Real Analysis

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

Last update: 20 July 2014

Lecture 32

Let X and Y be topological spaces.
Let f:X→Y be a continuous function.
Let E⊆X.

(a) If E is connected then f(E) is connected.
(b) If E is compact then f(E) is compact.

Let X=ℝ1. Let E⊆X.

(a) E is connected if and only if E is an interval.
(b) E is compact and connected if and only if there exist m,M∈ℝ such that E=[m,M].

(Intermediate value theorem) Let f:[a,b]→ℝ be continuous. Then there exist m,M∈ℝ such that f([a,b])= [m,M].

(Rolle's theorem) Let f:[a,b]→ℝ be continuous with f(a)=f(b). If f′:(a,b)→ℝ exists then there exists c∈(a,b) such that f′(c)=0.

Idea that makes it work: If M is the maximum value of f and f(c)=M then f′(c)=0.

(The mean value theorem) Let f:[a,b]→ℝ be a continuous function such that f′:(a,b)→ℝ exists. Then there exists c∈(a,b) such that f(b)=f(a)+ f′(c)(b-a).

(Taylor's theorem) Let f:[a,b]→ℝ be a function such that f(N):(a,b)→ℝ exists. Then there exists c∈(a,b) such that f(b) = f(a)+f′(a) (b-a)+12! f″(a)(b-a)2 +13!f(3) (a)(b-a)2+⋯ +1(N-1)! f(N-1)(a) (b-a)N-1 +1N!f(N) (c)(b-a)N.

If we let a=0 and b=x then Taylor's theorem says that f(x) can be expanded as a polynomial in x, f(x)=f(0) +f′(0)x+ 12!f″(0) x2+13!f(3) (0)x3+14! f(4)(0)x4+⋯ You need more and more terms for more and more accuracy in the expansion. Just like... You need more and more digits π=3.141592… for more and more accuracy in ℝ.

When does an infinite polynomial like 11-x=1+x+x2 +x3+x4+⋯ produce a good function? i.e. For which a∈ℂ does the series ∑n=0∞xn converge. In other words: Let s1=1, s2=1+a, s3=1+a+a2, s4=1+a+a2+a3,… For which a∈ℂ does the sequence (s1,s2,s3,…) converge? In other words: Does limn→∞sn exist? In other words: Does there exist ℓ∈ℂ such that
if ε∈ℝ>0 then there exists N∈ℤ>0 such that
if n∈ℤ>0 and n>N then d(sn,ℓ)<ε.
In this example: s1 = 1, s2 = 1+a, s3 = 1+a+a2=1-a31-a, s4 = 1+a+a2+a3= 1-a41-a, ⋮ sn = 1+a+a2+⋯+an-1 =1-an1-a. So limn→∞sn = limn→∞ 1-an1-a = 11-alimn→∞ 1-an = 11-a (1-limn→∞an), where we have used the limit theorems:

Let (an) and (bn) be sequences (in ℝ or in ℝn) and assume limn→∞an and limn→∞bn exist. Then

(a) limn→∞(an+bn)=limn→∞an+limn→∞bn,
(b) if c∈ℝ then limn→∞can=climn→∞an,
(c) limn→∞(anbn)=(limn→∞an)(limn→∞bn),
(d) if bn≠0 for all n∈ℤ>0 then limn→∞(anbn)= (limn→∞an) (limn→∞bn) .

Skills:

(1) Algebraic manipulations - Expressions
(2) Graphing
(3) Limits
(4) Sequences (functions ℤ>0→ℝ and ℤ>0→ℂ)
(5) Series (a special kind of sequence)
(6) Approximations - Taylor's theorem and Integral approximations
(7) Mathematical Language and Proof machine: Fields, Ordered fields, Open sets, Inductions, Definitions and Proof machine

Notes and References

These are notes from a 2010 course on Real Analysis 620-295. This page comes from 100524Lect32.pdf and was given on 24 May 2010.

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