Real Analysis

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

Last update: 8 July 2014

Lecture 4

Induction, sequences, bounds

Prove that if $n\in {\mathbb{Z}}_{>0}$ then $1+2+3+\cdots +n=\frac{n(n+1)}{2}\text{.}$

		Proof.
	Proof by induction. Base case: Assume $n=1\text{.}$ To show: $1=\frac{1(1+1)}{2}\text{.}$ Righthand side: $$\frac{1(1+1)}{2}=\frac{1·2}{2}=1\text{.}$$ Induction step: Let $N\in {\mathbb{Z}}_{>0}\text{.}$ Assume that if $n\in {\mathbb{Z}}_{>0}$ and $n<N$ then $$1+2+\cdots +n=\frac{n(n+1)}{2}\text{.}$$ To show: $1+2+\cdots +(N-1)+N=\frac{N(N+1)}{2}\text{.}$ Lefthand side: $$\begin{array}{ccc}{\displaystyle 1+2+\cdots +(N-1)+N}& =& {\displaystyle (1+2+\cdots +(N-1))+N}\\ & =& {\displaystyle \frac{(N-1)(N-1+1)}{2}+N,\text{by the induction hypothesis,}}\\ & =& {\displaystyle \frac{(N-1)N}{2}+\frac{2N}{2}}\\ & =& {\displaystyle \frac{N(N-1+2)}{2}}\\ & =& {\displaystyle \frac{N(N+1)}{2}\text{.}}\end{array}$$ $\square$

Some notation

Write out the first 10 terms of $$\sum _{n=4}^{100}{(-1)}^{n+1}\frac{2^n{x}^n}{{(n-3)}^2}$$

Solution $$\begin{array}{ccc}{\displaystyle \sum _{n=4}^{100}{(-1)}^{n+1}\frac{2^n{x}^n}{{(n-3)}^2}}& =& {\displaystyle {(-1)}^5\frac{2^4{x}^4}{{(4-3)}^2}+\frac{{(-1)}^6{2}^5{x}^5}{{(5-3)}^2}+\cdots +\frac{{(-1)}^{14}{2}^{13}{x}^{13}}{{(13-3)}^2}+\cdots }\\ & =& {\displaystyle -2^4{x}^4+\frac{2^5{x}^5}{2^2}-\frac{2^6{x}^6}{3^2}+\frac{2^7{x}^7}{4^2}-\frac{2^8{x}^8}{5^2}+\frac{2^9{x}^9}{6^2}}\\ & & {\displaystyle -\frac{2^{10}{x}^{10}}{7^2}+\frac{2^{11}{x}^{11}}{8^2}-\frac{2^{12}{x}^{12}}{9^2}+\frac{2^{13}{x}^{13}}{{10}^2}+\cdots \text{.}}\end{array}$$

Write $\text{sin}\hspace{0.17em}x$ as a series in sum notation. $$\begin{array}{ccc}\text{sin}\hspace{0.17em}x& =& {\displaystyle x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\frac{x^9}{9!}-\cdots }\\ & =& {\displaystyle \frac{x^{2·0+1}}{(2·0+1)!}+\frac{{(-1)}^1{x}^{2·1+1}}{(2·1+1)!}+\frac{{(-1)}^1{x}^{2·2+1}}{(2·2+1)!}+\frac{{(-1)}^1{x}^{2·3+1}}{(2·3+1)!}+\cdots }\\ & =& {\displaystyle \sum _{n=0}^{\infty }\frac{{(-1)}^n{x}^{2n+1}}{(2n+1)!}}\\ & =& {\displaystyle \sum _{n\in {\mathbb{Z}}_{\ge 0}}\frac{{(-1)}^n{x}^{2n+1}}{(2n+1)!}\text{.}}\end{array}$$

Bounds and intervals

${ℝ}_{\ge 0}$ is the set of decimal expansions.

$ℝ$ is the set of positive and negative decimal expansions

$$\begin{array}{ccc}0.9999\dots \in ℝ,& & 1.000\dots \in ℝ,\\ -0.9999\dots \in ℝ,& & -1.000\dots \in ℝ\text{.}\end{array}$$ Let $a,b\in ℝ\text{.}$ The sets $$\begin{array}{ccc}[a,b]& =& \{x\in ℝ\hspace{0.17em}|\hspace{0.17em}a\le x\le b\},\\ (a,b)& =& \{x\in ℝ\hspace{0.17em}|\hspace{0.17em}a<x<b\},\\ [a,b)& =& \{x\in ℝ\hspace{0.17em}|\hspace{0.17em}a\le x<b\},\\ (a,b]& =& \{x\in ℝ\hspace{0.17em}|\hspace{0.17em}a<x\le b\}\end{array}$$ are intervals in $ℝ\text{.}$ The set $(a,b)$ is an open interval in $ℝ\text{.}$

Graph $[2,3)\text{.}$ $$\begin{array}{c} 0 1 2 3 \end{array}$$

Graph $\left\{\frac{1}{n}\hspace{0.17em}\right|\hspace{0.17em}n\in {\mathbb{Z}}_{>0}\}\text{.}$ $$\begin{array}{c} 17 16 15 14 13 12 1 \end{array}$$

Let $S$ be a subset of $ℝ\text{.}$

An upper bound of $S$ in $ℝ$ is $b\in ℝ$ such that if $x\in S$ then $x\le b\text{.}$

A least upper bound of $S$ in $ℝ$ is $\text{sup}\left(s\right)\in ℝ$ such that

(a)	$\text{sup}\left(s\right)$ is an upper bound of $S$ in $ℝ,$
(b)	If $b$ is an upper bound of $S$ in $ℝ$ then $b\ge \text{sup}\left(S\right)\text{.}$

A maximum of $S$ is $m\in S$ such that if $x\in ℝ$ and $x>m$ then $x\notin S\text{.}$

Write $S=\{x\in ℝ\hspace{0.17em}|\hspace{0.17em}|x-2|<3\hspace{0.17em}\text{and}\hspace{0.17em}|x+1|<1\}\text{.}$

If $x\in ℝ$ then the absolute value of $x$ is $$\left|x\right|=\{\begin{array}{cc}x,& \text{if}\hspace{0.17em}x\ge 0,\\ -x,& \text{if}x<0\text{.}\end{array}$$ So $$|x-2|=\{\begin{array}{cc}x-2,& \text{if}\hspace{0.17em}x-2\ge 0\\ -(x-2),& \text{if}x-2<0\end{array}=\{\begin{array}{cc}x-2,& \text{if}\hspace{0.17em}x\ge 2,\\ 2-x,& \text{if}\hspace{0.17em}x<2\text{.}\end{array}$$ The graph of $y=x-2$ is $$\begin{array}{c} 1 2 x 1 2 y \end{array}$$ $$\begin{array}{ccc}{S}_1& =& \{x\in ℝ\hspace{0.17em}|\hspace{0.17em}|x-2|<3\}=(-1,5)\\ S_2& =& \{x\in ℝ\hspace{0.17em}|\hspace{0.17em}|x+1|<1\}\\ & =& \{x\in ℝ\hspace{0.17em}|\hspace{0.17em}(x+1\ge 0\hspace{0.17em}\text{and}\hspace{0.17em}x+1<1)\hspace{0.17em}\text{or}\hspace{0.17em}(x+1<0\hspace{0.17em}\text{and}\hspace{0.17em}-(x+1)<1)\}\\ & =& \{x\in ℝ\hspace{0.17em}|\hspace{0.17em}(x\ge -1\hspace{0.17em}\text{and}\hspace{0.17em}x<0)\hspace{0.17em}\text{or}\hspace{0.17em}(x<-1\hspace{0.17em}\text{and}\hspace{0.17em}x+1>-1)\}\\ & =& \{x\in ℝ\hspace{0.17em}|\hspace{0.17em}x\in [-1,0)\hspace{0.17em}\text{or}\hspace{0.17em}(x<-1\hspace{0.17em}\text{and}\hspace{0.17em}x>-2)\}\\ & =& \{x\in ℝ\hspace{0.17em}|\hspace{0.17em}x\in [-1,0)\hspace{0.17em}\text{or}\hspace{0.17em}x\in (-2,-1)\}\\ & =& [-1,0)\cup (-2,-1)\\ & =& (-2,0)\text{.}\end{array}$$

Graph $S=\{x\in ℝ\hspace{0.17em}|\hspace{0.17em}{x}^2<9\}\text{.}$ The graph of $y=x^2$ is $$\begin{array}{c} x y -1 1 1 \end{array}$$ So $$\begin{array}{ccc}S& =& \{x\in ℝ\hspace{0.17em}|\hspace{0.17em}x>-3\hspace{0.17em}\text{and}\hspace{0.17em}x<3\}\\ & =& (-3,3)\text{.}\end{array}$$

Notes and References

These are notes from a 2010 course on Real Analysis 620-295. This page comes from 100308Lect4.pdf and was given on 8 March 2010.

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