\documentclass[fullpage,10pt, openany]{book} \usepackage{amsmath,etex} \usepackage[usenames]{color} %%%%% Postscript drawing packages \usepackage{psboxit} \usepackage{pstricks} \usepackage{pstricks-add} \usepackage{fancybox} %%%%Note: Eventually I might convert all the pstricks drawings into metafont. Who knows. \usepackage{hangcaption} \usepackage[amsmath,thmmarks,standard,thref]{ntheorem} \usepackage[colorlinks=true,linkcolor=blue, pdftitle={David A. Santos Calculus Once Again Analysis Notes}, pdfauthor={david santos}, bookmarksopen, dvips]{hyperref} \usepackage{pdflscape} \usepackage{thumbpdf} %\usepackage[dvips]{graphicx} %%%%% FONTS \usepackage[latin1]{inputenc} % Fourier for math | Utopia (scaled) for rm | Helvetica for ss | Latin Modern for tt \usepackage{fourier} % math & rm \usepackage[scaled=0.875]{helvet} % ss \renewcommand{\ttdefault}{lmtt} %tt %\usepackage{marvosym} \usepackage{pifont} \usepackage{mathrsfs} \usepackage{stmaryrd} \usepackage{amssymb} \usepackage[tiling]{pst-fill} % PSTricks package for filling/tiling \usepackage{pst-text} % PSTricks package for character path \usepackage{pst-grad} % PSTricks package for gradient filling \usepackage{multicol} \everymath{\displaystyle} \mathversion{bold} \newcommand{\WriteBig}[5] {% \DeclareFixedFont{\bigsf}{T1}{phv}{b}{n}{#2}% \DeclareFixedFont{\smallrm}{T1}{ptm}{(m)}{n}{#3}% \psboxfill{\smallrm #1} % \centerline{% \pscharpath[fillstyle=gradient,% gradangle=-45,% gradmidpoint=0.5,% addfillstyle=boxfill,% gradbegin=#4,% gradend=#5,% fillangle=45,% fillsep=0.7mm]{\rput[b](0,0){\bigsf#1}}} } %%%%%FLOAT PLACEMENT \renewcommand{\textfraction}{0.05} \renewcommand{\topfraction}{0.95} \renewcommand{\bottomfraction}{0.95} \renewcommand{\floatpagefraction}{0.35} \setcounter{totalnumber}{5} %%%%%%%%%PRINTED AREA \topmargin -.6in \textheight 9.4in \oddsidemargin -.3in \evensidemargin -.3in \textwidth 7in %%%%% THEOREM-LIKE ENVIRONMENTS \newcommand{\proofsymbol}{\Pisymbol{pzd}{113}} \newtheorem{pro}{Problem}[section] \theorempreskipamount .5cm \theorempostskipamount .5cm \theoremstyle{change} \theoremheaderfont{\sffamily\bfseries} \theorembodyfont{\normalfont} \newtheorem{thm}{\textbf{\textsc{\red Theorem}}} \newtheorem{prop}[thm]{\textbf{\textsc{\magenta Definition-Proposition}}} \newtheorem{cor}[thm]{\textbf{\textsc{\blue Corollary}}} \newtheorem{df}[thm]{Definition} \newtheorem{axi}[thm]{Axiom} \newtheorem{exa}[thm]{Example} \newtheorem{lem}[thm]{\textbf{\textsc{\green Lemma}}} \newenvironment{pf}[0]{\itshape\begin{quote}{\bf Proof: \ }}{\proofsymbol\end{quote}} \newenvironment{f-pf}[0]{\itshape\begin{quote}{\bf First Proof: \ }}{\proofsymbol\end{quote}} \newenvironment{s-pf}[0]{\itshape\begin{quote}{\bf Second Proof: \ }}{\proofsymbol\end{quote}} \newenvironment{t-pf}[0]{\itshape\begin{quote}{\bf Third Proof: \ }}{\proofsymbol\end{quote}} \newenvironment{fo-pf}[0]{\itshape\begin{quote}{\bf Fourth Proof: \ }}{\proofsymbol\end{quote}} \newenvironment{fi-pf}[0]{\itshape\begin{quote}{\bf Fifth Proof: \ }}{\proofsymbol\end{quote}} \newenvironment{si-pf}[0]{\itshape\begin{quote}{\bf Sixth Proof: \ }}{\proofsymbol\end{quote}} \newenvironment{solu}[0]{\begin{quote}{\bf Solution: \ } \itshape }{\end{quote}} \newenvironment{rem}[0]{\begin{quote}{\huge\textcolor{red}{\Pisymbol{pzd}{43}}}\small\itshape }{\end{quote}} \newcommand{\QUOTEME}[2]{\begin{quote}{\it\textbf{#1}}\nolinebreak[1] \blue\hspace*{\fill} \mbox{-\textsl{#2}} \hspace*{\fill}\end{quote}} \usepackage{answers} \Newassociation{answer}{Answer}{calculillo} %%Note: I had to comment-out marvosym.sty on the line \def\Rightarrow{{\mvchr58}} %%because it wasn't giving the standard \implies command. Anyone knowing a better way %%of dealing with this problem, please email me. %%%%% Non-standard commands and symbols \newcommand{\BBZ}{\mathbb{Z}} \newcommand{\BBR}{\mathbb{R}} \newcommand{\BBN}{\mathbb{N}} \newcommand{\BBC}{\mathbb{C}} \newcommand{\BBQ}{\mathbb{Q}} \def\binom#1#2{{#1\choose#2}} \def\arc#1{{\wideparen{#1}}} \newcommand{\dis}{\displaystyle} \newcommand{\ngrows}{n\rightarrow +\infty} \newcommand{\unif}{\stackrel{\mathrm{\tiny unif}}{\longrightarrow} } \def\fun#1#2#3#4#5{\everymath{\displaystyle}{{#1} : \vspace{1cm} \begin{array}{ccc}{#4} & \rightarrow & {#5}\\ {#2} & \mapsto & {#3} \\ \end{array}}} \def\sgn#1{{\mathrm{signum}}\left(#1\right)} \def\d#1{\mathbf{d}#1} \def\acc#1{\mathbf{Acc}\left(#1\right)} \def\card#1{\mathrm{card}\left(#1\right)} \def\floor#1{\llfloor #1 \rrfloor} \def\ceil#1{\llceil #1 \rrceil} \def\norm#1{\Big|\Big| #1 \Big|\Big|} \def\absval#1{\left| #1 \right|} \def\N#1{\mathscr{N}_{#1}} \def\closure#1{\overline{#1}} \def\bdy#1{\operatorname{Bdy}\left(#1\right)} \def\interior#1{\widering{#1}} \def\interiorone#1{\mathring{#1}} \def\signum#1{\operatorname{signum}\left( #1 \right)} \def\card#1{\operatorname{card}\left( #1 \right)} \def\fracpart#1{\left\{#1\right\}} \def\gl#1#2{{\bf GL}_{#1}(#2)} \def\magma#1#2{\langle #1,#2\rangle} \def\ball#1#2{\mathscr{B}_{#1}\left(#2\right)} \def\field#1#2#3{\langle #1,#2, #3\rangle} \def\seq#1#2#3{\left\{#1\right\} _{#2} ^{#3}} \newcommand{\dint}{\displaystyle\int } \newcommand{\dsum}{\displaystyle\sum } \newcommand{\curlyP}{\mathscr{P} } \def\dom#1{{\mathbf{Dom}}\left(#1\right)} \def\target#1{{\mathbf{Target}}\left(#1\right)} \def\im#1{{\mathbf{Im}}\left(#1\right)} \newcommand{\idefun}{{\bf Id\ }} \def\supp#1{{\mathbf{supp}}\left(#1\right)} %%%%%%%%%%%%%%%%%INTERVALS %%%%%%%% lo= left open, rc = right closed, etc. \def\lcrc#1#2{{\bf\Big[ }#1 \ ; #2 {\bf\Big] }} \def\loro#1#2{{\bf\Big] }#1 \ ; #2 {\bf \Big[}} \def\lcro#1#2{{\bf \Big[} #1 \ ; #2 {\bf \Big[} } \def\lorc#1#2{\Big]#1 \ ; #2 \Big]} \newcommand{\dott}{{\scriptscriptstyle \stackrel{\bullet}{{}}}} \def\dotprod#1#2{\vec{#1} \dott \vec{#2}} \newcommand{\ngroes}{n\rightarrow +\infty} \newcommand{\sequ}{\left\{a_n\right\} _{n=1} ^{+\infty}} \def\soo#1{\mathit{o}\left(#1\right)} \def\meas#1{\mu\left(#1\right)} \def\outermeas#1{\overline{\mu}\left(#1\right)} \def\boo#1{\mathit{O}\left(#1\right)} \def\smallo#1#2{\mathit{o}_{\substack #1}\left(#2\right)} \def\asympto#1{\sim_{\substack #1}} \def\bigo#1#2{\mathit{O}_{\substack #1}\left(#2\right)} \def\ll#1#2{<<_{\substack #1}\left(#2\right)} \newcommand{\closR}{\closure{\closure{\BBR}}} %%%%Title Page \makeatletter \def\thickhrulefill{\leavevmode \leaders \hrule height 1pt\hfill \kern \z@} \renewcommand{\maketitle}{\begin{titlepage}% \let\footnotesize\small \let\footnoterule\relax \parindent \z@ \reset@font \null\vfil \begin{flushleft} \@title \end{flushleft} \par \hrule height 4pt \par \begin{flushright} \@author \par \end{flushright} \vskip 60\p@ \vspace*{\stretch{2}} \vskip 60\p@ \vspace*{\stretch{2}} \begin{center} \Large\textsf{\today\ Version} \end{center} \end{titlepage}% \setcounter{footnote}{0}% } \makeatother \title{\WriteBig{Calculus, once again}{1.5cm}{1mm}{green}{red}} \author{\textcolor{blue}{David A. SANTOS} \\ \href{mailto:dsantos@ccp.edu}{dsantos@ccp.edu}} %%%%%%% %%%%%%%%%%%%%%% %%%%%%%%% %%%%%%%%%%%%%% \usepackage{fancyhdr} %%%%for page headers and footers %%%%%%%%%%%%%%%%%% %%%%%% And voilą the document !!! \begin{document} \pagenumbering{roman} \maketitle \clearpage \begin{quote} Copyright \copyright{} 2007 David Anthony SANTOS. Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1.2 or any later version published by the Free Software Foundation; with no Invariant Sections, no Front-Cover Texts, and no Back-Cover Texts. A copy of the license is included in the section entitled ``GNU Free Documentation License''. \end{quote} \clearpage {\tiny\chapter*{\rlap{GNU Free Documentation License}} \phantomsection % so hyperref creates bookmarks \addcontentsline{toc}{chapter}{GNU Free Documentation License} %\label{label_fdl} \begin{center} Version 1.2, November 2002 Copyright \copyright{} 2000,2001,2002 Free Software Foundation, Inc. \bigskip 51 Franklin St, Fifth Floor, Boston, MA 02110-1301 USA \bigskip Everyone is permitted to copy and distribute verbatim copies of this license document, but changing it is not allowed. \end{center} \begin{center} {\bf\large Preamble} \end{center} The purpose of this License is to make a manual, textbook, or other functional and useful document ``free'' in the sense of freedom: to assure everyone the effective freedom to copy and redistribute it, with or without modifying it, either commercially or noncommercially. Secondarily, this License preserves for the author and publisher a way to get credit for their work, while not being considered responsible for modifications made by others. This License is a kind of ``copyleft'', which means that derivative works of the document must themselves be free in the same sense. It complements the GNU General Public License, which is a copyleft license designed for free software. We have designed this License in order to use it for manuals for free software, because free software needs free documentation: a free program should come with manuals providing the same freedoms that the software does. But this License is not limited to software manuals; it can be used for any textual work, regardless of subject matter or whether it is published as a printed book. We recommend this License principally for works whose purpose is instruction or reference. \begin{center} {\Large\bf 1. APPLICABILITY AND DEFINITIONS\par} \phantomsection \addcontentsline{toc}{section}{1. APPLICABILITY AND DEFINITIONS} \end{center} This License applies to any manual or other work, in any medium, that contains a notice placed by the copyright holder saying it can be distributed under the terms of this License. Such a notice grants a world-wide, royalty-free license, unlimited in duration, to use that work under the conditions stated herein. The ``\textbf{Document}'', below, refers to any such manual or work. Any member of the public is a licensee, and is addressed as ``\textbf{you}''. You accept the license if you copy, modify or distribute the work in a way requiring permission under copyright law. A ``\textbf{Modified Version}'' of the Document means any work containing the Document or a portion of it, either copied verbatim, or with modifications and/or translated into another language. 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The ``\textbf{Title Page}'' means, for a printed book, the title page itself, plus such following pages as are needed to hold, legibly, the material this License requires to appear in the title page. For works in formats which do not have any title page as such, ``Title Page'' means the text near the most prominent appearance of the work's title, preceding the beginning of the body of the text. A section ``\textbf{Entitled XYZ}'' means a named subunit of the Document whose title either is precisely XYZ or contains XYZ in parentheses following text that translates XYZ in another language. (Here XYZ stands for a specific section name mentioned below, such as ``\textbf{Acknowledgements}'', ``\textbf{Dedications}'', ``\textbf{Endorsements}'', or ``\textbf{History}''.) To ``\textbf{Preserve the Title}'' of such a section when you modify the Document means that it remains a section ``Entitled XYZ'' according to this definition. The Document may include Warranty Disclaimers next to the notice which states that this License applies to the Document. These Warranty Disclaimers are considered to be included by reference in this License, but only as regards disclaiming warranties: any other implication that these Warranty Disclaimers may have is void and has no effect on the meaning of this License. \begin{center} {\Large\bf 2. VERBATIM COPYING\par} \phantomsection \addcontentsline{toc}{section}{2. VERBATIM COPYING} \end{center} You may copy and distribute the Document in any medium, either commercially or noncommercially, provided that this License, the copyright notices, and the license notice saying this License applies to the Document are reproduced in all copies, and that you add no other conditions whatsoever to those of this License. You may not use technical measures to obstruct or control the reading or further copying of the copies you make or distribute. However, you may accept compensation in exchange for copies. If you distribute a large enough number of copies you must also follow the conditions in section~3. You may also lend copies, under the same conditions stated above, and you may publicly display copies. \begin{center} {\Large\bf 3. COPYING IN QUANTITY\par} \phantomsection \addcontentsline{toc}{section}{3. COPYING IN QUANTITY} \end{center} If you publish printed copies (or copies in media that commonly have printed covers) of the Document, numbering more than 100, and the Document's license notice requires Cover Texts, you must enclose the copies in covers that carry, clearly and legibly, all these Cover Texts: Front-Cover Texts on the front cover, and Back-Cover Texts on the back cover. Both covers must also clearly and legibly identify you as the publisher of these copies. The front cover must present the full title with all words of the title equally prominent and visible. You may add other material on the covers in addition. Copying with changes limited to the covers, as long as they preserve the title of the Document and satisfy these conditions, can be treated as verbatim copying in other respects. If the required texts for either cover are too voluminous to fit legibly, you should put the first ones listed (as many as fit reasonably) on the actual cover, and continue the rest onto adjacent pages. If you publish or distribute Opaque copies of the Document numbering more than 100, you must either include a machine-readable Transparent copy along with each Opaque copy, or state in or with each Opaque copy a computer-network location from which the general network-using public has access to download using public-standard network protocols a complete Transparent copy of the Document, free of added material. If you use the latter option, you must take reasonably prudent steps, when you begin distribution of Opaque copies in quantity, to ensure that this Transparent copy will remain thus accessible at the stated location until at least one year after the last time you distribute an Opaque copy (directly or through your agents or retailers) of that edition to the public. It is requested, but not required, that you contact the authors of the Document well before redistributing any large number of copies, to give them a chance to provide you with an updated version of the Document. \begin{center} {\Large\bf 4. MODIFICATIONS\par} \phantomsection \addcontentsline{toc}{section}{4. MODIFICATIONS} \end{center} You may copy and distribute a Modified Version of the Document under the conditions of sections 2 and 3 above, provided that you release the Modified Version under precisely this License, with the Modified Version filling the role of the Document, thus licensing distribution and modification of the Modified Version to whoever possesses a copy of it. In addition, you must do these things in the Modified Version: \begin{itemize} \item[A.] Use in the Title Page (and on the covers, if any) a title distinct from that of the Document, and from those of previous versions (which should, if there were any, be listed in the History section of the Document). You may use the same title as a previous version if the original publisher of that version gives permission. \item[B.] List on the Title Page, as authors, one or more persons or entities responsible for authorship of the modifications in the Modified Version, together with at least five of the principal authors of the Document (all of its principal authors, if it has fewer than five), unless they release you from this requirement. \item[C.] State on the Title page the name of the publisher of the Modified Version, as the publisher. \item[D.] Preserve all the copyright notices of the Document. \item[E.] Add an appropriate copyright notice for your modifications adjacent to the other copyright notices. \item[F.] Include, immediately after the copyright notices, a license notice giving the public permission to use the Modified Version under the terms of this License, in the form shown in the Addendum below. \item[G.] Preserve in that license notice the full lists of Invariant Sections and required Cover Texts given in the Document's license notice. \item[H.] Include an unaltered copy of this License. \item[I.] Preserve the section Entitled ``History'', Preserve its Title, and add to it an item stating at least the title, year, new authors, and publisher of the Modified Version as given on the Title Page. If there is no section Entitled ``History'' in the Document, create one stating the title, year, authors, and publisher of the Document as given on its Title Page, then add an item describing the Modified Version as stated in the previous sentence. \item[J.] Preserve the network location, if any, given in the Document for public access to a Transparent copy of the Document, and likewise the network locations given in the Document for previous versions it was based on. These may be placed in the ``History'' section. You may omit a network location for a work that was published at least four years before the Document itself, or if the original publisher of the version it refers to gives permission. \item[K.] For any section Entitled ``Acknowledgements'' or ``Dedications'', Preserve the Title of the section, and preserve in the section all the substance and tone of each of the contributor acknowledgements and/or dedications given therein. \item[L.] Preserve all the Invariant Sections of the Document, unaltered in their text and in their titles. Section numbers or the equivalent are not considered part of the section titles. \item[M.] Delete any section Entitled ``Endorsements''. Such a section may not be included in the Modified Version. \item[N.] Do not retitle any existing section to be Entitled ``Endorsements'' or to conflict in title with any Invariant Section. \item[O.] Preserve any Warranty Disclaimers. \end{itemize} If the Modified Version includes new front-matter sections or appendices that qualify as Secondary Sections and contain no material copied from the Document, you may at your option designate some or all of these sections as invariant. To do this, add their titles to the list of Invariant Sections in the Modified Version's license notice. These titles must be distinct from any other section titles. You may add a section Entitled ``Endorsements'', provided it contains nothing but endorsements of your Modified Version by various parties--for example, statements of peer review or that the text has been approved by an organization as the authoritative definition of a standard. You may add a passage of up to five words as a Front-Cover Text, and a passage of up to 25 words as a Back-Cover Text, to the end of the list of Cover Texts in the Modified Version. Only one passage of Front-Cover Text and one of Back-Cover Text may be added by (or through arrangements made by) any one entity. If the Document already includes a cover text for the same cover, previously added by you or by arrangement made by the same entity you are acting on behalf of, you may not add another; but you may replace the old one, on explicit permission from the previous publisher that added the old one. The author(s) and publisher(s) of the Document do not by this License give permission to use their names for publicity for or to assert or imply endorsement of any Modified Version. \begin{center} {\Large\bf 5. COMBINING DOCUMENTS\par} \phantomsection \addcontentsline{toc}{section}{5. COMBINING DOCUMENTS} \end{center} You may combine the Document with other documents released under this License, under the terms defined in section~4 above for modified versions, provided that you include in the combination all of the Invariant Sections of all of the original documents, unmodified, and list them all as Invariant Sections of your combined work in its license notice, and that you preserve all their Warranty Disclaimers. The combined work need only contain one copy of this License, and multiple identical Invariant Sections may be replaced with a single copy. If there are multiple Invariant Sections with the same name but different contents, make the title of each such section unique by adding at the end of it, in parentheses, the name of the original author or publisher of that section if known, or else a unique number. Make the same adjustment to the section titles in the list of Invariant Sections in the license notice of the combined work. In the combination, you must combine any sections Entitled ``History'' in the various original documents, forming one section Entitled ``History''; likewise combine any sections Entitled ``Acknowledgements'', and any sections Entitled ``Dedications''. You must delete all sections Entitled ``Endorsements''. \begin{center} {\Large\bf 6. COLLECTIONS OF DOCUMENTS\par} \phantomsection \addcontentsline{toc}{section}{6. COLLECTIONS OF DOCUMENTS} \end{center} You may make a collection consisting of the Document and other documents released under this License, and replace the individual copies of this License in the various documents with a single copy that is included in the collection, provided that you follow the rules of this License for verbatim copying of each of the documents in all other respects. You may extract a single document from such a collection, and distribute it individually under this License, provided you insert a copy of this License into the extracted document, and follow this License in all other respects regarding verbatim copying of that document. \begin{center} {\Large\bf 7. AGGREGATION WITH INDEPENDENT WORKS\par} \phantomsection \addcontentsline{toc}{section}{7. AGGREGATION WITH INDEPENDENT WORKS} \end{center} A compilation of the Document or its derivatives with other separate and independent documents or works, in or on a volume of a storage or distribution medium, is called an ``aggregate'' if the copyright resulting from the compilation is not used to limit the legal rights of the compilation's users beyond what the individual works permit. When the Document is included in an aggregate, this License does not apply to the other works in the aggregate which are not themselves derivative works of the Document. If the Cover Text requirement of section~3 is applicable to these copies of the Document, then if the Document is less than one half of the entire aggregate, the Document's Cover Texts may be placed on covers that bracket the Document within the aggregate, or the electronic equivalent of covers if the Document is in electronic form. Otherwise they must appear on printed covers that bracket the whole aggregate. \begin{center} {\Large\bf 8. TRANSLATION\par} \phantomsection \addcontentsline{toc}{section}{8. TRANSLATION} \end{center} Translation is considered a kind of modification, so you may distribute translations of the Document under the terms of section~4. Replacing Invariant Sections with translations requires special permission from their copyright holders, but you may include translations of some or all Invariant Sections in addition to the original versions of these Invariant Sections. You may include a translation of this License, and all the license notices in the Document, and any Warranty Disclaimers, provided that you also include the original English version of this License and the original versions of those notices and disclaimers. In case of a disagreement between the translation and the original version of this License or a notice or disclaimer, the original version will prevail. If a section in the Document is Entitled ``Acknowledgements'', ``Dedications'', or ``History'', the requirement (section~4) to Preserve its Title (section~1) will typically require changing the actual title. \begin{center} {\Large\bf 9. TERMINATION\par} \phantomsection \addcontentsline{toc}{section}{9. TERMINATION} \end{center} You may not copy, modify, sublicense, or distribute the Document except as expressly provided for under this License. Any other attempt to copy, modify, sublicense or distribute the Document is void, and will automatically terminate your rights under this License. However, parties who have received copies, or rights, from you under this License will not have their licenses terminated so long as such parties remain in full compliance. \begin{center} {\Large\bf 10. FUTURE REVISIONS OF THIS LICENSE\par} \phantomsection \addcontentsline{toc}{section}{10. FUTURE REVISIONS OF THIS LICENSE} \end{center} The Free Software Foundation may publish new, revised versions of the GNU Free Documentation License from time to time. Such new versions will be similar in spirit to the present version, but may differ in detail to address new problems or concerns. See http://www.gnu.org/copyleft/. Each version of the License is given a distinguishing version number. If the Document specifies that a particular numbered version of this License ``or any later version'' applies to it, you have the option of following the terms and conditions either of that specified version or of any later version that has been published (not as a draft) by the Free Software Foundation. If the Document does not specify a version number of this License, you may choose any version ever published (not as a draft) by the Free Software Foundation. } \QUOTEME{Que a quien robe este libro, o lo tome prestado y no lo devuelva, se le convierta en una serpiente en las manos y lo venza. Que sea golpeado por la par\'{a}lisis y todos sus miembros arruinados. Que languidezca de dolor gritando por piedad, y que no haya coto a su agon\'{\i}a hasta la \'{u}ltima disoluci\'{o}n. Que las polillas roan sus entra\~{n}as y, cuando llegue al final de su castigo, que arda en las llamas del Infierno para siempre.} {Maldici\'{o}n an\'{o}nima contra los ladrones de libros en el monasterio de San Pedro, Barcelona.} \clearpage \tableofcontents \renewcommand{\headrulewidth}{1pt} \renewcommand\footrulewidth{1pt} \lhead[\rm\thepage]{\it \rightmark} \rhead[\it \leftmark]{\rm\thepage} \cfoot{} \renewcommand{\sectionmark}{\markboth{\sectionname\ \thechapter}} \renewcommand{\sectionmark}{\markright} \chapter*{Preface} \markboth{}{} \addcontentsline{toc}{chapter}{Preface} \markright{Preface}For many years I have been lucky enough to have students ask for more: more challenging problems, more illuminating proofs to different theorems, a deeper look at various topics, etc. To those students I normally recommend the books in the bibliography. Some of the same students have complained of not finding the books or wanting to buy them, but being impecunious, not being able to afford to buy them. Hence I have decided to make this compilation. \bigskip Here we take a semi-rigorous tour through Calculus. We don't construct the real numbers, but we examine closer the real number axioms and some of the basic theorems of Calculus. We also consider some Olympiad-level problems whose solution can be obtained through Calculus. \bigskip The reader is assumed to be familiar with proofs using mathematical induction, proofs by contradiction, and the mechanics of differentiation and integration. \hfill{David A. SANTOS} \hfill{\href{mailto:dsantos@ccp.edu}{dsantos@ccp.edu}} \cfoot{}\rfoot{\psovalbox{\thepage}} \pagestyle{fancy} \fancyhead{} \renewcommand{\chaptermark}{\markboth{\chaptername\ \thechapter}} \renewcommand{\sectionmark}{\markright} \fancyhead[RO]{\itshape\leftmark} \fancyhead[RE]{\itshape\rightmark} \chapter{Preliminaries} \begin{center} \fcolorbox{blue}{yellow}{ \begin{minipage}{.90\linewidth} \noindent\textcolor{red}{\textbf{Why bother?}} We will use the language of set theory throughout these notes. There are various elementary results that pop up in later proofs, among them, the De Morgan Laws and the Monotonicity Reversing of Complementation Rule. \medskip The concept of a {\em function} lies at the core of mathematics. We will give a brief overview here of some basic properties of functions. \end{minipage}} \end{center} \pagenumbering{arabic} \setcounter{page}{1} \Opensolutionfile{calculillo}[calculillo1] \section{Sets} This section contains some of the set notation to be used throughout these notes. The one-directional arrow $\implies$ reads ``implies'' and the two-directional arrow $\iff$ reads ``if and only if.'' \begin{df} We will accept the notion of {\em set} as a primitive notion, that is, a notion that cannot be defined in terms of more elementary notions. By a {\em set} we will understand a well-defined collection of objects, which we will call the {\em elements} of the set. If the element $x$ belongs to the set $S$ we will write $x\in S$, and in the contrary case we will write $x\not\in S$.\footnote{ Georg Cantor(1845-1918), the creator of set theory, said ``{\em A set is any collection into a whole of definite, distinguishable objects, called {\bf elements}, of our intuition or thought.''}} The {\em cardinality} of a set is the number of elements the set has. It can either be finite or infinite. We will denote the cardinality of the set $S$ by $\card{S}$. \end{df} \begin{rem} Some sets are used so often that merit special notation. We will denote by $$ \BBN = \{0,1,2,3,\ldots \} $$ the set of natural numbers, by $$\BBZ = \{\ldots, -3,-2,-1,0,1,2,3,\ldots \}\footnote{$\BBZ$ for the German word {\em Z\"{a}hlen} meaning ``integer.''} $$by $\BBQ$ the set of rational numbers\footnote{$\BBQ$ for ``quotients.''}, by $\BBR$ the real numbers, and by $\BBC$ the set of complex numbers. We will occasionally also use $\alpha \BBZ = \{\ldots,-3\alpha,-2\alpha,-\alpha,0,\alpha, 2\alpha, 3\alpha, \ldots\}$, etc. \bigskip We will also denote the empty set, that is, the set having no elements by $\varnothing$. \end{rem} \vspace{1cm} \begin{figure}[h] \begin{minipage}{5cm}$$\psset{unit=1pc} \pscircle[fillstyle=hlines, fillcolor=red](-1,0){2}\pscircle[fillstyle=hlines, fillcolor=red](1,0){2} \uput[d](-1,-2){A}\uput[d](1,-2){B} $$\vspace{1cm}\footnotesize\hangcaption{$A\cup B$} \label{fig:a_union_b} \end{minipage} \hfill \begin{minipage}{5cm}$$ \psset{unit=1pc} \pscircle(-1,0){2}\pscircle(1,0){2} \uput[d](-1,-2){A}\uput[d](1,-2){B} \pscustom[fillstyle=solid,fillcolor=green]{\psarc(1,0){2}{120}{240}\psarc(-1,0){2}{270}{60}} $$\vspace{1cm}\footnotesize\hangcaption{$A\cap B$} \label{fig:a_intersection_b} \end{minipage} \hfill \begin{minipage}{5cm}$$\psset{unit=1pc} \pscircle[fillstyle=solid,fillcolor=blue](-1,0){2}\pscircle(1,0){2} \uput[d](-1,-2){A}\uput[d](1,-2){B} \pscustom[fillstyle=solid,fillcolor=white]{\psarc(1,0){2}{120}{240}\psarc(-1,0){2}{270}{60}} $$\vspace{1cm}\footnotesize\hangcaption{$A\setminus B$} \label{fig:a_minus_b} \end{minipage} \end{figure} \begin{df} The {\em union} of two sets $A$ and $B$ is the set $$A\cup B = \{x:(x\in A)\ \mathrm{or}\ (x\in B)\}.$$ This is read ``$A$ union $B$.'' See figure \ref{fig:a_union_b}. The {\em intersection} of two sets $A$ and $B$ is $$A\cap B = \{x:(x\in A)\ \mathrm{and} \ (x\in B)\}.$$ This is read ``$A$ intersection $B$.'' See figure \ref{fig:a_intersection_b}. The {\em set difference} of two sets $A$ and $B$ is $$A\setminus B = \{x:(x\in A)\ \mathrm{and} (x\not\in B)\}.$$ This is read ``$A$ set minus $B$.'' See figure \ref{fig:a_minus_b}. \end{df} \begin{df} Two sets $A$ and $B$ are {\em disjoint} if $A\cap B = \varnothing$. \end{df} \begin{exa} Write $A\cup B$ as the disjoint union of three sets. \end{exa} \begin{solu} Observe that $$ A\cup B = (A\setminus B) \cup (A\cap B) \cup (B\setminus A), $$and that the sets on the dextral side are disjoint. \end{solu} \begin{df} A {\em subset} $B$ of a set $A$ is a subcollection of $A$, and we denote this by $B \subseteqq A$. \footnote{There seems not to be an agreement here by authors. Some use the notation $\subset $ or $\subseteq$ instead of $ \subseteqq$. Some see in the notation $\subset $ the exclusion of equality. In these notes, we will always use the notation $\subseteqq $, and if we wished to exclude equality we will write $\subsetneqq$.} This means that $x\in B \implies x\in A$. \end{df} \begin{rem} $\varnothing$ and $A$ are always subsets of any set $A$. \end{rem} Observe that $$A= B \iff (A \subseteq B)\quad \mathrm{and}\quad (B\subseteq A).$$We use this observation on the next theorem. \begin{thm}[De Morgan Laws] \label{thm:DeMorgan}Let $A, B, C$ be sets. Then $$A \setminus (B\cap C) = (A\setminus B) \cup (A\setminus C), \qquad A \setminus (B\cup C) = (A\setminus B) \cap (A\setminus C). $$ \end{thm} \begin{pf} We have $$\begin{array}{lll} x \in A\setminus(B \cup C) & \iff & x \in A \quad \mathrm{and}\quad x\not \in (B \quad \mathrm{or} \quad C) \\ & \iff & (x \in A) \ \ \ \mathrm{and} \ \ \ ( (x \not \in B) \ \ \ \mathrm{and}\ \ \ (x \not \in C)) \\ & \iff & (x \in A \ \ \ \mathrm{and} \ \ \ x \not \in B) \ \ \ \mathrm{and} \ \ \ (x \in A \ \ \ \mathrm{and} \ \ \ x \not \in C) \\ & \iff & (x \in A\setminus B) \ \ \ \mathrm{and}\ \ \ (x \in A \setminus C) \\ & \iff & x \in (A\setminus B) \cap(A\setminus C). \end{array} $$ Also, $$\begin{array}{lll} x \in A\setminus(B \cap C) & \iff & x \in A \quad \mathrm{and}\quad x\not \in (B \quad \mathrm{and}\quad C) \\ & \iff & (x \in A) \ \ \ \mathrm{and} \ \ \ ( (x \not \in B) \ \ \ \mathrm{or}\ \ \ (x \not \in C)) \\ & \iff & (x \in A \ \ \ \mathrm{and} \ \ \ x \not \in B) \ \ \ \mathrm{or} \ \ \ (x \in A \ \ \ \mathrm{and} \ \ \ x \not \in C) \\ & \iff & (x \in A\setminus B) \ \ \ \mathrm{or}\ \ \ (x \in A \setminus C) \\ & \iff & x \in (A\setminus B) \cup(A\setminus C) \end{array} $$ \end{pf} \begin{thm}[Monotonicity Reversing of Complementation]\label{thm:mono-reversing} Let $A, B, X$ be sets. Then $$ A\subseteqq B \iff X\setminus B \subseteqq X\setminus A. $$ \end{thm} \begin{pf} We have $$ \begin{array}{lll} A\subseteqq B & \iff & (x\in A) \implies (x\in B) \\ & \iff & (x\not \in B)\implies (x\not\in A)\\ & \iff & (x\in X\quad \mathrm{and}\quad x\not \in B)\implies (x\in X\quad \mathrm{and}\quad x\not\in A)\\ & \iff & X\setminus B \subseteqq X\setminus A. \end{array}$$ \end{pf} \begin{df} Let $A_1, A_2, \ldots, A_n$, be sets. The {\em Cartesian Product} of these $n$ sets is defined and denoted by \index{sets!Cartesian Product} $$A_1\times A_2\times \cdots \times A_n = \{(a_1, a_2, \ldots , a_n): a_k \in A_k\},$$ that is, the set of all ordered $n$-tuples whose elements belong to the given sets. \end{df} \begin{rem} In the particular case when all the $A_k$ are equal to a set $A$, we write $$A_1\times A_2\times \cdots \times A_n = A^n.$$If $a\in A$ and $b\in A$ we write $(a,b)\in A^2.$\end{rem} \begin{exa} The Cartesian product is not necessarily commutative. For example, $(\sqrt{2}, 1)\in\BBR \times \BBZ$ but $(\sqrt{2}, 1)\not\in\BBZ \times \BBR$. Since $\BBR \times \BBZ$ has an element that $\BBZ \times \BBR$ does not, $\BBR \times \BBZ \neq \BBZ \times \BBR$. \end{exa} \begin{exa} Prove that if $X\times X = Y\times Y$ then $X=Y$. \end{exa} \begin{solu} Let $x\in X$. Then $(x,x)\in X\times X$, which gives $(x,x)\in Y\times Y$, so $y\in Y$. Hence $X\subseteq Y$. \bigskip Similarly, if $y\in Y$ then $(y,y)\in Y\times Y$, which gives $(y,y)\in X\times X$, so $y\in X$. Hence $Y\subseteq X$. \bigskip Thus $X\subseteq Y$ and $Y \subseteq X$ gives $X=Y$. \end{solu} \subsection*{Homework}\addcontentsline{toc}{subsection}{Homework} \begin{multicols}{2}\columnseprule 1pt \columnsep 25pt\multicoltolerance=900\small \begin{pro}For a fixed $n\in\BBN$ put $A_n =\{nk:k\in\BBN\} $. \begin{enumerate} \item Find $A_2\cap A_3$. \item Find $\bigcap _{n=1} ^\infty A_n$. \item Find $\bigcup _{n=1} ^\infty A_n$. \end{enumerate} \begin{answer}Observe that $A_n = \{0, n, 2n, 3n, \ldots \}$. \begin{enumerate} \item $A_6$. \item $\BBN $. \item $\{0\}$. \end{enumerate} \end{answer} \end{pro} \begin{pro} Prove the following properties of the empty set: $$ A\cap \varnothing = \varnothing ,\quad A\cup \varnothing = A. $$ \end{pro} \begin{pro} Prove the following commutative laws: $$ A\cap B = B \cap A,\quad A\cup B = B \cup A. $$ \end{pro} \begin{pro} Prove by means of set inclusion the following distributive law: $$(A \cup B) \cap C = (A \cap C) \cup (B \cap C).$$ \begin{answer} We have, $$\begin{array}{lll} x \in (A \cup B) \cap C & \iff & x\in (A \cup B) \quad \mathrm{and} \quad x\in C \\ & \iff & (x\in A \quad \mathrm{or} \quad x\in B) \quad \mathrm{and} \quad x\in C \\ & \iff & (x\in A \quad \mathrm{and} \quad x \in C) \quad \mathrm{or} \quad (x\in B \quad \mathrm{and} \quad x\in C) \\ & \iff & (x\in A\cap C) \quad \mathrm{or} \quad (x\in B \cap C)\\ & \iff & x \in (A \cap C) \cup (B \cap C), \end{array}$$which establishes the equality. \end{answer} \end{pro} \begin{pro} Prove the following associative laws: $$ A\cap (B \cap C) = (A\cap B)\cap C, \quad A \cup (B\cup C) = (A\cup B)\cup C. $$ \end{pro} \begin{pro} Prove that $$ A\cap B = A \iff A \subseteq B. $$ \end{pro} \begin{pro} Prove that $$ A\cup B = A \iff B \subseteq A. $$ \end{pro} \begin{pro} Prove that $$ A\subseteq B \implies A \cap C \subseteq B\cap C. $$ \end{pro} \begin{pro} Prove that $$ A\subseteq B \quad \mathrm{and}\quad C \subseteq B \implies A \cup C \subseteq B. $$ \end{pro} \begin{pro} Prove the following distributive laws: $$ A\cap (B\cup C) = (A\cap B)\cup (A\cap C), \qquad A\cup (B\cap C) = (A\cup B)\cap (A\cup C). $$ \end{pro} \begin{pro} Is there any difference between the sets $\varnothing$, $\{\varnothing\}$ and $\{\{\varnothing\}\}$? Explain. \end{pro} \begin{pro} Is the Cartesian product associative? Explain. \end{pro} \begin{pro} Let $A,B$, and $C$ be sets. Shew that $$A\times (B\setminus C) = (A\times B)\setminus (A\times C).$$ \begin{answer}We check the two statements $$x\in A\times (B\setminus C) \Longleftrightarrow x\in (A\times B)\setminus (A\times C).$$ Let us prove first $\Longrightarrow$. By definition of $\times$, $x = (a,b)$, where $a\in A, b\in B, b\notin C$. Thus $x\in A\times B$ but $x\notin A\times C$. By definition of $\setminus$ we are done. Now we prove the assertion $\Longleftarrow$. By definition of $\times$ and $\setminus$, $x=(a,b)$ where $a\in A, b\in B$. Since $x\notin A\times C$, we observe that $b\notin C$. Thus $a\in A, b\in B\setminus C$, and we gather that $x\in A\times (B\setminus C)$. \end{answer} \end{pro} \begin{pro} Prove that a set with $N\in\BBN$ elements has exactly $2^N$ subsets. \begin{answer} Attach a binary code to each element of the subset, $1$ if the element is in the subset and $0$ if the element is not in the subset. The total number of subsets is the total number of such binary codes, and there are $2^N$ in number. \end{answer} \end{pro} \end{multicols} \section{Numerical Functions} \begin{df} By a {\em (numerical) function}\index{function} $f:\dom{f}\rightarrow\target{f}$ we mean the collection of the following ingredients: \begin{dingautolist}{202} \item a {\em name} for the function. Usually we use the letter $f$. \item a set of real number inputs called the {\em domain} of the function. The domain of $f$ is denoted by $\dom{f}\subseteqq \BBR$. \item an {\em input parameter }, also called {\em independent variable} or {\em dummy variable}. We usually denote a typical input by the letter $x$.\index{function!domain} \item a set of possible real number outputs of the function, called the {\em target set} of the function. The target set of $f$ is denoted by $\target{f}\subseteqq \BBR$.\index{function!target set} \item an {\em assignment rule} or {\em formula}, assigning to {\bf every input} a {\bf unique} output. This assignment rule for $f$ is usually denoted by $x\mapsto f(x)$. The output of $x$ under $f$ is also referred to as the {\em image of $x$ under $f$}, \index{function!image} and is denoted by $f(x)$. \index{function!assingment rule} \end{dingautolist} \end{df} The notation\footnote{Notice the difference in the arrows. The straight arrow $\longrightarrow$ is used to mean that a certain set is associated with another set, whereas the arrow $\mapsto$ (read ``maps to'') is used to denote that an input becomes a certain output.} $$\fun{f}{x}{f(x)}{\dom{f}}{\target{f}} $$ read ``the function $f$, with domain $\dom{f}$, target set $\target{f}$, and assignment rule $f$ mapping $x$ to $f(x)$'' conveys all the above ingredients. \begin{rem} Oftentimes we will only need to mention the assignment rule of a function, without mentioning its domain or target set. In such instances we will sloppily say ``the function $f$'' or more commonly, ``the function $x\mapsto f(x)$'', e.g., the square function $x\mapsto x^2$.\footnote{This corresponds to the even sloppier American usage ``the function $f(x)=x^2$.''} \end{rem} \begin{df} The {\em image} $\im{f}$ of a function $f$ is its set of actual outputs. In other words, $$\im{f} = \{f(a): a\in\dom{f}\}.$$ Observe that we always have $\im{f} \subseteq \target{f}$. For a set $A$, we also define $$ f(A)=\{f(a): a\in A\}. $$ \end{df} \begin{thm} Let $f: X \rightarrow Y$ be a function and let $A\subseteqq X$, $A'\subseteqq X$. Then \begin{enumerate} \item $A \subseteqq A' \implies f(A) \subseteqq f(A')$ \item $f(A\cup A') =f(A)\cup f(A')$ \item $f(A\cap A') \subseteqq f(A)\cap f(A') $ \item $ f(A)\setminus f(A') \subseteqq f(A\setminus A') $ \end{enumerate} \end{thm} \begin{pf} \begin{enumerate} \item $x\in A \implies x\in A' $ and hence $f(x) \in f(A)\implies f(x)\in f(A') \implies f(A)\subseteqq f(A')$ \item Since $A \subseteqq A\cup A'$ and $A' \subseteqq A\cup A'$, we have $f(A) \subseteqq f(A\cup A')$ and $f(A') \subseteqq f(A\cup A')$, by part (1) and thus $f(A) \subseteqq f(A') \subseteqq f(A\cup A')$. Moreover, if $y\in f(A\cup A')$, then $\exists x\in A\cup A'$ such that $y=f(x)$. Then either $x\in A$ and so $f(x)\in f(A)$ or $x\in A'$ and so $f{x}\in f(A')$. Either way, $f(x)\in f(A)\cup f(A')$ and $$y\in f(A\cup A')\implies y\in f(A)\cup f(A')\implies f(A \cup A')\subseteqq f(A)\cup f(A').$$Hence $$f(A \cup A')=f(A)\cup f(A'). $$ \item Let $y\in f(A\cap A')$. Then $\exists x\in A\cap A'$ such that $f(x)=y$. Thus we have both $x\in A \implies f(x)\in f(A)$ and $x\in A' \implies f(x)\in f(A')$. Therefore $f(x)\in f(A)\cap f(A')$ and we conclude that $f(A\cap A') \subseteqq f(A)\cap f(A') $. \item Let $ y\in f(A)\setminus f(A')$. Then $y\in f(A)$ and $y\notin f(A')$. Thus $\exists x\in A$ such that $f(x)=y$. Since $y\notin f(A')$, then $x\notin A'$. Therefore $x\in A\setminus A'$ and finally, $y\in f(A\setminus A')$. This means that $ f(A)\setminus f(A') \subseteqq f(A\setminus A') $ as claimed. \end{enumerate} \end{pf} \subsection{Injective and Surjective Functions} \begin{df} A function is {\em injective} or {\em one-to-one} whenever two different values of its domain generate two different values in its image. A function is {\em surjective} or {\em onto} if every element of its target set is hit, that is, the target set is the same as the image of the function. A function is {\em bijective} if it is both injective and surjective. \end{df} \begin{exa} The function $$\fun{a}{x}{x^2}{\BBR}{\BBR} $$is neither injective nor surjective. \bigskip The function $$\fun{b}{x}{x^2}{\BBR}{\lcro{0}{+\infty}} $$is surjective but not injective. \bigskip The function $$\fun{c}{x}{x^2}{\lcro{0}{+\infty}}{\BBR} $$is injective but not surjective. \bigskip The function $$\fun{d}{x}{x^2}{\lcro{0}{+\infty}}{\lcro{0}{+\infty}} $$is a bijection. \end{exa} A bijection between two sets essentially tells us that the two sets have the same size. We will make this statement more precise now for finite sets. \begin{thm}\label{thm:size_domain_image_injections_surjections} Let $f:A\rightarrow B$ be a function, and let $A$ and $B$ be finite. If $f$ is injective, then $\card{A} \leq \card{B}$. If $f$ is surjective then $\card{B}\leq \card{A}$. If $f$ is bijective, then $\card{A} = \card{B}$. \end{thm} \begin{pf} Put $n = \card{A}$, $A = \{x_1,x_2, \ldots ,x_n\}$ and $m = \card{B}$, $B = \{y_1,y_2, \ldots ,y_m\}$. \bigskip If $f$ were injective then $f(x_1), f(x_2), \ldots , f(x_n)$ are all distinct, and among the $y_k$. Hence $n \leq m$. \bigskip If $f$ were surjective then each $y_k$ is hit, and for each, there is an $x_i$ with $f(x_i) = y_k$. Thus there are at least $m$ different images, and so $n \geq m$. \end{pf} \subsection{Algebra of Functions} \begin{df} Let $f: \dom{f}\rightarrow \target{f}$ and $g: \dom{g}\rightarrow \target{g}$. Then $\dom{f\pm g} = \dom{f}\cap \dom{g}$ and the sum (respectively, difference) function $f + g$ (respectively, $f - g$) is given by $$\fun{f \pm g}{x}{f(x) \pm g(x)}{\dom{f}\cap \dom{g}}{\target{f\pm g}}. $$ In other words, if $x$ belongs both to the domain of $f$ and $g$, then $$(f \pm g)(x) = f(x) \pm g(x).$$ \end{df} \begin{df} Let $f: \dom{f}\rightarrow \target{f}$ and $g: \dom{g}\rightarrow \target{g}$. Then $\dom{fg} = \dom{f}\cap \dom{g}$ and the product function $fg$ is given by $$\fun{fg}{x}{f(x)\cdot g(x)}{\dom{f}\cap \dom{g}}{\target{fg}}. $$ In other words, if $x$ belongs both to the domain of $f$ and $g$, then $$(fg)(x) = f(x)\cdot g(x).$$ \end{df} \begin{df} Let $g:\dom{g} \rightarrow \target{g}$ be a function. The {\em support} of $g$, denoted by $\supp{g}$ is the set of elements in $\dom{g}$ where $g$ does not vanish, that is $$\supp{g} = \{x\in\dom{g}: g(x) \neq 0\}.$$ \end{df} \begin{df} Let $f: \dom{f}\rightarrow \target{f}$ and $g: \dom{g}\rightarrow \target{f}$. Then $\dom{\dfrac{f}{g}} = \dom{f}\cap \supp{g}$ and the quotient function $\dfrac{f}{g}$ is given by $$\fun{\dfrac{f}{g}}{x}{\dfrac{f(x)}{g(x)}}{\dom{f}\cap \supp{g}}{\target{f/g}}. $$ In other words, if $x$ belongs both to the domain of $f$ and $g$ and $g(x) \neq 0$, then $\dfrac{f}{g}(x) = \dfrac{f(x)}{g(x)}.$ \end{df} \begin{df} Let $f: \dom{f} \rightarrow \target{f}$, $g: \dom{g} \rightarrow \target{g}$ and let $U = \{x\in\dom{g}: g(x) \in \dom{f}\}$. We define the {\em composition} function of $f$ and $g$ as \begin{equation} \fun{f\circ g}{x}{f(g(x))}{U}{\target{f\circ g}} .\end{equation}We read $f\circ g$ as ``$f$ {\em composed with} $g$.'' \end{df} \subsection{Inverse Image} \begin{df}Let $X$ and $Y$ be subsets of $\BBR$ and let $f:X\rightarrow Y$ be a function. Let $B\subseteqq Y$. The {\em inverse image of $B$ by $f$} is the set $$f^{-1}(B) =\{x\in X: f(x)\in B\}.$$ If $B=\{b\}$ consists of only one element, we write, abusing notation, $f^{-1}(\{b\})=f^{-1}(b)$. It is clear that $f^{-1}(Y)=X$ and $f^{-1}(\varnothing)=\varnothing$. \end{df} \begin{exa} Let $$\fun{f}{x}{x^2}{\{-2,-1,0,1,3\}}{\{0,1,4,5,9\}}. $$Then $f^{-1}(\{0,1\})=\{0,-1,1\}$, $f^{-1}(1)=\{-1,1\}$, $f^{-1}(5)=\varnothing$, $f^{-1}(4)=2$, $f^{-1}(0)=0$, etc. Notice that we have abused notation in all but the first example. \end{exa} \begin{thm} Let $f: X \rightarrow Y$ be a function and let $B\subseteqq Y$, $B'\subseteqq Y$. Then \begin{enumerate} \item $B \subseteqq B' \implies f^{-1}(B) \subseteqq f^{-1}(B')$ \item $f^{-1}(B\cup B') =f^{-1}(B)\cup f^{-1}(B')$ \item $f^{-1}(B\cap B') = f^{-1}(B)\cap f^{-1}(B') $ \item $ f^{-1}(B)\setminus f(B') = f^{-1}(B\setminus B') $ \end{enumerate} \end{thm} \begin{pf} \begin{enumerate} \item Assume $x\in f^{-1}(B)$. Then there is $y\in B\subseteqq B'$ such that $f(x)=y$. But $y$ is also in $B'$ so $ x\in f^{-1}(B')$. Thus $f^{-1}(B) \subseteqq f^{-1}(B')$. \item Since $B\subseteqq B\cup B'$ and $B'\subseteqq B\cup B'$, we have $f^{-1}(B)\subseteqq f^{-1}(B\cup B')$ and $f^{-1}(B')\subseteqq f^{-1}(B\cup B')$, by part (1). Thus $f^{-1}(B)\cup f^{-1}(B')\subseteqq f^{-1}(B\cup B')$. Now, let $x\in f^{-1}(B\cup B')$. There is $y\in B\cup B'$ such that $f(x)=y$. Either $y\in B$ and so $y\in B \implies x\in f^{-1}(B)$ or $y\in B'$ and so $y\in B \implies x\in f^{-1}(B')$. Either way, $x\in f^{-1}(B) \cup f^{-1}(B')$. Thus $f^{-1}(B\cup B')\subseteqq f^{-1}(B) \cup f^{-1}(B')$. We conclude that $f^{-1}(B\cup B')= f^{-1}(B) \cup f^{-1}(B')$. \item Let $x\in f^{-1}(B\cap B')$. Then $\exists y\in B\cap B'$ such that $f(x)=y$. Thus we have both $y\in B \implies x\in f^{-1}(B)$ and $y\in B' \implies x\in f^{-1}(B')$. Therefore $x\in f^{-1}(B)\cap f^{-1}(B')$ and we conclude that $f^{-1}(B\cap B') \subseteqq f^{-1}(B)\cap f^{-1}(B') $. Now, let $x\in f^{-1}(B)\cap f^{-1}(B') $. Then $x\in f^{-1}(B)$ and $x\in f^{-1}(B')$. Then $f(x)\in B$ and $f(x)\in B'$. Thus $f(x)\in B\cap B'$ and so $x\in f^{-1}(B\cap B')$. Hence $f^{-1}(B)\cap f^{-1}(B')\subseteqq f^{-1}(B\cap B')$ also, and we conclude that $f^{-1}(B)\cap f^{-1}(B')= f^{-1}(B\cap B')$. \item Let $ x\in f^{-1}(B)\setminus f^{-1}(B')$. Then $x\in f^{-1}(B)$ and $x\notin f^{-1}(B')$. Thus $f(x)\in B$ and $f(x)\notin B'$. Thus $f(x)\in B\setminus B'$ and therefore $x\in f^{-1}(B\setminus B')$, giving $f^{-1}(B)\setminus f^{-1}(B') \subseteqq f^{-1}(B\setminus B')$. Now, let $x\in f^{-1}(B\setminus B')$. Then $f(x)\in B\setminus B'$, which means that $f(x)\in B$ but $f(x)\notin B'$. Thus $x\in f^{-1}(B)$ but $x\notin f^{-1}(B')$, which gives $x\in f^{-1}(B)\setminus f^{-1}(B')$ and so $ f^{-1}(B\setminus B')\subseteqq f^{-1}(B)\setminus f^{-1}(B')$. This establishes the desired equality. \end{enumerate} \end{pf} \begin{thm} Let $f:X\rightarrow Y$ be a function. Let $A\times B\subseteqq X\times Y$. Then \begin{enumerate} \item $A \subseteqq (f^{-1}\circ f)(A)$ \item $(f\circ f^{-1})(B)\subseteqq B$ \end{enumerate} \end{thm} \begin{pf} We have \begin{enumerate} \item Let $x\in A$. Then $\exists y\in Y$ such that $y=f(x)$. Thus $y\in f(A)$. Therefore $x\in f^{-1}(f(A))$. \item $y\in (f\circ f^{-1})(B)$. Then $\exists x\in f^{-1}(B)$ such that $f(x) =y$. Thus $x\in f^{-1}(y)$. Hence $f(x)\in B$. Therefore $y\in B$. \end{enumerate} \end{pf} \subsection{Inverse Function} \begin{df} Let $A\times B \subseteqq \BBR^2$. A function $F: A \rightarrow B$ is said to be {\em invertible} if there exists a function $F^{-1}$ (called the {\em inverse} of $F$) such that $F\circ F^{-1} = \idefun _B$ and $F^{-1}\circ F = \idefun _A$. Here $\idefun _S$ is the identity on the set $S$ function with rule $\idefun _S(x) = x.$ \end{df} The central question is now: given a function $F: A \rightarrow B$, when is $F^{-1}: B \rightarrow A$ a function? The answer is given in the next theorem. \begin{thm}\label{thm:invertible<->bijective} Let $A\times B \subseteqq \BBR^2$. A function $f: A \rightarrow B$ is invertible if and only if it is a bijection. That is, $f^{-1}: B \rightarrow A$ is a function if and only if $f$ is bijective. \end{thm} \begin{pf} Assume first that $f$ is invertible. Then there is a function $f^{-1}: B \rightarrow A$ such that \begin{equation} f\circ f^{-1} = \idefun _B \ \ {\rm and} \ \ f^{-1}\circ f = \idefun _A.\end{equation} Let us prove that $f$ is injective and surjective. Let $s, t$ be in the domain of $f$ and such that $f(s) = f(t)$. Applying $f^{-1}$ to both sides of this equality we get $(f^{-1}\circ f)(s) = (f^{-1}\circ f)(t)$. By the definition of inverse function, $(f^{-1}\circ f)(s) = s$ and $(f^{-1}\circ f)(t) = t$. Thus $s = t.$ Hence $f(s) = f(t) \implies s = t$ implying that $f$ is injective. To prove that $f$ is surjective we must shew that for every $b \in f(A) \ \exists a \in A$ such that $f(a) = b.$ We take $a = f^{-1}(b)$ (observe that $f^{-1}(b) \in A$). Then $f(a) = f(f^{-1}(b)) = (f\circ f^{-1})(b) = b$ by definition of inverse function. This shews that $f$ is surjective. We conclude that if $f$ is invertible then it is also a bijection. \bigskip Assume now that $f$ is a bijection. For every $b\in B$ there exists a unique $a$ such that $f(a) = b$. This makes the rule $g: B \rightarrow A$ given by $g(b) = a$ a function. It is clear that $g\circ f = \idefun _A$ and $f\circ g = \idefun _B. $ We may thus take $f^{-1} = g.$ This concludes the proof. \end{pf} \subsection*{Homework}\addcontentsline{toc}{subsection}{Homework} \begin{multicols}{2}\columnseprule 1pt \columnsep 25pt\multicoltolerance=900\small \begin{pro}Find all functions with domain $\{a, b\}$ and target set $\{c, d\}$. \begin{answer} There are $2^2 = 4$ such functions, namely: \begin{dingautolist}{202} \item $f_1$ given by $f_1(a) = f_1(b) = c.$ Observe that $\im{f_1} = \{c\}$. \item $f_2$ given by $f_2(a) = f_2(b) = d.$ Observe that $\im{f_2} = \{d\}$. \item $f_3$ given by $f_3(a) = c, f_3(b) = d.$ Observe that $\im{f_3} = \{c, d\}$. \item $f_4$ given by $f_4(a) = d, f_4(b) = c.$ Observe that $\im{f_4} = \{c, d\}$. \end{dingautolist} \end{answer} \end{pro} \begin{pro} Let $A$, $B$ be finite sets with $\card{A} = n$ and $\card{B} = m$. Prove that \begin{itemize} \item The number of functions from $A$ to $B$ is $m^n$. \item If $n \leq m$, the number of injective functions from $A$ to $B$ is $m(m-1)(m-2)\cdots (m-n+1)$. If $n>m$ there are no injective functions from $A$ to $B$. \end{itemize} \begin{answer} Each of the $n$ elements of $A$ must be assigned an element of $B$, and hence there are $\underbrace{m\cdot m \cdots m}_{n\ \mathrm{factors}} = m^n$ possibilities, and thus $m^n$ functions.If a function from $A$ to $B$ is injective then we must have $n \leq m$ in view of Theorem \ref{thm:size_domain_image_injections_surjections}. If to different inputs we must assign different outputs then to the first element of $A$ we may assign any of the $m$ elements of $B$, to the second any of the $m-1$ remaining ones, to the third any of the $m-2$ remaining ones, etc., and so we have $m(m-1)\cdots (m-n+1)$ injective functions. \end{answer} \end{pro} \begin{pro} Let $A$ and $B$ be two finite sets with $\card{A} = n$ and $\card{B} = m$. If $n< m$ prove that there are no surjections from $A$ to $B$. If $n \geq m$ prove that the number of surjective functions from $A$ to $B$ is $$m^n - \binom{m}{1}(m-1)^n + \binom{m}{2}(m-2)^n - \binom{m}{3}(m-3)^n + \cdots + (-1)^{m-1}\binom{m}{m-1}(1)^{n}. $$ \end{pro} \begin{pro} Let $h:\BBR \rightarrow \BBR$ be given by $h(1 - x) = 2x$. Find $h(3x)$. \begin{answer} Rename the independent variable, say $h(1 - s) = 2s.$ Now, if $1 - s = 3x$ then $s = 1 - 3x.$ Hence $$h(3x) = h(1 - s) = 2s = 2(1 - 3x) = 2 - 6x.$$ \end{answer} \end{pro} \begin{pro} Consider the polynomial $$(1 - x^2 + x^4)^{2003} = a_0 + a_1x+a_2x^2 + \cdots + a_{8012}x^{8012}. $$Find \begin{dingautolist}{202} \item $a_0$ \item $a_0 + a_1+a_2 + \cdots + a_{8012}$ \item $a_0 - a_1+a_2 -a_3 + \cdots -a_{8011}+ a_{8012}$ \item $a_0+a_2 + a_4 + \cdots + a_{8010} + a_{8012}$ \item $a_1+a_3+ \cdots +a_{8009}+a_{8011}$ \end{dingautolist} \begin{answer} Put $$p(x) = (1 - x^2 + x^4)^{2003} = a_0 + a_1x+a_2x^2 + \cdots + a_{8012}x^{8012}.$$Then \begin{dingautolist}{202} \item $a_0 = p(0) = (1 - 0^2 + 0^4)^{2003} = 1.$ \item $a_0 + a_1+a_2 + \cdots + a_{8012} = p(1) = (1 - 1^2 + 1^4)^{2003} = 1.$ \item $$\begin{array}{lll}a_0 - a_1+a_2-a_3 + \cdots -a_{8011} + a_{8012} &= & p(-1)\\ & = & (1 - (-1)^2 + (-1)^4)^{2003}\\ & = & 1. \end{array}$$ \item The required sum is $\dfrac{p(1)+p(-1)}{2} = 1$. \item The required sum is $\dfrac{p(1)-p(-1)}{2} = 0$. \end{dingautolist} \end{answer} \end{pro} \begin{pro} Let $f:\BBR \rightarrow \BBR$, be a function such that $\forall x\in ]0;+\infty[$, $$[f(x^3 + 1)]^{\sqrt{x}} = 5,$$ find the value of $$\left[f\left(\frac{27 + y^3}{y^3}\right)\right]^{\sqrt{\frac{27}{y}}}$$ for $y\in ]0;+\infty[$. \begin{answer}We have $$\begin{array}{lll}\left[f\left(\frac{27 + y^3}{y^3}\right)\right]^{\sqrt{\frac{27}{y}}} & = & \left[f\left( \left(\frac{3}{y}\right)^3 + 1\right)\right]^{3\sqrt{\frac{3}{y}}} \\ & = & \left(\left[f\left( \left(\frac{3}{y}\right)^3 + 1\right)\right]^{\sqrt{\frac{3}{y}}}\right)^3 \\ & = & 5^3 \\ & = & 125. \end{array}$$ \end{answer} \end{pro} \begin{pro} Let $f$ satisfy $f(n + 1) = (-1)^{n + 1}n - 2f(n), n \geq 1$ If $f(1) = f(1001)$ find $$f(1) + f(2) + f(3) + \cdots + f(1000).$$ \begin{answer} We have \\ \begin{tabular}{lllll} $f(2)$ & = & $(-1)^21 - 2f(1)$ & = & $1 - 2f(1)$ \\ $f(3)$ & = & $(-1)^32 - 2f(2)$ & = & $-2 - 2f(2)$ \\ $f(4)$ & = & $(-1)^43 - 2f(3)$ & = & $3 - 2f(3)$ \\ $f(5)$ & = & $(-1)^54 - 2f(4)$ & = & $-4 - 2f(4)$ \\ $\vdots$ & $\vdots$ & $\vdots$ & $\vdots$ & $\vdots$ \\ $f(999)$ & = & $(-1)^{999}998 - 2f(998)$ & = & $-998 - 2f(998)$ \\ $f(1000)$ & = & $(-1)^{1000}999 - 2f(999)$ & = & $999 - 2f(999)$ \\ $f(1001)$ & = & $(-1)^{1001}1000 - 2f(1000)$ & = & $-1000 - 2f(1000)$ \\ \end{tabular} \bigskip Adding columnwise, $$f(2) + f(3) + \cdots + f(1001) = 1 - 2 + 3 - \cdots + 999 - 1000 - 2(f(1) + f(2) + \cdot + f(1000)).$$ This gives $$2f(1) + 3(f(2) + f(3) + \cdots + f(1000)) + f(1001) = - 500.$$ Since $f(1) = f(1001)$ we have $2f(1) + f(1001) = 3f(1).$ Therefore$$f(1) + f(2) + \cdots + f(1000) = - \frac{500}{3}.$$ \end{answer} \end{pro} \begin{pro} If $f(a)f(b) = f(a + b) \ \forall \ a, b \in \BBR$ and $f(x) > 0 \ \forall \ x \in \BBR$, find $f(0)$. Also, find $f(-a)$ and $f(2a)$ in terms of $f(a).$ \begin{answer} Set $a = b = 0.$ Then $(f(0))^2 = f(0)f(0) = f(0 + 0) = f(0)$. This gives $f(0)^2 = f(0).$ Since $f(0) > 0$ we can divide both sides of this equality to get $f(0) = 1$. \bigskip Further, set $b = -a.$ Then $f(a)f(-a) = f(a - a) = f(0) = 1.$ Since $f(a) \neq 0,$ may divide by $f(a)$ to obtain $f(-a) = \dis{\frac{1}{f(a)}}$. Finally taking $a = b$ we obtain $(f(a))^2 = f(a)f(a) = f(a + a) = f(2a).$ Hence $f(2a) = (f(a))^2$ \end{answer} \end{pro} \begin{pro} Prove that $\fun{f}{x}{\dfrac{x-1}{x+1}}{\BBR\setminus\{-1\}}{\BBR\setminus\{1\}}$ is a bijection and find $f^{-1}$. \begin{answer} To prove that $f$ is injective, we prove that $f(a)=f(b)\implies a=b$. We have $$ \begin{array}{lll} f(a)=f(b) & \implies & \dfrac{a-1}{a+1} = \dfrac{b-1}{b+1} \\ &\implies & (a-1)(b+1) = (a+1)(b-1)\\ &\implies & ab+a-b-1 = ab-a+b-1\\ & \implies & 2a=2b\\ &\implies & a=b,\\ \end{array}$$whence $f$ is injective. To prove that $f$ is surjective we must prove that any $y \in \BBR\setminus\{1\}$ has a pre-image $a\in \BBR\setminus \{-1\}$ such that $f(a)=y$. That is, $$\dfrac{a-1}{a+1}=y\implies a-1=ya+y \implies a-ya=1+y \implies a(1-y)=1+y \implies a=\dfrac{1+y}{1-y}.$$ Thus $f\left(\dfrac{1+y}{1-y}\right)=y$, and $f$ is surjective. This also serves to prove that $f^{-1}(x)=\dfrac{1+x}{1-x}$. \end{answer} \end{pro} \begin{pro} Let $f^{[1]}(x) = f(x) = x + 1, f^{[n + 1]} = f\circ f^{[n]}, n \geq 1.$ Find a closed formula for $f^{[n]}$ \begin{answer} We have $f^{[2]}(x) = f(x + 1) = (x + 1) + 1 = x + 2, f^{[3]}(x) = f(x + 2) = (x + 2) + 1 = x + 3$ and so, recursively, $f^{[n]}(x) = x + n.$ \end{answer} \end{pro} \begin{pro} Let $f, g:\lcrc{0}{1}\rightarrow\BBR$ be functions. Demonstrate that there exist $(a, b)\in \lcrc{0}{1}^2$ such that $\dfrac{1}{4}\leq \absval{f(a)+g(b)-ab}$. \end{pro} \begin{pro} Demonstrate that there is no function $f:\BBR\setminus \{1/2\}\rightarrow \BBR$ such that $$ x\in \BBR\setminus \{1/2\}\implies f(x)\left(f\left(\dfrac{x-1}{2x-1}\right)\right)=x^2+x+1$$ \end{pro} \begin{pro} Find all functions $f:\BBR\setminus \{-1,0\}\rightarrow \BBR$ such that $$ x\in \BBR\setminus \{-1,0\}\implies f(x)+f\left(\dfrac{-1}{x+1}\right)=3x+2.$$ \end{pro} \begin{pro} Let $f^{[1]}(x) = f(x) = 2x, f^{[n + 1]} = f\circ f^{[n]}, n \geq 1.$ Find a closed formula for $f^{[n]}$ \begin{answer} We have $f^{[2]}(x) = f(2x) = 2^2x, f^{[3]}(x) = f(2^2x) = 2^3x$ and so, recursively, $f^{[n]}(x) = 2^nx.$ \end{answer} \end{pro} \begin{pro} Find all functions $g:\BBR\rightarrow \BBR$ that satisfy $g(x + y) + g(x - y) = 2x^2 + 2y^2$. \begin{answer} Let $y = 0.$ Then $2g(x) = 2x^2$, that is, $g(x) = x^2.$ Let us check that $g(x) = x^2$ works. We have $$g(x + y) + g(x - y) = (x + y)^2 + (x - y)^2 = x^2 + 2xy + y^2 + x^2 - 2xy + y^2 = 2x^2 + 2y^2,$$ which is the functional equation given. Our choice of $g$ works. \end{answer} \end{pro} \begin{pro} Find all the functions $f:\BBR\rightarrow \BBR$ that satisfy $f(xy) = yf(x)$. \begin{answer} Let $x = 1$. Then $f(y) = yf(1)$. Since $f(1)$ is a constant, we may let $k = f(1).$ So all the functions satisfying the above equation satisfy $f(y) = ky.$ \end{answer} \end{pro} \begin{pro} Find all functions $f:\BBR\setminus \{0\}\rightarrow \BBR$ for which $$f(x) + 2f\left(\frac{1}{x}\right) = x.$$ \begin{answer} From $\dis{f(x) + 2f(\frac{1}{x}) = x}$ we obtain $\dis{f(\frac{1}{x}) = \frac{x}{2} - \frac{1}{2}f(x)}$. Also, substituting $1/x$ for $x$ on the original equation we get $$f(1/x) + 2f(x) = 1/x.$$Hence $$f(x) = \frac{1}{2x} - \frac{1}{2}f(1/x) = \frac{1}{2x} - \frac{1}{2}\left(\frac{x}{2} - \frac{1}{2}f(x)\right),$$ which yields $\dis{f(x) = \frac{2}{3x} - \frac{x}{3}}$. \\ \end{answer} \end{pro} \begin{pro}Find all functions $f:\BBR\setminus\{-1\}\rightarrow \BBR$ such that $$(f(x))^2\cdot f\left(\frac{1 - x}{1 + x}\right) = 64x.$$ \begin{answer} We have $$(f(x))^2\cdot f\left(\frac{1 - x}{1 + x}\right) = 64x, $$whence $$(f(x))^4\cdot \left(f\left(\frac{1 - x}{1 + x}\right))\right)^2 = 64^2x^2 \qquad (I) $$ Substitute $x$ by $\frac{1 - x}{1 + x}$. Then $$f\left(\frac{1 - x}{1 + x}\right) ^2 f(x) = 64\left(\frac{1 - x}{1 + x}\right) . \qquad (II) $$ Divide (I) by (II), $$ f(x)^3 = 64 x^2\left(\frac{1+x}{1-x}\right), $$ from where the result follows. \end{answer} \end{pro} \begin{pro} Let $f^{[1]} = f$ be given by $\dis{f(x) = \frac{1}{1 - x}}$. Find \\ (i) $f^{[2]}(x) = (f\circ f)(x)$, \\ (ii) $f^{[3]}(x) = (f\circ f\circ f)(x)$, and \\ (iii) $f^{[69]} = \underbrace{(f\circ f\circ \cdots f\circ f)}_{69 \ {\rm compositions \ with \ itself}}(x)$. \begin{answer} We have (i) $\dis{f^{[2]}(x) = (f\circ f)(x) = f(f(x)) = \frac{1}{1 - \frac{1}{1 - x}} = \frac{x - 1}{x}}$. \\ (ii) $\dis{f^{[3]}(x) = (f\circ f \circ f)(x) = f(f^{[2]}(x))) = f\left(\frac{x - 1}{x}\right) = \frac{1}{1 - \frac{x - 1}{x}} = x}$. \\ (iii) Notice that $f^{[4]}(x) = (f\circ f^{[3]})(x) = f(f^{[3]}(x)) = f(x) = f^{[1]}(x)$. We see that $f$ is cyclic of period 3, that is, $f^{[1]} = f^{[14]} = f^{[7]} = \ldots, f^{[2]} = f^{[5]} = f^{[8]} = \ldots, f^{[3]} = f^{[6]} = f^{[9]} = \ldots$. Hence $f^{[69]}(x) = f^{[3]}(x) = x.$ \end{answer} \end{pro} \begin{pro} Let $f:A\rightarrow B$ and $g:B\rightarrow C$ be functions. Shew that (i) if $g\circ f$ is injective, then $f$ is injective. (ii) if $g\circ f$ is surjective, then $g$ is surjective. \begin{answer} To see (i) observe that $$f(a) = f(b)\implies g(f(a))=g(f(b))\implies a=b, $$whence $f$ is injective. (The first implication is clear, the second implication follows because $g\circ f$ is injective.) \bigskip To see (ii), given $y\in C$, \quad $\exists x\in A$ such that $g(f(x))=y$, since $g\circ f$ is surjective. But then, letting $a=f(x)\in B$ we have $g(a)=y$ and $g$ is surjective. \end{answer} \end{pro} \end{multicols} \section{Countability} \begin{df} A set $X$ is countable if either it is finite or if there is a bijection $f : X \rightarrow \BBN$, that is, the set $X$ has as many elements as $\BBN$. \end{df} Any countable set can be thus enumerated a sequence $$ x_1, x_2, x_3, \ldots . $$ Thus the strictly positive integers can be enumerated as customarily: $$1,2,3,\ldots . $$ Another possible enumeration\footnote{Which is relevant in chaos theory, for {\em Sarkovkii's Theorem}.}is the following $$ 3, 5, 7, 9, \ldots, \qquad ,2\cdot 3, 2\cdot 5, 2\cdot 7,2\cdot 9, \ldots \qquad , 2^2\cdot 3, 2^2\cdot 5, 2^2\cdot 7, 2^2\cdot 9, \ldots \qquad, \ldots 2^4, 2^3, 2^2, 2, 1, $$that is, we start with the odd integers in increasing order, then $2$ times the odd integers, $2^2$ times the odd integers, etc., and at the end we put the powers of $2$ in decreasing order. \begin{lem}\label{lem:subsets-of-N-countable-are} Any subset $X \subseteqq \BBN$ is countable. \end{lem} \begin{pf}If $X$ is finite, then there is nothing to prove. If $X$ is infinite, we can arrange the elements of $X$ increasing order, say, $$ x_1< x_2< x_3< \cdots. $$ We then map the smallest element $x_1\in S$ to $1$, the next smallest $x_2$ to $2$, etc. \end{pf} \begin{rem} Hence, even though $2\BBN \subsetneqq \BBN$, the sets $2\BBN$ and $\BBN$ have the same number of elements. This can also be seen by noticing that $f:\BBN \rightarrow 2\BBN$ given by $x_n=2n$ is a bijection. \end{rem} \begin{lem} A set $X$ is countable if and only if there is an injection $f : X \rightarrow \BBN$. \end{lem} \begin{pf} The assertion is evident if $X$ is finite. Hence assume $X$ is infinite. If $f : X \rightarrow \BBN$ is an injection then $f(X)$ is an infinite subset of $\BBN$. Hence there is a bijection $g : f(X) \rightarrow \BBN$ by virtue of Lemma \ref{lem:subsets-of-N-countable-are}. Thus $(g\circ f) : X \rightarrow \BBN$ is a bijection. \end{pf} \begin{rem} An obvious consequence of the above lemma is that if $X'$ is countable and there is an injection $f : X \rightarrow X'$ then $X$ is countable. \end{rem} \begin{thm} $\BBZ$ is countable. \end{thm} \begin{pf} One can take, as a bijection between the two sets, for example, $f : \BBZ \rightarrow \BBN$, $$ f(x)= \begin{cases} 2 x + 1 & \text{if $x \ge 0$} \\ - 2 x & \text{if $x < 0$.} \end{cases} $$ \end{pf} \begin{thm} $\BBQ$ is countable. \end{thm} \begin{pf} Consider $f : \BBQ \rightarrow \BBN$ given $$ f \left(\frac{a}{b}\right)= 2^{|a|} 3^b 5^{1+\signum{a}}, $$ where $\dfrac{a}{b}$ is in least terms, and $b > 0$. By the uniqueness of the prime factorisation of an integer, $f$ is an injection. \end{pf} \begin{rem} The above theorem means that there as many rational numbers as natural numbers. Thus the rationals can be enumerated as $$q_1, q_2, q_3, \ldots , $$ \end{rem} \begin{thm}[Cantor's Diagonal Argument] $\BBR$ is uncountable. \end{thm} \begin{pf} Assume $\BBR$ were countable so that its complete set of elements may be enumerated, say, as in the list \begin{align*} r_1 &= n_1 . d_{11} d_{12} d_{13} \dots \\ r_2 &= n_2 . d_{21} d_{22} d_{13} \dots \\ r_3 &= n_3 . d_{31} d_{32} d_{33} \dots , \end{align*} where we have used decimal notation. Define the new real $r = 0.d_1 d_2 d_3 \dots$ by $d_i = 0$ if $d_{ii} \neq 0$ and $d_i = 1$ if $d_{ii} = 0$. This is real number (as it is a decimal), but it differs from $r_i$ in the $i^{\text{th}}$ decimal place. It follows that the list is incomplete and the reals are uncountable. \end{pf} \begin{thm}The interval $\loro{-1}{1}$ is uncountable. \end{thm} \begin{pf} Observe that the map $f:\loro{-1}{1}\rightarrow \BBR$ given by $f(x)=\tan \dfrac{\pi x}{2}$ is a bijection. \end{pf} \subsection*{Homework}\addcontentsline{toc}{subsection}{Homework} \begin{multicols}{2}\columnseprule 1pt \columnsep 25pt\multicoltolerance=900\small \begin{pro} Prove that there as many numbers in $[0;1]$ as in any interval $[a;b]$ with $a 0,$$whence the denominator never vanishes and since sums, multiplications and divisions of rational numbers are rational, $\dfrac{a + b}{1 + ab}$ is also rational. We must prove now that $-1 < \dfrac{a + b}{1 + ab} < 1$ for $(a, b)\in ]-1; 1[^2$. We have $$\begin{array}{lll} -1 < \dfrac{a + b}{1 + ab} < 1 & \Leftrightarrow & -1 - ab < a + b < 1 + ab \\ & \Leftrightarrow & -1 -ab - a - b < 0 < 1 + ab - a - b \\ & \Leftrightarrow & -(a + 1)(b + 1) < 0 < (a - 1)(b - 1). \end{array}$$ Since $(a, b)\in ]-1; 1[^2$, $(a + 1)(b + 1) > 0$ and so $-(a + 1)(b + 1) < 0$ giving the sinistral inequality. Similarly $a - 1 < 0$ and $b - 1 < 0$ give $(a - 1)(b - 1) > 0$, the dextral inequality. Since the steps are reversible, we have established that indeed $-1 < \dfrac{a + b}{1 + ab} < 1$. \item Since $a \otimes b = \dfrac{a + b}{1 + ab} = \dfrac{b + a}{1 + ba} = b \otimes a$, commutativity follows trivially. Now $$\begin{array}{ccc}a \otimes (b \otimes c) & = & a \left(\frac{b + c}{1 + bc}\right) \\ & = & \dfrac{a + \left(\dfrac{b + c}{1 + bc}\right)}{1 + a\left(\dfrac{b + c}{1 + bc}\right)}\\ & = & \dfrac{a(1 + bc) + b + c}{1 + bc + a(b + c)} = \dfrac{a + b + c + abc}{1 + ab + bc + ca}. \end{array}$$One the other hand, $$\begin{array}{ccc} (a \otimes b) \otimes c & = & \left(\dfrac{a + b}{1 + ab}\right) c \\ & = & \dfrac{\left(\dfrac{a + b}{1 + ab}\right) + c}{1 + \left(\dfrac{a + b}{1 + ab}\right)c} \\ & = & \dfrac{(a + b)+ c(1 + ab)}{1 + ab + (a + b)c}\\ & = & \dfrac{a + b + c + abc}{1 + ab + bc + ca},\end{array}$$ whence $ \otimes $ is associative. \item If $a \otimes e = a$ then $\dfrac{a + e}{1 + ae} = a$, which gives $a + e = a + ea^2$ or $e(a^2 - 1) = 0$. Since $a \neq \pm 1$, we must have $e = 0$. \item If $a \otimes b = 0$, then $\dfrac{a + b}{1 + ab} = 0$, which means that $b = -a$, that is, $a^{-1} = -a$. \end{enumerate} \end{answer} \end{pro} \begin{pro} Let $G$ be a group satisfying $(\forall a\in G)$$$a^2 = e.$$ Prove that $G$ is an abelian group. \begin{answer} We must shew that $\forall (a, b)\in G^2$ we have $a b = b a.$ But $$\begin{array}{lll} a b & = & e (a b) e \\ & = & (b^2) (a b) (a^2) \\ & = & b ((b a ) (b a)) a \\ & = & b (b a)^2 a \\ & = & b (e) a \\ & = & b a, \end{array}$$whence the result follows. \end{answer} \end{pro} \begin{pro} Let $G$ be a group where $(\forall (a, b)\in G^2)$ $$((a b)^3 = a^3 b^3) \quad \mathrm{and} \quad ((a b)^5 = a^5 b^5).$$ Shew that $G$ is abelian. \begin{answer} We have $$\begin{array}{lll}(a b)^3 = a^3 b^3 & \implies & a b (a b) a b = a (a^2 b^2) b \\ & \implies & b a b a = a^2 b^2 \\ & \implies & (b a)^2 = a^2 b^2. \end{array}$$ Similarly $$\begin{array}{lll}(a b)^5 = a^5 b^5 & \implies & (b a)^4 = a^4 b^4. \end{array}$$ But we also have $$ (b a)^4\ = ((b a)^2)^2 = (a^2 b^2)^2 = a^2 (b^2 a^2) b^2,$$ and so $$a^2 (b^2 a^2) b^2 = (b a)^4 = a^4 b^4 \implies b^2 a^2 = a^2 b^2.$$ We have shewn that $\forall (a, b)\in G^2$ $$((b a)^2 = a^2 b^2) \quad \mathrm{and} \quad (b^2 a^2 = a^2 b^2).$$Hence $$\begin{array}{lll} (b a)^2 = a^2 b^2 = b^2 a^2 & \implies & b a b a = b^2 a^2 \\ & \implies & a b = b a, \\ \end{array}$$ proving that the group is abelian. \end{answer} \end{pro} \begin{pro} Suppose that in a group $G$ there exists a pair $(a, b)\in G^2$ satisfying $$(a b)^k = a^k b^k$$ for three consecutive integers $k = i, i + 1, i + 2.$ Prove that $a b = b a$. \begin{answer} Since $$(a b)^{i + 2} = \underbrace{(a b) (a b) \cdots (a b)}_{ i + 2 \ \ {\rm times}} = a (b a)^{i + 1} b, $$ multiplying by $a^{-1}$ on the left and by $b^{-1}$ on the right the equality \begin{equation}(a b)^{i + 2} = a^{i + 2} b^{i + 2} \label{eq:group_5} \end{equation}we obtain \begin{equation}(b a)^{i + 1} = (a)^{i + 1} (b)^{i + 1}. \label{eq:group_6} \end{equation}By hypothesis \begin{equation}(a b)^{i + 1} = (a)^{i + 1} (b)^{i + 1}. \label{eq:group_7} \end{equation} Hence (\ref{eq:group_6}) and (\ref{eq:group_7}) yield \begin{equation}(a b)^{i + 1} = (b a)^{i + 1}. \label{eq:group_8} \end{equation} Similarly, from (\ref{eq:group_7}) we obtain \begin{equation}(a b)^{i} = (b a)^{i}, \label{eq:group_9} \end{equation} from which \begin{equation}(a b)^{-i} = (b a)^{-i}. \label{eq:group_10} \end{equation} Multiplying (\ref{eq:group_8}) and (\ref{eq:group_10}) together, we deduce $$a b = b a,$$which is what we wanted to shew. \end{answer} \end{pro} \end{multicols} \section{Addition and Multiplication in $\BBR$} Since $\BBR$ is a field, it satisfies the following list of axioms, which we list for future reference. \begin{axi}[Arithmetical Axioms of $\BBR$] $\field{\BBR}{\cdot}{+}$---that is, the set of real numbers endowed with multiplication $\cdot$ and addition $+$---is a field. This entails that $+$ and $\cdot$ verify the following properties. \begin{enumerate} \item[{\bf R1:}]\label{axi:r1} $+$ and $\cdot$ are closed binary operations, that is, $$\forall (a, b)\in \BBR^2, \ \ \ a + b \in \BBR, \quad \ \ \ a\cdot b \in \BBR,$$ \item[{\bf R2:}]\label{axi:r2} $+$ and $\cdot$ are associative binary operations, that is, $$ \forall(a, b, c)\in \BBR^3, \quad a +(b + c) = (a + b)+ c, \quad a \cdot (b \cdot c) = (a \cdot b)\cdot c$$ \item[{\bf R3:}]\label{axi:r3} $+$ and $\cdot$ are commutative binary operations, that is, $$ \forall(a, b)\in \BBR^2, \quad a +b = b+a, \quad a \cdot b = b \cdot a,$$ \item[{\bf R4:}]\label{axi:r4} $\BBR$ has an additive identity element $0$, and a multiplicative identity element $1$, with $0\neq 1$, such that $$\forall a\in \BBR, \quad 0 +a = a +0 = a,\quad 1\cdot a = a\cdot 1 = a,$$ \item[{\bf R5:}]\label{axi:r5} Every element of $\BBR$ has an additive inverse, and every element of $\BBR\setminus \{0\}$ has a multiplicative inverse, that is, $$\forall a\in \BBR, \quad \exists (-a)\in \BBR \ {\rm such\ that\ } a + (-a) = (-a) + a = 0,$$ $$\forall b\in \BBR\setminus \{0\}, \quad \exists b^{-1}\in \BBR\setminus \{0\} \ {\rm such\ that\ } b\cdot b^{-1} = b^{-1}\cdot b = 1,$$ \item[{\bf R6:}]\label{axi:r6} $+$ and $\cdot$ satisfy the following distributive law: $$\forall (a, b, c,)\in \BBR ^3, \quad a\cdot (b+c) = a\cdot b + a\cdot c.$$ \end{enumerate} \end{axi} Since $+$ and $\cdot$ are associative in $\BBR$, we may write a sum $a_1+a_2+\cdots + a_n$ or a product $a_1a_2\cdots a_n$ of real numbers without risking ambiguity. We often use the following shortcut notation. \begin{df} For real numbers $a_i$ we define $$a_1+a_2+\cdots + a_n=\sum _{k=1} ^n a_k \qquad \mathrm{and}\qquad a_1a_2\cdots a_n= \prod _{k=1} ^n a_k.$$ \end{df} \begin{rem} By convention $\sum _{k\in\varnothing} a_k=0$ and $\prod _{k\in\varnothing} a_k=1$ . \end{rem} \begin{thm}[Lagrange's Identity]\label{thm:lagranges-id} Let $a_k, b_k$ be real numbers. Then $$ \left(\sum _{k=1} ^na_kb_k\right)^2 =\left(\sum _{k=1} ^na_k ^2\right)\left(\sum _{k=1} ^nb_k ^2\right)-\sum _{1\leq k n$ we take $ \binom{n}{k}=0$. \end{df} \begin{lem}[Pascal's Identity]\label{lem:pascal-id} For $n\geq 1$ and $1\leq k \leq n$, $$\binom{n}{k} = \binom{n-1}{k}+\binom{n-1}{k-1}. $$ \end{lem} \begin{pf} We have $$\begin{array}{lll}\binom{n-1}{k}+\binom{n-1}{k-1} & = & \dfrac{(n-1)!}{k!(n-1-k)!}+ \dfrac{(n-1)!}{(k-1)!(n-k)!}\\ & = & \dfrac{(n-1)!}{(k-1)!(n-1-k)}\left(\dfrac{1}{k}+\dfrac{1}{n-k}\right)\\ &= &\dfrac{(n-1)!}{(k-1)!(n-1-k)}\left(\dfrac{n}{k(n-k)}\right)\\ & = &\dfrac{n!}{k!(n-k)!} = \binom{n}{k}.\end{array} $$ \end{pf} Using Pascal's Identity we obtain {\em Pascal's Triangle.} \renewcommand{\arraystretch}{1.2} $$\begin{array}{ccccccccccc} & & & & & \binom{0}{0} & & & & & \\ & & & & \binom{1}{0} & & \binom{1}{1} & & & & \\ & & & \binom{2}{0} & & \binom{2}{1} & & \binom{2}{2} & & & \\ & & \binom{3}{0} & & \binom{3}{1} & & \binom{3}{2} & & \binom{3}{3} & & \\ & \binom{4}{0} & & \binom{4}{1} & & \binom{4}{2} & & \binom{4}{3} & & \binom{4}{4} & \\ \binom{5}{0} & & \binom{5}{1} & & \binom{5}{2} & & \binom{5}{3} & & \binom{5}{4} & & \binom{5}{5} \\ \vdots & & \vdots& & \vdots & & \vdots & & \vdots & & \vdots \\ \end{array}$$ When the numerical values are substituted, the triangle then looks like this. $$ \begin{array}{ccccccccccc} & & & & & 1 & & & & & \\ & & & & 1 & & 1 & & & & \\ & & & 1 & & 2 & & 1 & & & \\ & & 1 & & 3 & & 3 & & 1 & & \\ & 1 & & 4 & & 6 & & 4 & & 1 & \\ 1 & & 5 & & 10 & & 10 & & 5 & & 1 \\ \vdots & & \vdots& & \vdots & & \vdots & & \vdots & & \vdots \\ \end{array}$$ We see from Pascal's Triangle that binomial coefficients are symmetric. This symmetry is easily justified by the identity $\binom{n}{k} = \binom{n}{n - k}$. We also notice that the binomial coefficients tend to increase until they reach the middle, and that then they decrease symmetrically. \begin{thm}[Binomial Theorem] For $n \in\BBN,$ $$(x + y)^n = \sum _{k = 0} ^n \binom{n}{k}x^ky^{n-k} .$$ \end{thm} \begin{pf} The theorem is obvious for $n=0$ (defining $(x+y)^0=1$), $n=1$ (as $(x+y)^1=x+y$), and $n=2$ (as $(x+y)^2=x^2+2xy+y^2$). Assume $n\geq 3$. The induction hypothesis is that $(x + y)^n = \sum _{k = 0} ^n \binom{n}{k}x^ky^{n-k} .$ Then we have $$\begin{array}{lll}(x+y)^{n+1} & = & (x+y)(x+y)^n\\ & = & (x+y)\left(\sum _{k = 0} ^n \binom{n}{k}x^ky^{n-k}\right)\\ & = & \sum _{k = 0} ^n \binom{n}{k}x^{k+1}y^{n-k}+ \sum _{k = 0} ^n \binom{n}{k}x^ky^{n-k+1}\\ & = & x^{n+1} +\sum _{k = 0} ^{n-1} \binom{n}{k}x^{k+1}y^{n-k} + \sum _{k = 1} ^n \binom{n}{k}x^ky^{n-k+1} + y^{n+1}\\ & = & x^{n+1} +\sum _{k = 1} ^{n} \binom{n}{k-1}x^{k}y^{n-k+1} + \sum _{k = 1} ^n \binom{n}{k}x^ky^{n-k+1} + y^{n+1}\\ & = & x^{n+1} +\sum _{k = 1} ^{n} \left(\binom{n}{k-1}+\binom{n}{k}\right)x^{k}y^{n-k+1}+ y^{n+1}\\ & = & x^{n+1} +\sum _{k = 1} ^{n} \binom{n+1}{k}x^{k}y^{n-k+1}+ y^{n+1}\\ & = & \sum _{k = 0} ^{n+1} \binom{n+1}{k}x^{k}y^{n-k+1},\\ \end{array}$$proving the theorem. \end{pf} \begin{lem}\label{lem:finite-geom-sum} If $a\in\BBR$, $a\neq 1$ and $n\in\BBN\setminus \{0\}$, then $$1 + a + a^2 + \cdots a^{n-1} = \dfrac{1 - a^{n}}{1 -a}. $$ \end{lem} \begin{pf} For, put $S = 1 + a + a^2 + \cdots + a^{n-1}.$ Then $aS = a + a^2 + \cdots + a^{n-1} + a^n.$ Thus $$S - aS = (1 + a + a^2 + \cdots +a^{n-1}) - (a + a^2 + \cdots + a^{n-1} + a^n) = 1 - a^n,$$ and from $(1-a)S = S-aS=1 - a^n $ we obtain the result. \end{pf} \begin{thm} Let $n$ be a strictly positive integer. Then $$y^n - x^n = (y - x)(y^{n-1} + y^{n-2}x + \cdots + yx^{n-2} + x^{n-1}).$$ \label{thm:diffbinom}\end{thm} \begin{pf} By making the substitution $a = \frac{x}{y}$ in Lemma \ref{lem:finite-geom-sum} we see that $$1 + \frac{x}{y} + \left(\frac{x}{y}\right)^2 + \cdots + \left(\frac{x}{y}\right)^{n-1} = \dfrac{1- \left(\frac{x}{y}\right)^n}{1-\frac{x}{y}} $$ we obtain $$\left(1-\frac{x}{y}\right)\left(1 + \frac{x}{y} + \left(\frac{x}{y}\right)^2 + \cdots + \left(\frac{x}{y}\right)^{n-1}\right) = 1- \left(\frac{x}{y}\right)^n, $$ or equivalently, $$\left(1-\frac{x}{y}\right)\left(1 + \frac{x}{y} + \frac{x^2}{y^2} + \cdots +\frac{x^{n-1}}{y^{n-1}}\right) = 1- \frac{x^n}{y^n}. $$ Multiplying by $y^n$ both sides, $$y\left(1-\frac{x}{y}\right) y^{n-1}\left(1 + \frac{x}{y} + \frac{x^2}{y^2} + \cdots +\frac{x^{n-1}}{y^{n-1}}\right) = y^n\left(1- \frac{x^n}{y^n}\right), $$ which is $$y^n - x^n = (y-x)(y^{n-1} + y^{n-2}x + \cdots + yx^{n-2} + x^{n-1}), $$ yielding the result. \end{pf} \begin{thm}\label{thm:sum-of-first-n-integers} $1 + 2 + \cdots + n= \dfrac{n(n+1)}{2}$. \end{thm} \begin{f-pf} Observe that $$ k^2 - (k - 1)^2 = 2k - 1.$$ From this $$ \begin{array}{lcl} 1^2 - 0^2 & = & 2\cdot 1 - 1 \\ 2^2 - 1^2 & = & 2\cdot 2 - 1 \\ 3^2 - 2^2 & = & 2\cdot 3 - 1 \\ \vdots & \vdots & \vdots \\ n^2 - (n - 1)^2 & = & 2\cdot n - 1 \end{array} $$ Adding both columns, $$n^2 - 0^2 = 2(1 + 2 + 3 + \cdots + n) - n.$$ Solving for the sum, $$1 + 2 + 3 + \cdots + n = n^2/2 + {n}/{2} = \frac{n(n + 1)}{2}.$$ \end{f-pf} \begin{s-pf} We may utilise Gauss' trick: If $$ A_n = 1 + 2 + 3 + \cdots + n $$ then$$ A_n = n + (n - 1) + \cdots + 1.$$Adding these two quantities, $$ \begin{array}{lcccccccc} A_n & = & 1 & + & 2 & + & \cdots & + & n \\ {A_n} & {=} & n & + & (n - 1) & + & \cdots & + & 1 \\ \hline 2A_n & = & (n + 1) & + & (n + 1) & + & \cdots & + & (n + 1) \\ & = & n(n + 1), & & & & & \end{array}$$since there are $n$ summands. This gives $A_n = \dis{\frac{n(n + 1)}{2}}$, that is, $$ 1 + 2 + \cdots + n = \frac{n(n + 1)}{2}.$$Applying Gauss's trick to the general arithmetic sum $$ (a) + (a + d) + (a + 2d) + \cdots + (a + (n - 1)d) $$we obtain \begin{equation} (a) + (a + d) + (a + 2d) + \cdots + (a + (n - 1)d) = \frac{n(2a + (n - 1)d)}{2} \end{equation} \end{s-pf} \begin{thm}\label{thm:sum-of-first-squares} $1^2 + 2^2 + 3^2 + \cdots + n^2=\frac{n(n + 1)(2n + 1)}{6}.$ \end{thm}\begin{pf} Observe that $$ k^3 - (k - 1)^3 = 3k^2 - 3k + 1.$$ Hence $$ \begin{array}{lcl} 1^3 - 0^3 & = & 3\cdot 1^2 - 3\cdot 1 + 1 \\ 2^3 - 1^3 & = & 3\cdot 2^2 - 3\cdot 2 + 1 \\ 3^3 - 2^3 & = & 3\cdot 3^2 - 3\cdot 3 + 1 \\ \vdots & \vdots & \vdots \\ n^3 - (n - 1)^3 & = & 3\cdot n^2 - 3\cdot n + 1 \end{array} $$ Adding both columns, $$n^3 - 0^3 = 3(1^2 + 2^2 + 3^2 + \cdots + n^2) - 3(1 + 2 + 3 + \cdots + n) + n.$$ From the preceding example $1 + 2 + 3 + \cdots + n = \cdot n^2/2 + {n}/{2} = \frac{n(n + 1)}{2}$ so $$n^3 - 0^3 = 3(1^2 + 2^2 + 3^2 + \cdots + n^2) - \frac{3}{2}\cdot n(n + 1) + n.$$ Solving for the sum, $$1^2 + 2^2 + 3^2 + \cdots + n^2 = \frac{n^3}{3} + \frac{1}{2}\cdot n(n + 1) - \frac{n}{3}.$$ After simplifying we obtain $$ 1^2 + 2^2 + 3^2 + \cdots + n^2 = \frac{n(n + 1)(2n + 1)}{6}.$$ \end{pf} \subsection*{Homework}\addcontentsline{toc}{subsection}{Homework} \begin{multicols}{2}\columnseprule 1pt \columnsep 25pt\multicoltolerance=900\small \begin{pro} Prove that for $n\geq 1$, $$2^n = \sum _{k=0} ^n \binom{n}{k}; \qquad 0 = \sum _{k=0} ^n (-1)^k\binom{n}{k}, \qquad 2^{n-1} = \sum _{\substack{0 \leq k \leq n\\ k \ \mathrm{even}}} \binom{n}{k} = \sum _{\substack{1 \leq k \leq n \\ k \ \mathrm{odd}}} \binom{n}{k}. $$ \begin{answer} The first two follow immediately from the Binomial Theorem, the first by putting $x=y=1$ and then $x=-y=1$. The third follows by adding the first two and dividing by $2$. The fourth follows by subtracting the second from the first and then dividing by $2$. \end{answer} \end{pro} \begin{pro} Given that $1002004008016032$ has a prime factor $p > 250000,$ find it. \begin{answer} If $a = 10^3 , b = 2$ then $$ 1002004008016032 = a^5 + a^4 b + a^3 b^2 + a^2 b^3 + ab^4 + b^5 = \frac{ a^6 - b^6 }{ a - b }. $$ This last expression factorises as $$ {\everymath{\displaystyle} \begin{array}{lcl}\frac{a^6 - b^6}{a - b} & = & (a + b)(a^2 + ab + b^2)(a^2 - ab + b^2 ) \\ & = & 1002\cdot 1002004\cdot 998004 \\ & = & 4\cdot 4\cdot 1002\cdot 250501 \cdot k,\end{array} }$$where $k < 250000$. Therefore $p = 250501$. \end{answer} \end{pro} \begin{pro} Prove that $(a+b+c)^2 = a^2+b^2+c^2+2ab+2bc+2ca$. \end{pro} \begin{pro} Let $a, b,c$ be real numbers. Prove that $$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca).$$ \begin{answer} From the Binomial Theorem, $$ (A+B)^3 = A^3+3A^2B+3AB^2+B^3\implies A^3+B^3=(A+B)^3-3AB(A+B). $$Then $$ \begin{array}{lll} a^3 + b^3 + c^3 - 3abc & = & (a + b)^3 + c^3 - 3ab(a + b) - 3abc \\ & = & (a + b + c)^3 - 3(a + b)c(a + b + c) - 3ab(a + b + c) \\ & = & (a + b + c)((a + b + c)^2 - 3ac - 3bc - 3ab) \\ & = & (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca). \end{array} $$ \end{answer} \end{pro} \begin{pro}Prove that $$ \binom{n}{k} = \dfrac{n}{k}\binom{n-1}{k-1}.$$ \begin{answer} $$ \binom{n}{k} = \dfrac{n!}{k!(n-k)!} = \dfrac{n}{k}\cdot \dfrac{(n-1)!}{(k-1)!(n-k)!} = \dfrac{n}{k}\binom{n-1}{k-1}.$$ \end{answer} \end{pro} \begin{pro}Prove that $$ \binom{n}{k} = \dfrac{n}{k}\cdot \dfrac{n-1}{k-1} \cdot\binom{n-2}{k-2}.$$ \begin{answer} $$ \binom{n}{k} = \dfrac{n!}{k!(n-k)!} = \dfrac{n(n-1)}{k(k-1)}\cdot \dfrac{(n-2)!}{(k-2)!(n-k)!} = \dfrac{n}{k}\cdot \dfrac{n-1}{k-1} \cdot\binom{n-2}{k-2}.$$ \end{answer} \end{pro} \begin{pro}Prove that $$\sum _{k = 1} ^n k\binom{n}{k}p^{k}(1 - p)^{n - k} = np. $$ \begin{answer} We use the identity $k\binom{n}{k} = n\binom{n - 1}{k - 1}$. Then $$\begin{array}{lll}\sum _{k = 1} ^n k\binom{n}{k}p^{k}(1 - p)^{n - k} & = &\sum _{k = 1} ^n n\binom{n - 1}{k - 1}p^{k}(1 - p)^{n - k} \\ & = & \sum _{k = 0} ^{n-1} n\binom{n - 1}{k}p^{k+1}(1 - p)^{n-1 - k} \\ & = & np\sum _{k = 0} ^{n-1} \binom{n - 1}{k}p^{k}(1 - p)^{n-1 - k} \\ & = & np(p + 1-p)^{n - 1} \\ & = & np. \end{array}$$ \end{answer} \end{pro} \begin{pro}Prove that $$\sum _{k = 2} ^n k(k-1)\binom{n}{k}p^{k}(1 - p)^{n - k} = n(n - 1)p^2. $$ \begin{answer} We use the identity $$k(k-1)\binom{n}{k} = n(n-1)\binom{n - 2}{k - 2}.$$ Then $$\begin{array}{lll}\sum _{k = 2} ^n k(k-1)\binom{n}{k}p^{k}(1 - p)^{n - k} & = &\sum _{k = 2} ^n n(n-1)\binom{n - 2}{k - 2}p^{k}(1 - p)^{n - k} \\ & = & \sum _{k = 0} ^{n-2} n(n-1)\binom{n - 2}{k}p^{k+2}(1 - p)^{n-1 - k} \\ & = & n(n-1)p^2\sum _{k = 0} ^{n-2} \binom{n - 1}{k}p^{k}(1 - p)^{n-2 - k} \\ & = & n(n-1)p^2(p + 1-p)^{n - 2} \\ & = & n(n-1)p^2. \end{array}$$ \end{answer} \end{pro} \begin{pro}Demonstrate that $$\sum _{k = 0} ^n (k - np)^2\binom{n}{k}p^{k}(1 - p)^{n - k} = np(1- p). $$ \begin{answer} We use the identity $$(k - np)^2 = k^2 -2knp + n^2p^2 = k(k-1) + k(1-2np) + n^2p^2.$$ Then $$\begin{array}{lll} \sum _{k = 0} ^n (k - np)^2\binom{n}{k}p^{k}(1 - p)^{n - k} & = & \sum _{k = 0} ^n (k(k-1) + k(1-2np)\\ & & \quad + n^2p^2)\binom{n}{k}p^{k}(1 - p)^{n - k} \\ & = & \sum _{k = 0} ^n k(k-1)\binom{n}{k}p^{k}(1 - p)^{n - k}\\ & & + (1 - 2np)\sum _{k = 0} ^n k\binom{n}{k}p^{k}(1 - p)^{n - k}\\ & & + n^2p^2\sum _{k = 0} ^n \binom{n}{k}p^{k}(1 - p)^{n - k} \\ & = & n(n-1)p^2 + np(1 -2np) + n^2p^2\\ & = & np(1-p). \end{array}$$ \end{answer} \end{pro} \begin{pro} Let $x\in\BBR\setminus\{1\}$ and let $n\in\BBN\setminus\{0\}$. Prove that $$ \sum _{k=0} ^n \dfrac{2^k}{x^{2^k}+1}=\dfrac{1}{x-1}-\dfrac{2^{n+1}}{x^{2^{n+1}}+1}. $$ \end{pro} \begin{pro} Consider the $n^k$ $k$-tuples $(a_1, a_2, \ldots , a_k)$ which can be formed by taking $a_i\in\{1,2,\ldots , n\}$, repetitions allowed. Demonstrate that $$ \sum _{a_i\in \{1,2,\ldots , n\}} \min (a_1, a_2, \ldots ,a_k) = 1^k + 2^k + \cdots + n^k. $$ \begin{answer} Observe that the number of $k$-tuples with $\min ((a_1, a_2, \ldots , a_k))=t$ is $(n-t+1)^k-(n-t)^k$. \end{answer} \end{pro} \end{multicols} \section{Order Axioms} \begin{rem} \fcolorbox{red}{cyan}{ \begin{minipage}{.90\linewidth} \noindent\textcolor{red}{\textbf{Vocabulary Alert!}} We will call a number $x$ {\em positive} if $x\geq 0$ and {\em strictly positive} if $x>0$. Similarly, we will call a number $y$ {\em negative} if $y\leq 0$ and {\em strictly negative} if $y<0$. This usage differs from most Anglo-American books, who prefer such terms as {\em non-negative} and {\em non-positive}. \end{minipage}} \end{rem} We assume $\BBR$ endowed with a relation $>$ which satisfies the following axioms. \begin{axi}[Trichotomy Law]\label{axi:trichotomy} $\forall (x,y)\in\BBR ^2$ exactly one of the following holds: $$ x> y, \quad x=y, \quad \mathrm{or}\quad y>x.$$ \end{axi} \begin{axi}[Transitivity of Order]\label{axi:transitivity} $\forall (x,y, z)\in\BBR ^3$, $$\mathrm{if}\quad x> y \quad \mathrm{and}\quad y>z \quad \mathrm{then}\quad x>z.$$ \end{axi} \begin{axi}[Preservation of Inequalities by Addition]\label{axi:preservation-of-ineqs-by-addition} $\forall (x,y, z)\in\BBR ^3$, $$\mathrm{if} \quad x> y \quad \mathrm{then}\quad x+z>y+z.$$ \end{axi} \begin{axi}[Preservation of Inequalities by Positive Factors]\label{axi:preservation-of-ineqs} $\forall (x,y, z)\in\BBR ^3$, $$\mathrm{if} \quad x> y \quad \mathrm{and}\quad z>0 \quad \mathrm{then}\quad xz>yz.$$ \end{axi} \begin{rem} $xx$. $x\leq y$ means that either $y>x$ or $y=x$, etc. \end{rem} \begin{thm}\label{thm:square-of-a-real} The square of any real number is positive, that is, $\forall a\in \BBR$, $\qquad a^2 \geq 0$. In fact, if $a\neq 0$ then $a^2>0$. \end{thm} \begin{pf} If $a=0$, then $0^2= 0$ and there is nothing to prove. Assume now that $a\neq 0$. By trichotomy, either $a>0$ or $a<0$. Assume first that $a>0$. Applying Axiom \ref{axi:preservation-of-ineqs} with $x=z=a$ and $y=0$ we have $$aa>a0\implies a^2>0,$$so the theorem is proved if $a>0$. \bigskip If $a<0$ then $-a>0$ and we apply the result just obtained: $$ -a>0 \implies (-a)^2>0 \implies 1\cdot a^2>0 \implies a^2>0, $$so the result is true regardless the sign of $a$. \end{pf} Theorem \ref{thm:square-of-a-real} will prove to be extremely powerful and will be the basis for many of the classical inequalities that follow. \begin{thm}If $(x, y)\in\BBR^2$, $$ x>y \iff x-y > 0. $$ \end{thm} \begin{pf} This is a direct consequence of Axiom \ref{axi:preservation-of-ineqs-by-addition} upon taking $z=-y$. \end{pf} \begin{thm}If $(x, y, a, b)\in\BBR^4$, $$ x>y \quad \mathrm{and} \quad a \geq b \implies x+a > y + b. $$ \end{thm} \begin{pf} We have $$x>y \implies x+a>y+a, \quad y+a\geq y+b, $$by Axiom \ref{axi:preservation-of-ineqs-by-addition} and so by Axiom \ref{axi:transitivity} $x+a>y+b$. \end{pf} \begin{thm}If $(x, y, a, b)\in\BBR^4$, $$ x>y >0\quad \mathrm{and} \quad a \geq b>0 \implies xa > yb. $$ \end{thm} \begin{pf} Indeed $$x>y \implies xa>ya, \quad ya\geq yb, $$by Axiom \ref{axi:preservation-of-ineqs} and so by Axiom \ref{axi:transitivity} $xa>yb$. \end{pf} \begin{thm}\label{thm:0<1} $1>0$. \end{thm} \begin{pf} By definition of $\BBR$ being a field $0\neq 1$. Assume that $1<0$ then $1^2>0$ by Theorem \ref{thm:square-of-a-real}. But $1^2 = 1$ and so $1>0$, a contradiction to our original assumption. \end{pf} \begin{thm} $x>0 \implies -x<0\quad \mathrm{and}\quad x^{-1}>0$. \end{thm} \begin{pf} Indeed, $-1<0$ since $-1\neq 0$ and assuming $-1>0$ would give $0=-1+1>1$, which contradicts Theorem \ref{thm:0<1}. Thus $$ -x=-1\cdot x < 0. $$Similarly, assuming $x^{-1}<0$ would give $1=x^{-1}x<0$. \end{pf} \begin{thm} $x>1 \implies x^{-1}<1$. \end{thm} \begin{pf} Since $x^{-1}\neq 1$, assuming $x^{-1}>1$ would give $1=xx^{-1}>1\cdot 1=1$, a contradiction. \end{pf} \subsection{Absolute Value} \begin{df}[The Signum (Sign) Function] Let $x$ be a real number. We define $ \signum{x}= \left\{ \begin{tabular}{cl} $-1$ & \rm{if} \ $x < 0$, \\ $0$ & \rm{if} \ $x = 0$,\\ $+1$ & \rm{if} \ $x > 0$. \\ \end{tabular} \right. $ \end{df} \begin{lem}\label{lem:signum-is-multiplicative} The signum function is multiplicative, that is, if $(x,y)\in\BBR^2$ then $\signum{x\cdot y}=\signum{x}\signum{y}$. \end{lem} \begin{pf} Immediate from the definition of signum. \end{pf} \begin{df}[Absolute Value] Let $x\in\BBR$. The {\em absolute value} of $x$ is defined and denoted by $$ \absval{x}=\signum{x}x. $$ \end{df} \begin{thm}\label{thm:absval-properties} Let $x\in\BBR$. Then \begin{enumerate} \item $ |x| = \left\{ \begin{array}{ll} -x & \mathrm{if} \ x < 0, \\ x & \mathrm{if} \ x \geq 0. \end{array} \right.$ \item $\absval{x}\geq 0$, \item $\absval{x}=\max (x,-x)$, \item $\absval{-x} = \absval{x}$, \item \label{eq:abs_val_interval} $-\absval{x} \leq x \leq \absval{x}$. \item $\sqrt{x^2} = |x| $ \item $|x|^2 = |x^2| = x^2$ \item $x = \signum{x}\absval{x}$ \end{enumerate} \end{thm} \begin{pf} These are immediate from the definition of $\absval{x}$. \end{pf} \begin{thm} $(\forall (x, y)\in \BBR^2)$, $$\absval{xy} =\absval{x}\absval{y}. $$ \end{thm} \begin{pf} We have $$\absval{xy} = \signum{xy}xy = \left(\signum{x}x\right) \left(\signum{y}y\right) = \absval{x}\absval{y}, $$where we have used Lemma \ref{lem:signum-is-multiplicative}. \end{pf} \begin{thm}\label{thm:|x|within_t}Let $t\geq 0$. Then $$|x| \leq t \iff -t \leq x \leq t. $$ \end{thm} \begin{pf} Either $\absval{x} =x$ or $\absval{x}=-x$. If $\absval{x} =x$, $$\absval{x}\leq t \iff x\leq t \iff -t \leq 0\leq x \leq t. $$ If $\absval{x} =-x$, $$\absval{x}\leq t \iff -x\leq t \iff -t \leq x \leq 0 \leq t. $$ \end{pf} \begin{thm}If $(x,y)\in\BBR^2$, $\max (x, y) = \dfrac{x+y+\absval{x-y}}{2}$ and $\min (x, y) = \dfrac{x+y-\absval{x-y}}{2}$. \end{thm} \begin{pf}Observe that $\max (x, y) + \min (x, y) = x+y$, since one of these quantities must be the maximum and the other the minimum, or else, they are both equal. \bigskip Now, either $\absval{x-y} = x-y$, and so $x\geq y$, meaning that $\max (x, y)-\min (x, y) = x-y$, or $\absval{x-y} = -(x-y)=y-x$, which means that $y\geq x$ and so $\max (x, y)-\min (x, y) = y-x$. In either case we get $\max(x,y)-\min (x, y)=\absval{x-y}$. Solving now the system of equations $$\begin{array}{lll}\max (x, y) + \min (x, y) & = & x+y\\ \max(x,y)-\min (x, y) & = &\absval{x-y},\\ \end{array} $$for $\max(x,y)$ and $\min (x,y)$ gives the result. \end{pf} \subsection*{Homework}\addcontentsline{toc}{subsection}{Homework} \begin{multicols}{2}\columnseprule 1pt \columnsep 25pt\multicoltolerance=900\small \begin{pro} Let $x, y$ be real numbers. Then $$ 0 \leq x < y \iff x^2 < y^2. $$ \end{pro} \begin{pro}Let $t \geq 0$. Prove that $$|x| \geq t \iff (x \geq t) \quad \mathrm{or}\quad (x\leq -t). $$ \end{pro} \begin{pro} Let $(x, y)\in \BBR^2$. Prove that $\max (x, y)=-\min (-x, -y)$. \end{pro} \begin{pro} Let $x, y , z$ be real numbers. Prove that $$ \max (x, y, z) = x + y + z - \min (x, y) - \min (y, z) - \min (z, x) + \min (x, y , z). $$ \end{pro} \begin{pro} Let $a 0$ for $1\leq i \leq n$. Then $$\min \left(\dfrac{a_1}{b_1},\dfrac{a_2}{b_2},\ldots , \dfrac{a_n}{b_n} \right) \leq \dfrac{a_1+a_2+\cdots + a_n}{b_1+b_2+\cdots + b_n}\leq \max \left(\dfrac{a_1}{b_1},\dfrac{a_2}{b_2},\ldots , \dfrac{a_n}{b_n} \right) . $$ \end{thm} \begin{pf} For every $k$, \quad $1\leq k \leq n$, $$\min \left(\dfrac{a_1}{b_1},\dfrac{a_2}{b_2},\ldots , \dfrac{a_n}{b_n} \right) \leq \dfrac{a_k}{b_k} \leq \max \left(\dfrac{a_1}{b_1},\dfrac{a_2}{b_2},\ldots , \dfrac{a_n}{b_n} \right) \implies b_k\min \left(\dfrac{a_1}{b_1},\dfrac{a_2}{b_2},\ldots , \dfrac{a_n}{b_n} \right) \leq a_k \leq b_k\max \left(\dfrac{a_1}{b_1},\dfrac{a_2}{b_2},\ldots , \dfrac{a_n}{b_n} \right). $$Adding all these inequalities for $1\leq k\leq n$, $$(b_1+b_2+\cdots + b_n)\min \left(\dfrac{a_1}{b_1},\dfrac{a_2}{b_2},\ldots , \dfrac{a_n}{b_n} \right) \leq a_1+a_2+\cdots + a_n \leq (b_1+b_2+\cdots + b_n)\max \left(\dfrac{a_1}{b_1},\dfrac{a_2}{b_2},\ldots , \dfrac{a_n}{b_n} \right), $$from where the result is obtained.\end{pf} \subsection{Bernoulli's Inequality} \begin{thm} If $0 \leq a < b, \ n\geq 1 \in \BBN $ $$na^{n - 1} < \frac{b^{n} - a^{n}}{b - a} < nb^{n - 1}.$$ \label{thm:ineq_bin}\end{thm} \begin{pf} By Theorem \ref{thm:diffbinom}, $$\begin{array}{lll} \dfrac{b^{n} - a^{n}}{b - a} & = & b^{n - 1} + b^{n - 2}a + b^{n - 3}a^2 + \cdots + b^2a^{n - 3} + ba^{n - 2} + a^{n - 1} \\ & < & b^{n - 1} + b^{n - 1} + \cdots + b^{n - 1} + b^{n - 1} \\ & = & nb^{n - 1}, \end{array} $$from where the dextral inequality follows. The sinistral inequality can be established similarly. \end{pf} \begin{thm}[Bernoulli's Inequality] If $x > -1, x \neq 0,$ and if $n \in\BBN\setminus \{0\}$ then $$(1 + x)^n > 1 + nx.$$ \label{thm:bernoulli} \end{thm} \begin{pf} Set $b = 1 + x, a = 1$ in Theorem \ref{thm:ineq_bin} and use the sinistral inequality. \end{pf} \begin{rem} If $x> 0$ then Bernoulli's Inequality is an easy consequence of the Binomial Theorem, as $$(1+x)^n = 1 + \binom{n}{1}x+ \binom{n}{2}x^2+ \cdots > 1 + \binom{n}{1}x = 1+nx. $$ \end{rem} \subsection{Rearrangement Inequality} \begin{df} Given a set of real numbers $\{x_1, x_2, \ldots , x_n\}$ denote by $$\check{x}_1\geq \check{x}_2\geq \cdots \geq \check{x}_n$$ the decreasing rearrangement of the $x_i$ and denote by $$\hat{x}_1\leq \hat{x}_2\leq \cdots \leq \hat{x}_n$$the increasing rearrangement of the $x_i$. \end{df} \begin{df} Given two sequences of real numbers $\{x_1, x_2, \ldots , x_n\}$ and $\{y_1, y_2, \ldots , y_n\}$ of the same length $n$, we say that they are {\em similarly sorted} if they are both increasing or both decreasing, and {\em differently sorted} if one is increasing and the other decreasing.. \end{df} \begin{exa} The sequences $1 \leq 2 \leq \cdots \leq n$ and $1^2 \leq 2^2 \leq \cdots \leq n^2$ are similarly sorted, and the sequences $\dfrac{1}{1^2} \geq \dfrac{1}{2^2} \geq \cdots \geq \dfrac{1}{n^2}$ and $1^3 \leq 2^3 \leq \cdots \leq n^3$ are differently sorted. \end{exa} \begin{thm}[Rearrangement Inequality]\label{thm:rearrangement-ineq} Given sets of real numbers $\{a_1, a_2, \ldots , a_n\}$ and $\{b_1, b_2, \ldots , b_n\}$ we have $$ \sum _{1\leq k\leq n}\check{a}_k\hat{b}_k \leq \sum _{1\leq k \leq n} a_kb_k \leq \sum _{1\leq k \leq n}\hat{a}_k\hat{b}_k. $$Thus the sum $ \sum _{1\leq k \leq n} a_kb_k $ is minimised when the sequences are differently sorted, and maximised when the sequences are similarly sorted. \end{thm} \begin{rem} Observe that $$\sum _{1\leq k \leq n}\check{a}_k\hat{b}_k = \sum _{1\leq k \leq n}\hat{a}_k\check{b}_k\qquad\mathrm{and}\qquad \sum _{1\leq k \leq n}\hat{a}_k\hat{b}_k = \sum _{1\leq k \leq n}\check{a}_k\check{b}_k. $$ \end{rem} \begin{pf} Let $\{\sigma (1),\sigma(2), \ldots , \sigma(n)\}$ be a reordering of $\{1,2, \ldots , n\}$. If there are two sub-indices $i, j$, such that the sequences pull in opposite directions, say, $a_i >a_j$ and $b_{\sigma (i)}< b_{\sigma (j)}$, then consider the sums $$ \begin{array}{lll}S & = & a_1b_{\sigma (1)}+a_2b_{\sigma (2)}+\cdots +a_ib_{\sigma (i)}+\cdots +a_jb_{\sigma (j)}+\cdots + a_nb_{\sigma (n)}\\ S' & = & a_1b_{\sigma (1)}+a_2b_{\sigma (2)}+\cdots +a_ib_{\sigma (j)}+\cdots +a_jb_{\sigma (i)}+\cdots + a_nb_{\sigma (n)}\\ \end{array}$$ Then $$ S'-S = (a_i-a_j)(b_{\sigma(j)}-b_{\sigma (i)}) >0. $$This last inequality shews that the closer the $a$'s and the $b$'s are to pulling in the same direction the larger the sum becomes. This proves the result. \end{pf} \subsection{Arithmetic Mean-Geometric Mean Inequality} \begin{thm}[Arithmetic Mean-Geometric Mean Inequality]\label{thm:AMGM-ineq} Let $a_{1}, \dots, a_{n}$ be positive real numbers. Then their geometric mean is at most their arithmetic mean, that is, $$ \sqrt[n]{a_{1}\cdots a_{n}} \leq \dfrac{a_{1} + \cdots + a_{n}}{n}, $$ with equality if and only if $a_{1} = \cdots = a_{n}$. \end{thm} We will provide multiple proofs of this important inequality. Some other proofs will be found in latter chapters. \begin{f-pf} Our first proof uses the Rearrangement Inequality (Theorem \ref{thm:rearrangement-ineq}) in a rather clever way. We may assume that the $a_k$ are strictly positive. Put $$x_1=\dfrac{a_1}{(a_1a_2\cdots a_n)^{1/n}}, \quad x_2=\dfrac{a_1a_2}{(a_1a_2\cdots a_n)^{2/n}},\quad \ldots ,\quad x_n=\dfrac{a_1a_2\cdots a_n}{(a_1a_2\cdots a_n)^{n/n}}=1, $$ and$$ y_1 = \dfrac{1}{x_1}, \quad y_2 = \dfrac{1}{x_2},\quad \ldots ,\quad y_n = \dfrac{1}{x_n}=1. $$ Observe that for $2 \leq k \leq n$, $$x_ky_{k-1} = \dfrac{a_1a_2\cdots a_k}{(a_1a_2\cdots a_n)^{k/n}}\cdot \dfrac{(a_1a_2\cdots a_n)^{(k-1)/n}} {a_1a_2\cdots a_{k-1}} = \dfrac{a_k}{(a_1a_2\cdots a_n)^{1/n}}.$$ The $x_k$ and $y_k$ are differently sorted, so by virtue of the Rearrangement Inequality we gather $$\begin{array}{lll} 1+1+\cdots + 1 & = & x_1y_1+x_2y_2+\cdots + x_ny_n \\ & \leq & x_1y_n+x_2y_{1}+\cdots + x_ny_{n-1} \\ & = & \dfrac{a_1}{(a_1a_2\cdots a_n)^{1/n}}+ \dfrac{a_2}{(a_1a_2\cdots a_n)^{1/n}} + \cdots + \dfrac{a_n}{(a_1a_2\cdots a_n)^{1/n}}, \end{array}$$ or $$ n \leq \dfrac{a_1+a_2+\cdots + a_n}{(a_1a_2\cdots a_n)^{1/n}}, $$from where we obtain the result. \end{f-pf} \begin{s-pf} This second proof is a clever induction argument due to Cauchy. It proves the inequality first for powers of $2$ and then interpolates for numbers between consecutive powers of $2$. \bigskip Since the square of a real number is always positive, we have, for positive real numbers $a, b$ $$ (\sqrt{a}-\sqrt{b})^2 \geq 0 \implies \sqrt{ab} \leq \dfrac{a+b}{2}, $$proving the inequality for $k=2$. Observe that equality happens if and only if $a=b$. Assume now that the inequality is valid for $k=2^{n-1}>2$. This means that for any positive real numbers $x_1, x_2, \ldots , x_{2^{n-1}} $ we have \begin{equation}\label{eq:amgm-n-1} \left(x_1x_2\cdots x_{2^{n-1}}\right)^{1/2^{n-1}} \leq \dfrac{x_1+x_2+\cdots +x_{2^{n-1}}}{2^{n-1}}. \end{equation} Let us prove the inequality for $2k=2^n$. Consider any any positive real numbers $y_1, y_2, \ldots , y_{2^{n}}$. Notice that there are $2^n-2^{n-1} = 2^{n-1}(2-1) = 2^{n-1}$ integers in the interval $\lcrc{2^{n-1}+1}{2^n}$. We have $$\begin{array}{lll}\left(y_1y_2\cdots y_{2^{n}}\right)^{1/2^{n}} & = & \sqrt{\left(y_1y_2\cdots y_{2^{n-1}}\right)^{1/2^{n-1}}\left(y_{2^{n-1}+1}\cdots y_{2^{n}}\right)^{1/2^{n-1}}} \\ & \leq & \dfrac{\left(y_1y_2\cdots y_{2^{n-1}}\right)^{1/2^{n-1}}+\left(y_{2^{n-1}+1}\cdots y_{2^{n}}\right)^{1/2^{n-1}}}{2}\\ & \leq & \dfrac{\dfrac{y_1+y_2+\cdots +y_{2^{n-1}}}{2^{n-1}}+\dfrac{y_{2^{n-1}+1}+\cdots +y_{2^{n}}}{2^{n-1}}}{2}\\ & = & \dfrac{y_1 + \cdots + y_{2^n}}{2^n}, \end{array}$$ where the first inequality follows by the Case $n=2$ and the second by the induction hypothesis (\ref{eq:amgm-n-1}). The theorem is thus proved for powers of $2$. \bigskip Assume now that $2^{n-1}A$ then $\dfrac{a_1+a_2+\cdots +a_n}{n}>\dfrac{nA}{n}=A$, impossible. Similarly, the $a_i$ cannot be all $ a_ia_j $$since $a_{j}-A>0$ and $A-a_i>0$. \bigskip This change has replaced one of the $a$'s by a quantity equal to the arithmetic mean, has not changed the arithmetic mean, and made the geometric mean larger. Since there at most $n$ $a$'s to be replaced, the procedure must eventually terminate when all the $a$'s are equal (to their arithmetic mean). Strict inequality then holds when at least two of the $a$'s are unequal. \end{fo-pf} \subsection{Cauchy-Bunyakovsky-Schwarz Inequality} \begin{thm}[Cauchy-Bunyakovsky-Schwarz Inequality]\label{thm:CBS-ineq} Let $x_k, y_k$ be real numbers, $1 \leq k \leq n$. Then $$\absval{\sum _{k = 1} ^n x_ky_k} \leq \left(\sum _{k = 1} ^n x_k ^2 \right)^{1/2}\left(\sum _{k = 1} ^n y_k ^2 \right)^{1/2}, $$ with equality if and only if $$(a_1,a_2,\ldots , a_n)=t(b_1,b_2,\ldots , b_n) $$for some real constant $t$. \end{thm} \begin{f-pf}The inequality follows at once from Lagrange's Identity $$ \left(\sum _{k=1} ^nx_ky_k\right)^2 =\left(\sum _{k=1} ^nx_k ^2\right)\left(\sum _{k=1} ^ny_k ^2\right)-\sum _{1\leq k 0$ and $y>0$ the following inequality holds: $$\dfrac{a^2}{x}+\dfrac{b^2}{y}\geq \dfrac{(a+b)^2}{x+y}. $$Equality holds if and only if $\dfrac{a}{x} = \dfrac{b}{y}$. \end{lem} \begin{pf} Since the square of a real number is always positive, we have $$\begin{array}{lll}(ay-bx)^2\geq 0& \implies & a^2y^2-2abxy+b^2x^2\geq 0\\ & \implies & a^2y(x+y)+b^2x(x+y) \geq (a+b)^2xy\\ & \implies & \dfrac{a^2}{x}+\dfrac{b^2}{y}\geq \dfrac{(a+b)^2}{x+y}.\\ \end{array}$$ Equality holds if and only if the first inequality is $0$.\end{pf} \begin{rem} Iterating the result on Lemma \ref{lem:for-CBS}, $$\dfrac{a_1 ^2}{b_1} + \dfrac{a_2 ^2}{b_2} + \cdots + \dfrac{a_n ^2}{b_n} \geq \dfrac{(a_1+a_2+\cdots + a_n)^2}{b_1+b_2+\cdots + b_n}, $$ with equality if and only if $\dfrac{a_1}{b_1}=\dfrac{a_2}{b_2}=\cdots = \dfrac{a_n}{b_n}.$ \end{rem} \begin{t-pf} By the preceding remark, we have $$\begin{array}{lll}x_1 ^2 + x_2 ^2 + \cdots + x_n ^2 & = & \dfrac{x_1 ^2 y_1 ^2}{y_1 ^2}+ \dfrac{x_2 ^2 y_2 ^2}{y_2 ^2}+\cdots + \dfrac{x_n ^2 y_n ^2}{y_n ^2} \\ & \geq & \dfrac{(x_1y_1+x_2y_2+\cdots + x_ny_n)^2}{y_1 ^2+y_2 ^2 + \cdots + y_n ^2}, \end{array} $$and upon rearranging, CBS is once again obtained.\end{t-pf} \subsection{Minkowski's Inequality} \begin{thm}[Minkowski's Inequality]\label{thm:minkowski-ineq} Let $x_k, y_k$ be any real numbers. Then $$ \left(\sum _{k=1} ^n (x_k+y_k)^2\right)^{1/2}\leq \left(\sum _{k=1} ^n x_k ^2\right)^{1/2} + \left(\sum _{k=1} ^n y_k ^2\right)^{1/2}. $$ \end{thm} \begin{pf} We have $$ \begin{array}{lll}\sum _{k=1} ^n (x_k+y_k)^2 & = & \sum _{k=1} ^n x_k ^2+2\sum _{k=1} ^n x_ky_k +\sum _{k=1} ^ny_k ^2\\ & \leq & \sum _{k=1} ^n x_k ^2+2\left(\sum _{k=1} ^n x_k ^2\right)^{1/2}\left(\sum _{k=1} ^ny_k ^2\right)^{1/2} +\sum _{k=1} ^ny_k ^2\\ & = & \left(\left(\sum _{k=1} ^nx_k ^2\right)^{1/2} +\left(\sum _{k=1} ^ny_k ^2\right)^{1/2} \right)^2,\end{array}$$where the inequality follows from the CBS Inequality.\end{pf} \subsection*{Homework}\addcontentsline{toc}{subsection}{Homework} \begin{multicols}{2}\columnseprule 1pt \columnsep 25pt\multicoltolerance=900\small \begin{pro} Let $(a, b, c, d)\in \BBR^4$. Prove that $$ \absval{\absval{a-c}-\absval{b-c}} \leq \absval{a-b} \leq \absval{a-c}+\absval{b-c}. $$ \end{pro} \begin{pro}Let $(x_1,x_2, \ldots, x_n)\in\BBR^n$ be such that $$ x_1 ^2+x_2 ^2+\cdots +x_n ^2= x_1 ^3+x_2 ^3+\cdots +x_n ^3=x_1 ^4+x_2 ^4+\cdots +x_n ^4. $$Prove that $x_k\in \{0,1\}$. \begin{answer} The given equalities entail t $$ \sum _{k=1} ^n (x_k ^2-x_k)^2=0. $$A sum of squares is $0$ if and only if every term is $0$. This gives the result. \end{answer} \end{pro} \begin{pro}Let $n \geq 2$ an integer. Let $(x_1,x_2, \ldots, x_n)\in\BBR^n$ be such that $$ x_1 ^2+x_2 ^2+\cdots +x_n ^2=x_1x_2+x_2x_3+\cdots + x_{n-1}x_n+x_nx_1 . $$Prove that $x_1=x_2=\cdots=x_n$. \begin{answer} The given equality entails that $$ \dfrac{1}{2}\left((x_1-x_2)^2+(x_2-x_3)^2+\cdots + (x_{n-1}-x_n)^2+(x_n-x_1)^2\right)=0. $$A sum of squares is $0$ if and only if every term is $0$. This gives the result. \end{answer} \end{pro} \begin{pro}\label{pro:mediant} If $b>0$ and $B>0$ prove that $$ \dfrac{a}{b}<\dfrac{A}{B}\implies \dfrac{a}{b}<\dfrac{a+A}{b+B}<\dfrac{A}{B}. $$Further, if $p$ and $q$ are positive integers such that $$ \dfrac{7}{10} < \dfrac{p}{q} < \dfrac{11}{15}, $$what is the least value of $q$? \\ \begin{answer} Since $aB \dfrac{11}{15} $ and that $\dfrac{4}{6} < \dfrac{7}{10}$. Thus by considering the cases with denominators $q = 1, 2, 3, 4, 5, 6$, we see that no such fraction lies in the desired interval. The smallest denominator is thus $7$. \end{answer} \end{pro} \begin{pro} Prove that if $r \geq s \geq t$ then $$ r^2 - s^2 + t^2 \geq (r - s + t)^2. $$ \begin{answer} We have $$ (r - s + t)^2 - t^2 = (r - s + t - t)(r - s + t + t) = (r - s)(r - s + 2t).$$ Since $t - s \leq 0,$ $r - s + 2t = r + s + 2(t - s) \leq r + s$ and so $$(r - s + t)^2 - t^2 \leq (r - s)(r + s) = r^2 - s^2$$which gives $$(r - s + t)^2 \leq r^2 - s^2 + t^2.$$ \end{answer} \end{pro} \begin{pro} Assume that $a_k, b_k, c_k, k = 1, \ldots, n$, are positive real numbers. Shew that $$\left(\sum _{k = 1} ^n a_kb_kc_k\right)^{4} \leq \left(\sum _{k = 1} ^n a_k ^4\right)\left(\sum _{k = 1} ^n b_k ^4\right) \left(\sum _{k = 1} ^n c_k ^2\right)^{2}.$$\begin{answer}Using the CBS Inequality (Theorem \ref{thm:CBS-ineq}) on $\sum _{k = 1} ^n (a_kb_k)c_k$ once we obtain $$\sum _{k = 1} ^n a_kb_kc_k \leq \left(\sum _{k = 1} ^n a_k ^2b_k ^2\right)^{1/2} \left(\sum _{k = 1} ^n c_k ^2\right)^{1/2}. $$Using CBS again on $\left(\sum _{k = 1} ^n a_k ^2b_k ^2\right)^{1/2}$ we obtain $$ \begin{array}{lll} \sum _{k = 1} ^n a_kb_kc_k & \leq & \left(\sum _{k = 1} ^n a_k ^2 b_k ^2\right)^{1/2} \left(\sum _{k = 1} ^n c_k ^2\right)^{1/2} \\ & \leq & \left(\sum _{k = 1} ^n a_k ^4\right)^{1/4} \left(\sum _{k = 1} ^n b_k ^4\right)^{1/4} \left(\sum _{k = 1} ^n c_k ^2\right)^{1/2}, \\ \end{array} $$which gives the required inequality. \end{answer} \end{pro} \begin{pro} Prove that for integer $n>1$, $$ n!<\left(\dfrac{n+1}{2}\right)^n. $$ \begin{answer} This follows directly from the AM-GM Inequality applied to $1,2,\ldots , n$: $$ n!^{1/n} (1\cdot 2 \cdots n)^{1/n}< \dfrac{1+2+\cdots + n}{n} = \dfrac{n+1}{2}, $$where strict inequality follows since the factors are unequal for $n>1$. \end{answer} \end{pro} \begin{pro}\label{pro:factorial-n^n/2} Prove that for integer $n>2$, $$ n^{n/2}1(n-k)+k= n$. Thus $$ n!^2 = (1\cdot n)(2\cdot (n-1))(3\cdot (n-2))\cdots ((n-1)\cdot 2)(n\cdot 1)>n\cdot n\cdot n\cdots n=n^n. $$ \end{answer} \end{pro} \begin{pro} Prove that for all integers $n\geq 0$ the inequality $n(n-1)<2^{n+1}$ is verified. \begin{answer} From the Binomial Theorem, for $n\geq 2$, $$2^n = (1+1)^n = \binom{n}{0}+\binom{n}{1}+\binom{n}{2}+\cdots +\binom{n}{n}>\binom{n}{2}=\dfrac{n(n-1)}{2}\implies 2^{n+1}>n(n-1). $$ This establishes the inequality for $n\geq 2$. For $n=0$, $0=0(0-1)<2^{0+1}$ and for $n=1$, $0=1(1-1)<2^{1+1}$, so the inequality is true for all natural numbers. \end{answer} \end{pro} \begin{pro}\label{pro:sum-squares-ineq} Prove that $\forall (a, b, c)\in\BBR^3$, $$ a^2+b^2+c^2\geq ab + bc +ca. $$ \begin{answer} Assume without loss of generality that $a\geq b \geq c$. Then $a\geq b \geq c$ is similarly sorted as itself, so by the Rearrangement Inequality $$ a^2 + b^2 + c^2=aa+bb+cc \geq ab + bc + ca. $$ This also follows directly from the identity $$ a^2+b^2+c^2-ab-bc-ca = \left(a-\dfrac{b+c}{2}\right)^2+\dfrac{3}{4}\left(b-c\right)^2. $$ One can also use the AM-GM Inequality thrice: $$a^2+b^2\geq 2ab; \quad b^2+c^2\geq 2bc;\quad c^2+a^2\geq 2ca, $$and add. \end{answer} \end{pro} \begin{pro} Prove that $\forall (a, b, c)\in\BBR^3$, with $a\geq 0$, $b \geq 0$, $c\geq 0$, the following inequalities hold: $$ a^3+b^3+c^3\geq \max(a^2b + b^2c +c^2a, a^2c+b^2a+c^2b), $$ $$ a^3+b^3+c^3\geq 3abc, $$ $$ a^3+b^3+c^3\geq \dfrac{1}{2}\left(a^2(b+c) + b^2(c+a) + c^2(a+b)\right). $$ \begin{answer} Assume without loss of generality that $a\geq b \geq c$. Then $a\geq b \geq c$ is similarly sorted as $a^2\geq b^2 \geq c^2$, so by the Rearrangement Inequality $$ a^3+b^3+c^3= aa^2 + bb^2 + cc^2 \geq a^2b + b^2c + c^2a,$$ and $$ a^3+b^3+c^3= aa^2 + bb^2 + cc^2 \geq a^2c + b^2a + c^2b.$$ Upon adding $$ a^3+b^3+c^3= aa^2 + bb^2 + cc^2 \geq \dfrac{1}{2}\left(a^2(b+c) + b^2(c+a) + c^2(a+b)\right).$$Again, if $a\geq b \geq c$ then $$ab\geq ac \geq bc, $$ thus $$ a^3+b^3+c^3= \geq a^2b + b^2c + c^2a = (ab)a+(bc)b+(ac)c\geq (ab)c+(bc)a+(ac)b=3abc. $$ This last inequality also follows directly from the AM-GM Inequality, as $$ (a^3b^3c^3)^{1/3}\leq \dfrac{a^3+b^3+c^3}{3}, $$or from the identity $$ a^3+b^3+c^3-3abc= (a+b+c)(a^2+b^2+c^2-ab-bc-ca), $$and the inequality of problem \ref{pro:sum-squares-ineq}. \end{answer} \end{pro} \begin{pro}[Chebyshev's Inequality] Given sets of real numbers $\{a_1, a_2, \ldots , a_n\}$ and $\{b_1, b_2, \ldots , b_n\}$ prove that $$\dfrac{1}{n} \sum _{1\leq k\leq n}\check{a}_k\hat{b}_k \leq \left(\dfrac{1}{n}\sum _{1\leq k \leq n} a_k\right)\left(\dfrac{1}{n}\sum _{1\leq k \leq n}b_k \right)\leq \dfrac{1}{n}\sum _{1\leq k \leq n}\hat{a}_k\hat{b}_k. $$ \begin{answer} We apply $n$ times the Rearrangement Inequality $$\begin{array}{lllll}\check{a}_1\hat{b}_1+ \check{a}_2\hat{b}_2+\cdots + \check{a}_n\hat{b}_n & \leq & a_1b_1+a_2b_2+ \cdots + a_nb_n & \leq & \hat{a}_1\hat{b}_1+ \hat{a}_2\hat{b}_2+\cdots + \hat{a}_n\hat{b}_n\\ \check{a}_1\hat{b}_1+ \check{a}_2\hat{b}_2+\cdots + \check{a}_n\hat{b}_n & \leq & a_1b_2+a_2b_3+ \cdots + a_nb_1 & \leq & \hat{a}_1\hat{b}_1+ \hat{a}_2\hat{b}_2+\cdots + \hat{a}_n\hat{b}_n\\ \check{a}_1\hat{b}_1+ \check{a}_2\hat{b}_2+\cdots + \check{a}_n\hat{b}_n & \leq & a_1b_3+a_2b_4+ \cdots + a_nb_2 & \leq & \hat{a}_1\hat{b}_1+ \hat{a}_2\hat{b}_2+\cdots + \hat{a}_n\hat{b}_n\\ & & \vdots & & \\ \check{a}_1\hat{b}_1+ \check{a}_2\hat{b}_2+\cdots + \check{a}_n\hat{b}_n & \leq & a_1b_n+a_2b_1+ \cdots + a_nb_{n-1} & \leq & \hat{a}_1\hat{b}_1+ \hat{a}_2\hat{b}_2+\cdots + \hat{a}_n\hat{b}_n\\ \end{array}$$ Adding we obtain the desired inequalities. \end{answer} \end{pro} \begin{pro} If $x > 0$, from $$\sqrt{x + 1} - \sqrt{x} = \frac{1}{\sqrt{x + 1} + \sqrt{x}},$$prove that $$\frac{1}{2\sqrt{x + 1}} < \sqrt{x + 1} - \sqrt{x} < \frac{1}{2\sqrt{x}}.$$ Use this to prove that if $n > 1$ is a positive integer, then $$2\sqrt{n + 1} - 2 < 1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \cdots + \frac{1}{\sqrt{n}} < 2\sqrt{n} - 1$$ \end{pro} \begin{pro} If $0 < a \leq b$, shew that $$\frac{1}{8}\cdot\frac{(b - a)^2}{b} \leq \frac{a + b}{2} - \sqrt{ab} \leq \frac{1}{8}\cdot\frac{(b - a)^2}{a} $$ \begin{answer} Use the fact that $(b - a)^2 = (\sqrt{b} - \sqrt{a})^2(\sqrt{b} + \sqrt{a})^2$. \end{answer} \end{pro} \begin{pro} Shew that $$\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdots \frac{9999}{10000} < \frac{1}{100}.$$ \begin{answer} Let $$A = \frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdots \frac{9999}{10000} $$ and $$ B =\frac{2}{3}\cdot\frac{4}{5}\cdot\frac{6}{7}\cdots\frac{10000}{10001}.$$ \bigskip Clearly, $x^2 - 1 < x^2$ for all real numbers $x$. This implies that $$\frac{x - 1}{x} < \frac{x}{x + 1}$$ whenever these four quantities are positive. Hence $${\everymath{\displaystyle}\begin{array}{ccc} {1}/{2} & < & {2}/{3} \\ {3}/{4} & < & {4}/{5} \\ {5}/{6} & < & {6}/{7} \\ \vdots & \vdots & \vdots \\ {9999}/{10000} & < & {10000}/{10001} \\ \end{array} }$$ As all the numbers involved are positive, we multiply both columns to obtain $$\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdots \frac{9999}{10000} < \frac{2}{3}\cdot\frac{4}{5}\cdot\frac{6}{7}\cdots\frac{10000}{10001},$$ or $A < B.$ This yields $A^2 = A\cdot A < A\cdot B.$ Now $$A\cdot B = \frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot\frac{4}{5}\cdot\frac{5}{6}\cdot\frac{6}{7} \cdot\frac{7}{8}\cdots\frac{9999}{10000}\cdot\frac{10000}{10001} = \frac{1}{10001},$$ and consequently, $A^2 < A\cdot B = 1/10001.$ We deduce that $A < 1/\sqrt{10001} < 1/100.$ \end{answer} \end{pro} \begin{pro} Prove that for all $x>0$, $$\sum _{k=1} ^n \dfrac{1}{(x+k)^2}<\dfrac{1}{x}-\dfrac{1}{x+n}. $$ \begin{answer} Observe that for $k\geq 1$, $(x+k)^2>(x+k)(x+k-1)$ and so $$ \dfrac{1}{(x+k)^2} <\dfrac{1}{(x+k)(x+k-1)} = \dfrac{1}{x+k-1}-\dfrac{1}{x+k}. $$Hence $$\begin{array}{lll}\dfrac{1}{(x+1)^2} + \dfrac{1}{(x+2)^2} + \dfrac{1}{(x+3)^2} + \cdots + \dfrac{1}{(x+n-1)^2} + \dfrac{1}{(x+n)^2} & < & \dfrac{1}{x(x+1)} + \dfrac{1}{(x+1)(x+2)}+\dfrac{1}{(x+2)((x+3))} + \cdots + \dfrac{1}{(x+n-2)(x+n-1)} + \dfrac{1}{(x+n-1)(x+n)}\\ & = & \dfrac{1}{x}-\dfrac{1}{x+1}+\dfrac{1}{x+1}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-\dfrac{1}{x+3}+\cdots +\dfrac{1}{x+n-2}-\dfrac{1}{x+n-1}+\dfrac{1}{x+n-1}-\dfrac{1}{x+n}\\ & = & \dfrac{1}{x}-\dfrac{1}{x+n}. \end{array}$$ \end{answer} \end{pro} \begin{pro} Let $x_i\in\BBR$ such that $\sum _{i=1}\absval{x_i}=1$ and $\sum _{i=1}x_i=0$. Prove that $$ \absval{\sum _{i=1} ^n \dfrac{x_i}{i}} \leq \dfrac{1}{2}\left(1-\dfrac{1}{n}\right). $$ \begin{answer} For $1 \leq i \leq n$, we have $$\absval{\dfrac{2}{i}-1-\dfrac{1}{n}} \leq 1 - \dfrac{1}{n} \iff \left(\dfrac{2}{i}-\left(1+\dfrac{1}{n}\right)\right)^2\leq \left(1-\dfrac{1}{n}\right)^2 \iff \dfrac{4}{i^2}-\dfrac{4}{i}\left(1+\dfrac{1}{n}\right)+\dfrac{4}{n}\leq 0 \iff \dfrac{(i-n)(i-1)}{i^2n}\leq 0.$$Thus $$\absval{\sum _{i=1} ^n \dfrac{x_i}{i}} = \dfrac{1}{2}\absval{\sum _{i=1} ^n \left(\dfrac{2}{i}-\left(1+\dfrac{1}{n}\right)\right)x_i}, $$as $\sum _{i=1}x_i=0$. Now $$ \absval{\sum _{i=1} ^n \left(\dfrac{2}{i}-\left(1+\dfrac{1}{n}\right)\right)x_i} \leq \sum _{i=1} ^n \absval{ \dfrac{2}{i}-1-\dfrac{1}{n}}\absval{x_i}\leq \left(1-\dfrac{1}{n}\right)\sum _{i=1} ^n\absval{x_i} = \left(1-\dfrac{1}{n}\right). $$ \end{answer} \end{pro} \begin{pro} Let $n$ be a strictly positive integer. Let $x_i \geq 0$. Prove that $$ \prod _{k=1} ^n (1+x_k) \geq 1 + \sum _{k=1} ^n x_k. $$When does equality hold? \begin{answer} Expanding the product $$\prod _{k=1} ^n (1+x_k) = 1 + \sum _{k=1} ^n x_k + \sum _{1 \leq i 1$ equality is achieved when $\sum _{1 \leq i 0$. Use mathematical induction to prove that $$ \sqrt{a+\sqrt{a+\sqrt{a+\cdots + \sqrt{a}}}} < \dfrac{1+\sqrt{4a+1}}{2}, $$ where the left member contains an arbitrary number of radicals. \begin{answer} Let $$P(n): \ \ \ \underbrace{\sqrt{a+\sqrt{a+\sqrt{a+\cdots + \sqrt{a}}}}}_{n \ \mathrm{radicands}} < \dfrac{1+\sqrt{4a+1}}{2}. $$Let us prove $P(1)$, that is $$\forall a > 0, \ \ \ \sqrt{a} < \dfrac{1+\sqrt{4a+1}}{2}.$$ To get this one, let's work backwards. If $a>\dfrac{1}{4}$ $$ \begin{array}{lll}\sqrt{a} < \dfrac{1+\sqrt{4a+1}}{2} & \iff & 2\sqrt{a} < 1 + \sqrt{4a + 1} \\ & \iff & 2\sqrt{a} -1 < \sqrt{4a + 1} \\ & \iff & (2\sqrt{a} -1)^2 < (\sqrt{4a + 1})^2 \\ & \iff & 4a - 4\sqrt{a} + 1 < 4a + 1 \\ & \iff & -2\sqrt{a} < 0. \end{array}$$ all the steps are reversible and the last inequality is always true. If $a\leq \dfrac{1}{4}$ then trivially $2\sqrt{a} -1 < \sqrt{4a + 1}$. Thus $P(1)$ is true. Assume now that $P(n)$ is true and let's derive $P(n + 1)$. From $$\underbrace{\sqrt{a+\sqrt{a+\sqrt{a+\cdots + \sqrt{a}}}}}_{n \ \mathrm{radicands}} < \dfrac{1+\sqrt{4a+1}}{2} \implies \underbrace{\sqrt{a+\sqrt{a+\sqrt{a+\cdots + \sqrt{a}}}}}_{n+1 \ \mathrm{radicands}} < \sqrt{a+ \dfrac{1+\sqrt{4a+1}}{2}}. $$ we see that it is enough to shew that $$\sqrt{a+ \dfrac{1+\sqrt{4a+1}}{2}} = \dfrac{1+\sqrt{4a+1}}{2}. $$ But observe that $$\begin{array}{lll} (\sqrt{4a + 1} + 1)^2 = 4a + 2\sqrt{4a + 1} + 2 & \implies & \dfrac{1+\sqrt{4a+1}}{2}=\sqrt{a+ \dfrac{1+\sqrt{4a+1}}{2}}, \end{array}$$proving the claim. \end{answer} \end{pro} \begin{pro} Let $a, b, c$ be positive real numbers. Prove that $$ (a+b)(b+c)(c+a)\geq 8abc. $$ \begin{answer} From the AM-GM Inequality, $$ a+b \geq 2\sqrt{ab}; \quad b+c \geq 2\sqrt{bc}; c+a \geq 2\sqrt{ca}, $$and the desired inequality follows upon multiplication of these three inequalities. \end{answer} \end{pro} \begin{pro}[IMO, 1978]Let $a_k$ be a sequence of pairwise distinct positive integers. Prove that $$\sum _{k=1} ^n \dfrac{a_k}{k^2}\geq \sum _{k=1} ^n \dfrac{1}{k}. $$ \begin{answer} By the Rearrangement inequality $$\sum _{k=1} ^n \dfrac{a_k}{k^2}\geq \sum _{k=1} ^n \dfrac{\check{a}_k}{k^2}\geq \sum _{k=1} ^n \dfrac{1}{k}, $$as $\check{a}_k \geq k$, the $a$'s being pairwise distinct positive integers. \end{answer} \end{pro} \begin{pro}[Harmonic Mean-Geometric Mean Inequality] Let $x_i>0$ for $1 \leq i \leq n$. Then $$ \dfrac{n}{\dfrac{1}{x_1}+\dfrac{1}{x_2}+\cdots + \dfrac{1}{x_n}} \leq (x_1x_2\cdots x_n)^{1/n}, $$ with equality iff $x_1=x_2=\cdots =x_n$. \begin{answer}By the AM-GM Inequality, $$ \left(\dfrac{1}{x_1}\dfrac{1}{x_2}\cdots \dfrac{1}{x_n}\right)^{1/n}\leq \dfrac{\dfrac{1}{x_1}+\dfrac{1}{x_2}+\cdots + \dfrac{1}{x_n}}{n}, $$ whence the inequality. \end{answer} \end{pro} \begin{pro}[Arithmetic Mean-Quadratic Mean Inequality] Let $x_i\geq 0$ for $1 \leq i \leq n$. Then $$ \dfrac{x_1+x_2+\cdots + x_n}{n} \leq \left(\dfrac{x_1 ^2+x_2 ^2+\cdots + x_n ^2}{n}\right)^{1/2}, $$ with equality iff $x_1=x_2=\cdots =x_n$. \begin{answer}By the CBS Inequality, $$ \left(1\cdot x_1 + 1\cdot x_2 + \cdots + 1\cdot x_n\right)^2\leq \left(1^2+1^2+ \cdots + 1^2\right)\left(x_1 ^2 + x_2 ^2 + \cdots + x_n ^2\right), $$which gives the desired inequality. \end{answer} \end{pro} \begin{pro} Given a set of real numbers $\{a_1, a_2, \ldots , a_n\}$ prove that there is an index $m\in \{0,1,\ldots , n\}$ such that $$ \absval{\sum _{1\leq k \leq m} a_k - \sum _{m \max _{1\leq k \leq n}\absval{a_k}$ and $\absval{T_p }> \max _{1\leq k \leq n}\absval{a_k}$, then $2|a_p|=|T_{p-1}-T_p|>2\max _{1\leq k \leq n}\absval{a_k}$, a contradiction. \end{answer} \end{pro} \begin{pro} Give a purely geometric proof of Minkowski's Inequality for $n=2$. That is, prove that if $(a, b), (c, d) \in \BBR^2$, then $$\sqrt{(a + c)^2 + (b + d)^2} \leq \sqrt{a^2 + b^2} + \sqrt{c^2 + d^2}.$$Equality occurs if and only if $ad = bc.$ \label{minkowski} \begin{answer} It is enough to prove this in the case when $a, b, c, d$ are all positive. To this end, put $O = (0,0)$, $L = (a, b)$ and $M = (a+c, b+d)$. By the triangle inequality $OM \leq OL + LM$, where equality occurs if and only if the points are collinear. But then $$\sqrt{(a+c)^2 + (b+d)^2} = OM \leq OL + LM = \sqrt{a^2+b^2}+ \sqrt{c^2+d^2},$$and equality occurs if and only if the points are collinear, that is $\dfrac{a}{b} = \dfrac{c}{d}$. \end{answer} \end{pro} \begin{pro} Let $x_k\in \lcrc{0}{1}$ for $1\leq k \leq n$. Demonstrate that $$\min \left(\prod _{k=1} ^n x_k,\prod _{k=1} ^n (1-x_k) \right) \leq \dfrac{1}{2^n}. $$ \end{pro} \begin{pro} If $n>0$ is an integer and if $a_k>0$, $1\leq k \leq n$ are real numbers, demonstrate that $$ \left(\sum _{k=1}^n\dfrac{a_k}{k}\right)^2 \leq \sum _{j=1}^n\sum _{k=1}^n \dfrac{a_ja_k}{j+k-1}. $$ \end{pro} \begin{pro} Let $n$ be a strictly positive integer, let $a_k\geq 0$, $1\leq k \leq n$ be real numbers such that $a_1\geq a_2\geq \cdots \geq a_n$, and let $b_k$, $1\leq k \leq n$ be real numbers. Assume that for all indices $k\in\{1,2,\ldots , n\}$, $$\sum _{i=1} ^ka_i\leq \sum _{i=1} ^kb_i.$$ Prove that $$\sum _{i=1} ^na_i ^2\leq \sum _{i=1} ^nb_i ^2$$ \end{pro} \begin{pro} Let $n\geq 2$ an integer and let $a_k$, $1\leq k \leq n$ be real numbers such that $a_1\leq a_2\leq \cdots \leq a_n$. Prove that there is an index $k\in\{1,2,\ldots , n\}$ such that $$(a_{k+1}-a_k)^2 \leq \dfrac{12}{n(n^2-1)}(a_1 ^2 + a_2 ^2 + \cdots a_n ^2). $$ \end{pro} \begin{pro}[AIME 1991] Let $P = \{a_1, a_2, \ldots, a_n\}$ be a collection of points with $$0 < a_1 < a_2 < \cdots < a_n < 17.$$ Consider $$S_n = \min _P \sum _{k = 1} ^n\sqrt{(2k - 1)^2 + a_k ^2},$$where the minimum runs over all such partitions $P$. Shew that exactly one of $S_2, S_3, \ldots , S_n, \ldots$ is an integer, and find which one it is. \begin{answer} Use Minkowski's Inequality and the fact that $17^2 + 144^2= 145^2$. The desired value is $S_{12}$. \end{answer} \end{pro} \end{multicols} \section{Completeness Axiom} \begin{center} \fcolorbox{blue}{yellow}{ \begin{minipage}{.90\linewidth} \noindent\textcolor{red}{\textbf{Why bother?}}We saw that both $\BBQ$ and $\BBR$ are fields, and hence they both satisfy the same arithmetical axioms. Why the need then for $\BBR$? In this section we will study a property of $\BBR$ that is not shared with $\BBQ$, that of completeness. It essentially means that there are no `holes' on the real line. \end{minipage}} \end{center} \begin{df}A number $u$ is an {\em upper bound} for a set of numbers $A\subseteq \BBR$ if for all $a\in A$ we have $a \leq u$. The smallest such upper bound is called the {\em supremum or least upper bound} of the set $A$, and is denoted by $\sup A$. If $\sup A\in A$ then we say that $A$ has a {\em maximum} and we denote it by $\max A (=\sup A)$. Similarly, a number $l$ is a {\em lower bound} for a set of numbers $B \subseteq \BBR$ if for all $b\in B$ we have $l \leq b$. The largest such lower bound is called the {\em infimum or greatest lower bound} of the set $B$, and is denoted by $\inf B$. If $\sup B\in B$ then we say that $B$ has a {\em minimum} and we denote it by $\inf B (=\inf B)$. \end{df} \begin{rem} We define $\inf (\BBR) = -\infty$, $\sup (\BBR) = +\infty$, $\inf (\varnothing) = +\infty$ and $\sup (\varnothing)=-\infty$. \end{rem} \begin{df} A set of numbers $A$ is said to be {\em complete} if every non-empty subset of $A$ which is bounded above has a supremum lying in $A$. \end{df} \begin{axi}[Completeness of $\BBR$]\label{axi:completeness-of-R} Any non-empty set of real numbers which is bounded above has a supremum. Any non-empty set of real numbers which is bounded below has a infimum. \end{axi} \begin{thm}[Approximation Property of the Supremum and Infimum]\label{thm:approx-sup-inf} Let $A\neq \varnothing$ be a set of real numbers possessing a supremum $\sup A$. Then $$\forall \varepsilon >0 \quad \exists a\in A \quad \mathrm{such\ that}\quad \sup A-\varepsilon \leq a.$$ \bigskip Let $B\neq \varnothing$ be a set of real numbers possessing an infimum $\inf B$. Then $$\forall \varepsilon >0 \quad \exists b\in B \quad \mathrm{such\ that}\quad \inf A+\varepsilon \geq b.$$ \end{thm} \begin{pf} If $\forall a\in A,\quad \sup A-\varepsilon > a$ then $\sup A-\varepsilon$ would be an upper bound smaller than the least upper bound, a contradiction to the definition of $\sup A$. Hence there must be a rogue $a\in A$ such that $\sup A-\varepsilon \leq a$. \bigskip If $\forall b\in A,\quad \inf B+\varepsilon < b$ then $\inf B+\varepsilon$ would be a lower bound greater than the greatest lower bound, a contradiction to the definition of $\inf B$. Hence there must be a rogue $b\in B$ such that $\inf B+\varepsilon \geq b$. \end{pf} \begin{rem} The above result should be intuitively clear. $\sup A$ sits on the fence, just to the right of $A$, so that going just a bit to the left should put $\sup A-\varepsilon$ within $A$, etc. \end{rem} \begin{thm}[Monotonicity Property of the Supremum and Infimum]\label{thm:monotonicity-sup-inf} Let $\varnothing \subsetneqq A \subseteq B \subseteqq \BBR$ and suppose that both $A$ and $B$ have a supremum and an infimum. Then $\sup A \leq \sup B$ and $\inf B \leq \inf A$. \end{thm} \begin{pf} Assume $B$ is bounded above with supremum $\sup B$. Suppose $x\in A$. Then $x\in B$ and so $x\leq \sup B$. Thus $\sup B$ is an upper bound for the elements of $A$, and so $A$ and so by definition, $\sup A \leq \sup B$. \bigskip Assume $B$ is bounded below with infimum $\inf B$. Suppose $x\in A$. Then $x\in B$ and so $x\geq \inf B$. Thus $\inf B$ is a lower bound for the elements of $A$ and so by definition, $\inf A \geq \inf B$. \end{pf} \begin{lem}\label{lem:a+epsilon 0$ the following inequality holds: $$ a < b + \varepsilon .$$ Then $a \leq b.$ \end{lem} \begin{pf} Assume contrariwise that $a > b.$ Hence $\dis{\frac{a - b}{2} > 0}$. Since the inequality $a < b + \varepsilon$ holds for every $\varepsilon > 0$ in particular it holds for $\varepsilon = \dfrac{a - b}{2}$. This implies that $$a < b + \frac{a - b}{2} \quad \mathrm{or} \quad a < b.$$Thus starting with the assumption that $a > b$ we reach the incompatible conclusion that $a < b.$ The original assumption must be wrong. We therefore conclude that $a \leq b.$ \end{pf} \begin{thm}[Additive Property of the Supremum]\label{thm:additive-sup-inf} Let $\varnothing \subsetneqq A \subseteq \BBR$, and $B \subseteqq \BBR$. Put $$ A+B = \{x+y: (x, y)\in A\times B\} $$ and suppose that both $A$ and $B$ have a supremum. Then $A+B$ has also a supremum and $$\sup (A+B) = \sup A+\sup B.$$ \end{thm} \begin{pf} If $t\in A+B$ then $t= x+y$ with $(x, y)\in A\times B$. Then $t= x+y \leq \sup A + \sup B$, and so $\sup A + \sup B$ is an upper bound for $A+B$. By the Completeness Axiom, $A+B$ is bounded. Thus $\sup (A+B)\leq \sup A + \sup B$. \bigskip We now prove that $\sup A + \sup B \leq \sup (A+B)$. By the approximation property, $\forall \varepsilon > 0$ $\exists a\in A$ and $b\in B$ such that $\sup A - \dfrac{\varepsilon}{2}0$, then there exists a natural number $n$ such that $nx>y$. \end{thm} \begin{pf} Consider the set $$ A=\{nx: n\in\BBN\}. $$Since $1\cdot x\in A$, $A$ is non-empty. If $\forall n\in\BBN$ we had $nx\leq y$, then $A$ would be bounded above by $y$. By the Completeness Axiom, $A$ would have a supremum $\sup A$. Thus $\forall n\in\BBN, \quad nx\leq \sup A$. Since $(n+1)x\in A$, we would also have $$(n+1)x\leq \sup A\implies nx\leq \sup A-x.$$This means that $\sup A-x$ is an upper bound for $A$ which is smaller than its supremum, a contradiction Thus there must be an $n$ for which $nx>y$. \end{pf} \begin{cor}$\BBN$ is unbounded above. \end{cor} \begin{pf} This follows by taking $x=1$ in Theorem \ref{thm:archimedean-prop}. \end{pf} The Completeness Axioms tells us, essentially, that there are no ``holes'' in the real numbers. We will see that this property distinguishes the reals from the rational numbers. \begin{lem}\label{lem:root-2-irrational}[Hipassos of Metapontum]$\sqrt{2}$ is irrational. \end{lem} \begin{pf} Assume there is $s\in \BBQ$ such that $s^2=2$. We can find integers $m, n\neq 0$ such that $s=\dfrac{m}{n}$. The crucial part of the argument is that we can choose $m, n$ such that this fraction be in least terms, and hence, $m, n$ must not be both even. Now, $m^2s^2=n^2$, that is $2m^2=n^2$. This means that $n^2$ is even. But then $n$ itself must be even, since the product of two odd numbers is odd. Thus $n=2a$ for some non-zero integer $a$ (since $n\neq 0$). This means that $2m^2 = (2a)^2=4a^2 \implies m^2=2a^2$. This means once again that $m$ is even. But then we have a contradiction, since $m$ and $n$ were not both even. \end{pf} \begin{thm} $\BBQ$ is not complete. \label{thm:Q-is-incomplete} \end{thm} \begin{pf} We must shew that there is a non-empty set of rational numbers which is bounded above but that does not have a supremum in $\BBQ$. Consider the set $A=\{r\in \BBQ: r^2\leq 2\}$ of rational numbers. This set is bounded above by $u=2$. For assume that there were a rogue element of $A$, say $r_0$ such that $r_0>2$. Then $r_0 ^2 > 4$ and so $r_0$ would not belong to $A$, a contradiction. Thus $r\leq 2$ for every $r\in A$ and so $A$ is bounded above. Suppose that $A$ had a supremum $s$, which must satisfy $s\leq 2$. Now, by Lemma \ref{lem:root-2-irrational} we cannot have $s^2 = 2$ and thus $s^2<2$. By Theorem \ref{thm:archimedean-prop} there is an integer $n$ such that $2-s^2>\dfrac{1}{10^n}$. Put $t=s+\dfrac{1}{10^{n-1}}$, a rational number and observe that since $s\leq 2$ one has $$t^2=s^2+\dfrac{2s}{10^{n-1}}+ \dfrac{1}{10^{2n-2}} s$, that is $t$ is an element of $A$ larger than its least upper bound, a contradiction. Hence $A$ does not have a least upper bound. \end{pf} \subsection{Greatest Integer Function} \begin{thm}\label{thm:existence-of-floor} Given $y\in \BBR$ there exists a unique integer $n$ such that $$ n\leq y 0$, $x$ is farther from $\sqrt{5}$ than $\dfrac{2x+5}{x+2}$ is. \begin{answer} It needs to be proved that $$ \absval{\dfrac{2x+5}{x+2}-\sqrt{5}}<\absval{x-\sqrt{5}}. $$ \end{answer} \end{pro} \begin{pro}[Existence of $n$-th Roots] Let $a>0$ and let $n\in \BBR$, $n\geq 2$. Prove that there is a unique $b\in \BBR$ such that $b^n = a$. \begin{answer} Consider the set $E=\{x: x>0, x^na$. In the first case it may be advantageous to prove $\left(b+\dfrac{a-b^n}{N}\right)^na$, for integral $M$ sufficiently large, again using the Binomial Theorem to establish the inequality. \end{answer} \end{pro} \end{multicols} \chapter{Topology of $\BBR$} \section{Intervals} \begin{center} \fcolorbox{blue}{yellow}{ \begin{minipage}{.90\linewidth} \noindent\textcolor{red}{\textbf{Why bother?}} In this section we give a more precise definition of what an interval is, and establish the interesting property that between any two real numbers there is always a rational number. \end{minipage}} \end{center} \begin{df} An {\em interval}\index{interval} $I$ is a subset of the real numbers with the following property: if $s\in I$ and $t\in I$, and if $s < x < t$, then $x\in I$. In other words, intervals are those subsets of real numbers with the property that every number between two elements is also contained in the set. Since there are infinitely many decimals between two different real numbers, intervals with distinct endpoints contain infinitely many members. \end{df} \begin{rem} The empty set $\varnothing$ is trivially an interval. \end{rem} We will now establish that there are nine types of intervals. \renewcommand{\arraystretch}{2} \begin{table}[h]\begin{center}\begin{tabular}{llc}{\bf Interval Notation} & {\bf Set Notation} & {\bf Graphical Representation} \\ $[a; b]$ & $\{x\in \BBR: a \leq x \leq b\}$\footnote{This is ``the set of all real numbers $x$ such that $a$ is less than or equal to $x$, and $x$ is less than or equal to $b$.''} & $ \psset{unit=1pc} \psline[linewidth=1.5pt, linecolor=red]{*-*}(-5,0)(5,0)\uput[d](-5,0){a}\uput[d](5,0){b} $ \\ $]a; b[$ & $\{x\in \BBR: a < x < b\}$ & $ \psset{unit=1pc} \psline[linewidth=1.5pt, linecolor=red]{o-o}(-5,0)(5,0)\uput[d](-5,0){a}\uput[d](5,0){b} $ \\ $[a; b[$ & $\{x\in \BBR: a \leq x < b\}$ & $ \psset{unit=1pc} \psline[linewidth=1.5pt, linecolor=red]{*-o}(-5,0)(5,0)\uput[d](-5,0){a}\uput[d](5,0){b} $ \\ $]a; b]$ & $\{x\in \BBR: a < x \leq b\}$ & $ \psset{unit=1pc} \psline[linewidth=1.5pt, linecolor=red]{o-*}(-5,0)(5,0)\uput[d](-5,0){a}\uput[d](5,0){b} $ \\ $]a; +\infty[$ & $\{x\in \BBR: x>a\}$ & $ \psset{unit=1pc} \psline[linewidth=1.5pt, linecolor=red]{o->}(-5,0)(5,0)\uput[d](-5,0){a}\uput[d](5,0){+\infty} $ \\ $[a; +\infty[$ & $\{x\in \BBR: x\geq a\}$ & $ \psset{unit=1pc} \psline[linewidth=1.5pt, linecolor=red]{*->}(-5,0)(5,0)\uput[d](-5,0){a}\uput[d](5,0){+\infty} $ \\ $]-\infty; b[$ & $\{x\in \BBR: x< b\}$ & $ \psset{unit=1pc} \psline[linewidth=1.5pt, linecolor=red]{<-o}(-5,0)(5,0)\uput[d](-5,0){-\infty}\uput[d](5,0){b} $ \\ $]-\infty; b]$ & $\{x\in \BBR: x\leq b\}$ & $ \psset{unit=1pc} \psline[linewidth=1.5pt, linecolor=red]{<-*}(-5,0)(5,0)\uput[d](-5,0){-\infty}\uput[d](5,0){b} $ \\ $]-\infty; +\infty[$ & $\BBR$ & $ \psset{unit=1pc} \psline[linewidth=1.5pt, linecolor=red]{<->}(-5,0)(5,0)\uput[d](-5,0){-\infty}\uput[d](5,0){+\infty} $ \\ \end{tabular}\footnotesize\hangcaption{Types of Intervals. Observe that we indicate that the endpoints are included by means of shading the dots at the endpoints and that the endpoints are excluded by not shading the dots at the endpoints.} \label{tab:intervals}\end{center}\end{table} \begin{rem} If $x\in \BBR$, then $\{x\}=\lcrc{x}{x}$. \end{rem} \begin{thm} The only kinds of intervals are those sets shewn in Table \ref{tab:intervals}, and conversely, all sets shewn in this table are intervals. \end{thm} \begin{pf} The converse is easily established, so assume that $I \subseteqq \BBR$ possesses the property that $\forall (a, b)\in I^2, \quad \lcrc{a}{b}\subseteqq I$. Since $\varnothing$ is an interval one may assume that $I\neq \varnothing$. Let $a\in I$ be a fixed element of $I$ and put $M_a = \{x\in I: x\leq a\} = \lorc{-\infty}{a}\cap I$ and $N_a = \{x\in I: x\geq a\} = \lcro{a}{+\infty}\cap I$. \bigskip If $N_a$ is not bounded above, then $\forall b\in \lcro{a}{+\infty}, \quad \exists c\in N_a$ such that $b\leq c$. Since $a\leq b \leq c$, this entails that $b\in N_a$. Thus $N_a = \lcro{a}{+\infty}$. \bigskip If $N_a$ is bounded above, then it has supremum $s=\sup (N_a)$ and $N_a \subseteqq \lcrc{a}{s}$. By Theorem \ref{thm:approx-sup-inf}, $\forall b\in \lcro{a}{s}, \quad c\in N_a$ such that $b \leq c$, and since $a\leq b \leq c$, this entails that $b\in N_a$. Thus $$\lcro{a}{s}\subseteqq N_a \subseteqq \lcrc{a}{s},$$ and so $N_a= \lcro{a}{s}$ or $N_a=\lcrc{a}{s}$. \bigskip Thus $N_a$ is one among three possible forms: $\lcro{a}{+\infty}$, $\lcrc{a}{s}$, or $\lcro{a}{s}$. Applying a similar reasoning, one obtains gathers that $M_a$ is of one of the forms $\lorc{-\infty}{a}$, $\lorc{l}{a}$, or $\lcrc{l}{a}$, where $l=\inf (M_a)$. Since $I = M_a\cup N_a$, there are $3$ choices for $M_a$ and $3$ for $N_a$, hence there are $3\cdot 3 = 9$ choices for $I$. The result is established. \end{pf} \begin{exa} Determine $\bigcap _{k=1} ^\infty \lcrc{1-\dfrac{1}{2^k}}{1+\dfrac{1}{k}}$. \end{exa} \begin{solu} Observe that the intervals are, in sequence, $$\lcrc{\frac{1}{2}}{2}; \quad \lcrc{\frac{3}{4}}{\frac{3}{2}}; \quad \lcrc{\frac{7}{8}}{\frac{4}{3}};\quad \ldots . $$ We claim that $\bigcap _{k=1} ^\infty \lcrc{1-\dfrac{1}{2^k}}{1+\dfrac{1}{k}}=1$. For we see that $$\forall k\geq 1, \quad \dfrac{1}{2}\leq 1-\dfrac{1}{2^k}< 1 < 1+\dfrac{1}{k}\leq 2,$$ so $1$ is in every interval. Could this intersection contain a number smaller than $1$? No, for if $\dfrac{1}{2}\leq a<1$, then we can take $k$ large enough so that $$a< 1-\dfrac{1}{2^k}, $$for example $$ a<1-\dfrac{1}{2^k} \implies k>-\log_2(1-a), $$ so taking $k\geq \floor{-\log_2(1-a)}+1$ will work. Could the intersection contain a number $b$ larger than $1$? No, for if $1< b<2$, then we can take $k$ large enough so that $$ 1+\dfrac{1}{k}\dfrac{1}{b-1}, $$ so taking $k\geq \floor{\dfrac{1}{b-1}}+1$ will work. Hence the only number in the intersection is $1$. \end{solu} \section{Dense Sets} \begin{df} A set $B\subseteqq \BBR$ is {\em dense in $A\subseteqq \BBR$} if $\forall (a_1, a_2)\in A^2$,\quad $a_1 \dfrac{1}{y-x}$ by the Archimedean Property of $\BBR$. Consider the rational number $r = \dfrac{m}{n}$, where $m$ is the least natural number with $m>nx$. This means that $$m > nx \geq m-1.$$ We claim that $x<\dfrac{m}{n} \dfrac{1}{n} \implies x> \dfrac{m}{n}-\dfrac{1}{n} \mathrm{and}\ y>x+ \dfrac{1}{n} \implies y > \dfrac{m}{n}-\dfrac{1}{n} + \dfrac{1}{n} = \dfrac{m}{n}. $$Thus $\dfrac{m}{n}$ is a rational number between $x$ and $y$. \end{pf} \begin{thm}\label{thm:irrationals-are-dense} $\BBR\setminus \BBQ$ is dense in $\BBR$. \end{thm} \begin{pf} Let $a\dfrac{1}{b-a}$. Thus $$0 < \dfrac{1}{2^n}<\dfrac{1}{n}0$} is the set$$\ball{x_0}{\varepsilon}=\loro{x_0-\varepsilon}{x_0+\varepsilon}. $$ \end{df} \begin{df} A set $\N{x_0}\subseteqq \BBR$ is an {\em open neighbourhood of a point $x_0$} if $\exists \varepsilon >0$ such that $\ball{x_0}{r}\subseteqq \N{x_0}$, that is, there is a sufficiently small open ball containing $x_0$ completely contained in $\N{x_0}$. \end{df} \begin{df} A set $U\subseteqq\BBR$ is said to be {\em open in $\BBR$} if $\forall\ x\in U$ there is an open neighbourhood $\N{x_0}$ such that $\N{x_0}\subseteqq U$. A set $F\subseteq \BBR$ is said to be {\em closed in $\BBR$} if its complement $U=\BBR\setminus F$ is open in $\BBR$. \end{df} \begin{thm}\label{thm:open-balls-are-open} Every open ball is open. \end{thm} \begin{pf} Let $\ball{x_0}{r}$ with $r>0$ be an open ball and let $x\in \ball{x_0}{r}$. We must shew that there is a sufficiently small neighbourhood of $x$ completely within $\ball{x_0}{r}$ . That is, we search for $\varepsilon > 0$ such that $y\in \ball{x}{\varepsilon}\implies y\in \ball{x_0}{r}$. Now, $$y\in \ball{x}{\varepsilon}\implies y \in\ball{x_0}{r} \iff |y-x|<\varepsilon \implies |y-x_0|0$ small enough. But then $$\loro{x-\varepsilon}{x+\varepsilon}\subseteqq U_N\subseteqq \bigcup _{n=1} ^\infty U_n, $$and so given an arbitrary point of the union, there is a small enough open neighbourhood enclosing the point and within the union, meaning that the union is open. \bigskip If $\bigcap _{n=1} ^\infty F_n$ is an arbitrary intersection of closed sets, then there are open sets $U_n = \BBR\setminus F_n$. By the De Morgan Laws, $$\bigcap _{n=1} ^\infty F_n =\bigcap _{n=1} ^\infty (\BBR\setminus U_n) = \BBR \setminus \bigcup _{n=1} ^\infty U_n,$$and since $\bigcup _{n=1} ^\infty U_n$ is open by the above paragraph, $\bigcap _{n=1} ^\infty F_n$ is the complement of an open set, that is, it is closed. \bigskip Let $U_1, U_2, \ldots, U_L$ be a sequence of open sets in $\BBR$ and consider $x\in \bigcap _{n=1} ^L U_n$. Then $x$ belongs to each of the $U_k$ and so there are $\varepsilon _k > 0$ such that $x\in \loro{x-\varepsilon_k}{x+\varepsilon_k}\subseteqq U_k$. Let $\varepsilon = \min _{1\leq k \leq L} \varepsilon _k$ be the smallest one of such. But then for all $k$, $$\loro{x-\varepsilon}{x+\varepsilon}\subseteqq \loro{x-\varepsilon_k}{x+\varepsilon_k} \subseteqq U_k, \implies \loro{x-\varepsilon}{x+\varepsilon}\subseteqq \bigcap _{n=1} ^L U_n, $$and so given an arbitrary point of the intersection, there is a small enough open neighbourhood enclosing the point and within the intersection, meaning that the intersection is open. \bigskip Using the De Morgan Laws and the preceding paragraph, the remaining statement can be proved. \end{pf} \begin{exa} The intersection of an infinite number of open sets may not be open. For example $$ \bigcap _{k=1} ^\infty \loro{1-\dfrac{1}{n+1}}{2-\dfrac{1}{n+1}} = \lcro{1}{2},$$ which is neither open nor closed.\end{exa} \begin{thm}[Characterisation of the Open Sets of $\BBR$]\label{thm:character-open-in-R} A set $A\subseteqq \BBR$ is open if an only if it is the countable union of open sets of $\BBR$. \end{thm} \section{Interior, Boundary, and Closure of a Set} \begin{df} Let $A\subseteqq \BBR$. The {\em interior} of $A$ is defined and denoted by $$\interiorone{A} = \bigcup _{\substack{\Omega \subseteqq A\\ \Omega\ \mathrm{open}}} \Omega, $$that is, the largest open set inside $A$. The points of $\interiorone{A}$ are called the {\em interior points of $A$}. \end{df} \begin{df} Let $A\subseteqq \BBR$. The {\em closure} of $A$ is defined and denoted by $${\closure{A}} = \bigcup _{\substack{\Omega \supseteqq A\\ \Omega\ \mathrm{closed}}}\Omega, $$that is, the smallest closed set containing $A$. The points of $\closure{A}$ are called the {\em adherence points of $A$}. \end{df} \begin{rem} One always has $\interiorone{A}\subseteqq A\subseteqq \closure{A}$. A set $U$ is open if and only if $U=\interiorone{U}$. A set $F$ is closed if and only if $F=\closure{F}$. \end{rem} \begin{df} Let $A\subseteqq \BBR$. The {\em boundary} of $A$ is defined and denoted by $$\bdy{A} = \closure{A}-\interiorone{A}. $$The elements of $\bdy{A}$ are called the {\em boundary points} of $A$. \end{df} \begin{exa}We have \begin{enumerate} \item $\interior{\lorc{0}{1}} = \loro{0}{1}$, \quad $\closure{\lorc{0}{1}} = \lcrc{0}{1}$, \quad $\bdy{\lorc{0}{1}} = \{0,1\}$ \item $\interior{\{0,1\}} = \varnothing$, \quad $\closure{\{0,1\}} = \{0,1\}$, \quad $\bdy{\{0,1\}} = \{0,1\}$ \item $\interiorone{\BBQ} = \varnothing$, \quad $\closure{\BBQ} = \BBR$, \quad $\bdy{\BBQ} = \BBR$ \end{enumerate} \end{exa} \begin{thm}\label{thm:demorgan-interior-closure} Let $A\subseteqq \BBR$. Then $$ \BBR\setminus \interiorone{A} = \closure{\BBR\setminus A}, \qquad \BBR\setminus \closure{A} = \interior{\BBR\setminus A}. $$ \end{thm} \begin{pf} The theorem follows from the De Morgan Laws, as $$\BBR\setminus \interiorone{A} = \BBR \setminus \bigcup _{\substack{\Omega \subseteqq A\\ \Omega\ \mathrm{open}}} \Omega = \bigcap _{\substack{\Omega \subseteqq A\\ \Omega\ \mathrm{open}}} \left(\BBR\setminus \Omega \right) =\bigcap _{\substack{\BBR\setminus A \subseteqq \BBR\setminus \Omega\\ \Omega\ \mathrm{open}}} \left(\BBR\setminus \Omega \right) =\bigcap _{\substack{\BBR\setminus A \subseteqq F\\ F\ \mathrm{closed}}} F = \closure{\BBR\setminus A}, $$and $$\BBR\setminus \closure{A} = \BBR \setminus \bigcap _{\substack{F \supseteqq A\\ F\ \mathrm{closed}}} F = \bigcup _{\substack{F \supseteqq A\\ F\ \mathrm{closed}}} \left(\BBR\setminus F \right) =\bigcup _{\substack{\BBR\setminus A \supseteqq \BBR\setminus F\\ F\ \mathrm{closed}}} \left(\BBR\setminus F \right) =\bigcap _{\substack{\BBR\setminus A \supseteqq \Omega\\ \Omega\ \mathrm{open}}} \Omega = \interior{\BBR\setminus A}. $$ \end{pf} \begin{thm}\label{thm:belonging-to-closure} $x\in\closure{A} \iff \forall \N{x}$,\quad $\N{x}\cap A \neq \varnothing$. That is, $x$ is an adherent point if and only if every neighbourhood of $x$ has a nonempty intersection with $A$. \end{thm} \begin{pf} Assume $x\in\closure{A}$ and let $r>0$. If $\loro{x-r}{x+r}\cap A = \varnothing$, then $\loro{x-r}{x+r}\subseteqq \BBR\setminus A$. Since $\loro{x-r}{x+r}$ is open, we have---in particular--- $\loro{x-r}{x+r}\subseteqq \interior{\BBR\setminus A} = \BBR\setminus \closure{A}$ by Theorem \ref{thm:demorgan-interior-closure}. This means that $x\not\in \closure{A}$, a contradiction. \bigskip Conversely, assume that for all neighbourhoods $\N{x}$ of $x$ we have $\N{x}\cap A \neq \varnothing$. If $x\not\in \closure{A}$ then $x\in \BBR\setminus \closure{A} = \interior{\BBR\setminus A}$. Since $\interior{\BBR\setminus A}$ is open there is an $r'>0$ such that $\loro{x-r'}{x+r'}\subseteqq \interior{\BBR\setminus A}\subseteqq \BBR\setminus A$. But then $\loro{x-r'}{x+r'}\cap A = \varnothing$, a contradiction. \end{pf} \begin{thm}\label{thm:sup-belongs-to-closed} Let $\varnothing \varsubsetneqq A\subseteqq \BBR$ be bounded above. Then $\sup {A}\in \closure{A}$. If, moreover, $A$ is closed then $\sup (A)\in A$. \end{thm} \begin{pf} Let $r>0$. By Theorem \ref{thm:approx-sup-inf}, there exists $a\in A$ such that $\sup (A)-r0$ might be and, hence, $\sup (A)\in \closure{A}$ by Theorem \ref{thm:belonging-to-closure}. If $A$ is closed, then $A=\closure{A}$. \end{pf} \begin{df} Let $A\subseteqq \BBR$. A point $x\in A$ is called an {\em isolated point of $A$} if there exists an $r>0$ such that $\ball{x}{r}\cap A = \{x\}$. The set of isolated points of $A$ is denoted by $A^*$. \bigskip A point $y \in \BBR$ is called an {\em accumulation point of $A$ in $\BBR$} if $$\forall \N{x}, \quad (\N{x}\setminus \{x\})\cap A \neq \varnothing , $$that is, if any neighbourhood of $x$ meets $A$ at a point different than $x$. The set of accumulation points of $A$ is called the {\em derived set of $A$} and is denoted by $\acc{A}$.\end{df} \begin{exa}We have \begin{enumerate} \item $0$ is an isolated point of the set $A = \{0\}\cup \lcrc{1}{2}$. \item Every point of the set $A = \left\{1, \dfrac{1}{2}, \dfrac{1}{3}, \ldots \right\}$ is isolated. This is because we may take $r = \dfrac{1}{2^{n+2}}$ in the definition of isolated point, and then $\loro{\dfrac{1}{n}-\dfrac{1}{2^{n+2}}}{\dfrac{1}{n}+\dfrac{1}{2^{n+2}}} \cap A = \left\{\dfrac{1}{n}\right\}$. Observe that $\dfrac{1}{n}-\dfrac{1}{n+1} = \dfrac{1}{n(n+1)}$ and $\dfrac{1}{n-1}-\dfrac{1}{n} = \dfrac{1}{n(n-1)}$ and that $2^{n+2}> \max (n(n+1), n(n-1))$. \item $0$ is an accumulation point of $A = \left\{1, \dfrac{1}{2}, \dfrac{1}{3}, \ldots \right\}$. \end{enumerate} \end{exa} \begin{thm}\label{thm:accu-points-infinite-meet} $x$ is an accumulation point of $A$ if and only if every neighbourhood of $x$ in $\BBR$ has an infinite number of points of $A$. \end{thm} \begin{pf} Suppose $x\in A'$. Suppose a neighbourhood of $x$ had only finitely many elements of $A$, say $\{y_1, y_2, \ldots , y_n\}$. Take $2r=\min _{1 \leq k \leq n} \absval{y_k-x}$. Then $\left(\loro{x-r}{x+r}\setminus \{x\}\right)\cap A = \varnothing$ contradicting the fact that every neighbourhood of $x$ meets $A$ at a point different from $x$. \bigskip Conversely if every neighbourhood of $x$ in $\BBR$ has an infinite number of points of $A$, then a fortiori, any intersection of such a neighbourhood with $A$ will contain a point different from $x$, and so $x\in \acc{A}$. \end{pf} \begin{thm} A set is closed if and only if it contains all its accumulation points. \end{thm} \begin{pf} If $A$ is closed then $\BBR \setminus A$ is open. If $c\in \BBR \setminus A$ then there exists $r>0$ such that $\loro{c-r}{c+r}\subseteqq \BBR \setminus A$, a neighbourhood that clearly does not contain any points of $A$, which means $c\not\in \acc{A}$. \bigskip Conversely, suppose a set $\acc{A}\subseteqq A$. We will prove that $\BBR\setminus A$ is open. If $x\in \BBR\setminus A$, then {\em a fortiori}, $x\not\in \acc{A}$. This means that there is an $r>0$ such that $\loro{x-r}{x+r}\cap A = \varnothing$. Hence $\loro{x-r}{x+r}\subseteqq \BBR\setminus A$, and so $\BBR\setminus A$ is open. \end{pf} \begin{rem} One has $$A^* \subseteqq A, \quad \closure{A}-A \subseteqq \acc{A}, \quad A^* \cap \acc{A} = \varnothing, \quad A^* \cup \acc{A} = \closure{A}. $$ \end{rem} \section{Connected Sets} \begin{df} A set $X\subseteqq \BBR$ is connected if, given open sets $U, V$ of $\BBR$ with $U\cup V = X$, $U\cap V = \varnothing$, either $U=\varnothing$ or $V=\varnothing.$ \end{df} \begin{thm} If $X\subseteqq \BBR$ is connected, and if $(a, c)\in X^2$, $b\in\BBR$, are such that $a0$ such that $\loro{\sup E-\varepsilon}{\sup E + \varepsilon}\subseteqq U_s$. By Theorem \ref{thm:approx-sup-inf} there is $x\in E$ such that $\sup E - \varepsilon < x \leq \sup E$. Thus there is a finite subcover from the $U_i$, say, $U_{p_1}$ $U_{p_2}$, \ldots , $U_{p_n}$ such that $\lcrc{a}{x}\subseteqq \bigcup _{i=1} ^n U_{k_i}$. \bigskip We thus have $$ \lcrc{a}{\sup E}\subseteqq \lcrc{a}{x}\bigcup \loro{\sup E-\varepsilon}{\sup E + \varepsilon}\subseteqq \left(\bigcup _{i=1} ^n U_{k_i}\right) \cup U_s, $$a finite subcover. This means that $\sup E \in E$. \bigskip Suppose now that $\sup E < b$, and consider $y = \sup E + \dfrac{1}{2}\min (b - \sup E, \varepsilon)$. Then $$ \sup E < y, \qquad \lcrc{a}{y}=\lcrc{a}{\sup E}\cup \lcrc{\sup E}{y} \subseteqq \left(\bigcup _{i=1} ^n U_{k_i}\right) \cup U_s,$$ whence $y\in E$, contradicting the definition of $\sup E$. This proves that $\sup E = b$ and finishes the proof of the theorem. \end{pf} \begin{thm}[Heine-Borel]\label{thm:heine-borel} A set $A$ of $\BBR$ is closed and bounded if and only if it is compact. \end{thm} \begin{pf} Let $A$ be closed and bounded in $\BBR$, and let $U_1, U_2, \ldots ,$ be an open cover for $A$. There exist $(a, b)\in \BBR^2$, $a\leq b$, such that $A\subseteqq \lcrc{a}{b}$. Since $$\lcrc{a}{b}\subseteqq (\BBR \setminus A)\cup\bigcup _{i=1} ^\infty U_i, $$by Theorem \ref{thm:[ab]-is-compact} there is a finite subcover of the $U_i$, say, $U_{k_i}$ such that $$\lcrc{a}{b}\subseteqq (\BBR \setminus A)\cup\bigcup _{i=1} ^\infty U_{k_i}. $$Therefore $$A=A\cap \lcrc{a}{b}\subseteqq \lcrc{a}{b}\subseteqq \bigcup _{i=1} ^\infty U_{k_i}, $$and so $A$ admits an open subcover. \bigskip Conversely, suppose that every open cover of $A$ admits a finite subcover. The open cover $\loro{-n}{n}, n\in\BBR$ of $A$ must admit a finite subcover by our assumption, hence there is $N\in \BBN$ such that $A\subseteqq \loro{-N}{N}$, meaning that $A$ is bounded. Let us shew now that $\BBR \setminus A$ is open. \bigskip Let $x\in \BBR\setminus A$. We have $$ \bigcup _{n\geq 1} \left(\BBR\setminus \lcrc{x-\frac{1}{n}}{x+\frac{1}{n}}\right) = \BBR \setminus \bigcap _{n\geq 1} \lcrc{x-\frac{1}{n}}{x+\frac{1}{n}} =\BBR\setminus \{x\} \supseteqq A,$$since $x\not\in A$. By hypothesis there is $N\in\BBN$ and $n_1, n_2, \ldots , n_N$ such that $$A\subseteqq \bigcup _{k=1} ^N \left(\BBR\setminus \lcrc{x-\frac{1}{n_k}}{x+\frac{1}{n_k}}\right) \subseteqq \BBR\setminus \lcrc{x-\frac{1}{n_m}}{x+\frac{1}{n_m}},$$where $m=\max (n_1, n_2, \ldots , n_N)$. This gives $ \lcrc{x-\frac{1}{n_m}}{x+\frac{1}{n_m}}\subseteqq \BBR\setminus A$, meaning that $\BBR\setminus A$ is open, whence $A$ is closed. \end{pf} \begin{cor}[Cantor's Intersection Theorem] Let $$\lcrc{a_1}{b_1}\supseteqq\lcrc{a_2}{b_2}\supseteqq \lcrc{a_3}{b_3}\supseteqq\ldots$$be a sequence of non-empty, bounded, nested closed intervals. Then $$ \bigcap _{j=1} ^\infty \lcrc{a_j}{b_j} \neq \varnothing . $$ \end{cor} \begin{pf}Assume that $\lcrc{a_1}{b_1}\cap \bigcap _{j=2} ^\infty \lcrc{a_j}{b_j} =\varnothing$. Then $$\lcrc{a_1}{b_1}\subseteqq \BBR\setminus \bigcap _{j=2} ^\infty \lcrc{a_j}{b_j} =\bigcup _{j=2} ^\infty \left(\BBR\setminus \lcrc{a_j}{b_j}\right).$$ The $\BBR\setminus \lcrc{a_j}{b_j}$ for an open cover for $\lcrc{a_1}{b_1}$, which is closed and bounded. By Theorem \ref{thm:mono-reversing} we have $$\lcrc{a_{j}}{a_{j}} \subseteqq \lcrc{a_{i}}{b_{i}} \implies \BBR\setminus \lcrc{a_{i}}{b_{i}}\subseteqq \BBR\setminus \lcrc{a_{j}}{b_{j}}.$$ By the Heine-Borel Theorem \ref{thm:heine-borel} there is a finite subcover, say $$\lcrc{a_1}{b_1}\subseteqq\bigcup _{j=1} ^N \left(\BBR\setminus \lcrc{a_{n_j}}{b_{n_j}}\right)\subseteqq \BBR\setminus \lcrc{a_{n_N}}{b_{n_N}}.$$But then $ \lcrc{a_{n_N}}{b_{n_N}}\subseteqq \BBR\setminus \lcrc{a_{1}}{b_{1}}$, which contradicts $ \lcrc{a_{n_N}}{b_{n_N}}\subseteqq\lcrc{a_{1}}{b_{1}}$, and the proof is complete.\end{pf} \begin{thm}[Bolzano-Weierstrass]\label{thm:bolzano-weierstrass} Ever bounded infinite set of $\BBR$ has at least one accumulation point. \end{thm} \begin{pf} Let $A$ be a bounded set of $\BBR$ with $\acc{A}=\varnothing$. Then $A^* = A = \closure{A}$. Notice that then every element of $A$ is an isolated point of $A$, and hence, $$\forall a\in A, \quad \exists r_a>0, \quad \mathrm{such\ that}\ \quad \loro{a-r_a}{a+r_a} \cap A = \{a\}. $$ Observe that $$A\subseteqq \bigcup _{a\in A} \loro{a-r_a}{a+r_a}, $$and so the $ \loro{a-r_a}{a+r_a}$ form an open cover for $A$. Since $A=\closure{A}$, $A$ is closed. By the Heine-Borel Theorem \ref{thm:heine-borel} $A$ has a finite subcover from among the $ \loro{a-r_a}{a+r_a}$ and so there exists an integer $N>0$ and $a_i$ such that $$ A\subseteqq \bigcup _{i=1} ^N \loro{a_i-r_{a_i}}{a_i+r_{a_i}}. $$ Since $$A=A\cap \bigcup _{i=1} ^N \loro{a_i-r_{a_i}}{a_i+r_{a_i}} = \bigcup _{i=1} ^N \{a_i\}, $$$A$ has only $N$ elements and thus it is finite. \end{pf} \begin{thm}Let $X\subseteqq \BBR$. Then the following are equivalent. \begin{enumerate} \item $X$ is compact. \item $X$ is closed and bounded. \item every infinite set of $X$ has an accumulation point. \item every infinite sequence of $X$ has a converging subsequence in $X$. \end{enumerate} \label{thm:equivalent-statements-for-compactness} \end{thm} \subsection*{Homework}\addcontentsline{toc}{subsection}{Homework} \begin{multicols}{2}\columnseprule 1pt \columnsep 25pt\multicoltolerance=900\small \begin{pro} Give an example shewing that the union of an infinite number of closed sets is not necessarily closed. \end{pro} \begin{pro} Prove that a set $A\subseteqq \BBR$ is dense if and only if $\closure{A}=\BBR$. \end{pro} \begin{pro} For any set $A\subseteqq \BBR$ prove that $\bdy{A} = \bdy{\BBR\setminus A}$. \end{pro} \begin{pro} Let $A\neq \varnothing$ be a subset of $\BBR$. Assume that $A$ is bounded above. Prove that $\sup (A) = \sup (\closure{A})$. \end{pro} \begin{pro} Demonstrate that the only subsets of $\BBR$ which are simultaneously open and closed in $\BBR$ are $\varnothing$ and $\BBR$. One codifies this by saying that $\BBR$ is {\em connected}. \end{pro} \begin{pro}\label{pro:additive-closed-groups} Prove that the closed additive subgroups of the real numbers are (i) just zero; or (ii) all integral multiples of a fixed non-zero number (which may be assumed positive); or (iii) or all reals. \begin{answer} For the proof of this let $G$ be such a set (so that $x + y$ is in $G$ if $x, y$ are, and $G$ is closed), and suppose that we are not in cases (i) or (ii). Then it is enough to show that $G$ contains arbitrarily small positive numbers, for then multiples of these will be dense in $\BBR$ , but $G$ being closed forces $G = \BBR$. To achieve this let $\mathscr{I} = \inf \{x:x\in G, x > 0\}$. If $\mathscr{I} = 0$ we are done; but if $\mathscr{I} > 0$ there cannot be numbers $x\in G$ arbitrarily close to and greater than $\mathscr{I}$, for then $x - \mathscr{I}$ would run through small positive members of $G$, in particular smaller than $\mathscr{I}$, contradicting its definition. This means that $\mathscr{I}$ belongs itself to $G$, and from there it is easy to see that we are in case (ii) contrary to the assumption. Hence indeed $\mathscr{I} = 0$, $G = \BBR$. \end{answer} \end{pro} \begin{pro} Let $A\in \BBR$. Prove the following \begin{multicols}{2} \begin{enumerate} \item $\closure{\closure{A}} = \closure{A}$ \item $\interiorone{\interiorone{A}} = \interiorone{A}$ \item $A\subseteqq B \implies \closure{A}\subseteqq \closure{B}$ \item $A\subseteqq B \implies \interiorone{A}\subseteqq \interiorone{B}$ \item $\closure{A\cup B} =\closure{A}\cup \closure{B}$ \item $\closure{A\cap B} \subseteqq\closure{A}\cap \closure{B}$ \item $\interiorone{A}\cup \interiorone{B} \subseteqq \interior{A\cup B}$ \item $\interior{A\cap B} =\interiorone{A}\cap \interiorone{B}$ \end{enumerate} \end{multicols} \end{pro} \end{multicols} \section{$\closure{\closure{\BBR}}$} \begin{center} \fcolorbox{blue}{yellow}{ \begin{minipage}{.90\linewidth} \noindent\textcolor{red}{\textbf{Why bother?}} The algebraic rules introduced here will simplify some computations and statements in subsequent chapters. \end{minipage}} \end{center} Geometrically, each real number can be viewed as a point on a straight line. We make the convention that we orient the real line with $0$ as the origin, the positive numbers increasing towards the right from $0$ and the negative numbers decreasing towards the left of $0$, as in figure \ref{fig:the_real_line}. \vspace{1cm} \begin{figure}[h] $$\psset{unit=.5} \renewcommand{\pshlabel}[1]{{\tiny #1}} \psaxes[yAxis=false](0, 0)(-7,0)(7, 0) \psline[linestyle=dotted]{->}(7,0)(11,0) \psline[linestyle=dotted]{->}(-7,0)(-11,0) \uput[dr](11,0){+\infty}\uput[dl](-11,0){-\infty} $$\vspace{1cm} \footnotesize\hangcaption{The Real Line.} \label{fig:the_real_line} \end{figure} We append the object $+\infty$, which is larger than any real number, and the object $-\infty$, which is smaller than any real number. Letting $x\in\BBR$, we make the following conventions. \begin{equation} (+\infty) + (+\infty) = +\infty \end{equation} \begin{equation} (-\infty) + (-\infty) = -\infty \end{equation} \begin{equation} x + (+\infty) = +\infty \end{equation} \begin{equation} x + (-\infty) = -\infty \end{equation} \begin{equation} x(+\infty) = +\infty\ \ \ \mathrm{if} \ x> 0 \end{equation} \begin{equation} x(+\infty) = -\infty\ \ \ \mathrm{if} \ x< 0 \end{equation} \begin{equation} x(-\infty) = -\infty\ \ \ \mathrm{if} \ x> 0 \end{equation} \begin{equation} x(-\infty) = +\infty\ \ \ \mathrm{if} \ x< 0 \end{equation} \begin{equation} \label{eq:1/big}\dfrac{x}{\pm \infty} = 0 \end{equation} Observe that we leave the following undefined: $$\dfrac{\pm \infty}{\pm \infty}, \ \ \ \ (+\infty) + (-\infty), \ \ \ 0(\pm \infty). $$ \begin{df} We denote by $\closure{\closure{\BBR}}=\lcrc{-\infty}{+\infty}$ the set of real numbers such with the two symbols $-\infty$ and $+\infty$ appended, obeying the algebraic rules above. Observe that then every set in $\closure{\BBR}$ has a supremum (it may as well be $+\infty$ if the set is unbounded by finite numbers) and an infimum (which may be $-\infty$). \end{df} \section{Lebesgue Measure} \begin{df}Let $(a, b)\in\BBR^2$. The {\em measure} of the open interval $\loro{a}{b}$ is $b-a$. We denote this by $\meas{\loro{a}{b}} = b-a$. If $G = \bigcup _{k=1} ^\infty \loro{a_k}{b_k}$ is a union of disjoint, bounded, open intervals, then $\meas{G} = \sum _{k=1} ^\infty (b_k-a_k)$. \end{df} \begin{df} Let $E\subseteqq \BBR$ be a bounded set. The {\em outer measure of $E$} is defined and denoted by $$\outermeas{E} = \inf _{\substack{E \subseteqq O\\ O\ \mathrm{open}}} \meas{O}. $$ \end{df} \begin{df} A set $E\subseteqq \BBR$ is said to be {\em Lebesgue measurable} if $\forall \varepsilon > 0$, $\exists G\supseteqq E$ open such that $\outermeas{G\setminus E}<\varepsilon$. In this case $\meas{E} = \outermeas{E}$. \end{df} \section{The Cantor Set} \begin{df}[The Cantor Set] The Cantor set $C$ is the canonical example of an uncountable set of measure zero. We construct $C$ as follows. Begin with the unit interval $C_0 = \lcrc{0}{1}$, and remove the middle third open segment $R_1 := \loro{\frac{1}{3}}{\frac{2}{3}}$. Define $C_1$ as \begin{equation} C_1 := C_0 \setminus R_1 = \lcrc{0}{\frac{1}{3}} \bigcup \lcrc{\frac{2}{3}}{1} \end{equation} Iterate this process on each remaining segment, removing the open set \begin{equation} R_2 := \loro{\frac{1}{9}}{\frac{2}{9}} \bigcup \loro{\frac{7}{9}}{\frac{8}{9}} \end{equation} to form the four-interval set \begin{equation} C_2 := C_1 \setminus R_2 = \lcrc{0}{\frac{1}{9}} \bigcup \lcrc{\frac{2}{9}}{\frac{1}{3}} \bigcup \lcrc{\frac{2}{3}}{\frac{7}{9}} \bigcup \lcrc{\frac{8}{9}}{1} \end{equation} Continue the process, forming $C_3, C_4, \ldots$ Note that $C_k$ has $2^k$ pieces. At each step, the endpoints of each closed segment will remain in the set. See figure \ref{fig:cantor-set}. The \emph{Cantor set} is defined as \begin{equation} C := \bigcap_{k=1}^{\infty} C_k = C_0 \setminus \bigcup_{n=1}^{\infty}R_n \end{equation} \label{df:cantor-set} \end{df} \vspace{1cm} \begin{figure}[h] $$\psset{unit=1pc} \begin{array}{ll} \uput[l](-16,0){C_0} & \rput(-13.5,0){\psline{|-|}(0,0)(27,0) \uput[d](0,0){0}\uput[d](27,0){1}} \\ \uput[l](-16,0){C_1} & \rput(-13.5,0){\psline{|-|}(0,0)(9,0) \psline{|-|}(18,0)(27,0) \uput[d](0,0){0}\uput[d](27,0){1}\uput[d](9,0){\frac{1}{3}}\uput[d](18,0){\frac{2}{3}}}\\ \uput[l](-16,0){C_2} & \rput(-13.5,0){\psline{|-|}(0,0)(3,0)\psline{|-|}(6,0)(9,0) \psline{|-|}(18,0)(21,0)\psline{|-|}(24,0)(27,0) \uput[d](0,0){0}\uput[d](27,0){1}\uput[d](9,0){\frac{1}{3}}\uput[d](18,0){\frac{2}{3}}\uput[d](3,0){\frac{1}{9}}\uput[d](6,0){\frac{2}{9}}\uput[d](21,0){\frac{7}{9}}\uput[d](24,0){\frac{8}{9}}}\\ \uput[l](-16,0){\vdots} & \rput(0,0){\vdots} \end{array}$$ \vspace{1cm} \hangcaption{Construction of the Cantor Set.}\label{fig:cantor-set} \end{figure} \begin{thm}[Cardinality of the Cantor Set]The Cantor Set is uncountable. \end{thm} \begin{pf} Starting with the two pieces of $C_1$, we mark the sinistral segment ``0'' and the dextral segment ``1''. We then continue to $C_2$, and consider only the leftmost pair. Again, mark the segments ``0'' and ``1'', and do the same for the rightmost pair. Successively then, mark the $2^{k-1}$ leftmost segments of $C_k$ ``0'' and the $2^{k-1}$ rightmost segments ``1.'' The elements of the Cantor Set are those with infinite binary expansions. Since there uncountable many such expansions, the Cantor Set in uncountable.\end{pf} \begin{thm}[Measure of the Cantor Set] The Cantor Set has (Lebesgue) measure $0$. \end{thm} \begin{pf}Using the notation of Definition \ref{df:cantor-set}, observe that \begin{align} \mu(R_1) &= \frac{2}{3} - \frac{1}{3} = \frac{1}{3}\\ \mu(R_2) &= \left(\frac{2}{9} - \frac{1}{9}\right) + \left(\frac{8}{9} - \frac{7}{9}\right) = \frac{2}{9}\\ &\ \vdots\\ \mu(R_k) &= \sum_{n=1}^{k}\frac{2^{n-1}}{3^n} \end{align} Since the $R$'s are disjoint, the measure of their union is the sum of their measures. Taking the limit as $k \to \infty$, \begin{equation} \mu\left(\bigcup_{n=1}^{\infty}R_n\right) = \sum_{n=1}^{\infty} \frac{2^{n-1}}{3^n} = 1. \end{equation} Since clearly $\mu(C_0) = 1$, we then have \begin{equation} \mu(C) = \mu\left(C_0 \setminus \bigcup_{n=1}^{\infty}R_n\right) = \mu(C_0) - \sum_{n=1}^{\infty} \frac{1}{2^n} = 1-1 = 0. \end{equation} \end{pf} \begin{thm}The Cantor set is closed and its interior is empty. \end{thm} \begin{pf}Each of $C_0, C_1, C_2, \ldots$, is closed, being the union of a finite number of closed intervals. Thus the Cantor Set is closed, as it is the intersection of closed sets. \bigskip Now, let $I$ be an open interval. Since the numbers of the form $\dfrac{m}{3^n}$, $(m, n)\in \BBZ$ are dense in the reals, there is exists a rational number $\dfrac{m}{3^n}\in I$. Expressed in ternary, this rational number has a finite expansion. If this expansion contains the digit ``1'', then this number does not belong to Cantor Set, and we are done. If not, since $I$ is open, there must exist a number $k > n$ such that $\dfrac{m}{3^n} + \dfrac{1}{3^k} \in I$. By construction, the last digit of the ternary expansion of this number is also ``1'', and hence this number does not belong to the Cantor Set either.\end{pf} \chapter{Sequences} \section{Limit of a Sequence} \begin{center} \fcolorbox{blue}{yellow}{ \begin{minipage}{.90\linewidth} \noindent\textcolor{red}{\textbf{Why bother?}} The {\em limit} concept is at the centre of calculus. We deal with discrete quantities first, that is, with limits of sequences. \end{minipage}} \end{center} \begin{df} A {\em (numerical) sequence} is a function $a: \BBN \rightarrow \BBR$. We usually denote $a(n)$ by $a_n$.\footnote{It is customary to start at $n=1$ rather than $n=0$. We won't be too fuzzy about such complications, but we will be careful to write sense.} \end{df} \begin{rem} We will use the notation $\seq{a_n}{n=k}{l}$ to denote the sequence $a_k, a_{k+1}\ldots, a_l$. For example $$\seq{a_n}{n=0}{10} = \{a_0, a_1, a_2, \ldots ,a_{10}\}, $$ $$\seq{b_n}{n=4}{6} = \{b_4, b_5, b_{6}\}, $$ $$\seq{\left(1+\dfrac{1}{n}\right)^n}{n=1}{+\infty} = \{2, \dfrac{9}{4},\dfrac{64}{27}, \ldots ,\}, $$etc. \end{rem} \begin{exa} The {\em Harmonic sequence} is $$1, \quad \dfrac{1}{2},\quad \dfrac{1}{3}, \quad \ldots, $$or $a_n = \dfrac{1}{n}$ for $n\geq 1$. \end{exa} \begin{df} A sequence $\seq{a_n}{n=1}{+\infty}$ is {\em bounded} if there exists a constant $K>0$ such that $\forall n, \absval{a_n}\leq K$. It is {\em increasing} if for all $n>0$, $a_{n}\leq a_{n+1}$ and {\em decreasing} if for all $n\geq 0$, $a_{n}\geq a_{n+1}$. \end{df} \section{Convergence of Sequences} \begin{df} A sequence $\seq{a_n}{n=1}{+\infty}$ is said to {\em converge} if $$\exists L\in \BBR, \forall \varepsilon > 0, \quad \exists N>0\quad \mathrm{such\ that }\quad \forall n\in \BBN, \quad n\geq N \implies \absval{a_n-L}<\varepsilon .$$ In other words, eventually\footnote{A good word to use in informal speech ``eventually'' will mean ``for large enough values'' or in the case at hand $\forall n \geq N$ for some strictly positive integer $N$.} the differences $$ \absval{a_n-L},\absval{a_{n+1}-L}, \absval{a_{n+2}-L}, \ldots $$ remain smaller that an arbitrarily prescribed small quantity. We denote the fact that the sequence $\seq{a_n}{n=1}{+\infty}$ converges to $L$ as $n\rightarrow +\infty$ by $$ \lim _{n\rightarrow +\infty}a_n = L, \quad \mathrm{or\ by}\quad a_n \rightarrow L \quad \mathrm{as}\quad n\rightarrow +\infty .$$ A sequence that does not converge is said to {\em diverge}. Thus a sequence diverges if $$\forall L\in \BBR, \exists \varepsilon >0, \forall N\in\BBN ,\exists n\in\BBN \quad \mathrm{such\ that}\quad n > N \quad \mathrm{and}\quad \absval{a_n-L}\geq \varepsilon. $$ \end{df} \begin{rem}Given a sequence sequence $\seq{a_n}{n=1}{+\infty}$ and $L\in \BBR$, $$ a_n \rightarrow L \quad \mathrm{as} \quad \ngrows \quad\mathrm{if and only if} \quad \lim \inf a_n = \lim \sup a_n = \lim a_n =L.$$ \end{rem} \begin{df} A sequence $\seq{b_n}{n=1}{+\infty}${\em diverges to plus infinity} if $\forall M>0, \quad \exists N>0$ such that $\forall n\geq N, \quad b_n>M$. A sequence $\seq{c_n}{n=1}{+\infty}${\em diverges to minus infinity} if $\forall M>0, \quad \exists N>0$ such that $\forall n\geq N, \quad c_n<-M$. A sequence that diverges to plus or minus infinity is said to {\em properly diverge}. Otherwise it is said to {\em oscillate}. \end{df} \begin{df} Given a sequence $\seq{a_n}{n=1}{+\infty}$, we say that $\lim _{\ngrows}a_n$ {\em exists} it is either convergent or properly divergent. \end{df} \begin{exa} The constant sequence $$1,1,1,1, \ldots $$converges to $1$. Similarly, if a sequence is eventually stationary, that is, constant, it converges to that constant. \end{exa} \begin{exa} Consider the sequence $$1, \dfrac{1}{2}, \dfrac{1}{3}, \ldots, \frac{1}{n}, \ldots, $$ We claim that $\dfrac{1}{n}\rightarrow 0$ as $n\rightarrow+\infty$. Suppose we wanted terms that get closer to $0$ by at least $.00001 = \dfrac{1}{10^5}$. We only need to look at the $100000$-term of the sequence: $\dfrac{1}{100000} = \dfrac{1}{10^5}$. Since the terms of the sequence get smaller and smaller, any term after this one will be within $.00001$ of $0$. We had to wait a long time---till after the $100000$-th term---but the sequence eventually did get closer than $.00001$ to $0$. The same argument works for any distance, no matter how small, so we can eventually get arbitrarily close to $0$. A rigorous proof is as follows. If $\varepsilon > 0$ is no matter how small, we need only to look at the terms after $N = \floor{ \frac{1}{\varepsilon} + 1}$ to see that, indeed, if $n > N$, then $$s_n = \frac{1}{n} < \frac{1}{N} = \frac{1}{\floor{\frac{1}{\varepsilon} + 1} } < \varepsilon.$$ Here we have used the inequality $$ t - 1 < \floor{t} \leq t, \ \ \forall t\in \BBR.$$ \end{exa} \begin{exa} The sequence $$0, 1,4,9,16, \ldots ,n^2, \ldots$$ diverges to $+\infty$, as the sequence gets arbitrarily large. A rigorous proof is as follows. If $M > 0$ is no matter how large, then the terms after $N = \floor{ \sqrt{M}} + 1$ satisfy ($n > N$) $$t_n = n^2 > N^2 = (\floor{ \sqrt{M}} + 1)^2 > M. $$ \end{exa} \begin{exa} The sequence $$1,-1,1,-1,1,-1,\ldots, (-1)^n,\ldots$$ has no limit (diverges), as it bounces back and forth from $-1$ to $+1$ infinitely many times. \end{exa} \begin{exa} The sequence $$0, -1,2,-3,4,-5,\ldots, (-1)^nn, \ldots , $$ has no limit (diverges), as it is unbounded and alternates back and forth positive and negative values.. \end{exa} \bigskip We will now see some properties of limits of sequences. \begin{thm}[Uniqueness of Limits] If $a_n\rightarrow L$ and $a_n\rightarrow L'$ as $n\rightarrow +\infty$ then $L=L'.$ \end{thm} \begin{pf}The statement only makes sense if both $L$ and $L'$ are finite, so assume so. If $L\neq L'$, take $\varepsilon = \dfrac{|L-L'|}{2}>0$ in the definition of convergence. Now $$\lim _{n\rightarrow +\infty} a_n = L \implies \exists N_1> 0, \quad \forall n\geq N_1 \absval{a_n-L}<\varepsilon, $$ $$\lim _{n\rightarrow +\infty} a_n = L' \implies \exists N_2> 0, \quad \forall n\geq N_2 \absval{a_n-L'}<\varepsilon. $$If $n>\max (N_1, N_2)$ then by the Triangle Inequality (Theorem \ref{tri_ineq}) then $$ \absval{L-L'}\leq \absval{L-a_n}+\absval{a_n-L'}< 2\varepsilon = \absval{L-L'}, $$a contradiction, so $L=L'$. \end{pf} \begin{thm}\label{thm:conve-seq-bounded-be} Every convergent sequence is bounded. \end{thm} \begin{pf} Let $\seq{a_n}{n=1}{+\infty}$ converge to $L$. Using $\varepsilon = 1$ in the definition of convergence, $\exists N>0$ such that $$ n\geq N \implies \absval{a_n-L}<1 \implies L-1}(-10,0)(10,0) \rput(-7,0){|} \rput(-3,0){|} \rput(0,0){|} \rput(2.9,0){|} \rput(4,0){|} \rput(5,0){|}\rput(6.5,0){|}\uput[d](-7,0){x_0} \uput[d](-3,0){x_1}\uput[d](0,0){{\tiny x_2}}\uput[d](2.9,0){\ddots }\uput[d](4,0){{\tiny x_n}}\uput[d](5,0){\ddots}\uput[d](6.5,0){{\tiny s}} $$ \footnotesize\hangcaption{Theorem \ref{thm:conve_seq}.} \label{fig:thm_conve_seq} \end{figure} When is it guaranteed that a sequence of real numbers has a limit? We have the following result. \begin{thm}\label{thm:bounded-increasing-seqs-convergent-be}Every bounded increasing sequence $\{a_n\}_{n = 0} ^{+\infty}$ of real numbers converges to its supremum. Similarly, every bounded decreasing sequence of real numbers converges to its infimum. \end{thm} \begin{pf} The idea of the proof is sketched in figure \ref{fig:thm_conve_seq}. By virtue of Axiom \ref{axi:completeness-of-R}, the sequence has a supremum $s$. Every term of the sequence satisfies $a_n \leq s$. We claim that eventually all the terms of the sequence are closer to $s$ than a preassigned small distance $\varepsilon > 0$. Since $s - \varepsilon$ is not an upper bound for the sequence, there must be a term of the sequence, say $a_{n_0}$ with $s - \varepsilon \leq a_{n_0}$ by virtue of the Approximation Property Theorem \ref{thm:approx-sup-inf}. Since the sequence is increasing, we then have $$ s-\varepsilon \leq a_{n_0} \leq a_{n_0+1} \leq a_{n_0+2} \leq a_{n_0+2} \leq \ldots \leq s, $$ which means that after the $n_0$-th term, we get to within $\varepsilon$ of $s$. \bigskip To obtain the second half of the theorem, we simply apply the first half to the sequence $\{-a_n\}_{n=0} ^{+\infty}$. \end{pf} \begin{thm}[Order Properties of Sequences]\label{thm:order-prop-seq} Let $\seq{a_n}{n=1}{+\infty}$ be a sequence of real numbers converging to the real number $L$. Then \begin{enumerate} \item If $a0$ such that $$\forall n \geq N_1, \quad \absval{a_n-L}0$ such that $$\forall n \geq N_2, \quad \absval{a_n-L}a$, then by part (1) we will eventually have $a_n>a$, a contradiction. \item If, to the contrary, $L0$ there are $N_1>0$, $N_2>0$ such that $$\forall n\geq \max (N_1, N_2), \quad \absval{u_n-L}<\varepsilon, \quad \absval{v_n-L}<\varepsilon \implies -\varepsilon < u_n-L < \varepsilon, \quad -\varepsilon < v_n-L < \varepsilon. $$ Thus for such $n$, $$-\varepsilon < u_n -L \leq a_n - L \leq v_n - L < \varepsilon, \implies -\varepsilon < a_n-L < \varepsilon \implies \absval{a_n-L}<\varepsilon, $$ from where $\seq{a_n}{n=1}{+\infty}$ converges to $L$. \end{pf} \begin{thm} Let $\seq{a_n}{n=1}{+\infty}$ be a sequence of real numbers such that $a_n\rightarrow L$. Then $\absval{a_n}\rightarrow \absval{L}$. \end{thm} \begin{pf} From Corollary \ref{cor:triangle-ineq}, we have the inequality $\absval{|a_n|-|L|}\leq \absval{a_n-L}$ from where the result follows. \end{pf} \begin{thm}\label{thm:abs-val-conver-seq} Let $\seq{a_n}{n=1}{+\infty}$ be a sequence of real numbers such that $a_n\rightarrow 0$, and let $\seq{b_n}{n=1}{+\infty}$ be a bounded sequence. Then $a_nb_n\rightarrow 0$. \end{thm} \begin{pf}Eventually $\absval{a_n}<\varepsilon$. Assume that eventually $|b_n|\leq U$. Then $$\absval{a_nb_n} \leq U\absval{a_n}0$ in the definition of convergence, we have that eventually $$\absval{|b_n|-|l|}<\dfrac{|l|}{2}\implies |l|-\dfrac{|l|}{2} <\absval{b_n} < |l|+\dfrac{|l|}{2} \implies \dfrac{|l|}{2}<\absval{b_n}, $$That is, eventually $|b_n|$ is strictly positive and so $\dfrac{1}{b_n}$ makes sense. Also, eventually, $\dfrac{1}{|b_n|}<\dfrac{2}{|l|}$ Now, for sufficiently large $n$, $$\absval{\dfrac{1}{b_n}-\dfrac{1}{l}} = \absval{\dfrac{l-b_n}{|b_n||l|}} = \dfrac{\absval{b_n-l}}{\absval{b_n}\absval{l}}<\dfrac{2\varepsilon}{\absval{l}\absval{l}}, $$ from where the result follows.\end{pf} \begin{thm}[Algebraic Properties of Sequences]\label{thm:algebra-of-seq-limits} Let $k\in \BBR$. If $\seq{a_n}{n=1}{+\infty}$ converges to $L$ and $\seq{b_n}{n=1}{+\infty}$ converges to $L'$ then $$ \lim _{n\rightarrow +\infty} (ka_n+ b_n ) = kL+ L', \qquad \lim _{n\rightarrow +\infty} (a_nb_n ) = LL'. $$ Moreover, if $L'\neq 0$ then $$ \lim _{n\rightarrow +\infty} \left(\dfrac{a_n}{b_n} \right) = \dfrac{L}{L'}. $$ \end{thm} \begin{pf} The trick in all these proofs is the following observation: If one multiplies a bounded quantity by an arbitrarily small quantity, one gets an arbitrarily small quantity. Hence once considers the absolute value of the desired terms of the sequence from the expected limit. \bigskip Given $\varepsilon>0$ there are $N_1>0$ and $N_2>0$ such that $\absval{a_n-L}<\varepsilon$ and $\absval{b_n-L'}<\varepsilon$. Then $$\absval{(ka_n+b_n)-(kL+L')} = \absval{(ka_n-kL)+(b_n-L')} \leq \absval{k}\absval{a_n-L}+\absval{b_n-L'}<\varepsilon (k+1), $$and so the sinistral side is arbitrarily close to $0$, establishing the first assertion. \bigskip For the product, observe that by Theorem \ref{thm:conve-seq-bounded-be} there exists a constant $K>0$ such that $\absval{b_n}0, \quad a_n>0$ and $\seq{a_n}{n=1}{+\infty}$ converges to $L$ must it be the case that $L>0$? \begin{answer}No. Take $a_n = \dfrac{1}{n}$. Then $a_n > 0$ always, but $L=0$. \end{answer} \end{pro} \begin{pro} Prove that if $a_n \rightarrow +\infty $ and if $\seq{b_n}{n=1}{+\infty}$ is bounded, then $a_n+b_n \rightarrow +\infty$. \end{pro} \begin{pro} Prove that if $a_n \rightarrow +\infty $ and $b_n \rightarrow +\infty $ is bounded, then $a_n+b_n \rightarrow +\infty$. \end{pro} \begin{pro} Prove that if $a_n \rightarrow +\infty $ and if there exists $K>0$ such that eventually $b_n\geq K$, then $a_nb_n \rightarrow +\infty$. \end{pro} \begin{pro} Prove that if $a_n \rightarrow +\infty $ and $b_n \rightarrow +\infty $ is bounded, then $a_nb_n \rightarrow +\infty$. \end{pro} \begin{pro} Prove that if $a_n \rightarrow +\infty $ and if $\seq{b_n}{n=1}{+\infty}$ is bounded, then $a_n+b_n \rightarrow +\infty$. \end{pro} \begin{pro} Prove that if $a_n \rightarrow +\infty $ then $\dfrac{1}{a_n}\rightarrow 0$. \end{pro} \begin{pro} Prove that if $a_n \rightarrow 0 $ and if eventually $a_n>0$, then $\dfrac{1}{a_n}\rightarrow +\infty$. \end{pro} \begin{pro} Prove that $\sum _{i=1} ^n \dfrac{n}{n^2+i} \rightarrow 1$ as $\ngrows$. \begin{answer} We have for $n>1$,$$\dfrac{n^2}{n^2+n} = \underbrace{\dfrac{n}{n^2+n}+\cdots + \dfrac{n}{n^2+n}}_{n\ \mathrm{times}}< \sum _{i=1} ^n \dfrac{n}{n^2+i}<\underbrace{\dfrac{n}{n^2+1}+ \cdots + \dfrac{n}{n^2+1}}_{n\ \mathrm{times}} =\dfrac{n^2}{n^2+1}, $$and the result follows by the Sandwich Theorem since each of the sequences on the extremes converges to $1$. \end{answer} \end{pro} \begin{pro} Prove that $\dfrac{1}{(n!)^{1/n}} \rightarrow 0$. \begin{answer} Evidently $n! \leq n^n$. By problem \ref{pro:factorial-n^n/2}, if $n>2$ then $n^{n/2}\leq n!$. Thus $$\dfrac{1}{n}\leq \dfrac{1}{(n!)^{1/n}}\leq \dfrac{1}{n^{1/2}}$$and the result follows by the Sandwich Theorem. \end{answer} \end{pro} \begin{pro} Prove that $\dfrac{2^n}{n!}\rightarrow 0$. \begin{answer} For $n \geq 2$ we have $$ \dfrac{2^n}{n!} = \dfrac{2}{1}\cdot \dfrac{2}{2}\cdot \dfrac{2}{3}\cdots \dfrac{2}{n} \leq 2\cdot 1 \cdot 1 \cdots 1 \cdot \dfrac{2}{n} = \dfrac{4}{n} \rightarrow 0.$$ \end{answer} \end{pro} \begin{pro} Let $x_1 , x_2 , \ldots$ be a {\em bounded} sequence of real numbers, and put $s_n = x_1 + x_2 + \cdots + x_n$. Suppose that $\dfrac{s_{n^2}}{n^2}\rightarrow 0$. Prove that $\dfrac{s_n}{n} \rightarrow 0$. \begin{answer} There is a positive integer $m$ with $m^2 \leq n < (m + 1)^2$. Consider $$ \absval{\dfrac{s_{m^2}}{m^2} - \dfrac{s_{n}}{n}}. $$\end{answer} \end{pro} \begin{pro} Prove rigorously that the sequence $\seq{\sin n}{n=0}{+\infty}$ is divergent. \begin{answer} Since $-1\leq \sin n \leq 1$, any possible limit must be finite. By way of contradiction assume that $\sin n \rightarrow a$ as $\ngrows$. Then $$\lim _{\ngrows} \sin n = a \implies \lim _{\ngrows} \sin (n+2) = a, $$ whence $$\lim _{\ngrows} (\sin (n+2)-\sin n) = a-a=0. $$Now, $$\sin (n+2)-\sin n=2(\sin 1)\cos (n+1) \implies \cos (n+1)\rightarrow 0, \quad \mathrm{as}\ \ngrows. $$ From $$\cos (n+1) = \cos n \cos 1 -\sin n\sin 1 $$we obtain $$\sin n = \dfrac{1}{\sin 1}\left(\cos n\cos 1 - \cos (n+1)\right) \rightarrow \dfrac{1}{\sin 1}\left(0\cdot \cos 1 - 0\right) = 0, $$and so $a=0$. But then $$ 1 = \sin ^2 n + \cos ^2n\rightarrow 0^2 + 0^2 = 0, $$a contradiction. \end{answer} \end{pro} \begin{pro} Prove that $(n!)^{1/n}\rightarrow +\infty$ as $\ngrows$. \begin{answer} By problem \ref{pro:factorial-n^n/2}, $(n!)^{1/n}>\sqrt{n}$ for $n\geq 3$. Hence, for all $M>0$, as long as $n>M^2$ we will have $$ (n!)^{1/n}>\sqrt{n}>M, $$giving the result. \end{answer} \end{pro} \begin{pro} A sequence of real numbers $a_1, a_2, \ldots $ satisfies, for all $m, n$, the inequality $$ \absval{a_m+a_n-a_{m+n}} \leq \dfrac{1}{m+n}. $$Prove that this sequence is an arithmetic progression. \end{pro} \begin{pro} Prove rigorously that $\sqrt{n+1}-\sqrt{n}\rightarrow 0$ as $\ngrows$. \begin{answer} We have $$\sqrt{n+1}-\sqrt{n} = \dfrac{n+1-n}{\sqrt{n+1}+\sqrt{n}} = \dfrac{1}{\sqrt{n+1}+\sqrt{n}}<\dfrac{1}{2\sqrt{n}}. $$ Hence, as long as $\dfrac{1}{2\sqrt{n}}< \varepsilon$ that is, as long as $n>\dfrac{1}{4\varepsilon ^2} $ we will have $$\absval{\sqrt{n+1}-\sqrt{n}} < \dfrac{1}{2\sqrt{n}}<\varepsilon .$$ \end{answer} \end{pro} \begin{pro} Prove that the sequence $H_n = 1 + \dfrac{1}{2}+\dfrac{1}{3}+\cdots + \dfrac{1}{n}$ diverges to $+\infty$. \begin{answer} Write $$ \sum_{n=1}^{2^M} \frac{1}{n} = \sum_{m=1}^M \sum_{n=2^{m-1}+1}^{2^m} \frac{1}{n}. $$ Since $1/n \geq 1/N$ when $n \leq N$, we gather that $$ \sum_{n=2^{m-1}+1}^{2^m} \frac{1}{n} \geq \sum_{n=2^{m-1}+1}^{2^m} 2^{-m} = (2^m - 2^{m-1}) 2^{-m} = \frac{1}{2}.$$ Thus $$ \sum_{n=1}^{2^M} \frac{1}{n} \geq \frac{M}{2} $$ and the sequence can be made arbitrarily large. \end{answer} \end{pro} \begin{pro} Find $$ \lim _{K\rightarrow +\infty} \sum _{n=1} ^K \dfrac{\sqrt{(n-1)!}}{(1+\sqrt{1})(1+\sqrt{2})(1+\sqrt{3}) \cdots (1+\sqrt{n})}. $$ \begin{answer} Observe that for $n \geq 2$, $$\begin{array}{l}\dfrac{\sqrt{(n-1)!}}{(1+\sqrt{1})(1+\sqrt{2})(1+\sqrt{3}) \cdots (1+\sqrt{n-1})} - \dfrac{\sqrt{(n)!}}{(1+\sqrt{1})(1+\sqrt{2})(1+\sqrt{3}) \cdots (1+\sqrt{n})} \\ = \dfrac{\sqrt{(n-1)!}}{(1+\sqrt{1})(1+\sqrt{2})(1+\sqrt{3}) \cdots (1+\sqrt{n-1})}\left(1-\dfrac{\sqrt{n}}{1+\sqrt{n}}\right) \\ = \dfrac{\sqrt{(n-1)!}}{(1+\sqrt{1})(1+\sqrt{2})(1+\sqrt{3}) \cdots (1+\sqrt{n})}. \end{array}$$Therefore $$\sum _{n=1} ^K \dfrac{\sqrt{(n-1)!}}{(1+\sqrt{1})(1+\sqrt{2})(1+\sqrt{3}) \cdots (1+\sqrt{n})}=1 -\frac{\sqrt{K!}}{(1+\sqrt{1})(1+\sqrt{2})\cdots (1+\sqrt{K})}.$$ Now prove that $u_K=\frac{\sqrt{K!}}{(1+\sqrt{1})(1+\sqrt{2})\cdots (1+\sqrt{K})}$ decreases to $0$. \end{answer} \end{pro} \begin{pro} What reasonable meaning can be given to $$ \sqrt{1+\sqrt{1+\sqrt{1+\sqrt{\cdots}}}}\qquad ? $$ \begin{answer} Put $x_1 = 1, \quad x_{n+1} = \sqrt{1+x_n}, n\geq 0$. We claim that the sequence $\seq{x_n}{n=1}{+\infty}$ is increasing and bounded above. By Theorem \ref{thm:bounded-increasing-seqs-convergent-be} the sequence must have a limit $L$. To prove that the sequence is increasing consider $x_{n+1}-x_n$ (fill in this gap). To prove that the sequence is bounded, we claim that for all $n\geq 1$, $x_n < 4$. For this is clearly true for $n=1$. So assume that $x_n<4$. Then $$x_{n+1} = \sqrt{1+x_n} < \sqrt{1+4}=\sqrt{5}<4, $$ and so the assertion follows by induction. \bigskip Since we have shewn that $L$ exists we now may compute $$L= \lim _{\ngrows} x_{n+1} = \lim _{\ngrows} \sqrt{1+x_n} = \sqrt{1+L}\implies L=\sqrt{1+L} \implies L^2-L-1=0 \implies L= \dfrac{1+\sqrt{5}}{2}, $$where we have chosen the positive root as the sequence is clearly strictly positive. \end{answer} \end{pro} \begin{pro} Prove that $$\dfrac{1+2+\cdots + n}{n^2}\rightarrow \dfrac{1}{2}, \quad \mathrm{as}\ \ngroes.$$ \begin{answer} By Theorem \ref{thm:sum-of-first-n-integers}, $1+2+\cdots + n = \dfrac{n^2+n}{2}$, and the desired result follows. \end{answer} \end{pro} \begin{pro} Calculate the following limits: \begin{enumerate} \item $\lim _{\ngroes}\left(\dfrac{1^2}{n^2}+ \dfrac{2^2}{n^2}+ \cdots + \dfrac{(n-1)^2}{n^2}\right)$, \item $\lim _{\ngroes}\left(\dfrac{1}{1\cdot 2}+ \dfrac{1}{2\cdot 3}+ \cdots + \dfrac{1}{n(n+1)}\right)$, \item $\lim _{\ngroes}\left(\dfrac{1}{1\cdot 2\cdot 3}+ \dfrac{1}{2\cdot 3\cdot 4}+ \cdots + \dfrac{1}{n(n+1)(n+2)}\right)$,\end{enumerate} \begin{answer} $\dfrac{1}{3}$; $1$; $\dfrac{1}{4}$. \end{answer} \end{pro} \begin{pro} What reasonable meaning can be given to $$ \dfrac{1}{1+\dfrac{1}{1+\dfrac{1}{1+\dfrac{1}{\vdots}}}} \qquad ? $$ \begin{answer} Put $x_1 = 1, \quad x_{n+1} = \dfrac{1}{1+x_n}, n\geq 0$. We claim that the sequence $\seq{x_n}{n=1}{+\infty}$ is increasing and bounded above. By Theorem \ref{thm:bounded-increasing-seqs-convergent-be} the sequence must have a limit $L$. To prove that the sequence is increasing consider $x_{n+1}-x_n$ (fill in this gap). To prove that the sequence is bounded, we claim that for all $n\geq 1$, prove by induction that $x_n < 4$ (fill in this gap). \bigskip Since we have shewn that $L$ exists we now may compute $$L= \lim _{\ngrows} x_{n+1} = \lim _{\ngrows} \dfrac{1}{1+x_n} = \dfrac{1}{1+L}\implies L=\dfrac{1}{1+L} \implies L^2+L-1=0 \implies L= \dfrac{\sqrt{5}-1}{2}, $$where we have chosen the positive root as the sequence is clearly strictly positive. \end{answer} \end{pro} \begin{pro} Let $K\in\BBN\setminus \{0\}$, and let $a_1, \ldots , a_K, \lambda _1 , \ldots , \lambda _K$ be strictly positive real numbers. Prove that $$ \lim _{\ngrows} \left(\sum _{k=1} ^K \lambda _k a_k ^n\right)^{1/n} = \max _{1\leq k \leq K} a_k, \qquad \lim _{\ngrows} \left(\sum _{k=1} ^K \lambda _k a_k ^{-n}\right)^{-1/n} = \min _{1\leq k \leq K} a_k. $$ \end{pro} \begin{pro} Prove that if $\seq{\dfrac{a_n}{b_n}}{n=1}{+\infty}$ is a monotonic sequence, then the $\seq{\dfrac{a_1+a_2+\cdots + a_n}{b_1+b_2+\cdots + b_n}}{n=1}{+\infty}$ is also monotonic in the same sense. \begin{answer} Assume that $\seq{\dfrac{a_n}{b_n}}{n=1}{+\infty}$ is increasing. Then $$\dfrac{a_1}{b_1}\leq \dfrac{a_2}{b_2} \leq \cdots \leq \dfrac{a_n}{b_n} \leq \dfrac{a_{n+1}}{b_{n+1}}. $$ Using Theorem \ref{thm:min-max-fractions}, $$ \dfrac{a_1+a_2+\cdots + a_n}{b_1+b_2+\cdots + b_n}\leq \dfrac{a_n}{b_n} \leq \dfrac{a_{n+1}}{b_{n+1}} \implies \dfrac{a_1+a_2+\cdots + a_n}{b_1+b_2+\cdots + b_n} \leq \dfrac{a_1+a_2+\cdots + a_{n+1}}{b_1+b_2+\cdots + b_{n+1}} \leq \dfrac{a_{n+1}}{b_{n+1}}, $$proving that $\seq{\dfrac{a_1+a_2+\cdots + a_n}{b_1+b_2+\cdots + b_n}}{n=1}{+\infty}$ is also increasing. If $\seq{\dfrac{a_n}{b_n}}{n=1}{+\infty}$ were decreasing, $\seq{-\dfrac{a_n}{b_n}}{n=1}{+\infty}$ is increasing and we apply what we just have proved. \end{answer} \end{pro} \begin{pro} Let $a, b, c$ be real numbers such that $b^2-4ac<0$. Let $\seq{X_n}{n=1}{+\infty}$, $\seq{Y_n}{n=1}{+\infty}$ be sequences of real numbers such that $$ aX^2 _n + bX_nY_n + cY_n ^2 \rightarrow 0, \quad \mathrm{as}\ \ngrows. $$Prove that $X_n\rightarrow 0$ and $Y_n \rightarrow 0$ as $\ngrows$. \end{pro} \begin{pro}[Gram's Product]Prove that $$ \lim _{\ngrows}\prod _{k=2} ^n \dfrac{k^3-1}{k^3+1} =\dfrac{2}{3}.$$ \begin{answer} We have $$\prod _{k=2} ^n \dfrac{k^3-1}{k^3+1} = \prod _{k=2} ^n \dfrac{k-1}{k+1} \prod _{k=2} ^n \dfrac{k^2+k+1}{k^2-k+1}. $$ Now $$\prod _{k=2} ^n \dfrac{k-1}{k+1} = \dfrac{(n-1)!}{\frac{(n+1)!}{2}} = \dfrac{2}{n(n+1)}. $$ By observing that $(k+1)^2-(k+1)+1 = k^2+k+1$, we gather that $$ \prod _{k=2} ^n \dfrac{k^2-k+1}{k^2+k+1} = \dfrac{3^2+3+1}{2^2-2+1}\cdot \dfrac{4^2+4+1}{3^2+3+1}\cdot \dfrac{5^2+5+1}{4^2+4+1} \cdots \dfrac{n^2+n+1}{(n-1)^2+(n-1)+1} =\dfrac{n^2+n+1}{3}. $$ Thus $$ \prod _{k=2} ^n \dfrac{k^3-1}{k^3+1} = \dfrac{2}{3} \cdot \dfrac{n^2+n+1}{n(n+1)} \rightarrow \dfrac{2}{3},$$as $\ngrows$. \end{answer} \end{pro} \begin{pro} Prove that the sequence $\seq{x_n}{n=1}{+\infty}$ with $x_n = 1 + \dfrac{1}{2^2} + \dfrac{1}{3^2} + \cdots +\dfrac{1}{n^2}$ satisfies $x_n \leq 2-\dfrac{1}{n}$ for $n\geq 1$. Hence deduce that it converges. \begin{answer} Clearly $x_n 0, b>0$. Also, find its limit. \end{pro} \begin{pro} Prove the convergence of the sequence, $x_1=a$, $x_{n+1} = \dfrac{1}{2}\left(x_n+\dfrac{b}{x_n}\right)$, $n \geq 1$ and $(a, b)\in\BBR^2$, $a<0, b>0$. Also, find its limit. \end{pro} \begin{pro} Let $(a, b)\in \BBR^2$, $a>b>0$. Set $a_1 = \dfrac{a+b}{2}$, $b_1 = \sqrt{ab}$. If for $n>1$, $$a_{n+1} = \dfrac{a_n+b_n}{2}, \qquad b_n = \sqrt{a_nb_n}, $$Prove that \begin{enumerate} \item $\seq{a_n}{n=1}{+\infty}$ is monotonically decreasing, \item $\seq{b_n}{n=1}{+\infty}$ is monotonically inreasing, \item both sequences converge, \item their limits are equal. \end{enumerate} \end{pro} \end{multicols} \section{Classical Limits of Sequences} \begin{center} \fcolorbox{blue}{yellow}{ \begin{minipage}{.90\linewidth} \noindent\textcolor{red}{\textbf{Why bother?}} In this section we will obtain various classical limits. In particular, we define the constant $e$ and obtain a few interesting results about it. \end{minipage}} \end{center} \begin{thm}\label{thm:geom-sequence-to-zero-goes} Let $r\in\BBR$ be fixed. If $\absval{r}<1$ then $r^n \rightarrow 0$ as $n\rightarrow +\infty$. If $\absval{r}>1$ then $r^n \rightarrow +\infty$ as $n\rightarrow +\infty$. \end{thm} \begin{pf} Taking $x = \absval{\frac{1}{r}} - 1$ in Bernoulli's Inequality (Theorem \ref{thm:bernoulli}), we find $$\left|\frac{1}{r}\right|^n > 1 + n\left(\left|\frac{1}{r}\right| - 1\right) > n\left(\left|\frac{1}{r}\right| - 1\right). $$Therefore $$|r|^n < \frac{|r|}{n(1 - |r|)} \rightarrow 0,$$as $n \rightarrow +\infty$, since $\dfrac{1}{n}\rightarrow 0$ as $\ngrows$. \bigskip If $|r| > 1$, again by Bernoulli's Inequality $$|r|^n = (1 + |r| - 1)^n > 1 + n(|r| - 1),$$ and the dextral side can be made arbitrarily large.\end{pf} \begin{thm}\label{thm:geometric-sum} Let $|r|<1$. Then $$1+r+r^2 + \cdots + r^n \rightarrow \dfrac{1}{1-r},\quad \mathrm{as}\quad \ngrows. $$ \end{thm} \begin{pf} If $S_n = 1+r+r^2 + \cdots + r^n $ then $rS_n = r+r^2+r^3+ + \cdots + r^{n+1}$ and $$S_n-rS_n = 1-r^{n+1} \implies S_n = \dfrac{1-r^{n+1}}{1-r}. $$Then apply Theorem \ref{thm:geom-sequence-to-zero-goes}. \end{pf} \begin{rem} An estimating trick that we will use often is the following. If $0 < r<1$ then the truncated sum is smaller than the infinite sum and so, for all positive integers $k$: $$1+r+r^2 + \cdots + r^k < 1+r+r^2 + \cdots + \cdots = \dfrac{1}{1-r}. $$ \end{rem} \begin{thm}\label{thm:root-n-of-a-to-1-goes} Let $a\in \BBR$, $a>0$, be fixed. Then $a^{1/n} \rightarrow 1$ as $n\rightarrow +\infty$. \end{thm} \begin{pf} If $a>1$ then $a^{1/n}>1$ and by Bernoulli's Inequality, $$a = (1+(a^{1/n}-1))^n >1+n(a^{1/n}-1) \implies 0\leq a^{1/n}-1 <\dfrac{a-1}{n},$$whence $a^{1/n}-1\rightarrow 0$ as $\ngrows$. \bigskip If $0 < a < 1$ then $b=\dfrac{1}{a}>1$ and so by what we just proved, $$b^{1/n}\rightarrow 1 \implies \dfrac{1}{a^{1/n}}\rightarrow 1 \implies a^{1/n}\rightarrow 1, $$ proving the theorem.\end{pf} \begin{thm}\label{thm:exps-are-faster-than-powers} Let $a\in\BBR$, $a>1$, $k\in\BBN\setminus\{0\}$, be fixed. Then $\dfrac{a^n}{n^k} \rightarrow +\infty$ as $n\rightarrow +\infty$. \end{thm} \begin{pf} Observe that $a^{1/k}>1$. We have, using the Binomial Theorem, $$\left(a^{1/k}\right)^n = \left(1+(a^{1/k}-1)\right)^n = \sum _{i=0} ^n \binom{n}{i}(a^{1/k}-1)^i.$$Since each term of the above expansion is $\geq 0$, we gather that $$ \left(a^{1/k}\right)^n \geq \dfrac{n(n-1)}{2}(a^{1/k}-1)^2 \implies \dfrac{\left(a^{1/k}\right)^n }{n} \geq \dfrac{(n-1)}{2}(a^{1/k}-1)^2 \implies \dfrac{\left(a^{1/k}\right)^n }{n}\rightarrow +\infty \implies \dfrac{a^n}{n^k} \rightarrow +\infty ,$$as desired. \end{pf} \begin{rem} In particular $\dfrac{2^n}{n}\rightarrow +\infty$ as $\ngrows$. \end{rem} \begin{thm}\label{thm:factorials-are-fasters-than-exps} Let $a\in\BBR$, , be fixed. Then $\dfrac{a^n}{n!} \rightarrow 0$ as $n\rightarrow +\infty$. \end{thm} \begin{pf}Put $N=\floor{\absval{a}}+1$ and let $n\geq N$. Then $$\absval{\dfrac{a^n}{n!}} = \left(\dfrac{|a|}{1}\cdot \dfrac{|a|}{2}\cdots \dfrac{|a|}{N}\right) \left(\dfrac{|a|}{N+1}\cdot \dfrac{|a|}{N+2}\cdots \dfrac{|a|}{n}\right)\leq \left(\dfrac{|a|^N}{N!}\right)\left(1\cdot 1 \cdots 1\cdot\dfrac{|a|}{n}\right) \rightarrow 0, $$as $\ngrows$. \end{pf} \begin{thm}\label{thm:e} The sequence $$e_n = \left(1 + \frac{1}{n}\right)^n, n = 1, 2, \ldots $$ is a bounded increasing sequence, and hence it converges to a limit, which we call $e$. Also, for all strictly positive integers $n$, $\left(1 + \frac{1}{n}\right)^n e$. \end{thm} \begin{pf} By Theorem \ref{thm:ineq_bin} $$ \dfrac{b^{n + 1} - a^{n + 1}}{b - a} \geq (n + 1)a^{n}. $$Putting $a = 1 + \dfrac{1}{n + 1}$, $b = 1 + \dfrac{1}{n}$ we obtain $$ \left(1 + \frac{1}{n}\right)^{n+1}>\left(1 + \frac{1}{n+1}\right)^{n+2}\left(\dfrac{n^3+4n^2+4n+1}{n(n+2)^2}\right). $$The result will follow as long as $\left(\dfrac{n^3+4n^2+4n+1}{n(n+2)^2}\right)> 1$. But $$n(n+2)^2=n(n^2+4n+4)=n^3+4n^2+4n< n^3+4n^2+4n+1 \implies \dfrac{n^3+4n^2+4n+1}{n(n+2)^2}> 1.$$ Thus the sequence is a sequence of strictly decreasing sequence of real numbers. Putting $a=1$, $b=1+\dfrac{1}{n}$ in $ \dfrac{b^{n + 1} - a^{n + 1}}{b - a} \geq (n + 1)a^{n}$ we get $$ \left(1+\dfrac{1}{n}\right)^{n+1} > 1+ n(n+1)>2, $$ so the sequence is bounded below. In view of Theorem \ref{thm:bounded-increasing-seqs-convergent-be} the sequence converges to a limit $L$. To see that $L=e$ observe that $$\left(1 + \frac{1}{n}\right)^{n+1} = \left(1 + \frac{1}{n}\right)^{n}\left(1 + \frac{1}{n}\right)\rightarrow e\cdot 1 = e. $$ It follows also from this proof and from Theorem \ref{thm:order-prop-seq} that for all strictly positive integers $n$, $\left(1 + \frac{1}{n}\right)^{n+1} >e$. \end{pf} \begin{rem} The inequality $\left(1+\dfrac{1}{n+1}\right)^{n+2}<\left(1+\dfrac{1}{n}\right)^{n+1}$ can be obtained by the Harmonic Mean-Geometric Mean Inequality by putting $x_1=1, x_2=x_2=\cdots =x_{n+2} = 1+\dfrac{1}{n}$ $$ \dfrac{n+2}{\frac{1}{x_1}+\frac{1}{x_2}+\cdots +\frac{1}{x_{n+2}}}\leq (x_1x_2\cdots x_{n+2})^{1/(n+2)} \implies \dfrac{n+2}{1+(n+1)\left(\dfrac{n}{n+1}\right)} < \left(1+\dfrac{1}{n}\right)^{(n+1)/(n+2)}.$$ \end{rem} \begin{thm}\label{thm:2y_k$ so that $\seq{y_k}{k=1}{+\infty}$ is an increasing sequence. We will prove that it is bounded above with supremum $e$. By the Binomial Theorem $$ \left(1+\dfrac{1}{n}\right)^n= \sum _{j=0} ^n \binom{n}{j}\cdot \dfrac{1}{n^j} = 1 + \binom{n}{1}\dfrac{1}{n} + \cdots + \binom{n}{k}\dfrac{1}{n^k} + \cdots + \binom{n}{n}\dfrac{1}{n^n} \geq 1 + \binom{n}{1}\dfrac{1}{n} + \cdots + \binom{n}{k}\dfrac{1}{n^k}, $$for $0 < k 4^{1/4}>5^{1/5}>\cdots . $$ Clearly, if $n>1$ then $n^{1/n}>1^{1/n}=1$. Also, by the Binomial Theorem, again, if $n>1$, $$ \left(1+\sqrt{\dfrac{2}{n}}\right)^n = 1^n + \binom{n}{1}\left(\sqrt{\dfrac{2}{n}}\right)^1 + \binom{n}{2}\left(\sqrt{\dfrac{2}{n}}\right)^2 + \cdots > 1 + \binom{n}{2}\left(\sqrt{\dfrac{2}{n}}\right)^2 = 1 + \dfrac{n(n-1)}{2}\left(\dfrac{2}{n}\right) = n. $$ We then conclude that $$ 1 < n^{1/n}< 1+\sqrt{\dfrac{2}{n}},$$and that $n^{1/n}\rightarrow 1$ follows from the Sandwich Theorem. \end{pf} \begin{rem} $2^{1/2} = 4^{1/4}$. \end{rem} \subsection*{Homework}\addcontentsline{toc}{subsection}{Homework} \begin{multicols}{2}\columnseprule 1pt \columnsep 25pt\multicoltolerance=900\small \begin{pro} What's wrong with the following? Since the product of the limits is the limit of the product, $$ e=\lim _{\ngroes} \left(1+\dfrac{1}{n}\right)^n = \underbrace{\left(\lim _{\ngroes} 1+\dfrac{1}{n}\right)\cdot \left(\lim _{\ngroes} 1+\dfrac{1}{n}\right) \cdots \left(\lim _{\ngroes} 1+\dfrac{1}{n}\right) }_{n\ \mathrm{times}} = \underbrace{1\cdot 1 \cdots 1}_{n\ \mathrm{times}} =1.$$ \begin{answer} The product rule for limits only applies to a finite number of factors. Here the number of factors grows with $n$. \end{answer} \end{pro} \begin{pro} Demonstrate that for all strictly positive integers $n$: $$ \cos \dfrac{\pi}{2^{n+1}} = \dfrac{1}{2}\underbrace{\sqrt{2+\sqrt{2+\sqrt{2+\cdots +\sqrt{2} } } }}_{n\ \mathrm{radicands}}, $$ $$ \sin \dfrac{\pi}{2^{n+1}} = \dfrac{1}{2}\underbrace{\sqrt{2-\sqrt{2+\sqrt{2+\cdots +\sqrt{2} } } }}_{n\ \mathrm{radicands}}. $$ Hence deduce {\em Vi\`{e}te's Formula for $\pi$:} $$ \pi = \lim _{\ngroes} 2^n\underbrace{\sqrt{2-\sqrt{2+\sqrt{2+\cdots +\sqrt{2} } } }}_{n\ \mathrm{radicands}}. $$ \end{pro} \begin{pro}\label{pro:converging-to-log2} Prove that the sequence $\seq{\sum _{k=n} ^{2n}\dfrac{1}{k}}{n=1}{+\infty}$ converges to $\log 2$. \begin{answer} From Theorem \ref{thm:e}, and since $x\mapsto \log x$ is increasing, $$ \left(1+\dfrac{1}{k+1}\right)^{k+1} < e < \left(1+\dfrac{1}{k}\right)^{k+1} \implies (k+1)\log \left(1+\dfrac{1}{k+1}\right) <1 < (k+1)\log \left(1+\dfrac{1}{k}\right). $$Rearranging, $$ \log \dfrac{k+2}{k+1}<\dfrac{1}{k+1}< \log \dfrac{k+1}{k}. $$ Summing from $k=n-1$ to $k=2n-1$, $$\begin{array}{lll} \sum _{k=n-1} ^{2n-1}\log \dfrac{k+2}{k+1}< \sum _{k=n-1} ^{2n-1}\dfrac{1}{k+1}< \sum _{k=n-1} ^{2n-1} \log \dfrac{k+1}{k} & \implies & \log \dfrac{2n+1}{n}< \dfrac{1}{n}+\dfrac{1}{n+1} + \cdots + \dfrac{1}{2n}<\log \dfrac{2n}{n-1}\\ & \implies & \log \left(2+\dfrac{1}{n}\right)<\dfrac{1}{n}+\dfrac{1}{n+1} + \cdots + \dfrac{1}{2n} <\log \left(2+\dfrac{2}{n-1}\right)\end{array}$$and the result follows from the Sandwich Theorem. \end{answer} \end{pro} \begin{pro} Prove that the sequence $\seq{1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots + \frac{1}{2n - 1} - \frac{1}{2n}}{n=1}{+\infty}$ converges to $\log 2$. \begin{answer} Observe that $${\everymath{\displaystyle}\begin{array}{lcl} \left( 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \cdots + \frac{1}{2n - 1} + \frac{1}{2n}\right) & & \\ \qquad - 2\left(\frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \cdots + \frac{1}{2n}\right) & & \\ & = & \left( 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \cdots + \frac{1}{2n - 1} + \frac{1}{2n}\right) \\ & & \qquad - 2\cdot\frac{1}{2}\left(1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \cdots + \frac{1}{n}\right) \\ & = & \left( 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \cdots + \frac{1}{2n - 1} + \frac{1}{2n}\right) \\ & & \qquad - \left(1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \cdots + \frac{1}{n}\right) \\ & = & \frac{1}{n + 1} + \frac{1}{n + 2} + \cdots + \frac{1}{2n}, \end{array}}$$and use the result of problem \ref{pro:converging-to-log2}. \end{answer} \end{pro} \begin{pro} Let $n$ be a strictly positive integer and let $x_n$ denote the unique real solution of the equation $x^n+x+1$. Prove that $x_n \rightarrow 1$ as $\ngroes$. \end{pro} \begin{pro} Let $$a_n = \sqrt{n+ \sqrt{(n-1)+ \sqrt{(n-2)+\cdots + \sqrt{2+\sqrt{1}}}}}, $$for $n\geq 1$. Prove that $a_n-\sqrt{n}\rightarrow \dfrac{1}{2}$. \end{pro} \begin{pro} Prove that $e$ is not a quadratic irrational. \begin{answer} We begin by looking at the Taylor series for $e^x$: \begin{equation*} e^x=\sum_{k=0}^{\infty}\frac{x^k}{k!}. \end{equation*} This converges for every $x\in\mathbb{R}$, so $e=\sum_{k=0}^{\infty}\frac{1}{k!}$ and $e^{-1}=\sum_{k=0}^{\infty}(-1)^k\frac{1}{k!}$. Arguing by contradiction, assume $ae^2+be+c=0$ for integers $a$, $b$ and $c$. That is the same as $ae+b+ce^{-1}=0$. Fix $n>\absval{a}+\absval{c}$, then $a,c\mid n!$ and $\forall k\le n$, $k!\mid n!\;$. Consider \begin{align*} 0=n!(ae+b+ce^{-1})&=an!\sum_{k=0}^{\infty}\frac{1}{k!}+b+ cn!\sum_{k=0}^{\infty}(-1)^k\frac{1}{k!}\\ &=b+\sum_{k=0}^n (a+c(-1)^k)\frac{n!}{k!}+\sum_{k=n+1}^\infty (a+c(-1)^k)\frac{n!}{k!} \end{align*} Since $k!\mid n!$ for $k\le n$, the first two terms are integers. So the third term should be an integer. However, \begin{align*} \absval{\sum_{k=n+1}^\infty (a+c(-1)^k)\frac{n!}{k!}}&\le (\absval{a}+\absval{c})\sum_{k=n+1}^\infty \frac{n!}{k!}\\ &=(\absval{a}+\absval{c})\sum_{k=n+1}^\infty \frac{1}{(n+1)(n+2)\dotsb k}\\ &\le (\absval{a}+\absval{c})\sum_{k=n+1}^\infty (n+1)^{n-k}\\ &=(\absval{a}+\absval{c})\sum_{t=1}^\infty (n+1)^{-t}\\ &=(\absval{a}+\absval{c})\frac{1}{n} \end{align*} is less than $1$ by our assumption that $n>\absval{a}+\absval{c}$. Since there is only one integer which is less than $1$ in absolute value, this means that $\sum_{k=n+1}^\infty (a+c(-1)^k)\frac{1}{k!}=0$ for every sufficiently large $n$ which is not the case because \begin{equation*} \sum_{k=n+1}^\infty (a+c(-1)^k)\frac{1}{k!}-\sum_{k=n+2}^\infty (a+c(-1)^k)\frac{1}{k!}=(a+c(-1)^{n+1})\frac{1}{(n+1)!} \end{equation*} is not identically zero. The contradiction completes the proof. \end{answer} \end{pro} \begin{pro} Find $\lim _{\ngrows} \prod _{k=1} ^n \left(1+\dfrac{k}{n}\right)$. \end{pro} \begin{pro} \label{pro:quadratic_integers} A {\em quadratic integer} is any number $x$ that satisfies an equation $$x^2 + mx + n = 0, \ \ (m,n)\in \BBZ^2.$$ Prove that the real quadratic integers are dense in the reals. \begin{answer} Apply Problem \ref{pro:additive-closed-groups} We can apply this to the stated problem by observing that for a fixed $d$, a positive integer without square factors, the numbers $a + b\sqrt{d}$ are quadratic integers if $a, b$ are rational integers, and that the set of such numbers is an additive group of reals. Clearly the closure of this group (it, together with its set of limit points) is a group too, for if $x_n \rightarrow x$ and $y_n \rightarrow y$ then $x_n+y_n \rightarrow x+y$. The new group is not of form (i) or (ii), hence must be all reals, and the proof (of a slightly stronger theorem) is complete. \end{answer} \end{pro} \end{multicols} \section{Averages of Sequences} \begin{center} \fcolorbox{blue}{yellow}{ \begin{minipage}{.90\linewidth} \noindent\textcolor{red}{\textbf{Why bother?}} In this section we will examine some classical results that allow us to compute more complicated limits. Had we the language of matrices, most results here could be deduced from a classical result of Toeplitz. Since we don't, we will develop ad hoc methods which are interesting by themselves. \end{minipage}} \end{center} We start with the following discrete analogues of L'H\^{o}pital's Rule. \begin{thm}Let $\seq{x_n}{n=1}{+\infty}$, $\seq{y_n}{n=1}{+\infty}$, be two sequences of real numbers such that $x_n \rightarrow 0$, $y_n\rightarrow 0$. Suppose, moreover, that the $x_n$ are eventually strictly decreasing. Then $$\lim _{\ngrows}\dfrac{x_n-x_{n-1}}{y_n-y_{n-1}} = \lim _{\ngrows} \dfrac{x_n}{y_n}, $$ provided the sinistral limit exists (be it finite or $+\infty$). \end{thm} \begin{pf} Assume first that $\dfrac{x_{n-1}-x_n}{y_{n-1}-y_n}=\dfrac{x_n-x_{n-1}}{y_n-y_{n-1}} \rightarrow L$, a finite real number. Then, given $\varepsilon >0$ we can find $N>0$ such that for $n> N$, $$ L-\varepsilon <\dfrac{x_{n-1}-x_{n}}{y_{n-1}-y_{n}} 0$ we can find $N'>0$ such that for all $n\geq N'$, $$\dfrac{x_{n-1}-x_{n}}{y_{n-1}-y_{n}} >M \implies x_{n-1}-x_{n}>M(y_{n-1}-y_{n}). $$ Reasoning as above, for positive integers $m\geq 0$, $$ x_{n}-x_{m+n}>M(y_{n}-y_{m+n}). $$Taking the limit as $m\rightarrow +\infty$, $$ x_n \geq My_n \implies \dfrac{x_n}{y_n}\geq M \implies \dfrac{x_n}{y_n}\rightarrow +\infty . $$ \end{pf} \begin{thm}[Stolz's Theorem]\label{thm:stolz} Let $\seq{a_n}{n=1}{+\infty}$, $\seq{b_n}{n=1}{+\infty}$, be two sequences of real numbers. Suppose that $\seq{b_n}{n=1}{+\infty}$ is strictly increasing for sufficiently large $n$ and that $b_n\rightarrow +\infty$ as $\ngrows$. Then $$\lim _{\ngrows}\dfrac{a_n-a_{n-1}}{b_n-b_{n-1}} = \lim _{\ngrows} \dfrac{a_n}{b_n}, $$ provided the sinistral side exists (be it finite or infinite). \end{thm} \begin{pf} Assume first that $\dfrac{a_n-a_{n-1}}{b_n-b_{n-1}}\rightarrow L$, finite. Then for every $\varepsilon > 0$ there is $N>0$ such that $(\forall) n \geq N$, $$ L-\varepsilon < \dfrac{a_{n+1}-a_n}{b_{n+1}-b_n} < L + \varepsilon , \qquad b_{n+1}>b_n.$$ This means that $$ (L-\varepsilon)(b_{n+1}-b_n) < a_{n+1}-a_n < (L+\varepsilon)(b_{n+1}-b_n) $$ By iterating the above relation for $N+1, N+2, \ldots , m+N$ we obtain $$\begin{array}{lllll} (L-\varepsilon)(b_{N+1}-b_N) & < & a_{N+1}-a_N & < & (L+\varepsilon)(b_{N+1}-b_N), \\ (L-\varepsilon)(b_{N+2}-b_{N+1}) & < & a_{N+2}-a_{N+1} & < & (L+\varepsilon)(b_{N+2}-b_{N+1}), \\ & & \vdots & & \\ (L-\varepsilon)(b_{m+N}-b_{m+N-1}) & < & a_{m+N}-a_{m+N-1} & < & (L+\varepsilon)(b_{m+N}-b_{m+N-1}). \end{array} $$ Adding columnwise, $$ (L-\varepsilon)(b_{m+N}-b_N) < a_{m+N}-a_N < (L+\varepsilon)(b_{m+N}-b_N) \implies \absval{\dfrac{a_{m+N}-a_N}{b_{m+N}-b_N}-L}<\varepsilon . $$ Now, $$\dfrac{a_{m+N}}{b_{m+N}} -L =\dfrac{a_{N}-Lb_N}{b_{m+N}} + \left(1-\dfrac{b_{N}}{b_{m+N}}\right)\left(\dfrac{a_{m+N}-a_N}{b_{m+N}-b_N}-L\right), $$ so by the Triangle Inequality $$\absval{\dfrac{a_{m+N}}{b_{m+N}} -L} \leq \absval{\dfrac{a_{N}-Lb_N}{b_{m+N}}} + \absval{1-\dfrac{b_{N}}{b_{m+N}}} \absval{\dfrac{a_{m+N}-a_N}{b_{m+N}-b_N}-L}. $$Since $N$ is fixed, $\dfrac{a_{N}-Lb_N}{b_{m+N}} \rightarrow 0$ and $\dfrac{b_{N}}{b_{m+N}}\rightarrow 0$ as $m\rightarrow +\infty$ Thus the dextral side is arbitrarily small, proving that $\dfrac{a_m}{b_m}\rightarrow L$ as $m\rightarrow +\infty$. \bigskip Assume now that $\dfrac{a_n-a_{n-1}}{b_n-b_{n-1}}\rightarrow +\infty$. For sufficiently large $n$ thus $a_n-a_{n-1}>b_n-b_{n-1}$. Thus the $a_n$ are eventually increasing and $a_n\rightarrow +\infty$ as $\ngrows$. Applying the results above to the $\dfrac{b_n}{a_n}$ we obtain $$ \lim _{\ngrows} \dfrac{b_n}{a_n} = \lim _{\ngrows} \dfrac{b_n-b_{n-1}}{a_n-a_{n-1}}=0$$ and so $ \lim _{\ngrows} \dfrac{a_n}{b_n}= +\infty$ too. \end{pf} \begin{thm}[C\`{e}saro]\label{thm:cesaro} If a sequence of real numbers converges to a number, then its sequence of arithmetic means converges to the same number, that is, if $x_n \rightarrow a$ then $\dfrac{x_1+x_2+\cdots + x_n}{n}\rightarrow a$. \end{thm} \begin{f-pf} Let $a_n= x_1+x_2+\cdots + x_n$ and $b_n=n$ in Stolz's Theorem. \end{f-pf} \begin{s-pf}It is instructive to give an ad hoc proof of this result. Given $\varepsilon > 0$ there exists $N>0$ such that if $n\geq N$ then $\absval{x_n-a}$. Then $$\absval{\dfrac{x_1+x_2+\cdots + x_n}{n}-a} = \absval{\dfrac{(x_1-a)+(x_2-a)+ \cdots +(x_n-a)}{n}}\leq \dfrac{\absval{(x_1-a)}+\absval{(x_2-a)}+ \cdots +\absval{(x_n-a)}}{n}. $$Now we run into a slight problem. We can control the differences $|x_k-a|$ after a certain point, but the early differences need to be taken care of. To this end we consider the differences $x_k-a$ with $k\leq \floor{\sqrt{n}}$ or $k>\floor{n}$ where $n$ is so large that $\floor{\sqrt{n}}\geq N$. We have $$\begin{array}{lll}\dfrac{\absval{(x_1-a)}+\absval{(x_2-a)}+ \cdots +\absval{(x_n-a)}}{n} & = & \dfrac{\absval{(x_1-a)}+\absval{(x_2-a)}+ \cdots +\absval{(x_{\floor{\sqrt{n}}}-a)}}{n}\\ & & + \dfrac{\absval{(x_{\floor{\sqrt{n}}+1}-a)}+\absval{(x_2-a)}+ \cdots +\absval{(x_n-a)}}{n} \\ & < &\dfrac{\floor{\sqrt{n}}\max _{1 \leq k \leq \floor{\sqrt{n}}}\absval{x_k-a}}{n} + \dfrac{(n-\floor{\sqrt{n}})\varepsilon}{n}. \end{array} $$ These two last quantities tend to $0$ as $n\rightarrow +\infty$, from where the result follows. \end{s-pf} \begin{exa} Since $n^{1/n}\rightarrow 1$, $\dfrac{1+2^{1/2}+3^{1/3}+\cdots + n^{1/n}}{n}\rightarrow 1$. \end{exa} \begin{exa} Since $\dfrac{1}{n}\rightarrow 0$, $\dfrac{1+\dfrac{1}{2}+\dfrac{1}{3}+\cdots + \dfrac{1}{n}}{n}\rightarrow 0$. \end{exa} \begin{exa} Since $\left(1+\dfrac{1}{n}\right)^n\rightarrow e$, $\dfrac{\left(1+\dfrac{1}{1}\right)^1+\left(1+\dfrac{1}{2}\right)^2+\left(1+\dfrac{1}{3}\right)^3+\cdots + \left(1+\dfrac{1}{n}\right)^n}{n}\rightarrow e$. \end{exa} \begin{exa} The converse of C\`{e}saro's Theorem is false. For, the sequence $a_n= (-1)^n$ oscillates and does not converge. But its sequence of averages is $b_n = \dfrac{1-1+1-1+\cdots + (-1)^n}{n} \rightarrow 0 $ as $\ngroes$ since the numerator is either $0$ or $-1$. \end{exa} \begin{thm}\label{thm:cesaro-2} If a sequence of positive real numbers converges to a number, then its sequence of geometric means converges to the same number, that is, if $\forall n>0, \quad x_n \geq 0$ and $x_n \rightarrow a$ then $(x_1x_2\cdots x_n)^{1/n}\rightarrow a$. \end{thm} \begin{pf}The proof mimics C\`{e}saro's Theorem \ref{thm:cesaro}. Since $x_n\rightarrow a$, for all $\varepsilon >0$ there is $N>0$ such that for all $n \geq N$, $$\absval{x_n-a}<\varepsilon \implies a-\varepsilon 0$. Prove that $$\lim _{\ngrows} \dfrac{a_n}{a_{n-1}} = \lim _{\ngrows} \sqrt[n]{a_n}. $$ \end{pro} \begin{pro} Let $x_n\rightarrow a$ and $y_n\rightarrow b$. Prove that $\dfrac{x_1y_n+x_2y_{n-1}+\cdots + x_ny_1}{n}\rightarrow ab$. \end{pro} \begin{pro} Prove that $\lim _{\ngroes} \left(\binom{2n}{n}\right)^{1/n} = 4$. \end{pro} \begin{pro} Prove that $\lim _{\ngroes} \dfrac{1}{n}\left(n(n+1)\cdots (n+n)\right)^{1/n} = 4e$. \end{pro} \begin{pro} Prove that $\lim _{\ngroes} \dfrac{1}{n}\left(1\cdot 3\cdot 5\cdots (2n-1)\right)^{1/n} = \dfrac{2}{e}$. \end{pro} \begin{pro} Prove that $\lim _{\ngroes} \dfrac{1}{n^2}\left(\dfrac{(3n)!}{n!}\right)^{1/n} = \dfrac{2}{e}$. \end{pro} \end{multicols} \section{Orders of Infinity} \begin{center} \fcolorbox{blue}{yellow}{ \begin{minipage}{.90\linewidth} \noindent\textcolor{red}{\textbf{Why bother?}} It is clear that the sequences $\seq{n}{n=1}{+\infty}$ and $\seq{n^2}{n=1}{+\infty}$ both tend to $+\infty$ as $\ngrows$. We would like now to refine this statement and compare one with the other. In other words, we will examine their speed towards $+\infty$. \end{minipage}} \end{center} \begin{df} We write $a_n = \boo{b_n}$ if $\exists C>0$, $\exists N>0$ such that $\forall n\geq N$ we have $\absval{a_n}\leq C\absval{b_n}$. We then say that $a_n$ is {\em Big Oh} of $b_n$, or that $a_n$ {\em is of order at most $b_n$} as $\ngroes$. Observe that this means that $\absval{\dfrac{a_n}{b_n}}$ is bounded for sufficiently large $n$. The notation $a_n << b_n$, due to Vinogradov, is often used as a synonym of $a_n = \boo{b_n}$. \end{df} \begin{rem} A sequence $\seq{a_n}{n=1}{+\infty}$ is bounded if and only if $a_n<<1$. \end{rem} An easy criterion to identify Big Oh relations is the following. \begin{thm}\label{thm:sufficient-conds-for-big-oh} If $\lim _{\ngroes} \dfrac{a_n}{b_n} =c\in\BBR$, then $a_n<0$. This proves the theorem. \end{pf} \begin{rem} The $=$ in the relation $a_n = \boo{b_n}$ is not a true equal sign. For example $n^2 = \boo{n^3}$ since $\lim _{\ngroes} \dfrac{n^2}{n^3} = 0$ and so $n^2<0$, $n^3>Mn^2$, meaning that $n^3\neq \boo{n^2}$. Thus the Big Oh relation is not symmetric.\footnote{One should more properly write $a_n\in \boo{b_n}$, as $\boo{b_n}$ is the set of sequences growing to infinity no faster than $b_n$, but one keeps the $=$ sign for historical reasons.} \end{rem} \begin{thm}[Lexicographic Order of Powers]\label{thm:lexico-powers-big-oh} Let $(\alpha, \beta)\in\BBR$ and consider the sequences $\seq{n^\alpha}{n=1}{+\infty}$ and $\seq{n^\beta}{n=1}{+\infty}$. Then $n^\alpha << n^\beta \iff \alpha \leq \beta$. \end{thm} \begin{pf} If $\alpha \leq \beta$ then $\lim_{\ngroes} \dfrac{n^\alpha}{n^\beta} $ is either $1$ (when $\alpha = \beta$) or $0$ (when $\alpha < \beta$), hence $n^{\alpha}<0$. If $\alpha >\beta$ then this would mean that for all large $n$ we would have $n^{\alpha-\beta}\leq C$, which is absurd, since for a strictly positive exponent $\alpha-\beta$, $n^{\alpha-\beta}\rightarrow +\infty$ as $\ngroes$.\end{pf} \begin{exa} As $\ngroes$, $$ n^{1/10}<0$ and $N>0$ such that $\absval{b_n} \leq C\absval{ca_n}$ whenever $n \geq N$. Therefore, $\absval{b_n} \leq C'\absval{a_n}$ whenever $n \geq N$, where $C' = C \absval{c}$, meaning that $b_n=\boo{a_n}$. Similarly, if $b_n= \boo{a_n}$ the there are constants $C_1>0$ and $N_1>0$ such that $\absval{b_n} \leq C_1\absval{a_n}$ whenever $n \geq N_1$. Since $c\neq 0$ this is equivalent to $\absval{b_n} \leq \dfrac{C_1}{c}\left(c\absval{a_n}\right)=C''\left(c\absval{a_n}\right)$ whenever $n \geq N_1$, where $C'' = \dfrac{C_1}{c}$, meaning that $b_n=\boo{ca_n}$. Therefore, $\boo{a_n} = \boo{ca_n}$. \end{pf} \begin{exa} As $\ngroes$, $$ \boo{n^3}=\boo{\dfrac{n^3}{3}}=\boo{4n^3}. $$ \end{exa} \begin{thm}[Sum Rule]\label{thm:sum-rule-big-oh}Let $a_n = \boo{x_n}$ and $b_n = \boo{y_n}$. Then $a_n + b_n = O(\max(\absval{x_n}, \absval{y_n}))$.\end{thm} \begin{pf} There exist strictly positive constants $C_1, N_1, C_2, N_2$ such that $$n\geq N_1, \implies \absval{a_n} \leq C_1\absval{x_n}\qquad \mathrm{and} \qquad n\geq N_2, \implies \absval{b_n} \leq C_2\absval{y_n}.$$ Let $N' = \max(N_1, N_2)$. Then for $n \geq N$, by the Triangle inequality $$\absval{a_n + b_n} \leq \absval{a_n} + \absval{b_n} \leq C_1\absval{x_n} + C_2\absval{y_n}.$$ Let $C' = \max(C_1,C_2)$. Then $$ \absval{a_n + b_n} \leq C'(\absval{x_n} + \absval{y_n}) \leq 2C' \max(\absval{x_n}, \absval{y_n}), $$ whence the theorem follows. \end{pf} \begin{cor}\label{cor:leading-term-dominates}Let $a_n= k_0n^m+k_1n^{m-1}+k_2n^{m-2}+\cdots + k_{m-1}n+k_n$ be a polynomial of degree $m$ in $n$ with real number coefficients. The $a_n=\boo{n^m}$, that is, $a_n$ is of order at most its leading term.\end{cor} \begin{pf} By the Sum Rule Theorem \ref{thm:sum-rule-big-oh} the leading term dominates.\end{pf} \begin{thm}[Transitivity Rule]\label{thm:transitivity-big-oh} If $a_n = O(b_n)$ and $b_n = O(c_n)$, then $a_n = \boo{c_n}$.\end{thm} \begin{pf} There are strictly positive constants $C_1, C_2, N_1, N_2$ such that $$n\geq N_1, \implies \absval{a_n} \leq C_1\absval{b_n}\qquad \mathrm{and} \qquad n\geq N_2, \implies \absval{b_n} \leq C_2\absval{c_n}.$$ If $n\geq \max (N_1,N_2)$, then $\absval{a_n} \leq C_1\absval{b_n} \leq C_1C_2\absval{c_n}=C\absval{c_n}$ , with $C=C_1C_2$. This gives $a_n = \boo{c_n}$. \end{pf} \begin{exa} By Corollary \ref{cor:leading-term-dominates}, $ 5n^4-2n^2+100n-8 = \boo{5n^4}$. By Theorem \ref{thm:constants-matter-not-big-oh}, $\boo{5n^4}=\boo{n^4}$. Hence $$5n^4-2n^2+100n-8=\boo{n^4}.$$ \end{exa} \begin{thm}[Multiplication Rule]\label{thm:multiply-big-oh} If $a_n = O(x_n)$ and $b_n = O(y_n)$, then $a_nb_n = \boo{x_ny_n}$.\end{thm} \begin{pf} There are strictly positive constants $C_1, C_2, N_1, N_2$ such that $$n\geq N_1, \implies \absval{a_n} \leq C_1\absval{x_n}\qquad \mathrm{and} \qquad n\geq N_2, \implies \absval{b_n} \leq C_2\absval{y_n}.$$ If $n\geq \max (N_1,N_2)$, then $\absval{a_nb_n} \leq C_1C_2\absval{x_ny_n} = C\absval{x_ny_n}$, with $C=C_1C_2$. This gives $a_nb_n = \boo{x_ny_n}$. \end{pf} \begin{thm}[Lexicographic Order of Exponentials]\label{thm:lexico-exps-big-oh} Let $(a, b)\in\BBR$, $a>1$, $b>1$, and consider the sequences $\seq{a^n}{n=1}{+\infty}$ and $\seq{b^n}{n=1}{+\infty}$. Then $a^n << b^n \iff a \leq b$. \end{thm} \begin{pf} Put $r=\dfrac{a}{b}$, and use Theorems \ref{thm:geom-sequence-to-zero-goes} and \ref{thm:sufficient-conds-for-big-oh}. \end{pf} \begin{exa} $\dfrac{1}{2^n}<< 1 << 2^n << e^n << 3^n$. \end{exa} \begin{lem}\label{lem:exps-are-faster-than-powers-big-oh} Let $a\in\BBR$, $a>1$, $k\in\BBN\setminus \{0\}$. Then $n^k << a^n$. \end{lem} \begin{pf} By Theorem \ref{thm:exps-are-faster-than-powers}, $\lim _{\ngroes} \dfrac{n^k}{a^n}=0$. Now apply Theorem \ref{thm:sufficient-conds-for-big-oh}. \end{pf} \begin{thm}[``Exponentials are faster than powers'']\label{thm:exps-are-faster-than-powers-big-oh} Let $a\in\BBR$, $a>1$, $\alpha\in\BBR$. Then $n^\alpha << a^n$. \end{thm} \begin{pf}Put $k=\max (1, \floor{\alpha}+1)$. Then by Theorem \ref{thm:lexico-powers-big-oh}, $n^\alpha << n^k$. By Lemma \ref{lem:exps-are-faster-than-powers-big-oh}, $n^k<0$. Then $ (\log n)^{\beta}<< n^\alpha$. \end{thm} \begin{pf}If $\beta \leq 0$, then $(\log n)^\beta <<1$ and the assertion is evident, so assume $\beta >0$. For $x>0$, then $\log x < x$. Putting $x=n^{\alpha /\beta}$, we get $$ \log n^{\alpha /\beta} < n^{\alpha/\beta} \implies \log n < \dfrac{\beta n^{\alpha/ \beta}}{\alpha} \implies (\log n)^\beta < \dfrac{\beta^\beta n^\alpha}{\alpha^\beta },$$ whence $(\log n)^\beta << n^\alpha$. \end{pf} By the Multiplication Rule (Theorem \ref{thm:multiply-big-oh}) and Theorems \ref{thm:lexico-powers-big-oh}, \ref{thm:exps-are-faster-than-powers-big-oh}, \ref{thm:logs-are-slower-than-powers-big-oh}, in order to compare two expressions of the type $a^nn^b(\log )^c$ and $u^nn^{v}(\log )^{w}$ we simply look at the lexicographic order of the exponents, keeping in mind that logarithms are slower than powers, which are slower than exponentials. \begin{exa} In increasing order of growth we have $$\dfrac{1}{e^n} << {\dfrac{1}{2^n}} <<{\dfrac{1}{n^2}} = {\dfrac{1}{\log n}} << {1} <<{(\log\log n)^{10}} <<{\sqrt{\log n}} <<{\dfrac{n}{\log n}}<< {n} <<{n\log n} <<{e^n}. $$ \end{exa} \begin{exa} Decide which one grows faster as $\ngroes$: $n^{\log n}$ or $(\log n)^n$. \end{exa} \begin{solu} Since $n^{\log n} = e^{(\log n)^2}$ and $(\log n)^n=e^{n\log\log n}$, and since $(\log n)^2<1$ $\lim _{\ngroes} \dfrac{n^\alpha}{a^n} = 0$, and so $n^\alpha = \soo{a^n}$. Also, for $\gamma >0$, $\lim _{\ngroes} \dfrac{(\log n)^\beta}{n^\gamma} = 0$, and so $(\log n)^\beta = \soo{n^\gamma}$. \end{rem} \begin{df} We write $a_n \sim b_n$ if $\dfrac{a_n}{b_n}\rightarrow 1$ as $n\rightarrow +\infty$, and say that $a_n$ is {\em asymptotic} to $b_n$. \end{df} Asymptotic sequences are thus those that grow at the same rate as the index increases. \vspace{2cm} \begin{figure}[h] $$\psset{unit=1pc} \pscircle(-2,0){4}\pscircle(2,0){4} \pscircle(0,0){1}\rput(0,0){f\sim g} \pscircle(-3.8,-1){1.7}\rput(-3.8,-1){f=\soo{g}} \pscircle(3.8,-1){1.7}\rput(3.8,-1){g=\soo{f}} \rput(-3.3,2.5){f=\boo{g}}\rput(3.3,2.5){g=\boo{f}} $$\vspace{1cm}\footnotesize\hangcaption{Diagram of $O$ relations.} \label{fig:O-relations} \end{figure} \begin{exa} The sequences $\seq{n^2-n\sin n}{n=1}{+\infty}$, $\seq{n^2 + n-1}{n=1}{+\infty}$ are asymptotic since $$ \dfrac{n^2-n\sin n}{n^2+n-1} = \dfrac{1-\dfrac{\sin n}{n}}{1+\dfrac{1}{n}-\dfrac{1}{n^2}}\rightarrow 1, $$as $\ngrows$. \end{exa} \begin{thm}Let $\seq{a_n}{n=1}{+\infty}$ and $\seq{b_n}{n=1}{+\infty}$ be two properly diverging sequences. Then $a_n\sim b_n \iff a_n = b_n(1+\soo{1})$. \end{thm} \begin{pf}Since the limit is $1>0$, either both diverge to $+\infty$ or both to $-\infty$. Assume the former, and so, eventually, $b_n$ will be strictly positive. Now, $$ \begin{array}{lll}\lim _{\ngroes} \dfrac{a_n}{b_n} = 1 & \iff & \forall \varepsilon >0, \exists N>0, 1-\varepsilon<\dfrac{a_n}{b_n}< 1+\varepsilon\\ & \iff & b_n - b_n\varepsilon < a_n < b_n + b_n\varepsilon\\ & \iff & \absval{a_n-b_n}< b_n\varepsilon\\ & \iff & a_n-b_n = \soo{b_n}. \end{array} $$ \end{pf} The relationship between the three symbols is displayed in figure \ref{fig:O-relations}. \subsection*{Homework}\addcontentsline{toc}{subsection}{Homework} \begin{multicols}{2}\columnseprule 1pt \columnsep 25pt\multicoltolerance=900\small \begin{pro} Prove that $e^n<0$, be a constant. Prove that $(a_n+b_k)^k < 0, \quad \exists N>0, \quad \mathrm{such\ that} \quad \forall n,m \geq N \quad \absval{a_{n}-a_{m}}< \varepsilon . $$ \end{df} \begin{thm}\label{thm:Cauchy-are-bounded} Cauchy sequences are bounded. \end{thm}\begin{pf} Let $ \seq{a_n}{n=1}{+\infty} $ be Cauchy. Take $ N>0 $ such that for all $ n\geq N $, $\absval{a_{n}-a_{N}}<1$ . Then $ a_{n}$ is bounded by $$ \max \left( |a_{1}|, \absval{a_{2}}, \ldots, \absval{a_{N}}\right)+1.$$ \end{pf} \begin{lem} \label{thm:Cauchy-seq-convers-if-a-subseq-does} If a Cauchy sequence of real numbers has a convergent subsequence, then the parent sequence converges, and it does so to the same limit as the subsequence. \end{lem} \begin{pf}Let $\seq{a_n}{n=1}{+\infty}$ be a Cauchy sequence of real numbers, and suppose that its subsequence $\seq{a_{n_k}}{k=1}{+\infty}$ converges to the real number $a$. Given $\varepsilon > 0 $, take $ N>0$ sufficiently large such that $$\forall m,n, n_k \geq N,\quad \absval{a_{n}-a_{m}}< \varepsilon, \qquad \mathrm{and}\quad \absval{a_{n_k}-a} < \varepsilon .$$ By the Triangle Inequality,$$ \absval{a_{n}-a} \leq \absval{a_{n}-a_{n_k}} + \absval{a_{n_k}-a} < \varepsilon + \varepsilon = 2 \varepsilon, $$whence $a_n\rightarrow a$.\end{pf} \begin{thm}[General Principle of Convergence] A sequence of real numbers converges if and only if it is Cauchy. \end{thm} \begin{pf} \begin{enumerate} \item[$ (\Rightarrow) $] If $ a_{n} \to a $, given $ \varepsilon>0 $, choose $ N >0$ such that $ \absval{a_{n}-a}< \varepsilon $ for all $ n \geq N $. Then if $ m,n \geq N $, $$ \absval{a_{n}-a_{m}} \leq \absval{a_{n}-a} + \absval{a_{m}-a} \leq \varepsilon + \varepsilon = 2 \varepsilon. $$ Since $ 2 \varepsilon > 0$ can be made arbitrarily small, $ a_{n} $ is Cauchy. \item[$ (\Leftarrow) $] Suppose $ a_{n} $ is Cauchy. By virtue of Theorem \ref{thm:Cauchy-are-bounded} it is bounded, say that for all $n>0$, $a_{n} \in \lcrc{\alpha}{\beta}$. Put $$ \mathscr{S} = \{ s: a_{n} \geq s \text{ for infinitely many $n$} \}. $$As $ \alpha \in \mathscr{S} $, $ \mathscr{S} \neq \varnothing $. $ \mathscr{S} $ is bounded above by $ \beta$. By the Completeness Axiom, $ \mathscr{S} $ has a supremum, $ a = \sup \mathscr{S} $. Given $ \varepsilon > 0 $, $ a - \varepsilon < a $ and so there is $ s \in \mathscr{S} $ such that $a - \varepsilon < s$. By definition of $ \mathscr{S}$, there are infinitely many $ n $ with $a_{n}\geq s > a-\varepsilon$. $ a+ \varepsilon > a $, so that $a+ \varepsilon \notin \mathscr{S}$ and so there are only finitely many $ n $ for which $a_{n} \geq a+ \varepsilon$. Thus there are infinitely many $ n $ with $a_{n} \in (a - \varepsilon, a+\varepsilon)$. Choose $ N>0 $ such that $\absval{a_{n}-a_{m}} < \varepsilon$ for all $ m,n \geq N $. We can find $ m \geq N $ with $a_{m}\in (a-\varepsilon, a + \varepsilon)$ ie $\absval{a_{m}-a}<\varepsilon$. Then if $ n \geq N $, $$ \absval{a_{n}-a} \leq \absval{a_{n}-a_{m}} + \absval{a_{m}-a} < \varepsilon + \varepsilon = 2 \varepsilon $$ As $ 2 \varepsilon $ can be made arbitrarily small this shews $ a_{n} \to a $. \end{enumerate} \end{pf} \subsection*{Homework}\addcontentsline{toc}{subsection}{Homework} \begin{multicols}{2}\columnseprule 1pt \columnsep 25pt\multicoltolerance=900\small \end{multicols} \section{Topology of sequences. Limit Superior and Limit Inferior} \begin{thm}\label{thm:dense-and-sequences}A set $X\subseteqq \BBR$ is dense in $\BBR$ is and only if for every $x\in \BBR$ there is a sequence $\seq{x_n}{n=1}{+\infty}$ of elements of $X\setminus \{x\}$ that converges to $x$. \end{thm} \begin{pf} \begin{description} \item[$\implies$] For each positive integer $n$, since $X$ is dense in $\BBR$, there exists $x_n\in X\setminus \{x\}$ such that $\absval{x_n -x}< \dfrac{1}{2^n}$. But then $x_n\to x$ as $\ngroes$. \item[$\Leftarrow$] Let $x\in \BBR $ and let $\seq{x_n}{n=1}{+\infty}$ of elements of $X\setminus \{x\}$ that converges to $x$. Then $\forall \varepsilon > 0$, $\exists N\in \BBN$ such that $\forall n\geq N$, $\absval{x_n-x}<\varepsilon$. But then we have found elements of $X\setminus \{x\}$ which are arbitrarily close to $x$, meaning that $X$ is dense in $\BBR$. \end{description} \end{pf} \begin{thm}\label{thm:accu-points-and-sequences} Let $X\subseteqq \BBR$. A point $x\in \BBR$ is an accumulation point of $X$ if and only if there exists a sequence of elements of $X\setminus \{x\}$ converging to $x$. \end{thm} \begin{pf} \begin{description} \item[$\implies$]If $x$ is an accumulation point of $X$, every closed interval $I_n := [x-1/n; x+1/n]$, $n\in\BBN$, satisfies $I_n \cap (X\setminus \{x\}) \neq \varnothing$, thus $\forall n\in \BBN$, $\exists x_n \in I_n \cap (X\setminus \{x\})$. Since $\absval{x_n-x}<\dfrac{1}{n}$, we conclude that $\lim x_n = x$. \item[$\Leftarrow$]Suppose now that $\seq{x_n}{1}{+\infty}$ is an infinite sequence of points of $X\setminus \{x\}$ converging to $x$. If $x\notin \acc{X}$, then $x\notin\acc{x_1,x_2,\ldots}$. Thus there is a neighbourhood of $x$, $\N{x}$ such that $\N{x}\cap \{x_1,x_2,\ldots\}$. Thus there is a $\varepsilon > 0$ such that $]x - \varepsilon; x + \varepsilon[ \subseteqq \N{x}$. For this $\varepsilon$ and for none of the $x_n$ it is true then that $\absval{ x_n - x } < \varepsilon$, contradicting the fact that $\lim _{\ngroes} x_n = x$. \end{description} \end{pf} \begin{df} Given a sequence $\seq{a_n}{1}{+\infty}$, the new sequence $$b_k = \inf _{n\geq k} a_n = \inf \{a_k, a_{k+1}, a_{k+2,} \}, \quad k \geq 1, $$satisfies $b_k\leq b_{k+1}$, that is, it is increasing, and hence it converges to its supremum. We then put $$ \lim \inf _{n\rightarrow +\infty} =\sup _{n\geq 1}\inf _{k\geq n} a_k. $$Similarly, the new sequence $$c_k = \sup _{n\geq k} a_n = \sup \{a_k, a_{k+1}, a_{k+2,} \}, \quad k \geq 1, $$satisfies $c_k\geq c_{k+1}$, that is, it is decreasing, and hence it converges to its infimum. We then put $$ \lim \sup _{n\rightarrow +\infty} =\inf _{n\geq 1}\sup _{k\geq n} a_k. $$ \end{df} We now prove the following theorem for future reference. \begin{thm}\label{thm:ratio-is-inferior-to-root} For any sequence $\seq{a_n}{n=0}{+\infty}$ of strictly positive real numbers $$ \lim _{\ngroes}\inf \dfrac{a_{n+1}}{a_n}\leq \lim _{\ngroes}\inf \sqrt[n]{a_n}\leq \lim _{\ngroes}\sup \sqrt[n]{a_n} \leq \lim _{\ngroes}\sup \dfrac{a_{n+1}}{a_n}. $$ \end{thm} \begin{pf} We will prove the last inequality. The first is quite similar, and the two middle ones are obvious. Put $r= \lim _{\ngroes}\sup \dfrac{a_{n+1}}{a_n}$. If $r=+\infty$ then there is nothing to prove. For $r<+\infty$ choose $r'>r$. There is $N\in \BBN$ such that $$\forall n \geq N, \qquad \dfrac{a_{n+1}}{a_n}\leq r'.$$ Hence, $$a_{N+1} \leq r'a_N, \quad a_{N+2} \leq r'a_{N+1}, \quad a_{N+3} \leq r'a_{N+2}, \ldots \quad a_{N+t} \leq r'a_{N+t-1}, $$ and so, upon multiplication and cancelling, $$a_{N+t}\leq a_N (r')^t, $$and putting $n=N+t$, $$a_n\leq a_N(r')^{-N} (r')^n \implies \sqrt[n]{a_n} \leq r' \sqrt[n]{a_N(r')^{-N}} \implies \lim _{\ngroes} \sup \sqrt[n]{a_n} \leq r', $$ since $a_N(r')^{-N}$ is a fixed real number (does not depend on $n$), and so, $\sqrt[n]{a_N(r')^{-N}} \to 1$ by Theorem \ref{thm:root-n-of-a-to-1-goes}. \end{pf} The following theorem is an easy exercise left to the reader. \begin{thm} Let $\seq{a_n}{1}{+\infty}$ be a sequence of real numbers. Then \begin{enumerate} \item if $\limsup _{n\rightarrow +\infty} a_n = +\infty$, then $\seq{a_n}{1}{+\infty}$ has a subsequence converging to $+\infty$. \item if $\limsup _{n\rightarrow +\infty}a_n = -\infty$, then $\lim _{n\rightarrow +\infty} a_n = -\infty$. \item if $\limsup _{n\rightarrow +\infty} a_n = a\in\BBR $, then $$ \forall\,\epsilon>0, \; \exists\, n_0\ \hbox{such\ that } a_n < a + \epsilon\hbox{ whenever } n \geq n_0 $$and also, there are infinitely many $a_n$ such that $a - \epsilon < a_n$. \item if $\liminf _{n\rightarrow +\infty} a_n = -\infty$, then $\seq{a_n}{1}{+\infty}$ has a subsequence converging to $-\infty$. \item if $\liminf _{n\rightarrow +\infty} a_n = +\infty$, then $\lim _{n\rightarrow +\infty}a_n = +\infty$. \item if $\liminf _{n\rightarrow +\infty} a_n = a\in\BBR $, then $$ \forall\,\epsilon>0, \; \exists\, n_0\ \hbox{ such \ that } a - \epsilon < a_n \hbox{ whenever } n \geq n_0 $$ and there are infinitely many $a_n$ such that $a_n < a + \epsilon$. \item $\liminf _{n\rightarrow +\infty} a_n \leq \limsup _{n\rightarrow +\infty} a_n$ is always verified, and furthermore, $\liminf _{n\rightarrow +\infty} a_n = \limsup _{n\rightarrow +\infty} a_n$ if and only if $\lim _{n\rightarrow +\infty} a_n$ exists, in which case $\liminf _{n\rightarrow +\infty} a_n =\lim _{n\rightarrow +\infty} a_n = \limsup _{n\rightarrow +\infty} a_n$. \end{enumerate} \end{thm} \subsection*{Homework}\addcontentsline{toc}{subsection}{Homework} \begin{multicols}{2}\columnseprule 1pt \columnsep 25pt\multicoltolerance=900\small \begin{pro} Identify the set of accumulation points of the set $\{\sqrt{a}-\sqrt{b}: (a, b)\in \BBN^2\}$. \end{pro} \begin{pro} Consider the following enumeration of the proper fractions $$ \dfrac{0}{1}, \dfrac{1}{1}, \dfrac{0}{2},\dfrac{1}{2},\dfrac{2}{2},\dfrac{0}{3},\dfrac{1}{3},\dfrac{2}{3}, \dfrac{3}{3}\ldots .$$ Clearly, the fraction $\dfrac{a}{b}$ in this enumeration occupies the $a+ \dfrac{b(b+1)}{2}$-th place. For each integer $k\geq 1$, cover the $k$-th fraction $\dfrac{a}{b}$ by an interval of length $2^{-k}$ centred at $\dfrac{a}{b}$. Shew that the point $\dfrac{\sqrt{2}}{2}$ does not belong to any interval in the cover. \end{pro} \end{multicols} \chapter{Series} \section{Convergence and Divergence of Series} \begin{df}Let $\{a_n\} _{n=1} ^{+\infty}$ be a sequence of real numbers. A {\em series} is the sum of a sequence. We write $$s_n= a_1 + a_2 + \cdots + a_n = \sum _{k =1} ^n a_k. $$ Here $s_n$ is the $n$-th {\em partial sum}. Observe in particular that $$ a_n = s_n - s_{n-1}. $$ \end{df} \begin{df} If the sequence $\{s_n\} _{n=1} ^{+\infty}$ has a finite limit $S$, we say that the series converges to $S$ and write $$\sum _{k =1} ^{+\infty} a_k = \lim _{\ngroes} s_n = S. $$Otherwise we say that the series {\em diverges}. \end{df} Observe that $\sum _{n=1} ^{+\infty} a_n$ converges to $S$ if $\forall \varepsilon > 0$, $\exists N$ such that $\forall n \geq N$, $$\absval{\left(\sum _{k \leq n} a_k\right) -S} =|s_n-S| < \varepsilon .$$ Now, since $$\left(\sum _{k \leq n} a_k\right) -S = \left(\sum _{k \leq n} a_k\right) - \left(\sum _{k \geq 1} a_k\right) =\sum _{k > n} a_k, $$we see that a series converges if and only if its ``tail'' can be made arbitrarily small. Hence, the reader should notice that adding or deleting a finite amount of terms to a series does not affect its convergence or divergence. Furthermore, since the sequence of partial sums of a convergent series must be a Cauchy sequence we deduce that a series is convergent if and only if $\forall \varepsilon > 0$, $\exists N>0$ such that $\forall m\geq N, n \geq N$, $m \leq n$, \begin{equation}\absval{s_n-s_m} = \absval{\sum _{k=m} ^n a_k}<\varepsilon .\label{eq:Cauchy-seq-ser} \end{equation} \begin{thm}[$n$-th Term Test for Divergence] If $\sum _{n=1} ^\infty a_n$ converges, then $a_n\rightarrow 0$ as $\ngroes$. \label{thm:n-th-term-test}\end{thm} \begin{pf} Put $s_n = \sum _{k=1} ^n a_k$. Then $$ \lim _{\ngroes} s_n = S \implies a_n = s_n -s_{n-1} \rightarrow S-S = 0. $$ \end{pf} In general, the problem of determining whether a series converges or diverges requires some work and it will be dealt with in the subsequent sections. We continue here with some other examples. \begin{exa}\label{exa:harmonic-series-diverges} The series $\sum _{n =1} ^{+\infty} \left(1+\dfrac{2}{n}\right)^n$ diverges, since its $n$-th term $\left(1+\dfrac{2}{n}\right)^n \rightarrow e^2$. \end{exa} \begin{exa} We will prove that the {\em harmonic series} $\sum _{n=1} ^{+\infty} \dfrac{1}{n}$ diverges, even though $\dfrac{1}{n}\rightarrow 0$ as $\ngroes$. Thus the condition in Theorem \ref{thm:n-th-term-test} though necessary for convergence is not sufficient. The divergence of the harmonic series was first demonstrated by Nicole d'Oresme (ca. 1323-1382), but his proof was mislaid for several centuries. The result was proved again by Pietro Mengoli in 1647, by Johann Bernoulli in 1687, and by Jakob Bernoulli shortly thereafter. Write the partial sums in dyadic blocks, $$ \sum_{n=1}^{2^M} \frac{1}{n} = \sum_{m=1}^M \sum_{n=2^{m-1}+1}^{2^m} \frac{1}{n}. $$ As $1/n \ge 1/N$ when $n \le N$, we deduce that $$ \sum_{n=2^{m-1}+1}^{2^m} \frac{1}{n} \ge \sum_{n=2^{m-1}+1}^{2^m} 2^{-m} = (2^m - 2^{m-1}) 2^{-m} = \frac{1}{2} $$ Hence, $$ \sum_{n=1}^{2^M} \frac{1}{n} \ge \frac{M}{2} $$ so the series diverges in the limit $M \to +\infty$. \end{exa} The following theorem says that linear combinations of convergent series converge. \begin{thm}\label{thm:line-comb-conv} Let $\sum _{n=1} ^{+\infty} a_n = A$ and $\sum _{n=1} ^{+\infty} b_n=B$ be convergent series and let $\gamma \in \BBR$ be a real number. Then the series \mbox{$\sum _{n=1} ^{+\infty} (a_n + \gamma b_n)$} converges to $A+\gamma B$. \end{thm} \begin{pf} For all $\varepsilon > 0$ there exist $N, N'$ such that for all $n \geq \max (N,N')$, $$ \absval{\sum _{k \leq n} a_k-A} < \dfrac{\varepsilon}{2} , \qquad \absval{\sum _{k \leq n} b_k-B} < \dfrac{\varepsilon}{2(|\gamma| + 1)}. $$ Hence, by the triangle inequality and by the obvious inequality $\dfrac{|\gamma|}{|\gamma |+1}\leq 1$, we have $$\absval{\left(\sum _{k \leq n} a_k+ \gamma b_k \right) - (A+\gamma B)} \leq \absval{\sum _{k \leq n} a_k - A} + | \gamma |\absval{\sum _{k \leq n} b_k -B} \leq \dfrac{\varepsilon}{2}+ |\gamma |\dfrac{\varepsilon}{2(|\gamma| + 1)} < \dfrac{\varepsilon}{2}+\dfrac{\varepsilon}{2} = \varepsilon . $$ \end{pf} \begin{df} A {\em geometric series} with common ratio $r$ and first term $a$ is one of the form $$a + ar + ar^2 + ar^3 + \cdots = \sum _{n=0} ^{+\infty} ar^{n}. $$ \end{df} By Theorem \ref{thm:geometric-sum}, if $|r|<1$ then the series converges and we have $$a + ar + ar^2 + ar^3 + \cdots = \sum _{n=0} ^{+\infty} ar^{n} = \dfrac{a}{1-r}. $$ \begin{exa} A fly starts at the origin and goes $1$ unit up, $1/2$ unit right, $1/4$ unit down, $1/8$ unit left, $1/16$ unit up, etc., {\em ad infinitum.} In what coordinates does it end up? \end{exa} \begin{solu} Its $x$ coordinate is $$\frac{1}{2} - \frac{1}{8} + \frac{1}{32} - \cdots = \frac{\frac{1}{2}}{1 - \frac{-1}{4}} = \frac{2}{5}.$$ Its $y$ coordinate is $$1 - \frac{1}{4} + \frac{1}{16} - \cdots = \frac{1}{1 - \frac{-1}{4}} = \frac{4}{5}.$$Therefore, the fly ends up in $$\left(\frac{2}{5}, \frac{4}{5}\right).$$ Here we have used the fact the sum of an infinite geometric progression with common ratio $r$, with $|r|<1$ and first term $a$ is $$ a + ar + ar^2 + ar^3 + \cdots = \dfrac{a}{1-r}. $$ \end{solu} \begin{df}A {\em telescoping sum} is a sum where adjacent terms cancel out. That is, $\sum _{n=0} ^Na_n$ is a telescoping sum if we can write $a_n = b_{n+1}-b_n$ and then $$\sum _{n=0} ^Na_n = a_0 + a_1 + \cdots + a_N = (b_{1}-b_0) + (b_{2}-b_1) + \cdots + (b_{N+1}-b_N) = b_{N+1}-b_0. $$ \end{df} \begin{exa}\label{exa:simple-telescope} We have $$\sum _{n=1} ^N \dfrac{1}{n(n+1)} = \sum _{n=1} ^N \left(\dfrac{1}{n}-\dfrac{1}{n+1}\right) = \left(\dfrac{1}{1}-\dfrac{1}{2}\right) + \left(\dfrac{1}{2}-\dfrac{1}{3}\right) + \cdots + \left(\dfrac{1}{N}-\dfrac{1}{N+1}\right) = 1-\dfrac{1}{N+1}. $$Thus $$ \sum _{n=1} ^{+\infty} \dfrac{1}{n(n+1)} =\lim _{N\to+\infty} \sum _{n=1} ^N \dfrac{1}{n(n+1)} = \lim _{N\to+\infty}\left(1-\dfrac{1}{N+1}\right) =1. $$ \end{exa} \begin{exa} We have $$\begin{array}{lll}\sum _{n=1} ^N \dfrac{1}{n(n+1)(n+2)} & = & \dfrac{1}{2}\sum _{n=1} ^N \left(\dfrac{1}{n(n+1)}-\dfrac{1}{(n+1)(n+2)}\right) = \dfrac{1}{2}\left(\left(\dfrac{1}{1\cdot 2}-\dfrac{1}{2\cdot 3}\right) + \left(\dfrac{1}{2\cdot 3}-\dfrac{1}{3\cdot 4}\right) + \cdots + \left(\dfrac{1}{N(N+1)}-\dfrac{1}{(N+1)(N+2)}\right)\right) \\ & = & \dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{(N+1)(N+2)}\right). \end{array}$$Thus $$ \sum _{n=1} ^{+\infty} \dfrac{1}{n(n+1)(n+2)} =\lim _{N\to+\infty} \sum _{n=1} ^N \dfrac{1}{n(n+1)(n+2)} = \lim _{N\to+\infty}\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{(N+1)(N+2)}\right) =\dfrac{1}{4}. $$ \end{exa} \subsection*{Homework}\addcontentsline{toc}{subsection}{Homework} \begin{multicols}{2}\columnseprule 1pt \columnsep 25pt\multicoltolerance=900\small \begin{pro} Find the sum of $\sum _{n=3} ^\infty \dfrac{2^n}{e^{n+1}} $. \begin{answer} This is a geometric series with common ratio $|r|=\frac{2}{e} <1$, so it converges. We have $$ \sum _{n=3} ^\infty \dfrac{2^n}{e^{n+1}} = \dfrac{2^3}{e^4} + \dfrac{2^4}{e^5} + \cdots = \dfrac{\frac{2^3}{e^4}}{1 - \frac{2}{e}} = \dfrac{8}{e^4-2e^3}. $$ \end{answer} \end{pro} \begin{pro} Find the sum of the series $\sum _{n=2} ^{+\infty} \dfrac{1}{4n^2-1}$. \begin{answer}Observe that $$\dfrac{1}{4n^2-1} = \dfrac{1}{2(2n-1)} -\dfrac{1}{2(2n+1)}.$$ Hence $$\sum _{n=2} ^{+\infty} \dfrac{1}{4n^2-1} = \left(\dfrac{1}{2(1)} -\dfrac{1}{2(3)}\right) + \left(\dfrac{1}{2(3)} -\dfrac{1}{2(5)}\right) + \left(\dfrac{1}{2(5)} -\dfrac{1}{2(7)}\right) + \cdots =\dfrac{1}{2(1)} = \dfrac{1}{2}. $$ \end{answer} \end{pro} \begin{pro} Find the exact numerical value of the sum $\sum _{n=0} ^{+\infty}\arctan \dfrac{1}{n^2+n+1}$. \begin{answer} Since $\tan (x-y) = \dfrac{\tan x - \tan y}{1+\tan x\tan y}$, observe that $\arctan \dfrac{1}{n^2+n+1} = \arctan (n+1) - \arctan n$. Hence the series telescopes to $\lim _{\ngroes}\arctan (n+1)-\arctan 1 = \dfrac{\pi}{4} .$ \end{answer} \end{pro} \begin{pro} Find the exact numerical value of the infinite sum $$ \sum _{n=1} ^{+\infty} \dfrac{\sqrt{(n-1)!}}{(1+\sqrt{1})\cdots (1+\sqrt{n})}.$$ \end{pro} \begin{pro} Shew that $$\sum _{k = 1} ^n \frac{k}{k^4 + k^2 + 1} = \frac{1}{2}\cdot\frac{n^2 + n}{n^2 + n + 1},$$and thus prove that $\sum _{k = 1} ^n \frac{k}{k^4 + k^2 + 1}$ converges. \end{pro} \begin{pro} Let $b(n)$ denote the number of ones in the binary expansion of the positive integer $n$, for example $b(3) = b(11_2)=2$. Prove that $\sum _{n=1} \dfrac{b(n)}{n(n+1)} = \log 4$. \end{pro} \begin{pro} Find $$1 + \dfrac{1}{2} +\dfrac{1}{3} + \dfrac{1}{6}+\dfrac{1}{8} + \dfrac{1}{9}+\dfrac{1}{12} +\dfrac{1}{16}+\dfrac{1}{18}+\cdots ,$$ which is the sum of the reciprocals of all positive integers of the form $2^n3^m$ for integers $n \geq 0, m\geq 0$. \begin{answer} By unique factorisation of the integers, the desired sum is $$\left(1+\dfrac{1}{2} + \dfrac{1}{2^2}+\dfrac{1}{2^3}+\cdots \right)\left(1+\dfrac{1}{3} + \dfrac{1}{3^2}+\dfrac{1}{3^3}+\cdots \right) = \dfrac{1}{1-\dfrac{1}{2}}\cdot\dfrac{1}{1-\dfrac{1}{3}} = 3. $$ \end{answer} \end{pro} \begin{pro} The {\em Fibonacci Numbers} $f_n$ are defined recursively as follows: $$f_0 =1,\quad f_1=1, \quad f_{n+2} = f_n + f_{n+1}, \quad n\geq 0. $$ Prove that $\sum _{n=1} ^{+\infty} \dfrac{f_n}{3^n}=\dfrac{3}{5}$. \end{pro} \begin{pro} Let $\sum _{n \geq 0} a_n $ be a convergent series and let $\sum _{n \geq 0} b_n $ be a divergent series. Prove that $\sum _{n \geq 0} (a_n+ b_n)$ diverges. \begin{answer}Since the sum of two convergent series is convergent by Theorem \ref{thm:line-comb-conv}, if $\sum _{n \geq 0} (a_n+ b_n)$ then from the identity $b_n = (a_n+b_n)-a_n$ we would deduce that $\sum _{n \geq 0} b_n $ converges, a contradiction. \end{answer} \end{pro} \begin{pro} Prove that if $\sum _{n\geq 1} a_n$ is a series of positive terms and that its partial sums are bounded, then $\sum _{n\geq 1} a_n$ converges. Shew that this is not necessarily true if $\sum _{n\geq 1} a_n$ is not a series of positive terms. \begin{answer}Put $s_N = \sum _{1 \leq n \leq N} a_n$. There is a positive constant $M $ such that $\forall N>0$, $s_N\leq M$. Observe that because the terms are positive $$s_{N+1} = s_N + a_{N+1} \geq s_N, $$and so the sequence $\seq{s_N}{N=1}{+\infty}$ is a monotonically increasing bounded above sequence and so it converges by Theorem \ref{thm:bounded-increasing-seqs-convergent-be}. This is not necessarily true if the series does not have positive terms. For example, the series $\sum _{n\geq 1} (-1)^{n+1}$ has bounded partial sums, in fact they are either $1$ or $0$. But the sequence of partial sums then is $$ 1,0,1,0,1,0, \ldots $$which does not converge. \end{answer} \end{pro} \end{multicols} \section{Convergence and Divergence of Series of Positive Terms} We have several tools to establish convergence and divergence of series of positive terms. We will start with some simple comparison tests. \begin{thm}[Direct Comparison Test] \label{thm:direct-comparison} Let $\seq{a_n}{n=0}{+\infty}$, $\seq{b_n}{n=0}{+\infty}$, $\seq{c_n}{n=0}{+\infty}$, be sequences of positive real numbers. Suppose that eventually $a_n \leq b_n$, that is, that $\exists N \geq 0$ such that $\forall n \geq N$ there holds $a_n\leq b_n$. If $\sum _{n \geq 0} b_n $ converges, then $\sum _{n \geq 0} a_n$ converges. If eventually $a_n \geq c_n$, and $\sum _{n \geq 0} c_n $ diverges, then $\sum _{n \geq 0} a_n$ also must diverge. \end{thm} \begin{pf}The theorem is clear from the inequalities $$ \sum _{n \geq N} a_n \leq \sum _{n \geq N} b_n , \qquad \sum _{n \geq N} a_n \geq \sum _{n \geq N} c_n. $$ If $\sum _{n \geq 0} b_n $ converges, then its tail can be made as small as we please, and so the tail of $\sum _{n \geq 0} a_n $ can be made as small as we please. Similarly if $\sum _{n \geq 0} c_n $ diverges, because it is a series of positive terms, its tail grows without bound and so the tail of $\sum _{n \geq 0} a_n $ grows without bound. \end{pf} \begin{rem} Call a divergent series of positive terms a ``giant'' and a converging series of positive terms a ``midget.'' The comparison tests say that if a series is bigger than a giant it must be a giant, and if a series is smaller than a midget, it must be a midget. \end{rem} \begin{exa} From example \ref{exa:simple-telescope}, $\sum _{n \geq 1} \dfrac{1}{n(n+1)}$ converges. Since for $n\geq 1$, $$n(n+1)<(n+1)^2\implies \dfrac{1}{(n+1)^2} < \dfrac{1}{n(n+1)},$$we deduce that the series $$\sum _{n \geq 1} \dfrac{1}{(n+1)^2} = \sum _{n \geq 2} \dfrac{1}{n^2} $$ converges. Since adding a finite amount of terms to a series does not affect convergence, we deduce that $1+ \sum _{n \geq 2} \dfrac{1}{n^2} = \sum _{n \geq 1} \dfrac{1}{n^2} $ converges. \end{exa} \begin{exa} $\sum _{n=1} ^{+\infty} \dfrac{1}{n^n}$ converges. For $n \geq 2$ we have $\dfrac{1}{n^n} \leq \dfrac{1}{n^2}$ and the series converges by direct comparison with $\sum _{n=1} ^{+\infty} \dfrac{1}{n^2}$. \end{exa} \begin{exa} From example \ref{exa:harmonic-series-diverges}, $\sum _{n\geq 1}\dfrac{1}{n}$ diverges. Since for $n\geq 1$, $\log n < n$, we deduce that $\sum _{n\geq 2} \dfrac{1}{\log n}$ diverges. Notice that here we start the sum at $n=2$ since the logarithm vanishes at $n=1$. \end{exa} \begin{exa} Prove that $$ \sum _{\stackrel{p}{p \ {\rm prime} }} \dfrac{1}{p}$$diverges.\end{exa} \begin{solu} We will prove this by contradiction. Let $p_1=2$, $p_2=3$, $p_3=5$, \ldots be the sequence of primes in ascending order and assume that the series converges. Then there exists an integer $K$ such that $$ \sum _{m \geq K+1} \dfrac{1}{p_m}< \dfrac{1}{2}. $$ Let $P = p_1p_2 \cdots p_K$ and consider the numbers $1+nP$ for $n=1,2,3,\ldots$. None of these numbers has a prime divisor in the set $\{p_1,p_2, \ldots , p_K\}$ and hence all the prime divisors of the $1+nP$ belong to the set $\{p_{K+1}, p_{K+2}, \ldots \}$. This means that for each $t\geq 1$, $$ \sum _{n=1} ^t \dfrac{1}{1+nP} \leq \sum _{s\geq 1} \left(\sum _{m \geq K+1} \dfrac{1}{p_m}\right)^s \leq \sum _{s\geq 1}\dfrac{1}{2^s} = 1, $$that is, $ \sum _{n=1} ^t \dfrac{1}{1+nP}$, a series of positive terms, has bounded partial sums and so it converges. But since $1+nP \sim nP$ as $n\to +\infty$ and $\dfrac{1}{P}\sum _{n\geq 1}\dfrac{1}{n}$ diverges, we obtain a contradiction. \end{solu} Since the convergent behaviour of a series depends of its tail, the following asymptotic comparison tests should be clear, and its proof follows the same line of reasoning as Theorem \ref{thm:direct-comparison}. \begin{thm}[Asymptotic Comparison Test] \label{thm:asymp-comparison} Let $\seq{a_n}{n=0}{+\infty}$, $\seq{b_n}{n=0}{+\infty}$, $\seq{c_n}{n=0}{+\infty}$, be sequences of real numbers which are eventually positive. Suppose that $a_n <1$ then $\zeta (p) = \sum _{n=1} ^{+\infty} \dfrac{1}{n^p}$ converges, but diverges when $p \leq 1$. \label{thm:p-series-test} \end{thm} \begin{pf} If $p\leq 0$, divergence follows from Theorem \ref{thm:n-th-term-test}. If $p > 0$, then using the fact that $x\mapsto x^p$ is monotonically increasing, we may use Theorem \ref{thm:condensed-cauchy}. Since $$ \sum _{k\geq 0} \dfrac{2^k}{2^{pk}} = \sum _{k \geq 0} \left(2^{(1-p)}\right)^k $$ is a geometric series with ratio $2^{1-p}$, it converges by Theorem \ref{thm:geometric-sum} when $$2^{1-p}< 1 \implies (1-p)\log _2 2 < \log _2 1 \implies 1-p < 0 \implies p>1, $$and diverges for $p>1$. The case $p=1$ has been shewn to diverge in example \ref{exa:harmonic-series-diverges}. \end{pf} \begin{exa} Since $\sqrt{2}>1$, the series $\sum _{n=1} ^{+\infty} \dfrac{1}{n^{\sqrt{2}}}$ converges. \end{exa} \begin{exa} Since $$ \dfrac{n^{\sqrt{2}}+(\log\log n)^{2007}}{n^3 + n(\log n)^5 + 1}\sim \dfrac{n^{\sqrt{2}}}{n^3}=\dfrac{1}{n^{3-\sqrt{2}}} $$and $3-\sqrt{2}>1$, the series \mbox{$\sum _{n\geq 1} \dfrac{n^{\sqrt{2}}+(\log\log n)^{2007}}{n^3 + n(\log n)^5 + 1}$} converges. \end{exa} \begin{cor}[De Morgan's Logarithmic Scale] If $p>1$ then all of $$\sum _{n=1} ^{+\infty} \dfrac{1}{n^p}; \quad \sum _{n\geq e} ^{+\infty} \dfrac{1}{n(\log n)^p}; \quad \sum _{n\geq e^e} ^{+\infty} \dfrac{1}{n(\log n)(\log\log n)^p}; \quad \sum _{n\geq e^{e^e}} ^{+\infty} \dfrac{1}{n(\log n)(\log\log n)(\log\log\log n)^p}; \quad \ldots $$ converge, but diverge when $p \leq 1$. \label{cor:demorgan-log-scale} \end{cor} \begin{pf} The theorem is proved inductively by successive applications of Cauchy's Condensation Test. We will prove how the case for $\sum _{n\geq e} ^{+\infty} \dfrac{1}{n(\log n)^p}$ follows from the case $\sum _{n=1} ^{+\infty} \dfrac{1}{n^p}$ and leave the rest to the reader. We see that $$ \sum _{k \geq 1} \dfrac{2^k}{2^k(\log 2^k)^p} = \dfrac{1}{(\log 2)^p}\sum _{k \geq 1} \dfrac{1}{k^p}, $$ and so this case follows from Theorem \ref{thm:p-series-test}. \end{pf} \begin{exa} Determine whether $\sum _{n=4} ^{+\infty} \dfrac{(\log n)^{100}}{n^{3/2}\log \log n}$ converges. \end{exa} \begin{solu}Since $(\log n)^{100}<< n^{1/4} $, eventually $\dfrac{(\log n)^{100}}{n^{1/4}}<<1$. We have $\dfrac{(\log n)^{100}}{n^{3/2}\log \log n} << \dfrac{(\log n)^{100}}{n^{1/4}}\cdot \dfrac{1}{n^{5/4}\log\log n}$ and since $\sum _{n=4} ^{+\infty} \dfrac{1}{n^{5/4}\log\log n}< +\infty$, we have $\sum _{n=4} ^{+\infty}\dfrac{(\log n)^{100}}{n^{3/2}\log \log n} <+\infty$, that is, the series converges. \end{solu} The reader should be aware that the value of the exponent in Theorems \ref{thm:p-series-test} and \ref{cor:demorgan-log-scale} is fixed. The following examples should dissuade him that ``having an exponent higher than $1$'' implies convergence. \begin{exa} Test $\dis{\sum _{n = 1} ^\infty \dfrac{1}{n^{1+1/n}}}$ for convergence by comparing it to a suitable $p$-series. Use the direct comparison test. \end{exa} \begin{solu} By induction $n<2^n \implies n^{1/n}<2$ and so $n^{1+1/n}<2n \implies \dfrac{1}{2n}< \dfrac{1}{n^{1+/1n}}$. So the series diverges by direct comparison to $\dis{\sum _{n = 1} ^\infty \dfrac{1}{2n}}$. \end{solu} \begin{exa} Test $\dis{\sum _{n = 2} ^\infty \dfrac{1}{n^{1+1/\log n}}}$ for convergence by comparing it to a suitable $p$-series. Use the direct comparison test \end{exa} \begin{solu} We have $n = e^{\log n} \implies n^{\frac{1}{\log n}} =e$ and so $n^{1+1/\log n} =en, \qquad n>1.$. So the series diverges by direct comparison to $\dis{\sum _{n = 2} ^\infty \dfrac{1}{en}}$. \end{solu} \begin{exa} Test $\dis{\sum _{n = 2} ^\infty \dfrac{1}{n^{1+1/\log \log n}}}$ for convergence by comparing it to a suitable $p$-series. Use the direct comparison test. \end{exa} \begin{solu} By considering the monotonicity of $f(x) = e^x -\dfrac{x^2}{2}$ (see Theorem \ref{thm:x^2\dfrac{x^2}{2}$ for $x>0$. Now, $$ n^{1/\log \log n} = e^{\log n^{1/\log \log n}} = e^{\frac{\log n}{\log \log n}} > \dfrac{(\log n)^2}{2(\log \log n)^2}. $$This gives $$\dfrac{2(\log \log n)^2}{n(\log n)^2}> \dfrac{1}{n^{1+\frac{1}{\log \log n}}}. $$Now, $$\sum _{n=2} ^{+\infty}\dfrac{2(\log \log n)^2}{n(\log n)^2} $$ can be shewn to converge by comparing to a series in the De Morgan logarithmic scale. \end{solu} \begin{exa} Prove that the series $\sum _{n=1} ^{+\infty} \dfrac{1}{n^{1.8+\sin n}}$ diverges. \end{exa} \begin{solu} For $k\in \BBZ$, the interval $I_k =\lcrc{(2k+\frac{4}{3})\pi}{(2k+\frac{5}{3})\pi}$ has length $\dfrac{\pi}{3}>1$ and $x\in I_k \implies \sin x \leq -\dfrac{\sqrt{3}}{2}$. The gap between $I_k$ and $I_{k+1}$ is $<\dfrac{5\pi}{3}<6$. Hence, among any seven consecutive integers, at least one must fall into $I_k$ and for this value of $n$ we must have $1.8+\sin n < 1-\dfrac{\sqrt{3}}{2}<1$. This means that $$\sum _{n=1} ^{+\infty} \dfrac{1}{n^{1.8+\sin n}} = \sum _{m=0} ^{+\infty}\qquad \sum _{n=7m+1} ^{n=7m+7}\dfrac{1}{n^{1.8+\sin n}} \geq \sum _{m=0} ^{+\infty} \dfrac{1}{7m+7}, $$and since the rightmost series diverges, the original series diverges by the direct comparison test. \end{solu} The following result puts the harmonic series at the ``frontier'' of convergence and divergence for series with monotonically decreasing positive terms. \begin{thm}[Pringsheim's Theorem] Let $\sum _{n \geq 1} a_n$ be a converging series of positive terms of monotonically decreasing terms. Then $a_n =\soo{\dfrac{1}{n}}$. \end{thm} \begin{pro}Since the series converges, its sequence of partial sums is a Cauchy sequence and by \ref{eq:Cauchy-seq-ser}, given $\varepsilon > 0$, $\exists m>0$, such that $\forall n \geq m$, $$ \sum _{k=m+1} ^{n} a_k < \varepsilon . $$ Because the series decreases monotonically, each of $a_{m+1}, a_{m+2}, \ldots , a_{n}$ is at least $a_n$ and thus $$ (n-m)a_{n} \leq \sum _{k=m+1} ^{n} a_k < \varepsilon.$$ Again, since the series converges, $a_n \to 0$ as $\ngroes$ we may choose $n$ large enough so that $a_n < \dfrac{\varepsilon}{m}$. In this case $$ (n-m)a_{n}<\varepsilon \implies na_n < \varepsilon + ma_n < 2\varepsilon \implies a_n < \dfrac{ 2\varepsilon}{n},$$ which proves the theorem. \end{pro} The disadvantage of the comparison tests is that in order test for convergence, we must appeal to the behaviour of an auxiliary series. The next few tests provide a way of testing the series against its own terms. \begin{thm}[Root Test] Let $\sum _{n=1} ^{+\infty} a_n$ be a series of positive terms. Put $r=\lim \sup (a_n)^{1/n}$. Then the series converges if $r <1$ and diverges if $r >1$. The test is inconclusive if $r =1$. \end{thm} \begin{pf} If $r<1$ choose $r'$ with $r1$ then there is a sequence $\seq{n_k}{k=1}{+\infty}$ of positive integers such that $$ \sqrt[n_k]{a_{n_k}} \to r. $$ This means that $a_n$ will be $>1$ for infinitely many values of $n$, and so, the condition $a_n \to 0$ necessary for convergence, does not hold. By considering $\sum _{n=1} ^{+\infty} \dfrac{1}{n}$, which diverges, and $\sum _{n=1} ^{+\infty} \dfrac{1}{n^2}$, which converges, one sees that $r=1$ may appear in series of different conditions. \end{pf} \begin{thm}[Ratio Test] Let $\sum _{n=1} ^{+\infty} a_n$ be a series of strictly positive terms. Put $r=\lim \sup \dfrac{a_{n+1}}{a_n} $. Then the series converges if $r <1$ and diverges if $r >1$. The test is inconclusive if $ r =1$. \end{thm} \begin{pf} If $r<1$, then there exists $N\in \BBN$ such that $$a_{N+1} \leq ra_{N}, \quad a_{N+2} \leq ra_{N+1}, \quad a_{N+3} \leq ra_{N+2}, \ldots \quad a_{N+t} \leq ra_{N+t-1}. $$Multiplying all these inequalities together, $$a_{N+t}\leq a_N r^{t}. $$ Putting $N+t =n$ we deduce that $$a_n \leq a_N r^{-N}r^{n}. $$ Since $ a_N r^{-N}$ is a constant, we may use direct comparison between $\sum _{n=1} ^{+\infty} a_n$ and the converging geometric series $a_N r^{-N}\sum _{n=1} ^{+\infty}r^{n}$, concluding that $\sum _{n=1} ^{+\infty} a_n$ converges. If $r>1$ then $a_{n+1}\geq a_n \geq a_N$ for all $n \geq N$. This means that the condition $a_n \to 0$, necessary for convergence, does not hold. By considering $\sum _{n=1} ^{+\infty} \dfrac{1}{n}$, which diverges, and $\sum _{n=1} ^{+\infty} \dfrac{1}{n^2}$, which converges, one sees that $r=1$ may appear in series of different conditions. \end{pf} \begin{rem} The root test is more general than the ratio test, as can be seen from Theorem \ref{thm:ratio-is-inferior-to-root}. \end{rem} \begin{exa} Since $$ \dfrac{\frac{(n+1)!}{(n+1)^{n+1}}}{\frac{n!}{n^n}} = \dfrac{1}{\left(1+\dfrac{1}{n}\right)^n} \rightarrow \dfrac{1}{e}<1$$the series $\sum _{n=1} ^{+\infty} \dfrac{n!}{n^n}$ converges. \end{exa} \begin{exa} Since $$ \left(\dfrac{(n!)^n}{n^{n^2}}\right)^{1/n} = \dfrac{n!}{n^n} \rightarrow 0$$the series $\sum _{n=1} ^{+\infty} \dfrac{(n!)^n}{n^{n^2}}$ converges. \end{exa} \subsection*{Homework}\addcontentsline{toc}{subsection}{Homework} \begin{multicols}{2}\columnseprule 1pt \columnsep 25pt\multicoltolerance=900\small \begin{pro}True or False: If the infinite series $\dsum _{n=1} ^{+\infty} a_n$ of strictly positive terms, converges, then $\dsum _{n=1} ^{+\infty} a_n ^2$ must necessarily converge. \begin{answer}True. For, we must have $a_n\to 0$ and so eventually $0 < a_n \leq 1 $. This means that eventually $a_n ^2 \leq a_n$ and the series of squares converges by direct comparison to the original series. \end{answer} \end{pro} \begin{pro}True or False: If the infinite series $\dsum _{n=1} ^{+\infty} a_n$ of strictly positive terms converges, then $\dsum _{n=1} ^{+\infty} \sin (a_n)$ must necessarily converge. \begin{answer}True. Since $a_n\to 0$, we must have $\sin a_n \to a_n$ and so the series converges by asymptotic comparison to the original series. (Recall that $\lim _{x\to 0} \dfrac{\sin x}{x}=1$.) \end{answer} \end{pro} \begin{pro}True or False: If the infinite series $\dsum _{n=1} ^{+\infty} a_n$ of strictly positive terms converges, then $\dsum _{n=1} ^{+\infty} \tan (a_n)$ must necessarily converge. \begin{answer}True. Since $a_n\to 0$, we must have $\tan a_n \to a_n$ and so the series converges by asymptotic comparison to the original series. (Recall that $\lim _{x\to 0} \dfrac{\tan x}{x}=1$.) \end{answer} \end{pro} \begin{pro}True or False: If the infinite series $\dsum _{n=1} ^{+\infty} a_n$ converges, then $\dsum _{n=1} ^{+\infty} \cos (a_n)$ must necessarily converge. \begin{answer}False. Since $a_n\to 0$, we must have $\cos a_n \to 1$ and so the series diverges by the $n$-th Term Test. \end{answer} \end{pro} \begin{pro} Use the comparison tests to shew that if $a_n > 0$ and $\dis{\sum _{n = 1} ^\infty a_n}$ converges, then $\dis{\sum _{n = 1} ^\infty \frac{a_n}{n}}$ converges. \\ \begin{answer} Only the fact that $\dfrac{a_n}{n} \leq a_n$ is needed here. \end{answer} \end{pro} \begin{pro}Give an example of a series converging to $1$ with $n$-th term $a_n>0$ satisfying $a_n <<\dfrac{1}{n^2}$. (That is, the $n$-th term goes to zero faster than the reciprocal of a square.) \begin{answer} Take $a_n = \dfrac{1}{2^n}$. Then $a_n << \dfrac{1}{n^2}$ and $\dsum _{n=1} ^{+\infty}\dfrac{1}{2^n}=1$. \end{answer} \end{pro} \begin{pro}Give an example of a converging series of strictly positive terms $\dsum _{n=1} ^{+\infty} a_n$ such that $\dsum _{n=1} ^{+\infty} (a_n)^{1/n}$ also converges. \begin{answer} Take $a_n = \dfrac{1}{2^{n^2}}$ or $a_n=\dfrac{1}{n^{2n}}$. \end{answer} \end{pro} \begin{pro} Give an example of a converging series of strictly positive terms $\dsum _{n=1} ^{+\infty} a_n$ such that $\dsum _{n=1} ^{+\infty} (a_n)^{1/n}$ diverges. \begin{answer} Take $a_n = \dfrac{1}{2^n}$ or $a_n=\dfrac{1}{n^n}$. \end{answer} \end{pro} \begin{pro}Give an example of a converging series of strictly positive terms $a_n$ such that $\lim _{n\to +\infty} (a_n)^{1/n}$ does not exist. \begin{answer} For even $n\geq 0$ take $a_n = \dfrac{1}{2^n}$ and for odd $n\geq 1$ take $a_n = \dfrac{1}{3^n}$. Then $$ \sum _{n=0} ^{+\infty} a_n = \sum _{n=0} ^{+\infty} \dfrac{1}{2^{2n}} + \sum _{n=1} ^{+\infty} \dfrac{1}{3^{2n-1}}, $$ and both series on the right are geometric convergent series. However if $n$ is even, $ (a_n)^{1/n} = \dfrac{1}{2}$ and if $n$ is odd $ (a_n)^{1/n} = \dfrac{1}{3}$ meaning that $\lim _{\ngrows} (a_n)^{1/n}$ does not exist. \end{answer} \end{pro} \begin{pro}Test $\dis{\sum _{n = 1} ^\infty \frac{3^n}{n^{2n}}}$ using both direct comparison and the root test. \\ \begin{answer} By the root test $$a_n ^{1/n}=\left(\frac{3^n}{n^{2n}}\right)^{1/n} = \dfrac{3}{n}\rightarrow 0 <1, $$and the series converges. By direct comparison, for $n \geq 3$ we have $$ \dfrac{3^n}{n^{2n}} = \dfrac{3^n}{n^n}\cdot \dfrac{1}{n^n} \leq 1 \dfrac{1}{n^n} \leq \dfrac{1}{n^3}, $$and the series converges by direct comparison to $\dis{\sum _{n = 1} ^\infty \frac{1}{n^3}}$. \end{answer} \end{pro} \begin{pro}Let $\mathscr{S}$ be the set of positive integers none of whose digits in its decimal representation is a $0$: $$\mathscr{S} = \{1,2,3,4,5,6,7,8,9,11,12,13,14,15,16,17,18,19,21,\cdots\}. $$ Prove that the series $\sum _{n\in \mathscr{S}}\dfrac{1}{n}$ converges. \begin{answer} We divide the sum into decimal blocks. There are $9^k$ $k$-digit integers in the interval $[10^k; 10^{k+1}[$ that do not have a $0$ in their decimal representation. Thus $$\sum _{n\in \mathscr{S}}\dfrac{1}{n} = \sum _{k=0} ^{+\infty} \sum _{n\in [10^k; 10^{k+1}[\cap \mathscr{S}} \dfrac{1}{n} \leq \sum _{k=0} ^{+\infty} 9^k\left(\dfrac{1}{10^k}\right) = 10.$$ \end{answer} \end{pro} \begin{pro} Let $d(n)$ be the number of strictly positive divisors of the integer $n$. Prove that $d(n)\leq 2\sqrt{n}$. Use this to prove that $$ \sum _{n\geq 1} \dfrac{d(n)}{n^2} $$converges. \end{pro} \begin{pro} Let $p_n$ be the $n$-th prime. Thus $p_1=2$, $p_2=3$, $p_3=5$, etc. Put $a_1=p_1$ and $a_{n+1} = p_1p_2\cdots p_n+1$ for $n\geq 1$. Find $$\sum _{n=1} ^{+\infty}\dfrac{1}{a_n} . $$ \end{pro} \begin{pro} Determine whether $\sum _{n\geq 2}a_n$ converges, when $a_n$ is given as below. \begin{multicols}{2} \begin{enumerate} \item $\left(1+\dfrac 1n\right)^n - e$. \item $\cosh^\alpha n - \sinh^\alpha n$. \item \mbox{$\log \dfrac{(n^3+1)^2}{(n^2+1)^3} $.} \item $\sqrt[n]{n+1} - \sqrt[n]{n}$. \item $\arccos\left(\dfrac {n^3+1}{n^3+2} \right)$. \item $\dfrac {a^n}{1+a^{2n}}$. \item $\dfrac{2\cdot 4\cdot 6\cdots(2n)}{n^n}$. \item $\dfrac{1!+2!+\dots+n!}{(n+2)!}$. \item $\dfrac{1!-2!+\dots\pm n!}{(n+1)!}$. \item $\dfrac{(\log n)^n}{n^{\log n}}$. \item $\dfrac1{(\log n)^{\log n}}$. \end{enumerate} \end{multicols} \begin{answer} \noindent \begin{multicols}{2} \begin{enumerate} \item $a_n \sim -\dfrac e{2n} \implies $ diverges. \item $a_n \sim \dfrac \alpha{2^{\alpha-1}}e^{n(\alpha-2)} \implies $ converges iff $\alpha < 2$. \item $a_n \sim -\dfrac 3{n^2} \implies $ converges. \item $a_n \sim \dfrac 1{n^2} \implies $ converges. \item $a_n \sim \sqrt{\dfrac 2{n^3}} \implies $ converges. \item converges iff $|a| \ne 1$. \item Converges. \item $a_n \le \dfrac{(n-1)(n-1)!+n!}{(n+2)!} \le \dfrac 2{(n+1)(n+2)} \implies $ converges. \item Converges. \item $a_n \not\longrightarrow 0\implies $ diverges. \item $a_n= \dfrac1{n^{\log\log n}} \implies $ converges. \end{enumerate} \end{multicols} \end{answer} \end{pro} \end{multicols} \section{Summation by Parts} In this section we consider series whose terms have arbitrary signs. We first need the following result. \begin{thm}[Summation by Parts] Let $A_n = \sum _{0\leq k \leq n} a_k$,\ $A_{-1}=0$. Then for $ p \leq q$, $$ \sum _{p \leq k \leq q} a_kb_k = \sum _{p \leq k \leq q-1}A_k(b_{k}-b_{k+1}) +A_qb_q-A_{p-1}b_p. $$ \end{thm} \begin{pf}Changing a subindex, $$\begin{array}{lll} \sum _{p \leq k \leq q} a_kb_k & = & \sum _{p \leq k \leq q} (A_k-A_{k-1})b_k\\ & = & \sum _{p \leq k \leq q}A_kb_k-\sum _{p \leq k \leq q}A_{k-1}b_k \\ & = & \sum _{p \leq k \leq q-1}A_kb_k-\sum _{p \leq k \leq q-1}A_{k}b_{k+1} +A_{q}b_q -A_{p-1}b_p. \end{array}$$ giving the result. \end{pf} \begin{rem} An alternative and more symmetric formulation will be given once we introduce Riemann-Stieltjes integration. \end{rem} We now obtain a convergence test. \begin{thm}[Dirichlet's Test] The series $\sum _{k\geq 0}a_kb_k$ converges if \begin{enumerate} \item the partial sums $A_n = \sum _{k=0} ^n a_k$ are bounded \item $b_{n}\geq b_{n+1}$ \item $b_n\to 0$ as $\ngroes$ \end{enumerate} \end{thm} \begin{pf} \end{pf} \section{Alternating Series} A series of the form $\sum _{n=1} ^{+\infty} (-1)^na_n$ where the all the $a_n\geq 0$ is called an {\em alternating series.} \begin{thm}[Leibniz's Alternating Series Test] The alternating series $\sum _{n=1} ^{+\infty} (-1)^na_n$ converges if all the following conditions are met \begin{itemize} \item the $a_n$ eventually decrease, that is, $a_{n+1} \leq a_n$ for all $n\geq N$. \item $a_n\rightarrow 0$ \end{itemize} \end{thm} \begin{exa} The series $\sum _{n=1} ^{+\infty} (-1)^{n+1}\dfrac{1}{n}$ converges by Leibniz's Test. In fact, one can prove that it equals $\log 2$. \end{exa} \section{Absolute Convergence} If $\sum _{n=1} ^{+\infty} |a_n|$ converges then $\sum _{n=1} ^{+\infty} a_n$ converges. The converse is not true. \begin{exa} Since $\Big|\dfrac{\sin n}{n^2}\Big| \leq \dfrac{1}{n^2}$, $\sum _{n=1} ^{+\infty} |\dfrac{\sin n}{n^2}\Big|$ converges by the comparison test. Thus $\sum _{n=1} ^{+\infty} \dfrac{\sin n}{n^2}$ converges absolutely and so it converges.\end{exa} \begin{exa} Determine whether the following two infinite series converge: (I) $\displaystyle{\sum_{n=2}^\infty {(-1)^n\sin(3n)\over n^2}}$, \quad (II) $\displaystyle{\sum_{n=1}^\infty {(-1)^nn\over n^2+2}}$.\end{exa} \begin{solu} We have $$ \Big|(-1)^n\dfrac{\sin 3n}{n^2}\Big| \leq \dfrac{1}{n^2}, $$so (I) converges absolutely. As for number (II), $f(x) = \dfrac{x}{x^2+2}$ is decreasing (take the first derivative) $\dfrac{n}{n^2+2} \rightarrow 0,$ so it converges by Leibniz's Test. \end{solu} \chapter{Real Functions of One Real Variable} \section{Limits of Functions} \begin{prop}[Cauchy-Heine, Sinistral Limit]\label{prop:sinistral-limit} Let $f:\loro{a}{b}\rightarrow \BBR$ and let $x_0\in\loro{a}{b}$. The following are equivalent. \begin{enumerate} \item $\forall \varepsilon > 0$, $\exists \delta >0$ such that $$ x_0-\delta 0$, $\exists \delta >0$ such that $$ x_0-\delta 0$ such that for all $\delta >0$ we can find $x$ such that $$0<\absval{x-x_0}<\delta \implies \absval{f(x)-L}\geq \varepsilon_0.$$In particular, for each strictly positive integer $n$ we can find $x_n$ satisfying $$0<\absval{x_n-x_0}<\dfrac{1}{n} \implies \absval{f(x_n)-L}\geq \varepsilon_0,$$a contradiction to the fact that $f(x_n)\rightarrow L$. \end{enumerate} \end{pf} In an analogous manner, we have the following. \begin{prop}[Cauchy-Heine, Dextral Limit] \label{prop:dextral-limit} Let $f:\loro{a}{b}\rightarrow \BBR$ and let $x_0\in\loro{a}{b}$. The following are equivalent. \begin{enumerate} \item For each sequence $\seq{x_n}{n=1}{+\infty}$ of points in the interval $\loro{a}{b}$ with $x_n>x_0$, $x_n\rightarrow x_0 \implies f(x_n)\rightarrow L$. \item $\forall \varepsilon > 0$, $\exists \delta >0$ such that $$ x_0 0$, $\exists \delta >0$ such that $$ 0 <\absval{x-x_0}<\delta \implies \absval{f(x)-L}<\varepsilon .$$ \end{enumerate} If either condition is fulfilled we say that {\em $f$ has a (two-sided) limit $L$ as $x$ decreases towards $x_0$} and we write $$L=\lim _{x\rightarrow x_0}f(x). $$ \end{prop} We now prove analogues of the theorems that the proved for limits of sequences. \begin{thm}[Uniqueness of Limits] Let $X\subseteqq \BBR$, $a\in\BBR$, and $f:X\rightarrow \BBR$. If $\lim _{x\rightarrow a}f(x)=L$ and $\lim _{x\rightarrow a}f(x)=L'$ then $L=L'$. \end{thm} \begin{pf}If $L\neq L'$ then take $2\varepsilon =\absval{L-L'}$ in the definition of limit. There is $\delta >0$ such that $$0<\absval{x-a}<\delta \implies \absval{f(x)-L}<\dfrac{\absval{L-L'}}{2}, \quad \absval{f(x)-L'}<\dfrac{\absval{L-L'}}{2}. $$By the Triangle Inequality $$\absval{L-L'}\leq \absval{L-f(x)}+\absval{f(x)-L'}< \dfrac{\absval{L-L'}}{2}+\dfrac{\absval{L-L'}}{2} = \absval{L-L'},$$ but $\absval{L-L'}<\absval{L-L'}$ is a contradiction. \end{pf} \begin{thm}[Local Boundedness]\label{thm:local-boundedness} Let $X\subseteqq \BBR$, $a\in\BBR$, and $f:X\rightarrow \BBR$. If $\lim _{x\rightarrow a}f(x)=L$ exists and is finite, then $f$ is bounded in a neighbourhood of $a$. \end{thm} \begin{pf} Take $\varepsilon = 1$ in the definition of limit. Then there is a $\delta >0$ such that $$0<\absval{x-a}<\delta \implies \absval{f(x)-L}<1 \implies \absval{f(x)}<1+\absval{L},$$and so $f$ is bounded on this neighbourhood. \end{pf} \begin{thm}[Order Properties of Limits]\label{thm:order-prop-func-limits} Let $X\subseteqq \BBR$, $a\in\BBR$, and $f:X\rightarrow \BBR$. Let $\lim _{x\rightarrow a}f(x)=L$ exist and be finite. Then \begin{enumerate} \item If $s0$ in the definition of limit. There is $\delta >0$ such that $$ 0<\absval{x-a}<\delta \implies \absval{f(x)-L}0$ in the definition of limit. There is $\delta >0$ such that $$ 0<\absval{x-a}<\delta \implies \absval{f(x)-L}s$ then (1) asserts that there is a neighbourhood of $\mathscr{N}'_{a}\subseteqq \N{a}$ such that $f(x)>s$, a contradiction to the assumption that $\forall x\in\N{a}$, $s\leq f(x)$. \item If on the said neighbourhood $\N{a}$ we had, on the contrary, $L0$ there is $\delta >0$ such that $$0<\absval{x-x_0}<\delta \implies \absval{a(x)-L}<\varepsilon \quad \mathrm{and}\quad \absval{c(x)-L}<\varepsilon \implies L-\varepsilon 0$ there are $\delta _1>0$ and $\delta _2>0$ such that $$ 0<\absval{x-a}<\delta _1\implies \absval{f(x)-L}<\varepsilon, \quad \mathrm{and} \quad 0<\absval{x-a}<\delta _2\implies \absval{g(x)-L'}<\varepsilon . $$Take $\delta = \min (\delta _1, \delta _2)$. Then $$ 0<\absval{x-a}<\delta \implies \absval{f(x)+\lambda g(x) - (L+\lambda L')}\leq \absval{f(x)- L}+ \absval{\lambda}\absval{g(x) - L'} <(1+\absval{\lambda})\varepsilon . $$Since the dextral side can be made arbitrarily small, the assertion follows. \item For all $\varepsilon >0$ there are $\delta _1>0$ and $\delta _2>0$ such that $$ 0<\absval{x-a}<\delta _1\implies \absval{f(x)-L}<\varepsilon, \quad \mathrm{and} \quad 0<\absval{x-a}<\delta _2\implies \absval{g(x)-L'}<\varepsilon . $$Also, by Theorem \ref{thm:local-boundedness}, $g$ is locally bounded and so there exists $B>0$, and $\delta _3>0$ such that $$0<\absval{x-a}<\delta _3 \implies \absval{g(x)}0$ there are $\delta _1>0$, $B>0$, and $\delta _2>0$ such that $$ 0<\absval{x-a}<\delta _1\implies \absval{f(x)}<\varepsilon, \quad \mathrm{and} \quad 0<\absval{x-a}<\delta _2\implies \absval{g(x)}0$ there is a sufficiently small $\delta'>0$ such that $$ \absval{\absval{g(x)}-\absval{L'}} < \dfrac{\absval{L'}}{2} \implies \absval{L'}- \dfrac{\absval{L'}}{2}<\absval{g(x)}<\absval{L'}+ \dfrac{\absval{L'}}{2} \implies \dfrac{\absval{L'}}{2}<\absval{g(x)}< \dfrac{3\absval{L'}}{2},$$that is, $\absval{g(x)}$ is bounded away from $0$ $x$ sufficiently close to $a$. Now, for all $\varepsilon >0$ there are $\delta _1>0$ and $\delta _2>0$ such that $$ 0<\absval{x-a}<\delta _1\implies \absval{f(x)-L}<\varepsilon, \quad \mathrm{and} \quad 0<\absval{x-a}<\delta _2\implies \absval{g(x)-L'}<\varepsilon . $$For $\delta = \min (\delta _1, \delta _2, \delta' )$, $$0 <\absval{x-a}<\delta \implies L-\varepsilon 0$, $\exists M$, $M>\max (0,a)$, such that $$ x\geq M \implies \absval{f(x)-L}<\varepsilon .$$ \end{enumerate} If either condition is fulfilled we say that {\em $f$ has a limit $L$ as $x$ tends towards $+\infty$} and we write $$L=\lim _{x\rightarrow +\infty}f(x). $$ \end{prop} \begin{prop}[Cauchy-Heine, Limit at $-\infty$]\label{prop:limit-infty} Let $f:\loro{-\infty}{a}\rightarrow \BBR$ The following are equivalent. \begin{enumerate} \item For each sequence $\seq{x_n}{n=1}{+\infty}$ of points in the interval $\loro{-\infty}{a}$, $$x_n\rightarrow -\infty \implies f(x_n)\rightarrow L.$$ \item $\forall \varepsilon > 0$, $\exists M$, $M<\min (0,a)$, such that $$ x\leq M \implies \absval{f(x)-L}<\varepsilon .$$ \end{enumerate} If either condition is fulfilled we say that {\em $f$ has a limit $L$ as $x$ tends towards $-\infty$} and we write $$L=\lim _{x\rightarrow -\infty}f(x). $$ \end{prop} \begin{df} We write $\lim _{x\rightarrow a+} f(x)=+\infty$ or $\lim _{x\searrow a} f(x)=+\infty$ if $\forall M>0$, $\exists \delta > 0$ such that $$x\in \loro{a}{a+\delta}\implies f(x)>M. $$Similarly, we write $\lim _{x\rightarrow a-} f(x)=+\infty$ or $\lim _{x\nearrow a} f(x)=+\infty$ if $\forall M>0$, $\exists \delta > 0$ such that $$x\in \loro{a-\delta}{a}\implies f(x)>M. $$Finally, we write $\lim _{x\rightarrow a} f(x)=+\infty$ if $\forall M>0$, $\exists \delta > 0$ such that $$x\in \loro{a-\delta}{a+\delta}\implies f(x)>M. $$ \end{df} \begin{df} We write $\lim _{x\rightarrow a+} f(x)=-\infty$ or $\lim _{x\searrow a} f(x)=-\infty$ if $\forall M<0$, $\exists \delta > 0$ such that $$x\in \loro{a}{a+\delta}\implies f(x) 0$ such that $$x\in \loro{a-\delta}{a}\implies f(x) 0$ such that $$x\in \loro{a-\delta}{a+\delta}\implies f(x)0$, then $\lim _{x\rightarrow a} (f(x)g(x))= +\infty$. \end{enumerate} \end{pro} \begin{pro}[Cauchy Criterion for Functional Limits] Let $X\subseteqq \BBR$, $a\in X$, and $f:X\rightarrow \BBR$. Prove that $f$ has a limit at $a$ (finite or infinite) if and only if for all $\varepsilon >0$ there is a $\delta > 0$ such that $\absval{x'-x''}<\delta$ implies $\absval{f(x')-f(x'')}<\varepsilon$. \end{pro} \end{multicols} \section{Continuity} \begin{df} A function $f:\loro{a}{b}\rightarrow \BBR$ is said to be {\em continuous at the point $x_0\in\loro{a}{b}$}, if we can exchange limiting operations, as in $$\lim _{x\rightarrow x_0}f(x) = f\left(\lim _{x\rightarrow x_0}x\right) \qquad (=f(x_0)). $$In other words, a function is continuous at the point $x_0$ if $$ \forall \varepsilon >0, \exists \delta >0, \quad \mathrm{such\ that} \quad \absval{x-x_0}<\delta \implies \absval{f(x)-f(x_0)}<\varepsilon. $$ \end{df} \begin{df} A function $f:\lcrc{a}{b}\rightarrow \BBR$ is said to be {\em right continuous at $a$}, if $$f(a) =f(a+).$$It is said to be {\em left continuous at $b$}, if $$f(b) =f(b-).$$ \end{df} In view of the above definitions and Proposition \ref{prop:limit}, we have the following \begin{thm} The following are equivalent. \begin{enumerate} \item The function $f:\loro{a}{b}\rightarrow \BBR$ is continuous at the point $x_0\in\loro{a}{b}$. \item $f(x_0-)=f(x_0)=f(x_0+)$. \item If $\seq{x_n}{n=1}{+\infty}$, and for all $n$, $x_n\in \loro{a}{b}$, then $x_n\rightarrow x_0 \implies f(x_n)\rightarrow f(x_0)$. \end{enumerate} \end{thm} \begin{exa} What are the points of discontinuity of the function $$\fun{f}{x}{\left\{\begin{array}{ll} \dfrac{1}{p+q} & \mathrm{if}\ x\in\BBQ\cap \lcro{0}{+\infty}, x=\dfrac{p}{q}, \ \mathrm{in\ lowest\ terms}\\ 0 & \mathrm{if}\ x\in \lcro{0}{+\infty}\setminus \BBQ\\ \end{array}\right.}{\lcro{0}{+\infty}}{\BBR}?$$ \end{exa} \begin{solu} Let $a\in\BBQ$. Since $\lcro{0}{+\infty}\setminus \BBQ$ is dense in $\lcro{0}{+\infty}$, there exists a sequence $\seq{a_n}{n=1}{+\infty}$ of points in $\lcro{0}{+\infty}\setminus \BBQ$ such that $a_n\rightarrow a$ as $\ngroes$. Observe that $f(a_n)=0$ but $f(a)\neq 0$. Hence $a_n\rightarrow a$ does not imply $f(a_n)\rightarrow f(a)$ and $f$ is not continuous at $a$. On the other hand, let $n\in \lcro{0}{+\infty}\setminus \BBQ$. Then $f(b)=0$. Let $\seq{b_n}{n=1}{+\infty}$ be a sequence in $\lcro{0}{+\infty}\cap \BBQ$ converging to $b$, $b_n = \dfrac{p_n}{q_n}$ in lowest terms. By Dirichlet's Approximation Theorem we must have $p_n\rightarrow +\infty$ and $q_n\rightarrow +\infty$. Hence $\dfrac{1}{p_n+q_n}\rightarrow 0$. So $f$ is continuous at $b$. In conclusion, $f$ is continuous at every irrational in $\lcro{0}{+\infty}$ and discontinuous at every rational in $\lcro{0}{+\infty}$. \end{solu} \begin{prop}[Oscillation of a function at a point] Let $f$ be bounded. The function $\omega: \dom{f} \to [0;+\infty[$, called the {\em oscillation of $f$ at $x$} and given by $$\omega (f, x) = \lim _{\delta \to 0+} \sup \{\absval{f(a)-f(b)}: \absval{a-x}< \delta, \absval{b-x}< \delta\} $$is well-defined. Moreover, $f$ is continuous at $x$ if and only if $\omega (f, x)=0$. \label{prop:oscillation-at-a-point}\end{prop} \begin{pf} Observe that in fact $$\omega (f, x) = \lim _{\delta \to 0+} \sup \{\absval{f(a)-f(b)}: \absval{a-x}< \delta, \absval{b-x}< \delta\} = \inf _{\delta > 0} \sup \{\absval{f(a)-f(b)}: \absval{a-x}< \delta, \absval{b-x}< \delta\} \leq \absval{f(a)-f(b)} \leq 2\absval{f} < +\infty. $$This says that $\omega (f, x)$ is well-defined. \bigskip \end{pf} \begin{df}We say that a function $f$ is continuous on the closed interval $\lcrc{a}{b}$ if it is continuous everywhere on $\loro{a}{b}$, continuous on the right at $a$ and continuous on the left at $b$. If $X\subseteqq \BBR$, then $f:X\rightarrow \BBR$ is said to be {\em continuous on $X$} (or {\em continuous}) if it is continuous at every element of $X$. \end{df} \begin{thm} \label{thm:continuous-iff-open-sets}Let $X\subseteqq \BBR$. A function $f:X\rightarrow \BBR$ is continuous if and only if the the inverse image of an open set is open in $X$. \end{thm} \begin{pf} \begin{enumerate} \item[$\implies$] Let $A \subseteqq \BBR$ be an open set. We must shew that $f^{-1}(A)$ is open in $X$. Let $a\in f^{-1}(A)$. Since $f(a)\in A$ and $A$ is open in $\BBR$, there exists an $r>0$ such that $\loro{f(a)-r}{f(a)+r}\subseteqq A$. Since $f$ is continuous at $a$, there exists a $\delta >0$ such that $$\begin{array}{lll} \absval{x-a}<\delta \implies \absval{f(x)-f(a)} 0$, we must find a $\delta > 0$ such that for all $a\in X$, $$\absval{x-a}<\delta \implies \absval{f(x)-f(a)}< \varepsilon. $$Now $$ \absval{f(x)-f(a)}< \varepsilon \implies f(x)\in \loro{f(a)-\varepsilon}{f(a)+\varepsilon} \implies x\in f^{-1}\left(\loro{f(a)-\varepsilon}{f(a)+\varepsilon}\right).$$Now, $\loro{f(a)-\varepsilon}{f(a)+\varepsilon}\subseteqq \BBR$ is open in $\BBR$, and so, by assumption, so is $f^{-1}\left(\loro{f(a)-\varepsilon}{f(a)+\varepsilon}\right)$. This means that if $t\in f^{-1}\left(\loro{f(a)-\varepsilon}{f(a)+\varepsilon}\right)$ then there is a $r >0$ such that $$\loro{t-r}{t+r}\subseteqq f^{-1}\left(\loro{f(a)-\varepsilon}{f(a)+\varepsilon}\right).$$ But clearly $a\in f^{-1}\left(\loro{f(a)-\varepsilon}{f(a)+\varepsilon}\right)$, and hence there is a $\delta > 0$ such that $$\loro{a-\delta}{a+\delta}\subseteqq f^{-1}\left(\loro{f(a)-\varepsilon}{f(a)+\varepsilon}\right).$$ Thus $$x\in\loro{a-\delta}{a+\delta}\implies x\in f^{-1}\left(\loro{f(a)-\varepsilon}{f(a)+\varepsilon}\right),$$ or equivalently, $$ \absval{x-a}<\delta \implies f(x)\in \loro{f(a)-\varepsilon}{f(a)+\varepsilon}, $$ that is, $$ \absval{x-a}<\delta \implies \absval{f(x)-f(a)}< \varepsilon, $$ as we needed to shew. \end{enumerate}\end{pf} \begin{thm}Let $X\subseteqq \BBR$. A function $f:X\rightarrow \BBR$ is continuous if and only if the the inverse image of a closed set is closed in $X$. \label{thm:continuous-iff-closed-sets} \end{thm} \begin{pf} Let $F\subseteqq \BBR$ be a closed set. Then $\BBR \setminus F$ is open. By Theorem \ref{thm:continuous-iff-open-sets} $f^{-1}(\BBR \setminus F)$ is open in $X$, and so $X\setminus f^{-1}(\BBR \setminus F)$ is closed in $X$. But $X\setminus f^{-1}(\BBR \setminus F) = f^{-1}(F)$, proving the theorem. \end{pf} \begin{thm}If two continuous functions agree on a dense set of the reals, then they are identical. That is, if $X\subseteqq \BBR$ is dense in $\BBR$ and if $f:\BBR\rightarrow \BBR$ and $g:\BBR\rightarrow \BBR$ satisfy $f(x)=g(x)$ for all $x\in X$, then $f(x)=g(x)$ for all $x\in\BBR$.\label{thm:fun-agree-in-dense-set} \end{thm} \begin{pf} Let $a\in\BBR\setminus X$. Since $X$ is dense in $\BBR$, there is a sequence $\seq{x_n}{n=1}{+\infty}\subseteqq X$ such that $x_n\rightarrow a$ as $\ngroes$. Notice that since $x_n\in X$, we have $f(x_n)=g(x_n)$. By continuity $$f(a) = f\left(\lim _{\ngroes}x_n\right) = \lim _{\ngroes} f(x_n) = \lim _{\ngroes} g(x_n) = g\left(\lim _{\ngroes}x_n\right) =g(a), $$proving the theorem. \end{pf} \begin{thm}[Cauchy's Functional Equation]\label{thm:cauchy-functional-equation} Let $f$ be a continuous function defined over the real numbers that satisfies the {\em Cauchy functional equation}: $$\forall (x,y)\in \BBR^2, \quad f(x+y)=f(x)+f(y). $$Then $f$ is linear, that is, there is a constant $c$ such that $f(x)=cx$. \end{thm} \begin{pf} Our method of proof is as follows. We first prove the assertion for positive integers $n$ using induction. We then extend our result to negative integers. Thence we extend the result to reciprocals of integers and after that to rational numbers. Finally we extend the result to all real numbers by means of Theorem \ref{thm:fun-agree-in-dense-set}. \bigskip We prove by induction that for integer $n \geq 0$, $f(nx)=nf(x)$. Using the functional equation, $$f(0\cdot x) =f(0\cdot x+ 0\cdot x) = f(0\cdot x)+f(0\cdot x)\implies f(0\cdot x)=0f(x),$$ and the assertion follows for $n=0$. Assume $n\geq 1$ is an integer and that $f((n-1)x) = (n-1)f(x)$. Then $$f(nx) = f((n-1)x+x) = f((n-1)x)+f(x) = (n-1)f(x)+ f(x) = nf(x), $$ proving the assertion for all strictly positive integers. \bigskip Let $m<0$ be an integer. Then $-m>0$ is a strictly positive integer, for which the result proved in the above paragraph holds, and thus and by the above paragraph, $f(-mx) = -mf(x)$. Now, $$0=f(0) \implies 0=f(mx+(-mx)) = f(mx)+f(-mx)\implies f(mx)=-f(-mx)=-(-mf(x))=mf(x), $$and the assertion follows for negative integers. We have thus proved the theorem for all integers. \bigskip Assume now that $x=\dfrac{a}{b}$, with $a\in \BBZ$ and $b\in \BBZ\setminus \{0\}$. Then $f(a) =f(a\cdot 1)= af(1)$ and $f(a) = f\left(b\dfrac{a}{b}\right) = bf\left(\dfrac{a}{b}\right)$ by the result we proved for integers and hence $$af(1)= bf\left(\dfrac{a}{b}\right) \implies = f\left(\dfrac{a}{b}\right) =f(1)\left(\dfrac{a}{b}\right).$$ We have established that for all rational numbers $x\in \BBQ$, $f(x) = xf(1)$. \bigskip We have not used the fact that the function is continuous so far. Since the rationals are dense in the reals the extension of the result now follows from Theorem \ref{thm:fun-agree-in-dense-set}.\end{pf} \subsection*{Homework}\addcontentsline{toc}{subsection}{Homework} \begin{multicols}{2}\columnseprule 1pt \columnsep 25pt\multicoltolerance=900\small \begin{pro} Find all functions $f:\BBR\rightarrow \BBR$, continuous at $x=0$ such that $\forall x\in \BBR$, $f(x)=f(3x)$. \end{pro} \begin{pro} Find all functions $f:\BBR\rightarrow \BBR$, continuous at $x=0$ such that $\forall x\in \BBR$, $f(x)=f\left(\dfrac{x}{1+x^2}\right)$. \end{pro} \begin{pro} Determine the set of points of discontinuity of the function $f:\BBR\rightarrow \BBR$, $f:x\mapsto \floor{x}+ \sqrt{x-\floor{x}}$. \end{pro} \begin{pro} What are the points of discontinuity of the function $\fun{f}{x}{\left\{\begin{array}{ll} x & \mathrm{if}\ x\in \BBQ\\ 0 & \mathrm{if}\ x\in \BBR\setminus \BBQ\\ \end{array}\right.}{\BBR}{\BBR}$? \end{pro} \begin{pro} What are the points of discontinuity of the function $\fun{f}{x}{\left\{\begin{array}{ll} 0 & \mathrm{if}\ x\in \BBQ\\ x & \mathrm{if}\ x\in \BBR\setminus \BBQ\\ \end{array}\right.}{\BBR}{\BBR}$? \end{pro} \begin{pro} What are the points of discontinuity of the function $\fun{f}{x}{\left\{\begin{array}{ll} 0 & \mathrm{if}\ x\in \BBQ\\ 1 & \mathrm{if}\ x\in \BBR\setminus \BBQ\\ \end{array}\right.}{\BBR}{\BBR}$? \end{pro} \begin{pro} What are the points of discontinuity of the function $\fun{f}{x}{\left\{\begin{array}{ll} \cos x & \mathrm{if}\ x\in \BBQ\\ \sin x & \mathrm{if}\ x\in \BBR\setminus \BBQ\\ \end{array}\right.}{\BBR}{\BBR}$? \end{pro} \begin{pro} Find all functions $f:\BBR\rightarrow \BBR$, continuous at $x=1$ such that $\forall x\in \BBR$, $f(x)=-f(x^2)$. \end{pro} \begin{pro} Let $a\in \BBR$ be fixed. Find all functions $f:\BBR\rightarrow \BBR$, continuous everywhere such that $\forall (x,y)\in \BBR^2$, $f(x-y)=f(x)-f(y)+axy$. \end{pro} \begin{pro} Let $f:\lcro{0}{+\infty}\rightarrow \lcro{0}{+\infty}$, $x\mapsto \sqrt{x+\sqrt{x+\sqrt{x+\cdots}}}$. Is $f$ right-continuous at $0$? \begin{answer} $f(0)=0$, but for $x>0$, $f(x)=\dfrac{1+\sqrt{1+4x}}{2}$, so $f$ is not right-continuous at $x=0$. \end{answer} \end{pro} \end{multicols} \section{Algebraic Operations with Continuous Functions} \begin{thm}[Algebra of Continuous Functions]\label{thm:algebra-of-cont-fun} Let $f, g:\loro{a}{b}\rightarrow \BBR$ be continuous a the point $x_0\in \loro{a}{b}$. Then \begin{enumerate} \item $f+g$ is continuous at $x_0$. \item $fg$ is continuous at $x_0$. \item if $g(x_0)\neq 0$, $\dfrac{f}{g}$ is continuous at $x_0$. \end{enumerate} \end{thm} \begin{pf} This follows directly from Theorem \ref{thm:algebra-of-fun-limits}. \end{pf} \begin{thm}\label{thm:continuity-composition} Let $X, Y$ be subsets of $\BBR$, $a\in X$ and $b\in Y$, $f:X\rightarrow \BBR$, $g:Y\rightarrow \BBR$ such that $f(X)\subseteqq Y$. If $f$ is continuous at $a$ and $g$ is continuous at $f(a)$, then $g\circ f$ is continuous at $a$. \end{thm} \begin{pf} This follows at once from Theorem \ref{thm:limit-composition}. \end{pf} \begin{thm} Let $f:I\rightarrow \BBR$ be a monotone function, where $I \subseteqq \BBR$ is a non-empty interval. Then the set of points of discontinuity of $f$ is either finite or countable. \end{thm} With Theorems \ref{thm:algebra-of-cont-fun} and \ref{thm:continuity-composition} we can now demonstrate the \section{Monotonicity and Inverse Image} \begin{df}Let $X$ and $Y$ be subsets of $\BBR$. Let $f:X\rightarrow Y$, and assume that $X$ has at least two elements. Then $f$ is said to be \begin{itemize} \item {\em increasing} if $\forall (a,b)\in X^2$, $a0$. \item {\em decreasing} if $\forall (a,b)\in X^2$, $a f(b)$. Equivalently, if the ratio $\dfrac{f(b)-f(a)}{b-a} <0$. \end{itemize} $f$ is said to be {\em monotonic} if it is either increasing or decreasing, and {\em strictly monotonic} if it is either strictly increasing or strictly decreasing. \end{df} \begin{rem} Observe that if $f$ is increasing, then $-f$ is decreasing, and conversely. Similarly for strictly monotonic functions. \end{rem} \begin{thm}\label{thm:monotone->injective} Let $X\subseteqq \BBR$ and let $f:X\rightarrow \BBR$ be strictly monotone. Then $f$ is injective. \end{thm} \begin{pf} Recall that $f$ is injective if $x\neq y\implies f(x)\neq f(y)$. If $f$ is strictly increasing then $xf(y)$. In either case, the condition for injectivity is fulfilled. \end{pf} \begin{thm}Let $I\subseteqq \BBR$ be an interval and let $f:I\rightarrow f(I)$ be strictly monotone. Then $f^{-1}$ is strictly monotone in the same sense as $f$. \end{thm} \begin{pf}Assume first that $f$ is strictly increasing and put $x=f^{-1}(a)$, $y=f^{-1}(b)$ and that $a0$. We must shew that there is $\delta >0$ such that $$ \absval{y-b}<\delta \implies \absval{f^{-1}(y)-b}<\varepsilon . $$ If $a$ is not an endpoint of $I$, there is an $\alpha >0$ such that $\loro{a-\alpha}{a+\alpha}\subseteqq I$. Put $\varepsilon ' = \min (\varepsilon , \alpha)$. Since both $f$ and $f^{-1}$ are both strictly monotone $$ \absval{f^{-1}(y)-b}<\varepsilon ' \implies b-\varepsilon ' < f^{-1}(y)< b+\varepsilon ' \implies f(b-\varepsilon ') < f(f^{-1}(y))< f(b+\varepsilon ') \implies f(b-\varepsilon ') < y< f(b+\varepsilon '). $$Since $f$ is strictly increasing and $a-\varepsilon '0$ such that $f(a-\varepsilon ') = b-\eta f(b)$ the argument is similar. We would like to shew that if $a' 0$, then by the Intermediate Value Theorem there must be an $s\in\loro{0}{1}$ such that $g(s) = 0$. This entails $$(1 - s)a + sa' = (1-s)b +sb' \implies 0 > (1-s)(a -b) = s(b' -a') > 0,$$absurd. This entails that $g(1) < 0 \implies f(a')injective}, and thus $f$ is invertible, by Theorem \ref{thm:invertible<->bijective}. \end{enumerate} \end{pf} \section{Convex Functions} \begin{df}Let $A\times B\subseteqq \BBR^2$. A function $f: A \rightarrow B$ is {\em convex} in $A$ if $\forall (a, b, \lambda)\in A^2 \times [0; 1]$, $$ f(\lambda a + (1 - \lambda)b) \leq f(a)\lambda + (1 - \lambda)f(b).$$ It is {\em strictly convex} if the inequality above is strict. Similarly, a function $g: A \rightarrow B$ is {\em concave} in $A$ if $\forall (a, b, \lambda)\in A^2 \times [0; 1]$, $$ g(\lambda a + (1 - \lambda)b) \geq g(a)\lambda + (1 - \lambda)g(b).$$It is {\em strictly concave} if the inequality above is strict. \end{df} \subsection{Graphs of Functions} \begin{df} Given a function $f$, its {\em graph} is the set on the plane $$\Gamma _f = \{(x,y)\in\BBR^2: y = f(x)\}.$$ \end{df} \begin{exa} Figures \ref{fig:x} through \ref{fig:floorx} shew the graphs of a few standard functions, with which we presume the reader to be familiar. \end{exa} \vspace{1cm} \section{Classical Functions} \subsection{Affine Functions} \begin{df} An {\em affine function} is one with assignment rule of the form $x\mapsto ax+b$, where $a, b$ are real constants. \end{df} \begin{thm} The graph of an affine function is a line on the plane. Conversely, any non-vertical straight line on the plane is the graph on an affine function. \end{thm} \subsection{Quadratic Functions} \subsection{Polynomial Functions} \subsection{Exponential Functions} \begin{prop}\label{prop:e^x} Let $x\in \BBR$ be fixed. The sequence $\seq{\left(1+\dfrac{x}{n}\right)^n}{n>-x}{+\infty}$ is bounded and strictly increasing. Thus it converges and we define {\em the natural exponential function} by $$ \exp : \BBR\rightarrow \BBR, \qquad \exp(x) := \lim _{\ngroes} \left(1+\dfrac{x}{n}\right)^n. $$ \end{prop} \begin{pf} Observe that $1+\dfrac{x}{n}>0$ for $n>-x$. Using the AM-GM Inequality with $x_1 =1, x_2 = \cdots =x_{n+1}=1+\dfrac{x}{n}$ $$ \left(1+\dfrac{x}{n}\right)^{n/(n+1)} < \dfrac{1 + n\left(1+\dfrac{x}{n}\right)}{n+1} = 1+\dfrac{x}{n+1} \implies \left(1+\dfrac{x}{n}\right)^{n} < \left(1+\dfrac{x}{n+1}\right)^{n+1},$$whence the sequence is increasing. \bigskip For $0 < x \leq 1$ then $ \left(1+\dfrac{x}{n}\right)^{n}\leq \left(1+\dfrac{1}{n}\right)^{n}1$ then by the already proved monotonicity, $$ \left(1+\dfrac{x}{n}\right)^{n} \leq \left(1+\dfrac{\floor{x}+1}{n}\right)^{n} <\left(1+\dfrac{\floor{x}+1}{n(\floor{x}+1)}\right)^{n(\floor{x}+1)} < e^{\floor{x}+1}. $$ \bigskip If $x\leq 0$ then $1+\dfrac{x}{n} \leq 1$ and so $\left(1+\dfrac{x}{n}\right)^{n} \leq 1$. \end{pf} \begin{rem} By Theorem \ref{thm:e}, $\exp (1) = e$. We will later prove, in ????, that for all $x\in \BBR$, $\exp (x) = e^x$. \end{rem} \subsection{Logarithmic Functions} \subsection{Trigonometric Functions} \begin{thm} Let $x\in \loro{0}{\frac{\pi}{2}}$. Then $\sin x < x < \tan x$. \end{thm} \begin{pf} \end{pf} \subsection*{Homework}\addcontentsline{toc}{subsection}{Homework} \begin{multicols}{2}\columnseprule 1pt \columnsep 25pt\multicoltolerance=900\small \begin{pro} How many solutions does the equation $$ \sin x = \dfrac{x}{100} $$ have? \end{pro} \begin{pro} Prove that $$ \frac{2}{\pi}x\leq\sin(x)\leq x, \forall\;x\in \lcrc{0}{\frac{\pi}{2}}. $$ \begin{answer} Consider a unit circle and take any point $P$ on the circumference of the circle. Drop the perpendicular from $P$ to the horizontal line, $M$ being the foot of the perpendicular and $Q$ the reflection of $P$ at $M$. (refer to figure) Let $x = \angle POM.$ For $x$ to be in $[0,\frac{\pi}{2}]$, the point $P$ lies in the first quadrant, as shown. The length of line segment $PM$ is $\sin(x)$. Construct a circle of radius $MP$, with $M$ as the center. Length of line segment $PQ$ is $2\sin(x)$. Length of arc $PAQ$ is $2x$. Length of arc $PBQ$ is $\pi\sin(x)$. Since $PQ \leq$ length of arc $PAQ$ (equality holds when $x = 0$) we have $2\sin(x) \leq 2x$. This implies $$\sin(x) \leq x$$ Since length of arc $PAQ$ is $\leq $ length of arc $PBQ$ (equality holds true when $x = 0$ or $x = \frac{\pi}{2}$), we have $2x \leq \pi\sin(x)$. This implies $$\frac{2}{\pi}x \leq \sin(x)$$ Thus we have $$ \frac{2}{\pi}x\leq\sin(x)\leq x, \forall\;x\in [0,\frac{\pi}{2}] $$ \end{answer} \end{pro} \begin{pro} How many solutions does the equation $$ \sin x = \log x $$ have? \end{pro} \begin{pro} How many solutions does the equation $$ \sin (\sin (\sin (\sin (\sin (x))))) = \dfrac{x}{3} $$ have? \end{pro} \begin{pro}[Chebyshev Polynomials] \end{pro} \begin{pro}[Cardano's Formula] \end{pro} \end{multicols} \subsection{Inverse Trigonometric Functions} \section{Continuity of Some Standard Functions.} \subsection{Continuity Polynomial Functions} \begin{lem}\label{lem:constant-fun-continuous-ist} Let $K\in \BBR$ be a constant. The constant function $f:\BBR\rightarrow \BBR$, $f(x)=K$ is everywhere continuous. \end{lem} \begin{pf} Given $a\in \BBR$ and $\varepsilon >0$, take $\delta = \varepsilon$. Then clearly $$ \absval{x-a}<\delta \implies \absval{f(x)-f(a)}<\varepsilon ,$$since $f(x)=f(a)=K$ and the quantity after the implication is $0<\varepsilon$ and we obtain a tautology. \end{pf} \begin{lem}\label{lem:ide-fun-continuous-ist} The identity function $f:\BBR\rightarrow \BBR$, $f(x)=x$ is everywhere continuous. \end{lem} \begin{pf} Given $a\in \BBR$ and $\varepsilon >0$, take $\delta = \varepsilon$. Then clearly $$ \absval{x-a}<\delta \implies \absval{f(x)-f(a)}<\varepsilon ,$$since the quantity after the implication is $\absval{x-a}<\delta$ and we obtain a tautology. \end{pf} \begin{lem} \label{lem:power-fun-continuous-ist} Given a strictly positive integer $n$, the power function $f:\BBR\rightarrow \BBR$, $f(x)=x^n$ is everywhere continuous. \end{lem} \begin{pf} By Lemma \ref{lem:ide-fun-continuous-ist}, the function $x\mapsto x$ is continuous. Applying this Lemma and the product rule from Theorem \ref{thm:algebra-of-cont-fun} $n$ times, we obtain the result. \end{pf} \begin{thm}[Continuity of Polynomial Functions]Let $n$ be a fixed positive integer. Let $a_k\in \BBR$, $0 \leq k \leq n$ be constants. Then the polynomial function $f:\BBR \rightarrow \BBR$, $f(x)=a_0 + a_1x + a_2x^2 + \cdots + a_nx^n$ is everywhere continuous. \end{thm} \begin{pf} This follows from Lemma \ref{lem:power-fun-continuous-ist} and the sum rule from Theorem \ref{thm:algebra-of-cont-fun} applied $n+1$ times. \end{pf} \subsection{Continuity of the Exponential and Logarithmic Functions} \begin{lem}\label{lem:exp-func-cont-at-0} Let $a>1$. The exponential function $\BBR\rightarrow \BBR$, $x\mapsto a^x$ is continuous at $x=0$. \end{lem} \begin{pf} For integral $n>0$ we know that $\lim _{\ngroes }a^{1/n}=1$ by virtue of Theorem \ref{thm:root-n-of-a-to-1-goes}. We wish to shew that $a^x \rightarrow 1$ as $x\rightarrow 0$. Observe first that $\lim _{\ngroes }a^{-1/n}= \lim _{\ngroes }\dfrac{1}{a^{1/n}}=1$ also. Thus given $\varepsilon >0$, and since $a>1$, there is $N>0$ such that $$1-\varepsilon < a^{-1/N}< a^{1/N}<1+ \varepsilon. $$If $x\in \loro{-\dfrac{1}{N}}{\dfrac{1}{N}}$ then, $$ a^{-1/N}0$, $a\neq 1$. The exponential function $f:\BBR\rightarrow \loro{0}{+\infty}$, $x\mapsto a^x$ is everywhere continuous. \end{thm} \begin{pf} Assume first that $a>1$. Let us shew that it is continuous at an arbitrary $u\in \BBR$. If $x\rightarrow u$ then $x-u \rightarrow 0$. Thus $$\lim _{x\rightarrow u} a^x =a^u\lim _{x\rightarrow u} a^{x-u} = a^u\lim _{x-u\rightarrow 0} a^{x-u}=a^u\lim _{t\rightarrow 0} a^{t} = a^u\cdot 1 = a^u, $$by Lemma \ref{lem:exp-func-cont-at-0}, and so the continuity is established for $a>1$. \bigskip If $01$ and by what we have proved, $x\mapsto \dfrac{1}{a^x}$ is continuous. Then $$\lim _{x\rightarrow u} a^x = \lim _{x\rightarrow u} \dfrac{1}{\dfrac{1}{a^x}} = \dfrac{1}{\dfrac{1}{a^u}} = a^u, $$ proving continuity in the case $00$, $a\neq 1$. Then $\loro{0}{+\infty}\rightarrow \BBR$, $x\mapsto \log _ax$ is everywhere continuous. \end{lem} \begin{pf} Its inverse function $\BBR\rightarrow \loro{0}{+\infty}$, $x\mapsto a^x$, is everywhere continuous and strictly monotone. The result then follows from Theorem \ref{thm:monotone-from-interval-has-cont-inverse}. \end{pf} \subsection{Continuity of the Power Functions} \begin{thm}\label{thm:cont-of-power-fun} Let $p\in \BBR$. Then $\loro{0}{+\infty}\rightarrow \loro{0}{+\infty}$, $x\mapsto x^p$ is everywhere continuous. \end{thm} \begin{pf} This follows by the continuity of compositions: $x^p = e^{p\log x}$. \end{pf} \subsection*{Homework}\addcontentsline{toc}{subsection}{Homework} \begin{multicols}{2}\columnseprule 1pt \columnsep 25pt\multicoltolerance=900\small \begin{pro} Prove the continuity of the function $\BBR \rightarrow \lcrc{-1}{1}$, $x\mapsto \sin x$. \end{pro} \begin{pro} Prove the continuity of the function $\lcrc{-1}{1}\rightarrow \lcrc{-\frac{\pi}{2}}{\frac{\pi}{2}}$, $x\mapsto \arcsin x$. \end{pro} \begin{pro} Prove the continuity of the function $\BBR \rightarrow \lcrc{-1}{1}$, $x\mapsto \cos x$. \end{pro} \begin{pro} Prove the continuity of the function $\lcrc{-1}{1}\rightarrow \lcrc{0}{\pi}$, $x\mapsto \arccos x$. \end{pro} \begin{pro} Prove the continuity of the function $\BBR \setminus (2\BBZ+1)\dfrac{\pi}{2} \rightarrow \BBR $, $x\mapsto \tan x$. \end{pro} \begin{pro} Prove the continuity of the function $\BBR\rightarrow \loro{-\frac{\pi}{2}}{\frac{\pi}{2}}$, $x\mapsto \arctan x$. \end{pro} \end{multicols} \section{Inequalities Obtained by Continuity Arguments} The technique used Theorem \ref{thm:cauchy-functional-equation}, of proving results in a dense set of the real numbers and extending the result by continuity can be exploited in a variety of situations. We now use it to give a generalisation of Bernoulli's Inequality. \begin{thm}[Generalisation of Bernoulli's Inequality]\label{thm:generalisation-bernoulli-ineq}Let $(\alpha, x) \in \BBR^2$ with $x\geq -1$. If $0 < \alpha < 1$ then $$ (1+x)^\alpha \leq 1 + \alpha x. $$ If $\alpha \in \loro{-\infty}{0}\cup \loro{1}{+\infty}$ then $$ (1+x)^\alpha \geq 1 + \alpha x. $$Equality holds in either case if and only if $x=0$. \end{thm} \begin{pf} Let $\alpha \in \BBQ$, $0<\alpha <1$. Then $\alpha = \dfrac{m}{n}$ for integers $m, n$ with $1 \leq m < n$. Since $x+1\geq 0$, we may use the AM-GM Inequality to obtain $$\begin{array}{lll} (1+x)^{\alpha} & = & (1+x)^{m/n}\\ & = & \left((1+x)^m\cdot 1^{n-m}\right)^{1/n}\\ & \leq & \dfrac{m(1+x)+(n-m)\cdot 1}{n}\\ & = & \dfrac{n+mx}{n}\\ & = & 1+\dfrac{m}{n}x\\ & = & 1 + \alpha x. \end{array}$$ Equality holds when are the factors are the same, that is, when $1+x=1 \implies x=0$. \bigskip Assume now that $\alpha \in \BBR\setminus \BBQ$ with $0 <\alpha <1$. We can find a sequence of rational numbers $\seq{a_n}{n=1}{+\infty}\subseteqq \BBQ$ such that $a_n\rightarrow \alpha$ as $\ngroes$. Then $$ (1+x)^{a_n}\leq 1 + a_nx, $$whence by the continuity of the power functions (Theorem \ref{thm:cont-of-power-fun}), $$ (1+x)^\alpha = \lim _{\ngroes}(1+x)^{a_n}\leq \lim _{\ngroes} (1 + a_nx) = 1 + \alpha x,$$giving the result for all real numbers $\alpha$ with $0 < \alpha < 1$, except that we need to prove that equality holds only for $x=0$. Take a rational number $r$ with $0 < \alpha < r < 1$, and recall that we are assuming that $\alpha$ is irrational. Then $$ (1+x)^{\alpha} = (1+x)^{\alpha /r})^r \leq \left(1 + \dfrac{\alpha}{r}x\right)^r . $$Since the exponent on the right is rational, by what we have proved above $\left(1 + \dfrac{\alpha}{r}x\right)^r \leq 1 + x$ with equality if and only if $x=0$. Hence the full result has been proved for the case $\alpha\in\BBR$ with $0 < \alpha < 1$. \bigskip Let $\alpha > 1$. If $1+\alpha x < 0$, then obviously $(1+x)^\alpha >0 > 1+ \alpha x$, and there is nothing to prove. Hence we will assume that $\alpha x \geq -1$. By the first part of the theorem, since $0<\dfrac{1}{\alpha}<1$, $$ (1+\alpha x)^{1/\alpha} \leq 1 + \dfrac{1}{\alpha}\cdot \alpha x = 1+x \implies 1+\alpha x\leq (1+x)^\alpha, $$with equality only if $x=0$. The theorem has been proved for $\alpha > 1$. \bigskip Finally, let $\alpha < 0$. Again, if $1+\alpha x < 0$, then obviously $(1+x)^\alpha >0 > 1+ \alpha x$, and there is nothing to prove. Assume thus $\alpha x \geq -1$. Choose a strictly positive integer $n$ satisfying $0<-\alpha < n$. Now, $$1 \geq 1-\dfrac{\alpha ^2}{n^2}x^2 = \left(1 - \dfrac{\alpha}{n}x\right) \left(1 + \dfrac{\alpha}{n}x\right) \implies \dfrac{1}{1 - \dfrac{\alpha}{n}x} \geq 1 + \dfrac{\alpha}{n}x,$$and so by the first pat of the theorem $$\begin{array}{lll} (1+x)^{-\alpha /n} \leq 1 - \dfrac{\alpha}{n}x & \implies & (1+x)^{\alpha /n} \geq \dfrac{1}{1 - \dfrac{\alpha}{n}x} \\ & \implies & (1+x)^{\alpha /n} \geq 1 + \dfrac{\alpha}{n}x \\ & \implies & (1+x)^{\alpha } \geq \left(1 + \dfrac{\alpha}{n}x\right)^n, \\ \end{array}$$and since $n$ is a positive integer, $ \left(1 + \dfrac{\alpha}{n}x\right)^n \geq 1 + n\cdot \dfrac{\alpha}{n}x = 1+ \alpha x$ and so $(1+x)^\alpha \geq 1+\alpha x$ also when $\alpha < 0$. This finishes the proof of the theorem. \end{pf} \begin{thm}[Monotonicity of Power Means]\label{thm:monotonicity-power-means} Let $a_1, a_2, \ldots , a_n$ be strictly positive real numbers and let $(\alpha, \beta )\in\BBR^2$ be such that $\alpha \cdot \beta \neq 0 $ and $\alpha < \beta$. Then $$ \left(\dfrac{a_1 ^\alpha + a_2 ^\alpha + \cdots + a_n ^\alpha}{n}\right)^{1/\alpha} \leq \left(\dfrac{a_1 ^\beta + a_2 ^\beta + \cdots + a_n ^\beta}{n}\right)^{1/\beta}, $$ with equality if and only is $a_1=a_2=\cdots = a_n$. \end{thm} \begin{pf}Assume first that $0 < \alpha < \beta$. Put $c_\alpha = \left(\dfrac{a_1 ^\alpha + a_2 ^\alpha + \cdots + a_n ^\alpha}{n}\right)^{1/\alpha}$ and $d_k = \left(\dfrac{a_k}{c_\alpha}\right)^\alpha$. Observe that $$ \dfrac{c_\beta}{c_\alpha} = \left(\dfrac{\left(\dfrac{a_1}{c_\alpha}\right)^\beta+\left(\dfrac{a_2}{c_\alpha}\right)^\beta+\cdots + \left(\dfrac{a_n}{c_\alpha}\right)^\beta}{n}\right)^{1/\beta}= \left(\dfrac{d_1 ^{\beta/\alpha}+d_2 ^{\beta/\alpha}+\cdots + d_n ^{\beta/\alpha}}{n}\right)^{1/\beta}, $$and that $$\left(\dfrac{d_1 +d_2 +\cdots + d_n }{n}\right)^{1/\alpha} = \dfrac{1}{c_\alpha}\left(\dfrac{a_1 ^\alpha + a_2 ^\alpha + \cdots + a_n ^\alpha}{n}\right)^{1/\alpha}=1 \implies d_1 +d_2 +\cdots + d_n=n. $$Put $d_k = 1+x_k$. Then $x_1+x_2+\cdots + x_n = 0$. By Theorem \ref{thm:generalisation-bernoulli-ineq}, \begin{equation}\label{eq:d's-in-mono-means} d_k ^{\beta/\alpha} = (1+x_k)^{\beta/\alpha} \geq 1 + \dfrac{\beta}{\alpha}x_k. \end{equation} Letting $k$ run from $1$ through $n$ and adding, $$ d_1 ^{\beta/\alpha}+d_2 ^{\beta/\alpha}+\cdots + d_n ^{\beta/\alpha} \geq n + \dfrac{\beta}{\alpha}(x_1+x_2+\cdots + x_n) = n. $$ Hence $$ \dfrac{d_1 ^{\beta/\alpha}+d_2 ^{\beta/\alpha}+\cdots + d_n ^{\beta/\alpha}}{n}\geq 1 \implies \dfrac{c_\beta}{c_\alpha}\geq 1, $$proving the theorem when $0<\alpha < \beta$. \bigskip If $\alpha < \beta < 0$, then $0 < \dfrac{\beta}{\alpha} < 1$. The inequality in (\ref{eq:d's-in-mono-means}) is reversed, giving $\dfrac{d_1 ^{\beta/\alpha}+d_2 ^{\beta/\alpha}+\cdots + d_n ^{\beta/\alpha}}{n}\leq 1$, and since $\beta < 0$, $$\dfrac{c_\beta}{c_\alpha} = \left( \dfrac{d_1 ^{\beta/\alpha}+d_2 ^{\beta/\alpha}+\cdots + d_n ^{\beta/\alpha}}{n}\right)^{1/\beta}\geq 1^{1/\beta} = 1, $$proving the theorem when $\alpha < \beta <0$. \bigskip Finally, we tackle the case $\alpha<0 < \beta$. By the AM-GM Inequality, putting $G = (a_1a_2\cdots a_n)^{1/n}$ $$G^\alpha = (a_1 ^\alpha a_2 ^\alpha\cdots a_n ^\alpha)^{1/n} \leq \dfrac{a_1 ^\alpha + a_2 ^\alpha + \cdots + a_n ^\alpha}{n}.$$ Raising the quantities at the extreme of the inequalities to the power $-1/\alpha$ and remembering that $-1/\alpha > 0$, we gather that $$ \left(\dfrac{a_1 ^\alpha + a_2 ^\alpha + \cdots + a_n ^\alpha}{n}\right)^{1/\alpha} \leq G. $$ In a similar manner, $$G^\beta = (a_1 ^\beta a_2 ^\beta\cdots a_n ^\beta)^{1/n} \leq \dfrac{a_1 ^\beta + a_2 ^\beta + \cdots + a_n ^\beta}{n},$$ and $$ G\leq \left( \dfrac{a_1 ^\beta + a_2 ^\beta + \cdots + a_n ^\beta}{n}\right)^{1/\beta},$$since $\beta >0$. This finishes the proof. \end{pf} \begin{lem}\label{lem:homogenised-youngs} Let $\alpha, a, x$ be real numbers with $\alpha > 1$, $a>0$, and $x\geq 0$. Then $$ x^\alpha -ax \geq (1-\alpha) \left(\dfrac{a}{\alpha}\right)^{\alpha/(\alpha -1)}. $$ \end{lem} \begin{pf} By Theorem \ref{thm:generalisation-bernoulli-ineq}, since $\alpha >1$, $$ (1+z)^\alpha \geq 1 + \alpha z, \qquad z \geq -1, $$with equality only is $z=0$. Putting $z = 1 + y$, $$ y^\alpha \geq 1 + \alpha (y-1) \implies y^\alpha - \alpha y \geq 1-\alpha, \qquad y \geq 0, $$with equality only if $y=1$. Let $c>0$ be a constant. Multiplying the above inequality by $c^\alpha$ we obtain $$(cy)^\alpha - \alpha c^{\alpha -1} (cy) \geq (1-\alpha)c^{\alpha}, \qquad \mathrm{for} \qquad y\geq 0. $$ Putting $x=cy$ and $a=\alpha c^{\alpha -1}$, we get $$ x^\alpha - ax \geq (1-\alpha)\left(\dfrac{a}{\alpha}\right)^{\alpha /(\alpha -1)}, $$with equality if and only if $x=c=\left(\dfrac{a}{\alpha}\right)^{\alpha /(\alpha -1)}$. \end{pf} \begin{thm}[Young's Inequality]\label{thm:youngs-ineq} Let $p > 1$ and put $\dfrac{1}{p} + \dfrac{1}{q} = 1$. Then for $(x, y)\in ([0;+\infty[)^2$ we have $$ xy \leq \frac{x^p}{p} + \frac{y^q}{q}. $$ \end{thm} \begin{pf} Put $\alpha = p$, $a = py$ in Lemma \ref{lem:homogenised-youngs}, obtaining $$ x^p - (py)x \geq (1-p)\left(\dfrac{py}{p}\right)^{p/(p-1)} = (1-p)y^{p/(p-1)}. $$ Now, $\dfrac{1}{q} = \dfrac{p-1}{p} \implies q = \dfrac{p}{p-1}$ and $p-1 = \dfrac{p}{q}$. Hence $$ x^p - (py)x \geq (1-p)y^{p/(p-1)} \implies (1-p)y^{p/(p-1)} \geq -\dfrac{p}{q}y^q,$$and rearranging gives the result sought. \end{pf} We now derive a generalisation of the Cauchy-Bunyakovsky-Schwarz Inequality. \begin{thm}[H\"{o}lder Inequality] \label{thm:holders-ineq} Let $x_j ,y_k$, $1 \leq j, k \leq n$, be real numbers. Let $p > 1$ and put $\dfrac{1}{p} + \dfrac{1}{q} = 1$. Then $$ \absval{\sum _{k=1} ^n x_ky_k}\leq \left(\sum _{k=1} ^n \absval{x_k} ^p\right) ^{1/p}\left(\sum _{k=1} ^n \absval{y_k} ^q\right) ^{1/q}. $$ \end{thm} \begin{pf} If either $\sum _{k=1} ^n \absval{x_k} ^p = 0$ or $\sum _{k=1} ^n \absval{y_k} ^q = 0$ there is nothing to prove, so assume otherwise. From Young's Inequality we have $$ \frac{|x_k|}{ \left(\sum _{k=1} ^n \absval{x_k} ^p\right) ^{1/p}}\frac{|y_k|}{{ \left(\sum _{k=1} ^n \absval{y_k} ^q\right) ^{1/q}}} \leq \dfrac{|x_k| ^p}{ \left(\sum _{k=1} ^n \absval{x_k} ^p\right) p} + \dfrac{|y_k| ^q}{ \left(\sum _{k=1} ^n \absval{y_k} ^q\right)q}. $$Adding, we deduce $$ \begin{array}{lll}\sum _{k = 1} ^n \dfrac{|x_k|}{ \left(\sum _{k=1} ^n \absval{x_k} ^p\right) ^{1/p}}\dfrac{|y_k|}{{ \left(\sum _{k=1} ^n \absval{y_k} ^q\right) ^{1/q}}} & \leq& \dfrac{1}{ \left(\sum _{k=1} ^n \absval{x_k} ^p\right)p}\sum _{k = 1} ^n |x_k| ^p + \dfrac{1}{ \left(\sum _{k=1} ^n \absval{y_k} ^q\right)q}\sum _{k = 1} ^n |y_k| ^q \\ & = & \dfrac{\sum _{k=1} ^n \absval{x_k} ^p}{ \left(\sum _{k=1} ^n \absval{x_k} ^p\right)p} + \dfrac{ \left(\sum _{k=1} ^n \absval{y_k} ^q\right)q}{ \left(\sum _{k=1} ^n \absval{y_k} ^q\right)q}\\ & = & \dfrac{1}{p} + \dfrac{1}{q}\\ & = & 1. \end{array} $$ This gives $$ \sum _{k = 1} ^n |x_ky_k| \leq \left(\sum _{k=1} ^n \absval{x_k} ^p\right) ^{1/p} \left(\sum _{k=1} ^n \absval{y_k} ^q\right) ^{1/q}. $$The result follows by observing that $$ \left| \sum _{k = 1} ^n x_ky_k \right|\leq\sum _{k = 1} ^n |x_ky_k| \leq \left(\sum _{k=1} ^n \absval{x_k} ^p\right) ^{1/p} \left(\sum _{k=1} ^n \absval{y_k} ^q\right) ^{1/q}. $$ \end{pf} Finally, we derive a generalisation of Minkowski's Inequality. \begin{thm}[Generalised Minkowski Inequality] Let $p \in ]1; +\infty[$. Let $x_j ,y_k$, $1 \leq j, k \leq n$, be real numbers. Then the following inequality holds $$ \left(\sum _{k=1} ^n \absval{x_k + y_k} ^p\right) ^{1/p}\leq \left(\sum _{k=1} ^n \absval{x_k} ^p\right) ^{1/p} + \left(\sum _{k=1} ^n \absval{y_k} ^p\right) ^{1/p}. $$ \label{thm:minkowski_inequality-general}\end{thm} \begin{pf}From the triangle inequality for real numbers $$|x_k + y_k|^p = |x_k + y_k||x_k + y_k|^{p - 1} \leq \left(|x_k| + |y_k|\right)|x_k + y_k|^{p - 1}.$$ Adding \begin{equation} \sum _{k = 1} ^n |x_k + y_k|^p \leq \sum _{k = 1} ^n |x_k||x_k + y_k|^{p - 1} + \sum _{k = 1} ^n |y_k||x_k + y_k|^{p - 1}.\label{eq:minkowski_1}\end{equation} By the H\"{o}lder Inequality \begin{equation}\begin{array}{lll} \sum _{k = 1} ^n |x_k||x_k + y_k|^{p - 1} & \leq & \left(\sum _{k = 1} ^n |x_k|^p\right)^{1/p}\left(\sum _{k = 1} ^n|x_k + y_k|^{(p - 1)q}\right)^{1/q} \\ & = & \left(\sum _{k = 1} ^n |x_k|^p\right)^{1/p}\left(\sum _{k = 1} ^n|x_k + y_k|^{p}\right)^{1/q}\\ \end{array}\label{eq:minkowski_2}\end{equation} In the same manner we deduce \begin{equation}\sum _{k = 1} ^n |y_k||x_k + y_k|^{p - 1} \leq \left(\sum _{k=1} ^n \absval{y_k} ^p\right) ^{1/p} \left(\sum _{k=1} ^n \absval{x_k + y_k} ^p\right) ^{1/q} \label{eq:minkowski_3}.\end{equation} Hence (\ref{eq:minkowski_1}) gives $$\begin{array}{lll} \sum _{k = 1} ^n |x_k + y_k|^p & \leq & \left(\sum _{k=1} ^n \absval{x_k} ^p\right) ^{1/p}\left(\sum _{k=1} ^n \absval{x_k + y_k} ^p\right) ^{1/q} + \left(\sum _{k=1} ^n \absval{y_k} ^q\right) ^{1/q}\left(\sum _{k=1} ^n \absval{x_k + y_k} ^p\right) ^{1/q}\\ & = & \left(\left(\sum _{k=1} ^n \absval{x_k} ^p\right) ^{1/p} + \left(\sum _{k=1} ^n \absval{y_k} ^p\right) ^{1/p}\right)\left(\sum _{k=1} ^n \absval{x_k + y_k} ^p\right) ^{1/q}\\ , \end{array}$$from where we deduce the result. \end{pf} \subsection*{Homework}\addcontentsline{toc}{subsection}{Homework} \begin{multicols}{2}\columnseprule 1pt \columnsep 25pt\multicoltolerance=900\small \begin{pro} Prove that if $\alpha >0$ and $n>0$ an integer then $$ \dfrac{n^{1+\alpha}-(n-1)^{1+\alpha}}{1+\alpha} t\right\}= \loro{b}{+\infty}\cup f^{-1}\left(\loro{t}{+\infty} \cap \loro{a}{b}\right). $$ Then $U, V$ are open sets of $\BBR$ by virtue of Theorem \ref{thm:continuous-iff-open-sets}. But then $\BBR = U \cup V$ and $U\cap V = \varnothing$, $U\neq \varnothing$, $V\neq \varnothing$, contradicting the fact that $\BBR$ is connected. Thus there must exist a $c$ such that $f(c)=t$. \end{pf} \begin{cor} A continuous function defined on an interval maps that interval into an interval. \end{cor} \begin{pf} This follows at once from the Intermediate Value Theorem and the definition of an interval. \end{pf} \begin{thm}[Bolzano's Theorem]If $f:\lcrc{u}{v}\rightarrow \BBR$ is continuous and $f(u)f(v)<0$, then there is a $w\in \loro{u}{v}$ such that $f(w)=0$. \end{thm} \begin{pf} This follows at once from the Intermediate Value Theorem by putting $a=\min (f(u), f(v))<0$ and $b=\max (f(u), f(v))>0$ . \end{pf} \begin{cor} Every polynomial $p(x)\in \BBR[x]$ with real coefficients and odd degree has at least one real root. \end{cor} \begin{pf}Let $p(x) = a_0 + a_1x+ a_2x^2+\cdots + a_nx^n$, with $a_n\neq 0$ and $n$ odd. Since $p$ has odd degree, $\lim _{x\rightarrow -\infty}p(x) = (-\infty)\signum{a_n}$ and $\lim _{x\rightarrow +\infty}p(x) = (+\infty)\signum{a_n}$, which are of opposite sign. The polynomial must then attain positive and negative values and between values of opposite sign, it will have a real root. \end{pf} \begin{cor}\label{cor:retention-of-sign-cont} If $f$ is continuous at the point $a$ and $f(a)\neq 0$, then there is a neighbourhood of $a$ where $f(x)$ has the same sign as $f(a)$. \end{cor} \begin{pf} Take $\varepsilon = \dfrac{\absval{f(a)}}{2}>0 $ in the definition of continuity. There is a $\delta > 0$ such that $$ \absval{x-a}<\delta \implies \absval{f(x)-f(a)}< \dfrac{\absval{f(a)}}{2} \implies f(a)- \dfrac{\absval{f(a)}}{2} a$ and $f(b)0$ and $g(b)<0$. By Bolzano's Theorem, there must be a $c\in \loro{a}{b}$ such that $g(c)=0$, that is, $f(c)-c=0$, finishing the proof. \end{pf} \subsection*{Homework}\addcontentsline{toc}{subsection}{Homework} \begin{multicols}{2}\columnseprule 1pt \columnsep 25pt\multicoltolerance=900\small \begin{pro} Let $p(x), q(x)$ be polynomials with real coefficients such that $$ p(x^2+x+1) = p(x)q(x).$$Prove that $p$ must have even degree. \begin{answer} If $p$ had odd degree, then, by the Intermediate Value Theorem it would have a real root. Let $\alpha$ be its largest real root. Then $$ 0=p(\alpha )q(\alpha )=p(\alpha ^2+\alpha+1)$$meaning that $\alpha ^2+\alpha+1>\alpha$ is a real root larger than the supposedly largest real root $\alpha$, a contradiction. \end{answer} \end{pro} \begin{pro} A function $f$ defined over all real numbers is continuous and for all real $x$ satisfies $$\left(f(x)\right) \cdot \left((f\circ f)(x)\right) =1. $$Given that $f(1000)=999$, find $f(500)$. \begin{answer} Observe that $f(1000)f(f(1000)) = 1 \implies f(999) = \dfrac{1}{999}$. So the range of $f$ include all numbers from $\dfrac{1}{999}$ to $999$. By the intermediate value theorem, there is a real number $a$ such that $f(a) = 500$. Thus $$f(a)f(f(a)) = 1 \implies f(500) = \dfrac{1}{500}. $$ \end{answer} \end{pro} \begin{pro} Let $f:\BBR\rightarrow \BBR$ be a continuous function such that $\lim _{x\rightarrow -\infty} f(x)=0=\lim _{x\rightarrow +\infty} f(x)$. If $f$ is strictly negative somewhere on $\BBR$ then $f$ attains a finite absolute minimum on $\BBR$. If $f$ is strictly positive somewhere on $\BBR$ then $f$ attains a finite absolute maximum on $\BBR$. \end{pro} \begin{pro} Let $f:\lcrc{0}{1}\rightarrow \lcrc{0}{1}$ be continuous. Prove that there is no $c\in \lcrc{0}{1}$ such that $f^{-1}(\{c\})$ has exactly two elements. \end{pro} \begin{pro} Let $f, g$ be continuous functions from $\lcrc{0}{1}$ to $\lcrc{0}{1}$ such that $$\forall x\in \lcrc{0}{1} \qquad f(g(x)) = g(f(x)).$$Prove that $f$ and $g$ have a common fixed point in $\lcrc{0}{1}$. \begin{answer} \end{answer} \end{pro} \begin{pro} A continuous function $f:\BBR \rightarrow \BBR$ satisfies $$\forall x\in\BBR \qquad f(x+f(x))=f(x). $$Prove that $f$ is constant. \end{pro} \begin{pro} Let $I$ be a closed and bounded interval on the line and let $f$ be continuous on $I$. Suppose that for each $x \in I$, there exists a $y \in I$ such that $$ |f(y)| \leq \frac{1}{2}|f(x)|.$$ Prove the existence of a $t \in I$ such that $f(t ) = 0.$ \end{pro} \begin{pro} Find all continuous functions that satisfy the functional equation $$ f(x) + f(y) = f\left(\frac{x + y}{1 - xy}\right),$$ for all $-1 < x, y < 1.$ \end{pro} \begin{pro}[Putnam 1947] A real valued continuous function satisfies for all real $x, y$ the functional equation $$ f(\sqrt{x^2 + y^2}) = f(x)f(y) .$$ Prove that $f(x) = (f(x))^{x^2}$. \end{pro} \begin{pro}Suppose that $f:\lcrc{0}{1} \rightarrow \lcrc{0}{1}$ is continuous. Prove that there is a number $c$ in $\lcrc{0}{1}$ such that $f(c) = 1 - c$. \begin{answer} If either $f(0)=1$ or $f(1)=0$, we are done. So assume that $0\geq f(0)<1$ and $00$. By Bolzano's Theorem there is a $c\in \loro{0}{1}$ such that $g(c)=0$, that is, $f(c)+c-1=0$, as required. \end{answer} \end{pro} \begin{pro}[Universal Chord Theorem]\label{pro:universal-chord-thm} Suppose that $f$ is a continuous function of $\lcrc{0}{1}$ and that $f(0) = f(1)$. Let $n$ be a strictly positive integer. Prove that there is some number $x \in \lcrc{0}{1}$ such that $f(x) = f(x + 1/n).$ \\ \begin{answer} Consider $g(x) = f(x) - f(x + 1/n)$, which is clearly continuous. If $g$ is never $0$ in $\lcrc{0}{1}$ then by Corollary \ref{cor:retention-of-sign-cont} $g$ must be either strictly positive or strictly negative. But then $$ 0=f(0)-f(1)= \left(f(0)-f\left(\dfrac{1}{n}\right)\right) + \left(f\left(\dfrac{1}{n}\right)-f\left(\dfrac{2}{n}\right)\right) + \left(f\left(\dfrac{2}{n}\right)-f\left(\dfrac{3}{n}\right)\right)+ \cdots + \left(f\left(\dfrac{n-1}{n}\right)-f\left(\dfrac{n}{n}\right)\right). $$The sum of each parenthesis on the right is strictly positive or strictly negative and hence never $0$, a contradiction. \end{answer} \end{pro} \begin{pro} Under the same conditions of problems \ref{pro:universal-chord-thm} prove that there are no universal chords of length $a, 0 < a < 1, a \neq 1/n.$ \begin{answer} Consider the function $f:\lcrc{0}{1}\rightarrow \lcrc{0}{1}$, $x \mapsto \dfrac{\sin \frac{2\pi x}{a}}{\sin \frac{2\pi}{a}}-x$. \end{answer} \end{pro} \end{multicols} \section{Variation of a Function and Uniform Continuity} \begin{df} A {\em partition} $\curlyP$ of the interval $\lcrc{a}{b}$ is any finite set of points $x_0, x_1, \ldots , x_n$ such that $$ a=x_0 0$ there exists a partition of $\lcrc{a}{b}$ into a finite number of subintervals of equal length such that the oscillation of $f$ on each of these subintervals is at most $\varepsilon$. \end{thm} \begin{pf}Let $P_\varepsilon$ mean the following: there is an $\varepsilon >0$ such that for all partitions of $\lcrc{a}{b}$ into a finite number of intervals of equal length, the oscillation of $f$ is $\geq \varepsilon$. By bisecting $\lcrc{a}{b}$, at least one of the halves must have property $P_\varepsilon$, say $\lcrc{a_1}{b_1}$. If $\lcrc{a}{b}$ we to have property $P_\varepsilon$, then by bisecting $\lcrc{a_1}{b_1}$, at least one of the halves must have property $P_\varepsilon$, say $\lcrc{a_2}{b_2}$. Continuing in this way we have constructed a sequence of imbricated intervals $$\lcrc{a}{b} \supseteqq \lcrc{a_1}{b_1} \supseteqq \lcrc{a_2}{b_2} \supseteqq \cdots \supseteqq \lcrc{a_n}{b_n}\supseteqq \cdots $$ where the length of $ \lcrc{a_n}{b_n}$ is $b_n-a_n=\dfrac{b-a}{2^n} \rightarrow 0$ as $\ngroes$. By the Cantor Intersection Theorem, there is a point $c\in \bigcap _{n=1} ^\infty \lcrc{a_n}{b_n}$. Moreover, we have $\omega (f, \lcrc{a_n}{b_n}) \geq \varepsilon$. Since $c\in \lcrc{a}{b}$, $f$ is continuous at $c$. Hence there is a $\delta > 0$ such that $$x\in \loro{c-\delta}{c+\delta}\implies \absval{f(x)-f(c)}<\dfrac{\varepsilon}{2}$$. Taking $(x',x'')\in \loro{c-\delta}{c+\delta}^2$ we have $$ \absval{f(x')-f(x'')} \leq \absval{f(x')-f(c)} + \absval{f(c)-f(x'')} < \varepsilon, $$ whence $$\omega (f, \lcrc{a}{b}\cap \loro{c-\delta}{c+\delta}) < \varepsilon. $$ Now, if there was an $\varepsilon > 0$ such that for all partitions of $\lcrc{a}{b}$ into a finite number of intervals of equal length, the oscillation of $f$ is $\geq \varepsilon$, then by taking $n$ large enough above we could find one of the $\lcrc{a_n}{b_n}$ completely inside one of the subintervals of the partition. By the above, the oscillation there would be $<\varepsilon$, a contradiction. \end{pf} \begin{thm}\label{thm:oscillation-2} Let $f:\lcrc{a}{b} \rightarrow \BBR$ be a continuous function. Given $\varepsilon > 0$ there exists a $\delta >0$ such that on any subinterval $I\subseteqq \lcrc{a}{b}$ having length $<\delta$ the oscillation of $f$ on $I$ is $<\varepsilon$. \end{thm} \begin{pf}Let $\delta = \dfrac{b-a}{n}$. By Theorem \ref{thm:oscillation-1} we may choose $n$ so large that the oscillation of $f$ on each of \begin{equation}\lcrc{a}{a+\delta}, \quad \lcrc{a+\delta}{a+2\delta}, \quad \ldots \quad , \quad \lcrc{a+(n-1)\delta}{b}, \label{eq:subintervals-oscillation} \end{equation} is $<\dfrac{\varepsilon}{2}$. Let $I\subseteqq \lcrc{a}{b}$ be any subinterval of length $<\delta$ and let $x'\in I$ be the point where $f$ achieves its largest value and $x''\in I$ be the point where $f$ achieves its smallest value. Then $x'$ and $x''$ either belong to the same interval in \ref{eq:subintervals-oscillation}---in which case $\absval{ f(x')-f(x'')}<\dfrac{\varepsilon}{2}$---or since $I$ has length smaller than $\delta$, to two consecutive subintervals $$\lcrc{a+(j-1)\delta}{a+j\delta}, \lcrc{a+j\delta}{a+(j+1)\delta}. $$ In this case $$ f(x')-f(x'') = (f(x')-f(a+j\delta)) + (f(a+j\delta)-f(x''))< \dfrac{\varepsilon}{2} + \dfrac{\varepsilon}{2} = \varepsilon.$$ The theorem now follows.\end{pf} \begin{df} A function $f$ is said to be {\em uniformly continuous} on $\lcrc{a}{b}$ if $\forall \varepsilon > 0$ there exists $\delta > 0$ depending only on $\varepsilon$ such that for any $(u, v)\in \lcrc{a}{b}^2 $, $$ \absval{u-v}<\delta \implies \absval{f(u)-f(v)}<\varepsilon . $$ \end{df} \begin{thm}\label{thm:semi-heine} If $f:\lcrc{a}{b}\rightarrow \BBR$ is continuous, then $f$ is uniformly continuous. \end{thm} \begin{pf} This follows from Theorem \ref{thm:oscillation-2}. \end{pf} \begin{thm}[Heine's Theorem] \label{thm:heine} If $f:X\rightarrow \BBR$ is continuous and $X$ is compact, then $f$ is uniformly continuous. \end{thm} \begin{pf} This follows from Theorem \ref{thm:semi-heine}. \end{pf} \begin{thm}\label{thm:-x+y-mono} Let $f$ be an increasing function on an open interval $\loro{a}{b}$. Then, for any $x$ satisfying $a 0$ such that $a0$ be an integer, and let $$ \mathscr{S}_m = \left\{x\in \loro{a}{b}: f(x+)-f(x-)\geq \dfrac{1}{m}\right\}. $$If $x_10$ such that $$\sum _{k=1} ^n \absval{f(x_k)-f(x_{k-1})} \leq V $$for all partitions of $\lcrc{a}{b}$, the we say that {\em $f$ is of bounded variation on $\lcrc{a}{b}$.} \end{df} \begin{thm} If $f$ is monotonic on $\lcrc{a}{b}$, then $f$ is bounded variation on $\lcrc{a}{b}$. \end{thm} \begin{pf} Let $$ a=x_00$ such that $$\absval{f(a)-f(x)}+\absval{f(x)-f(b)}\leq V. $$ But then $$\absval{f(x)} \leq \absval{f(x)-f(a)} + \absval{f(a)} \leq V + \absval{f(a)}. $$and so $f$ is bounded by the constant quantity $V + \absval{f(a)}$. \end{pf} \subsection*{Homework}\addcontentsline{toc}{subsection}{Homework} \begin{multicols}{2}\columnseprule 1pt \columnsep 25pt\multicoltolerance=900\small \begin{pro} Shew that $\loro{0}{+\infty} \rightarrow \loro{0}{+\infty}$, $x \mapsto x^2$, is not uniformly continuous. \end{pro} \end{multicols} \section{Classical Limits} \begin{lem}\label{lem:ineqs-for-exp} If $0 < x \leq 1$ then $$1 \leq \dfrac{\exp (x)-1}{x} \leq 1 + x(e-2).$$ If $-\dfrac{1}{2}\leq x < 0$ then $$ 1 + x\leq \dfrac{\exp (x)-1}{x} \leq 1 + \dfrac{x}{4}. $$ \end{lem} \begin{pf} Since $\left(1+\dfrac{x}{n}\right)^n\leq \exp (x)$ for $n>-x$ by Proposition \ref{prop:e^x}, we have $1+x\leq \exp (x)$ for all $x>-1$. Now, for $n \geq 2$ and $0-2$, $1+x+\dfrac{x^2}{4}=\left(1+\dfrac{x}{2}\right)^2\leq \exp (x)$ by Proposition \ref{prop:e^x}. Now we assume that $-\dfrac{1}{2}\leq x \leq 0$. As before, $$\begin{array}{lll} \left(1 + \dfrac{x}{n}\right)^n & = & 1 + x +x^2\left(\dfrac{1}{2!}\left(\dfrac{1}{n}\right)\left(1-\dfrac{1}{n}\right) +\dfrac{1}{3!}\left(\dfrac{1}{n}\right)\left(1-\dfrac{1}{n}\right) \left(1-\dfrac{2}{n}\right)x +\cdots +\dfrac{1}{n!}\left(\dfrac{1}{n}\right)\left(1-\dfrac{1}{n}\right) \left(1-\dfrac{2}{n}\right)\cdots \left(1-\dfrac{n-1}{n}\right) x^{n-2}\right).\\ \end{array}$$ Since $x^k \leq 0$ for odd $k$ and $x^k \leq \dfrac{1}{2^k}$ for even $k$ we may delete the odd terms from the dextral side and so $$\begin{array}{lll} \left(1 + \dfrac{x}{n}\right)^n & \leq & 1 + x +x^2\left(\dfrac{1}{2!}\left(\dfrac{1}{n}\right)\left(1-\dfrac{1}{n}\right) +0 +\cdots +\dfrac{1}{n!}\left(\dfrac{1}{n}\right)\left(1-\dfrac{1}{n}\right) \left(1-\dfrac{2}{n}\right)\cdots \left(1-\dfrac{2k-1}{n}\right) x^{2k}+ \cdots\right)\\ & \leq & 1 + x + x^2 \left(\dfrac{1}{2}+\dfrac{1}{2^2} + \cdots \right)\\ & \leq & 1 + x + x^2. \end{array}$$On taking limits $\exp (x)\leq 1 + x + x^2$ for $-\dfrac{1}{2}\leq x\leq 0$. Thus we have $$-\dfrac{1}{2}\leq x< 0\implies 1+x+\dfrac{x^2}{4}\leq \exp (x) \leq 1 + x + x^2 \implies 1+x\leq \dfrac{\exp (x)-1}{x}\leq 1 + \dfrac{x}{4}, $$since division by negative $x$ reverses the sense of the inequalities. \end{pf} \begin{thm}$\lim _{x\rightarrow 0} \dfrac{\exp (x)-1}{x}=1$. \label{thm:limit(expx-1)/x}\end{thm} \begin{pf} We prove that $\lim _{x\rightarrow 0+} \dfrac{\exp (x)-1}{x}=1$ and that $\lim _{x\rightarrow 0-} \dfrac{\exp (x)-1}{x}=1$. Let us start with the first assertion. For $00$, $x\mapsto \dfrac{x^2}{1+x}$ is strictly increasing, $\dfrac{x^2(e-2)}{1+x} <\dfrac{e-2}{2}<1$ for $0}(0,0)(-1.5,-1.5)(1.5,1.5) $$\vspace{2cm}\hangcaption{Theorem \ref{thm:limit-sinx/x}.}\label{fig:limit-sinx/x} \end{figure} \begin{thm}If $a\in \BBR$, then $\lim _{x\rightarrow 0} \dfrac{(1+x)^a-1}{x}=a$. \label{thm:limit((1+x)a-1)/x} \end{thm} \begin{pf}This is evident for $a=0$. Assume now $a\neq 0$. Since $x\mapsto \exp (x)$ is continuous and since $a\log (1+x)\to 0$ as $x\to 0$, by Theorems \ref{thm:limit(expx-1)/x} and \ref{thm:limit(log(1+x)-x/x}, $$ \lim _{x\rightarrow 0} \dfrac{(1+x)^a-1}{x} = a\lim _{x\to 0} \dfrac{\exp (a\log (1+x))-1}{a\log (1+x)}\cdot \lim _{x\to 0} \dfrac{\log (1+x)}{x} = a\cdot 1 \cdot 1 =a. $$ \end{pf} \begin{thm}$\lim _{\theta\rightarrow 0} \dfrac{\sin \theta}{\theta}=1$.\label{thm:limit-sinx/x} \end{thm} \begin{pf} We first prove that $\lim _{\theta\to 0+}\dfrac{\sin \theta}{\theta}=1$. Since $\theta\mapsto \dfrac{\sin \theta}{\theta}$ is an even function it will also follow that $\lim _{\theta\to 0-}\dfrac{\sin \theta}{\theta}=1$. \bigskip Assume $0<\theta <\dfrac{\theta}{2}$ and consider $\triangle OAB$ right-angled at $A$, with $OA =1$ and $\angle BOA =\theta$. $C$ is the point where line $OB$ meets the unit circle with centre at $O$ and $D$ is its perpendicular projection. The area of $\triangle OAC$ is smaller than the area of the circular sector $OAC$, which is smaller than the area of $\triangle OAB$. Hence $$\dfrac{1}{2}\sin \theta < \dfrac{\theta }{2}< \dfrac{1}{2}\tan \theta \implies \dfrac{1}{\cos \theta} <\dfrac{\sin \theta}{\theta} < 1 \implies \lim _{\theta \to 0+}\dfrac{\sin\theta}{\theta}=1$$by the Sandwich Theorem, proving the theorem.\end{pf} \chapter{Differentiable Functions} \section{Derivative at a Point} \begin{df} Let $I$ be an interval, $a\in \interiorone{I}$, and $f:I\rightarrow \BBR$. We say that $f$ is differentiable at $a$ if the limit $$\lim _{x\rightarrow a} \dfrac{f(x)-f(a)}{x-a}=\lim _{h\rightarrow 0} \dfrac{f(a+h)-f(a)}{h} $$exists and is finite. In such a case we denote this limit by $f'(a)$, $Df(a)$, or $\dfrac{\d{f}}{\d{x}}(a)$ and we call this quantity {\em the derivative of $f$ at $a$.} \end{df} \begin{df} Let $I$ be an interval, $a\in \interiorone{I}$, and $f:I\rightarrow \BBR$. If $$\lim _{x\rightarrow a+} \dfrac{f(x)-f(a)}{x-a}=\lim _{h\rightarrow 0+} \dfrac{f(a+h)-f(a)}{h} $$exists and is finite we say that {\em $f$ is differentiable at $a$ on the right} and write $f_+ '(a)$ for this limit. If $$\lim _{x\rightarrow a-} \dfrac{f(x)-f(a)}{x-a}=\lim _{h\rightarrow 0-} \dfrac{f(a+h)-f(a)}{h} $$exists and is finite we say that {\em $f$ is differentiable at $a$ on the left} and write $f_- '(a)$ for this limit. \end{df} \begin{thm} Let $I$ be an interval, $a\in \interiorone{I}$, and $f:I\rightarrow \BBR$. Then $f$ is differentiable at $a$ if and only if both $f_+(a)$ and $f_-(a)$ exist and are equal. In this case $f_+(a) = f'(a)=f_-(a)$. \end{thm} \begin{pf} Obvious. \end{pf} \begin{thm} Let $I$ be an interval, $a\in \interiorone{I}$, and $f:I\rightarrow \BBR$. If $f$ is differentiable at $a$ then it is continuous at $a$. \end{thm} \begin{pf} We have $$ \lim _{h\rightarrow 0} f(a+h)-f(a) = \lim _{h\rightarrow 0} \left(\dfrac{f(a+h)-f(a)}{h}\right)h = \left(\lim _{h\rightarrow 0} \dfrac{f(a+h)-f(a)}{h}\right)\left(\lim _{h\rightarrow 0} h\right) = f'(a)\cdot 0 = 0. $$ Thus $\lim _{h\rightarrow 0} f(a+h)-f(a) = 0 \implies \lim _{h\rightarrow 0} f(a+h)=f(a)$ and so $f$ is continuous. \end{pf} \begin{thm} Let $I\subseteqq \BBR$ be an interval. If $f:I\rightarrow \BBR$ is identically constant, then $f'(I) = 0$. \label{thm:derivative-of-constant}\end{thm} \begin{pf}Assume that $f(I)=K$, a constant. Let $c\in\interiorone{I}$. Then $f'(c) = \lim _{x\rightarrow c} \dfrac{f(x)-f(c)}{x-c} =\lim _{x\rightarrow c} \dfrac{K-K}{x-c}=0.$ If $c$ is an endpoint of $I$, then the argument is modified to be either the left or right derivative. \end{pf} \subsection*{Homework}\addcontentsline{toc}{subsection}{Homework} \begin{multicols}{2}\columnseprule 1pt \columnsep 25pt\multicoltolerance=900\small \begin{pro} Let $f:\BBR\rightarrow \BBR$, $$f(x) \left\{ \begin{array}{ll} x+1 & \mathrm{if}\ x\in \BBQ \\ 2-x & \mathrm{if}\ x\in \BBR\setminus \BBQ\end{array}\right. $$ Prove that $f$ is nowhere differentiable. \end{pro} \begin{pro} Let $f:\BBR \rightarrow \BBR$, $x\mapsto \absval{x}$. Prove that $f$ is not differentiable at $x=0$ and that for $x\neq 0$, $f'(x) = \signum{x}$. \end{pro} \begin{pro} Let $f:\BBR \rightarrow \BBR$, $x\mapsto x\absval{x}$. Determine whether $f'(0)$ exists. \end{pro} \end{multicols} \section{Differentiation Rules} \begin{thm}\label{thm:differentiation-rules} Let $I$ be an interval, $a\in \interiorone{I}$, $\lambda \in \BBR$ a constant, and $f, then , g:I\rightarrow \BBR$. If $f$ and $g$ are differentiable at $a$ then \begin{enumerate} \item {\bf (Linearity Rule)} $f+\lambda g$ is differentiable at $a$ and $(f+\lambda g)'(a)=f'(a) + \lambda g'(a)$ \item {\bf (Product Rule)} $fg$ is differentiable at $a$ and $(fg)'(a)=f'(a)g(a) + f(a)g'(a)$ \item if $g(a)\neq 0$, $\dfrac{1}{g}$ is differentiable at $a$ and $\left(\dfrac{1}{g}\right)'(a)=-\dfrac{g'(a)}{(g(a))^2}$ \item {\bf (Quotient Rule)} if $g(a)\neq 0$, $\dfrac{f}{g}$ is differentiable at $a$ and $\left(\dfrac{f}{g}\right)'(a)=\dfrac{f'(a)g(a)-f(a)g'(a)}{(g(a))^2}$ \end{enumerate} \end{thm} \begin{pf} \begin{enumerate} \item This follows by the linearity of limits. \item We have $$\begin{array}{lll}(fg)'(a) & = & \lim _{h\rightarrow 0} \dfrac{(fg)(a+h)-(fg)(a)}{h}\\ & = & \lim _{h\rightarrow 0} \dfrac{g(a+h)(f(a+h)-f(a))+f(a)(g(a+h)-g(a))}{h}\\ & = & \lim _{h\rightarrow 0}g(a+h)\lim _{h\rightarrow 0} \dfrac{(f(a+h)-f(a))}{h}+ \lim _{h\rightarrow 0}f(a)\lim _{h\rightarrow 0}\dfrac{(g(a+h)-g(a))}{h}\\ & = & g(a)f'(a)+f(a)g'(a), \end{array}$$as desired. \item We have $$\begin{array}{lll}\left(\dfrac{1}{g}\right)'(a) & = & \lim _{h\rightarrow 0} \dfrac{\dfrac{1}{g(a+h)}-\dfrac{1}{g(a)}}{h}\\ & = & \lim _{h\rightarrow 0} \dfrac{\dfrac{g(a)-g(a+h)}{g(a+h)g(a)}}{h}\\ & = & \lim _{h\rightarrow 0} \dfrac{g(a)-g(a+h)}{h}\lim _{h\rightarrow 0} \dfrac{1}{g(a+h)g(a)}\\ & = & \left(-g'(a)\right) \left(\dfrac{1}{g(a)g(a)}\right)\\ & = & -\dfrac{g'(a)}{g(a)^2},\\ \end{array}$$as desired. \item Using (2) and (3), $$\begin{array}{lll}\left(\dfrac{f}{g}\right)'(a) & = & f'(a)\left(\dfrac{1}{g}\right)(a) + f(a)\left(\dfrac{1}{g}\right)'(a)\\ & = & \dfrac{f'(a)}{g(a)} -\dfrac{f(a)g'(a)}{g(a)^2}\\ & = & \dfrac{f'(a)g(a)-f(a)g'(a)}{(g(a))^2}, \end{array} $$as desired. \end{enumerate} \end{pf} \begin{thm}[Chain Rule] Let $I, J$ be intervals of $\BBR$, with $a\in I$. Let $f:I\rightarrow \BBR$ and $g:J\rightarrow \BBR$ be such that $f(I)\subseteqq J$. If $f$ is differentiable at $a$ and $g$ is differentiable at $f(a)$, then $g\circ f$ is differentiable at $a$ and $(g\circ f)' = g'(f(a))f'(a)$.\label{thm:chain-rule} \end{thm} \begin{pf}Put $b = f(a)$, and $$\varphi(y) = \left \{ \begin{array}{ll} \frac{g(y) - g(b)}{y-b} & \textrm{if $y \neq b$} \\ g'(b) & \textrm{if $y = b$} \end{array} \right. $$ Since $g$ is differentiable at $b$, $\varphi$ is continuous at $y=b$. Now, for $x \neq a$, $$\dfrac{g(f(x))-g(f(a))}{x-a} = \varphi(f(x)) \dfrac{f(x) - f(a)}{x-a}. $$ (If $f(x) \neq f(a)$ this follows directly from the definition of $\varphi$. If $f(x) = f(a)$, both sides of the equality are $0$.) By the continuity of $f$ at $a$ and of $\varphi$ at $b$, $$\lim_{x \to a} \varphi(f(x)) = \varphi(f(a)) = g'(f(a)), $$ whence \begin{eqnarray*} (g \circ f)'(a) &=& \lim_{x\to a} \frac{g(f(x))-g(f(a))}{x-a} \\ &=& \lim_{x\to a} \varphi(f(x)) \frac{f(x) - f(a)}{x-a} \\ &=& g'(f(a))f'(a), \end{eqnarray*}as desired. \end{pf} \begin{thm}[Inverse Function Rule] Let $I$ be an interval of $\BBR$, with $a\in I$. Let $f:I\rightarrow \BBR$ be strictly monotonic and continuous over $I$. If $f$ is differentiable at $a$ and $f'(a) \neq 0$, then the inverse $f^{-1}: f(I)\rightarrow \BBR$ is differentiable at $f(a)$ and $$ (f^{-1})'(f(a))=\dfrac{1}{f'(a)}. $$ \end{thm} \begin{pf} Put $b= f(a)$. Observe that $ \lim _{ y\rightarrow b} f^{-1}(y) = a$, and by the composition rule for limits, $$ \lim _{ y\rightarrow b} \dfrac{f^{-1}(y)-f^{1}(b)}{y-a} = \lim _{ y\rightarrow b} \dfrac{f^{-1}(y)-a}{f(f^{-1}(y))-a} = \dfrac{1}{f'(a)},$$proving the theorem. \end{pf} \begin{rem} Once it is known that $ (f^{-1})'$ exists, we may proceed as follows. Since $f^{-1}(f(x))=x$, differentiating on both sides, using the Chain Rule on the sinistral side, $$ (f^{-1})'(f(x))f'(x) =1, $$from where the result follows. \end{rem} \begin{df} Let $I$ be an interval of $\BBR$. Let $f:I\rightarrow \BBR$ be differentiable at every point of $I$. The function $f':I\rightarrow \BBR$, $x\mapsto f'(x)$ is called the {\em derivative function} or {\em derivative} of the function $f$. \end{df} \begin{thm} Let $n\geq 0$ be an integer. Let $f:\BBR\rightarrow \BBR$, $x\mapsto x^n$. Then $f$ is everywhere differentiable and $f':\BBR\rightarrow \BBR$ is given by $x\mapsto nx^{n-1}$. \label{thm:power-rule-1} \end{thm} \begin{pf} Assume first $n$ is strictly positive. By Theorem \ref{thm:diffbinom}, $$\begin{array}{lll}\lim _{x\rightarrow a} \dfrac{x^n-a^n}{x-a} & = & \lim _{x\rightarrow a} \dfrac{(x-a)(x^{n-1}+ax^{n-2}+a^2x^{n-3}+\cdots + a^{n-2}x + a^{n-1})}{x-a} \\ & = & \lim _{x\rightarrow a} (x^{n-1}+ax^{n-2}+a^2x^{n-3}+\cdots + a^{n-2}x + a^{n-1}) \\ & = & na^{n-1}.\end{array} $$Observe that this is true for all $a\in \BBR$. \bigskip If $n=0$ then $f$ is constant, say $f(x)=K$ for all $x$ and so $$ \lim _{x\rightarrow a} \dfrac{f(x)-f(a)}{x-a} = \lim _{x\rightarrow a} \dfrac{K-K}{x-a} = 0. $$ \end{pf} \begin{thm}\label{thm:power-rule-2} Let $n>0$ be an integer and $f:\loro{0}{+\infty} \rightarrow \loro{0}{+\infty}$, $x\mapsto \dfrac{1}{x^n}$. Then $f'$ exists everywhere in $\loro{0}{+\infty}$ and $f':\loro{0}{+\infty} \rightarrow \loro{0}{+\infty}$ is given by $f'(x) = -\dfrac{n}{x^{n+1}}$. \end{thm} \begin{pf} We use the result above, part (3) of Theorem \ref{thm:differentiation-rules}, and the Chain Rule, to get $$\dfrac{\d{}}{\d{x}}\dfrac{1}{x^{n}} = -\dfrac{nx^{n-1}}{(x^n)^2} = -\dfrac{n}{x^{n+1}}, $$ and the theorem follows.\end{pf} \begin{lem}\label{lem:power-rule} Let $q\in\BBZ$, $q>0$ be an integer, and $f:\loro{0}{+\infty} \rightarrow \loro{0}{+\infty}$, $x\mapsto x^{1/q}$. Then $f'$ exists everywhere in $\loro{0}{+\infty}$ and $f':\loro{0}{+\infty} \rightarrow \loro{0}{+\infty}$ is given by $f'(x) = \dfrac{x^{1/q-1}}{q}$. \end{lem} \begin{pf} We have $(f(x))^q=x$. Using the Chain Rule $qf'(x)(f(x))^{q-1}=1$. Since $f(x)\neq 0$, $$ f'(x) = \dfrac{1}{q(f(x))^{q-1}} = \dfrac{1}{q(x^{1/q})^{q-1}} = \dfrac{1}{q}x^{1/q-1}. $$ \end{pf} \begin{thm}\label{thm:power-rule-3} Let $r\in \BBQ$ and let $f:\loro{0}{+\infty} \rightarrow \loro{0}{+\infty}$, $x\mapsto x^r$. Then $f'$ exists everywhere in $\loro{0}{+\infty}$ and $f':\loro{0}{+\infty} \rightarrow \loro{0}{+\infty}$ is given by $f'(x) = rx^{r-1}$. \end{thm} \begin{pf} Let $r=\dfrac{a}{b}$, where $a, b$ are integers, with $b>0$. We use the Chain Rule, Lemma \ref{lem:power-rule}, and Theorem \ref{thm:power-rule-2}. Then $$\dfrac{\d{}}{\d{x}}x^{a/b} = \dfrac{\d{}}{\d{x}}(x^{1/b})^a = a(x^{1/b})^{a-1}\cdot \dfrac{1}{b}x^{1/b-1} = \dfrac{a}{b}x^{a/b-1} = rx^{r-1}, $$ proving the theorem. \end{pf} \begin{thm}[Derivative of the Exponential Function] Let $\exp :\BBR\rightarrow \BBR$, $x\mapsto e^x$. Then $\exp$ is everywhere differentiable and $\exp ':\BBR\rightarrow \BBR$ is given by $x\mapsto e^x$. \label{thm:derivative-of-exp} \end{thm} \begin{pf} Using Theorem \ref{thm:limit(expx-1)/x}, we have, with $h=x-a$, $$\begin{array}{lll} \lim _{x\rightarrow a}\dfrac{e^x-e^a}{x-a} & = & e^a\lim _{x\rightarrow a}\dfrac{e^{x-a}-1}{x-a}\\ &= & e^a \lim _{h\rightarrow 0} \dfrac{e^h-1}{h}\\ & = & e^a\cdot 1\\ & =& e^a. \end{array}$$ \end{pf} \begin{thm}[Derivative of the Logarithmic Function]\label{thm:derivative-of-log} Let $f:\loro{0}{+\infty} \rightarrow \loro{-\infty}{+\infty}$, $x\mapsto \log x$. Then $f'$ exists everywhere in $\loro{0}{+\infty}$ and $f':\loro{0}{+\infty} \rightarrow \BBR\setminus \{0\}$ is given by $f'(x) = \dfrac{1}{x}$. \end{thm} \begin{pf} Let $a>0$. Then, with $h=\dfrac{x}{a}-1$, and using Theorem \ref{thm:limit(log(1+x)-x/x}, $$\begin{array}{lll} \lim _{x\rightarrow a}\dfrac{\log x-\log a}{x-a} & = & \lim _{x\rightarrow a}\dfrac{\log \dfrac{x}{a}}{x-a}\\ & = & \dfrac{1}{a}\cdot \lim _{x\rightarrow a}\dfrac{\log \left(1+\dfrac{x}{a}-1\right)}{\dfrac{x}{a}-1}\\ &= & \dfrac{1}{a}\cdot \lim _{h\rightarrow 0} \dfrac{\log (1+h)}{h}\\ & = & \dfrac{1}{a}\cdot 1\\ & =& \dfrac{1}{a}. \end{array}$$ \end{pf} \begin{thm}[Power Rule]\label{thm:power-rule-4} Let $t\in \BBR$ and let $f:\loro{0}{+\infty} \rightarrow \loro{0}{+\infty}$, $x\mapsto x^t$. Then $f'$ exists everywhere in $\loro{0}{+\infty}$ and $f':\loro{0}{+\infty} \rightarrow \loro{0}{+\infty}$ is given by $f'(x) = tx^{t-1}$. \end{thm} \begin{pf} Using the Chain Rule, $$ \dfrac{\d{}}{\d{x}}x^t = \dfrac{\d{}}{\d{x}}\left(\exp (t\log x)\right) = \dfrac{t}{x}\cdot \left(\exp (t\log x)\right) = \dfrac{t}{x}\cdot x^t = tx^{t-1}. $$ \end{pf} \begin{thm}[Derivative of $\sin$]. Let $\sin :\BBR\rightarrow \BBR$, $x\mapsto \sin x$. Then $\sin$ is everywhere differentiable and $\sin ':\BBR\rightarrow \BBR$ is given by $x\mapsto \cos x$. \label{thm:derivative-of-sin} \end{thm} \begin{pf} We make a change of variables, and use Theorem \ref{thm:limit-sinx/x}, $$\begin{array}{lll} \lim _{x\rightarrow a}\dfrac{\sin x-\sin a}{x-a} & = & \lim _{x\rightarrow a}\dfrac{\sin (x-a+a)-\sin a}{x-a}\\ & = & \lim _{x\rightarrow a}\dfrac{\sin (x-a)\cos a+\cos(x-a)\sin a-\sin a}{x-a}\\ & = & (\cos a)\lim _{x\rightarrow a}\dfrac{\sin (x-a)}{x-a} + (\sin a)\lim _{x\rightarrow a}\dfrac{\cos(x-a)-1}{x-a}\\ & = & (\cos a)\lim _{h\rightarrow 0}\dfrac{\sin h}{h} + (\sin a)\lim _{h\rightarrow 0}\dfrac{\cos h-1}{h}\\ & = & (\cos a)\cdot 1 + (\sin a)\lim _{h\rightarrow 0}\dfrac{\cos^2 h-1}{h(\cos h +1)}\\ & = & (\cos a)\cdot 1 + (\sin a)\lim _{h\rightarrow 0}\dfrac{-\sin^2h}{h(\cos h +1)}\\ & = & (\cos a) + (\sin a)\lim _{h\rightarrow 0}\dfrac{\sin h}{h}\cdot \lim _{h\rightarrow 0}\dfrac{-\sin h}{\cos h +1}\\ & = & \cos a, \end{array} $$and the theorem follows. \end{pf} \begin{thm}[Derivatives of the Goniometric Functions]\label{thm:derivatives-goniometric} $$ \begin{array}{lllll}1. & \dfrac{\d{}}{\d{x}} \sin x & = & \cos x & x\in \BBR \\ 2. & \dfrac{\d{}}{\d{x}} \cos x & = & \sin x & x\in \BBR \\ 3. & \dfrac{\d{}}{\d{x}} \tan x & = & \sec^2 x & x\in \BBR \setminus (2\BBZ+1)\dfrac{\pi}{2}\\ 4. & \dfrac{\d{}}{\d{x}} \sec x & = & \sec x\tan x & x\in \BBR \setminus (2\BBZ+1)\dfrac{\pi}{2}\\ 5. & \dfrac{\d{}}{\d{x}} \csc x & = & -\csc x\cot x & x\in \BBR \setminus \BBZ\pi\\ 6. & \dfrac{\d{}}{\d{x}} \cot x & = & -\csc^2x & x\in \BBR \setminus \BBZ\pi\\ \end{array}$$ \end{thm} \begin{pf} (1) is Theorem \ref{thm:derivative-of-sin}. To prove (2), observe that $$\dfrac{\d{}}{\d{x}} \cos x = \dfrac{\d{}}{\d{x}} \sin \left(\dfrac{\pi}{2}-x\right) = -\cos \left(\dfrac{\pi}{2}-x\right) = -\sin x. $$ To prove (3), we use the Quotient Rule, $$\dfrac{\d{}}{\d{x}} \tan x = \dfrac{\d{}}{\d{x}} \dfrac{\sin x}{\cos x} = \dfrac{(\cos x)(\cos x)-(-\sin x)(\sin x)}{\cos^2x} =\dfrac{1}{\cos^2x}=\sec^2x. $$ To prove (4), we use once again the Quotient Rule, $$\dfrac{\d{}}{\d{x}} \sec x = \dfrac{\d{}}{\d{x}} \dfrac{1}{\cos x} = \dfrac{(0)(\cos x)-(-\sin x)(1)}{\cos^2x} =\dfrac{\sin x}{\cos^2x}=\sec x\tan x. $$ To prove (5), observe that $$\dfrac{\d{}}{\d{x}} \csc x = \dfrac{\d{}}{\d{x}} \sec \left(\dfrac{\pi}{2}-x\right) = -\sec \left(\dfrac{\pi}{2}-x\right)\tan \left(\dfrac{\pi}{2}-x\right) = -\csc x\cot x. $$ To prove (6), observe that $$\dfrac{\d{}}{\d{x}} \cot x = \dfrac{\d{}}{\d{x}} \tan \left(\dfrac{\pi}{2}-x\right) = -\sec ^2 \left(\dfrac{\pi}{2}-x\right) = -\csc^2 x. $$ \end{pf} \begin{df}[Higher Order Derivatives] Let $I$ be an interval of $\BBR$ and let $f:I\rightarrow \BBR$. For $a\in I$ we define the successive derivatives of $f$ at $a$, inductively. Put $f(a) = f^{(0)}(a)$. If $n \geq 1$, $$ f^{(n)}(a) = f'(f^{(n-1)}(a)), $$provided $f$ is differentiable at $f^{(n-1)}(a)$. \end{df} \begin{rem} We usually write $f''$ instead of $f^{(2)}$. \end{rem} \begin{thm}[Leibniz's Rule]Let $n$ be a positive integer. $$ (fg)^{(n)} = \sum _{k=0} ^n \binom{n}{k}f^{(k)}g^{(n-k)} $$ \end{thm} \begin{pf} This is a generalisation of the Product Rule. The proof is by induction on $n$. For $n=0$ and $n=1$ the assertion is obvious. Assume that $ (fg)^{(n)} = \sum _{k=0} ^n \binom{n}{k}f^{(k)}g^{(n-k)} $. Observe that $$\begin{array}{lll}(fg)^{(n+1)} & = & ((fg)^{(n)})'\\ & = & \left(\sum _{k=0} ^n \binom{n}{k}f^{(k)}g^{(n-k)}\right)'\\ & = & \sum _{k=0} ^n \binom{n}{k}(f^{(k+1)}g^{(n-k)}+f^{(k)}g^{(n-k+1)})\\ & = & \sum _{k=0} ^n \binom{n}{k}f^{(k+1)}g^{(n-k)}+\sum _{k=0} ^n \binom{n}{k}f^{(k)}g^{(n-k+1)}\\ & = & f^{(0)}g^{(n+1)}+ \sum _{k=0} ^n \left(\binom{n}{k} +\binom{n}{k+1} \right)f^{(k)}g^{(n+1-k)} + f^{(n+1)}g^{(0)}\\ & = & \sum _{k=0} ^{n+1} \binom{n+1}{k}f^{(k)}g^{(n+1-k)}, \end{array} $$proving the statement. \end{pf} \subsection*{Homework}\addcontentsline{toc}{subsection}{Homework} \begin{multicols}{2}\columnseprule 1pt \columnsep 25pt\multicoltolerance=900\small \begin{pro} Prove that $$ \dfrac{2}{x^2 -1} = \dfrac{1}{x-1} - \dfrac{1}{x+1} $$and use this result to find the $100$th derivative of $f(x) = \dfrac{2}{x^2 -1}$. \begin{answer} Observe that that $$ \dfrac{1}{x-1} - \dfrac{1}{x+1} = \dfrac{(x+1) - (x-1)}{(x-1)(x+1)}= \dfrac{2}{x^2 -1}. $$ If $f(x) = (x-1)^{-1}$ then $$f'(x) = -1(x-1)^{-2}; f''(x) = (-1)(-2)(x-1)^{-3}; (-1)(-2)(-3)(x-1)^{-4}; \ldots ; f^{(100)}(x) = 100!(x-1)^{-101}.$$Similarly, if $g(x) = (x+1)^{-1}$ then $$g'(x) = -1(x+1)^{-2}; g''(x) = (-1)(-2)(x+1)^{-3}; (-1)(-2)(-3)(x+1)^{-4}; \ldots ; g^{(100)}(x) = 100!(x+1)^{-101}.$$Hence $$ \dfrac{\mathrm{d}^{100}}{\d{x}^{100}} \ \dfrac{2}{x^2 -1} = f^{(100)}(x) - g^{(100)}(x) = 100!(x-1)^{-101} - 100!(x+1)^{-101}. $$ \end{answer} \end{pro} \begin{pro} Find the $100$-th derivative of $x\mapsto x^{2}\sin x$. \begin{answer} We use Leibniz's Rule and the observation that the third derivative of $x\mapsto x^{2}$ is $0$. Also $(\sin x)^{(4n)} = \sin x$, $(\sin x)^{(4n+2)} = -\sin x$, $(\sin x)^{(4n+1)} = \cos x$, and $(\sin x)^{(4n+3)} = -\cos x$, Then $$ \dfrac{\d{}^{100}}{\d{x^{100}}}x^{2}\sin x = \binom{100}{0}x^2(\sin x)^{(100)} + \binom{100}{1}(x^2)'(\sin x)^{(99)}+ \binom{100}{2}(x^2)''(\sin x)^{(98)} = x^2\sin x-200x\cos x-9900\sin x. $$ \end{answer} \end{pro} \begin{pro} Demonstrate that the polynomial $p(x)\in\BBR [x]$ has a zero at $x=a$ of multiplicity $k$ if and only if $$ p(a) = p'(a) =\cdots = p^{(k-1)}(a) = 0. $$ \end{pro} \begin{pro} Demonstrate that if for all $x\in \BBR$ there holds the identity $$ \sum _{k=0} ^n a_k(x-a)^k=\sum _{k=0} ^n b_k(x-b)^k, $$then $a_k = \sum _{j=k} ^n \binom{n}{j}b_j(a-b)^{j-k}.$ \end{pro} \begin{pro} Let $p$ be a polynomial of degree $r$ and consider the polynomial $F$ with $$F(x) = p(x)+p'(x)+p''(x) + \cdots + p^{(r)}(x). $$Prove that $$ \dfrac{\d{\left(F(x)\exp (-x)\right)}}{\d{x}} = -\exp (-x)p(x). $$ \end{pro} \end{multicols} \section{Rolle's Theorem and the Mean Value Theorem} \begin{thm}[Rolle's Theorem] Let $(a, b)\in \BBR^2$ such that $a0$ such that $\loro{c-\varepsilon}{c+\varepsilon}\subseteqq I$ and $\forall x\in \loro{c-\varepsilon}{c+\varepsilon}, f'(x) =0 $. By Theorem \ref{thm:derivative-0-gives-const-fun}, $f$ must be constant on $\loro{c-\varepsilon}{c+\varepsilon}$ and so it is not strictly increasing, a contradiction. If $f$ is strictly decreasing, we apply what has been proved to $-f$. \item[$\Leftarrow$] Conversely, suppose that $\forall x\in I$, $f'(x) \geq 0$. and the set $\interior{\{x\in I^\circ : f'(x) =0\}} = \varnothing$. From Theorem \ref{thm:increasing-iff-firstderiv-pos}, $f$ is increasing on $I$. Suppose that there exist $(a, b)\in I^2$, $a 0$ for $x \in \loro{0}{1}.$ Is there a number $d \in \loro{0}{1}$ such that $$ \frac{2f'(c)}{f(c)} = \frac{f'(1 - c)}{f(1 - c)} ?$$ \begin{answer} Set $g(x) = f(x)^2f(1-x)$. Since $g(0) = g(1)=0$, $g$ satisfies the hypotheses of Rolle's Theorem. There is a $c\in\loro{0}{1}$ such that $$ g'(c)= 0 \implies 2f'(c)f(c)f(1-c) - f(c)^2f'(1-c)=0. $$Since by assumption $f(c)f(1-c) \neq 0$ we must have, upon dividing by every term by $f(c)^2f(1-c)$, the assertion. \end{answer} \end{pro} \begin{pro} Let $n\geq 1$ be an integer and let $f:[0;1]\rightarrow \BBR$ be differentiable and such that $f(0)=0$ and $f(1)=1$. Prove that there exist distinct points $0 < a_0< a_2< \cdots < a_{n-1} < 1$ such that $$ \sum _{k=0} ^{n-1} f'(a_k)=n. $$ \begin{answer} For $0\leq k \leq n-1$, consider the interval $\lcrc{\dfrac{k}{n}}{\dfrac{k+1}{n}}$. By the Mean Theorem, there are $a_k\in\loro{\dfrac{k}{n}}{\dfrac{k+1}{n}}$ such that $$ f'(a_k) = \dfrac{f\left(\dfrac{k+1}{n}\right)-f\left(\dfrac{k}{n}\right)}{\dfrac{1}{n}} = n\left(f\left(\dfrac{k+1}{n}\right)-f\left(\dfrac{k}{n}\right)\right). $$Summing from $k=0$ to $k=n-1$ and noting that the dextral side telescopes, $$ \sum _{k=0} ^{n-1}f'(a_k) = n\sum _{k=0} ^{n-1} \left(f\left(\dfrac{k+1}{n}\right)-f\left(\dfrac{k}{n}\right)\right) =n(f(1)-f(0)) =n. $$ \end{answer} \end{pro} \begin{pro} Let $n\geq 1$ be an integer and let $f:[0;1]\rightarrow \BBR$ be differentiable and such that $f(0)=0$ and $f(1)=1$. Prove that there exist distinct points $0 < a_0< a_2< \cdots < a_{n-1} <1$ such that $$ \sum _{k=0} ^{n-1 } \dfrac{1}{f'(a_k)}=n. $$ \begin{answer} Let $k_i\in \lcrc{0}{1}$ be the smallest number such that $f(k_i) = \dfrac{i}{n}$, $1\leq i \leq n-1$. Put $k_0=0, k_n=1$. The existence of the $k_i$ is guaranteed by the Intermediate Value Theorem. Moreover, since the $k_i$ are chosen to be the first time $f$ is $\dfrac{i}{n}$, once again, by the Intermediate Value Theorem we must have $$0f(a)$.\item We say that $f$ has a {\em local extremum at $a$} if $f$ has either a local maximum or a local minimum at $a$.\item We say that $f$ has a {\em strict local extremum at $a$} if $f$ has either a strict local maximum or a strict local minimum at $a$. The plural of extremum is {\em extrema}. \end{enumerate} \end{df} \begin{thm} If $f:I\rightarrow \BBR$ is continuous on the interval $I$, differentiable on $\interiorone{I}$, and if $f$ has a local extremum at $a\in \interiorone{I}$, then $f'(a) = 0$. \end{thm} \begin{pf} Suppose $f$ admits a local maximum at $a$. Let $h\neq 0$ be so small that $a+h\in I$. Now $$ h>0 \implies \dfrac{f(a+h)-f(a)}{h} \leq 0, \qquad h< 0 \implies \dfrac{f(a+h)-f(a)}{h} \geq 0. $$ Upon taking limits as $h\rightarrow 0$, $f'(a)\leq 0$ and $f'(a)\geq 0$, whence $f'(a)=0$. \end{pf} \begin{df} Let $f:I\rightarrow \BBR$. The points $x\in I$ where $f'(x)=0$ are called {\em critical points or stationary points} of $f$. \end{df} \begin{thm} Let $f:\lcrc{a}{b}\rightarrow \BBR$ be a twice differentiable function having a critical point at $c\in\loro{a}{b}$. If $f''(c)<0$ then $f$ has a relative maximum at $x=c$, and if $f''(c)>0$ then $f$ has a relative minimum at $x=c$. \end{thm} \begin{pf} Assume that $f'(c)=0 $ and $f''(c)<0$. Since $$\lim _{x\rightarrow c} \dfrac{f'(x)}{x-c} = \lim _{x\rightarrow c} \dfrac{f'(x)-f'(c)}{x-c} = f''(c)<0, $$ there exists $\delta >0$ such that $f'(x)>0$ when $c-\delta 0$ when $c0$ then we apply what has been proved to $-f$. \end{pf} \begin{thm}[Darboux's Theorem]\label{thm:darboux} Let $f$ be differentiable on $\lcrc{a}{b}$ and suppose that $f'(a)0$. Find the minimum value of $f$. \begin{answer} We have $f'(x) = x^x(\log x + 1)$ whence $f'(x) = 0 \implies x = e^{-1}$. Since $f'(x) < 0 $ for $0 < x < e^{-1}$ and $f'(x) > 0$ for $x> e^{-1}$, $x = e^{-1}$ is a local (relative) minimum. Thus $f(x) \geq f(e^{-1}) = \left(\dfrac{1}{e}\right)^{1/e}$. \end{answer} \end{pro} \end{multicols} \section{Convex Functions} \begin{df} Let $I \subseteqq \BBR$ be an interval. A function $f:I\rightarrow \BBR$ is said to be {\em convex} if $$\forall (a, b)\in I^2, \forall \lambda \in \lcrc{0}{1}, f(\lambda a + (1-\lambda)b)\leq \lambda f(a) + (1-\lambda)f(b). $$ We say that $f$ is {\em concave} if $-f$ is convex. \end{df} \begin{rem}$f$ is convex if given any two points on its graph, the straight line joining these two points lies above the graph of $f$. See figure \ref{fig:convex}. \end{rem} \vspace{1cm} \begin{figure}[h] \centering \begin{minipage}{7cm} $$ \psset{unit=.8pc} \rput(0,-4){\parabola[linewidth=2pt,linecolor=red]{<->}(2.25,5.0625)(0,0) \psline[linewidth=2pt,linecolor=blue](2,4)(-1,1)\psdots[dotscale=1,dotstyle=*](2,4)(-1,1)(.5,2.5)(.5,.25) } $$\vspace{1cm}\footnotesize \hangcaption{ A convex curve }\label{fig:convex}\end{minipage} \begin{minipage}{7cm} $$ \psset{unit=.8pc} \parabola[linewidth=2pt,linecolor=red]{<->}(-2.25,-5.0625)(0,0) \psline[linewidth=2pt,linecolor=blue](2,-4)(-1,-1)\psdots[dotscale=1,dotstyle=*](2,-4)(-1,-1)(.5,-2.5)(.5,-.25) $$\vspace{1cm}\footnotesize \hangcaption{ A concave curve. }\label{fig:concave}\end{minipage} \end{figure} \begin{df} Let $(x_1, x_2, \ldots , x_n)\in \BBR^n$ and let $\lambda _k\in \lcrc{0}{1}$ be such that $\sum _{k=1} ^n \lambda _k =1$. The sum $$ \sum _{k=1} ^n \lambda _k x_k $$ is called a {\em convex combination} of the $x_k$. \end{df} \begin{thm}\label{thm:convex-combinations-in-intervals} If $(x_1, x_2, \ldots , x_n)\in \lcrc{a}{b}^n$, then any convex combination of the $x_k$ also belongs to $\lcrc{a}{b}$. \end{thm} \begin{pf} Assume $\lambda _k\in \lcrc{0}{1}$ be such that $\sum _{k=1} ^n \lambda _k =1$. Since the $\lambda _k \geq 0$ we have $$a\leq x_k \leq b \implies \lambda _ka \leq \lambda _kx_k \leq \lambda _kb. $$ Adding, and bearing in mind that $\sum _{k=1} ^n \lambda _k =1$, $$ \left(\sum _{k=1} ^n \lambda _k\right)a \leq \sum _{k=1} ^n \lambda _kx_k \leq \left(\sum _{k=1} ^n \lambda _k\right)b \implies a \leq \sum _{k=1} ^n \lambda _kx_k \leq b, $$ proving the theorem. \end{pf} \begin{thm}[Jensen's Inequality]\label{thm:Jensen}Let $I \subseteqq \BBR$ be an interval and let $f:I\rightarrow \BBR$ be a convex function. Let $n\geq 1$ be an integer, $x_k\in I$, and $\lambda _k\in \lcrc{0}{1}$ be such that $\sum _{k=1} ^n \lambda _k =1$. Then $$f\left(\sum _{k=1} ^n \lambda _kx_k\right) \leq \sum _{k=1} ^n \lambda _kf(x_k). $$ \end{thm} \begin{pf} The proof is by induction on $n$. For $n=2$ we must shew that given $(x_{1}, x_{2})\in \lcrc{a}{b}^2$, $$ f\left( \lambda _{1}x_{1}+\lambda _{2}x_{2}\right) \leq \lambda _{1}f(x_{1})+\lambda _{2}f(x_{2}).$$ As $\lambda _{1}+\lambda _{2} =1$, we may put $\lambda = \lambda _{2}=1-\lambda _{1}$ and so the above inequality becomes $$f\left( \lambda x_{1}+\left( 1-\lambda \right) x_{2}\right) \leq \lambda f(x_{1})+\left( 1-\lambda \right) f(x_{2}),$$retrieving the definition of convexity. \bigskip Assume now that $f\left( \sum_{k=1}^{n-1}\mu _{k}x_{k}\right) \leq \sum_{k=1}^{n-1}\mu _{k}f\left( x_{k}\right) $, when $\sum_{k=1}^{n-1}\mu _{k}=1$, $\mu _k\in\loro{0}{1}$. We must prove that $f\left( \sum_{k=1}^{n}\lambda _{k}x_{k}\right) \leq \sum_{k=1}^{n}\lambda _{k}f\left( x_{k}\right) $, when $\sum_{k=1}^{n}\lambda _{k}=1$, $\lambda _k\in\loro{0}{1}$. \bigskip If $\lambda _n = 1$ the assertion is trivial, since then $\lambda_1 = \cdots = \lambda _{n-1}=0$. So assume that $\lambda _n \neq 1$. Observe that $\sum_{k=1}^{n-1}\frac{\lambda _{k}}{1-\lambda _{n}}=\frac{\left( \sum_{k=1}^{n}\lambda _{k}\right) -\lambda _{n}}{1-\lambda _{n}}=\frac{% 1-\lambda _{n}}{1-\lambda _{n}}=1$ so that $\sum_{k=1}^{n-1}\frac{\lambda _{k}% }{1-\lambda _{n}}x_{k}$ is a convex combination of the $x_k$ and hence also belongs to $\lcrc{a}{b}$, by Theorem \ref{thm:convex-combinations-in-intervals}. Since $f$ is convex, $$\begin{array}{lll}f\left( \sum_{k=1}^{n}\lambda _{k}x_{k}\right) & = & f\left( \sum_{k=1}^{n-1}\lambda _{k}x_{k}+\lambda _{n}x_{n}\right)\\ & = & f\left( \left( 1-\lambda _{n}\right) \sum_{k=1}^{n-1}\frac{\lambda _{k}}{1-\lambda _{n}} x_{k}+\lambda _{n}x_{n}\right)\\ & \leq & \left( 1-\lambda _{n}\right) f\left( \sum_{k=1}^{n-1}\frac{\lambda _{k}% }{1-\lambda _{n}}x_{k}\right) +\lambda _{n}f\left( x_{n}\right) \end{array}$$ By the inductive hypothesis, with $\mu_k =\frac{\lambda _{k}}{1-\lambda _{n}}=1$, $$f\left( \sum_{k=1}^{n-1}\frac{\lambda _{k}}{1-\lambda _{n}}x_{k}\right) \leq \sum_{k=1}^{n-1}\frac{\lambda _{k}}{1-\lambda _{n}}f\left( x_{k}\right). $$Finally, we gather, $$\begin{array}{lll}f\left( \sum_{k=1}^{n}\lambda _{k}x_{k}\right) &\leq & \left( 1-\lambda _{n}\right) f\left( \sum_{k=1}^{n-1}\frac{\lambda _{k}}{1-\lambda _{n}} x_{k}\right) +\lambda _{n}f\left( x_{n}\right) \\ & \leq & \left( 1-\lambda _{n}\right) \sum_{k=1}^{n-1}\frac{\lambda _{k}}{% 1-\lambda _{n}}f\left( x_{k}\right) +\lambda _{n}f\left( x_{n}\right)\\ & = & \sum_{k=1}^{n-1}\lambda _{k}f\left( x_{k}\right) +\lambda _{n}f\left( x_{n}\right)\\ & = & \sum_{k=1}^{n}\lambda _{k}f\left( x_{k}\right),\end{array} $$ proving the theorem. \end{pf} \begin{thm}\label{thm:convex-iff-Av-Ra-Ch-increasing} Let $I \subseteqq \BBR$ be an interval and let $f:I\rightarrow \BBR$. For $a\in I$ we put $$\fun{T_a}{x}{\dfrac{f(x)-f(a)}{x-a}}{I\setminus\{a\}}{\BBR}.$$Then $f$ is convex if and only if $\forall a\in I$, $T_a$ is increasing over $I\setminus \{a\}$. \end{thm} \begin{pf}Let $a0$. Then $\dfrac{x^2}{2} < \exp (x)$.\label{thm:x^20$, $f''(x)>0$ and so $f'$ is strictly increasing. Thus $f'(x)>f'(0)=1>0$ and so $f$ is increasing. Thus $$f(x)>f(0)\implies \exp (x)-\dfrac{x^2}{2}>0,$$proving the theorem.\end{pf} \begin{thm}\label{lem:limit-x/exp(x)}$\lim _{x\to +\infty} \dfrac{x}{\exp (x)}=0$. \end{thm} \begin{pf} From Theorem \ref{thm:x^20$, $$ 0<\dfrac{x}{\exp (x)}< \dfrac{2}{x} \implies 0 \leq \lim _{x\to +\infty}\dfrac{x}{\exp (x)} \leq \lim _{x\to +\infty}\dfrac{2}{x} =0,$$ and the theorem follows from the Sandwich Theorem. \end{pf} \begin{thm}\label{thm:limit-xa/exp(x)} Let $\alpha\in \BBR$. Then $\lim _{x\to +\infty} \dfrac{x^\alpha}{\exp (x)}=0$. \end{thm} \begin{pf} If $\alpha <1$ then $$\dfrac{x^\alpha}{\exp (x)} = \dfrac{x}{\exp (x)}\cdot x^{\alpha -1} \to 0 \cdot 0, $$by Lemma \ref{lem:limit-x/exp(x)}. If $\alpha \geq 1$ then $$\dfrac{x^\alpha}{\exp (x)} = \alpha ^{-\alpha}\left(\dfrac{\alpha x}{\exp (\alpha x)}\right)^\alpha \to \alpha ^{-\alpha}\cdot 0^\alpha=0, $$by continuity and by Lemma \ref{lem:limit-x/exp(x)}. \end{pf} \begin{thm} Let $x>0$. Then $\log x < x$.\label{thm:logx1$, $f'(x)>0$, which means that $f$ has a minimum at $x=1$. Thus $$f(x)>f(1) \implies x-\log x> 1. $$Since $x-\log x>1$ then {\em a fortiori} we must have $x-\log x>0$ and the theorem follows.\end{pf} \begin{lem}\label{lem:limit-logx/x} $\lim _{x\to +\infty} \dfrac{\log x}{x}=0$. \end{lem} \begin{pf} From Theorem \ref{thm:logx1$, $\log x>0$ and hence, $$x>1 \implies 0<\dfrac{\log x}{x} < \dfrac{1}{2x}, $$whence $\lim _{x\to +\infty} \dfrac{\log x}{x}=0$ by the Sandwich Theorem. \end{pf} \begin{thm}\label{thm:limit-logx/xa} Let $\alpha\in \loro{0}{+\infty}$. Then $\lim _{x\to +\infty} \dfrac{\log x}{x^\alpha}=0$. \end{thm} \begin{pf} If $\alpha >1$ then $$\dfrac{\log x}{x^\alpha} = \dfrac{\log x}{x}\cdot x^{1-\alpha } \to 0 \cdot 0, $$by Lemma \ref{lem:limit-logx/x}. If $0<\alpha \leq 1$ then $$\dfrac{\log x}{x^\alpha} =\dfrac{\log x^\alpha }{\alpha x^\alpha} \to \dfrac{1}{\alpha}\cdot 0=0, $$by continuity and by Lemma \ref{lem:limit-logx/x}. \end{pf} \begin{thm}For $x\in \loro{0}{\frac{\pi}{2}}$, $\sin x < x < \tan x$. \end{thm} \begin{pf} Observe that we gave a geometrical argument for this inequality in Theorem \ref{thm:limit-sinx/x}. First, let f(x)= $\sin x-x$. Then $f'(x) = \cos x-1< 0$, since for $x\in \loro{0}{\frac{\pi}{2}}$, the cosine is strictly positive. This means that $f$ is strictly decreasing. Thus for all $x\in \loro{0}{\frac{\pi}{2}}$, $$f(0)> f(x) \implies 0 > \sin x-x \implies \sin x < x,$$giving the first half of the inequality. \bigskip For the second half, put $g(x)= \tan x -x$. Then $g'(x) = \sec ^2 x -1$. Now, since $\absval{\cos x} < 1$ for $x\in \loro{0}{\frac{\pi}{2}}$, $\sec^2 x > 1$. Hence $g'(x)>0$, and so $g$ is strictly increasing. This gives $$ g(0) 0$ and $x-\tan x < 0$ for $x\in \loro{0}{\frac{\pi}{2}}$. Thus $f$ is strictly decreasing for $x\in \loro{0}{\frac{\pi}{2}}$ and so $$f(x)> f\left(\frac{\pi}{2}\right) \implies \dfrac{\sin x}{x} > \dfrac{2}{\pi}, $$ proving the theorem.\end{pf} \begin{df} If $w_1,w_2,\ldots,w_n$ are positive real numbers such that $w_1+w_2+\cdots+w_n=1$, we define the \emph{$r$-th weighted power mean} of the $x_i$ as: $$M_w^r(x_1,x_2,\ldots,x_n)=\left({w_1x_1^r+w_2x_2^r+\cdots+w_nx_n^r}\right)^{1/r}.$$ When all the $w_i=\frac{1}{n}$ we get the standard power mean. The weighted power mean is a continuous function of $r$, and taking limit when $r\to0$ gives us $$M_w^0=x_1^{w_1}x_2^{w_2}\cdots w_n^{w_n}.$$ \end{df} \begin{thm}[Generalisation of the AM-GM Inequality]\label{thm:AMGM-ineq-generalised1} If $r1$ and $y_i=x_i^r$ for $1\leq i\leq n$. This implies that $y_i^t=x_i^s$. The function $f:]0;+\infty[\rightarrow ]0;+\infty[$, $f(x)=x^t$ is strictly convex, since its second derivative is $f''(x)=\frac{1}{t(t-1)}x^{t-2}>0$ for all $x\in ]0;+\infty[$. By Jensen's inequality, \begin{eqnarray*} (w_1y_1+w_2y_2+\cdots+w_ny_n)^t&=&f(w_1y_1+w_2y_2+\cdots+w_ny_n)\\ &\le&w_1f(y_1)+w_2f(y_2)+\cdots+w_nf(y_n)\\ &=&w_1y_1^t+w_2y_2^t+\cdots+w_ny_n^t. \end{eqnarray*} with equality if and only if $y_1=y_2=\cdots=y_n$. By substituting $t=\frac{s}{r}$ and $y_i=x_i^r$ back into this inequality, we get $$ (w_1x_1^r+w_2x_2^r+\cdots+w_nx_n^r)^{s/r}\le w_1x_1^s+w_2x_2^s+\cdots+w_nx_n^s $$ with equality if and only if $x_1=x_2=\cdots=x_n$. Since $s$ is positive, the function $x\mapsto x^{1/s}$ is strictly increasing, so raising both sides to the power $1/s$ preserves the inequality: $$ (w_1x_1^r+w_2x_2^r+\cdots+w_nx_n^r)^{1/r}\le (w_1x_1^s+w_2x_2^s+\cdots+w_nx_n^s)^{1/s}, $$ which is the inequality we had to prove. Equality holds if and only if all the $x_i$ are equal. \bigskip The cases $r<0 0, \exists \delta > 0$ such that $$x\in \loro{a-\delta}{a+\delta} \implies \absval{\alpha (x)} \leq \varepsilon\absval{\beta (x)}. $$ We express the condition above with the notation $ \alpha (x) = \smallo{x\to a}{\beta (x)}$ (read ``$\alpha$ of $x$ is small oh of $\beta$ of $x$ as $x$ tends to $a$''). \bigskip Finally, we say that {\em $\alpha$ is Big Oh of $\beta$ around $x=a$}---written $\alpha (x) = \bigo{x\to a}{\beta (x)}$, or $\alpha (x)\ll{x\to a}{\beta (x)}$---if $\exists C>0$ and $\exists \delta>0$ such that $\forall x\in \loro{a-\delta}{a+\delta}$, $\absval{\alpha (x)}\leq C\absval{\beta (x)}$. \end{df} \begin{rem} Notice that $a$ above may be finite or $\pm \infty$. If $a$ is understood, we prefer to write $\alpha (x)=\soo{\beta (x)}$ rather $\alpha (x)=\smallo{x\to a}{\beta (x)}$. Also $$\alpha = \smallo{x\to a}{\beta} \iff \lim _{x \rightarrow a} \dfrac{\alpha (x)}{\beta (x)} = 0 \quad \mathrm{and} \quad \beta (a) = 0 \implies \alpha (a) =0.$$ \end{rem} \begin{exa} $\sin :\BBR \rightarrow [-1; 1]$ is infinitesimal as $x \rightarrow 0$, since $\lim_{x\rightarrow 0} \sin x = 0$. \end{exa} \begin{exa} $f:\BBR\setminus\{0\} \rightarrow \BBR, x \mapsto \dfrac{1}{x}$ is infinitesimal as $x \rightarrow +\infty$, since $\lim_{x\rightarrow +\infty} \dfrac{1}{x} = 0$. \end{exa} \begin{exa} We have $x^2 = \soo{x}$ as $x\rightarrow 0$ since $$\lim _{x\rightarrow 0} \dfrac{x^2}{x} = \lim _{x\rightarrow 0} x = 0.$$ \end{exa} \begin{exa} We have $x = \soo{x^2}$ as $x\rightarrow +\infty$ since $$\lim _{x\rightarrow +\infty} \dfrac{x}{x^2} = \lim _{x\rightarrow +\infty} \dfrac{1}{x} = 0.$$ \end{exa} \begin{df} We write $ \alpha (x) = \gamma (x) + \soo{\beta(x)}$ as $x \rightarrow a$ if $ \alpha (x) - \gamma (x) = \soo{\beta(x)}$ as $x \rightarrow a$. Similarly, $ \alpha (x) = \gamma (x) + \boo{\beta(x)}$ as $x \rightarrow a$ means that $ \alpha (x) - \gamma (x) = \boo{\beta(x)}$ as $x \rightarrow a$. \end{df} \begin{exa} We have $\sin x = x + \soo{x}$ as $x \rightarrow 0$ since $$\lim _{x\rightarrow 0} \dfrac{\sin x - x}{x} = \lim _{x\rightarrow 0} \dfrac{\sin x}{x } - \lim _{x\rightarrow 0} 1 = 1 - 1 = 0.$$ \end{exa} \begin{thm}\label{thm:properties-soo-boo} Let $f, g, \alpha , \beta, u, v$ be real-valued functions defined on an interval containing $a\in \closR$. Let $\lambda \in \BBR$ be a constant. Let $h$ be a real valued function defined on an interval containing $b\in \closR$. Then \begin{enumerate} \item $f=\soo{g} \implies f=\boo{g}$. \item $f=\soo{\alpha} \implies \lambda f=\soo{\alpha}$. \item $f=\soo{\alpha}, g=\soo{\alpha} \implies f+g=\soo{\alpha}$. \item $f=\soo{\alpha}, g=\soo{\beta} \implies fg=\soo{\alpha\beta}$. \item $f=\boo{\alpha} \implies \lambda f=\boo{\alpha}$. \item $f=\boo{\alpha}, g=\boo{\alpha} \implies f+g=\boo{\alpha}$. \item $f=\boo{\alpha}, g=\boo{\beta} \implies fg=\boo{\alpha\beta}$. \item $f=\boo{\alpha}, g=\soo{\beta} \implies fg=\soo{\alpha\beta}$. \item $f=\boo{\alpha}, \alpha=\boo{\beta} \implies f=\boo{\beta}$. \item $f=\soo{\alpha}, \alpha=\boo{\beta} \implies f=\soo{\beta}$. \item $f=\boo{\alpha}, \alpha=\soo{\beta} \implies f=\soo{\beta}$. \item $f=\soo{\alpha}, \lim _{x\to b}h(x) = a \implies f\circ h=\smallo{x\to b}{\alpha \circ h}$. \item $f=\boo{\alpha}, \lim _{x\to b}h(x) = a \implies f\circ h=\bigo{x\to b}{\alpha \circ h}$. \end{enumerate} \end{thm} \begin{pf} These statements follow directly from the definitions. \begin{enumerate} \item If $f=\soo{g}$ then $\forall \varepsilon >0$ there exists $\delta > 0$ such that $$x\in \loro{a-\delta}{a+\delta}\implies \absval{\dfrac{f(x)}{g(x)}-0}<\varepsilon \implies \absval{f(x)}<\varepsilon \absval{g(x)} \implies f=\boo{g},$$using $C=\varepsilon$ in the definition of Big Oh. \item This follows by Theorem \ref{thm:algebra-of-fun-limits}. \item This follows by Theorem \ref{thm:algebra-of-fun-limits}. \item Both $\lim _{x\to a}\dfrac{f(x)}{\alpha (x)} = 0$ and $\lim _{x\to a}\dfrac{g(x)}{\beta (x)} = 0$. Hence $\lim _{x\to a}\dfrac{f(x)g(x)}{\alpha (x)\beta (x)}=\lim _{x\to a}\dfrac{f(x)}{\alpha (x)} \cdot \lim _{x\to a}\dfrac{g(x)}{\beta (x)} =0 \implies fg=\soo{\alpha\beta}$. \item If $f=\boo{\alpha}$ then there is $\delta > 0$ and $C>0$ such that $$x\in \loro{a-\delta}{a+\delta} \implies \absval{f(x)}\leq C\absval{g(x)}\implies \absval{\lambda f(x)}\leq C\absval{\lambda}\cdot \absval{g(x)} \implies\lambda f=\boo{\alpha}$$ \item There exists $\delta _1>0, \delta _2 > 0$ and $C_1>0$, $C_2>0$ such that $$ x\in \loro{a-\delta _1}{a+\delta _1} \implies \absval{f(x)} \leq C_1\absval{\alpha (x)} \qquad \mathrm{and} \qquad x\in \loro{a-\delta _2}{a+\delta _2} \implies \absval{g(x)} \leq C_2\absval{\alpha (x)} .$$Thus if $\delta = \min (\delta _1, \delta _2)$, $$x\in \loro{a-\delta}{a+\delta} \implies \absval{f(x)+g(x)}\leq \absval{f(x)}+\absval{g(x)} \leq C_1\alpha (x) +C_2\alpha (x) = (C_1+C_2)\alpha (x)\implies f+g=\boo{\alpha}.$$ \item There exists $\delta _1>0, \delta _2 > 0$ and $C_1>0$, $C_2>0$ such that $$ x\in \loro{a-\delta _1}{a+\delta _1} \implies \absval{f(x)} \leq C_1\absval{\alpha (x)} \qquad \mathrm{and} \qquad x\in \loro{a-\delta _2}{a+\delta _2} \implies \absval{g(x)} \leq C_2\absval{\beta (x)} .$$Thus if $\delta = \min (\delta _1, \delta _2)$, $$x\in \loro{a-\delta}{a+\delta} \implies \absval{f(x)g(x)} =\absval{f(x)}\absval{g(x)} \leq C_1\absval{\alpha (x)}\cdot C_2\absval{\beta (x)} = (C_1C_2)\absval{\alpha (x)\beta (x)}\implies fg=\boo{\alpha\beta}.$$ \item There exists $\delta _1>0, \delta _2 > 0$ and $C_1>0$, such that $\forall \varepsilon > 0$ $$ x\in \loro{a-\delta _1}{a+\delta _1} \implies \absval{f(x)} \leq C_1\absval{\alpha (x)} \qquad \mathrm{and} \qquad x\in \loro{a-\delta _2}{a+\delta _2} \implies \absval{g(x)} \leq \varepsilon\absval{\beta (x)} .$$Thus if $\delta = \min (\delta _1, \delta _2)$, $$x\in \loro{a-\delta}{a+\delta} \implies \absval{f(x)g(x)} =\absval{f(x)}\absval{g(x)} \leq C_1\absval{\alpha (x)}\cdot \varepsilon\absval{\beta (x)} = \varepsilon (C_1)\absval{\alpha (x)\beta (x)}\implies fg=\soo{\alpha\beta}.$$ \item There exists $\delta _1>0, \delta _2 > 0$ and $C_1>0$, $C_2>0$ such that $$ x\in \loro{a-\delta _1}{a+\delta _1} \implies \absval{f(x)} \leq C_1\absval{\alpha (x)} \qquad \mathrm{and} \qquad x\in \loro{a-\delta _2}{a+\delta _2} \implies \absval{\alpha (x)} \leq C_2\absval{\beta (x)} .$$Thus if $\delta = \min (\delta _1, \delta _2)$, $$x\in \loro{a-\delta}{a+\delta} \implies \absval{f(x)}\leq C_1\absval{\alpha (x)} \leq C_1C_2\absval{\beta (x)} \implies f=\boo{\beta}.$$ \item There exists $\delta _1>0, \delta _2 > 0$ and $C>0$, such that $\forall \varepsilon > 0$ $$ x\in \loro{a-\delta _1}{a+\delta _1} \implies \absval{f(x)} \leq \varepsilon\absval{\alpha (x)} \qquad \mathrm{and} \qquad x\in \loro{a-\delta _2}{a+\delta _2} \implies \absval{\alpha (x)} \leq C\absval{\beta (x)} .$$Thus if $\delta = \min (\delta _1, \delta _2)$, $$x\in \loro{a-\delta}{a+\delta} \implies \absval{f(x)}\leq \varepsilon\absval{\alpha (x)} \leq C\varepsilon\absval{\beta (x)} \implies f=\soo{\beta}.$$ \item There exists $\delta _1>0, \delta _2 > 0$ and $C>0$, such that $\forall \varepsilon > 0$ $$ x\in \loro{a-\delta _1}{a+\delta _1} \implies \absval{f(x)} \leq C\absval{\alpha (x)} \qquad \mathrm{and} \qquad x\in \loro{a-\delta _2}{a+\delta _2} \implies \absval{\alpha (x)} \leq \varepsilon\absval{\beta (x)} .$$Thus if $\delta = \min (\delta _1, \delta _2)$, $$x\in \loro{a-\delta}{a+\delta} \implies \absval{f(x)}\leq C\absval{\alpha (x)} \leq C\varepsilon\absval{\beta (x)} \implies f=\soo{\beta}.$$ \item There exists $\delta _1>0, \delta _2 > 0$ such that $\forall \varepsilon > 0$ $$ x\in \loro{a-\delta _1}{a+\delta _1} \implies \absval{f(x)} \leq \varepsilon \absval{\alpha (x)} \qquad \mathrm{and} \qquad x\in \loro{b-\delta _2}{b+\delta _2} \implies \absval{h(x)-a} \leq \varepsilon \implies h(x) \in \loro{a-\varepsilon}{a+\varepsilon} .$$Thus if $\delta = \min (\delta _1, \delta _2, \varepsilon)$, $$x\in \loro{b-\delta}{b+\delta} \implies \absval{(f\circ h)(x)}\leq \varepsilon\absval{(\alpha\circ h) (x)} \implies f\circ h=\smallo{x\to b}{\alpha \circ h}.$$ \item There exists $\delta _1>0, \delta _2 > 0, C>0$ such that $\forall \varepsilon > 0$ $$ x\in \loro{a-\delta _1}{a+\delta _1} \implies \absval{f(x)} \leq C \absval{\alpha (x)} \qquad \mathrm{and} \qquad x\in \loro{b-\delta _2}{b+\delta _2} \implies \absval{h(x)-a} \leq \varepsilon \implies h(x) \in \loro{a-\varepsilon}{a+\varepsilon} .$$Thus if $\delta = \min (\delta _1, \delta _2, \varepsilon)$, $$x\in \loro{b-\delta}{b+\delta} \implies \absval{(f\circ h)(x)}\leq C\absval{(\alpha\circ h) (x)} \implies f\circ h=\bigo{x\to b}{\alpha \circ h}.$$ \end{enumerate} \end{pf} \begin{rem} In the above theorem, (8), (10), and (11) essentially say that $\boo{\mathit{o}} =\soo{\mathit{O}}=\soo{\mathit{o}} = \mathit{o} $ and (9) says that $\boo{\mathit{O}}=\mathit{O}$. \end{rem} The following corollary is immediate. \begin{cor} Let $\alpha$ and $\beta$ be infinitesimal functions as $x\rightarrow a$. Then the following hold. \begin{enumerate} \item The sum of two infinitesimals is an infinitesimal: $$\soo{\beta(x)} + \soo{\beta(x)} = \soo{\beta(x)}. $$ \item The difference of two infinitesimals is an infinitesimal: $$\soo{\beta(x)} - \soo{\beta(x)} = \soo{\beta(x)}. $$ \item $\forall c\in \BBR\setminus \{0\}, \soo{c\beta(x)} = \soo{\beta(x)}. $ \item $\forall n\in\BBN, n \geq 2, 1 \leq k \leq n - 1, \ \soo{(\beta(x))^n} = \soo{(\beta(x))^k}$. \item $\soo{\soo{\beta(x)}} = \soo{\beta(x)}$. \item $\forall n\in\BBN, n \geq 1, \ (\beta(x))^n\soo{\beta(x)} = \soo{(\beta(x))^{n + 1}}$. \item $\forall n\in\BBN, n \geq 2, \dfrac{\soo{(\beta(x))^n}}{\beta(x)} = \soo{(\beta(x))^{n - 1}}$. \item $\dfrac{\soo{\beta(x)}}{\beta(x)} = \soo{1}$. \item If $c_k$ are real numbers, then $\soo{\sum_{k = 1} ^n c_k(\beta(x))^k} = \soo{\beta(x)}$. \item $(\alpha\beta)(x) = \soo{\alpha(x)}$ and $(\alpha\beta)(x) = \soo{\beta(x)}$. \item If $\alpha \sim \beta,$ then $(\alpha - \beta)(x) = \soo{\alpha(x)}$ and $(\alpha - \beta)(x) = \soo{\beta(x)}$. \end{enumerate} \label{cor:small-oh-properties} \end{cor} \begin{thm}[Canonical small oh Relations] The following relationships hold \begin{enumerate} \item $\forall (\alpha , \beta)\in \BBR^2$, $x^\alpha = \smallo{x\to +\infty}{x^\beta} \iff \alpha < \beta$. \item $\forall (\alpha , \beta)\in \BBR^2$, $x^\alpha = \smallo{x\to 0+}{x^\beta} \iff \alpha > \beta$. \item $\log x = \smallo{x\to +\infty}{x}$. \item $\forall (\alpha , \beta)\in \BBR^2, \beta > 0$, $(\log x)^\alpha = \smallo{x\to +\infty}{x^\beta}$. \item $\forall (\alpha , \beta)\in \BBR^2, \beta < 0$, $\absval{\log x}^\alpha = \smallo{x\to 0+}{x^\beta}$. \item $\forall (\alpha , a)\in \BBR^2, a >1$, $x^\alpha = \smallo{x\to +\infty}{a^x}$ \item $\forall (\alpha , a)\in \BBR^2, a >1$, $a^x = \smallo{x\to -\infty}{\absval{x}^\alpha}$ \end{enumerate} \label{thm:canonical-oh-relations} \end{thm} \begin{pf} \begin{enumerate} \item Immediate. \item Immediate. \item This follows from Lemma \ref{lem:limit-logx/x}. \item If $\alpha = 0$ then eventually $(\log x)^\alpha =1$ and so the assertion is immediate. If $\alpha < 0$ the assertion is also immediate, since then $(\log x)^\alpha \to 0$ as $x\to +\infty$. If $\alpha > 0$, by Theorem \ref{thm:limit-logx/xa}, $$ \dfrac{\log x}{x^{\beta /\alpha}} \to 0,$$ whence $$ \dfrac{(\log x)^\alpha}{x^\beta} = \left( \dfrac{\log x}{x^{\beta /\alpha}} \right)^\alpha \to 0^\alpha = 0. $$ \item If $x\to 0+$ then $\dfrac{1}{x}\to +\infty$. Hence by the preceding part and by continuity, as $x\to 0+$ and for $\gamma > 0$, $$ \dfrac{\left(\absval{\log \dfrac{1}{x}}\right)^\alpha}{\left(\dfrac{1}{x}\right)^\gamma} \to 0. $$ But $$\dfrac{\left(\absval{\log \dfrac{1}{x}}\right)^\alpha}{\left(\dfrac{1}{x}\right)^\gamma} = \dfrac{\left(\absval{-\log x}\right)^\alpha}{\left(\dfrac{1}{x}\right)^\gamma} = x^\gamma \absval{\log x}^\alpha, $$and so $ \absval{\log x}^\alpha = \smallo{x\to 0+}{x^{-\gamma}}$, and so putting $\beta = -\gamma < 0$ we have $ \absval{\log x}^\alpha= \smallo{x\to 0+}{x^{\beta}}$. \item For $\alpha < 1$ we have $$\dfrac{x^\alpha}{a^x} =\dfrac{x\log a}{\exp \left(x\log a\right)}\cdot \dfrac{x^{\alpha-1}}{\log a}\to 0 \cdot 0, $$ since $\dfrac{x\log a}{\exp \left(x\log a\right)} \to 0$ by continuity and Theorem \ref{thm:limit-xa/exp(x)}, and $\dfrac{x^{\alpha-1}}{\log a} \to 0$ since $\alpha-1<0$. If $\alpha > 1$ then $$ \dfrac{x^\alpha}{a^x} = \left(\dfrac{x}{\left(a^{1/\alpha}\right)^x}\right)^\alpha =\dfrac{\alpha ^\alpha}{(\log a)^\alpha } \cdot \left(\dfrac{x\dfrac{\log a}{\alpha}}{\exp \left( x\dfrac{\log a}{\alpha}\right)}\right)^\alpha \to \dfrac{\alpha ^\alpha}{(\log a)^\alpha } \cdot 0^\alpha = 0, $$by continuity and Theorem \ref{thm:limit-xa/exp(x)}. \item If $\alpha > 0$, $a>1$ then $\absval{x}^\alpha \to +\infty$ but $a^x\to 0$ as $x\to -\infty$, hence there is nothing to prove. If $\alpha =0$, again the result is obvious. Assume $\alpha < 0$. If $x\to -\infty$ then $-x\to +\infty$ and so by the preceding part $$ \dfrac{\absval{x}^{-\alpha}}{a^{-x}} \to 0$$since the above result is valid regardless of the sign of $\alpha$. Now $$ \dfrac{a^x}{\absval{x}^\alpha}=\dfrac{\absval{x}^{-\alpha}}{a^{-x}},$$proving the result. \end{enumerate} \end{pf} \begin{exa} In view of Corollary \ref{cor:small-oh-properties} and Theorem \ref{thm:canonical-oh-relations}, we have $$\soo{-2x^3 + 8x^2} = \soo{x},$$as $x\rightarrow 0.$ \end{exa} \begin{exa} In view of Corollary \ref{cor:small-oh-properties} and Theorem \ref{thm:canonical-oh-relations}, we have $$\soo{-2x^3 + 8x^2} = \soo{x^4},$$as $x\rightarrow +\infty.$ \end{exa} \subsection*{Homework}\addcontentsline{toc}{subsection}{Homework} \begin{multicols}{2}\columnseprule 1pt \columnsep 25pt\multicoltolerance=900\small \begin{pro} Which one is faster as $x\to +\infty$, $(\log\log x)^{\log x}$ or $(\log x)^{\log\log x}$? \begin{answer} $(\log\log x)^{\log x} = \exp ((\log x)(\log\log\log x)) $ and $(\log x)^{\log\log x} = \exp ((\log\log x)^2)$. Now, lexicographically, $$ (\log\log x)^2 <<(\log x)(\log\log\log x) \implies \exp((\log\log x)^2) <<\exp ((\log x)(\log\log\log x))$$ and thus $(\log\log x)^{\log x}$ is faster. \end{answer} \end{pro} \end{multicols} \section{Asymptotic Equivalence} \begin{df} Let $I\subseteqq \closR$ be an interval, and let $a\in I$. We say that $\alpha$ is {\em asymptotic} to a function $\beta:I \rightarrow \BBR$ as $x\rightarrow a$, and we write $\alpha\sim \beta$, if $\alpha \sim \beta \iff \alpha - \beta = \smallo{a}{\beta}$. \end{df} \begin{rem} If in a neighbourhood $\N{a}$ of $a$ $\beta \neq 0$ then $$ \alpha \sim \beta \iff \left\{\begin{array}{l} \dfrac{\alpha}{\beta}\sim 1 \\ \beta (a)= 0 \implies \alpha (a) = 0\end{array}\right. $$ \end{rem} \begin{exa} We have $\sin x \sim x$ as $x \rightarrow 0$, since $\lim_{x\rightarrow 0} \dfrac{\sin x}{x} = 1$. \end{exa} \begin{exa} We have $x^2 + x \sim x$ as $x \rightarrow 0$, since $\lim_{x\rightarrow 0} \dfrac{x^2 + x}{x} = 1$. \end{exa} \begin{exa} We have $x^2 + x \sim x^2$ as $x \rightarrow +\infty$, since $\lim_{x\rightarrow +\infty} \dfrac{x^2 + x}{x^2} = 1$. \end{exa} \begin{thm}\label{thm:asymptotic-means-big-oh} $$\alpha \sim \beta \implies \left\{\begin{array}{l} \alpha = \boo{\beta} \\ \beta = \boo{\alpha}\end{array}\right. $$ \end{thm} \begin{pf} If $\alpha - \beta = \soo{\beta}$ there is a neighbourhood $\N{a}$ of $a$ such that $$\forall \varepsilon > 0, x\in\N{a} \implies \absval{\alpha (x)-\beta (x)} \leq \varepsilon \absval{\beta (x)}. $$In particular, for $\varepsilon = \dfrac{1}{2}$, we have $$x\in\N{a} \implies \absval{\alpha (x)-\beta (x)} \leq \dfrac{1}{2} \absval{\beta (x)}. $$Hence $$ x\in\N{a} \implies \absval{\alpha (x)} = \absval{\alpha (x)-\beta (x)+\beta (x)} \leq \absval{\alpha (x)-\beta (x)} + \absval{\beta (x)} \leq \dfrac{3}{2} \absval{\beta (x)} \implies \alpha = \boo{\beta}, $$and $$ x\in\N{a} \implies \absval{\beta (x)} = \absval{\beta (x)-\alpha (x)+\alpha (x)} \leq \absval{\beta (x)-\alpha (x)} + \absval{\alpha (x)} \leq \dfrac{1}{2} \absval{\beta (x)} + \absval{\alpha (x)} \implies \absval{\beta (x)} \leq 2\absval{\alpha (x) } \implies \beta = \boo{\alpha} . $$ \end{pf} \begin{thm} The relation of asymptotic equivalence $\sim$ is an equivalence relation on the set of functions defined on a neighbourhood of $a$. \end{thm} \begin{pf} We have \begin{enumerate} \item[{\bf Reflexivity}] $\alpha - \alpha = 0 = \soo{\alpha}$. \item[{\bf Symmetry}] $\alpha - \beta = \soo{\beta} \implies \beta = \boo{\alpha}$ by Theorem \ref{thm:asymptotic-means-big-oh}. Now by (10) of Theorem \ref{thm:properties-soo-boo}, $$\alpha - \beta = \soo{\beta} \quad \mathrm{and} \beta = \boo{\alpha} \implies \alpha - \beta = \soo{\alpha} \implies \beta -\alpha = \soo{\alpha}, $$whence $\beta \sim \alpha$. \item[{\bf Transitivity}] Assume $\alpha -\beta = \soo{\beta}$ and $\beta - \gamma = \soo{\gamma}$. Then by Theorem \ref{thm:asymptotic-means-big-oh} we also have $\beta = \boo{\gamma}$. Hence $\alpha - \beta = \soo{\gamma}$ by (10) of Theorem \ref{thm:properties-soo-boo}. Finally $\alpha - \beta = \soo{\gamma}$ and $\beta - \gamma = \soo{\gamma}$ give $\alpha - \gamma = \soo{\gamma}$ by (3) of Theorem \ref{thm:properties-soo-boo}. \end{enumerate} \end{pf} The relationship between $\mathit{o}, \mathit{O},$ and $\sim$ is displayed in figure \ref{fig:O-relations2}. \vspace{2cm} \begin{figure}[h] $$\psset{unit=1pc} \pscircle(-2,0){4}\pscircle(2,0){4} \pscircle(0,0){1}\rput(0,0){f\sim g} \pscircle(-3.8,-1){1.7}\rput(-3.8,-1){f=\soo{g}} \pscircle(3.8,-1){1.7}\rput(3.8,-1){g=\soo{f}} \rput(-3.3,2.5){f=\boo{g}}\rput(3.3,2.5){g=\boo{f}} $$\vspace{1cm}\footnotesize\hangcaption{Diagram of Big Oh relations.} \label{fig:O-relations2} \end{figure} \begin{thm} The relation of asymptotic equivalence $\sim$ possesses the following properties. \begin{enumerate} \item $ \left\{\begin{array}{l} \alpha \sim {\beta} \\ \gamma \sim {\delta}\end{array}\right. \implies \alpha \gamma \sim \beta\delta$. \item $\left\{\begin{array}{l} \alpha \sim {\beta} \\ n\in \BBN \setminus \{0\}\end{array}\right. \implies \alpha ^n \sim \beta ^n$ \item if $\alpha \sim \beta$ and if there is a neighbourhood $\N{a}$ of $a$ where $\forall x\in \N{a}\setminus \{a\}, \beta (a)\neq 0$, then $\dfrac{1}{\alpha}$ and $\dfrac{1}{\beta}$ are defined on $\N{a}\setminus \{a\}$ and $\dfrac{1}{\alpha} \asympto{a} \dfrac{1}{\beta}$. \item $\left\{\begin{array}{l} \alpha = \soo{\beta} \\ \beta \sim \gamma \end{array}\right. \implies \alpha = \soo{\gamma}$. \item $\left\{\begin{array}{l} \alpha \sim \beta \\ \beta =\soo{\gamma} \end{array}\right. \implies \alpha =\soo{\gamma}$. \item if $\alpha \sim \beta$ and if there is a neighbourhood $\N{a}$ of $a$ where $\forall x\in \N{a}\setminus \{a\}, \beta (a) > 0$, and if $r\in \BBR$ then $\alpha ^r \asympto{a} \beta ^r$. \item {\bf (Dextral Composition)} If $\alpha \asympto{a} \beta$ and if $\lim _{x\to b} \gamma (x)=a$, then $\alpha \circ \gamma \asympto{a} \beta \circ \gamma$. \end{enumerate} \end{thm} \begin{pf} We prove the assertions in the given order. \begin{enumerate} \item Since $\alpha - \beta = \soo{\beta}$ and $\gamma - \delta = \soo{\delta}$ then $\alpha = \boo{\beta}$, and so $$\alpha \gamma - \beta \delta = \alpha (\gamma - \delta) - \delta (\beta - \alpha) = \boo{\beta} \soo{\delta} -\delta \soo{\beta}=\soo{\beta\delta}.$$ \item This follows upon applying the preceding product rule $n-1$ times, using $\gamma = \alpha$ and $\delta =\beta$. \item Observe that $$\dfrac{1}{\alpha} - \dfrac{1}{\beta} = \dfrac{\beta - \alpha}{\alpha \beta} =\dfrac{\soo{\alpha}}{\alpha \beta} = \soo{\dfrac{1}{\beta}}, $$upon using $\beta - \alpha = \soo{\alpha}$ and (8) of Corollary \ref{cor:small-oh-properties}. \item We have $\alpha = \soo{\beta}$ and $\beta - \gamma = \soo{\gamma}$. This last implies that $\beta = \boo{\gamma}$ by Theorem \ref{thm:asymptotic-means-big-oh}. Hence $$\alpha = \soo{\beta} = \soo{\boo{\gamma}} = \soo{\gamma}. $$ \item We have $\alpha - \beta = \soo{\beta}$ and $\beta = \soo{\gamma}$. This last implies that $\alpha = \boo{\beta}$ by Theorem \ref{thm:asymptotic-means-big-oh}. Hence $$\alpha = \boo{\beta} = \boo{\soo{\gamma}} = \soo{\gamma}. $$ \item Since $\beta$ is eventually strictly positive, so is $\alpha$. Hence $\alpha \sim \beta \iff \dfrac{\alpha}{\beta}(x)\to 1$ as $x\to a$. Since the function $x\mapsto x^r$ is continuous in $\loro{0}{+\infty}$, $$\dfrac{\alpha}{\beta}(x)\to 1 \implies \dfrac{\alpha ^r}{\beta ^r}(x)\to 1 \implies \alpha ^r \sim \beta ^r.$$ \item We have $\dfrac{\alpha (x)-\beta (x)}{\beta (x)}\to 0$ as $x\to a$. Now if $\gamma (x)\to a$ as $x\to b$ then as $x\to b$, $$\dfrac{\alpha (\gamma(x))-\beta (\gamma (x))}{\beta (\gamma (x))}\to 0. $$ \end{enumerate} \end{pf} \begin{thm}[Exponential Composition] $\exp (\alpha) \asympto{a} \exp (\beta)\iff \alpha-\beta \asympto{a} 0$. \end{thm} \begin{pf} We have $$\begin{array}{lll}\exp (\alpha) \asympto{a} \exp (\beta) & \iff & \exp (\alpha) - \exp (\beta) =\soo{\exp (\beta)} \\ & \iff & \left(\exp (-\beta)\right)\left(\exp (\alpha) - \exp (\beta)\right) = \left(\exp (-\beta)\right)\soo{\exp (\beta)} \\ & \iff & \exp (\alpha - \beta) - 1 = \soo{1}\\ & \iff & \alpha - \beta = \soo{0}. \end{array}$$ \end{pf} \begin{rem} The above theorem {\em does not say} that $\alpha \sim \beta \implies \exp (\alpha)\sim \exp (\beta)$. That this last assertion is false can be seen from the following counterexample: $x+1\sim x$ as $x\to 0$, but $\exp (x+1) = e\exp (x)$ is not asymptotic to $\exp (x)$. \end{rem} \begin{thm}[Logarithmic Composition] Suppose there is a neighbourhood of $a$ $\N{a}$ such that $\forall x\in \N{a}\setminus \{a\}, \beta (x)>0 $. Suppose, moreover, that $\alpha \asympto{a} \beta$ and that $\lim _{x\to a} \beta (x)=l$ with $l\in \lcrc{0}{+\infty}\setminus \{1\}$. Then $\log \circ \alpha \asympto{a}\log \circ \beta$. \end{thm} \begin{pf} Either $l\in \loro{0}{+\infty}\setminus \{1\} $ or $l=+\infty$ or $l=0$. \bigskip In the first case, $\log \alpha(x)\to \log l$ and $\log \beta (x) \to \log l$ as $x\to a$ hence $$\log \alpha \sim \log l \sim \log \beta, \qquad \mathrm{as}\ x\to a. $$ \bigskip In the second case $\beta (x)>1$ eventually, and thus $\log \beta (x) \neq 0$. Hence $$ \dfrac{\log \alpha (x)}{\log \beta (x)}-1 = \dfrac{\log \alpha (x) - \log \beta (x)}{\log \beta (x)} = \dfrac{\log \dfrac{\alpha (x)}{\beta (x)} }{\log \beta (x)} \to \dfrac{\log 1}{+\infty} = 0, $$ since $\dfrac{\alpha (x)}{\beta (x)} \to 1$ and $\log \beta (x) \to +\infty$ as $x\to +\infty$. \bigskip The third case becomes the second case upon considering $\dfrac{1}{\alpha}$ and $\dfrac{1}{\beta}$. \end{pf} \begin{thm}[Addition of Positive Terms] If $\alpha \sim \beta$ and $\gamma \sim \delta$ and there exists a neighbourhood of $a$ $\N{a}$ such that $\forall x\in \N{a}\setminus \{a\}, \beta (x)>0, \delta (x) >0$ then $$\alpha + \gamma \sim \beta + \delta . $$ \end{thm} \begin{pf} We have $\alpha - \beta = \soo{\beta}$ and $\gamma - \delta = \soo{\delta}$. Hence $$\begin{array}{lll} (\alpha +\gamma) - (\beta + \delta) & = & (\alpha - \beta) + (\gamma - \delta) \\ & = & \soo{\beta} + \soo{\delta}\\ & = & \soo{\beta + \delta}, \end{array}$$which means $\alpha + \gamma \sim \beta + \delta$. \end{pf} \begin{thm} The following asymptotic expansions hold as $x\to 0:$ \begin{enumerate} \item \label{property:first}$\exp (x)-1 \sim x$ and thus $\exp (x)=1+ x+ \soo{x}$ \item $\log (1+x) \sim x$ and thus $\log (1+x)=x+ \soo{x}$ \item $\sin x \sim x$ and thus $\sin (x)=x+ \soo{x}$ \item $\tan x \sim x$ and thus $\tan (x)= x+ \soo{x}$ \item $\arcsin x \sim x $ and thus $\arcsin (x)= x+ \soo{x}$ \item $\arctan x \sim x $ and thus $\tan (x)= x+ \soo{x}$ \item \label{property:penultimate} for $\alpha \in \BBR$ constant, $(1+x)^\alpha -1 \sim \alpha x $ and thus $(1+x)^\alpha =1+ \alpha x+ \soo{x}$ \item \label{property:last} $1-\cos x \sim \dfrac{x^2}{2} $ and thus $\cos (x)=1- \dfrac{x^2}{2}+ \soo{x^2}$ \end{enumerate} \label{thm:baby-asymp-dev}\end{thm} \begin{pf} Results \ref{property:first}---\ref{property:penultimate} follow from the fact that $$f'(a) \neq 0, \quad \dfrac{f(x)-f(a)}{x-a} \to f'(a) \implies f(x)-f(a)\sim f'(a)(x-a).$$Property \ref{property:last} follows from the identity $1-\cos x =2\sin ^2\dfrac{x}{2}$. \end{pf} \begin{exa} Since $\tan x = x + \soo{x}$, we have $$\tan \dfrac{x^2}{2} = \dfrac{x^2}{2} + \soo{\dfrac{x^2}{2}} = \dfrac{x^2}{2} + \soo{x^2}, $$as $x\rightarrow 0.$ Also, $$(\tan x)^{3} = (x + \soo{x})^3 = x^3 + 3x^2\soo{x} + 3x\soo{x^2} + (\soo{x})^3 = x^3 + \soo{x^3}.$$ \end{exa} \begin{exa} Since $\cos x = 1 - \dfrac{x^2}{2} + \soo{x^2}$, we have $$\cos 3x^2 = 1 - \dfrac{9x^4}{2} + \soo{x^4}.$$ \end{exa} \begin{exa} Find an asymptotic expansion of $\cot^2x$ of type $\soo{x^{-2}}$ as $x\rightarrow 0$. \label{exa:cotangent}\end{exa} \begin{solu} Since $\tan x \sim x$ we have $$\cot^2x \sim \dfrac{1}{x^2}.$$We can write this as $\cot^2x = \dfrac{1}{x^2} + \soo{\dfrac{1}{x^2}}$. \end{solu} \begin{exa} Calculate $$\lim_{x\rightarrow 0} \dfrac{\sin\sin\tan \dfrac{x^2}{2}}{\log \cos 3x}.$$ \end{exa} \begin{solu} We use theorems \ref{thm:baby-asymp-dev} and \ref{cor:small-oh-properties}. $$\begin{array}{lll} \sin\sin\tan \dfrac{x^2}{2} & = & \sin\sin\left(\dfrac{x^2}{2} + \soo{x^2}\right) \\ & = & \sin\left(\dfrac{x^2}{2} + \soo{x^2} + \soo{\dfrac{x^2}{2} + \soo{x^2}}\right) \\ & = & \sin \left(\dfrac{x^2}{2} + \soo{x^2}\right) \\ & = & \dfrac{x^2}{2} + \soo{x^2}, \end{array}$$and $$\begin{array}{ccc} \log \cos 3x & = & \log \left(1 - \dfrac{9x^2}{2} + \soo{x^2}\right) \\ & = & - \dfrac{9x^2}{2} + \soo{x^2} + \soo{- \dfrac{9x^2}{2!} + \soo{x^2}} \\ & = & - \dfrac{9x^2}{2} + \soo{x^2} \\ \end{array}$$The limit is thus equal to $$\lim_{x\rightarrow 0} \dfrac{\dfrac{x^2}{2} + \soo{x^2}}{- \dfrac{9x^2}{2} + \soo{x^2}} = \lim_{x\rightarrow 0} \dfrac{\dfrac{1}{2} + \soo{1}}{- \dfrac{9}{2} + \soo{1}} = -\dfrac{1}{9}.$$ \end{solu} \begin{exa} Find $\lim_{x\rightarrow 0} (\cos x)^{(\cot^2x)}$. \end{exa} \begin{solu} By example \ref{exa:cotangent}, we have $\cot^2x = \dfrac{1}{x^2} + \soo{\dfrac{1}{x^2}}.$ Also, $$\log \cos x = \log \left(1 - \dfrac{x^2}{2} + \soo{x^2}\right) = - \dfrac{x^2}{2} + \soo{x^2}. $$ Hence $$\begin{array}{ccc}(\cos x)^{\cot^2x}& = & \exp\left((\cot^2x)\log \cos x\right) \\ & = & \exp\left(\left(\dfrac{1}{x^2} + \soo{\dfrac{1}{x^2}}\right)\left(- \dfrac{x^2}{2} + \soo{x^2}\right)\right)\\ & = & \exp(-\dfrac{1}{2} + \soo{1}) \\ & \rightarrow & e^{-1/2}, \end{array}$$as $x\rightarrow 0.$ \end{solu} \subsection*{Homework}\addcontentsline{toc}{subsection}{Homework} \begin{multicols}{2}\columnseprule 1pt \columnsep 25pt\multicoltolerance=900\small \begin{pro} Prove that $\dfrac{\log (1+2\tan x)}{\sin x} \to 2$ as $x\to 0$. \end{pro} \begin{pro} Prove that $\left(1+\dfrac{1}{x}\right)^x \to e$ as $x\to +\infty$. \end{pro} \begin{pro} Prove that $(\tan x)^{\cot 4x} \to e^{1/2}$ as $x\to \dfrac{\pi}{4}$. \end{pro} \end{multicols} \section{Asymptotic Expansions} \begin{df} Let $n\in \BBN$ and let $f:\N{0}\to \BBR$ where $\N{0}$ is a neighbourhood of $0$. We say that $f$ admits an {\em asymptotic expansion} of order $n$ about $x=0$ if there exists a polynomial $p$ of degree $n$ such that $$ \forall x\in\N{0}, \quad f(x)=p(x)+\smallo{0}{x^n}. $$The polynomial $p$ is called the {\em regular part of the asymptotic expansion about $x=0$ of $f$}. \end{df} \begin{thm} If $f$ admits an asymptotic expansion about $0$, its regular part is unique. \end{thm} \begin{pf} Assume $ f(x)=p(x)+\smallo{0}{x^n}$ and $ f(x)=q(x)+\smallo{0}{x^n}$, where $p(x) =p_nx^n+\cdots + p_1x+p_0$ and $q(x) =q_nx^n+\cdots + q_1x+q_0$ are polynomials of degree $n$. If $p\neq q$ let $k$ be the largest $k$ for which $p_k\neq q_k$. Then subtracting both equivalencies, as $x\to 0$, $$p(x)-q(x)= \soo{x^n} \implies (p_n-q_n)x^n + (p_{n-1}-q_{n-1})x^{n-1}+\cdots + (p_1-q_1)x =\soo{x^n} \implies (p_k-q_k)x^k +\cdots + \cdots =\soo{x^n} . $$ But $(p_k-q_k)x^k +\cdots + \cdots \boo{x^k} $ as $x\to 0$, a contradiction, since $k \leq n$. \end{pf} \begin{df} Let $n\in \BBN$, $a\in \BBR$, and let $f:\N{a}\to \BBR$ where $\N{a}$ is a neighbourhood of $a$. We say that $f$ admits an {\em asymptotic expansion} of order $n$ about $x=a$ if there exists a polynomial $p$ of degree $n$ such that $$ \forall x\in\N{a}, \quad f(x)=p(x-a)+\smallo{a}{(x-a)^n}. $$The polynomial $p$ is called the {\em regular part of the asymptotic expansion about $x=a$ of $f$}. \end{df} \begin{df} Let $n\in \BBN$, and let $f:\N{+\infty}\to \BBR$ where $\N{+\infty}$ is a neighbourhood of $+\infty$. We say that $f$ admits an {\em asymptotic expansion} of order $n$ about $+\infty$ if there exists a polynomial $p$ of degree $n$ such that $$ \forall x\in\N{a}\cap \loro{0}{+\infty}, \quad f(x)=p\left(\dfrac{1}{x}\right)+\smallo{+\infty}{\dfrac{1}{x^n}}. $$The polynomial $p$ is called the {\em regular part of the asymptotic expansion about $+\infty$ of $f$}. \end{df} \begin{thm} Let $f:\N{0}\rightarrow \BBR$ be a function with an asymptotic expansion $f(x)=p(x) + \smallo{0}{x^n}$, where $p$ is a polynomial. Then, if $f$ is even, then $p$ is even and if $f$ is odd, then $p$ is odd. \end{thm} \begin{pf} Let $f(x)=p(x)+\soo{x^n}$ as $x\to 0$, where $p$ is a polynomial of degree $n$. Then $f(-x)=p(-x) + \soo{x^n}$. If $f$ is even then $$p(x)+\soo{x^n}=f(x)=f(-x)=p(-x)+\soo{x^n},$$ and so by uniqueness of the regular part of an asymptotic expansion we must have $p(x)=p(-x)$, so $p$ is even. Similarly if $f$ is odd then $$-p(x)+\soo{x^n}=-f(x)=f(-x)=p(-x)+\soo{x^n},$$ and so by uniqueness of the regular part of an asymptotic expansion we must have $-p(x)=p(-x)$, so $p$ is odd. \end{pf} We want to expand the function $f$ in powers of $x-a$: $$f(x) = a_0 + a_1(x-a) + a_2(x-a)^2 + \cdots + a_n(x-a)^n + \cdots , $$ and that we will truncate at the $n$-th term, obtaining thereby a polynomial of degree $n$ in powers of $x-a$. We must determine what the coefficients $a_k$ are, and what the remainder $$f(x) - a_0 - a_1(x-a) - a_2(x-a)^2 - \cdots - a_n(x-a)^n = R(x) $$ is. We hope that this remainder is $\smallo{a}{(x-a)^n}$. The coefficients $a_k$ are easily found. For $0\leq k \leq n$ since $f$ is $n+1$ times differentiable, differentiating $k$ times, $$f^{(k)}(x) = k!a_k + ((k+1)(k)\cdots 2)a_{k+1}(x-a) + ((k+2)(k+1)\cdots 3)a_{k+2}(x-a)^2 + \cdots + R^{(k)}(x), \implies \dfrac{f^{(k)}(a)}{k!} = a_k,$$ as long as $R(a)=R'(a)=R''(a) = \cdots =R^{(n)}(a) = 0$. We write our ideas formally in the following theorems. \begin{thm}[Taylor-Lagrange Theorem] Let $I\subseteqq \BBR$, $I \neq \varnothing$ be an interval of $\BBR$ and let $f:I\rightarrow \BBR$ be $n+1$ times differentiable in $I$. Then if $(x, a)\in I^2, $, there exist $c$ with $\inf (x, a)a$. Consider the function $\phi : \lcrc{a}{x}\rightarrow \BBR$ with $$ \phi (t) = f(x)- \sum_{k=0}^{n} f^{(k)}(t)\, \dfrac{(x-t)^k}{k!} -R\dfrac{(x-t)^{n+1}}{(n+1)!}, $$ where $R$ is a constant. Observe that $\phi (x)=0$. We now choose the constant $R$ so that $\phi (a) =0$. Observe that $\phi$ is differentiable and that it satisfies the hypotheses of Rolle's Theorem on $\lcrc{a}{x}$. Therefore, there exists $c\in \loro{a}{x}$ such that $\phi ' (c) = 0$. Now $$\phi '(t)= - \sum_{k=1}^{n}\left( f^{(k+1)}(t) \dfrac{(x-t)^k}{k!} -f^{(k)}(t)\dfrac{(x-t)^{k-1}}{(k-1)!} \right) + R\dfrac{(x-t)^n}{n!} = -\dfrac{(x-t)^n}{n!}f^{(n+1)}(t) + R\dfrac{(x-t)^n}{n!}, $$from where we gather, that $R = f^{(n+1)}(c)$ and the theorem follows. \end{pf} \begin{cor}[Taylor-Young Theorem] Let $f:\N{a}\rightarrow \BBR$ be $n+1$ times differentiable in $\N{a}$. Then $f$ admits the asymptotic expansion of order $n$ about $a$: $$f(x) = f(a)+f'(a)(x-a) + \dfrac{f''(a)}{2!}(x-a)^2 + \dfrac{f^{(3)}(a)}{3!}(x-a)^3+\cdots + \dfrac{f^{(n)}(a)}{n!}(x-a)^n+\smallo{a}{(x-a)^n}. $$ \label{cor:Taylor-Young}\end{cor} \begin{pf}Follows at once from Theorem \ref{thm:Taylor-Lagrange}. \end{pf} The following theorem follows at once from Corollary \ref{cor:Taylor-Young}. \begin{thm} Let $x \rightarrow 0$. Then \begin{enumerate} \item $\sin x = x - \dfrac{x^3}{3!} + \dfrac{x^5}{5!} - \cdots + (-1)^n\dfrac{x^{2n + 1}}{(2n + 1)!} + \soo{x^{2n + 2}}$. \\ \item $\cos x = 1 - \dfrac{x^2}{2!} + \dfrac{x^4}{4!} - \cdots + (-1)^n\dfrac{x^{2n}}{(2n)!} + \soo{x^{2n + 1}}$.\item $\tan x = x + \dfrac{x^3}{3} + \dfrac{2x^5}{15} + \soo{x^5}$. \item $e^x = 1 + x + \dfrac{x^2}{2!} + \dfrac{x^3}{3!} + \cdots + \dfrac{x^{n}}{n!} + \soo{x^n}$ \item $\log (1 + x) = x - \dfrac{x^2}{2} + \dfrac{x^3}{3} - \cdots + (-1)^{n + 1}\dfrac{x^n}{n} + \soo{x^n}$. \item $(1 + x)^{\tau} = 1 + \tau x + \dfrac{\tau (\tau - 1)}{2}x^2 + \cdots + \dfrac{\tau (\tau - 1)(\tau - 2) (\tau - 3)\cdots (\tau - n + 1)}{n!}x^n + \soo{x^n}$. \end{enumerate} \label{thm:maclaurin}\end{thm} \begin{exa} Find an asymptotic development of $\log (2\cos x + \sin x)$ around $x = 0$ of order $\soo{x^4}$. \end{exa} \begin{solu} By theorem \ref{thm:maclaurin}, $$\begin{array}{ccc}2\cos x + \sin x & = & 2\left(1 - \dfrac{x^2}{2} + \dfrac{x^4}{24} + \soo{x^5}\right) + \left(x - \dfrac{x^3}{6} + \soo{x^4} \right) \\ & = & 2 + x - x^2 - \dfrac{x^3}{6} + \dfrac{x^4}{12} + \soo{x^4}\\ & = & 2\left(1 + \dfrac{x}{2} - \dfrac{x^2}{2} - \dfrac{x^3}{12} + \dfrac{x^4}{24} + \soo{x^4}\right) , \end{array}$$ and so, $$\begin{array}{cccl} \log (2\cos x + \sin x) & = & & \log 2\left(1 + \dfrac{x}{2} - \dfrac{x^2}{2} - \dfrac{x^3}{12} + \dfrac{x^4}{24} + \soo{x^4}\right) \\ & = & & \log 2 + \log \left(1 + \dfrac{x}{2} - \dfrac{x^2}{2} - \dfrac{x^3}{12} + \dfrac{x^4}{24} + \soo{x^4}\right) \\ & = & & \log 2 + \left(\dfrac{x}{2} - \dfrac{x^2}{2} - \dfrac{x^3}{12} + \dfrac{x^4}{24} + \soo{x^4}\right) \\ & & - & \dfrac{1}{2}\left(\dfrac{x}{2} - \dfrac{x^2}{2} - \dfrac{x^3}{12} + \dfrac{x^4}{24} + \soo{x^4}\right)^2 \\ & & + & \dfrac{1}{3}\left(\dfrac{x}{2} - \dfrac{x^2}{2} - \dfrac{x^3}{12} + \dfrac{x^4}{24} + \soo{x^4}\right)^3 \\ & & - & \dfrac{1}{4}\left(\dfrac{x}{2} - \dfrac{x^2}{2} - \dfrac{x^3}{12} + \dfrac{x^4}{24} + \soo{x^4}\right)^4 + \soo{x^4} \\ & = & & \log 2 + \left(\dfrac{x}{2} - \dfrac{x^2}{2} - \dfrac{x^3}{12} + \dfrac{x^4}{24} \right) - \dfrac{1}{2}\left(\dfrac{x^2}{4} - \dfrac{x^3}{2} + \dfrac{x^4}{6} \right) \\ & & + & \dfrac{1}{3}\left(\dfrac{x^3}{8} - \dfrac{3x^4}{8} \right) - \dfrac{1}{4}\cdot\dfrac{x^4}{16} + \soo{x^4} \\ & = & & \log 2 + \dfrac{x}{2} - \dfrac{5x^2}{8} + \dfrac{5x^3}{24} - \dfrac{35x^4}{192} + \soo{x^4} \end{array}$$ as $x \rightarrow 0$. \end{solu} \subsection*{Homework}\addcontentsline{toc}{subsection}{Homework} \begin{multicols}{2}\columnseprule 1pt \columnsep 25pt\multicoltolerance=900\small \begin{pro} Prove that the limit $$\lim _{n\to +\infty} \left(1+\dfrac{1}{2}+\dfrac{1}{3}+\cdots + \dfrac{1}{n}\right)-\log n, $$exists. The constant $$\gamma =\lim _{n\to +\infty} \left(1+\dfrac{1}{2}+\dfrac{1}{3}+\cdots + \dfrac{1}{n}\right)-\log n $$is called the {\em Euler-Mascheroni} constant. It is not known whether $\gamma$ is irrational. \end{pro} \end{multicols} \chapter{Integrable Functions} \section{The Area Problem} \begin{df} Let $f:\lcrc{a}{b}\rightarrow \BBR$ be bounded, say with $m \leq f(x)\leq M$ for all $x\in \lcrc{a}{b}$. Corresponding to each partition $\curlyP = \{x_0, x_1, \ldots , x_n\}$ of $\lcrc{a}{b}$, we define the {\em upper Darboux sum} $$U(f, \curlyP) = \sum _{k=1} ^n (\sup _{x_{k-1} \leq x\leq x_k}f(x))(x_k-x_{k-1}),$$ and the {\em lower Darboux sum} $$L(f, \curlyP) = \sum _{k=1} ^n (\inf _{x_{k-1} \leq x\leq x_k}f(x))(x_k-x_{k-1}).$$ Clearly $$L(f, \curlyP) \leq U(f, \curlyP). $$ Finally, we put $$ \overline{\dint} _a ^b f(x)\d{x} = \inf _{\curlyP\ \mathrm{is\ a \ partition\ of}\ \lcrc{a}{b}} U(f, \curlyP),$$which we call the {\em upper Riemann integral of $f$} and $$ \qquad \underline{\dint} _a ^b f(x)\d{x} = \sup _{\curlyP\ \mathrm{is\ a \ partition\ of}\ \lcrc{a}{b}} L(f, \curlyP). $$which we call the {\em lower Riemann integral of $f$}. \end{df} \begin{df} Let $f:\lcrc{a}{b}\rightarrow \BBR$ be bounded. We say that {\em $f$ is Riemann integrable} if $ \overline{\dint} _a ^b f(x)\d{x}=\underline{\dint\limits} _a ^b f(x)\d{x} $. In this case, we denote their common value by $\dint _a ^b f(x)\d{x} $ and call it the {\em Riemann integral of $f$ over $\lcrc{a}{b}$}. \end{df} \begin{thm}\label{thm:refinement-ineqs-upper-lower-sums} Let $f$ be a bounded function on $\lcrc{a}{b}$ and let $\curlyP\subseteqq \curlyP'$ be two partitions of $\lcrc{a}{b}$. Then $$ L(f, \curlyP )\leq L(f, \curlyP ')\leq U(f, \curlyP ' )\leq U(f, \curlyP ). $$ \end{thm} \begin{pf} Clearly is enough to prove this when $\curlyP '$ has exactly one more point than $\curlyP$. Let $$ \curlyP =\{x_0, x_1, \ldots , x_n\} $$with $a=x_0 0$, $\exists \curlyP$ a partition of $\lcrc{a}{b}$ such that $$ U(f, \curlyP)-L(f, \curlyP) < \varepsilon . $$ \end{thm} \begin{pf} \begin{description} \item[$\Leftarrow$] If for all $\varepsilon > 0$, $ U(f, \curlyP)-L(f, \curlyP) < \varepsilon $ then by Theorem \ref{thm:riemann-on-top-dominates-riemann-on-bottom}, $$ L(f, \curlyP)\leq \underline{\dint} _a ^b f(x)\d{x} \leq \overline{\dint} _a ^b f(x)\d{x} \leq U(f, \curlyP) \implies 0 \leq \overline{\dint} _a ^b f(x)\d{x} - \underline{\dint} _a ^b f(x)\d{x} <\varepsilon, $$ and so $\overline{\dint} _a ^b f(x)\d{x} = \underline{\dint} _a ^b f(x)\d{x}$, which means that $f$ is Riemann-integrable. \item[$\implies$] Suppose $f$ is Riemann integrable. By the Approximation property of the supremum and infimum, for all $\varepsilon > 0$ there exist partitions $\curlyP_1$ and $\curlyP_2$ such that $$ U(f, \curlyP_2) -\int _a ^b f(x)\d{x}<\dfrac{\varepsilon}{2}, \qquad \int _a ^b f(x)\d{x}-L(f, \curlyP_1)<\dfrac{\varepsilon}{2}. $$ Hence by taking $\curlyP = \curlyP_1 \cup \curlyP_2$ then $$U(f, \curlyP)\leq U(f, \curlyP_2) <\int _a ^b f(x)\d{x} + \dfrac{\varepsilon}{2} < L(f, \curlyP_1) + \varepsilon < L(f, \curlyP)+\varepsilon, $$ from where $ U(f, \curlyP)-L(f, \curlyP) < \varepsilon $. \end{description} \end{pf} \begin{exa} \begin{itemize} \item $ f(x)= \begin{cases} 0& \text{$ x $ irrational,} \\ 1& \text{$ x $ rational.} \end{cases} \quad x \in [0;1] $ Then $ U(f, \curlyP)=1, L(f, \curlyP)=0 $, for any partition $\curlyP$, and so $ f $ is not Riemann integrable. \item $ f(x)= \begin{cases} 0& \text{$ x $ irrational,} \\ \frac{1}{q}& \text{$ x $ rational = $ \frac{p}{q} $ in lowest terms.} \end{cases} \quad x \in [0;1] $ is Riemann integrable with $$ \int_{0}^{1}f(x)\,\d{x} =0 $$ \end{itemize} \end{exa} \begin{df} Let $f$ be a bounded function on $\lcrc{a}{b}$ and let $\curlyP = \{x_0, x_1, \ldots , x_n\}$ be a partition of $\lcrc{a}{b}$. If $t_k$ are selected so that $x_{k-1}\leq t_k\leq x_k$, put $$ S(f, \curlyP) = \sum _{k=1} ^n f(t_k)(x_k-x_{k-1}), $$is the {\em Riemann sum of $f$ associated with $\curlyP$.}\end{df} \begin{thm}\label{thm:integrable-if-dominated-by-linear-combination} Let $f_1, f_2, \ldots , f_m$ be Riemann integrable over $\lcrc{a}{b}$, and let $f:\lcrc{a}{b}\rightarrow \BBR$. If for any subinterval $I \subseteqq \lcrc{a}{b}$ there exists strictly positive numbers $a_1, a_2, \ldots , a_m$ such that $$ \omega (f, I) \leq a_1\omega (f_1, I)+ a_2 \omega (f_2, I)+ \cdots + a_m\omega (f_m,I), $$then $f$ is also Riemann integrable. \end{thm} \begin{pf} Let $\curlyP = \{a=x_00$, also $\dfrac{1}{g}$ \item provided $\inf _{x\in [a;b]} \absval{g(x)}>0$, also $\dfrac{f}{g}$ \end{enumerate}\label{thm:algebra-of-int-functions} \end{thm} \begin{pf} Since $$\absval{f(x)+\lambda g(x) -f(t)-\lambda g(t)} \leq \absval{f(x)-f(t)}+\absval{\lambda}\absval{g(x)-g(t)}, \qquad \mathrm{and} \qquad \absval{\absval{f(x)-\absval{f(t)}}} \leq \absval{f(x)-f(t)}, $$we have $$ \omega (f+\lambda g, I) \leq \omega (f, I)+\absval{\lambda} \omega (g, I) \qquad \mathrm{and} \qquad \omega (\absval{f}, I) \leq \omega (f, I),$$ from where the first two assertions follow, upon appealing to Theorem \ref{thm:integrable-if-dominated-by-linear-combination}. \bigskip To prove the third assertion, put $a_1=\sup _{u\in [a;b]}\absval{f(u)}$ and $a_2=\sup _{u\in [a;b]}\absval{g(u)}$ $$\begin{array}{lll} \absval{f(x)g(x)-f(t)g(t)} & = & \absval{f(x)(g(x)-g(t))+g(t)(f(x)-f(t))} \\ & \leq & \absval{f(x)}\absval{g(x)-g(t)}+\absval{g(t)}\absval{f(x)-f(t)} \\ & \leq & \left(\sup _{u\in [a;b]}\absval{f(u)}\right)\absval{g(x)-g(t)} + \left(\sup _{u\in [a;b]}\absval{g(u)}\right) \absval{f(x)-f(t)}\\ & = & a_1\absval{g(x)-g(t)}+a_2\absval{f(x)-f(t)},\\ \end{array}$$ which gives $$\omega (fg, I) \leq a_1\omega (f, I) + a_2\omega (g, I), $$and so the third assertion follows from Theorem \ref{thm:integrable-if-dominated-by-linear-combination}. \bigskip To prove the fourth assertion, with $a= \inf _{x\in [a;b]} \absval{g(x)}>0$, observe that we have $$\begin{array}{lll} \absval{\dfrac{1}{g(x)}-\dfrac{1}{g(t)}} & = & \dfrac{1}{\absval{g(x)g(t)}}\absval{g(x)-g(t)} \\ & \leq & \dfrac{1}{a^2}\absval{g(x)-g(t)}, \\ \end{array}$$and this gives $\omega (\dfrac{1}{g}, I)\leq \dfrac{1}{a^2}\omega (g, I)$. The fourth assertion now follows by again appealing to Theorem \ref{thm:integrable-if-dominated-by-linear-combination}. \bigskip The fifth assertion follows from the third and the fourth.\end{pf} \begin{thm}\label{thm:additivity-of-integrals} Let $f$ and $g$ be Riemann integrable functions on $\lcrc{a}{b}$ and let $\lambda \in \BBR$ be a constant. Then $$ \dint _a ^b (f(x) + \lambda g(x))\d{x} = \dint _a ^b f(x) \d{x}+ \lambda\dint _a ^bg(x) \d{x}. $$ \end{thm} \begin{pf} Let $\curlyP = \{a=x_0 0$ there exist $\delta >0$ and $\delta'>0$ such that $$\absval{\sum _{k=1} ^n f(t_k)(x_k-x_{k-1}) -\int _a ^b f(x)\d{x}}<\dfrac{\varepsilon}{2} \qquad \mathrm{if}\quad \norm{\curlyP}<\delta, $$ $$\absval{\lambda\sum _{k=1} ^n g(t_k)(x_k-x_{k-1}) -\lambda \int _a ^b g(x)\d{x}}<\dfrac{\varepsilon}{2} \qquad \mathrm{if}\quad \norm{\curlyP}<\delta'. $$ Hence, if $\norm{\curlyP}<\min (\delta , \delta')$, $$\begin{array}{l}\absval{\sum _{k=1} ^n \left(f(t_k)+\lambda g(t_k)\right)(x_k-x_{k-1}) -\int _a ^b f(x)\d{x}-\lambda \int _a ^b g(x)\d{x}} \\ \qquad \leq \absval{\sum _{k=1} ^n f(t_k)(x_k-x_{k-1}) -\int _a ^b f(x)\d{x}} +\absval{\lambda \sum _{k=1} ^n g(t_k)(x_k-x_{k-1}) -\lambda\int _a ^b g(x)\d{x}} \\ \qquad < \varepsilon \end{array}$$proving the theorem.\end{pf} \begin{thm}\label{thm:monotonicity-of-integral-operator} Let $f$ and $g$ be Riemann integrable functions on $\lcrc{a}{b}$ with $f(x)\leq g(x)$ for all $x\in \lcrc{a}{b}$. Then $$ \dint _a ^b f(x)\d{x} \leq \dint _a ^bg(x) \d{x}. $$ \end{thm} \begin{pf} The function $h=g-f$ is positive for all $x\in [a;b]$ and hence $L(h, \curlyP)\geq 0$ for all partitions $\curlyP$. It is also integrable by Theorem \ref{thm:additivity-of-integrals}. Thus $$ \int _a ^b h(x)\d{x} = \underline{\int} _a ^b h(x)\d{x}\geq 0.$$But $$ \int _a ^b h(x)\d{x} \geq 0 \implies 0\leq \int _a ^b (g(x)-f(x))\d{x}=\int _a ^b g(x)\d{x}-\int _a ^b f(x)\d{x},$$ and so $\dint _a ^b f(x)\d{x} \leq \dint _a ^bg(x) \d{x}$, as claimed. \end{pf} \begin{thm}[Triangle Inequality for Integrals] Let $f$ be a Riemann integrable function on $\lcrc{a}{b}$. Then $$ \absval{\dint _a ^b f(x)\d{x}} \leq \dint _a ^b\absval{f(x)} \d{x}. $$ \end{thm} \begin{pf} By Theorem \ref{thm:algebra-of-int-functions}, $\absval{f}$ is integrable. Now, since $-\absval{f}\leq f \leq \absval{f}$ we just need to apply Theorem \ref{thm:monotonicity-of-integral-operator} twice. \end{pf} \begin{thm}[Chasles' Rule]\label{thm:chasles} Let $f$ be a Riemann integrable function on $\lcrc{a}{b}$ and let $c\in \loro{a}{b}$. Then $f$ is Riemann integrable function on $\lcrc{a}{c}$ and $\lcrc{c}{b}$. Moreover, $$ \dint _a ^b f(x)\d{x} = \dint _a ^c f(x)\d{x} + \dint _c ^b f(x)\d{x}. $$ Conversely, if $c\in \loro{a}{b}$ and $f$ is Riemann integrable on $\lcrc{a}{c}$ and $\lcrc{c}{b}$ then $f$ is Riemann integrable on $\lcrc{a}{b}$ and $$ \dint _a ^b f(x)\d{x} = \dint _a ^c f(x)\d{x} + \dint _c ^b f(x)\d{x}. $$ \end{thm} \begin{pf} Consider the partitions $$\curlyP = \{a=x_00$, we choose $\curlyP$ so that $$ U(f, \curlyP)-L(f, \curlyP)<\varepsilon. $$It follows that $$ \left( U(f, \curlyP')-L(f, \curlyP')\right)+ \left( U(f, \curlyP'')-L(f, \curlyP'')\right)= U(f, \curlyP)-L(f, \curlyP)<\varepsilon.$$ Hence $f$ is Riemann-integrable over both $[a;c]$ and $[c;b]$. Observe that $$0 \leq U(f, \curlyP') -\int _a ^c f(x)\d{x} <\varepsilon, \qquad 0 \leq U(f, \curlyP'') -\int _c ^b f(x)\d{x} <\varepsilon ,$$ $$0 \leq \int _a ^c f(x)\d{x} -L(f, \curlyP') <\varepsilon, \qquad 0 \leq \int _c ^b f(x)\d{x}- L(f, \curlyP'') <\varepsilon ,$$ and upon addition, $$ 0 \leq U(f, \curlyP)- \left(\int _a ^c f(x)\d{x}+\int _c ^b f(x)\d{x}\right) < 2\varepsilon, $$ $$ 0 \leq \left(\int _a ^c f(x)\d{x}+\int _c ^b f(x)\d{x}\right)-L(f, \curlyP) < 2\varepsilon, $$ whence $$ \dint _a ^b f(x)\d{x} = \dint _a ^c f(x)\d{x} + \dint _c ^b f(x)\d{x},$$as required. \end{pf} \begin{thm}[Converse of Chasles' Rule]\label{thm:chasles-converse} Let $f$ be a function defined on the interval $[a;b]$ and let $c\in]a;b[$. If $f$ is Riemann-integrable on $[a;c]$ and $[c;b]$ then it is also Riemann integrable in $[a;b]$ and $$ \dint _a ^b f(x)\d{x} = \dint _a ^c f(x)\d{x} + \dint _c ^b f(x)\d{x}. $$ \end{thm} \begin{pf} Since $f$ is Riemann-integrable on both subintervals, it is bounded there, and so it is bounded on the larger subinterval. By Theorem \ref{thm:riemann-int-if-upper-lower-sums-small}, given $\varepsilon > 0$ there exist partitions $\curlyP'$ and $\curlyP''$ such that $$U_{[a;c]}(f, \curlyP')-L_{[a;c]}(f, \curlyP')<\varepsilon, \qquad U_{[c;b]}(f, \curlyP'')-L_{[c;b]}(f, \curlyP'')<\varepsilon. $$ The above inequalities also hold in the refinement $\curlyP = \curlyP' \cup \curlyP''$, and $$U(f, \curlyP)= U_{[a;c]}(f, \curlyP)+U_{[c;b]}(f, \curlyP), \qquad L(f, \curlyP)= L_{[a;c]}(f, \curlyP)+L_{[c;b]}(f, \curlyP). $$ We then deduce that $$\begin{array}{lll}U(f, \curlyP)-L(f, \curlyP) & = & \left(U_{[a;c]}(f, \curlyP)+U_{[c;b]}(f, \curlyP)\right)- \left(L_{[a;c]}(f, \curlyP)+L_{[c;b]}(f, \curlyP)\right)\\ & = & \left(U_{[a;c]}(f, \curlyP)-L_{[a;c]}(f, \curlyP)\right)- \left(U_{[a;c]}(f, \curlyP)-L_{[c;b]}(f, \curlyP)\right)\\ & < & 2\varepsilon, \end{array}$$and so $f$ is Riemann integrable in $[a;b]$ by virtue of Theorem \ref{thm:riemann-int-if-upper-lower-sums-small}. Now $$ \begin{array}{lll}\int _a ^b f(x)\d{x} & \leq & U(f, \curlyP) \\ & < & L(f, \curlyP)+\varepsilon\\ & = & L_{[a;c]}(f, \curlyP)+L_{[c;b]}(f, \curlyP)+\varepsilon\\ & \leq & \int _a ^c f(x)\d{x}+\int _c ^b f(x)\d{x}+\varepsilon, \\ \end{array}$$ and similarly $$ \begin{array}{lll}\int _a ^b f(x)\d{x} & \geq & L(f, \curlyP) \\ & > & U(f, \curlyP)-\varepsilon\\ & = & U_{[a;c]}(f, \curlyP)+U_{[c;b]}(f, \curlyP)-\varepsilon\\ & \geq & \int _a ^c f(x)\d{x}+\int _c ^b f(x)\d{x}-\varepsilon, \\ \end{array}$$ hence $$\int _a ^c f(x)\d{x}+\int _c ^b f(x)\d{x}-\varepsilon \leq \int _a ^b f(x)\d{x}\leq \int _a ^c f(x)\d{x}+\int _c ^b f(x)\d{x}+\varepsilon $$ giving the desired equality between integrals.\end{pf} \begin{thm} Let $f$ be Riemann integrable over $[a;b]$ and let $g:\lcrc{\inf _{u\in [a;b]}f(u)}{\sup _{u\in [a;b]}f(u)}\rightarrow \BBR$ be continuous. Then $g\circ f$ is Riemann integrable on $[a;b]$. \end{thm} \begin{pf} Since $g$ is uniformly continuous on the compact interval $\lcrc{\inf _{u\in [a;b]}f(u)}{\sup _{u\in [a;b]}f(u)}$, for given $\varepsilon > 0$ we may find $\delta'$ such that $$ (s,t)\in \lcrc{\inf _{t\in [a;b]}f(t)}{\sup _{u\in [a;b]}f(u)}^2; \quad \absval{s-t}<\delta' \implies \absval{f(s)-f(t)}<\varepsilon. $$Let $\delta = \min (\delta', \varepsilon)$. Since $f$ is Riemann-integrable, we may choose a partition $\curlyP = \{a=x_0 0$ and $\delta > 0$, put $\varepsilon = \gamma \delta$. Let $f$ be Riemann integrable. There is a partition $\curlyP = \{a=x_0< x_1< \cdots < x_n=b\}$ such that $$ U(f, \curlyP)-L(f, \curlyP)< \varepsilon. $$Let $x\in\loro{x_i}{x_{i+1}}$ be such that $\omega (f, x)\geq \gamma$. Then $$\sup _{\loro{x_i}{x_{i+1}}}f(x) - \inf _{\loro{x_i}{x_{i+1}}}f(x) \geq \gamma. $$Now observe that $$\{x\in[a;b]: \omega (f, x)\geq \delta\} = \left( \bigcup _{\sup f-\inf f \geq \gamma} \loro{x_i}{x_{i+1}}\right) \cup \{x_0, x_1, \ldots , x_n\}. $$ Hence $$\begin{array}{lll} \mu \left(\{x\in[a;b]: \omega (f, x)\geq \gamma\}\right)& \leq & \sum _{\sup _{]x_i;x_{i+1}[}f(x) - \inf _{]x_i;x_{i+1}[}f(x) \geq \gamma}\absval{x_{i+1}-x_i}\\ & \leq & \dfrac{1}{\gamma}\sum _i \absval{x_{i+1}-x_i}\left(\sup _{\loro{x_i}{x_{i+1}}}f(x) - \inf _{\loro{x_i}{x_{i+1}}}f(x)\right) \\ & \leq & \dfrac{1}{\gamma} \left(U(f, \curlyP)-L(f, \curlyP)\right)\\ & < & \dfrac{\varepsilon}{\gamma}\\ & = & \delta. \end{array}$$ Letting $\delta \to 0+$ and $\gamma\to 0+$ we get $\mu \left(\{x\in[a;b]: \omega (f, x)\geq 0\}\right) = 0$, and in particular, $\mu \left(\{x\in[a;b]: \omega (f, x)> 0\}\right) = 0$ which means hat the set of discontinuities is a set of measure $0$. \item[$\Leftarrow$] Conversely, assume $\meas{\{x\in[a;b]: \omega (f, x)> 0\}} = 0$. We can write $$\{x\in[a;b]: \omega (f, x)> 0\} = \bigcup _{K \geq 1} \{x\in[a;b]: \omega (f, x)> \dfrac{1}{K}\}. $$ Fix $K$ large enough so that $\dfrac{1}{K}<\varepsilon$. Since $\meas{\{x\in[a;b]: \omega (f, x)\geq \dfrac{1}{K}\}}=0$, we can find open intervals $I_j(K)$ such that $$\{x\in[a;b]: \omega (f, x)\geq \dfrac{1}{K}\} \subseteqq \bigcup _{j\geq 1}I_j(K), \qquad \sum _{j\geq 1}\meas{I_j(K)}<\varepsilon. $$ It is easy to shew that $\left\{x\in[a;b]: \omega (f, x)> \dfrac{1}{K}\right\}$ is closed and bounded and hence compact, so we may find a finite subcover with $$ \{x\in[a;b]: \omega (f, x)> \dfrac{1}{K}\} \subseteqq I_{j_1}\cup I_{j_2}\cup \cdots \cup I_{j_N}. $$ Now $$\lcrc{a}{b}\setminus \left(I_{j_1}\cup I_{j_2}\cup \cdots \cup I_{j_N}\right) $$is a finite disjoint union of closed intervals, say $J_1\cup J_2\cup \cdots \cup J_M$. If $x\in J_i$ then $\omega (f, x)<\dfrac{1}{K}$. Thus on each of the $J_i$ we may find so fine a partition that $\omega (f, L)<\dfrac{1}{K}$ for every interval such partition. All these partitions and the endpoints of the $I_{j_k}$ form a partition, say $\curlyP$. Write ${\cal S} = S_1\cup S_2 \cup \cdots \cup S_M$ for the intervals of the partition $\curlyP$ that are not the $I_{j_k}$. Observe that $\omega (f, S_k) < \dfrac{1}{K}$. Then $$\begin{array}{lll} U(f, \curlyP)-L(f, \curlyP) & = & \sum _{I_{j_k}} \left(\sup _{I_{j_k}}f - \inf _{I_{j_k}}f\right)\left(\meas{I_{j_k}}\right) + \sum _{S_k} \left(\sup _{S_k}f - \inf _{S_k}f\right)\left(\meas{S_k}\right)\\ & \leq & 2\sup _{[a;b]} \absval{f}\sum _{k=1} ^N \meas{I_{j_k}}+ \dfrac{1}{K}\sum _{S_k} \meas{S_k} \\ & \leq & 2\sup _{[a;b]} \absval{f}\varepsilon + (b-a)\varepsilon\\ & = & \left(2\sup _{[a;b]} \absval{f} + (b-a)\right)\varepsilon. \\ \end{array}$$ This proves the theorem. \end{description} \end{pf} \begin{cor} Every continuous function $f$ on $[a;b]$ is Riemann integrable on $\lcrc{a}{b}$. \end{cor} \begin{pf} This is immediate from Theorem \ref{thm:riemann-int-iff-continuous-almost-everywhere}. \end{pf} \begin{cor} Every monotonic function $f$ on $\lcrc{a}{b}$ is Riemann integrable on $\lcrc{a}{b}$. \end{cor} \begin{pf} Since a countable set has measure $0$, and since the set of discontinuities of a monotonic function is countable (Theorem \ref{thm:discontinuities-of-monotone-functions}), the result is immediate.\end{pf} \begin{pspicture}[plotpoints=200]( -0.5 , -3)(10 ,2.5)% \psStep[algebraic , fillstyle=solid , fillcolor= yellow ](0.001,9.5){40}{2* sqrt(x)*cos(ln(x))*sin(x)}% \psStep[algebraic ,StepType= Riemann , fillstyle= solid , fillcolor=blue ](0.001 ,9.5) {40}{2*sqrt(x)* cos(ln(x))* sin(x)}% \psaxes{->}(0 ,0)(0 , -2.75)(10 ,2.5) \psplot [ algebraic , linecolor= white]{0.001}{9.75}{2* sqrt(x)* cos(ln(x))* sin(x)}% \uput [90](6 ,1.2){$f(x)=2\cdot\sqrt{x}\cdot\cos {(\log{x})}\cdot\sin{x}$}% \end{pspicture} \subsection*{Homework}\addcontentsline{toc}{subsection}{Homework} \begin{multicols}{2}\columnseprule 1pt \columnsep 25pt\multicoltolerance=900\small \begin{pro}\label{pro:epsilon-delta-riemann-integrability} Let $f$ be a bounded function on $\lcrc{a}{b}$. Then $f$ is Riemann integrable if and only if $\forall \varepsilon > 0$, $\exists \delta > 0$ such that for all partitions $\curlyP$ of $\lcrc{a}{b}$, $$ \norm{\curlyP}< \delta \implies U(f, \curlyP)-L(f, \curlyP) < \varepsilon . $$ \begin{answer} \begin{description} \item[$\Leftarrow$] This follows directly from Theorem \ref{thm:riemann-int-if-upper-lower-sums-small}. \item[$\implies$] If $f$ is Riemann integrable, let $\varepsilon > 0$ and let $\curlyP' = \{a=y_0< y_1 < \ldots < y_m=b\}$ be a partition with $m+1$ points such that $$U(f, \curlyP')-L(f, \curlyP')<\dfrac{\varepsilon}{2}. $$ As $f$ is bounded, there is $M>0$ such that $\forall x\in \lcrc{a}{b}, \quad\absval{f(x)}\leq M$. Take $\delta = \dfrac{\varepsilon}{8mM}$ and consider now an arbitrary partition $\curlyP = \{a=x_0< x_1 < \ldots < x_n=b\}$ with norm $\norm{\curlyP}<\delta$. Put $\curlyP'' = \curlyP \cup \curlyP'$. Arguing as in Theorem \ref{thm:refinement-ineqs-upper-lower-sums}, we obtain $$L(f, \curlyP'')-L(f, \curlyP)< 2mM\norm{\curlyP}<2mM\delta=\dfrac{\varepsilon}{4}. $$ Since by Theorem \ref{thm:upper-lower-riemann-sums-ineq} $L(f, \curlyP')\leq L(f, \curlyP'')$ we gather $$ L(f, \curlyP')-L(f, \curlyP)<\dfrac{\varepsilon}{4}. $$ In a similar fashion we establish that $$ U(f, \curlyP)-U(f, \curlyP')<\dfrac{\varepsilon}{4}, $$ and upon assembling the inequalities, $$U(f, \curlyP)-L(f, \curlyP)< U(f, \curlyP')-L(f, \curlyP')+\dfrac{\varepsilon}{2}<\varepsilon, $$since we had assumed that $ U(f, \curlyP')-L(f, \curlyP')<\dfrac{\varepsilon}{2}$. \end{description}\end{answer} \end{pro} \begin{pro} Let $f$ be a bounded function on $\lcrc{a}{b}$. Then $f$ is Riemann integrable on $\lcrc{a}{b}$ if and only if $$ \lim _{\norm{\curlyP}\rightarrow 0}S(f, \curlyP) $$exists and is finite. In this case we write $ \lim _{\norm{\curlyP}\rightarrow 0}S(f, \curlyP) = \dint _a ^b f(x)\d{x}$. \begin{answer} \begin{description} \item[$\implies$] Assume $f$ is Riemann-integrable. For $\varepsilon > 0$ let $\delta > 0$ be chosen so that the conditions of Theorem \ref{thm:epsilon-delta-riemann-integrability} be fulfilled. By definition of a Riemann sum, $$L(f, \curlyP)\leq S(f, \curlyP)\leq U(f, \curlyP), $$and therefore $$ U(f, \curlyP)< L(f, \curlyP)+\varepsilon \leq \underline{\int} _a ^b f(x)\d{x} + \varepsilon = \int _a ^b f(x)\d{x} + \varepsilon $$ and $$L(f, \curlyP) > U(f, \curlyP) - \varepsilon \geq \overline{\int} _a ^b f(x)\d{x} - \varepsilon =\int _a ^b f(x)\d{x} - \varepsilon . $$These inequalities give $$ \absval{S(f, \curlyP) - \int _a ^b f(x)\d{x}}< \varepsilon , $$whence $ \lim _{\norm{\curlyP}\rightarrow 0}S(f, \curlyP) = \int _a ^b f(x)\d{x}. $ \item[$\Leftarrow$] Suppose that $\lim _{\norm{\curlyP}\rightarrow 0}S(f, \curlyP) =L$, existing and finite. Given $\varepsilon > 0$ there is $\delta > 0$ such that $\norm{\curlyP}<\delta$ implies \begin{equation}\label{eq:ineq-lower-upper-darboux} L-\dfrac{\varepsilon}{3}< S(f, \curlyP)< A + \dfrac{\varepsilon}{3}. \end{equation} Now, choose $\curlyP = \{a=x_0 0, \varepsilon ' >0$ there is a partition $\curlyP$ of $\lcrc{a}{b}$ such that $$ \sum _{k=1} ^n (x_k-x_{k-1})\chi_{\{x\in[a;b]:\omega (f, [x_{k-1}; x_k]) \geq \varepsilon' \}}<\varepsilon . $$Here $\chi (.)$ is the {\em indicator function} defined on a set $E$ as $$ \chi _E(x) = \left\{\begin{array}{ll}1 & \mathrm{if}\ x\in E \\ 0 & \mathrm{if}\ x\not\in E \\ \end{array}\right. .$$ \begin{answer} \begin{description} \item[$\implies$] Let $\curlyP = \{a=x_0 0$, in view of Theorem \ref{thm:riemann-int-if-upper-lower-sums-small}, there is a partition $\curlyP = \{a=x_00$ such that $\forall x\in [a;b], \quad \absval{f(x)}\leq M$. Now, if $a \leq x < y \leq b$ with $\absval{x-y}<\dfrac{\varepsilon}{M}$, then $$ \absval{F(y)-F(x)} = \absval{\int _x ^y f(t)\d{t}} \leq \int _x ^y \absval{f(t)}\d{t}\leq \int _x ^y M\d{t} =M(y-x)<\varepsilon $$ Thus $F$ is continuous on $[a;b]$ and by Heine's Theorem, uniformly continuous on $[a;b]$. Now, take $u\in ]a;b[ $, and observe that $$x\neq u \implies \dfrac{F(x)-F(u)}{x-u} = \dfrac{1}{x-u} \int _u ^x f(t)\d{t}.$$ Moreover, $$f(u) = \dfrac{1}{x-u} \int _u ^x f(u)\d{t},$$and therefore, $$ \dfrac{F(x)-F(u)}{x-u} -f(u) = \dfrac{1}{x-u} \int _u ^x \left(f(t)-f(u)\right)\d{t}. $$ Since $f$ is continuous at $u$, there is $\delta > 0$ such that $$t\in [a;b], \absval{t-u}<\delta \implies \absval{f(t)-f(u)}<\varepsilon. $$ This gives $$\absval{\dfrac{F(x)-F(u)}{x-u} -f(u)}<\varepsilon $$ for $x\in ]a;b[$ with $\absval{x-u}< \delta$. From this it follows that $F'(u) = f(u)$.\end{pf} \begin{thm}[Young's Inequality for Integrals] Let $f$ be a strictly increasing continuous function on $\lcro{0}{+\infty}$ and let $f(0)=0$. If $A>0$ and $B>0$ then $$AB \leq \dint _0 ^A f(x)\d{x} + \dint _0 ^B f^{-1}(x)\d{x}. $$ \label{thm:young's-ineq-integrals} \end{thm} \begin{pf} The inequality is evident from Figure \ref{fig:young's-ineq-integrals}. The rectangle of area $AB$ fits nicely in the areas under the curves $y=f(x), x\in [0;A]$ and $x=f^{-1}(y), y\in [0;B]$. \end{pf} \begin{thm}[H\"{o}lder's Inequality for Integrals] Let $p>1$ and put $\dfrac{1}{q} = 1-\dfrac{1}{p}$. If $f$ and $g$ are Riemann integrable on $\lcrc{a}{b}$ then $$ \absval{\dint _a ^bf(x)g(x)\d{x}} \leq \left(\dint _a ^b \absval{f(x)}^p\d{x} \right)^{1/p} \left(\dint _a ^b \absval{g(x)}^q\d{x} \right)^{1/q}.$$ \end{thm} \begin{pf} First observe that all of $\absval{fg}, \absval{f}^p$ and $\absval{g}^q$ are Riemann-integrable, in view of Theorem \ref{thm:algebra-of-int-functions}. Now, with $f(x) = x^{p-1}$ in Young's Inequality (Theorem \ref{thm:young's-ineq-integrals}), we obtain, \begin{equation}\label{eq:young} AB \leq \dfrac{A^{p}}{p} + \dfrac{B^{1/(p-1)+1}}{1/(p-1)+1} = \dfrac{A^{p}}{p}+\dfrac{B^q}{q}.\end{equation} If any of the integrals in the statement of the theorem is zero, the result is obvious. Otherwise put $A^p=\int _a ^b \absval{f(x)}^p\d{x}$, $B^q=\int _a ^b \absval{g(x)}^p\d{x}$. Then by (\ref{eq:young}), $$ \dfrac{\absval{f(x)g(x)}}{AB} \leq \dfrac{A^{-p}\absval{f(x)}^p}{p}+ \dfrac{B^{-q}\absval{g(x)}^q}{q}. $$ Integrating throughout the above inequality, $$ \dfrac{1}{AB}\int _a ^b \absval{f(x)g(x)}\d{x}\leq \dfrac{1}{pA^p}\int _a ^b \absval{f(x)}^p\d{x} + \dfrac{1}{qB^q}\int _a ^b \absval{g(x)}^q\d{x} = \dfrac{1}{p}+\dfrac{1}{q}=1, $$whence the theorem follows. \end{pf} \vspace{2cm} \begin{figure}[h] \centering \psset{unit=1pc} \psaxes[labels=none,ticks=none]{->}(0,0)(0,0)(5,5) \uput[d](1,0){A} \uput[l](0,4){B} \psline[linestyle=dotted](1,0)(1,1.420735492) \psline[linestyle=dotted](0,4)(4.487398067,4) \psplot[linewidth=2pt,algebraic]{0}{4.487398067}{x + sin(x)/2} \pscustom[fillstyle=solid,fillcolor=red]{\psplot[algebraic]{0}{1}{x+sin(x)/2}\psline(1,1.420735492)(1,0)(0,0)} \pscustom[fillstyle=solid,fillcolor=blue]{\psplot[algebraic]{1}{4.487398067}{x+sin(x)/2}\psline(4.487398067,4)(0,4)(0,0)} \vspace{1cm} \hangcaption{Young's Inequality (Theorem \ref{thm:young's-ineq-integrals}).}\label{fig:young's-ineq-integrals} \end{figure} \vspace{1cm} \begin{thm} Let $f:\lcrc{a}{b}\rightarrow \BBR$. Then \begin{enumerate} \item If $f$ is continuous on $\lcrc{a}{b}$, $\forall x\in \lcrc{a}{b}, f(x)\geq 0$, $\exists c\in \lcrc{a}{b}$ with $f(c)>0$ then $\dint _a ^b f(x)\d{x}>0$. \item If $f,g$ are continuous on $\lcrc{a}{b}$, $\forall x\in \lcrc{a}{b}, f(x)\leq g(x)$, and $\exists c\in \lcrc{a}{b}$ with $f(c) 0$, such that $\forall x\in ]c-\delta; c+\delta[, f(x) \geq \dfrac{f(c)}{2}$. Therefore $$ \dint _a ^b f(x)\d{x}\geq \dint _{c-\delta} ^{c+\delta} f(x)\d{x}>\dint _{c-\delta} ^{c+\delta} \dfrac{f(c)}{2}\d{x} = \delta f(c)>0. $$ If $c=a$ then we consider a neighbourhood of the form $]a;a+\delta[$, and similarly if $c=b$, we consider a neighbourhood of the form $]b-\delta;b[$ \end{pf} \begin{thm}[First Mean Value Theorem for Integrals] Let $f, g$ be continuous on $\lcrc{a}{b}$, with $g$ of constant sign on $\lcrc{a}{b}$. Then there exists $c\in \loro{a}{b}$ such that $$\dint _a ^b f(x)g(x)\d{x} = f(c)\dint _a ^b g(x)\d{x}. $$ \label{thm:first-mvt-integrals} \end{thm} \begin{pf} If $g$ is identically $0$, there is nothing to prove. Similarly, if $f$ is constant in $[a;b]$ there is nothing to prove. Otherwise, $g$ is always strictly positive or strictly negative in the interval $[a;b]$. Let $$m =\inf _{x\in [a;b]}f(x); \quad M =\sup _{x\in [a;b]}f(x). $$ Then $$ m< \dfrac{\int _a ^b f(x)g(x)\d{x}}{\int _a ^bg(x)\d{x}} 0$. Find $\dint _0 ^a \dfrac{1}{f(x) + 1}\d{x}$. \begin{answer} Put $I = \int _0 ^a \frac{1}{f(x) + 1}{\rm d}x$. We have $$I = \int _0 ^a \frac{1}{f(u) + 1}{\rm d}u =\int _0 ^a \frac{f(u)f(a-u)}{f(u) + f(u)f(a-u)}{\rm d}u = \int _0 ^a \frac{f(a-u)}{1 + f(a-u)}{\rm d}u = -\int _a ^0 \frac{f(v)}{1 + f(v)}{\rm d}v = \int _0 ^a \frac{f(u)}{1 + f(u)}{\rm d}u ,$$whence $$2I = \int _0 ^a \frac{f(u)}{1+f(u)}{\rm d}u + \int _0 ^a \frac{f(a-u)}{1 + f(a-u)}{\rm d}u = \int _0 ^a \frac{2 + f(u) + f(a-u)}{2 + f(u) + f(a-u)}{\rm d}u = a, $$and so $I = \dfrac{a}{2}$. \end{answer} \end{pro} \begin{pro} Let $f$ be a Riemann integrable function over every bounded interval and such that $f(a+b)=f(a)+f(b)$ for all $(a, b)\in\BBR^2$. Demonstrate that $f(x)=xf(1)$. \begin{answer}Observe first that $f(0+0)=f(0)+f(0)$ and so $f(0)=0$. Integrate $f(u+y)=f(u)+f(y)$ for $u\in[0;x]$keeping $y$ constant, getting $$\int _0 ^{x}f(u+y)\d{u}=\int _0 ^{x}f(u)\d{u}+ \int _0 ^{x}f(y)\d{u} = \int _0 ^{x}f(u)\d{u}+xf(y). $$ Also, by substitution, $$\int _0 ^{x}f(u+y)\d{u} = \int _y ^{y+x}f(u)\d{u} = \int _0 ^{y+x}f(u)\d{u}-\int _0 ^yf(u)\d{u}.$$ Hence \begin{equation}\label{eq:Cauchy-int-equation} xf(y) = \int _0 ^{y+x}f(u)\d{u}-\int _0 ^yf(u)\d{u}- \int _0 ^{x}f(u)\d{u}.\end{equation}Exchanging $x$ and $y$: \begin{equation}\label{eq:Cauchy-int-equation2} yf(x) = \int _0 ^{y+x}f(u)\d{u}-\int _0 ^xf(u)\d{u}- \int _0 ^{y}f(u)\d{u}.\end{equation} From (\ref{eq:Cauchy-int-equation}) and (\ref{eq:Cauchy-int-equation2}) we gather that $xf(y)=xf(y)$. If $xy\neq 0$ then $\dfrac{f(x)}{x} = \dfrac{f(y)}{y}$. This means that for $\dfrac{f(x)}{x}$ is constant, and so for $x\neq 0$, $f(x)=cx$ for some constant $c$. Since $f(0)=0$, $f(x)=cx$ for all $x$. Taking $x=1$, $f(1)=c$. \end{answer} \end{pro} \begin{pro} Compute $\dint _0 ^2 x\floor{x^2}\d{x}$. \end{pro} \begin{pro}Find $\dint _{-1} ^2 |x^2-1| \ \ \d{x} $. \begin{answer}We have $$\begin{array}{lll} \dint _{-1} ^2 |x^2-1| \ \ \d{x} & = & \int _{-1} ^1 (1-x^2) \ \ \d{x} + \int _{1} ^2 (x^2-1) \ \ \d{x} \\ & = & (x - \frac{x^3}{3})\Big| _{-1} ^1 + (\frac{x^3}{3} - x)\Big| _1 ^2 \\ & = & 2(1 - \frac{1}{3}) + (\frac{8}{3} -2) - (\frac{1}{3}-1)\\ & = & \dfrac{4}{3} + \dfrac{2}{3} + \dfrac{2}{3} \\ & = & \dfrac{8}{3} \end{array}$$ \end{answer} \end{pro} \begin{pro} Let $n$ be a fixed integer. Let $f:\BBR \rightarrow \BBR$ be given by $$f(x)\quad = \quad \left\{\begin{array}{ll}x & \mathrm{if}\ x \leq 0\\ 2^n& \mathrm{if}\ 2^n-2^{n-2} 0$. Let $f$ be a continuous function on $\lcrc{0}{a}$ such that $f(x)+f(a-x)$ does not vanish on $\lcrc{0}{a}$. Evaluate $\dint _0 ^a\dfrac{f(x)\d{x}}{f(x)+f(a-x)}$. \end{pro} \begin{pro}Let $a>0$. Let $F$ be a differentiable function such that $\forall x\in \lcrc{0}{a}$ $F'(a-x)=F'(x)$. Evaluate $\dint _0 ^a F(x)\d{x}$. \end{pro} \begin{pro}Let $n \geq 0$ be an integer. Let $a$ be the unique differentiable function such that $\forall x\in \BBR$ $$(a(x))^{2n+1}+a(x)=x.$$ Evaluate $\dint _0 ^x a(t)\d{t}$. \end{pro} \begin{pro} Find $\dint _0 ^{\pi/2} \dfrac{\sin x\d{x}}{\sin x + \cos x}$. \end{pro} \begin{pro} Find $\dint _0 ^{\pi/2} \dfrac{1\d{x}}{1 + (\tan x)^{\sqrt{2}}}$. \end{pro} \begin{pro} Find $\dint \dfrac{1}{x\sqrt{x^2-1}} \d{x}$. \begin{answer} Put $u = \sqrt{x^2-1}; u^2 = x^2-1$ so that $2u\mathrm{d}u = 2x\d{x}$ and $\dfrac{\d{x}}{x} = \dfrac{x\d{x}}{x^2} = \dfrac{u\mathrm{d}u}{u^2 +1}$. Thus $$\dint \dfrac{1}{x\sqrt{x^2-1}} \d{x} = \dint \dfrac{u}{(u^2 + 1)u} \mathrm{d}u = \dint \dfrac{1}{u^2 + 1} \mathrm{d}u = \arctan u + C = \arctan\sqrt{x^2-1} + C. $$ \end{answer} \end{pro} \begin{pro} Find $\dis{\dint \dfrac{1}{1 + \sqrt{x + 1}} \d{x}}$. \begin{answer} Put $u = \sqrt{x+1}; u^2 = x + 1$; from where $\d{x} = 2u\mathrm{d}u$. Whence $$ \dint \dfrac{1}{1 + \sqrt{x + 1}} \d{x}= \dint \dfrac{2u}{1 + u} \ \mathrm{d}u = \dint \left(2- \dfrac{2}{1 + u}\right) \ \mathrm{d}u = 2u - 2\log|1 + u| + C = 2\sqrt{1+x}-2\log |1+\sqrt{1+x}| + C.$$ \end{answer} \end{pro} \begin{pro} Find $\dis{\dint \dfrac{x^{1/2}}{x^{1/2}-x^{1/3}} \d{x}}$. \begin{answer} Put $x = u^6; \d{x} =6u^5\mathrm{d}u$, giving $$ \begin{array}{lll} \dint \dfrac{x^{1/2}}{x^{1/2}-x^{1/3}} \d{x} & = & \dint \dfrac{(u^3)(6u^5)}{u^3-u^2} \mathrm{d}u\\ & = & \dint \dfrac{6u^6}{u-1} \mathrm{d}u \\ & = & 6\dint \left(u^5 + u^4 + u^3 + u^2 + u + 1 + \dfrac{1}{u-1}\right) \mathrm{d}u \\ & = & 6\left(\dfrac{u^6}{6}+\dfrac{u^5}{5}+\dfrac{u^4}{4}+\dfrac{u^3}{3}+\dfrac{u^2}{2}+u + \log|u-1|\right) + C \\ & = & x+\dfrac{6x^{5/6}}{5}+\dfrac{3x^{2/3}}{2}+2x^{1/2}+3x^{1/3}+6x^{1/6} + 6\log|x^{1/6}-1| + C. \end{array} $$ \end{answer} \end{pro} \begin{pro} Find $\dis{\dint \dfrac{a^{2x}}{\sqrt{a^x + 1}} \d{x}}$, $a> 0$. \begin{answer} Put $u^2 = a^x + 1; 2u\mathrm{d}u = (\log a)a^x \d{x}$ and so $$\dint \dfrac{a^{2x}}{\sqrt{a^x + 1}} \d{x} = \dint \dfrac{2u(u^2-1)}{u\log a} \mathrm{d}u =\dint \dfrac{2u^2-2}{\log a} \mathrm{d}u = \dfrac{2u^3}{3\log a} - \dfrac{2u}{\log a} + C = \dfrac{2(a^x+1)^{3/2}}{3\log a} - \dfrac{2(a^x+1)^{1/2}}{ \log a} + C.$$ \end{answer} \end{pro} \begin{pro} Find $\dis{\dint \dfrac{1}{(e^x - e^{-x})^2} \d{x}}$. \begin{answer} Observe that $(e^x - e^{-x})^2 = (e^{-x}(e^{2x} - 1))^2 = e^{-2x}(e^{2x} - 1)^2$, and so$$\dint \dfrac{1}{(e^x - e^{-x})^2} \d{x} = \dint \dfrac{e^{2x}}{(e^{2x} - 1)^2} \d{x} = \dint \dfrac{1}{2u^2} \mathrm{d}u = -\dfrac{1}{2u} + C = -\dfrac{1}{2(e^{2x}-1)} + C, $$on putting $u = e^{2x}-1$. \end{answer} \end{pro} \begin{pro} Prove that $\dis{\dint _1 ^{5} \dfrac{\lfloor x \rfloor}{x} \d{x}} = 4\log(5) - 3 \log(2) - \log(3)$. \begin{answer} We have $$\begin{array}{lll} \dint _1 ^{5} \dfrac{\lfloor x \rfloor}{x} \d{x} & = & \dint _1 ^{2} \dfrac{\lfloor x \rfloor}{x} \d{x} + \dint _2 ^{3} \dfrac{\lfloor x \rfloor}{x} \d{x} + \dint _3 ^{4} \dfrac{\lfloor x \rfloor}{x} \d{x} + \dint _4 ^5 \dfrac{\lfloor x \rfloor}{x} \d{x} \\ & = & \dint _1 ^{2} \dfrac{1}{x} \d{x} + \dint _2 ^{3} \dfrac{2}{x} \d{x} + \dint _3 ^{4} \dfrac{3}{x} \d{x} + \dint _4 ^5 \dfrac{4}{x} \d{x} \\ & = & (\log 2 - \log 1) + 2(\log 3 - \log 2) + 3(\log 4 - \log 3) + 4(\log 5 - \log 4) \\ & = & 4\log(5) - 3 \log(2) - \log(3) .\end{array} $$ \end{answer} \end{pro} \begin{pro} Find $\dint e^{e^x+x} \ \d{x}$. \begin{answer} Put $u= e^x$, etc. $$\dint e^{e^x+x}\d{x} = \dint e^x e^{e^x} \d{x}=\dint e^{e^x} \d{e^x} = e^{e^x}+C$$ \end{answer} \end{pro} \begin{pro} Find $\dis{\dint \tan x \log (\cos x) \ \d{x}}$.\\ \begin{answer}Put $u = \log (\cos x)$, etc. $$\dint \tan x \log (\cos x)\d{x} = \dint (\log (\cos x))\d{(-\log (\cos(x)))} = -\dfrac{(\log (\cos x))^2}{2}+C$$ \end{answer} \end{pro} \begin{pro} Find $\dis{\dint \dfrac{\log\log x}{x\log x} \ \d{x}}$.\\ \begin{answer} Put $u = \log\log x$, etc. $$\dint \dfrac{\log\log x}{x\log x} \d{x} = \dint \log\log x\d{\left(\log\log x\right)} = \dfrac{\log\log x}{2} +C $$ \end{answer} \end{pro} \begin{pro} Find $\dis{\dint \dfrac{x^{18}-1}{x^3-1} \ \d{x}}$.\\ \begin{answer} Carry out the long division. $$\dint \dfrac{x^{18}-1}{x^3-1} \d{x} = \dint (x^{15}+x^{12}+x^9+x^6+x^3+1)\d{x} = \dfrac{x^{16}}{16}+\dfrac{x^{13}}{13} + \dfrac{x^{10}}{10} +\dfrac{x^7}{7}+\dfrac{x^4}{4}+x+C $$ \end{answer} \end{pro} \begin{pro} Find $\dis{\dint \dfrac{1}{x^{8}+x} \ \d{x}}$.\\ \begin{answer}After an algebraic trick, put $u=1+x^{-7}$, etc. $$\dint \dfrac{1}{x^{8}+x}\d{x} = \dint \dfrac{x^{-8}}{1+x^{-7}}\d{x} = -\frac{1}{7}\dint \dfrac{\d{(1+x^{-7})}}{1+x^{-7}} = -\frac{1}{7}\log |1+x^{-7}|+C$$ \end{answer} \end{pro} \begin{pro} Find $\dis{\dint \dfrac{4^x}{2^x + 1} \ \d{x}}$.\\ \begin{answer}Put $u = 2^x+1$ $$ \dint \dfrac{2^x2^x}{2^x+1}\d{x} =\frac{1}{\log 2}\dint \dfrac{2^x}{2^x+1}\d{(2^x+1)} = \frac{1}{\log 2}\dint \dfrac{u-1}{u}\d{u} =\frac{1}{\log 2}\left(u - \log |u|\right) +C =\frac{1}{\log 2}\left(2^x+1 - \log |2^x+1|\right) +C $$ \end{answer} \end{pro} \begin{pro} Find $\dint \dfrac{x^2}{(x+1)^{10}} \ \d{x}$.\\ \begin{answer}Put $u=x+1$. Then $x^2= (u-1)^2 = u^2-2u+1$, and hence $$ \begin{array}{lll}\dint \dfrac{x^2}{(x+1)^{10}} \d{x} & =& \dint \dfrac{u^2-2u+1}{u^{10}} \d{u}\\ & = & \dint u^{-8} -2u^{-9} +u^{-10} \d{u} \\ & = & -\frac{u^{-7}}{7} +\frac{u^{-8}}{4}-\frac{u^{-9}}{9}+C\\ &= &-\frac{(x+1)^{-7}}{7} +\frac{(x+1)^{-8}}{4}-\frac{(x+1)^{-9}}{9}+C\\\end{array}$$ \end{answer} \end{pro} \begin{pro} Find $\dis{\dint \dfrac{1}{1 + e^x} \ \d{x}}$. \begin{answer}Algebraic trick, and then $u=e^{-x}+1$, etc. $$ \dint \dfrac{1}{1 + e^x} \d{x} = \dint \dfrac{e^{-x}}{e^{-x}+1}\d{x} = -\dint \dfrac{1}{e^{-x}+1} \d{(e^{-x}+1)} = -\log |e^{-x}+1|+C $$ \end{answer} \end{pro} \begin{pro} Find $\dis{\dint \dfrac{1}{1-\sin x} \ \d{x}}$.\\ \begin{answer} $$ \dint \dfrac{1}{1-\sin x} \d{x} = \dint \dfrac{1+\sin x}{1-\sin^2 x} \d{x} = \dint \dfrac{1+\sin x}{\cos^2 x} \d{x} =\dint \sec^2 x + \sec x\tan x \d{x} = \tan x +\sec x + C$$ \end{answer} \end{pro} \begin{pro} Find $\dis{\dint \sqrt{1+\sin 2x} \ \d{x}}$.\\ \begin{answer} $$ \begin{array}{lll}\dint \sqrt{1+\sin 2x}\d{x} & = & \dint \sqrt{\sin^2x+2\sin x\cos x+\cos^2x}\d{x} \\ &= & \dint \sqrt{(\sin x + \cos x)^2}\d{x}\\ & = & \dint |\sin x + \cos x|\d{x}\\ & = & \mp\cos x \pm\sin x +C \end{array}$$ \end{answer} \end{pro} \begin{pro} Find $\dis{\dint \dfrac{x}{\sqrt{1-x^4}} \ \d{x}}$. \begin{answer} Put $u =x^2$, etc. $$ \dint \dfrac{x}{\sqrt{1- (x^2)^2}} \d{x} = \frac{1}{2} \dint \dfrac{1}{\sqrt{1-u^2}}\d{u} = \frac{1}{2}\arcsin u +C = \frac{1}{2}\arcsin x^2 +C $$ \end{answer} \end{pro} \begin{pro} Find $\dint \sec^4x\d{x}$.\\ \begin{answer} We have $$ \begin{array}{lll}\dint \sec^4x\d{x} & = & \dint \sec^2x (\tan^2x+1)\d{x} \\ & = & \dint \sec^2x \tan^2x\d{x}+\dint \sec^2x \d{x} \\ & = & \dint (\tan x)^2\d{(\tan x)} + \dint \sec^2x\d{x} \\ & = & \dfrac{\tan^3x}{3} + \tan x +C. \end{array}. $$ \end{answer} \end{pro} \begin{pro} Find $\dint \sec^5x\d{x}$.\\ \begin{answer} We have $$ \begin{array}{lll}\dint \sec^5x \d{x} & = & \dint \sec^3x\sec^2x\d{x}\\ & = & \dint \sec^3x\d{(\tan x)} \\ & = & \sec^3x\tan x - \dint \tan x \d{(\sec^3x)} \\ & = & \sec^3x\tan x - 3\dint \tan^2 x\sec^2x\sec x \d{x} \\ & = & \sec^3x\tan x - 3\dint (\sec^2x-1)\sec^3x \d{x} \\ & = & \sec^3x\tan x - 3\dint \sec^5x\d{x}+3\dint \sec^3x \d{x} \\ \end{array} $$ The above implies that $$ \begin{array}{lll}\dint \sec^5x \d{x} & = & \dfrac{\tan x\sec^3x}{4} + \frac{3}{4}\dint \sec^3x \d{x} \\ & = & \dfrac{\tan x\sec^3x}{4} + \dfrac{3\tan x\sec x}{8} + \dfrac{3}{8}\log |\sec x + \tan x|+C, \\ \end{array} $$upon recalling from class that $$ \dint \sec^3x \d{x} = \dfrac{\tan x\sec x}{2} + \dfrac{1}{2}\log |\sec x + \tan x|+C $$ \end{answer} \end{pro} \begin{pro} Find $\dint e^{x^{1/3}}\d{x}$.\\ \begin{answer} First put $t = x^{1/3}$, then $t^3 = x \implies 3t^2\d{t} =\d{x}$. Thus $$ \begin{array}{lll}\dint e^{x^{1/3}}\d{x} & = & \dint 3t^2e^{t}\d{t}\\ & = & 3t^2e^t -6te^t - 6e^t +C \\ & = & 3x^{2/3}e^{x^{1/3}} -6x^{1/3}e^{x^{1/3}} -6e^{x^{1/3}} +C, \end{array} $$where the penultimate step results from tabular integration by parts. \end{answer} \end{pro} \begin{pro} Find $\dint \log (x^2+1)\d{x}$.\\ \begin{answer} We have $$ \begin{array}{lll}\dint \log (x^2+1)\d{x} & = & x\log (x^2+1) - \dint x\d{(\log(x^2+1))} \\ & = & x\log (x^2+1) - 2\dint \dfrac{x^2}{x^2+1}\d{x} \\ & = & x\log (x^2+1) - 2\dint \dfrac{x^2+1-1}{x^2+1}\d{x} \\ & = & x\log (x^2+1) - 2\dint \left(1-\dfrac{1}{x^2+1}\right)\d{x} \\ & = & x\log (x^2+1) - 2(x-\arctan x)+C \\ \end{array} $$ \end{answer} \end{pro} \begin{pro} \label{pro:xecos}Find $\dint xe^x\cos x\d{x}$.\\ \begin{answer} Put $$I = \dint xe^x\cos x := (Ax+B)e^x\cos x + (Cx+D)e^x\sin x+ K. $$Differentiating both sides, $$ xe^x\cos x = Ae^x\cos x + (Ax+B)e^x\cos x -(Ax+B)e^x\sin x + Ce^x\sin x + (Cx+D)e^x\sin x + (Cx+D)e^x\cos x. $$ Equating coefficients, $$\begin{array}{lll} xe^x\cos x & : & 1 = A+C \\ xe^x\sin x & : & 0 = -A+C \\ e^x\cos x & : & 0 = A+B+D \\ e^x\sin x & : & 0 = -B+C+D\\\end{array} $$ From the first two equations $C = \frac{1}{2}, A = \frac{1}{2}.$ Then the third and fourth equations become $-\frac{1}{2} = B+D; -\frac{1}{2} = -B+D$, whence $D=-\frac{1}{2}$, and $B=0$. We conclude that $$\dint xe^x\cos x = \dfrac{x}{2}e^x\cos x + \left(\dfrac{x-1}{2}\right)e^x\sin x+ K. $$ \end{answer} \end{pro} \begin{pro} Find $\dint x^{2/3}\log x\d{x}$.\\ \begin{answer} We will do this one two ways: first, by making the substitution $$t = \log x\implies e^t = x \implies e^t\d{t}=\d{x}.$$Observe also that $x^{2/3} = e^{2t/3}$. Then $$ \begin{array}{lll} \dint x^{2/3}\log x\d{x}& = & \dint te^{2t/3}e^t\d{t} \\ & = & \dfrac{3t}{5}e^{5t/3} - \dfrac{9}{25}e^{5t/3} + C \\ & = & \dfrac{3(\log x)}{5}x^{5/3} - \dfrac{9}{25}x^{5/3}+C. \end{array} $$ {\em Aliter:} By directly integrating by parts, $$ \begin{array}{lll} \dint x^{2/3}\log x\d{x}& = & \dint \log x\d{\left(\dfrac{3x^{5/3}}{5}\right)} \\ & = & \dfrac{3x^{5/3}}{5}\log x - \dfrac{3}{5}\dint x^{5/3}\d{(\log x)}\\ & = & \dfrac{3(\log x)}{5}x^{5/3} - \dfrac{3}{5}\dint x^{2/3}\d{ x}\\ & = & \dfrac{3(\log x)}{5}x^{5/3} - \dfrac{9}{25}x^{5/3}+C, \end{array} $$as before. \end{answer} \end{pro} \begin{pro} Find $\dint \sin (\log x)\d{x}$.\\ \begin{answer} This integral can be done multiple ways. For example, you may integrate by parts directly and then ``solve'' for the integral. Another way is the following. Start by putting $$t = \log x\implies e^t = x \implies e^t\d{t}=\d{x}.$$ Then $$ \dint \sin (\log x)\d{x} = \dint e^t\sin t\d{t},$$an integral that we found in class. We will find it again, using a method similar of problem \ref{pro:xecos}. Put $$I = \dint e^t\cos t\d{t} := Ae^t\cos t + Be^t\sin t + K.$$ Differentiating both sides $$ e^t\cos t = Ae^t\cos t -Ae^t\sin t + Be^t\sin t + Be^t\cos t. $$ Equating coefficients, $$\begin{array}{lll} e^t\cos t& : & 1 = A+B \\ e^t\sin t & : & 0 = -A + B\end{array} $$ and so $A = B = \frac{1}{2}$. We have thus $$ \begin{array}{lll}\dint \sin (\log x)\d{x} & = & \dint e^t\sin t\d{t} \\ & = & \frac{1}{2}e^t\cos t + \frac{1}{2}e^t\sin t +K \\ & = & \frac{1}{2}x\cos \log x + \frac{1}{2}x\sin \log x +K. \\ \end{array}$$ \end{answer} \end{pro} \begin{pro} Find $\dint \dfrac{\log\log x}{x}\d{x}$.\\ \begin{answer} Put $t = \log\log x \implies e^{e^t} =x \implies e^te^{e^t}\d{t} = \d{x}$. Hence $$ \begin{array}{lll}\dint \dfrac{\log\log x}{x}\d{x} & = & \dint \dfrac{te^te^{e^t}}{e^{e^t}}\d{t} \\ & = & te^t-e^t +C \\ & = & (\log x)(\log\log x) - (\log x) +C, \end{array} $$where the penultimate equality follows from a tabular integration by parts. \end{answer} \end{pro} \begin{pro}[$\dint \sec x\d{x}$ in three ways] A traditional indefinite integral is $$\dint \sec x \d{x} = \log (\tan x + \sec x) +C.$$Justify this formula. \bigskip Now, prove that $\dfrac{1}{\cos x} = \dfrac{\cos x}{2(1+\sin x)} + \dfrac{\cos x}{2(1-\sin x)}.$ Use this to find a second formula for $\dint \sec x\d{x}$. \bigskip A third way is as follows. Using $\sin 2\theta = 2\sin\theta \cos \theta$ shew that $\dint \csc x \d{x} = \log |\tan \frac{x}{2}| + C$. Now use $\csc (\frac{\pi}{2}+x) = \sec x$ to find yet another formula for $\dint \sec x\d{x}$. \begin{answer} Observe that $$\dint \sec x \d{x} =\dint \dfrac{\sec x \tan x + \sec^2x}{\tan x + \sec x} \d{x} = \dint \d{\left(\log (\tan x + \sec x)\right)} = \log (\tan x + \sec x) +C,$$ \bigskip For the second way, simple algebra will yield the identity. We have $$ \begin{array}{lll} \dint \sec x\d{x}& = & \dint \dfrac{\cos x}{2(1+\sin x)}\d{x} + \dint \dfrac{\cos x}{2(1-\sin x)}\d{x} \\ & = & \frac{1}{2}\log |1+\sin x| -\frac{1}{2}\log |1-\sin x| +C\\ & = & \dfrac{1}{2}\log \Big|\dfrac{1+\sin x}{1-\sin x}\Big| + C \end{array}. $$ \bigskip For the third way, we have $$\begin{array}{lll} \dint \csc x \d{x}& = & \dint \dfrac{1}{\sin x} \d{x}\\ & = & \dint \dfrac{1}{2\sin \frac{x}{2}\cos \frac{x}{2}} \d{x} \\ & = & \dint \dfrac{\cos \frac{x}{2}}{2\sin \frac{x}{2}\cos ^2 \frac{x}{2}} \d{x}\\ & = & \dint \dfrac{\sec^2 \frac{x}{2}}{2\tan \frac{x}{2}} \d{x}\\ & \stackrel{u = \tan \frac{x}{2}}{=} & \dint \dfrac{\d{u}}{u}\\ & = & \log |\tan \frac{x}{2}|+C. \end{array}$$ Thus $$\dint \sec x\d{x} = \dint \csc (\frac{\pi}{2}+x) \d{x} =\dint \csc (\frac{\pi}{2}+x) \d{(\frac{\pi}{2}+x)} = \log \Big|\tan (\frac{\pi}{4}+\frac{x}{2})\Big|+C. $$ \end{answer} \end{pro} \begin{pro} Find $\dint (\arcsin x)^2\d{x}$. \\ \begin{answer} Putting $t=\arcsin x$ we have $$ \sin t = x \implies \cos t \d{t} = \d{x}, $$whence $$\begin{array}{lll}\dint (\arcsin x)^2\d{x} & = & \dint t^2\cos t\d{t} \\ & = & t^2\sin t + 2t\cos t - 2\sin t + C \\ & = & (\arcsin x)^2x + 2(\arcsin x)\cos (\arcsin x) - 2x +C \\ & = & (\arcsin x)^2x + 2(\arcsin x)\sqrt{1-x^2} - 2x +C \\ \end{array} $$ \end{answer} \end{pro} \begin{pro} Find $\dint \dfrac{\d{x}}{\sqrt{x + 1} + \sqrt{x - 1}}$. \\ \begin{answer} We have $$ \begin{array}{lll} \dint \dfrac{\d{x}}{\sqrt{x + 1} + \sqrt{x - 1}}& = & \dint \dfrac{(\sqrt{x + 1} - \sqrt{x - 1})\d{x}}{2} \\ & = & \dfrac{1}{3}(x+1)^{3/2} - \dfrac{1}{3}(x-1)^{3/2}+C \end{array}. $$ \end{answer} \end{pro} \begin{pro} $\dint x\arctan x\d{x}$. \\ \begin{answer} We have $$ \begin{array}{lll}\dint x\arctan x\d{x} & = & \dint \arctan x\d{\left(\dfrac{x^2}{2}\right)} \\ & = &\dfrac{x^2}{2}\arctan x - \dint \dfrac{x^2}{2}\d{(\arctan x)} \\ & = &\dfrac{x^2}{2}\arctan x - \dint \frac{1}{2}\dfrac{x^2}{1 + x^2}\d{x} \\ & = &\dfrac{x^2}{2}\arctan x - \dint \frac{1}{2}\dfrac{x^2+1-1}{1 + x^2}\d{x} \\ & = &\dfrac{x^2}{2}\arctan x - \dfrac{x}{2} + \frac{1}{2}\arctan x+C \\ \end{array}. $$ \end{answer} \end{pro} \begin{pro} Find $\dint \sqrt{\tan x} \d{x} $.\\ \begin{answer} Put $u=\sqrt{\tan x}$ and so $u^2 = \tan x$, $2u\d{u} = \sec^2x\d{x} = (\tan^2x+1)\d{x} = (u^4+1)\d{x}$. Hence the integral becomes $$\dint \sqrt{\tan x} \d{x} = 2\dint \dfrac{u^2}{u^4+1}\d{u}. $$To decompose the above fraction into partial fractions observe (Sophie Germain's trick) that $u^4+1 = u^4+2u^2+1 -2u^2= (u^2+u\sqrt{2}+1)(u^2-u\sqrt{2}+1)$ and hence $$\begin{array}{lll} \dint \sqrt{\tan x} \d{x} & = & 2\dint \dfrac{u^2}{u^4+1}\d{u} \\ & = & -\dfrac{\sqrt{2}}{2}\dint \dfrac{u}{u^2+u\sqrt{2}+1}\d{u}+ \dfrac{\sqrt{2}}{2}\dint \dfrac{u}{u^2-u\sqrt{2}+1}\d{u}\\ & = & -\dfrac{\sqrt{2}}{4}\log (u^2+u\sqrt{2}+1) + \dfrac{\sqrt{2}}{4}\log (u^2-u\sqrt{2}+1) + \dfrac{\sqrt{2}}{2}\arctan (\sqrt{2}u+1)-\dfrac{\sqrt{2}}{2}\arctan (-\sqrt{2}u+1)+C\\ & = &-\dfrac{\sqrt{2}}{4}\log (\tan x+\sqrt{2\tan x}+1) + \dfrac{\sqrt{2}}{4}\log (\tan x-\sqrt{2\tan x}+1) \\ & & \qquad + \dfrac{\sqrt{2}}{2}\arctan (\sqrt{2\tan x}+1)-\dfrac{\sqrt{2}}{2}\arctan (-\sqrt{2\tan x}+1)+C \end{array}$$ \end{answer} \end{pro} \begin{pro} Find $\dint \dfrac{2x+1}{x^2(x-1)} \d{x}$.\\ \begin{answer} Put $$\dfrac{2x+1}{x^2(x-1)} = \dfrac{A}{x} + \dfrac{B}{x^2} + \dfrac{C}{x-1} \implies 2x+1 = Ax(x-1) + B(x-1) + Cx^2. $$Letting $x=1$ we get $3=C$. Letting $x=0$ we get $1 = -B\implies B=-1$. To get $A$ observe that equating the coefficients of $x^2$ on both sides we get $0 = A+C$, whence $A=-3$. Thus $$\begin{array}{lll}\dint \dfrac{2x+1}{x^2(x-1)} \d{x} & = &-3\dint \dfrac{1}{x}\d{x} - \dint \dfrac{1}{x^2}\d{x} +3\dint \dfrac{1}{x-1}\d{x}\\ & = & -3\log |x| +\dfrac{1}{x} + 3\log |x-1| +C \\ & = & 3\log \Big|\frac{x-1}{x}\Big| + \dfrac{1}{x}+C. \end{array}$$ \end{answer} \end{pro} \begin{pro} Find $\dint \log (x + \sqrt{x}) \d{x}$.\\ \begin{answer}Integrating by parts, $$\begin{array}{lll}\dint \log (x + \sqrt{x}) \d{x} & = &x\log (x + \sqrt{x})-\dint x \d{\log(x + \sqrt{x})} \\ & = & x\log (x + \sqrt{x}) -\dint \dfrac{x(1 + \dfrac{1}{2\sqrt{x}})}{x+\sqrt{x}} \d{x}\\ & = &x\log (x + \sqrt{x}) -\dint \left( 1 -\dfrac{1}{2}\cdot\dfrac{\sqrt{x}}{x+\sqrt{x}} \right) \d{x} \\ & = & x\log (x + \sqrt{x}) -x + \dfrac{1}{2}\dint \dfrac{\sqrt{x}}{x+\sqrt{x}} \d{x} \\ & \stackrel{u = \sqrt{x}}{=} & x\log (x + \sqrt{x}) -x + \dint \dfrac{u^2}{u^2+u} \d{u} \\ & \stackrel{u = \sqrt{x}}{=} & x\log (x + \sqrt{x}) -x + \dint 1-\dfrac{1}{u+1} \d{u} \\ & = & x\log (x + \sqrt{x}) -x + u - \log (u+1)+C \\ & = & x\log (x + \sqrt{x}) -x + \sqrt{x}-\log (\sqrt{x}+1)+C \\ \end{array}$$ \end{answer} \end{pro} \begin{pro} Find $\dint \dfrac{1}{x^4+1} \d{x}$.\\ \begin{answer} We use Sophie Germain's trick to factor $$x^4+1 = x^4+2x^2+1-2x^2 = (x^2+1)^2 - 2x^2 = (x^2-\sqrt{2}x+1)(x^2+\sqrt{2}x+1),$$and seek the partial fraction decomposition $$\dfrac{1}{x^4+1} = \dfrac{Ax+B}{x^2-\sqrt{2}x+1} +\dfrac{Cx+D}{x^2+\sqrt{2}x+1} \implies 1 = (Ax+B)(x^2+\sqrt{2}x+1) + (Cx+D)(x^2-\sqrt{2}x+1). $$ Equating coefficients $$\begin{array}{lll} x^3 & : & 0 = A +C \\ x^2 & : & 0 = B +D + \sqrt{2}(A -C) \\ x & : & 0 = A+ C+\sqrt{2}(B-D) \\ x^0 & : & 1 = B +D \\ \end{array}$$From the first and third equation it follows that $A=-C$ and that $B=D$. From the fourth equation $B=D = \frac{1}{2}$ and from the second equation $A = -\dfrac{1}{2\sqrt{2}} = -C$. Hence we must integrate $$\begin{array}{lll}\dint \dfrac{1}{x^4+1} \d{x} & = & \dint\dfrac{\sqrt{2}x+2}{4(x^2+\sqrt{2}x+1)}\d{x} - \dint\dfrac{\sqrt{2}x-2}{4(x^2-\sqrt{2}x+1)}\d{x} \\ & = & \dfrac{\sqrt{2}}{8}\dint\dfrac{2x+\sqrt{2}}{x^2+\sqrt{2}x+1}\d{x} + \dfrac{1}{4}\dint\dfrac{1}{x^2+\sqrt{2}x+1}\d{x} - \dfrac{\sqrt{2}}{8}\dint\dfrac{2x+\sqrt{2}}{x^2-\sqrt{2}x+1}\d{x} + \dfrac{1}{4}\dint\dfrac{1}{x^2-\sqrt{2}x+1}\d{x} \\ & = & \dfrac{\sqrt{2}}{8}\log (x^2 + x\sqrt{2}+1)-\dfrac{\sqrt{2}}{8}\log (x^2 - x\sqrt{2}+1) + \dfrac{1}{2}\dint \dfrac{\d{x}}{(x\sqrt{2}+1)^2+1} +\dfrac{1}{2}\dint \dfrac{\d{x}}{(-x\sqrt{2}+1)^2+1} \\ & = & \dfrac{\sqrt{2}}{8}\log (x^2 + x\sqrt{2}+1)-\dfrac{\sqrt{2}}{8}\log (x^2 - x\sqrt{2}+1) +\dfrac{\sqrt{2}}{4}\arctan (x\sqrt{2}+1)-\dfrac{\sqrt{2}}{4}\arctan (-x\sqrt{2}+1)+C \end{array}$$ \end{answer} \end{pro} \begin{pro} Find $\dint \dfrac{1}{x^3+1} \d{x}$.\\ \begin{answer} We begin by observing that $$\dfrac{1}{x^3+1} = \dfrac{A}{x+1} + \dfrac{Bx+C}{x^2-x+1} \implies 1 = A(x^2-x+1) + (Bx+C)(x+1). $$Letting $x=-1$ we obtain $1 = 3A \implies A = \frac{1}{3}$. Letting $x=0$ we obtain $1 = A + C \implies C = 1-A = \dfrac{2}{3}$. Finally, we must have $A+B = 0$, since the coefficient of $x^2$ must be zero. thus $B = -\frac{1}{3}$. We must then integrate $$\begin{array}{lll} \dint \dfrac{\d{x}}{3(x+1)} - \dint \dfrac{x-2}{3(x^2-x+1)}\d{x}& = & \frac{1}{3}\log |x+1| - \dint \dfrac{x-\frac{1}{2}}{3(x-\frac{1}{2})^2 + \frac{3}{4}} +\dfrac{1}{2}\dint \dfrac{1}{(x-\frac{1}{2})^2 + \frac{3}{4}}\\ & = & \frac{1}{3}\log |x+1| -\frac{1}{6}\log |(x-\frac{1}{2})^2 + \frac{3}{4}| +\dfrac{2}{3}\dint \dfrac{1}{\frac{4}{3}(x-\frac{1}{2})^2 + 1}\\ & = & \frac{1}{3}\log |x+1| -\frac{1}{6}\log |(x-\frac{1}{2})^2 + \frac{3}{4}| +\dfrac{2}{3}\cdot \dfrac{\sqrt{3}}{2}\arctan (x-\frac{1}{2}) \\ & = & \frac{1}{3}\log |x+1| -\frac{1}{6}\log |x^2-x+1| +\dfrac{\sqrt{3}}{3}\arctan \frac{2}{\sqrt{3}}(x-\frac{1}{2}) \\ \end{array}$$ \end{answer} \end{pro} \begin{pro} Demonstrate that for all strictly positive integers $n$, $$\left(1 + \frac{1}{n}\right)^{n}\left(1 + \frac{1}{4n}\right) 0$ $\exists N>0 $ (depending on $\varepsilon$ and on $x$) such that $$ n\geq N \implies \absval{f_n(x)-f(x)}<\varepsilon. $$ \end{df} \begin{exa} The sequence of functions $x\mapsto x^n, n=1,2,\ldots$ converges pointwise on the interval $\lcrc{0}{1}$ to the function $f:\lcrc{0}{1}\rightarrow \{0,1\}$ with $$ f(x) = \left\{\begin{array}{ll} 0 & \mathrm{if}\ x\in \lcro{0}{1}\\ 1 & \mathrm{if}\ x=1\\ \end{array}\right. $$ \end{exa} \section{Uniform Convergence} \begin{df} We say that a sequence of functions $\seq{f_n}{n=1}{+\infty}$ $f_n:I\rightarrow \BBR$ defined on an interval $I\subseteqq \BBR$ {\em converges uniformly to the function $f$} if $\forall x\in I$, $\forall \varepsilon >0$ $\exists N>0 $ (depending only on $\varepsilon$) such that $$ n\geq N \implies \absval{f_n(x)-f(x)}<\varepsilon. $$ In this case we write $f_n \unif f$. \end{df} \begin{thm} Let $\seq{f_n}{n=1}{+\infty}$ be a sequence of functions defined over a common domain $I$. If there exists a numerical sequence $\seq{a_n}{n=1}{+\infty}$ with $a_n\to 0$ as $\ngroes$, and a function $f$ defined over $I$ such that eventually$$ \absval{f_n(x)-f(x)}\leq a_n, $$then $f_n \unif f$. \end{thm} \begin{thm} If the sequence of continuous functions $\seq{f_n}{n=1}{+\infty}$ $f_n:I\rightarrow \BBR$ defined on an open interval $I\subseteqq \BBR$ converges uniformly to $f$ on $I$, then $f$ is continuous on $I$. Moreover, if $x_0\in I$ then we may exchange the limits, as in $$ \lim _{\ngroes} \left( \lim _{x\rightarrow x_0}f_n(x) \right) = \lim _{x\rightarrow x_0} \left( \lim _{\ngroes}f_n(x) \right) = \lim _{x\rightarrow x_0} f(x). $$ \end{thm} \begin{thm} If the sequence of integrable functions $\seq{f_n}{n=1}{+\infty}$ $f_n:I\rightarrow \BBR$ defined on an open interval $I\subseteqq \BBR$ converges uniformly to $f$ on $I$, then $f$ is integrable on $I$. Moreover, if $(a, b)\in I^2$ then we may exchange the limit with the integral, as in $$ \lim _{\ngroes} \left( \int _a ^b f_n(x)\d{x} \right) = \int _a ^b \left( \lim _{\ngroes}f_n(x) \right)\d{x} = \int _a ^b f(x)\d{x}. $$ \end{thm} \section{Integrals and Derivatives of Sequences of Functions} \section{Power Series} A {\em power series} about $x=a$ is a series of the form $$f(x) =\sum _{n=0} ^{+\infty} a_n(x-a)^n.$$This is a function of $x$, and truncating it gives polynomial approximations to $f$. The goal is to approximate ``decent'' functions about a given point $x=a$. \bigskip These expansions don't necessarily make sense for all $x$. The region where the power series converges is called the {\em interval of convergence.} \begin{exa} Find the interval of convergence of the series $\displaystyle{\sum_{n=1}^\infty \dfrac{2^n(x-3)^n}{\sqrt{n}}}$. \end{exa} \begin{solu} By the ratio test, the series will converge if $$\Big|\dfrac{2^{n+1}(x-3)^{n+1}}{\sqrt{n+1}}\cdot \dfrac{\sqrt{n}}{2^n(x-3)^n}\Big| = 2\sqrt{\dfrac{n}{n+1}}|x-3| \rightarrow r <1, $$ that is when $$ 2|x-3|<1 \implies \dfrac{5}{2}