\end_inset
,
torej
\begin_inset Formula $R=\frac{1}{\limsup_{j\to\infty}\sqrt[k]{\left|b_{k}\right|}}=1$
\end_inset
,
torej po zgornjem izreku vrsta konvergira za
\begin_inset Formula $x\in\left(-1,1\right)$
\end_inset
in divergira za
\begin_inset Formula $x\not\in\left[-1,1\right]$
\end_inset
.
Ročno lahko še preverimo,
da divergira tudi v
\begin_inset Formula $\left\{ -1,1\right\} $
\end_inset
.
\end_layout
\begin_layout Section
Zveznost
\begin_inset Note Note
status open
\begin_layout Plain Layout
TODO XXX FIXME PREVERI ŠE V profesrojevih PDFJIH,
recimo dodaj dokaz zveznosti x^2
\end_layout
\end_inset
\end_layout
\begin_layout Standard
Ideja:
Izdelati želimo formulacijo,
s katero preverimo,
če lahko z dovolj majhno spremembo
\begin_inset Formula $x$
\end_inset
povzročimo majhno spremembo funkcijske vrednosti.
\end_layout
\begin_layout Example*
Primer nezvezne funkcije je
\begin_inset Formula $f\left(x\right)=\begin{cases}
0 & ;0\leq x<1\\
1 & ;x=1
\end{cases}$
\end_inset
.
\end_layout
\begin_layout Definition*
Naj bo
\begin_inset Formula $D\subseteq\mathbb{R},a\in D$
\end_inset
in
\begin_inset Formula $f:D\to\mathbb{R}$
\end_inset
.
Pravimo,
da je
\begin_inset Formula $f$
\end_inset
zvezna v
\begin_inset Formula $a$
\end_inset
,
če
\begin_inset Formula $\forall\varepsilon>0\exists\delta>0\forall x\in D:\left|x-a\right|<\delta\Rightarrow\left|f\left(x\right)-f\left(a\right)\right|<\varepsilon$
\end_inset
.
\begin_inset Formula $f$
\end_inset
je zvezna na množici
\begin_inset Formula $x\subseteq D$
\end_inset
,
če je zvezna na vsaki točki v
\begin_inset Formula $D$
\end_inset
.
\end_layout
\begin_layout Theorem*
\begin_inset ERT
status open
\begin_layout Plain Layout
\backslash
hypertarget{kzzz}{Karakterizacija zveznosti z zaporedji}
\end_layout
\end_inset
.
Naj bodo
\begin_inset Formula $D,a,f$
\end_inset
kot prej.
Velja:
\begin_inset Formula $f$
\end_inset
zvezna v
\begin_inset Formula $a\Leftrightarrow\forall\left(a_{n}\right)_{n\in\mathbb{N}},a_{n}\in D:\lim_{n\to\infty}a_{n}=a\Rightarrow\lim_{n\to\infty}f\left(a_{n}\right)=f\left(a\right)$
\end_inset
ZDB
\begin_inset Formula $f$
\end_inset
je zvezna v
\begin_inset Formula $a$
\end_inset
,
če za vsako k
\begin_inset Formula $a$
\end_inset
konvergentno zaporedje na domeni velja,
da funkcijske vrednosti členov zaporedja konvergirajo k funkcijski vrednosti
\begin_inset Formula $a$
\end_inset
.
\end_layout
\begin_layout Proof
Dokazujemo ekvivalenco.
\end_layout
\begin_deeper
\begin_layout Labeling
\labelwidthstring 00.00.0000
\begin_inset Formula $\left(\Rightarrow\right)$
\end_inset
Predpostavimo,
da je
\begin_inset Formula $f$
\end_inset
zvezna v
\begin_inset Formula $a$
\end_inset
,
torej
\begin_inset Formula $\forall\varepsilon>0\exists\delta>0\forall x\in D:\left|x-a\right|<\delta\Rightarrow\left|f\left(x\right)-f\left(a\right)\right|<\varepsilon$
\end_inset
.
Naj bo
\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
\end_inset
poljubno zaporedje na
\begin_inset Formula $D$
\end_inset
,
ki konvergira k
\begin_inset Formula $a$
\end_inset
,
se pravi
\begin_inset Formula $\forall\varepsilon>0\exists n_{0}\in\mathbb{N}\forall n\in\mathbb{N}:n\geq n_{0}\Rightarrow\left|a-a_{n}\right|<\varepsilon$
\end_inset
.
Naj bo
\begin_inset Formula $\varepsilon$
\end_inset
poljuben.
Vsled zveznosti
\begin_inset Formula $f$
\end_inset
velja,
da je
\begin_inset Formula $\left|f\left(a_{n}\right)-f\left(a\right)\right|<\varepsilon$
\end_inset
za vse take
\begin_inset Formula $a_{n}$
\end_inset
,
da velja
\begin_inset Formula $\left|a_{n}-a\right|<\delta$
\end_inset
za neko
\begin_inset Formula $\delta\in\mathbb{R}$
\end_inset
.
Ker je zaporedje
\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
\end_inset
konvergentno k
\begin_inset Formula $a$
\end_inset
,
so vsi členi po nekem
\begin_inset Formula $n_{0}$
\end_inset
v
\begin_inset Formula $\delta-$
\end_inset
okolici
\begin_inset Formula $a$
\end_inset
,
torej velja pogoj
\begin_inset Formula $\left|a_{n}-a\right|<\delta$
\end_inset
,
torej velja
\begin_inset Formula $\left|f\left(a_{n}\right)-f\left(a\right)\right|<\varepsilon$
\end_inset
za vse
\begin_inset Formula $n\geq n_{0}$
\end_inset
.
\end_layout
\begin_layout Labeling
\labelwidthstring 00.00.0000
\begin_inset Formula $\left(\Leftarrow\right)$
\end_inset
PDDRAA
\begin_inset Formula $f$
\end_inset
ni zvezna v
\begin_inset Formula $a$
\end_inset
.
Da pridemo do protislovja,
moramo dokazati,
da
\begin_inset Formula $\exists\left(a_{n}\right)_{n\in\mathbb{N}},a_{n}\in D\ni:\lim_{n\to\infty}a_{n}=a$
\end_inset
,
a vendar
\begin_inset Formula $\lim_{n\to\infty}f\left(a_{n}\right)\not=f\left(a\right)$
\end_inset
.
Ker
\begin_inset Formula $f$
\end_inset
ni zvezna,
velja,
da
\begin_inset Formula $\exists\varepsilon>0\forall\delta>0\exists x\in D\ni:\left|x-a\right|<\delta\wedge\left|f\left(x\right)-f\left(a\right)\right|\geq\varepsilon$
\end_inset
.
Izberimo
\begin_inset Formula $\forall n\in\mathbb{N}:\delta_{n}\coloneqq\frac{1}{n}$
\end_inset
.
Tedaj
\begin_inset Formula $\forall n\in\mathbb{N}\exists\varepsilon>0,x\in D\eqqcolon x_{n}\ni:\left|x_{n}-a\right|<\frac{1}{n}\wedge\left|f\left(x_{n}\right)-f\left(a\right)\right|\geq\varepsilon$
\end_inset
.
S prvim argumentom konjunkcije smo poskrbeli za to,
da je naše konstruiramo zaporedje
\begin_inset Formula $\left(x_{n}\right)_{n\in\mathbb{N}}$
\end_inset
konvergentno k
\begin_inset Formula $a$
\end_inset
.
Konstruirali smo zaporedje,
pri katerem so funkcijske vrednosti za vsak
\begin_inset Formula $\varepsilon$
\end_inset
izven
\begin_inset Formula $\varepsilon-$
\end_inset
okolice
\begin_inset Formula $f\left(a\right)$
\end_inset
,
torej zaporedje ne konvergira k
\begin_inset Formula $f\left(a\right)$
\end_inset
.
\end_layout
\end_deeper
\begin_layout Theorem*
\begin_inset ERT
status open
\begin_layout Plain Layout
\backslash
hypertarget{kzsppom}{Karakterizacija zveznosti s pomočjo praslik odprtih množic}
\end_layout
\end_inset
.
Naj bo
\begin_inset Formula $f:D\to\mathbb{R}$
\end_inset
.
\begin_inset Formula $f$
\end_inset
je zvezna na
\begin_inset Formula $D\Leftrightarrow$
\end_inset
za vsako odprto množico
\begin_inset Formula $V\subset\mathbb{R}$
\end_inset
je
\begin_inset Formula $f^{-1}\left(V\right)$
\end_inset
spet odprta množica
\begin_inset Foot
status open
\begin_layout Plain Layout
Za funkcijo
\begin_inset Formula $f:D\to V$
\end_inset
za
\begin_inset Formula $X\subseteq V$
\end_inset
definiramo
\begin_inset Formula $f^{-1}\left(X\right)\coloneqq\left\{ x\in D;f\left(x\right)\in V\right\} \subseteq D$
\end_inset
.
\end_layout
\end_inset
.
\end_layout
\begin_layout Proof
Dokazujemo ekvivalenco.
\end_layout
\begin_deeper
\begin_layout Labeling
\labelwidthstring 00.00.0000
\begin_inset Formula $\left(\Leftarrow\right)$
\end_inset
Predpostavimo,
da za vsako odprto množico
\begin_inset Formula $V\subset\mathbb{R}$
\end_inset
je
\begin_inset Formula $f^{-1}\left(V\right)$
\end_inset
spet odprta množica.
Dokazujemo,
da je
\begin_inset Formula $f$
\end_inset
zvezna na
\begin_inset Formula $D$
\end_inset
.
Naj bosta
\begin_inset Formula $a\in D,\varepsilon>0$
\end_inset
poljubna.
Naj bo
\begin_inset Formula $V\coloneqq\left(f\left(a\right)-\varepsilon,f\left(a\right)+\varepsilon\right)$
\end_inset
odprta množica.
Po predpostavki sledi,
da je
\begin_inset Formula $f^{-1}\left(V\right)$
\end_inset
spet odprta.
Ker je
\begin_inset Formula $a\in f^{-1}\left(V\right)$
\end_inset
,
je
\begin_inset Formula $a\in V$
\end_inset
.
Ker je
\begin_inset Formula $V$
\end_inset
odprta,
\begin_inset Formula $\exists\delta>0\ni:\left(a-\delta,a+\delta\right)\in V$
\end_inset
.
Torej
\begin_inset Formula $\forall x\in D:\left|x-a\right|<\delta\Rightarrow\left|f\left(x\right)-f\left(a\right)\right|<\varepsilon$
\end_inset
,
torej je
\begin_inset Formula $f$
\end_inset
zvezna na
\begin_inset Formula $D$
\end_inset
.
\end_layout
\begin_layout Labeling
\labelwidthstring 00.00.0000
\begin_inset Formula $\left(\Rightarrow\right)$
\end_inset
Predpostavimo,
da je
\begin_inset Formula $f$
\end_inset
zvezna na
\begin_inset Formula $D$
\end_inset
,
to pomeni
\begin_inset Formula $\forall a\in D\forall\varepsilon>0\exists\delta>0\forall x\in D:\left|x-a\right|<\delta\Rightarrow\left|f\left(x\right)-f\left(a\right)\right|<\varepsilon$
\end_inset
.
Naj bo
\begin_inset Formula $V$
\end_inset
poljubna odprta podmnožica
\begin_inset Formula $\mathbb{R}$
\end_inset
in naj bo
\begin_inset Formula $a\in f^{-1}\left(V\right)$
\end_inset
poljuben (torej
\begin_inset Formula $f\left(a\right)\in V$
\end_inset
).
Ker je
\begin_inset Formula $f\left(a\right)\in V$
\end_inset
,
ki je odprta,
\begin_inset Formula $\exists\varepsilon>0\ni:\left(f\left(a\right)-\varepsilon,f\left(a\right)+\varepsilon\right)\subseteq V$
\end_inset
.
Ker je
\begin_inset Formula $f$
\end_inset
zvezna v
\begin_inset Formula $a$
\end_inset
,
\begin_inset Formula $\exists\delta>0\forall x\in D:\left|x-a\right|<\delta\Rightarrow\left|f\left(x\right)-f\left(a\right)\right|<\varepsilon$
\end_inset
,
torej je tudi neka odprta okolica
\begin_inset Formula $f\left(a\right)$
\end_inset
v
\begin_inset Formula $f^{-1}\left(V\right)$
\end_inset
.
Ker je bil
\begin_inset Formula $a$
\end_inset
poljuben,
je
\begin_inset Formula $f^{-1}\left(V\right)$
\end_inset
odprta,
ker je bila
\begin_inset Formula $V$
\end_inset
poljubna,
je izrek dokazan.
\end_layout
\end_deeper
\begin_layout Theorem*
Naj bosta
\begin_inset Formula $f,g:D\to\mathbb{R}$
\end_inset
zvezni v
\begin_inset Formula $a\in D$
\end_inset
.
Tedaj so v
\begin_inset Formula $a$
\end_inset
zvezne tudi funkcije
\begin_inset Formula $f+g,f-g,f\cdot g$
\end_inset
in
\begin_inset Formula $f/g$
\end_inset
,
slednja le,
če je
\begin_inset Formula $g\left(a\right)\not=0$
\end_inset
.
\end_layout
\begin_layout Proof
Ker je
\begin_inset Formula $f$
\end_inset
zvezna v
\begin_inset Formula $a$
\end_inset
po
\begin_inset ERT
status open
\begin_layout Plain Layout
\backslash
hyperlink{kzzz}{izreku o karakterizaciji zveznosti z zaporedji}
\end_layout
\end_inset
velja za vsako
\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}},\forall n\in\mathbb{N}:a_{n}\subset D,\lim_{n\to\infty}a_{n}=a$
\end_inset
tudi
\begin_inset Formula $\lim_{n\to\infty}f\left(a_{n}\right)=f\left(a\right)$
\end_inset
.
Po
\begin_inset ERT
status open
\begin_layout Plain Layout
\backslash
hyperlink{pmkdlim}{izreku iz poglavja o zaporedjih}
\end_layout
\end_inset
velja,
da
\begin_inset Formula $f\left(a_{n}\right)*g\left(a_{n}\right)\to\left(f*g\right)\left(a_{n}\right)$
\end_inset
za
\begin_inset Formula $*\in\left\{ +,-,\cdot,/\right\} $
\end_inset
.
Zopet uporabimo izrek o karakterizaciji zveznosti z zaporedji,
ki pove,
da so tudi
\begin_inset Formula $f*g$
\end_inset
za
\begin_inset Formula $*\in\left\{ +,-,\cdot,/\right\} $
\end_inset
zvezne v
\begin_inset Formula $a$
\end_inset
.
Pri deljenju velja omejitev
\begin_inset Formula $f\left(a\right)\not=0$
\end_inset
.
\end_layout
\begin_layout Theorem*
Če sta
\begin_inset Formula $D,E\subseteq\mathbb{R}$
\end_inset
in
\begin_inset Formula $f:D\to E$
\end_inset
in
\begin_inset Formula $g:E\to\mathbb{R}$
\end_inset
,
je
\begin_inset Formula $g\circ f:D\to\mathbb{R}$
\end_inset
.
Hkrati pa,
če je
\begin_inset Formula $f$
\end_inset
zvezna v
\begin_inset Formula $a$
\end_inset
in
\begin_inset Formula $g$
\end_inset
zvezna v
\begin_inset Formula $f\left(a\right)$
\end_inset
,
je
\begin_inset Formula $g\circ f$
\end_inset
zvezna v
\begin_inset Formula $a$
\end_inset
.
\begin_inset Foot
status open
\begin_layout Plain Layout
Velja
\begin_inset Formula $\left(g\circ f\right)\left(x\right)=g\left(f\left(x\right)\right)$
\end_inset
.
\end_layout
\end_inset
\end_layout
\begin_layout Proof
Vzemimo poljubno
\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}\subseteq D$
\end_inset
,
da
\begin_inset Formula $a_{n}\to a\in D$
\end_inset
.
Zopet uporabimo
\begin_inset ERT
status open
\begin_layout Plain Layout
\backslash
hyperlink{kzzz}{izrek o karakterizaciji zveznosti z zaporedji}
\end_layout
\end_inset
:
ker je
\begin_inset Formula $f$
\end_inset
zvezna,
velja
\begin_inset Formula $f\left(a_{n}\right)\to f\left(a\right)$
\end_inset
in ker je
\begin_inset Formula $g$
\end_inset
zvezna,
velja
\begin_inset Formula $g\left(f\left(a_{n}\right)\right)\to g\left(f\left(a\right)\right)$
\end_inset
.
Potemtakem
\begin_inset Formula $\left(g\circ f\right)\left(a_{n}\right)\to\left(g\circ f\right)\left(a\right)$
\end_inset
in po istem izreku je
\begin_inset Formula $g\circ f$
\end_inset
zvezna na
\begin_inset Formula $D$
\end_inset
.
\end_layout
\begin_layout Theorem*
Vsi polinomi so zvezni na
\begin_inset Formula $\mathbb{R}$
\end_inset
.
\end_layout
\begin_layout Proof
Vzemimo
\begin_inset Formula $p\left(x\right)=\sum_{k=0}^{n}a_{k}k^{k}$
\end_inset
.
Uporabimo prejšnji izrek.
Polinom je sestavljen iz vsote konstantne funkcije,
zmnožene z identiteto,
ki je s seboj
\begin_inset Formula $n-$
\end_inset
krat množena.
Ker vsota in množenje ohranjata zveznost,
je treba dokazati le,
da je
\begin_inset Formula $f\left(x\right)=x$
\end_inset
zvezna in da so
\begin_inset Formula $\forall c\in\mathbb{R}:f\left(x\right)=c$
\end_inset
zvezne.
\end_layout
\begin_deeper
\begin_layout Labeling
\labelwidthstring 00.00.0000
\begin_inset Formula $f\left(x\right)=x$
\end_inset
Ali velja
\begin_inset Formula $\forall\varepsilon>0\exists\delta>0\ni:\forall x\in\mathbb{R}:\left|x-a\right|<\delta\Rightarrow\left|f\left(x\right)-f\left(a\right)\right|<\varepsilon$
\end_inset
?
Da,
velja.
Vzamemo lahko katerokoli
\begin_inset Formula $\delta\in(0,\varepsilon]$
\end_inset
.
\end_layout
\begin_layout Labeling
\labelwidthstring 00.00.0000
\begin_inset Formula $f\left(x\right)=c$
\end_inset
Naj bo
\begin_inset Formula $c\in\mathbb{R}$
\end_inset
poljuben.
Tu je
\begin_inset Formula $\left|f\left(x\right)-f\left(a\right)\right|=\left|c-c\right|=0$
\end_inset
,
torej je desna stran implikacije vedno resnična,
torej je implikacija vedno resnična.
\end_layout
\end_deeper
\begin_layout Theorem*
Vse elementarne funkcije so na njihovih definicijskih območjih povsod zvezne.
To so:
polinomi,
potence,
racionalne funkcije,
koreni,
eksponentne funkcije,
logaritmi,
trigonometrične,
ciklometrične in kombinacije neskončno mnogo naštetih,
spojenih s
\begin_inset Formula $+,-,\cdot,/,\circ$
\end_inset
.
\end_layout
\begin_layout Proof
Tega izreka ne bomo dokazali.
\end_layout
\begin_layout Example*
\begin_inset Formula $f\left(x\right)\coloneqq\log\left(\sin^{3}x+\frac{1}{8}\right)+\frac{1}{\sqrt[4]{x-7}}$
\end_inset
je zvezna povsod,
kjer je definirana.
\end_layout
\begin_layout Definition*
Naj bo
\begin_inset Formula $\varepsilon>0,a\in\mathbb{R}$
\end_inset
in
\begin_inset Formula $f:\left(a-\varepsilon,a+\varepsilon\right)\setminus\left\{ a\right\} \to\mathbb{R}$
\end_inset
.
Pravimo,
da je
\begin_inset Formula $L\in\mathbb{R}$
\end_inset
limita
\begin_inset Formula $f$
\end_inset
v točki
\begin_inset Formula $a$
\end_inset
(zapišemo
\begin_inset Formula $L=\lim_{x\to a}f\left(x\right)$
\end_inset
),
če za vsako zaporedje
\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}},a_{n}\in\left(a-\varepsilon,a+\varepsilon\right)\setminus\left\{ a\right\} $
\end_inset
,
za katero velja
\begin_inset Formula $a_{n}\to a$
\end_inset
,
velja
\begin_inset Formula $f\left(a_{n}\right)\to L$
\end_inset
\end_layout
\begin_layout Definition*
ZDB če
\begin_inset Formula $\forall\varepsilon>0\exists\delta>0\ni:\left|x-a\right|<\delta\Rightarrow\left|f\left(x\right)-L\right|<\varepsilon$
\end_inset
\end_layout
\begin_layout Definition*
ZDB če za
\begin_inset Formula $\overline{f}:\left(a-\varepsilon,a+\varepsilon\right)\to\mathbb{R}$
\end_inset
s predpisom
\begin_inset Formula $\overline{f}\left(x\right)\coloneqq\begin{cases}
f\left(x\right) & ;x\in\left(a-\varepsilon,a+\varepsilon\right)\setminus\left\{ a\right\} \\
L & ;x\in a
\end{cases}$
\end_inset
velja,
da je zvezna v
\begin_inset Formula $a$
\end_inset
.
\end_layout
\begin_layout Note*
Vrednost
\begin_inset Formula $f\left(a\right)$
\end_inset
,
če sploh obstaja,
nima vloge pri vrednosti limite.
\end_layout
\begin_layout Corollary*
Naj bo
\begin_inset Formula $a\in D\subseteq\mathbb{R}$
\end_inset
in
\begin_inset Formula $f:D\to\mathbb{R}$
\end_inset
.
\begin_inset Formula $f$
\end_inset
je zvezna v
\begin_inset Formula $a\Leftrightarrow\lim_{x\to a}f\left(x\right)=f\left(a\right)$
\end_inset
.
\end_layout
\begin_layout Example*
Kvadratna funkcija
\begin_inset Formula $f\left(x\right)=x^{2}$
\end_inset
je zvezna.
Vzemimo poljuben
\begin_inset Formula $a\in\mathbb{R},\varepsilon>0$
\end_inset
.
Obstajati mora taka
\begin_inset Formula $\delta>0\ni:\forall x\in\mathbb{R}:\left|x-a\right|<\delta\Rightarrow\left|f\left(x\right)-f\left(a\right)\right|<\varepsilon$
\end_inset
.
\end_layout
\begin_layout Example*
Podan imamo torej
\begin_inset Formula $a$
\end_inset
in
\begin_inset Formula $\varepsilon$
\end_inset
,
želimo najti
\begin_inset Formula $\delta$
\end_inset
.
Želimo priti do neenakosti,
ki ima na manjši strani
\begin_inset Formula $\left|f\left(x\right)-f\left(a\right)\right|=\left|x^{2}-a^{2}\right|$
\end_inset
in na večji strani nek izraz z
\begin_inset Formula $\left|x-a\right|$
\end_inset
,
da ta
\begin_inset Formula $\left|x-a\right|$
\end_inset
nadomestimo z
\begin_inset Formula $\delta$
\end_inset
in nato večjo stran enačimo z
\begin_inset Formula $\varepsilon$
\end_inset
,
da izrazimo
\begin_inset Formula $\varepsilon$
\end_inset
v odvisnosti od
\begin_inset Formula $\delta$
\end_inset
in
\begin_inset Formula $a$
\end_inset
.
\end_layout
\begin_layout Example*
Računajmo:
\begin_inset Formula $\left|x^{2}-a^{2}\right|=\left|x-a\right|\left|x+a\right|$
\end_inset
.
Predelajmo izraz
\begin_inset Formula $\left|x+a\right|=\left|\left(x-a\right)+2a\right|\leq\left|x-a\right|+\left|2a\right|$
\end_inset
,
torej skupaj
\begin_inset Formula $\left|x^{2}-a^{2}\right|\leq\left|x-a\right|\left(\left|x-a\right|+\left|2a\right|\right)$
\end_inset
.
Sedaj nadomestimo
\begin_inset Formula $\left|x-a\right|$
\end_inset
z
\begin_inset Formula $\delta$
\end_inset
:
\begin_inset Formula $\left|x^{2}-a^{2}\right|\leq\delta\left(\delta+\left|2a\right|\right)$
\end_inset
.
Iščemo tak
\begin_inset Formula $\varepsilon$
\end_inset
,
da velja
\begin_inset Formula $\left|x^{2}-a^{2}\right|<\varepsilon$
\end_inset
,
zato enačimo
\begin_inset Formula $\delta\left(\delta+\left|2a\right|\right)=\varepsilon$
\end_inset
in dobimo kvadratno enačbo
\begin_inset Formula $\delta^{2}+\left|2a\right|\delta-\varepsilon=0$
\end_inset
,
ki jo rešimo z obrazcem za ničle:
\begin_inset Formula
\[
\delta_{1,2}=\frac{-2\left|a\right|\pm\sqrt{4\left|a\right|^{2}-4\varepsilon}}{2}=-\left|a\right|\pm\sqrt{\left|a\right|^{2}-\varepsilon}
\]
\end_inset
Toda ker iščemo le pozitivne
\begin_inset Formula $\delta$
\end_inset
,
je edina rešitev
\begin_inset Formula
\[
\delta=-\left|a\right|+\sqrt{\left|a\right|^{2}-\varepsilon}=\sqrt{\left|a\right|^{2}-\varepsilon}-\left|a\right|=\frac{\sqrt{\left|a\right|^{2}-\varepsilon}-\left|a\right|}{1}=\frac{\left(\sqrt{\left|a\right|^{2}-\varepsilon}-\left|a\right|\right)\left(\sqrt{\left|a\right|^{2}-\varepsilon}+\left|a\right|\right)}{\sqrt{\left|a\right|^{2}-\varepsilon}+\left|a\right|}=\frac{\varepsilon}{\sqrt{\left|a\right|^{2}-\varepsilon}+\left|a\right|}
\]
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Definition*
Naj bo
\begin_inset Formula $D\subset\mathbb{R},a\in\mathbb{R}\ni:\forall\varepsilon>0:D\cap\left(a,a+\varepsilon\right)\not=\emptyset$
\end_inset
.
Naj bo
\begin_inset Formula $f:D\to\mathbb{R}$
\end_inset
.
Število
\begin_inset Formula $L_{+}\in\mathbb{R}$
\end_inset
je desna limita funkcije
\begin_inset Formula $f$
\end_inset
v točki
\begin_inset Formula $a$
\end_inset
,
če
\begin_inset Formula $\forall\left(a_{n}\right)_{n\in\mathbb{N}}\subset D\cap\left(a,\infty\right):a_{n}\to a\Rightarrow f\left(a_{n}\right)\to L_{+}$
\end_inset
ZDB če za vsako k
\begin_inset Formula $a$
\end_inset
konvergentno zaporedje s členi desno od
\begin_inset Formula $a$
\end_inset
velja,
da funkcijske vrednosti členov konvergirajo k
\begin_inset Formula $L_{+}$
\end_inset
.
Oznaka
\begin_inset Formula $L_{+}=\lim_{x\to a^{+}}f\left(x\right)=\lim_{x\searrow a}f\left(x\right)=f\left(a+0\right)$
\end_inset
.
Podobno definiramo tudi levo limito
\begin_inset Formula $L_{-}=\lim_{x\to a^{-}}f\left(x\right)=\lim_{x\nearrow a}f\left(x\right)=f\left(a-0\right)$
\end_inset
.
\end_layout
\begin_layout Theorem*
Naj bo
\begin_inset Formula $D\subset\mathbb{R}$
\end_inset
in
\begin_inset Formula $a\in\mathbb{R}$
\end_inset
da velja
\begin_inset Formula $\forall\varepsilon>0:D\cap\left(a,a-\varepsilon\right)\not=\emptyset\wedge D\cap\left(a,a+\varepsilon\right)\not=\emptyset$
\end_inset
.
Naj bo
\begin_inset Formula $f:D\to\mathbb{R}$
\end_inset
.
Velja
\begin_inset Formula $\exists\lim_{x\to a}f\left(x\right)\Leftrightarrow\exists\lim_{x\nearrow a}f\left(x\right)\wedge\exists\lim_{x\searrow a}f\left(x\right)\wedge\lim_{x\nearrow a}f\left(x\right)=\lim_{x\searrow a}f\left(x\right)$
\end_inset
V tem primeru velja
\begin_inset Formula $\lim_{x\to a}f\left(x\right)=\lim_{x\nearrow a}f\left(x\right)=\lim_{x\searrow a}f\left(x\right)$
\end_inset
.
\end_layout
\begin_layout Definition*
Označimo
\begin_inset Formula $\lim_{x\searrow a}f\left(x\right)\eqqcolon f\left(a+0\right),\lim_{x\nearrow a}f\left(x\right)\eqqcolon f\left(a-0\right)$
\end_inset
.
Če
\begin_inset Formula $\exists f\left(a+0\right)$
\end_inset
in
\begin_inset Formula $\exists f\left(a-0\right)$
\end_inset
,
vendar
\begin_inset Formula $f\left(a+0\right)\not=f\left(a-0\right)$
\end_inset
,
pravimo,
da ima
\begin_inset Formula $f$
\end_inset
v točki
\begin_inset Formula $a$
\end_inset
\begin_inset Quotes gld
\end_inset
skok
\begin_inset Quotes grd
\end_inset
.
\end_layout
\begin_layout Example*
\begin_inset Formula $\lim_{x\to0}\frac{1}{1+e^{1/x}}$
\end_inset
ne obstaja.
Zakaj?
Izračunajmo levo in desno limito:
\begin_inset Formula
\[
\lim_{x\searrow0}\frac{1}{1+e^{1/x}}=0,\lim_{x\nearrow0}\frac{1}{1+e^{1/x}}=1
\]
\end_inset
Toda
\begin_inset Formula $\exists\lim_{x\to a}f\left(x\right)\Leftrightarrow\exists\lim_{x\nearrow a}f\left(x\right)\wedge\exists\lim_{x\searrow a}f\left(x\right)\wedge\lim_{x\nearrow a}f\left(x\right)=\lim_{x\searrow a}f\left(x\right)$
\end_inset
.
\end_layout
\begin_layout Definition*
Funkcija
\begin_inset Formula $f$
\end_inset
je na intervalu
\begin_inset Formula $D$
\end_inset
odsekoma zvezna,
če je zvezna povsod na
\begin_inset Formula $D$
\end_inset
,
razen morda v končno mnogo točkah,
v katerih ima skok.
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Example*
Naj bo
\begin_inset Formula $f:\mathbb{R}\setminus\left\{ 0\right\} \to\mathbb{R}$
\end_inset
s predpisom
\begin_inset Formula $x\mapsto\frac{\sin x}{x}$
\end_inset
.
Zanima nas,
ali obstaja
\begin_inset Formula $\lim_{x\to0}f\left(x\right)$
\end_inset
.
Grafični dokaz.
\end_layout
\begin_layout Example*
\begin_inset Float figure
placement document
alignment document
wide false
sideways false
status open
\begin_layout Plain Layout
TODO XXX FIXME SKICA S TKZ EUCLID,
glej ZVZ III/ANA1P1120/str.8
\end_layout
\begin_layout Plain Layout
\begin_inset Caption Standard
\begin_layout Plain Layout
Skica.
\end_layout
\end_inset
\end_layout
\end_inset
Očitno velja
\begin_inset Formula $\triangle ABD\subset$
\end_inset
krožni izsek
\begin_inset Formula $DAB\subset\triangle ABC$
\end_inset
,
torej za njihove ploščine velja
\begin_inset Formula
\[
\frac{\sin x}{2}\leq\frac{x}{2\pi}\cdot x=\frac{x}{2}\leq\frac{\tan x}{2}\quad\quad\quad\quad/\cdot\frac{2}{\sin x}
\]
\end_inset
\begin_inset Formula
\[
1\leq\frac{x}{\sin x}\leq\frac{1}{\cos x}\quad\quad\quad\quad/\lim_{x\to0}
\]
\end_inset
\begin_inset Formula
\[
\lim_{x\to0}1\leq\lim_{x\to0}\frac{x}{\sin x}\leq\lim_{x\to0}\frac{1}{\cos x}
\]
\end_inset
\begin_inset Formula
\[
1\leq\lim_{x\to0}\frac{x}{\sin x}\leq1
\]
\end_inset
\begin_inset Formula
\[
\lim_{x\to0}\frac{x}{\sin x}=1
\]
\end_inset
Da naš sklep res potrdimo,
je potreben spodnji izrek.
\end_layout
\begin_layout Theorem*
Če za
\begin_inset Formula $f,g,h:D\to\mathbb{R}$
\end_inset
velja za
\begin_inset Formula $a\in D$
\end_inset
:
\end_layout
\begin_deeper
\begin_layout Itemize
\begin_inset Formula $\exists\varepsilon>0\forall x\in\left(a-\varepsilon,a+\varepsilon\right)\setminus\left\{ a\right\} :f\left(x\right)\leq g\left(x\right)\leq h\left(x\right)$
\end_inset
in hkrati
\end_layout
\begin_layout Itemize
\begin_inset Formula $\exists\lim_{x\to a}f\left(x\right),\lim_{x\to a}h\left(x\right)$
\end_inset
in
\begin_inset Formula $\lim_{x\to a}f\left(x\right)=\lim_{x\to a}h\left(x\right)\eqqcolon L$
\end_inset
,
tedaj tudi
\begin_inset Formula $\exists\lim_{x\to a}g\left(x\right)$
\end_inset
in
\begin_inset Formula $\lim_{x\to a}g\left(x\right)=L$
\end_inset
.
\end_layout
\end_deeper
\begin_layout Proof
Naj bo
\begin_inset Formula $A=A\left(x\right)\coloneqq\max\left\{ \left|f\left(x\right)-L\right|,\left|h\left(x\right)-L\right|\right\} $
\end_inset
.
Velja
\end_layout
\begin_deeper
\begin_layout Itemize
\begin_inset Formula $g\left(x\right)-L\leq h\left(x\right)-L\leq\left|h\left(x\right)-L\right|\leq A\left(x\right)$
\end_inset
in
\end_layout
\begin_layout Itemize
\begin_inset Formula $L-g\left(x\right)\leq L-f\left(x\right)\leq\left|f\left(x\right)-L\right|\leq A\left(x\right)$
\end_inset
.
\end_layout
\end_deeper
\begin_layout Proof
Posledično
\begin_inset Formula $\left|g\left(x\right)-L\right|\leq A\left(x\right)$
\end_inset
.
Naj bo sedaj
\begin_inset Formula $\varepsilon>0$
\end_inset
poljuben.
Tedaj velja
\begin_inset Formula $\exists\delta_{1}>0\ni:\left|x-a\right|<\delta_{1}\Rightarrow\left|f\left(x\right)-L\right|<\varepsilon$
\end_inset
in
\begin_inset Formula $\exists\delta_{2}>0\ni:\left|x-a\right|<\delta_{2}\Rightarrow\left|h\left(x\right)-L\right|<\varepsilon$
\end_inset
.
Za
\begin_inset Formula $\delta\coloneqq\min\left\{ \delta_{1},\delta_{2}\right\} $
\end_inset
torej velja
\begin_inset Formula $\left|x-a\right|<\delta\Rightarrow\left|g\left(x\right)-L\right|<\varepsilon$
\end_inset
.
\end_layout
\begin_layout Subsection
Zvezne funkcije na kompaktnih množicah
\end_layout
\begin_layout Definition*
Množica
\begin_inset Formula $K\subseteq\mathbb{R}$
\end_inset
je kompaktna
\begin_inset Formula $\Leftrightarrow$
\end_inset
je zaprta in omejena ZDB je unija zaprtih intervalov.
\end_layout
\begin_layout Theorem*
Naj bo
\begin_inset Formula $K\subset\mathbb{R}$
\end_inset
kompaktna in
\begin_inset Formula $f:K\to\mathbb{R}$
\end_inset
zvezna.
Tedaj je
\begin_inset Formula $f$
\end_inset
omejena in doseže minimum in maksimum.
\end_layout
\begin_layout Example*
Primeri funkcij.
\end_layout
\begin_deeper
\begin_layout Enumerate
\begin_inset Formula $f_{1}\left(x\right)=\frac{1}{x}$
\end_inset
na
\begin_inset Formula $I_{1}=(0,1]$
\end_inset
.
\begin_inset Formula $f_{1}$
\end_inset
je zvezna in
\begin_inset Formula $\lim_{x\to0}f_{1}\left(x\right)=\infty$
\end_inset
,
torej ni omejena,
a
\begin_inset Formula $I_{1}$
\end_inset
ni zaprt.
\end_layout
\begin_layout Enumerate
\begin_inset Formula $f_{2}\left(x\right)=\begin{cases}
0 & ;x=0\\
\frac{1}{x} & ;x\in(0,1]
\end{cases}$
\end_inset
ni omejena in je definirana na kompaktni množici,
a ni zvezna.
\end_layout
\begin_layout Enumerate
\begin_inset Formula $f_{3}\left(x\right)=x$
\end_inset
na
\begin_inset Formula $x\in\left(0,1\right)$
\end_inset
.
Je omejena,
ne doseže maksimuma,
a
\begin_inset Formula $D_{f_{3}}$
\end_inset
ni kompaktna (ni zaprta).
\end_layout
\begin_layout Enumerate
\begin_inset Formula $f_{4}\left(x\right)=\begin{cases}
x & ;x\in\left(0,1\right)\\
\frac{1}{2} & ;x\in\left\{ 0,1\right\}
\end{cases}$
\end_inset
.
Velja
\begin_inset Formula $\sup f_{4}=1$
\end_inset
,
ampak ga ne doseže,
a ni zvezna
\end_layout
\end_deeper
\begin_layout Proof
Naj bo
\begin_inset Formula $K\subseteq\mathbb{R}$
\end_inset
kompaktna in
\begin_inset Formula $f:K\to\mathbb{R}$
\end_inset
zvezna.
\end_layout
\begin_deeper
\begin_layout Itemize
Omejenost navzgor:
PDDRAA
\begin_inset Formula $f$
\end_inset
ni navzgor omejena.
Tedaj
\begin_inset Formula $\forall n\in\mathbb{N}\exists x_{n}\in K\ni:f\left(x_{n}\right)\geq n$
\end_inset
(*).
Ker je
\begin_inset Formula $\left(x_{n}\right)_{n\in\mathbb{N}}$
\end_inset
omejeno zaporedje (vsi členi so na kompaktni
\begin_inset Formula $K$
\end_inset
),
ima stekališče,
recimo mu
\begin_inset Formula $s\in\mathbb{R}$
\end_inset
.
Vemo,
da tedaj obstaja podzaporedje
\begin_inset Formula $\left(x_{n_{k}}\right)_{k\in\mathbb{N}}\ni:s=\lim_{k\to\infty}x_{n_{k}}$
\end_inset
.
Ker je
\begin_inset Formula $K$
\end_inset
tudi zaprta,
sledi
\begin_inset Formula $s\in K$
\end_inset
.
Ker je
\begin_inset Formula $f$
\end_inset
zvezna na
\begin_inset Formula $K$
\end_inset
,
velja
\begin_inset Formula $f\left(s\right)=\lim_{k\to\infty}f\left(x_{n_{k}}\right)$
\end_inset
.
Toda po (*) sledi
\begin_inset Formula $\lim_{k\to\infty}f\left(x_{n_{k}}\right)=\infty$
\end_inset
,
zato
\begin_inset Formula $f\left(s\right)=\infty$
\end_inset
,
kar ni mogoče,
saj je
\begin_inset Formula $f\left(s\right)\in\mathbb{R}$
\end_inset
.
\begin_inset Formula $\rightarrow\!\leftarrow$
\end_inset
.
Torej je
\begin_inset Formula $f$
\end_inset
navzgor omejena.
\end_layout
\begin_layout Itemize
Omejenost navzdol:
PDDRAA
\begin_inset Formula $f$
\end_inset
ni navzdol omejena.
Tedaj
\begin_inset Formula $\forall n\in\mathbb{N}\exists x_{n}\in K\ni:f\left(x_{n}\right)\leq-n$
\end_inset
(*).
Ker je
\begin_inset Formula $\left(x_{n}\right)_{n\in\mathbb{N}}$
\end_inset
omejeno zaporedje (vsi členi so na kompaktni
\begin_inset Formula $K$
\end_inset
),
ima stekališče,
recimo mu
\begin_inset Formula $s\in\mathbb{R}$
\end_inset
.
Vemo,
da tedaj obstaja podzaporedje
\begin_inset Formula $\left(x_{n_{k}}\right)_{k\in\mathbb{N}}\ni:s=\lim_{k\to\infty}x_{n_{k}}$
\end_inset
.
Ker je
\begin_inset Formula $K$
\end_inset
tudi zaprta,
sledi
\begin_inset Formula $s\in K$
\end_inset
.
Ker je
\begin_inset Formula $f$
\end_inset
zvezna na
\begin_inset Formula $K$
\end_inset
,
velja
\begin_inset Formula $f\left(s\right)=\lim_{k\to\infty}f\left(x_{n_{k}}\right)$
\end_inset
.
Toda po (*) sledi
\begin_inset Formula $\lim_{k\to\infty}f\left(s_{n_{k}}\right)=-\infty$
\end_inset
,
zato
\begin_inset Formula $f\left(s\right)=-\infty$
\end_inset
,
kar ni mogoče,
saj je
\begin_inset Formula $f\left(s\right)\in\mathbb{R}$
\end_inset
.
\begin_inset Formula $\rightarrow\!\leftarrow$
\end_inset
.
Torej je
\begin_inset Formula $f$
\end_inset
navzgor omejena.
\end_layout
\begin_layout Itemize
Doseže maksimum:
Označimo
\begin_inset Formula $M\coloneqq\sup_{x\in K}f\left(x\right)$
\end_inset
.
Ravnokar smo dokazali,
da
\begin_inset Formula $M<\infty$
\end_inset
.
Po definiciji supremuma
\begin_inset Formula $\forall n\in\mathbb{N}\exists t_{n}\in K\ni:f\left(t_{n}\right)>M-\frac{1}{n}$
\end_inset
.
Ker je
\begin_inset Formula $K$
\end_inset
omejena,
ima
\begin_inset Formula $\left(t_{n}\right)_{n}$
\end_inset
stekališče in ker je zaprta,
velja
\begin_inset Formula $t\in K$
\end_inset
,
zato
\begin_inset Formula $\exists$
\end_inset
podzaporedje
\begin_inset Formula $\left(t_{n_{j}}\right)_{j\in\mathbb{N}}\ni:t=\lim_{j\to\infty}t_{n_{j}}$
\end_inset
.
Ker je
\begin_inset Formula $f$
\end_inset
zvezna,
velja
\begin_inset Formula $f\left(t\right)=\lim_{j\to\infty}f\left(t_{n_{j}}\right)$
\end_inset
.
Toda ker
\begin_inset Formula $f\left(t_{n_{j}}\right)>M-\frac{1}{n_{j}}\geq M-\frac{1}{j}$
\end_inset
,
velja
\begin_inset Formula $f\left(t\right)\geq M$
\end_inset
.
Hkrati po definiciji
\begin_inset Formula $M$
\end_inset
velja
\begin_inset Formula $f\left(t\right)\leq M$
\end_inset
.
Sledi
\begin_inset Formula $M=f\left(t\right)$
\end_inset
in zato
\begin_inset Formula $M=\max_{x\in K}f\left(x\right)$
\end_inset
.
\end_layout
\begin_layout Itemize
Doseže minimum:
Označimo
\begin_inset Formula $M\coloneqq\inf_{x\in K}f\left(x\right)$
\end_inset
.
Ko smo dokazali omejenost,
smo dokazali,
da
\begin_inset Formula $M>-\infty$
\end_inset
.
Po definiciji infimuma
\begin_inset Formula $\forall n\in\mathbb{N}\exists t_{n}\in K\ni:f\left(t_{n}\right)<M+\frac{1}{n}$
\end_inset
.
Ker je
\begin_inset Formula $K$
\end_inset
omejena,
ima
\begin_inset Formula $\left(t_{n}\right)_{n}$
\end_inset
stekališče in ker je zaprta,
velja
\begin_inset Formula $t\in K$
\end_inset
,
zato
\begin_inset Formula $\exists$
\end_inset
podzaporedje
\begin_inset Formula $\left(t_{n_{j}}\right)_{j\in\mathbb{N}}\ni:t=\lim_{j\to\infty}t_{n_{j}}$
\end_inset
.
Ker je
\begin_inset Formula $f$
\end_inset
zvezna,
velja
\begin_inset Formula $f\left(t\right)=\lim_{j\to\infty}f\left(t_{n_{j}}\right)$
\end_inset
.
Toda ker
\begin_inset Formula $f\left(t_{n_{j}}\right)<M-\frac{1}{n_{j}}\leq M-\frac{1}{j}$
\end_inset
,
velja
\begin_inset Formula $f\left(t\right)\leq M$
\end_inset
.
Hkrati po definiciji
\begin_inset Formula $M$
\end_inset
velja
\begin_inset Formula $f\left(t\right)\geq M$
\end_inset
.
Sledi
\begin_inset Formula $M=f\left(t\right)$
\end_inset
in zato
\begin_inset Formula $M=\min_{x\in K}f\left(x\right)$
\end_inset
.
\end_layout
\end_deeper
\begin_layout Theorem*
Naj bo
\begin_inset Formula $f:\left[a,b\right]\to\mathbb{R}$
\end_inset
zvezna in
\begin_inset Formula $f\left(a\right)f\left(b\right)<0$
\end_inset
.
Tedaj
\begin_inset Formula $\exists\xi\in\left(a,b\right)\ni:f\left(\xi\right)=0$
\end_inset
.
\end_layout
\begin_layout Proof
Interval
\begin_inset Formula $I_{0}=\left[a,b\right]$
\end_inset
razpolovimo.
To pomeni,
da pogledamo levo in desno polovico intervala
\begin_inset Formula $I_{0}$
\end_inset
,
torej
\begin_inset Formula $\left[a,\frac{a+b}{2}\right]$
\end_inset
in
\begin_inset Formula $\left[\frac{a+b}{2},b\right]$
\end_inset
.
Če je
\begin_inset Formula $f\left(\frac{a+b}{2}\right)=0$
\end_inset
,
smo našli iskano točko
\begin_inset Formula $\xi$
\end_inset
,
sicer z
\begin_inset Formula $I_{1}$
\end_inset
označimo katerokoli izmed polovic,
ki ima
\begin_inset Formula $f$
\end_inset
v krajiščih različno predznačene funkcijske vrednosti.
Torej
\begin_inset Formula $I_{1}=\begin{cases}
\left[a,\frac{a+b}{2}\right] & ;f\left(a\right)f\left(\frac{a+b}{2}\right)<0\\
\left[\frac{a+b}{2},b\right] & ;f\left(\frac{a+b}{2}\right)f\left(b\right)<0
\end{cases}$
\end_inset
.
S postopkom nadaljujemo.
Če v končno mnogo korakih najdemo
\begin_inset Formula $\xi$
\end_inset
,
da je
\begin_inset Formula $f\left(\xi\right)=0$
\end_inset
,
fino,
sicer pa dobimo zaporedje intervalov
\begin_inset Formula $I_{n}=\left[a_{n},b_{n}\right]\subset\left[a,b\right]=I_{0}\ni:$
\end_inset
\end_layout
\begin_deeper
\begin_layout Enumerate
\begin_inset Formula $\forall n\in\mathbb{N}:\left|I_{n}\right|=2^{-n}\left|I_{0}\right|$
\end_inset
in
\end_layout
\begin_layout Enumerate
\begin_inset Formula $\forall n\in\mathbb{N}:I_{n+1}\subset I_{n}$
\end_inset
,
torej
\begin_inset Formula $a_{n+1}\geq a_{n}\wedge b_{n+1}\leq b_{n}$
\end_inset
,
in
\end_layout
\begin_layout Enumerate
\begin_inset CommandInset label
LatexCommand label
name "enu:različni-predznaki-istoležnih-clenov"
\end_inset
\begin_inset Formula $\forall n\in\mathbb{N}:f\left(a_{n}\right)f\left(b_{n}\right)<0$
\end_inset
.
\end_layout
\end_deeper
\begin_layout Proof
Ker sta zaporedji
\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
\end_inset
in
\begin_inset Formula $\left(b_{n}\right)_{n\in\mathbb{N}}$
\end_inset
omejeni in monotoni,
imata po
\begin_inset ERT
status open
\begin_layout Plain Layout
\backslash
hyperlink{kmoz}{izreku o konvergenci monotonih in omejenih zaporedij}
\end_layout
\end_inset
limiti
\begin_inset Formula $\alpha\coloneqq\lim_{n\to\infty}a_{n}=\sup_{n\in\mathbb{N}}a_{n}$
\end_inset
in
\begin_inset Formula $\beta\coloneqq\lim_{n\to\infty}=\sup_{n\in\mathbb{N}}b_{n}$
\end_inset
in
\begin_inset Formula $\alpha,\beta\in I_{0}$
\end_inset
,
ker je
\begin_inset Formula $I_{0}$
\end_inset
zaprt.
\end_layout
\begin_layout Proof
Sledi
\begin_inset Formula $\forall n\in\mathbb{N}:\left|\alpha-\beta\right|=\beta-\alpha\leq b_{n}-a_{n}=\left|I_{n}\right|=2^{-n}\left|I_{0}\right|$
\end_inset
,
torej
\begin_inset Formula $\lim_{n\to\infty}\left|\alpha-\beta\right|=0\Rightarrow\alpha-\beta=0$
\end_inset
.
\end_layout
\begin_layout Proof
Ker je
\begin_inset Formula $f$
\end_inset
zvezna in
\begin_inset Formula $a_{n},b_{n},\xi\in I_{0}$
\end_inset
,
sledi
\begin_inset Formula $\lim_{n\to\infty}f\left(a_{n}\right)=f\left(\alpha\right)=f\left(\xi\right)=f\left(\beta\right)=\lim_{n\to\infty}f\left(b_{n}\right)$
\end_inset
.
\end_layout
\begin_layout Proof
Po točki
\begin_inset CommandInset ref
LatexCommand ref
reference "enu:različni-predznaki-istoležnih-clenov"
plural "false"
caps "false"
noprefix "false"
nolink "false"
\end_inset
velja
\begin_inset Formula $f\left(\alpha\right)f\left(\beta\right)\leq0$
\end_inset
.
Ker pa
\begin_inset Formula $f\left(\alpha\right)=f\left(\beta\right)$
\end_inset
,
velja
\begin_inset Formula $f\left(\alpha\right)=f\left(\beta\right)=f\left(\xi\right)=0$
\end_inset
.
\end_layout
\begin_layout Corollary*
Naj bo
\begin_inset Formula $I=\left[a,b\right]$
\end_inset
omejen zaprt interval
\begin_inset Formula $\in\mathbb{R}$
\end_inset
in
\begin_inset Formula $f:I\to\mathbb{R}$
\end_inset
zvezna.
Tedaj
\begin_inset Formula $\exists x_{-},x_{+}\in I\ni:\forall x\in I:f\left(x\right)\in\left[f\left(x_{-}\right),f\left(x_{+}\right)\right]$
\end_inset
in
\begin_inset Formula $\forall y\in\left[f\left(x_{-}\right),f\left(x_{+}\right)\right]\exists x\in I\ni:y=f\left(x\right)$
\end_inset
ZDB
\begin_inset Formula $f\left(I\right)=\left[f\left(x_{-}\right),f\left(x_{+}\right)\right]$
\end_inset
ZDB zvezna funkcija na zaprtem intervalu
\begin_inset Formula $\left[a,b\right]$
\end_inset
doseže vse funkcijske vrednosti na intervalu
\begin_inset Formula $\left[f\left(a\right),f\left(b\right)\right]$
\end_inset
.
\end_layout
\begin_layout Proof
Dokaz posledice.
Naj bo
\begin_inset Formula $y$
\end_inset
poljuben.
Če je
\begin_inset Formula $y=f\left(x_{-}\right)$
\end_inset
,
smo našli
\begin_inset Formula $x=x_{-}$
\end_inset
.
Če je
\begin_inset Formula $y=f\left(x_{+}\right)$
\end_inset
,
smo našli
\begin_inset Formula $x=x_{+}$
\end_inset
.
Sicer pa je
\begin_inset Formula $f\left(x_{-}\right)<y<f\left(x_{+}\right)$
\end_inset
.
Oglejmo si funkcijo
\begin_inset Formula $g\left(x\right)\coloneqq f\left(x\right)-y$
\end_inset
.
Ker je
\begin_inset Formula $g\left(x_{-}\right)=f\left(x_{-}\right)-y<0$
\end_inset
in
\begin_inset Formula $g\left(x_{+}\right)=f\left(x_{+}\right)-y>0$
\end_inset
in
\begin_inset Formula $g$
\end_inset
zvezna na
\begin_inset Formula $\left[x_{-}-y,x_{+}-y\right]$
\end_inset
,
torej po prejšnjem izreku
\begin_inset Formula $\exists x\in\left[x_{-}-y,x_{+}-y\right]\ni:g\left(x\right)=0$
\end_inset
,
kar pomeni ravno
\begin_inset Formula $f\left(x\right)=y$
\end_inset
.
\end_layout
\begin_layout Theorem*
Naj bo
\begin_inset Formula $I$
\end_inset
poljuben interval med
\begin_inset Formula $a,b\in\mathbb{R}\cup\left\{ -\infty,\infty\right\} $
\end_inset
in
\begin_inset Formula $f:I\to\mathbb{R}$
\end_inset
zvezna in strogo monotona.
Tedaj je
\begin_inset Formula $f\left(I\right)$
\end_inset
interval med
\begin_inset Formula $f\left(a+0\right)$
\end_inset
in
\begin_inset Formula $f\left(a-0\right)$
\end_inset
.
Inverzna funkcija
\begin_inset Formula $f^{-1}$
\end_inset
je definirana na
\begin_inset Formula $f\left(I\right)$
\end_inset
in zvezna.
\end_layout
\begin_layout Example*
\begin_inset Formula $f\coloneqq\arctan$
\end_inset
,
\begin_inset Formula $I\coloneqq\left(-\infty,\infty\right)$
\end_inset
,
zvezna.
Naj bo
\begin_inset Formula $y\in f\left(I\right)$
\end_inset
poljuben.
Tedaj
\begin_inset Formula $\exists!x\in I\ni:y=f\left(x\right)$
\end_inset
in definiramo
\begin_inset Formula $x\coloneqq f^{-1}\left(x\right)$
\end_inset
.
\begin_inset Formula $f^{-1}$
\end_inset
obstaja in je spet zvezna.
\end_layout
\begin_layout Proof
Ne bomo dokazali.
\begin_inset Note Note
status open
\begin_layout Plain Layout
Označimo
\begin_inset Formula $g=f^{-1}:f\left(I\right)\to\mathbb{R}$
\end_inset
.
Uporabimo
\begin_inset ERT
status open
\begin_layout Plain Layout
\backslash
hyperlink{kzsppom}{karakterizacijo zveznosti s pomočjo praslik odprtih množic}
\end_layout
\end_inset
.
Dokazujemo torej,
da
\begin_inset Formula $\forall V^{\text{odp.}}\subset\mathbb{R}:g^{-1}\left(V\right)$
\end_inset
je zopet odprta množica
\begin_inset Formula $\subseteq f\left(I\right)$
\end_inset
.
\end_layout
\begin_layout Proof
Velja
\begin_inset Formula $g^{-1}\left(V\right)=\left\{ x\in f\left(I\right);g\left(x\right)\in V\right\} =\left\{ x\in f\left(I\right):\exists v\in V\cap I\ni:x=f\left(v\right)\right\} =f\left(V\cap I\right)$
\end_inset
.
\end_layout
\begin_layout Proof
Torej dokazujemo
\begin_inset Formula $\forall V^{\text{odp.}}\subset\mathbb{R}:f\left(I\cap V\right)$
\end_inset
je spet zopet odprta
\begin_inset Formula $\subseteq f\left(I\right)$
\end_inset
,
kar je ekvivalentno
\begin_inset Formula
\[
\forall y\in f\left(I\cap V\right)\exists\delta>0\ni:\left(y-\delta,y+\delta\right)\cap f\left(I\right)\subset f\left(I\cap V\right).
\]
\end_inset
Pišimo
\begin_inset Formula $y=f\left(x\right),x\in I\cap V$
\end_inset
.
Privzemimo,
da
\begin_inset Formula $f$
\end_inset
narašča (če pada,
ravnamo podobno).
Ker jer
\begin_inset Formula $ $
\end_inset
\end_layout
\end_inset
\end_layout
\begin_layout Subsection
Enakomerna zveznost
\end_layout
\begin_layout Definition*
\begin_inset Formula $f:I\to\mathbb{R}$
\end_inset
je
\begin_inset ERT
status open
\begin_layout Plain Layout
\backslash
hypertarget{ez}{enakomerno zvezna}
\end_layout
\end_inset
na
\begin_inset Formula $I$
\end_inset
,
če
\begin_inset Formula
\[
\forall\varepsilon>0\exists\delta>0\forall x,y\in I:\left|x-y\right|<\delta\Rightarrow\left|f\left(x\right)-f\left(y\right)\right|<\varepsilon.
\]
\end_inset
\end_layout
\begin_layout Note*
Primerjajmo to z definicijo
\begin_inset Formula $f:I\to\mathbb{R}$
\end_inset
je (nenujno enakomerno) zvezna na
\begin_inset Formula $I$
\end_inset
,
če
\begin_inset Formula
\[
\forall\varepsilon>0,a\in I\exists\delta>0\forall x\in I:\left|x-a\right|<\delta\Rightarrow\left|f\left(x\right)-f\left(y\right)\right|<\varepsilon.
\]
\end_inset
Pri slednji definiciji je
\begin_inset Formula $\delta$
\end_inset
odvisna od
\begin_inset Formula $\varepsilon$
\end_inset
in
\begin_inset Formula $a$
\end_inset
,
pri enakomerni zveznosti pa le od
\begin_inset Formula $\varepsilon$
\end_inset
.
\end_layout
\begin_layout Example*
\begin_inset Formula $f\left(x\right)=\frac{1}{x}$
\end_inset
ni enakomerno zvezna,
ker je
\begin_inset Formula $\delta$
\end_inset
odvisen od
\begin_inset Formula $a$
\end_inset
.
Če pri fiksnem
\begin_inset Formula $\varepsilon$
\end_inset
pomaknemo tisto pozitivno točko,
v kateri preizkušamo zveznost,
bolj v levo,
bo na neki točki potreben ožji,
manjši
\begin_inset Formula $\delta$
\end_inset
\end_layout
\begin_layout Theorem*
Zvezna funkcija na kompaktni množici je enakomerno zvezna.
\end_layout
\begin_layout Proof
Naj bo
\begin_inset Formula $f:K\to\mathbb{R}$
\end_inset
zvezna,
kjer je
\begin_inset Formula $K$
\end_inset
kompaktna podmnožica
\begin_inset Formula $\mathbb{R}$
\end_inset
.
PDDRAA
\begin_inset Formula $f$
\end_inset
ni enakomerno zvezna.
Zanikajmo definicijo
\begin_inset ERT
status open
\begin_layout Plain Layout
\backslash
hyperlink{ez}{enakomerne zveznosti}
\end_layout
\end_inset
:
\begin_inset Formula $\exists\varepsilon>0\forall\delta>0\exists x_{\delta},y_{\delta}\in I:\left|x_{\delta}-y_{\delta}\right|<\delta\wedge\left|f\left(x_{\delta}\right)-f\left(y_{\delta}\right)\right|\geq\varepsilon$
\end_inset
.
\begin_inset Formula $x,y$
\end_inset
sta seveda lahko odvisna od
\begin_inset Formula $\delta$
\end_inset
in
\begin_inset Formula $\varepsilon$
\end_inset
,
zato v subskriptu pišemo
\begin_inset Formula $\delta$
\end_inset
,
ki ji pripadata.
Ker smo dejali,
da to velja,
si oglejmo
\begin_inset Formula $\forall n\in\mathbb{N}:\delta_{n}\coloneqq\frac{1}{n}$
\end_inset
in pripadajoči zaporedji
\begin_inset Formula $\left(x_{1/n}\right)_{n\in\mathbb{N}}$
\end_inset
in
\begin_inset Formula $\left(y_{1/n}\right)_{n\in\mathbb{N}}$
\end_inset
.
Ker je
\begin_inset Formula $K$
\end_inset
kompaktna,
ima zaporedje
\begin_inset Formula $\left(x_{1/n}\right)_{n\in\mathbb{N}}$
\end_inset
stekališče v
\begin_inset Formula $x\in K$
\end_inset
,
torej obstaja podzaporede
\begin_inset Formula $\left(x_{1/n_{k}}\right)_{k\in\mathbb{N}}$
\end_inset
,
ki konvergira k
\begin_inset Formula $x$
\end_inset
.
Podobno obstaja podzaporedje
\begin_inset Formula $\left(y_{1/n_{k_{l}}}\right)_{l\in\mathbb{N}}$
\end_inset
,
ki konvergira k
\begin_inset Formula $y\in K$
\end_inset
.
Pišimo sedaj
\begin_inset Formula $x_{l}\coloneqq x_{1/n_{k_{l}}}$
\end_inset
in
\begin_inset Formula $y_{l}\coloneqq y_{1/n_{k_{l}}}$
\end_inset
.
\end_layout
\begin_layout Proof
Velja torej
\begin_inset Formula $x_{l}\to x$
\end_inset
in
\begin_inset Formula $y_{l}\to y$
\end_inset
.
Sledi
\begin_inset Formula $\left|x-y\right|\leq\lim_{l\to\infty}\left(\left|x-x_{l}\right|+\left|x_{l}-y_{l}\right|+\left|y_{l}-y\right|\right)$
\end_inset
.
Levi in desni člen sta v limiti enaka 0 zaradi konvergence zaporedja,
srednji pa je manjši od
\begin_inset Formula $\frac{1}{j}$
\end_inset
zaradi naše predpostavke (PDDRAA),
potemtakem je
\begin_inset Formula $x=y$
\end_inset
.
\end_layout
\begin_layout Proof
Zato
\begin_inset Formula $\lim_{l\to\infty}\left(f\left(x_{l}\right)-f\left(y_{l}\right)\right)=\lim_{l\to\infty}\left[\left(f\left(x_{l}\right)-f\left(x\right)\right)+\left(f\left(x\right)-f\left(y\right)\right)+\left(f\left(y\right)-f\left(y_{l}\right)\right)\right]$
\end_inset
.
Levi in desni člen sta v limiti enaka 0 zaradi konvergence zaporedja in zveznosti
\begin_inset Formula $f$
\end_inset
,
srednji pa je tudi 0,
ker
\begin_inset Formula $x=y$
\end_inset
,
potemtakem
\begin_inset Formula $f\left(x_{l}\right)-f\left(y_{l}\right)\to0$
\end_inset
,
kar je v protislovju z
\begin_inset Formula $\left|f\left(x_{l}\right)-f\left(y_{l}\right)\right|\geq\varepsilon$
\end_inset
za fiksen
\begin_inset Formula $\varepsilon$
\end_inset
in
\begin_inset Formula $\forall l\in\mathbb{N}$
\end_inset
.
\begin_inset Formula $\rightarrow\!\leftarrow$
\end_inset
,
\begin_inset Formula $f$
\end_inset
je enakomerno zvezna.
\end_layout
\begin_layout Corollary*
En zaprt interval
\begin_inset Formula $\frac{1}{x}$
\end_inset
bo enakomerno zvezen,
\begin_inset Formula $\frac{1}{x}$
\end_inset
sama po sebi kot
\begin_inset Formula $\left(0,\infty\right)\to\mathbb{R}$
\end_inset
pa ni definirana na kompaktni množici.
Prav tako
\begin_inset Formula $\arcsin$
\end_inset
in
\begin_inset Formula $x\mapsto\sqrt{x}$
\end_inset
.
\end_layout
\begin_layout Section
Odvod
\end_layout
\begin_layout Standard
Najprej razmislek/ideja.
Odvod je hitrost/stopnja,
s katero se v danem trenutku neka količina spreminja.
\end_layout
\begin_layout Standard
\begin_inset Float figure
placement document
alignment document
wide false
sideways false
status open
\begin_layout Plain Layout
TODO XXX FIXME SKICA S TKZ EUCLID (ali pa —
bolje —
s čim drugim),
glej PS zapiski/ANA1P FMF 2023-12-04.pdf
\end_layout
\begin_layout Plain Layout
\begin_inset Caption Standard
\begin_layout Plain Layout
Skica.
\end_layout
\end_inset
\end_layout
\end_inset
\end_layout
\begin_layout Standard
Radi bi določili naklon sekante,
torej naklon premice,
določene z
\begin_inset Formula $x$
\end_inset
in neko bližnjo točko
\begin_inset Formula $x+h$
\end_inset
na grafu funkcije,
ki je odvisen le od
\begin_inset Formula $x$
\end_inset
,
ne pa tudi od izbire
\begin_inset Formula $h$
\end_inset
.
Bližnjo točko pošljemo proti začetni —
\begin_inset Formula $h$
\end_inset
pošljemo proti 0.
Naklon izračunamo s izrazom
\begin_inset Formula $\frac{f\left(x+h\right)-f\left(x\right)}{h}$
\end_inset
.
\end_layout
\begin_layout Definition*
Odvod funkcije
\begin_inset Formula $f$
\end_inset
v točki
\begin_inset Formula $x$
\end_inset
označimo
\begin_inset Formula $f'\left(x\right)\coloneqq\lim_{h\to0}\frac{f\left(x+h\right)-f\left(x\right)}{h}$
\end_inset
.
Če limita obstaja v točki
\begin_inset Formula $x$
\end_inset
,
pravimo,
da je funkcija odvedljiva v
\begin_inset Formula $x$
\end_inset
.
Pravimo,
da je
\begin_inset Formula $f$
\end_inset
odvedljiva na množici
\begin_inset Formula $I\subseteq\mathbb{R}$
\end_inset
,
če je odvedljiva na vsaki
\begin_inset Formula $t\in I$
\end_inset
.
\end_layout
\begin_layout Example*
Primeri odvodov preprostih funkcij.
\end_layout
\begin_deeper
\begin_layout Labeling
\labelwidthstring 00.00.0000
\begin_inset Formula $f\left(x\right)=c,c\in\mathbb{R}$
\end_inset
\begin_inset Formula $f'\left(x\right)=\lim_{h\to0}\frac{\cancelto{c}{f\left(x+h\right)}-\cancelto{c}{f\left(x\right)}}{h}=0$
\end_inset
\end_layout
\begin_layout Labeling
\labelwidthstring 00.00.0000
\begin_inset Formula $f\left(x\right)=x$
\end_inset
\begin_inset Formula $f'\left(x\right)=\lim_{h\to0}\frac{f\left(x+h\right)-f\left(x\right)}{h}=\lim_{h\to0}\frac{x+h-x}{h}=1$
\end_inset
\end_layout
\begin_layout Labeling
\labelwidthstring 00.00.0000
\begin_inset Formula $f\left(x\right)=x^{2}$
\end_inset
\begin_inset Formula $f'\left(x\right)=\lim_{h\to0}\frac{f\left(x+h\right)-f\left(x\right)}{h}=\lim_{h\to0}\frac{x^{2}+2xh+h^{2}-x^{2}}{h}=\lim_{h\to0}\frac{2xh+h^{2}}{h}=\lim_{h\to0}2x+h=2x$
\end_inset
\end_layout
\end_deeper
\begin_layout Claim*
Za poljuben
\begin_inset Formula $n\in\mathbb{N}$
\end_inset
so funkcije
\begin_inset Formula $f\left(x\right)=x^{n}$
\end_inset
odvedljive na
\begin_inset Formula $\mathbb{R}$
\end_inset
in velja
\begin_inset Formula $f'\left(x\right)=nx^{n-1}$
\end_inset
.
\end_layout
\begin_layout Proof
\begin_inset Formula
\[
\lim_{h\to0}\frac{f\left(x+h\right)-f\left(x\right)=\left(x+h\right)^{n}-x^{n}=\sum_{k=0}^{n}\binom{n}{k}h^{k}x^{n-k}-x^{n}=\cancel{x^{n}}+nhx^{n-1}+\sum_{k=2}^{n}\binom{n}{k}h^{k}x^{n-k}\cancel{-x^{n}}}{h}=
\]
\end_inset
\begin_inset Formula
\[
=\lim_{h\to0}\frac{nhx^{n-1}+\sum_{k=2}^{n}\binom{n}{k}h^{k}x^{n-k}}{h}=\lim_{h\to0}\frac{\cancel{h}\left(nx^{n-1}+\sum_{k=2}^{n}\binom{n}{k}h^{k-1}x^{n-k}\right)}{\cancel{h}}=\lim_{h\to0}\left(nx^{n-1}+\sum_{k=2}^{n}\binom{n}{k}h^{k-1}x^{n-k}\right)=
\]
\end_inset
\begin_inset Formula
\[
=nx^{n-1}+\cancel{\lim_{h\to0}\sum_{k=2}^{n}\binom{n}{k}\cancelto{0}{h^{k-1}}x^{n-k}}=nx^{n-1}
\]
\end_inset
\end_layout
\begin_layout Claim*
\begin_inset Formula $\sin'=\cos$
\end_inset
,
\begin_inset Formula $\cos'=-\sin$
\end_inset
\end_layout
\begin_layout Proof
Najprej dokažimo
\begin_inset Formula $\sin'=\cos$
\end_inset
.
\begin_inset Formula
\[
\lim_{h\to\infty}\frac{\sin\left(x+h\right)-\sin\left(x\right)=\sin x\cos h+\sin h\cos x-\sin x=\sin x\left(\cos h-1\right)+\sin h\cos x}{h}=
\]
\end_inset
\begin_inset Formula
\[
=\lim_{h\to0}\left(\sin x\frac{\cos h-1}{h}+\cos x\frac{\sin h}{h}\right)=\lim_{h\to0}\left(\sin x\frac{\left(\cos h-1\right)\left(\cos h+1\right)=\cos^{2}h-1=-\sin^{2}h}{h\left(\cos h+1\right)}+\cos x\frac{\sin h}{h}\right)=
\]
\end_inset
\begin_inset Formula
\[
=\lim_{h\to0}\left(\sin x\frac{-\sin h}{h}\cdot\frac{\sin h}{\cos h+1}+\cos x\frac{\sin h}{h}\right)=\lim_{h\to0}\cancelto{1}{\frac{\sin h}{h}}\left(\cos x-\cancel{\sin x\frac{\cancelto{0}{\sin h}}{\cos h+1}}\right)=\cos x
\]
\end_inset
Sedaj dokažimo še
\begin_inset Formula $\cos'=-\sin$
\end_inset
.
\begin_inset Formula
\[
\lim_{h\to0}\frac{\cos\left(x+h\right)-\cos\left(x\right)=\cos x\cos h-\sin x\sin h-\cos x=\cos x\left(\cos h-1\right)-\sin x\sin h}{h}=
\]
\end_inset
\begin_inset Formula
\[
=\lim_{h\to0}\left(\cos x\frac{\cos h-1}{h}-\sin x\frac{\sin h}{h}\right)=\lim_{h\to0}\left(\cos x\frac{\left(\cos h-1\right)\left(\cos h+1\right)=\cos^{2}h-1=-\sin^{2}h}{h\left(\cos h+1\right)}-\sin x\frac{\sin h}{h}\right)=
\]
\end_inset
\begin_inset Formula
\[
=\lim_{h\to0}\left(\cos x\frac{-\sin h}{h}\cdot\frac{\sin^{}h}{\cos h+1}-\sin x\frac{\sin h}{h}\right)=\lim_{h\to0}\left(\cancelto{1}{\frac{\sin h}{h}}\left(\cancel{-\cos x\frac{\cancelto{0}{\sin h}}{\cos h+1}}-\sin x\right)\right)=-\sin x
\]
\end_inset
\end_layout
\begin_layout Fact*
Od prej vemo
\begin_inset Formula $\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^{n}=e$
\end_inset
(limita zaporedja).
Velja tudi
\begin_inset Formula $\lim_{x\to\infty}\left(1+\frac{1}{x}\right)^{x}=e$
\end_inset
(funkcijska limita).
Ne bomo dokazali.
\end_layout
\begin_layout Claim*
Naj bo
\begin_inset Formula $a>0$
\end_inset
in
\begin_inset Formula $f\left(x\right)=a^{x}$
\end_inset
.
Tedaj je
\begin_inset Formula $f'\left(x\right)=a^{x}\ln a$
\end_inset
.
\end_layout
\begin_layout Proof
\begin_inset Formula
\[
\lim_{h\to0}\frac{f\left(x+h\right)-f\left(x\right)=a^{x}a^{h}-a^{x}}{h}=\lim_{h\to0}a^{x}\frac{a^{h}-1}{h}=\cdots
\]
\end_inset
Sedaj pišimo
\begin_inset Formula $\frac{1}{z}\coloneqq a^{h}-1$
\end_inset
.
Ulomek
\begin_inset Formula $\frac{a^{h}-1}{h}$
\end_inset
namreč ni odvisen od
\begin_inset Formula $x$
\end_inset
.
Sedaj
\begin_inset Formula
\[
a^{h}-1=\frac{1}{z}
\]
\end_inset
\begin_inset Formula
\[
a^{h}=\frac{1}{z}+1
\]
\end_inset
\begin_inset Formula
\[
h=\log_{a}\left(\frac{1}{z}+1\right)
\]
\end_inset
\begin_inset Formula
\[
h=\frac{\ln\left(\frac{1}{z}+1\right)}{\ln a}
\]
\end_inset
Nadaljujmo s prvotnim računom,
ločimo primere:
\end_layout
\begin_deeper
\begin_layout Labeling
\labelwidthstring 00.00.0000
\begin_inset Formula $a>1,h\searrow0$
\end_inset
Potemtakem
\begin_inset Formula $a^{h}-1\searrow0$
\end_inset
,
torej
\begin_inset Formula $\frac{1}{z}\searrow0$
\end_inset
,
sledi
\begin_inset Formula $z\nearrow\infty$
\end_inset
.
\begin_inset Formula
\[
\cdots=a^{x}\lim_{z\to\infty}\frac{\frac{1}{z}}{\frac{\ln\left(\frac{1}{z}+1\right)}{\ln a}}=a^{x}\lim_{z\to\infty}\frac{\frac{1}{z}\ln a}{\ln\left(\frac{1}{z}+1\right)}=a^{x}\lim_{z\to\infty}\frac{\ln a}{\ln\left(\frac{1}{z}+1\right)^{z}}=a^{x}\lim_{z\to\infty}\frac{\ln a}{\ln e}=a^{x}\ln a
\]
\end_inset
\end_layout
\begin_layout Labeling
\labelwidthstring 00.00.0000
\begin_inset Formula $a>1,h\nearrow0$
\end_inset
Potemtakem
\begin_inset Formula $a^{h}-1\nearrow0$
\end_inset
,
torej
\begin_inset Formula $\frac{1}{z}\nearrow0$
\end_inset
,
sledi
\begin_inset Formula $z\searrow-\infty$
\end_inset
.
\begin_inset Formula
\[
\cdots=a^{x}\lim_{z\to-\infty}\frac{\ln a}{\ln\left(\frac{1}{z}+1\right)^{z}}=a^{x}\lim_{z\to-\infty}\frac{\ln a}{\ln\cancelto{e}{\left(\frac{1}{z}+1\right)^{z}}}=a^{x}\ln a
\]
\end_inset
Kajti
\begin_inset Formula $\lim_{x\to\infty}\left(1+\frac{k}{x}\right)^{x}=e^{k}$
\end_inset
.
\end_layout
\begin_layout Labeling
\labelwidthstring 00.00.0000
\begin_inset Formula $a\in(0,1]$
\end_inset
Podobno kot zgodaj,
bodisi
\begin_inset Formula $z\nearrow\infty$
\end_inset
bodisi
\begin_inset Formula $z\searrow-\infty$
\end_inset
.
\end_layout
\end_deeper
\begin_layout Claim*
Če je
\begin_inset Formula $f$
\end_inset
odvedljiva v točki
\begin_inset Formula $x$
\end_inset
,
je tam tudi zvezna.
\end_layout
\begin_layout Proof
Predpostavimo,
da obstaja limita
\begin_inset Formula $\lim_{h\to0}\frac{f\left(x+h\right)-f\left(x\right)}{h}$
\end_inset
.
Želimo dokazati
\begin_inset Formula $f\left(x\right)=\lim_{t\to x}f\left(t\right)$
\end_inset
.
Računajmo:
\begin_inset Formula
\[
f\left(x\right)=\lim_{t\to x}f\left(t\right)
\]
\end_inset
\begin_inset Formula
\[
0=\lim_{t\to x}f\left(t\right)-f\left(x\right)
\]
\end_inset
\begin_inset Formula
\[
0=\lim_{h\to0}f\left(x+h\right)-f\left(x\right)
\]
\end_inset
\begin_inset Formula
\[
0=\lim_{h\to0}\left(f\left(x+h\right)-f\left(x\right)\right)=\lim_{h\to0}\left(\frac{f\left(x+h\right)-f\left(x\right)}{h}\cdot h\right)
\]
\end_inset
Limita obstaja,
čim obstajata
\begin_inset Formula $\lim_{h\to0}\frac{f\left(x+h\right)-f\left(x\right)}{h}$
\end_inset
,
ki obstaja po predpostavki,
in
\begin_inset Formula $\lim_{h\to0}h$
\end_inset
,
ki obstaja in ima vrednost
\begin_inset Formula $0$
\end_inset
.
\end_layout
\begin_layout Example*
\begin_inset Formula $f\left(x\right)=\left|x\right|=\sqrt{x^{2}}$
\end_inset
.
Je zvezna,
ker je kompozitum zveznih funkcij,
toda v
\begin_inset Formula $0$
\end_inset
ni odvedljiva,
kajti
\begin_inset Formula $\lim_{h\to0}\frac{f\left(0+h\right)-f\left(0\right)}{h}=\lim_{h\to0}\frac{\left|h\right|-0}{h}=\lim_{h\to0}\sgn h$
\end_inset
.
Limita ne obstaja,
ker
\begin_inset Formula $-1=\lim_{h\nearrow0}\sgn h\not=\lim_{h\searrow0}\sgn h=1$
\end_inset
.
\end_layout
\begin_layout Theorem*
Naj bosta
\begin_inset Formula $f,g$
\end_inset
odvedljivi v
\begin_inset Formula $x\in\mathbb{R}$
\end_inset
.
Tedaj so
\begin_inset Formula $f+g,f-g,f\cdot g,f/g$
\end_inset
(slednja le,
če
\begin_inset Formula $g\left(x\right)\not=0$
\end_inset
) in velja
\begin_inset Formula $\left(f\pm g\right)'=f'\pm g'$
\end_inset
,
\begin_inset Formula $\left(fg\right)'=f'g+fg'$
\end_inset
,
\begin_inset Formula $\left(f/g\right)'=\frac{f'g-fg'}{g^{2}}$
\end_inset
.
\end_layout
\begin_layout Proof
Dokažimo vse štiri trditve.
\end_layout
\begin_deeper
\begin_layout Labeling
\labelwidthstring 00.00.0000
\begin_inset Formula $f+g$
\end_inset
Velja
\begin_inset Formula $\left(f+g\right)\left(x\right)=f\left(x\right)+g\left(x\right)$
\end_inset
.
\begin_inset Formula
\[
\left(f+g\right)'\left(x\right)=\lim_{h\to0}\frac{\left(f+g\right)\left(x+h\right)-\left(f+g\right)\left(x\right)=f\left(x+h\right)+g\left(x+h\right)-f\left(x\right)-g\left(x\right)}{h}=
\]
\end_inset
\begin_inset Formula
\[
=\lim_{h\to0}\left(\frac{f\left(x+h\right)-f\left(x\right)}{h}+\frac{g\left(x+h\right)-g\left(x\right)}{h}\right)=f\left(x\right)'+g\left(x\right)'
\]
\end_inset
\end_layout
\begin_layout Labeling
\labelwidthstring 00.00.0000
\begin_inset Formula $-f$
\end_inset
Naj bo
\begin_inset Formula $g=-f$
\end_inset
.
\begin_inset Formula $g'\left(x\right)=\lim_{h\to0}\frac{g\left(x+h\right)-g\left(x\right)}{h}=\lim_{h\to0}\frac{-f\left(x+h\right)+f\left(x\right)}{h}=-\lim_{h\to0}\frac{f\left(x+h\right)-f\left(x\right)}{h}=-f\left(x\right)'$
\end_inset
,
zato
\begin_inset Formula
\[
\left(f-g\right)'\left(x\right)=\left(f+\left(-g\right)\right)'\left(x\right)=f'\left(x\right)+\left(-g\right)'\left(x\right)=f'\left(x\right)-g'\left(x\right)
\]
\end_inset
\end_layout
\begin_layout Labeling
\labelwidthstring 00.00.0000
\begin_inset Formula $f\cdot g$
\end_inset
Velja
\begin_inset Formula $\left(fg\right)\left(x\right)=f\left(x\right)g\left(x\right)$
\end_inset
.
Prištejemo in odštejemo isti izraz (v oglatih oklepajih).
\begin_inset Formula
\[
\left(fg\right)'\left(x\right)=\lim_{h\to0}\frac{\left(fg\right)\left(x+h\right)-\left(fg\right)\left(x\right)=f\left(x+h\right)g\left(x+h\right)-f\left(x\right)g\left(x\right)+\left[f\left(x\right)g\left(x+h\right)-f\left(x\right)g\left(x+h\right)\right]}{h}=
\]
\end_inset
\begin_inset Formula
\[
=\lim_{h\to0}\frac{g\left(x+h\right)\left(f\left(x+h\right)-f\left(x\right)\right)+f\left(x\right)\left(g\left(x+h\right)-g\left(x\right)\right)}{h}=
\]
\end_inset
\begin_inset Formula
\[
=\lim_{h\to0}\frac{f\left(x+h\right)-f\left(x\right)}{h}\cancelto{g\left(x\right)}{g\left(x+h\right)}+\lim_{h\to0}\frac{g\left(x+h\right)-g\left(x\right)}{h}f\left(x\right)=f'\left(x\right)g\left(x\right)+g'\left(x\right)f\left(x\right)
\]
\end_inset
\end_layout
\begin_layout Labeling
\labelwidthstring 00.00.0000
\begin_inset Formula $f/g$
\end_inset
Velja
\begin_inset Formula $\left(f/g\right)\left(x\right)=f\left(x\right)/g\left(x\right)$
\end_inset
.
Prištejemo in odštejemo isti izraz (v oglatih oklepajih).
\begin_inset Formula
\[
\left(f/g\right)'\left(x\right)=\lim_{h\to0}\frac{\left(f/g\right)\left(x+h\right)-\left(f/g\right)\left(x\right)=\frac{f\left(x+h\right)}{g\left(x+h\right)}-\frac{f\left(x\right)}{g\left(x\right)}=\frac{f\left(x+h\right)g\left(x\right)}{g\left(x+h\right)g\left(x\right)}-\frac{f\left(x\right)g\left(x+h\right)}{g\left(x\right)g\left(x+h\right)}=\frac{f\left(x+h\right)g\left(x\right)-f\left(x\right)g\left(x+h\right)}{g\left(x\right)g\left(x+h\right)}}{h}=
\]
\end_inset
\begin_inset Formula
\[
=\lim_{h\to0}\frac{f\left(x+h\right)g\left(x\right)-f\left(x\right)g\left(x+h\right)=f\left(x+h\right)g\left(x\right)-f\left(x\right)g\left(x+h\right)+\left[f\left(x\right)g\left(x\right)-f\left(x\right)g\left(x\right)\right]}{hg\left(x\right)g\left(x+h\right)}=
\]
\end_inset
\begin_inset Formula
\[
=\lim_{h\to0}\left(\frac{f\left(x+h\right)-f\left(x\right)}{h}\cdot\frac{g\left(x\right)}{g\left(x\right)g\left(x+h\right)}-\frac{g\left(x+h\right)-g\left(x\right)}{h}\cdot\frac{f\left(x\right)}{g\left(x\right)g\left(x+h\right)}\right)=
\]
\end_inset
\begin_inset Formula
\[
=\lim_{h\to0}\left(\left(\frac{1}{g\left(x\right)g\left(x+h\right)}\right)\left(\frac{f\left(x+h\right)-f\left(x\right)}{h}g\left(x\right)-\frac{g\left(x+h\right)-g\left(x\right)}{h}f\left(x\right)\right)\right)=
\]
\end_inset
\begin_inset Formula
\[
=\frac{1}{g^{2}\left(x\right)}\left(f'\left(x\right)g\left(x\right)-g'\left(x\right)f\left(x\right)\right)=\frac{f'\left(x\right)g\left(x\right)-f\left(x\right)g'\left(x\right)}{g^{2}\left(x\right)}
\]
\end_inset
\end_layout
\end_deeper
\begin_layout Example*
\begin_inset Formula $\tan'\left(x\right)=\left(\frac{\sin\left(x\right)}{\cos\left(x\right)}\right)'=\frac{\sin'\left(x\right)\cos\left(x\right)-\sin\left(x\right)\cos'\left(x\right)}{\cos^{2}\left(x\right)}=\frac{\cos^{2}\left(x\right)+\sin^{2}\left(x\right)}{\cos^{2}\left(x\right)}=\cos^{-2}\left(x\right)$
\end_inset
.
\end_layout
\begin_layout Theorem*
Naj bo
\begin_inset Formula $f$
\end_inset
odvedljiva v
\begin_inset Formula $x$
\end_inset
in
\begin_inset Formula $g$
\end_inset
odvedljiva v
\begin_inset Formula $f\left(x\right)$
\end_inset
.
Tedaj je
\begin_inset Formula $g\circ f$
\end_inset
odvedljiva v
\begin_inset Formula $x$
\end_inset
in velja
\begin_inset Formula $\left(g\circ f\right)\left(x\right)=g'\left(f\left(x\right)\right)\cdot f'\left(x\right)$
\end_inset
(opomba:
\begin_inset Formula $\left(g\circ f\right)\left(x\right)=g\left(f\left(x\right)\right)$
\end_inset
).
\end_layout
\begin_layout Proof
Označimo
\begin_inset Formula $a\coloneqq f\left(x\right)$
\end_inset
in
\begin_inset Formula $\delta_{h}\coloneqq f\left(x+h\right)-f\left(x\right)$
\end_inset
,
torej
\begin_inset Formula $f\left(x+h\right)\coloneqq a+\delta\left(h\right)$
\end_inset
.
\begin_inset Formula
\[
\lim_{h\to0}\frac{\left(g\circ f\right)\left(x+h\right)-\left(g\circ f\right)\left(x\right)=g\left(f\left(x+h\right)\right)-g\left(f\left(x\right)\right)=g\left(a+\delta_{h}\right)-g\left(a\right)}{h}=\lim_{h\to0}\frac{g\left(a+\delta_{h}\right)-g\left(a\right)}{\delta_{h}}\cdot\frac{\delta_{h}}{h}=
\]
\end_inset
\begin_inset Formula
\[
=\lim_{h\to0}\frac{g\left(a+\delta_{h}\right)-g\left(a\right)}{\delta_{h}}\cdot\frac{f\left(x+h\right)-f\left(x\right)}{h}=\cdots
\]
\end_inset
Ker je
\begin_inset Formula $f$
\end_inset
odvedljiva v
\begin_inset Formula $x$
\end_inset
,
je v
\begin_inset Formula $x$
\end_inset
zvezna,
zato sledi
\begin_inset Formula $h\to0\Rightarrow\delta_{h}\to0$
\end_inset
,
torej
\begin_inset Formula
\[
\cdots=g'\left(a\right)\cdot f'\left(x\right)=g'\left(f\left(x\right)\right)\cdot f'\left(x\right)
\]
\end_inset
\end_layout
\begin_layout Example*
\begin_inset Formula $\varphi\left(x\right)=\sin\left(x^{2}\right)=\left(g\circ f\right)\left(x\right),f\left(x\right)=x^{2},g\left(x\right)=\sin x$
\end_inset
in velja
\begin_inset Formula $\varphi'\left(x\right)=g'\left(f\left(x\right)\right)f'\left(x\right)=\sin'\left(x^{2}\right)\left(x^{2}\right)'=2x\cos\left(x^{2}\right)$
\end_inset
.
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Example*
\begin_inset Formula $\psi\left(x\right)=\sin^{2}\left(x\right)=\left(g\circ f\right)\left(x\right),f\left(x\right)=\sin,g\left(x\right)=x^{2}$
\end_inset
in velja
\begin_inset Formula $\psi'\left(x\right)=g'\left(f\left(x\right)\right)f'\left(x\right)=2\sin x\cos x=\sin2x$
\end_inset
(sinus dvojnega kota)
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Example*
\begin_inset Formula $\delta'\left(x\right)=\sin\left(e^{x^{2}}\right)=\sin\left(e^{\left(x^{2}\right)}\right)=\left(g\circ h\circ f\right)\left(x\right),g\left(x\right)=\sin x,h\left(x\right)=e^{x},f\left(x\right)=x^{2}$
\end_inset
.
\begin_inset Formula $\delta'\left(x\right)=\cos\left(e^{x^{2}}\right)e^{x^{2}}2x$
\end_inset
,
kajti
\begin_inset Formula $\left(e^{x}\right)'=e^{x}$
\end_inset
.
\end_layout
\begin_layout Definition*
Funkcija
\begin_inset Formula $f:I\subseteq\mathbb{R}\to\mathbb{R}$
\end_inset
je zvezno odvedljiva na
\begin_inset Formula $I$
\end_inset
,
če je na
\begin_inset Formula $I$
\end_inset
odvedljiva in je
\begin_inset Formula $f'$
\end_inset
na
\begin_inset Formula $I$
\end_inset
zvezna.
\end_layout
\begin_layout Example*
\begin_inset Formula $f\left(x\right)=\begin{cases}
x^{2}\sin\frac{1}{x} & ;x\not=0\\
0 & ;x=0
\end{cases}$
\end_inset
je na
\begin_inset Formula $\mathbb{R}$
\end_inset
odvedljiva,
a ne zvezno.
Odvedljivost na
\begin_inset Formula $\mathbb{R}\setminus\left\{ 0\right\} $
\end_inset
je očitna,
preverimo še odvedljivost v
\begin_inset Formula $0$
\end_inset
:
\begin_inset Formula
\[
f'\left(0\right)=\lim_{h\to0}\frac{f\left(h\right)-0}{h}=\lim_{h\to0}\cdots\text{NADALJUJEM JUTRI}
\]
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Corollary*
sssssssssss
\end_layout
\begin_layout Corollary*
sssssssssss
\end_layout
\begin_layout Corollary*
sssssssssss
\end_layout
\begin_layout Corollary*
sssssssssss
\end_layout
\begin_layout Corollary*
sssssssssss
\end_layout
\begin_layout Corollary*
sssssssssss
\end_layout
\begin_layout Corollary*
sssssssssss
\end_layout
\begin_layout Corollary*
sssssssssss
\end_layout
\end_body
\end_document
