introduction to real analysis math 2001 for...course description real analysis is the detailed study...
TRANSCRIPT
Introduction to Real Analysis MATH 2001
Juris Steprans
York University
September 6, 2012
Juris Steprans MATH 2001
Instructor
Instructor: Juris Steprans
My office is N530 in the Ross building.
Office hours are Thursdays from 4:30 to 6:00. You are alsowelcome to make an appointment to see me.
My telephone number is 736-5250 (ask for me) or 736-2100extension 55250. However, I check my email frequently andoften the best way of reaching me is sending an email [email protected]
You can find a link to my website on the Department site. It iswww.math.yorku.ca/Who/Faculty/Steprans/menu.html
Juris Steprans MATH 2001
Course description
Real analysis is the detailed study of the objects and argumentsused in elementary calculus. A first course in calculus is usuallyfocused on the computational techniques that make calculus souseful in analyzing many physical systems. Such a course mayoverlook several subtle points and the perceptive student will beleft wondering why these techniques work. Real analysis providesthe justification for the techniques and this analysis often providesprecise conditions under which the computational techniques canbe applied. Real Analysis I is an introduction the analysis of thearguments behind the computional techniques of Calculus.
Juris Steprans MATH 2001
Important dates
The class meets Thursdays from 7:00 to 10:00 in HNE B15.
There is a weekly tutorial on Thursdays from 6:00 to 7:00 inHNE B15.
The Mid-Term Examination will be held on Thursday,October 25 from 7:00 to 8:00.
There is no lecture on November 1.
The last date to drop this course without receiving a grade inNovember 19.
The last lecture is on Thursday, November 29.
The final examination will be held some time betweenDecember 5 and December 21.
Juris Steprans MATH 2001
Evaluation
The final grade will be determined on the basis of assigned work, amid-term examination and a final examination according to thefollowing scheme:
Graded assignments: 20%
Mid-term examination: 30%
Final examination: 50%
Students of this course will be able to find their scores for thevarious components of the final grade posted here as they becomeavailable.
Juris Steprans MATH 2001
The text
The text for the course is:
A Radical Approach to Real Analysis, second edition by David M.Bressoud, 2007. ISBN 0-88385-747-2. Mathematical Associationof America.
The course will cover Chapters 1 to 5 in detail and cover Chapter 6according to the time available. Additional resources can be foundon the course text web site maintained by David M. Bressoud, theauthor of the text.
Juris Steprans MATH 2001
How to read the text
The sections of the text will be covered in the same order aspresented in A Radical Approach to Real Analysis.
You will be told which sections will covered in upcominglectures and will be expected to have read those sectionsbefore the lecture.
You will also be expected to re-read them, with greaterunderstanding, after the lecture.
In other words, between any two lectures you will re-read thematerial of the previous lecture and read for the first time thematerial of the upcoming lecture.
Juris Steprans MATH 2001
Exercises
Exercises will be assigned after most lectures. These will alsobe listed on the course web site with due dates.
Not all exercises will be marked; but some will be collected formarking on a random basis.
The exercises form an essential part of the course and it is notpossible to successfully complete the course without seriouslyattempting all the exercises.
Some assigned exercises may return as questions on themid-term and final examinations.
Juris Steprans MATH 2001
From the preface to A Radical Approach to RealAnalysis
This course of analysis is radical; it returns to the roots of thesubject. It is not a history of analysis. It is rather an attempt tofollow the injunction of Henri Poincare to let history informpedagogy. It is designed to be a first encounter with real analysis,laying out its context and motivation in terms of the transitionfrom power series to those that are less predictable, especiallyFourier’s into which even great mathematicians have fallen.
Juris Steprans MATH 2001
More from the preface to A Radical Approach toReal Analysis
The traditional course begins with a discussion of properties of thereal numbers, moves on to continuity, then differentiability,integrability, sequences, and finally infinite series, culminating in arigorous proof of the properties of Taylor series and perhaps evenFourier series. This is the right way to build analysis, but it is notthe right way to teach it. It supplies little motivation for the earlydefinitions and theorems. Careful definitions mean nothing untilthe drawbacks of the geometric and intuitive understandings ofcontinuity, limits, and series are fully exposed. For this reason, thefirst part of this book follows the historical progression and movesbackwards. It starts with infinite series, illustrating the greatsuccesses that led the early pioneers onward as well as theobstacles that stymied even such luminaries as Euler . . . .
Juris Steprans MATH 2001
Even more from the preface to A RadicalApproach to Real Analysis
There is an intentional emphasis on the mistakes that have beenmade. These highlight difficult conceptual points. That Cauchyhad so much trouble proving the mean value theorem or coming toterms with the notion of uniform convergence should alert us tothe fact that these ideas are not easily assimilated. The studentneeds time with them. The highly refined proofs that we knowtoday leave the mistaken impression that the road of discovery inmathematics is straight and sure. It is not . . .
Juris Steprans MATH 2001
The content of a conventional course
A conventional course might begin with the following definitions:
Definition
If f : R→ R is a function and a ∈ R then limx→a f (x) = L if andonly if for all ε > 0 there is δ > 0 such that |f (z)− L| < ε for all zsuch that 0 < |z − a| < δ.
Definition
If f : R→ R is a function and a ∈ R then f is continuous at a ifand only if limx→a f (x) = f (a). The function f is said to becontinuous if and only if f is continuous at a for all a ∈ R.
Juris Steprans MATH 2001
Why are these natural definitions?
Where do they come from?
Is there not a more natural definition of continuitycorresponding to our intuition that a continuous function isone that we can draw without lifting our pencil from thepaper?
What are the defects of more natural definitions that preventthem from being useful?
Juris Steprans MATH 2001
The next lecture will look at the crisis in mathematics that mademathematicians realize that they had to look much more closely atthe assumptions and foundations of their subject. The subtletiesstemming from the study of the convergence of Fourier series couldnot be understood without providing precise definitions andproving facts that had, up to this point, seemed obvious. As weshall see, some of the ”obvious” facts were not only not obvious,but actually wrong.
Juris Steprans MATH 2001
The crisis in Mathematics: Chapter 1 of thetext
The crisis in mathematics that resulted in the modernunderstanding of analysis resulted from Joseph Fourier’s study ofthe distribution of heat in conductive plates. While it might seemthat there is some irony to be found in the fact that Fourier wasinterested in these questions for practical rather than theoreticalreasons and that his research led to a very abstract branch ofmathematics, the understanding of Fourier series still has practicalapplications in voice recognition, image compression, signalprocessing and many other areas.
Juris Steprans MATH 2001
Fourier and the heat problem
Juris Steprans MATH 2001
The heat equation
The key physical fact is that the rate of cooling (or heating)between two adjacent regions is proportional to the difference intheir temperatures.Let z(x ,w , t) denote the temperature at the point (x ,w) at timet. Given a square with bottom left hand corner at (x ,w) andheight and width h the amount of heat the square loses at the lefthand side from right to left is
h∂
∂xz(x ,w , t)
while the amount of heat the square loses at the right hand sidefrom left to right is
−h∂
∂xz(x + h,w , t)
noting the change in sign.
Juris Steprans MATH 2001
The heat equation continued
Hence, the total change of heat from left to right is:
h
(∂
∂xz(x ,w , t)− ∂
∂xz(x + h,w , t)
)where physical constants have been ignored.Similarly, the total change of heat from top to bottom is:
h
(∂
∂wz(x ,w , t)− ∂
∂wz(x ,w + h, t)
)
Juris Steprans MATH 2001
The heat equation continued
On the other hand, the total amount of heat lost or gained shouldbe proportional to the area of the square and the change intemperature
h2 ∂
∂tz(x ,w , t)
and so
h2 ∂
∂tz(x ,w , t) =
h
(∂
∂xz(x ,w , t)− ∂
∂xz(x + h,w , t) +
∂
∂wz(x ,w , t)− ∂
∂wz(x ,w + h, t)
)and so . . .
Juris Steprans MATH 2001
The heat equation continued
∂
∂tz(x ,w , t) =
∂
∂x
(z(x ,w , t)− z(x + h,w , t)
h
)+∂
∂w
(z(x ,w , t)− z(x ,w + h, t)
h
)Taking the limit of the right hand side as h goes to 0 yields:
∂
∂tz(x ,w , t) =
∂2
∂x2z(x ,w , t) +
∂2
∂w2z(x ,w , t)
Juris Steprans MATH 2001
The heat equation obtained
In the steady state, when the temperature is no longer changing itfollows that ∂
∂t z(x ,w , t) = 0 and hence the variable t can beignored and the heat equation becomes:
∂2
∂x2z(x ,w) +
∂2
∂w2z(x ,w) = 0 (1)
This is the equation in which Fourier was interested.
Juris Steprans MATH 2001
Solutions to the heat equation
Fourier was interested in finding solutions for z such thatz(0, x) = f (x) for some specified function f . The function fspecifies the temperature at which the bottom edge of Fourier’splate is maintained.
Juris Steprans MATH 2001
Solutions to the heat equation
Fourier’s intuition was that there should only be a single solutionto these equations because they model a real physical system.Hence he looked for the simplest type of solution and started byassuming that he would find a solution satisfying
z(w , x) = φ(w)ψ(x)
and in this case
∂2z(w , x)
∂w2= φ
′′(w)ψ(x) and
∂2z(w , x)
∂x2= φ(w)ψ
′′(x)
and soφ
′′(w)ψ(x) + φ(w)ψ
′′(x) = 0
Dividing by φ′′
(w)ψ′′
(x), assuming this is never 0 yields
φ(w)
φ′′(w)+
ψ(x)
ψ′′(x)= 0
Juris Steprans MATH 2001
Solutions to the heat equation
Since φ and ψ depend on different variables it must be thatφ(w)/φ′′(w) and ψ(x)/ψ′′(x) are both constant and, hence
φ′′(w) = Aφ(w)
ψ′′(x) = −Aψ(x)
and these equations should be familiar from first year calculus. Thefirst is the differential equation of exponential growth and thesecond for sin and cos. Hence . . .
Juris Steprans MATH 2001
φ(w) = C0e−√Aw + C1e
√Aw
ψ(x) = C2 cos(√
Ax) + C3 sin(√
Ax)
Fourier then observed that since ψ(x) is an even function for hisparticular problem the coefficient of sin(
√Ax) should be 0. Also,
since the temperature should not grow exponentially the furtheryou are from the heat source, the coefficient C1 should also be 0.Hence he has a solution
z(w , x) = C0e−√AwC2 cos(
√Ax) = Ce−
√Aw cos(
√Ax)
by combining constants.
Juris Steprans MATH 2001
Notice however, that when w = 0 this yields that
z(0, x) = Ce−√A0 cos(
√Ax) = C cos(
√Ax)
and so this would not model an arbitrary heat source at thebottom of the plate. Note though, that for each integer n, settingA = n2 yields the solution
z(w , x) = Ce−nw cos(nx)
which reduces to z(0, x) = C cos(nx) at the bottom of the plate.
Fourier’s brilliant idea was that, since each n gives a differentsolution the sum of the solutions should also be a solution to thedifferential equation. (Why?)
Juris Steprans MATH 2001
Hence, he reasoned, given an arbitrary heat source at the bottomof the plate defined by the function z(0, x) it should be possible tofind constants Cn such that
z(0, x) = C0+C1 cos(x)+C2 cos(2x)+C3 cos(3x)+C4 cos(4x)+. . .
and then
z(w , x) = C0+C1e−w cos(x)+C2e−2w cos(2x)+C3e−3w cos(3x) . . .
would be the solution he wanted.
Juris Steprans MATH 2001
By using arguments that we will examine at the end of the course,Fourier was able to determine the constants Ck . He reduced hisproblem to finding constants Ck such that
1 = C0 + C1 cos(x) + C2 cos(2x) + a3 cos(3x) + . . .
Integration yields that∫ 2π
01dx =
∫ 2π
0(C0 + C1 cos(x) + C2 cos(2x) + C3 cos(3x) + . . .) dx =
∫ 2π
0C0dx+
∫ 2π
0C1 cos(x)dx+
∫ 2π
0C2 cos(2x)dx+
∫ 2π
0C3 cos(3x)dx+. . .
=
∫ 2π
0C0dx = 2πC0
Juris Steprans MATH 2001
Fourier concluded that a the solution to his problem with aconstant heat source at the bottom edge is given by
z(x , 0) =
4
π
(cos(πx
2
)− 1
3cos
(3πx
2
)+
1
5cos
(5πx
2
)− 1
7cos
(7πx
2
). . .
)But does this it make sense to add together infinitely manyfunctions? What happens with the discontinuity at the corners?
Juris Steprans MATH 2001
Fourier realized there were some problems with this solution. Firstof all, it models the constant function 1 on the interval (−1, 1) butthe constant function −1 on the interval (1, 2). This sort ofbehaviour was not expected of functions in his time.
There were also flaws in the reasoning leading to Fourier’sdetermination of the constants 1/3, 1/5, 1/7. . . . In arriving at hisconclusion, Fourier needed to interchange integration with infinitesummation. Why is this justified? We shall see that it is notalways justified and that Fourier was somewhat lucky.
Finally, there is the observation that 1 + 1/3 + 1/5 + 1/7 . . . addsup to an infinite number. So how can Fourier’s series converge atevery point?
Juris Steprans MATH 2001
The next lecture will begin to look at the question of infinitesummation. But we will start with some clever arguments ofArchimedes that arrive at solutions to problems that seem torequire summing infinitely many quantities without having to doso. To prepare for this read Section 2.1 of the text.
Juris Steprans MATH 2001