MAT 5930. Analysis for Teachers

MAT 5930. Analysis for Teachers.

Wm C Bauldry

Summer, 2010

Wm C Bauldry ([email protected]) MAT 5930. Analysis for Teachers. Summer, 2010 1 / 139

Nordkapp, Norway.

Wm C Bauldry ([email protected]) MAT 5930. Analysis for Teachers. Summer, 2010 2 / 139 Introduction and Calculus Considered

1 Course Introduction (Class page, Course Info page, Syllabus)

2 Calculus Considered 1 A Standard Freshman Calculus Course B Refer to texts by Thomas (traditional), Stewart (popular, but. . . ), and Ostebee & Zorn (“reform”). MAT 1110 Topics List

2 An AP Calculus Course(AB,BC) 1 Functions, Graphs, and Limits Representations, one-to-one, onto, inverses, analysis of graphs, limits of functions, asymptotic behavior, continuity, uniform continuity 2 Derivatives Concept, deﬁnitions, interpretations, at a point, as a function, second derivative, applications, computation, numerical approx. 3 Integrals Concept, deﬁnitions, interpretations, properties, Fundamental Theorem, applications, techniques, applications, numerical approx.

Wm C Bauldry ([email protected]) MAT 5930. Analysis for Teachers. Summer, 2010 3 / 139

Calculus Considered

2 (Calculus Considered) 3 Calculus Problems §1 Precalculus material: function, induction, summation, slope, trigonometry §2 Limits and Continuity: Squeeze Theorem, discontinuity, removable discontinuity, different interpretations of limit expressions §3 Derivatives: trigonometric derivatives, power rule, indirect methods, Newton’s method, Mean Value Theorem,“Racetrack Principle” §4 Integration: Fundamental Theorem, Riemann sums, parts, multiple integrals §5 Inﬁnite Series: geometric, integrals and series, partial fractions, convergence tests (ratio, root, comparison, integral), Taylor& Maclaurin

Wm C Bauldry ([email protected]) MAT 5930. Analysis for Teachers. Summer, 2010 4 / 139 AP Calculus 2009 Results

AP Score Distributions, May 2009

n = 230,588 n = 72,965 x¯ = 2.99, s = 1.50x ¯ = 3.72, s = 1.40x ¯ = 4.14, s = 1.21

x ≥ 3 : 59.6% x ≥ 3 : 80.0% x ≥ 3 : 87.8% ASU: MAT 1110 credit (4 cr) MAT 1110 & 1120 (8) MAT 1110 (4) UNC-CH: MATH 231 (3 cr) MATH 231 & 232 (6) MATH 231 (3)

c 2010 The College Board.

Wm C Bauldry ([email protected]) MAT 5930. Analysis for Teachers. Summer, 2010 5 / 139

A Little Background

Our goal is A rigorous study of elementary calculus extending to a treatment of fundamental concepts of analysis involving functions of a real variable. (See “Course Objectives.”)

What this course is not A “refresher course” in calculus. A “how to” course for teaching AP calculus. What this course does Consider calculus from an advanced view—that is, as real analysis. Relate analysis concepts to the calculus classroom for depth.

Develop the profound understanding of fundamental mathematics . . . need[ed] to become accomplished mathematics teachers. — Liping Ma in Knowing and Teaching Elementary Mathematics.

Wm C Bauldry ([email protected]) MAT 5930. Analysis for Teachers. Summer, 2010 6 / 139 Notes to Analysis Students: Learning Mathematics

According to Paul Halmos, to learn mathematics, you have to know:

The ‘What’: Know the deﬁnitions, statements of theorems, and formulas What is an open set? When does a function have to have a maximum? What is the limit formula for the derivative?

The ‘How’: Apply techniques and use formulas to solve problems Find the minimum of the derivative on an interval. Determine where a function is continuous. Calculate the sum of the reciprocals of the squares.

The ‘Why’: Understand why a technique works, why a theorem is true Why does a bounded inﬁnite set have to have a limit point? Why does continuity imply integrability? Why can antiderivatives compute deﬁnite integrals?

Wm C Bauldry ([email protected]) MAT 5930. Analysis for Teachers. Summer, 2010 7 / 139

Notes to Analysis Students: Doing Mathematics

George Polya’s´ Four Steps in Problem Solving:

Step 1: Understand the Problem (“see clearly what is required”) What is the unknown? What are the data?

Step 2: Make a Plan (“the idea of the solution”) Do you know a related problem? How are the items (the data and the unknown) connected? Step 3: Carry Out the Plan Check each step. Can you see clearly and prove that the step is correct?

Step 4: Look Back at the Completed Solution (“review and discuss it”) Reconsider, reexamine the result & the path that led to it. Can you check the result? The argument?

For a complete version see “How to Solve It”

Wm C Bauldry ([email protected]) MAT 5930. Analysis for Teachers. Summer, 2010 8 / 139 Notes to Analysis Students: Teaching Mathematics

George Polya’s´ Ten Commandments for Teachers:

1 Be interested in your subject. 2 Know your subject. 3 Know about the ways of learning: The best way to learn anything is to discover it by yourself. 4 Try to read the faces of your students, try to see their expectations and difﬁculties, put yourself in their place. 5 Give them not only information, but “know-how,” attitudes of mind, the habit of methodical work. 6 Let them learn guessing. 7 Let them learn proving. 8 Look out for such features of the problem at hand as may be useful in solving the problems to come—try to disclose the general pattern that lies behind the present concrete situation. 9 Do not give away your whole secret at once—let the students guess before you tell it—let them ﬁnd out by themselves as much as is feasible. 10 Suggest it, do not force it down their throats.

Wm C Bauldry ([email protected]) MAT 5930. Analysis for Teachers. Summer, 2010 9 / 139

Fundamental Concept: Function

Who First used the term function? – Leibniz1 (1646 –1716) Introduced f(x) notation and originally thought a function must be a single algebraic expression? – Euler (1707–1783) Shocked the math world by presenting a bizarre function? – Weierstrass (1815 –1897) Introduced the formal deﬁnition of function in terms of sets of ordered pairs? – Homework! When did the function concept become central in school mathematics? – 1920’s (See: “The Function Concept in School Mathematics,” H. Hamley. Math. Gazette, 18(229), 169-179) 1 See: Earliest Known Uses of Some of the Words of Mathematics Wm C Bauldry ([email protected]) MAT 5930. Analysis for Teachers. Summer, 2010 10 / 139 The Deﬁnition of Function

Definition (Function) When two variables are so related that the value of the first variable is determined when the value of the second variable is given, then the first variable is said to be a function of the second. From Elements of the Differential and Integral Calculus by Granville, Smith, & Longley (1934).

Deﬁnition (Function) Let X and Y be nonempty sets. Let f be a collection of ordered pairs (x, y) with x X and y Y. Then f is a function from X to Y if to every x X there ∈ ∈ ∈ is assigned a unique y Y. From Calculus and Analytic Geometry by Thomas (1968). ∈ Deﬁnition (Function) A function is a rule for assigning to each member of one set, called the domain, one (and only one) member of another set, called the range. From Calculus from Graphical, Numeric, and Symbolic Points of View by Ostebee & Zorn (2002).

Wm C Bauldry ([email protected]) MAT 5930. Analysis for Teachers. Summer, 2010 11 / 139

Elementary Properties of Functions

Deﬁnition Let f :X Y be a function. → Well-deﬁned: If x1 = x2, then f(x1) = f(x2). One-to-one: If f(x ) = f(x ), then x = x ; or if x = x , then f(x ) = f(x ). 1 2 1 2 1 6 2 1 6 2 Onto: For each y, there is an x such that f(x) = y.

1 Inverse image: If B Y, then f − (B) = x X f(x) B . ⊆ { ∈ | ∈ }

Graph: The set of pairs (x, f(x)) x X X Y is the graph of f. { | ∈ } ⊂ ×

Even: If f( x) = +f(x) for each x, x X. − { − } ⊆ Odd: If f( x) = f(x) for each x, x X. − − { − } ⊆

Wm C Bauldry ([email protected]) MAT 5930. Analysis for Teachers. Summer, 2010 12 / 139 Interlude: Domains Matter

Example Is the statement

+1 = √+1 √+1 = √+1 +1 = ( 1) ( 1) = √ 1 √ 1 = 1 × × − × − − × − − correct? p If yes, then +1 = 1 −

If no, where’s the ﬂaw?

Wm C Bauldry ([email protected]) MAT 5930. Analysis for Teachers. Summer, 2010 13 / 139

The Principle of Mathematical Induction

Theorem (Mathematical Induction) Let P (n) be a statement about the natural number n. If P (1) is true, and whenever P (k) is true, then P (k + 1) must also be true, then P (n) is true for all n N. ∈

Theorem (Strong Induction) Let P (n) be a statement about the natural number n. If P (1) is true, and whenever P (k),P (k 1),P (k 2), ..., P (2), and P (1) are true, − − then P (k + 1) must also be true, then P (n) is true for all n N. ∈

Wm C Bauldry ([email protected]) MAT 5930. Analysis for Teachers. Summer, 2010 14 / 139 Summations and Induction, I

n n(n+1) Example ( k=1 k = 2 by Picture) P

Wm C Bauldry ([email protected]) MAT 5930. Analysis for Teachers. Summer, 2010 15 / 139

Summations and Induction, II

n n(n+1) Example (Demonstrate k=1 k = 2 by Induction) n n(n+1) Let P (n) be the proposition thatP k=1 k is equal to 2 . 1 P 1(1+1) Basis: Then P (1), which is k=1 k = 2 , is clearly true. Induction: Show that if P (n)Pis true, then P (n + 1) is true. Assume P (n) is true and add (n + 1) to both sides; i.e., n n(n + 1) (n + 1) + k = (n + 1) + . 2 k=1 n+1 X 1 Combine terms to see k=1 k = 2 ((2n + 2) + n(n + 1)) . Simplify: n+1 P (n + 1)(n + 2) k = 2 kX=1 which shows that P (n + 1) is true. By the Principle of Mathematical Induction, the result holds.

Wm C Bauldry ([email protected]) MAT 5930. Analysis for Teachers. Summer, 2010 16 / 139 Summations and Induction, III

Exercise Illustrate the statement with a picture and prove it using induction. n 2 1 2k = n + n k=1 X n 2 2k 1 = ? (HINT:ASTACK OF TRIANGLES) − k=1 X n 2 n(n + 1)(2n + 1) 3 k = 6 k=1 X 4 2n < n! for n = 4, 5, 6,...

2 5 2 (n n) for n N | − ∈

Wm C Bauldry ([email protected]) MAT 5930. Analysis for Teachers. Summer, 2010 17 / 139

Polya´ and Induction

George Polya´ constructed the example: Theorem All horses are of the same color. Proof. Basis. If there’s only one horse, there’s only one color. Induction. Suppose within a set of n horses, there is only one color. Look at any set of n + 1 horses. Number them: 1, 2, 3, . . . , n, n + 1. Consider the sets 1, 2, . . . , n and 2, 3, . . . , n + 1 . Each is a set of { } { } only n horses, therefore within each there is only one color. But the two sets overlap, so there is only one color among all n + 1 horses.

Homework: Find the ﬂaw.

Wm C Bauldry ([email protected]) MAT 5930. Analysis for Teachers. Summer, 2010 18 / 139 Important Inequalities ( 1.1) § Theorem

Let a, b, ai, and bi all be real numbers and n N. Then ∈ 1 a b if and only if (iff) b a b | | ≤ − ≤ ≤ 2 a + b a + b Triangle inequality | | ≤ | | | | n 2 n n 3 a b a2 b2 Cauchy-Bunyakovsky-Schwarz ineq. i i ≤ i · i "i=1 # "i=1 # "i=1 # X X X n n n 4 (a + b )2 (a )2 + (b )2 Minkowski inequality v i i ≤ v i v i ui=1 ui=1 ui=1 uX uX uX t t t I. II. 4 a a a and b b b 4 0 (a + b)2 = a2 + 2ab + b2 −| | ≤a ≤b | | a + b− | a| ≤+ b≤ | | ≤a 2 + 2 a b + b 2 = ( a + b )2 −|( a |+ −b | )| ≤ a + b ≤( | a| +| b| ) ≤ | |(a + b|)2| | |( a| +| b )2| | | | − | | | | ≤ ≤ | | | | ≤ | | | |

Wm C Bauldry ([email protected]) MAT 5930. Analysis for Teachers. Summer, 2010 19 / 139

Compare: Definitions of Limit ( 1.2) § “Ghosts of departed Quantities” Definition (First Limit (d’Alembert)) One quantity is the limit of another if the second can approach the first nearer than by a given quantity, so that the difference between them is absolutely inassignable. from Encyclopedie´ (1754).

Definition (Intuitive Limit (Cauchy)) When the successive values attributed to a variable approach indefinitely a fixed value so as to end by differing from it by as little as one wishes, this last is called the limit of all the others. from Cours d’analyse (1823).

Deﬁnition (Formal Limit (Weierstrass and Bolzano))

A function f(x) has a limit L at x = x0 if for any positive number , there exists a δ such that f(x) L < for all x in the deleted interval 0 < x x < δ. | − | | − 0| The notation “lim” was introduced by S.-A.-J. L’Huilier (1784) and “ lim ” is due to K. T. W. Weierstrass (1854). x→a

Wm C Bauldry ([email protected]) MAT 5930. Analysis for Teachers. Summer, 2010 20 / 139 Limit Proofs, I

Example (-δ Proof) Prove that lim 2x + 3 = 7 x→2 Proof. Let > 0. Then we need to ﬁnd a δ > 0 so that

f(x) L = (2x + 3) (7) < | − | | − | 2x 4 < | − | 2 x 2 < | − | Choosing δ > 0 so that δ < /2 yields that if 0 < x 2 < δ, then it | − | must follow that f(x) L = (2x + 3) (7) < . | − | | − |

Wm C Bauldry ([email protected]) MAT 5930. Analysis for Teachers. Summer, 2010 21 / 139

Limit Proofs, II

Example (-δ Proof) Prove: lim 4x2 1 = 35 x→3 − Proof. Let > 0. We need to ﬁnd a δ > 0 so that for 0 < x 3 < δ, then | − | f(x) L = (4x2 1) (35) < | − | − − 2 4x 36 = 2x + 6 2x 6 < | − | | | · | − | (4 x + 3 ) x 3 < | | · | − | Assume δ < 1. Then 1 < x 3 < 1 implies that 5 < x + 3 < 7, so that − − 20 < 4(x + 3) < 28. Choosing δ > 0 to be less than the minimum of /28 and 1 yields that if 0 < x 3 < δ, then it must follow that | − | f(x) L = (4x2 1) (35) < . | − | − −

Wm C Bauldry ([email protected]) MAT 5930. Analysis for Teachers. Summer, 2010 22 / 139 Limit Proofs, III

“For every > 0 there is a δ > 0 such that for all x, we see P (x) is true” negated becomes “There is an > 0 such that for every δ > 0 there is an xˆ with P (ˆx) false” Example (-δ “Non-Proof”) x Demonstrate that lim | | does not exist. x→0 x Proof. Suppose the limit is L. Let = 1 and let δ be any positive number. Choose any xp (0, δ). Then xp /xp = 1, so that f(x) L = 1 L . ∈ | | | − | | − | Choose any xn ( δ, 0). Then xn /xn = 1, so that ∈ − | | − f(x) L = 1 L . We have that | − | | − − | < 1 L < 1 < 1 L < 1 0 < L < 2 and − − ⇒ − − ⇒ < 1 L < 1 < 1 L < 1 2 < L < 0 which is− a contradiction.− − Therefore⇒ − there− − is no limit⇒L. −

Wm C Bauldry ([email protected]) MAT 5930. Analysis for Teachers. Summer, 2010 23 / 139

Limit Proofs, IV

Exercise Find the value and prove it correct for: 1 lim 4x 1 = x→3 − 2 lim 3x + 5 = x→2 − 2 3 lim 4 x = x→1 − 3 4 lim 2x + 1 = x→0 3 5 lim 3x + x + 1 = (STUCK?LOOK AT APROOF .) x→−1 6 Why won’t this approach work for lim ln(2x 1) ? x→1 − 7 Prove that lim 1 doesn’t exist. x→0 x

Wm C Bauldry ([email protected]) MAT 5930. Analysis for Teachers. Summer, 2010 24 / 139 Squeeze Theorem

Theorem (Squeeze Theorem or Sandwich Theorem) Suppose that m(x) f(x) M(x) on a deleted neighborhood 2of a and that ≤ ≤ lim m(x) = L = lim M(x). x→a x→a Then lim f(x) = L. x→a

1 Apply the theorem to f(x) = x2 sin(1/x) to determine a value for f(0) that makes f continuous everywhere. sin(x) 2 State an analogue of the theorem to use to determine lim . x→∞ x

2A deleted neighborhood of a is I0 = (a − δ, a + δ) − {a} for some δ > 0.

Wm C Bauldry ([email protected]) MAT 5930. Analysis for Teachers. Summer, 2010 25 / 139

Compare: Deﬁnitions of Continuity

Deﬁnition (Intuitive Continuity) The function f is continuous at x = a if and only if we can make f(x) arbitrarily close to f(a) whenever x is sufﬁciently close to a.

Deﬁnition (Formal Continuity—calculus level) Let f be deﬁned on an open interval containing a. Then f is I continuous at x = a if and only if for every > 0 there is a corresponding δ > 0 such that whenever x is within δ of a, then ∈ I f(x) must be within of L.

1 How do these definitions compare to the (calculus level) “limit definition” of continuity: Definition: f is continuous at a if and only if lim f(x) = f(a)? x→a

Wm C Bauldry ([email protected]) MAT 5930. Analysis for Teachers. Summer, 2010 26 / 139 Borzellino’s Functions, I

Example (f(x) = x sin(1/x) ) Does f have a limitp at x = 0? Is f continuous at x = 0?

Wm C Bauldry ([email protected]) MAT 5930. Analysis for Teachers. Summer, 2010 27 / 139

Borzellino’s Functions, II

Example (g(x) = (x4 x2)3/2 ) − Does g have a limit at x = 0? Is g continuous at x = 0?

Wm C Bauldry ([email protected]) MAT 5930. Analysis for Teachers. Summer, 2010 28 / 139 Types of Discontinuity

Deﬁnition (Four Principal Types of Discontinuity) Simple or First Kind Removable: The limit lim f(x) exists, but isn’t equal to f(a). x→a Jump: Both lim f(x) and lim f(x) exist, but have different x→a+ x→a− values.

Essential or Second Kind Inﬁnite: At least one of lim f(x) or lim f(x) is inﬁnite. x→a+ x→a− Oscillating: At least one of lim f(x) or lim f(x) doesn’t exist, but is x→a+ x→a− bounded.

1 Find and graph examples of each type of discontinuity.

Wm C Bauldry ([email protected]) MAT 5930. Analysis for Teachers. Summer, 2010 29 / 139

Continuous and Discontinuous a Lot!

Example (Thomae’s Function3) A function that is continuous at each irrational point and discontinuous at each rational point in [0, 1].

1/q if x=p/q, (p, q)=1 D(x) = (0 otherwise

Can a function be discontinuous only on the irrationals?

Compare to Baire’s theorem: Baire 1 functions are continuous except on a ﬁrst category set. 3 Also called Riemann’s function; Dirichlet’s function is χQ.

Wm C Bauldry ([email protected]) MAT 5930. Analysis for Teachers. Summer, 2010 30 / 139 Continuity Theorems

Theorem (Composition of Continuous Functions) The composition of continuous functions is a continuous function.

Theorem (Intermediate Value Theorem) Let f be continuous on the closed interval [a, b] and let yˆ be between f(a) and f(b). Then there is at least one xˆ [a, b] for which f(ˆx) =y. ˆ ∈ Bolzano’s original theorem: if f(a) f(b) < 0, then f must have a root. · Theorem (Extreme Value Theorem) Let f be continuous on the closed interval [a, b]. There are (at least) two values xm and xM in [a, b] such that f(xm) f(x) f(xM ) ≤ ≤ for all x [a, b]; i.e., f must take on minimum and maximum values. ∈

Wm C Bauldry ([email protected]) MAT 5930. Analysis for Teachers. Summer, 2010 31 / 139

Homework

1 Problems: 1.1 & 1.2; pg. 36: 1a,§ 2, 4, 6, 11, 13, 17, 18, 21, 23, 24, 27, 29, 30.

2 Read 1.3 §

3 Enjoy the The MacTutor History of Mathematics archive web page for the Mathematicians of the day.

4 Look at the Animated Function Families web page.

Wm C Bauldry ([email protected]) MAT 5930. Analysis for Teachers. Summer, 2010 32 / 139 Derivatives ( 1.3) § Deﬁnition (The Derivative Function) The derivative of f(x) is given by d f(x + h) f(x) f(x) = f 0(x) = lim − dx h→0 h if the limit exists.

Which came ﬁrst differentiation or integration? Proposition (Basic Rules of Differentiation) Results: Reductions: r r 1 1 (u )0 = r u − u0 1 (k u)0 = k (u0) · · · 2 u u u0 (u v)0 = (u0) (v0) 2 (e )0 = e u0, ln(u)0 = ± ± · u 3 (u v)0 = (u0) v + u (v0) · · · 3 sin(u)0 = + cos(u) u0, etc. · u (u0) v u (v0) 4 0 = · − · 1 u0 v v2 4 sin− (u)0 = , etc. √1 u2 5 (u(v))0 = u0(v) v0 Wm C Bauldry ([email protected]− ) MAT 5930. Analysis for Teachers. · Summer, 2010 33 / 139

Derivative People

Pierre de Fermat (1601-1665) Isaac Barrow (1630-1677)

Fermat: Solved extrema problems using tangents (derivatives) Barrow: “Differential triangle;” Newton attended Barrow’s lectures on geometry and optics

Wm C Bauldry ([email protected]) MAT 5930. Analysis for Teachers. Summer, 2010 34 / 139 The Chain Rule and Inverse Functions

Theorem 0 If f :[a, b] R is differentiable and f is not zero on [a, b], then, for → y = f(x), we have 0 1 f −1 (y) = f 0(x)

Proof. (Computation sketch.)

f f −1(y) = y −1 0 f f (y) = 1 0 −1 −10 f f (y) f (y) = 1 Chain Rule! · ← 0 −10 f (x) f (y) = 1 ·

Wm C Bauldry ([email protected]) MAT 5930. Analysis for Teachers. Summer, 2010 35 / 139

Properties of the Derivative

Theorem If f is differentiable at x = a, then f is continuous at x = a.

Proof. f(x) f(a) lim − lim(x a) = lim[f(x) f(a)] x→a x a × x→a − x→a − −

Theorem (Darboux’s Intermediate Value Theorem) If f is continuous on [a, b] and differentiable on (a, b), then f 0 has the intermediate value property.

But f 0 isn’t necessarily continuous!

Wm C Bauldry ([email protected]) MAT 5930. Analysis for Teachers. Summer, 2010 36 / 139 Newton’s Method

Example (A Functional Version of Newton’s Method) f(x) Deﬁne the function Newton(x) = x . − f 0(x) Maple TI f(x) := ... y1 := ... df := D(f) y2 := (y10) Newton(x) := x - f(x)/df(x) y3 := x - y1(x)/y2(x) Give an initial value and iterate: Maple TI 1.0 1.0 Newton(%) y3(ans) Newton(%) y3(ans) Et cetera.

1 Find all positive roots of f(x) = x7 1.4995x + 0.994. (Maple ﬁle.) −

Wm C Bauldry ([email protected]) MAT 5930. Analysis for Teachers. Summer, 2010 37 / 139

A Group Project

Project (The function M(x) = x f 0(x)/f 00(x)) − Determine what iterating f 0(x) M(x) = x − f 00(x) does. 1 Test the function f(x) = x4 8x3 + 19x2 12x with different − − starting values for x0.

1 x0 = 1.0 2 x0 = 2.0 3 x0 = 3.0 2 Do the same with g(x) = x + 2 sin(x). 3 What is your conjecture? Can you prove it?

Wm C Bauldry ([email protected]) MAT 5930. Analysis for Teachers. Summer, 2010 38 / 139 The Mean Value Theorem and . . .

Theorem (The Mean Value Theorem; Cauchy – 1823) Let f be differentiable on (a, b) and continuous on [a, b]. Then there is a f(b) f(a) value c (a, b) so that f 0(c) = − . ∈ b a − Somewhere the instantaneous rate of change = the average rate of change.

Theorem (The “Racetrack Principle;” Davis, Porta, Uhl – 1994) Suppose f(a) = g(a) and f 0(x) g0(x) for all x a. Then f(x) g(x) ≤ ≥ ≤ for all x a. ≥

Theorem (The “Speed Limit Law;” Ostebee and Zorn – 1997 ) 0 Let f be differentiable on [a, b] and M R. If f (x) M for all ∈ ≤ x [a, b], then f(x) f(a) M (x a). ∈ − ≤ · −

Wm C Bauldry ([email protected]) MAT 5930. Analysis for Teachers. Summer, 2010 39 / 139

Derivative Tests for Extrema

Proposition (First Derivative Test) Suppose f is differentiable on an interval containing c, a critical point. If the derivative f 0 changes sign from + below c to above c, then f(c) is a local maximum, − from below c to + above c, then f(c) is a local minimum. − If f 0 doesn’t change sign, then f(c) is a terrace point, not an extrema.

Proposition (Second Derivative Test) Suppose f is twice differentiable on an interval containing c, a critical point. If f 00(c) < 0, then f(c) is a local maximum, If f 00(c) > 0, then f(c) is a local minimum. If f 00(c) = 0, the test fails to classify the point.

Wm C Bauldry ([email protected]) MAT 5930. Analysis for Teachers. Summer, 2010 40 / 139 Homework

1 Problems: 1.3; pg. 36: 29,§ 34, 40b, 41, 44, 45, 46; Look at 39.

2 Read 1.4 §

3 Investigate the MAA Mathematical Sciences Digital Library

4 Explore the Mathworld website.

Wm C Bauldry ([email protected]) MAT 5930. Analysis for Teachers. Summer, 2010 41 / 139

Development of the Integral ( 1.4) § Hippocrates (430 BC): Quadrature of the Lune

Eudoxus (360 BC): Method of Exhaustion

Archimedes (250 BC): Vol(sphere) = 2/3 Vol(cyl), parabolic area ·

a an+1 Cavalieri (1635): xn dx = — without modern notation n + 1 Z0

a a(p/q)+1 Fermat (1679): xp/q dx = — without modern notation (p/q) + 1 Z0

Newton’s notation: x˙ & x Leibniz’ notation: dx & omn. (later dx) R Wm C Bauldry ([email protected]) MAT 5930. Analysis for Teachers. Summer, 2010 42 / 139 The Fundamental Theorem, I

Isaac Newton (1642-1723) Gottfried Leibniz (1646-1716)

Developed the Fundamental Theorem independently Used either antiderivatives or geometry to calculate integrals Newton did not use transcendental functions, preferring inﬁnite series; Leibniz avoided series

Wm C Bauldry ([email protected]) MAT 5930. Analysis for Teachers. Summer, 2010 43 / 139

The Fundamental Theorem, II

Theorem (Fundamental Theorem, Version 1) Suppose that f is integrable 4on [a, b] and, for x [a, b], deﬁne F (x) by ∈ x F (x) = f(t) dt Za Then F is continuous and if f is continuous at c, then F 0(c) = f(c).5

Theorem (Fundamental Theorem, Version 2) Let f be continuous on [a, b] with F 0(x) = f(x). Then

b f(x) dx = F (b) F (a) − Za

4 R R b Leibniz invented the · dx notation (1665); Fourier added the limits a · dx. (1822) 5Watch out for singular functions! Wm C Bauldry ([email protected]) MAT 5930. Analysis for Teachers. Summer, 2010 44 / 139 Deﬁnite Integral Deﬁnitions

b n Cauchy (1823) f(x) dx = lim f(xn−1) (xn xn−1) n · − Za k=1 Theorem: Continuous ⇒Xintegrable

b n Riemann (1854) f(x) dx = lim f(cn) (xn xn−1) n · − Za k=1 Theorem: Continuous exceptX on a “null set” ⇒ integrable

b n Stieltjes (1894) f(x) dg(x) = lim f(cn) (g(xn) g(xn−1)) n · − Za k=1 Theorem: Continuous and gXmonotone ⇒ integrable Lebesgue (1902) b −1 f(x) dµ(x) = lim f(cn) “length” f ([yk−1, yk]) · Za k Theorem: Bounded andX “measurable” ⇒ integrable

Wm C Bauldry ([email protected]) MAT 5930. Analysis for Teachers. Summer, 2010 45 / 139

Integral People

Augustin-L. Cauchy G. F. Bernhard Riemann Thomas J. Stieltjes Henri L. Lebesgue 1789-1857 1826-1866 1856-1894 1875-1941

Wm C Bauldry ([email protected]) MAT 5930. Analysis for Teachers. Summer, 2010 46 / 139 Riemann Sums

Deﬁnition (Riemann Sum) Let f be a bounded function on [a, b]. Let the partition be P = x0 = a < x1 < x2 < < xn = b and = ti be a collection of P { ··· } T { } points where ti [xi−1, xi] for each i = 1..n. The Riemann sum is ∈ n (f, , ) = f(tk) (xk xk−1) R P T · − k=1 X Deﬁnition (Riemann-Stieltjes Sum) Let f be bounded and g be increasing on [a, b]. Let the partition be P = x0 = a < x1 < x2 < < xn = b and = ti be a collection of P { ··· } T { } points where ti [xi−1, xi] for each i = 1..n. The Riemann-Stieltjes sum is ∈ n (f, g, , ) = f(tk) [g(xk) g(xk−1)] RS P T · − k=1 X

Wm C Bauldry ([email protected]) MAT 5930. Analysis for Teachers. Summer, 2010 47 / 139

Riemann’s Function (x) R From Riemann’s dissertation: Deﬁne 0.75

∞ ((kx)) 0.5 (x) = R k2 kX=1 0.25 6 where 0 0.25 0.5 0.75 1 1.25 1.5

 -0.25 x − bxc bxc ≤ x < bxc + 1  2 ((x)) = 0 x = bxc + 1 2 -0.5  1 x − bxc − 1 bxc + 2 ≤ x < bxc + 1 -0.75

(x) is discontinuous at all rationals x = p/q where q is even R 1 (x) is integrable What do you conjecture (x) =? R R Z0

6 Note: alternately ((x)) = (x − 1) − bx − 1/2c for x 6= bxc + 1/2. Wm C Bauldry ([email protected]) MAT 5930. Analysis for Teachers. Summer, 2010 48 / 139 A Function Deﬁned by an Integral

Example +1 x < 1.5 x Deﬁne: f(x) = and F (x) = f(t) dt. ( 1 otherwise) 0 − Z

Query: Where is F not 1 differentiable?

-1 0 1 2 3

-1

-2

Wm C Bauldry ([email protected]) MAT 5930. Analysis for Teachers. Summer, 2010 49 / 139

Functions Deﬁned by Integrals

Many important functions that have no elementary expressions are deﬁned by integrals. Deﬁnition (Several Special Functions) x 1 xn+1−1 ln(x) = dt Look at lim n+1 1 t n→−1 Z x 2 2 erf(x) = e−t dt,“error function” √π 0 ∞ Z Γ(x) = tx−1e−tdt, Generalized factorial as Γ(n) = (n 1)! − Z0 x sin(t) Si(x) = dt,“sine integral” 0 t Z x π 2 s(x) = sin t dt,“Fresnel sine integral” F 2 Z0 and hundreds more... See Maple’s Special Functions Explorer

Wm C Bauldry ([email protected]) MAT 5930. Analysis for Teachers. Summer, 2010 50 / 139 Which Witch is Which?

Example

-3 -2 -1 0 1 2 3

-1

-2

-3

The graph shows f(x), f 0(x), and F (x) = f(x) dx. Which is which? Z

Wm C Bauldry ([email protected]) MAT 5930. Analysis for Teachers. Summer, 2010 51 / 139

A Special Integral

∞ 2 The integral e−x dx is important in analysis and probability, but −∞ has no elementaryZ antiderivative — the Fundamental Theorem does not apply. Instead, consider ∞ ∞ ∞ ∞ 2 2 2 2 2 = e−x dx e−y dy = e−(x +y )dx dy. I Z0 Z0 Z0 Z0 Change to polar coordinates with

[x, y] [r, θ] = x2 + y2 , arctan(y/x) and dx dy r dr dθ. 7→ 7→ hp i The transformed integral is π/2 ∞ 2 2 = e−r r dr dθ, I Z0 Z0 which is not hard to evaluate. Homework!

Wm C Bauldry ([email protected]) MAT 5930. Analysis for Teachers. Summer, 2010 52 / 139 A Mean Value Theorem for Integrals

Deﬁnition (Average Value) Let f be an integrable function on [a, b]. The average value f¯ is 1 b f¯ = f(t) dt b a − Za Theorem (Integral Mean Value Theorem) Let f be a continuous on [a, b]. There is a point c (a, b) where f(c) = f.¯ ∈ Proof.

1 f is continuous on [a, b], so f has a max M and a min m by the EVT b 2 f is cont. so integrable. Then m(b a) f(x) dx M(b a) − ≤ a ≤ − b 3 R 1 By the IVT, there is a pt. c (a, b) so that f(c) = b a a f(x) dx ∈ − R

Wm C Bauldry ([email protected]) MAT 5930. Analysis for Teachers. Summer, 2010 53 / 139

Homework

1 Problems: 1.4; pg. 36: 48,§ 49, 53, 55, 56.

2 Read 1.5, 1.6 §

3 Investigate the Earliest Known Uses of Some of the Words of Mathematics website

4 Explore the Earliest Uses of Symbols of Calculus website.

Wm C Bauldry ([email protected]) MAT 5930. Analysis for Teachers. Summer, 2010 54 / 139 Sequential Limits ( 1.5) §

Deﬁnition

Let an be a sequence. Then lim an = L if and only if for every > 0 { } n→∞ there is an N N such that whenever n > N, then an L < . ∈ | − | Examples ( 1)n n + 1 Let an = − Let dn = n + 1 2n 1 2 − Let bn = √4n n + 3 2n Let en = cos(nπ/2) − − n Let cn = sin(2nπ) Let fn = n

Theorem (Bolzano-Weierstrass Thm for Sequences) A bounded sequence has a convergent subsequence.

Wm C Bauldry ([email protected]) MAT 5930. Analysis for Teachers. Summer, 2010 55 / 139

Series Limits

Deﬁnition ∞ Let an be a series. Then an = L if and only if for every > 0 k=0 X n thereP is an N N such that whenever n > N, then an L < . ∈ | k=0 − | P Examples ∞ ∞ 1 1 rk = when r < 1 = e 1 r | | k! k=0 − k=0 X∞ X∞ 1 ( 1)n+1 = − = ln(2) n ∞ n n=1 n=1 X∞ X∞ 1 π2 1 = = 1 j2 6 j (j + 1) j=1 j=1 X X ·

Wm C Bauldry ([email protected]) MAT 5930. Analysis for Teachers. Summer, 2010 56 / 139 Guido Grandi and Summing Series Divinely

Guido Grandi (1671-1742) discussed inﬁnite series. In his book Quadratura circula et hyperbolæper inﬁnitas hyperbolas geometrice exhibita (1703), he computes

0 = (1 +− 1) + (1 +− 1) + (1 +− 1) + (1 +− 1) + ··· 1 = 1 + (−1 + 1) + (−1 + 1) + (−1 + 1) + and writes: ··· By putting parentheses into the expression 1 1 + 1 1 + in different ways, I can, if I want, obtain 0 or 1. But− then− the idea··· of the creation ex nihilo is perfectly plausible.7

1 Although, he thought the true value of the series was 2 from setting x = 1 in 1 = 1 x + x2 x3 in the 2nd edition of Quadratura. 1+x − − ± · · ·

7 “Sed inquies: aggregatum ex infinitis differentiis infinitarum ipsi DV æqualium, sive continue,` sive alterne` sumptarum, est demum summa ex infinitis nullitatibus, seu 0, quomodo ergo quantitatem notabilem aggreget? At repono, eam Infiniti vim agnoscendam, ut etiam quod per se nullum est multiplicando, in aliquid commutet, sicuti finitam magnitudine´ dividendo, in nullam degenerare cogit: unde per infinitam Dei Creatoris potentiam omnia ex nihlo facta, omniaque in nihilum redigi posse: neque ade absurdum esse, quantitatem aliquam, ut ita dicam, creari per infinitam vel multiplicationem, vel additionem ipsius nihili, aut quodvis quantum infinita divisione, aut subductione in nihilum redigit.”

Wm C Bauldry ([email protected]) MAT 5930. Analysis for Teachers. Summer, 2010 57 / 139

Convergence Tests for Series

Theorem

Ratio Test: [D’Alembert (1761)] Let an be a positive term series an+1 and set r = lim . Then n an P 1 if 0 r < 1, the series converges. ≤ 2 if 1 < r , the series diverges. ≤ ∞ 3 if r = 1, the test fails.

Root Test: [Cauchy (1823)] Let an be a positive term series and n set ρ = lim √an. Then n P 1 if 0 ρ < 1, the series converges. ≤ 2 if 1 < ρ , the series diverges. ≤ ∞ 3 if ρ = 1, the test fails.

The root test is stronger than the ratio test.

Wm C Bauldry ([email protected]) MAT 5930. Analysis for Teachers. Summer, 2010 58 / 139 Very Useful Tests

Theorem

Limit Comparison: Let an and bn be positive series. Set r = lim an/bn. Then P P 1 if r = 0 and bn converges, then an converges. 2 if 0 < r < , the two series either converge or diverge together.∞P P 3 if r = and bn diverges, then an diverges. ∞ P P Integral Test: Let an = f(n) be positive terms. Then an and ∞ f(x) dx converge or diverge together. k P R For more tests, visit Mathworld’s Convergence Tests page.

Wm C Bauldry ([email protected]) MAT 5930. Analysis for Teachers. Summer, 2010 59 / 139

Group Project

Project Test each series for convergence.

∞ ∞ 2 + n3 n 1 5 ln 1 + 2n3 3n + 1 n=0 n=1 X∞ X∞ n 22n+1 2 6 ln(n) 5n n=2 n=0 X∞ X∞ ( 10)n √n + 1 √n 3 − 7 − n! n n=0 n=1 X X ∞ ∞ n n n−1 −3 ( 1) x 4 ( 1) n 8 − − 22n n! n=1 n=0 X X

Wm C Bauldry ([email protected]) MAT 5930. Analysis for Teachers. Summer, 2010 60 / 139 Taylor Polynomials and Series

Deﬁnition (Taylor Polynomial for f) Let f have n continuous derivatives. The Taylor polynomial for f of degree n is 00 (n) 0 f (a) 2 f (a) n Tn(x) = f(a) + f (a)(x a) + (x a) + + (x a) − 2! − ··· n! − (n+1) Set Mn+1 = max f (t) on the interval [a, b]. Then the error in approximating f by Tn for x [a, b] is bounded by ∈ Mn+1 n+1 Errn (x a) ≤ (n + 1)! −

Deﬁnition (Taylor Series for f) Let f have derivatives of all orders. The Taylor series for f is ∞ f (k)(a) T (x) = (x a)k k! − k=0 for x a < R where R 0 isX the radius of convergence. | − | ≥

Wm C Bauldry ([email protected]) MAT 5930. Analysis for Teachers. Summer, 2010 61 / 139

Taylor Comparison

Example

sin(x) & T5, T15, and T30. tan(sin(x)) & T5, T15, and T30

The Taylor series for sine works well, in contrast to the series for tan sin . ◦

Wm C Bauldry ([email protected]) MAT 5930. Analysis for Teachers. Summer, 2010 62 / 139 Complex Plots Show the Answer

Example

z = sin(x + iy) z = tan(sin(x + iy)) | | | | No poles Lots of poles

The radius of convergence of a Taylor series is the distance from its center to the nearest pole. Poles may lie in the plane, off the real axis.

Wm C Bauldry ([email protected]) MAT 5930. Analysis for Teachers. Summer, 2010 63 / 139

Homework

1 Problems: 1.5, 1.6; pg. 36: 36,§ 60, 61, 63, 66, 70, 72. Ponder the comment following #72

2 Read 2.1, 2.2 §

3 Investigate the Google Directory of Math Software

4 Explore uses of GeoGebra (free) software.

Wm C Bauldry ([email protected]) MAT 5930. Analysis for Teachers. Summer, 2010 64 / 139 Interior, Exterior, Boundary, & Limit Points ( 2.1) § Deﬁnition

Neighborhood: A (basic) neighborhood N(x) of x R is an open interval containing x; i.e., N(x) = (x δ , x +∈ δ ). − L R Deleted N’hood: A neighborhood N(x) minus x; i.e., N 0(x) = N(x) x . − { } Interior: The interior of a set is int(S) = x N(x) S for some n’hood N(x) . { | ⊆ } Exterior: The exterior of a set is ext(S) = x N(x) Sc for some n’hood N(x) . { | ⊆ } Boundary: The boundary of a set is c ∂(S) = int(S) ext(S) . ∪ Limit Point: The point x is a limit point or an accumulation point of S iff every neighborhood N(x) contains a point of S different from x; i.e., S N 0(x) = . ∩ 6 ∅

Wm C Bauldry ([email protected]) MAT 5930. Analysis for Teachers. Summer, 2010 65 / 139

Interior/Exterior in R2

Example

Exterior

Interior

Boundary

Wm C Bauldry ([email protected]) MAT 5930. Analysis for Teachers. Summer, 2010 66 / 139 Interior/Exterior Example

Example Identify the points A and B as interior, exterior, or boundary points.

A B

Wm C Bauldry ([email protected]) MAT 5930. Analysis for Teachers. Summer, 2010 67 / 139

Homework

A B

1 Problems: pg. xx. §

2 Read §

3 Investigate

4 Explore A B

(5) (4)

Wm C Bauldry ([email protected]) MAT 5930. Analysis for Teachers. Summer, 2010 68 / 139 (9)

(8)

A B

(5) (4) Limit and Continuity Deﬁnitions ( 2.2) § Deﬁnition (Limit) Let f : D R and let a be an accumulation point of Dlet a be an → accumulation point of D. Then lim f(x) = L if and only if for every x→a > 0 there is a δ > 0 such that if 0 < x a < δ, then f(x) L < . | − | | − |

Deﬁnition (Continuity) Let f : D R and let a D. Then f is continuous at x = a if and only → ∈ if for every > 0 there is a δ > 0 such that if x a < δ, then | − | f(x) f(a) < . ∀ > 0 ∃δ > 0 ∀x ∈ D [d(x, a) < δ =⇒ d(f(x), f(a)) < ]. | − | Deﬁnition (Continuity — General Topological Version) Let f : D R and let a D. Then f is continuous at x = a iff for every → ∈ −1 open set R, the inverse image f ( ) is an open set in D. O ⊆ O

Wm C Bauldry ([email protected]) MAT 5930. Analysis for Teachers. Summer, 2010 69 / 139

Limit and Continuity Example

Example

3 Consider

√x2 1 1 x 2 f(x) = − ≤ | | (0 x = 0 1

1 What is the lim f(x) ? x→0 The limit does not exist as 0 is -2 -1 0 1 2 not an accumulation point. -1 2 Is f continuous at x = 0? Yes!

Wm C Bauldry ([email protected]) MAT 5930. Analysis for Teachers. Summer, 2010 70 / 139 Uniform Continuity

Deﬁnition A function f is uniformly continuous on a set S iff for any > 0, there is a δ > 0 such that for any x1, x2 S with x1 x2 < δ, we ∈ | − | have f(x1) f(x2) < . | − | A function f is not uniformly continuous on a set S iff there is an > 0 for which any δ > 0 has points x1, x2 S with x1 x2 < δ, ∈ | − | but f(x1) f(x2) . | − | ≥ Theorem If f is continuous on a compact set, then f is uniformly continuous on it. If f is uniformly continuous on (a, b), then f can be extended to be (uniformly) continuous on [a, b].

If f 0 is bounded on (a, b), then f is uniformly continuous on (a, b).

Wm C Bauldry ([email protected]) MAT 5930. Analysis for Teachers. Summer, 2010 71 / 139

Homework

1 Problems: pg. xx. §

2 Read §

3 Investigate

4 Explore

Wm C Bauldry ([email protected]) MAT 5930. Analysis for Teachers. Summer, 2010 72 / 139 Derivatives ( 2.3) §

Wm C Bauldry ([email protected]) MAT 5930. Analysis for Teachers. Summer, 2010 73 / 139

Homework

1 Problems: 2.3 pg. 117. # 23,§ 24, 26, 29, 34, 35, 36

2 Read §

3 Investigate

4 Explore

Wm C Bauldry ([email protected]) MAT 5930. Analysis for Teachers. Summer, 2010 74 / 139 Riemann Integrals ( 2.4) §

Deﬁnitions

Partition: A set of n + 1 points where a = x0, b = xn, and P = x0 < x1 < x2 < < xn . Set k = [xk−1, xk]. { ··· } I Tagged Partition: A partition P along with a set of tags T = tk tk [xk−1, xk] for k = 1..n denoted by PT { | ∈ n } Riemann Sum: RS(PT , f) = f(tk)∆xk k=1 X n Darboux Upper Sum: U(P, f) = Mk ∆xk where Mk = sup f(x) k=1 x∈Ik Xn Darboux Lower Sum: L(P, f) = mk ∆xk where mk = inf f(x) x∈Ik k=1 X

Wm C Bauldry ([email protected]) MAT 5930. Analysis for Teachers. Summer, 2010 75 / 139

Riemann-Darboux Integrals

Deﬁnitions Upper Integral: b f(x) dx = inf U(P, f) a P R b Lower Integral: a f(x) dx = sup L(P, f) P Riemann-DarbouxR Integral: A function bounded on [a, b] is integrable iff b b b a f(x) dx = a f(x) dx. The common value is a f(x) dx R R R Proposition For P any partition, L(P, f) b f(x) dx b f(x) dx U(P, f) ≤ a ≤ a ≤ R R Theorem Suppose f is bounded on [a, b]. Then f is integrable iff for any > 0 there is a partition P ∗ such that U(P ∗, f) L(P ∗, f) < . −

Wm C Bauldry ([email protected]) MAT 5930. Analysis for Teachers. Summer, 2010 76 / 139 Integral Examples

Example Let D(x) be Dirichlet’s function. Then 1 1 D(x) dx = 0 = 1 = D(x) dx, 0 6 0 1 Z Z therefore D(x) dx DNE. Z0 Example Let T (x) be Thomae’s function. Then 1 1 T (x) dx = 0 = T (x) dx, 0 0 1 Z Z therefore T (x) dx = 0. Z0

Wm C Bauldry ([email protected]) MAT 5930. Analysis for Teachers. Summer, 2010 77 / 139

Integral Theorems

Theorem (Continuity Implies Integrability) If f is continuous on [a, b], then f is integrable on [a, b].

Proof.

U(P ) L(P ) = (Mk mk)∆xk = [f(xk,M ) f(xk,m)]∆xk • − − − f(xk,M ) f(xk,m) /(b a) and ∆xk = b a • − P≤ − P − P Theorem (Monotonicity Implies Integrability) If f is monotone on [a, b], then f is integrable on [a, b].

Proof.

U(P ) L(P ) = (Mk mk)∆xk = [f(xk) f(xk−1)]∆xk • − − − with ∆xk = (b a)/n and [f(xk) f(xk−1)] = f(b) f(a) • −P −P − P Wm C Bauldry ([email protected]) MAT 5930. Analysis for Teachers. Summer, 2010 78 / 139 A Series of Steps

Example

∞ 1 1 S(x)= U x k2 · − k k=1

Is S integrable on [0, 1]? Yes! 1 S(x) dx 0.44 0 ≈ Wm C Bauldry ([email protected]) MAT 5930.R Analysis for Teachers. Summer, 2010 79 / 139

Integral Mean Value Theorem

Theorem (Bonnet’s Mean Value Theorem) Suppose f is continuous and g is nonnegative and integrable on [a, b]. Then there is a value c (a, b) such that ∈ b b f(x)g(x) dx = f(c) g(x) dx Za Za Proof. f is continuous on a closed interval = f has max & min values ⇒ m g f g M g, then divide by g (nonzero else done) · ≤ ≤ · IVT = there is a c s.t. f(c) = ( f g)/( g) R ⇒ R R R R R Useful form (from the proof): b b b m g(x) dx f(x)g(x) dx M g(x) dx · ≤ ≤ · Za Za Za

Wm C Bauldry ([email protected]) MAT 5930. Analysis for Teachers. Summer, 2010 80 / 139 Fundamental Theorem of Calculus

Theorem (Fundamental Theorem of Calculus) If F is differentiable on [a, b] and F 0 = f is integrable on [a, b], then b f(x) dx = F (b) F (a). − Za Proof.

n n n 1 F (b) F (a) = F (xk) F (xk 1) = F 0(tk)∆xk = f(tk)∆xk − − − kX=1 kX=1 kX=1 n b 2 f integrable = there is a P s.t. f(tk)∆xk f(x) dx < ε ⇒ − a k=1 Z X b b 3 Thus [F (b) F (a)] f(x) dx < ε. So f(x) dx = F (b) F (a) − − a a − Z Z

Wm C Bauldry ([email protected]) MAT 5930. Analysis for Teachers. Summer, 2010 81 / 139

Fundamental Example

Example Problem. Suppose that f(x) 0 for all x [a, b], that f is continuous b ≥ ∈ on [a, b], and a f(x) dx = 0. Then f(x) = 0. Solution. AssumeR that at some c [a, b] we have f(c) > 0. ∈ Since f is continuous, there is a neighborhood Nδ(c) where f(x) f(c)/2 > 0 for all x Nδ(c). ≥ ∈ Every partition P has to include c in some interval k = [xk−1, xk]. I Reﬁne P until k Nδ(c). I ⊂ Then L(P, f) (f(c)/2) δ > 0. ≥ Hence b f(x) dx (f(c)/2) δ > 0 which is a contradiction. a ≥ R

Wm C Bauldry ([email protected]) MAT 5930. Analysis for Teachers. Summer, 2010 82 / 139 Homework

1 Problems: 2.4 pg. 117. # 39,§ 41, LA 42

2 Read 2.5 §

3 Investigate

4 Explore

Wm C Bauldry ([email protected]) MAT 5930. Analysis for Teachers. Summer, 2010 83 / 139

Sequence & Series ( 2.5) § Theorem The limit of a sequence is unique. Proof.

an L = an L < /2 eventually → ⇒ | − | an M = an M < /2 eventually → ⇒ | − | L M an L + an M < , so L = M | − | ≤ | − | | − |

Theorem A convergent sequence is bounded. Proof.

an an L + L < + L < 1 + L eventually | | ≤ | − | | | | | | | M max a1 , a2 ,..., an−1 , 1 + L ≥ {| | | | | | | |}

Wm C Bauldry ([email protected]) MAT 5930. Analysis for Teachers. Summer, 2010 84 / 139 A Sequence of Results

Theorem A bounded, increasing sequence of reals converges.

Theorem (Bolzano-Weierstrass Theorem) Every bounded, inﬁnite set of reals has an accumulation point.

Deﬁnition (Cauchy Sequence)

A real sequence an is Cauchy iff for every > 0 there is an N N { } ∈ such that if n, m > N, then an am < . | − |

Theorem A sequence of reals is Cauchy iff it converges.

These theorems all fail if we replace R with Q. Why?

Wm C Bauldry ([email protected]) MAT 5930. Analysis for Teachers. Summer, 2010 85 / 139

Upper and Lower Limits

Deﬁnition

For a sequence an , set { } lim sup an = inf sup ak n→∞ n→∞ k≥n

lim inf an = sup inf ak n→∞ n→∞ k≥n

Let S be the range of an . Then { } lim supn→∞ an is the largest accumulation point of S lim infn→∞ an is the smallest accumulation point of S

Theorem A sequence converges iff lim sup and lim inf are ﬁnite and equal.

Wm C Bauldry ([email protected]) MAT 5930. Analysis for Teachers. Summer, 2010 86 / 139 Series Convergence Tests, I

Let an and bn be series.

TheoremP (ComparisonP & Limit Comparison Tests)

Suppose 0 < an bn. Then ≤ if bn converges, so does an if P an diverges, so does Pbn. Suppose an 0 and bn > 0. Then P ≥ P if lim an/bn = L with 0 < L < , then an and bn converge or diverge together. ∞ P P

Theorem (Integral Test)

If a:[1, ) R is continuous, positive, and decreasing, then an and ∞ ∞ → a(x) dx converge or diverge together. 1 P R Wm C Bauldry ([email protected]) MAT 5930. Analysis for Teachers. Summer, 2010 87 / 139

Series Convergence Tests, II

Let an be a series with positive terms. (See Convergence Tests) TheoremP (Ratio & Root Tests)

n Set r = limn an+1/an. Then Set ρ = limn √an. Then →∞ →∞ if r < 1, an converges, if ρ < 1, an converges, if r = 1, Pthe test fails, if ρ = 1, Pthe test fails,

if r > 1, an diverges. if ρ > 1, an diverges. P P Theorem (Raabe’s Test & Cauchy’s Condensation Test)

an Set R = lim n 1 . Then If an is decreasing, then n · a − →∞ n+1 a converges if L > 1, then an converges, n iff if L = 1, the test fails, P n P 2 a2n converges. if L < 1, then an diverges. P P Wm C Bauldry ([email protected]) MAT 5930. Analysis for Teachers. Summer, 2010 88 / 139 Basic Convergence Test Flow Chart

Choosing a Convergence Test for Infinite Series Courtesy David J. Manuel

Do Series Diverges by the individual No terms approach 0? the Divergence Test.

Yes

Does Do Use Ratio Test the series Yes individual terms Yes alternate have factorials or (Ratio of Consecutive signs? exponentials? Terms) No No

Is Use Alternating individual term Yes Use Integral Test Series Test easy to integrate? (do absolute value of terms go to 0?) No

Do individual terms Use Comparison or involve fractions with Yes Limit Comp. Test powers of n? (Look at Dominating Terms) No

Wm C Bauldry ([email protected]) MAT 5930. Analysis for Teachers. Summer, 2010 89 / 139

Homework

1 Problems: 2.5 pg. 117. # 49,§ 52, 54, 57, 58, 62a,b.

2 Read 2.6 §

3 Investigate the “Interactive Real Analysis” web pages for convergence tests

TM 4 Explore The On-Line Encyclopedia of Integer Sequences

Wm C Bauldry ([email protected]) MAT 5930. Analysis for Teachers. Summer, 2010 90 / 139 Pointwise & Uniform Convergence ( 2.6) §

Let fn :D R for all n N. → ∈ Deﬁnition

Pointwise convergence: fn(x) or n fn(x) converges pointwise on D to f(x) iff for every{ } > 0, for every x D there is an P ∈ N = N(, x) N such that if n > N, then ∈

fn(x) f(x) < or fn(x) f(x) < | − | − n X

Uniform convergence: f (x) or f (x) converges uniformly on D to f(x) { n } n n iff for every > 0 there is an N = N() N such that for any x D if n > N, thenP ∈ ∈ fn(x) f(x) < or fn(x) f(x) < | − | − n X

PWC: “ x D N(, x) ... ” versus UC: “ N() x D... ” ∀ ∀ ∈ ∃ ∀ ∃ ∀ ∈

Wm C Bauldry ([email protected]) MAT 5930. Analysis for Teachers. Summer, 2010 91 / 139

A Sequence of Functions

Example

1/n fn(x)=n x 1 − f2 f3 f5 f15

f(x)=ln(x)

Is the convergence uniform on [1, 10]? On (0, 10)?

Wm C Bauldry ([email protected]) MAT 5930. Analysis for Teachers. Summer, 2010 92 / 139 Uniform Convergence and Continuity

Theorem

If fn is continuous on [a, b] for all n and fn f uniformly, then f is continuous and → lim lim fn(x) = lim lim fn(x). x→c n→∞ n→∞ x→c

Theorem

If fn is continuous on [a, b] for all n and fn converges uniformly, then the sum is continuous and ∞ ∞P lim fn(x) = lim fn(x). x→c x→c n=0 n=0 X X

1/n x Q Set Dn(x) = ∈ . Is limn Dn(x) continuous? Is Dn(x)? (0 x / Q ∈

Wm C Bauldry ([email protected]) MAT 5930. Analysis for Teachers. Summer, 2010 93 / 139

Uniform Convergence and Integration

Theorem

If fn is integrable on [a, b] for all n and fn f uniformly, then f is integrable and → b b lim fn(x)dx = lim fn(x)dx. n→∞ n→∞ Za Za Theorem

If fn is integrable on [a, b] for all n and fn converges uniformly, then the sum is integrable and P b ∞ ∞ b fn(x)dx = fn(x)dx. a n=0 n=0 a Z X X Z

Wm C Bauldry ([email protected]) MAT 5930. Analysis for Teachers. Summer, 2010 94 / 139 A Good Sequence of Functions

Example

0 x<0 1 1 fn(x)= x x 0 x  · n − ≤ ≤ n 0 1

Find fn, limn fn, limn fn, and limn fn.

Wm C Bauldry ([email protected] ) MATR 5930. Analysis forR Teachers. Summer, 2010 95 / 139

A Bad Sequence of Functions

Example

0 x<0 f4 2 1 2n x 0 x< 2n fn(x)= ≤ 2n(1 nx) 1 x< 1  2n n  − 1 ≤ 0 n x f3 ≤  

Find fn, limn fn, limn fn, and limn fn.

Wm C Bauldry ([email protected] ) MATR 5930. Analysis forR Teachers. Summer, 2010 96 / 139 Uniform Convergence and Differentiation

Differentiation does not behave as well as continuity and integration. Example 1 2 Let fn(x) = n sin n x . Show that

fn converges uniformly on R

fn0 doesn’t even converge pointwise anywhere.

Theorem

Suppose fn :[a, b] R. If → 1 each fn is continuously differentiable on [a, b], 2 f f pointwise on [a, b], and n → 3 fn0 converges uniformly on [a, b], { } d d then fn f uniformly, fn0 f 0 uniformly, and lim fn(x) = lim fn(x). n n → → dx →∞ →∞ dx

Wm C Bauldry ([email protected]) MAT 5930. Analysis for Teachers. Summer, 2010 97 / 139

The Best Uniform Convergence Test

The Weierstrass M–test provides a very useful method for testing uniform convergence. Theorem (The Weierstrass M–Test)

Let fn be a sequence of functions. If there is a sequence of constants Mn such that 1 fn(x) Mn for all n N and |∞ | ≤ ∈ 2 Mn converges, n=1 X∞ then fn converges uniformly (and absolutely). n=1 X

1 2 1 Does fn(x) converge where fn(x) = sin n x ? n n2 2 0 Why doesn’tP the test work for fn?

Wm C Bauldry ([email protected]) MAT 5930. Analysis for Teachers. Summer, 2010 98 / 139 A Series of Steps

Example

∞ 1 1 S(x)= U x k2 · − k k=1

Is the convergence uniform?

Wm C Bauldry ([email protected]) MAT 5930. Analysis for Teachers. Summer, 2010 99 / 139

Bernstein’s Polynomials

Theorem (Weierstrass)

Let f :[0, 1] R be continuous. For any > 0, there is a polynomial Bn(x) such that f→(x) B (x) < for all x [0, 1]. | − n | ∈ Proof.

Set n k n k n k B (x) = f x (1 x) − n n · k − kX=0

B1 = f(0) (1 − x) + f(1) x 2 B2 = f(0) (1 − x) + 2f(1/2) x(1 − x) B (x) 5 + f(1) x2 3 2 B20(x) B3 = f(0) (1 − x) + 3f(1/3) x(1 − x) 7 f(x)= x sin π√x 2 3 − 2 + 3f(2/3) x (1 − x) + f(1) x

Wm C Bauldry ([email protected]) MAT 5930. Analysis for Teachers. Summer, 2010 100 / 139 Powerful Series

Theorem

∞ Let R be the radius of convergence of the power series a (x c)k. k − Then the power series kX=0 1 converges absolutely for all x s.t. x c < R; | − | 2 converges uniformly for all x s.t. x c r < R; | − | ≤ 3 diverges for all x s.t. x c > R. | − | Example

∞ nxn Problem: Find the radius of convergence of . 2n+1 n=0 X (n+1) x n+1 n x n x Solution: Consider rn := un+1/un = 2n|+1| / |2n| |2| as n . Applying the ratio test gives convergence for x /2 1 or→ x < 2. So→ the ∞ radius of convergence is R = 2. | | ≤ | |

Wm C Bauldry ([email protected]) MAT 5930. Analysis for Teachers. Summer, 2010 101 / 139

Differentiating and Integrating Power Series

Theorem ∞ n Suppose that f(x) = anx has radius of convergence R > 0. Then n=0 X 1 f is continuous for x < R | | ∞ 0 n−1 2 f is differentiable for x < R and f (x) = n anx | | n=0 X x ∞ an n+1 3 f is integrable for x < R and f(t) dt = x | | n + 1 0 n=0 Z X

Theorem Series representation is unique on the interval of convergence.

Wm C Bauldry ([email protected]) MAT 5930. Analysis for Teachers. Summer, 2010 102 / 139 Taylor Series

Theorem (Taylor Series)

∞ n Suppose that f(x) = anx has radius of convergence R > 0. Then f is n=0 X inﬁnitely differentiable on x a < R and the coefﬁcients an are given by (n) | − | an = f (a)/n!.

Examples (Important Maclaurin Series)

∞ ∞ 2k+1 1 X k −1 X k x 1 = x for |x| < 1 4 tan (x) = (−1) ; x ∈ 1 − x 2k + 1 R n=0 n=0 ∞ 2k+1 ∞ k X k x x X x 2 sin(x) = (−1) ; x ∈ 5 e = ; x ∈ (2k + 1)! R k! R n=0 n=0 ∞ 2k ∞ k X k x X k−1 x 3 cos(x) = (−1) ; x ∈ 6 ln(1 + x) = (−1) ; |x| < 1 (2k)! R k n=0 n=1 See Wikipedia’s list.

Wm C Bauldry ([email protected]) MAT 5930. Analysis for Teachers. Summer, 2010 103 / 139

Maclaurin Series

A Maclaurin series is a Taylor series centered at a = 0. Alternate techniques can be useful for ﬁnding series expansions without ﬁnding a formula for the nth derivative. Example The series for sec(x) can be found from 1 1 = cos(x) x2 x4 x6 1 − 2! + 4! − 6! ± · · · 1 5 61 = 1 + x2 + x4 + x6 + ··· 2 24 720

To ﬁnd the inverse of the cosine series, use (see Landau symbols for “O(xn)” )

x2 x4 1 = 1 − + + O(x6) × a + a x + a x2 + a x3 + a x4 + O(x5) 2! 4! 0 1 2 3 4 a a a a 1 = a + a x + a − 0 x2 + a − 1 x3 + a − 2 + 0 x4 + O(x5) 0 1 2 2 3 2 4 2 24 1 5 ⇒ a = 1 ⇒ a = ⇒ a = ⇒ ... ; a = 0 ⇒ a = 0 ⇒ a = 0, &c 0 2 2 4 24 1 3 5

Wm C Bauldry ([email protected]) MAT 5930. Analysis for Teachers. Summer, 2010 104 / 139 Maclaurin Series, II

Example (Differentiation/Integration) The Maclaurin series for ln(1 + x) can be found as follows (subject to conditions): 1 ln(1 + x) = dx 1 + x Z 1 = 1 x + x2 x3 + x4 x5 1 + x − − − ± · · · x2 x3 x4 ln(1 + x) = x + ... − 2 3 − 4 ± We need to investigate what conditions are needed to be able to integrate term by term; i.e., when is ∞ ∞ an(x) dx = an(x) dx ? k=0 k=0 Z X X Z

Wm C Bauldry ([email protected]) MAT 5930. Analysis for Teachers. Summer, 2010 105 / 139

Off On a Tangent

What is the Maclaurin series for f(x) = tan(x)? Using the deﬁnition, we need the derivatives of tan(x): tan(0)(x) = tan(x) tan(1)(x) = sec2(x) tan(2)(x) = 2 sec2(x) tan(x) tan(3)(x) = 2 sec4(x) + 4 sec2(x) tan2(x) tan(4)(x) = 16 sec4(x) tan(x) + 8 sec2(x) tan3(x) tan(5)(x) = 16 sec6(x) + 88 sec4(x) tan2(x) + 16 sec2(x) tan4(x) tan(6)(x) = 272 sec6(x) tan(x) + 416 sec4(x) tan3(x) + 32 sec2(x) tan5(x) tan(7)(x) = 272 sec8(x) + 2880 sec6(x) tan2(x) + 1824 sec4(x) tan4(x) + 64 sec2(x) tan6(x) tan(8)(x) = 7936 sec8(x) tan(x) + 24576 sec6(x) tan3(x) + 7680 sec4(x) tan5(x) + 128 sec2(x) tan7(x) etc. (Volunteers?) tan(x) = x + 1 x3 + 2 x5 + 17 x7 + 62 x9 + 1382 x11 + 21844 x13 + 3 15 315 2835 155925 6081075 ···

Wm C Bauldry ([email protected]) MAT 5930. Analysis for Teachers. Summer, 2010 106 / 139 Patterns

Circular Function Hyperbolic Function

∞ 2n+1 ∞ 2n+1 X n x X x sin(x) = (−1) sinh(x) = (2n + 1)! (2n + 1)! n=0 n=0

∞ 2n ∞ 2n X n x X x cos(x) = (−1) cosh(x) = (2n)! (2n)! n=0 n=0

∞ 2n−1 ∞ 2n−1 X n n x X n n x tan(x) = B (−4) (1−4 ) tanh(x) = B 4 (1 − 4 ) 2n (2n)! 2n (2n)! n=0 n=0

∞ 2n+1 ∞ 2n+1 X (2n)! x X n (2n)! x arcsin(x) = arcsinh(x) = (−1) 4n(n!)2 (2n + 1) 4n(n!)2 (2n + 1) n=0 n=0

∞ 2n+1 ∞ 2n+1 X n x X x arctan(x) = (−1) arctanh(x) = (2n + 1) (2n + 1) n=0 n=0

Bn is the nth Bernoulli number; B0 = 1,B2 = 1/6,B4 = −1/30,B6 = 1/42,B8 = −1/30,... .

Wm C Bauldry ([email protected]) MAT 5930. Analysis for Teachers. Summer, 2010 107 / 139

Circular versus Hyperbolic

Wm C Bauldry ([email protected]) MAT 5930. Analysis for Teachers. Summer, 2010 108 / 139 Homework

1 Problems: 2.6 pg. 117; # 38,§ 65, 68, 71, 74.

2 Read 3.1 §

3 Investigate the HMC Math ‘Convergence Tests’ web site.

4 Explore the text Real Inﬁnite Series by Bonar and Khoury.

Wm C Bauldry ([email protected]) MAT 5930. Analysis for Teachers. Summer, 2010 109 / 139

In The Woods

Wm C Bauldry ([email protected]) MAT 5930. Analysis for Teachers. Summer, 2010 110 / 139 Algebras ( 3.1) §

Deﬁnition (Algebra of Sets) A collection of subsets of a set X is an algebra iff A 1 X ∈ A 2 if A , then Ac ∈ A ∈ A 3 if A, B , then A B ∈ A ∪ ∈ A Examples

,X A R either A or Ac is ﬁnite {∅ } { ⊆ | } c , 1, 2 , 3, 4 , 1, 2, 3, 4 A R A or A is countable {∅ { } { } { }} { ⊆ | } (R), the set of all subsets of R , ( , 0], (0, ), R P {∅ −∞ ∞ } (the “power set”)

Wm C Bauldry ([email protected]) MAT 5930. Analysis for Teachers. Summer, 2010 111 / 139

Duality, σ-Algebras

Theorem (De Morgan’s Laws) (A B)c = Ac Bc and (A B)c = Ac Bc ∪ ∩ ∩ ∪ Theorem If is an algebra, then A 1 ∅ ∈ A 2 if A, B , then A B ∈ A ∩ ∈ A Deﬁnition (σ-Algebra of Sets) An algebra of sets is a σ-algebra iff whenever A for k = 1, 2,... , then A k ∈ A ∞ ∞ the union A . (And the intersection A , too.) k ∈ A k ∈ A k[=1 k\=1

Wm C Bauldry ([email protected]) MAT 5930. Analysis for Teachers. Summer, 2010 112 / 139 Open Covers, Length, and Outer Measure

Definition A collection of open sets is an open cover of a set A iff A O. O ⊆ O∈O Define the length of an open interval I = (a, b) to be `(I) = b Sa. The “outer measure” of a set is the minimal length of open covers− of the set. Definition (Outer Measure) ∗ The outer measure of a set E R is µ (E) = inf `(I) where is ⊆ O O an open cover of by open intervals. I∈O E X Examples 1 The outer measure of a finite set is 0. 2 Outer measure of an open interval: I = (a, b) is µ∗(I) = b a. − 3 Outer measure of a closed interval: J = [a, b] is µ∗(J) = b a. −

Wm C Bauldry ([email protected]) MAT 5930. Analysis for Teachers. Summer, 2010 113 / 139

Properties

Proposition Outer measure is: ∗ nonnegative: for any set E R, µ (E) 0 ⊆ ≥ monotone: if A B, then µ∗(A) µ∗(B) ⊆ ≤ subadditive: for any two sets A and B, µ∗(A B) µ∗(A) + µ∗(B) ∪ ≤ countably subadditive: if En is a sequence of sets, then { } ∗ ∗ µ En µ (En) n ! ≤ n [ X Proof. (Countably Subadditivity).

Wolog µ∗(En) < for all n. Each En has a countable cover In,j of open ∞ n { } intervals s.t. j `(In,j) µ∗(En) + /2 . Then n In,j covers E = En. ≤ n { } So µ∗(E) `(In,j) (µ∗(En) + /2 ) = (µ∗(En)) + . ≤ Pn,j ≤ n S n S P P P Wm C Bauldry ([email protected]) MAT 5930. Analysis for Teachers. Summer, 2010 114 / 139 Lebesgue’s Measure (1902)

Goals for a set measure function m: 1 m(an interval) = it’s length 2 m(E + x) = m(E) (translation invariant ) countably 3 if En is pairwise disjoint, then m( En) = m(En) { } n n additive 4 every set can be measured by m. S P The continuum hypothesis implies that 1, 3 and 4 are incompatible. Additionally, 2 and 4 are incompatible. We insist on 1, 2, and (nearly) 3. So we give up 4: there will be sets that can’t be measured.

Deﬁnition (Caratheodory’s´ Condition) A set E is Lebesgue measurable iff for every ‘test set’ A R, ⊆ µ∗(A) = µ∗(A E) + µ∗(A Ec) ∩ ∩ Deﬁne M to be the family of all measurable sets of real numbers.

Wm C Bauldry ([email protected]) MAT 5930. Analysis for Teachers. Summer, 2010 115 / 139

Measuring Properties

Proposition If E measurable, then Ec is measurable; i.e., E M Ec M. ∈ ⇐⇒ ∈

Proposition If µ∗(E) = 0, then E is measurable.

Proof. 1 A = (A E) (A Ec), therefore µ∗(A) µ∗(A E) + µ∗(A Ec) ∩ ∪ ∩ ≤ ∩ ∩ 2 A E E, thus µ∗(A E) µ∗(E) = 0 ∩ ⊆ ∩ ≤ 3 A Ec A, thus µ∗(A Ec) µ∗(A) ∩ ⊆ ∩ ≤ 4 So µ∗(A) µ∗(A E) + µ∗(A Ec) 0 + µ∗(A). E M ≤ ∩ ∩ ≤ ∴ ∈

Wm C Bauldry ([email protected]) MAT 5930. Analysis for Teachers. Summer, 2010 116 / 139 Measure Creates a σ-Algebra

Proposition

If E1 and E2 M, then E1 E2 M. If En M, then En M. ∈ ∪ ∈ ∈ n ∈ Proof. S

1 c Let A be any set and use A E1 as a test-set for E2. So c ∩ c c c µ∗(A E ) = µ∗((A E ) E ) + µ∗((A E ) E ) ∩ 1 ∩ 1 ∩ 2 ∩ 1 ∩ 2 2 A (E E ) = (A E ) (A E Ec) ∩ 1 ∪ 2 ∩ 1 ∪ ∩ 2 ∩ 1 c 3 µ∗(A (E E )) + µ∗(A (E E ) ) ∩ 1 ∪ 2 ∩ 1 ∪ 2 c c = µ∗(A (E E )) +µ∗(A (E E )) ∩ 1 ∪ 2 ∩ 1 ∩ 2 | {zc } c c µ∗(A E ) + µ∗(A E E ) + µ∗(A E E ) ≤ ∩ 1 ∩ 2 ∩ 1 ∩ 1 ∩ 2 | {z } c c c µ∗(A E1) + µ∗((A E ) E2) + µ∗((A E ) E ) ≤ ∩ ∩ 1 ∩ ∩ 1 ∩ 2 c µ∗(A E ) + µ ∗(A E ) = µ∗(A) ≤ ∩ 1 ∩ 1 | {z c } 4 Then µ∗(A) µ∗(A (E E )) + µ∗(A (E E ) ) µ∗(A) ≤ ∩ 1 ∪ 2 ∩ 1 ∪ 2 ≤

Wm C Bauldry ([email protected]) MAT 5930. Analysis for Teachers. Summer, 2010 117 / 139

A Big σ-Algebra

Theorem M is a σ-algebra.

Deﬁnition Borel σ-algebra: the smallest σ-algebra that contains all open intervals

Proposition The Borel σ-algebra (R) is a subalgebra of M. Moreover, B (R) M (R). B $ $ P There are Lebesgue measurable sets that are not Borel sets and there are subsets of R that are not measurable. (Both are very complex.)

Wm C Bauldry ([email protected]) MAT 5930. Analysis for Teachers. Summer, 2010 118 / 139 Hm. . .

I wonder. . .

Wm C Bauldry ([email protected]) MAT 5930. Analysis for Teachers. Summer, 2010 119 / 139

The Measure

Deﬁnition (Lebesgue Measure)

For E M, deﬁne µ(E) = µ∗(E). That is, µ = µ∗ . ∈ M

Deﬁnition (Almost Everywhere) A property P holds almost everywhere, or a.e., iff the set on which P does not hold has measure 0.

Example 1 x χ Z The characteristic function of the integers is Z(x) = ∈ . Then (0 x / Z χ ∈ Z(x) = 0 a.e.

Wm C Bauldry ([email protected]) MAT 5930. Analysis for Teachers. Summer, 2010 120 / 139 Countable Additivity of Lebesgue Measure Theorem Let E be a countable sequence of pairwise-disjoint measurable sets. { n} Then µ( n∞=1 En) = n∞=1 µ(En) . Proof. S P I.FINITESEQUENCE. Basis: For n = 1, it’s trivially true. Induction: Assume the results holds for some value n 1. A ﬁnite union of sets in M is in M, so − n n n 1 µ( E ) = µ([ E ] E ) + µ([ E ] Ec ) Disjoint! k=1 k k=1 k ∩ n k=1 k ∩ n n 1 2 S S S = µ(En) + µ k=1− Ek 3 nS 1 n = µ(En) + k=1− µ(Ek) = k=1 µ(Ek) II.INFINITE SEQUENCEP . P n n n 1 E ∞ E , so µ( E ) = µ(E ) µ( ∞ E ) k=1 k ⊆ k=1 k k=1 k k=1 k ≤ k=1 k 2 TheS left sideS converges (bnddS & incr), P ∞ µ(Ek) µS( ∞ Ek) ∴ k=1 ≤ k=1 3 Opposite inequality follows from subadditivity.P S

Wm C Bauldry ([email protected]) MAT 5930. Analysis for Teachers. Summer, 2010 121 / 139

More Properties of µ

Proposition If A E and µ(E) = 0, then A M and µ(A) = 0. ⊆ ∈ Theorem ∞ Let En be a sequence of nested, measurable sets with { }n=1 µ(E1) < . Then ∞ ∞ µ En = lim µ(En) n→∞ n=1 ! \ Theorem (Regularity of µ) Let A M. Then for any > 0 there is an open set O and a closed set ∈ F such that i) F A O and ii) µ(O F ) < . ⊆ ⊆ −

Wm C Bauldry ([email protected]) MAT 5930. Analysis for Teachers. Summer, 2010 122 / 139 Homework

1 Problems: 3.1 pg. 166; # 1,§ 2, 5, 7, LA 8, 11.

2 Read 3.2 §

3 Investigate David Bressoud’s Engaging Students through the History of Mathematics talk (Powerpoint slides).

4 Explore the text A Garden of Integrals by Frank Burk.

Wm C Bauldry ([email protected]) MAT 5930. Analysis for Teachers. Summer, 2010 123 / 139

The Lebesgue Integral ( 3.2) § Deﬁnition (Measurable Function) Let f be an extended real-valued function on a measurable domain D. TFAE: For each r R, f −1((r, ∞))∈∈ M ⇔ f −1([r, ∞)) ∈ M ⇔ f −1((−∞, r)) ∈ M ⇔ f −1((−∞, r]) ∈ M A function f that satisﬁes this criteria is called a measurable function.

Definition (Step Function) A function φ:[a, b] R is a step function iff there is a partition of [a, b] such → n χ that φ is constant on each interval Ik = (xk 1, xk). Then φ(x) = ak Ik (x). − kX=1 Definition (Simple Function) A function ψ :[a, b] R is a simple function iff the range of ψ is a finite set → n n 1 a and each A = ψ− (a ) is measurable. Then ψ(x) = a χ (x). { k}k=1 k k k Ak kX=1

Wm C Bauldry ([email protected]) MAT 5930. Analysis for Teachers. Summer, 2010 124 / 139 Measurable Functions

Proposition A continuous function is measurable. Theorem (Algebra of Measurable Functions) Let f and g be measurable functions on a common domain D, and let 2 c R. Then f + c, cf, f g, f , and fg are all measurable. If a ∈ ± function h is equal to f a.e., then h is measurable. Deﬁnition (Convergence Almost Everywhere)

A sequence fn converges to f a.e., written fn f a.e., iff the set of { } → points where fn fails to converge has measure zero. Theorem A function f is measurable iff there is a sequence of simple functions ψn converging to f a.e. { }

Wm C Bauldry ([email protected]) MAT 5930. Analysis for Teachers. Summer, 2010 125 / 139

Lebesgue’s Diagram

From Lebesgue’s 1926 address to the “Societ´ e´ Mathematique”´ in Copenhagen.

Wm C Bauldry ([email protected]) MAT 5930. Analysis for Teachers. Summer, 2010 126 / 139 Integral of a Simple Function

Deﬁnition Let ψ be a simple function that is zero outside of a set of ﬁnite measure n χ −1 with ψ(x) = ak Ak (x) where Ak = ψ (ak). Then k=1 X n ψ(x) dµ = ak µ(Ak) k=1 Z X If E is a measurable set, then n ψ(x) dµ = ak µ(Ak E) ∩ E k=1 Z X Examples 1 Find the integral of Dirichlet’s function on the interval [0, 1]. X 2 Find the integral of Thomae’s function on the interval [0, 1]. How?

Wm C Bauldry ([email protected]) MAT 5930. Analysis for Teachers. Summer, 2010 127 / 139

Integral of a Bounded Function

Deﬁnition Let f be a bounded measurable function on a set E having ﬁnite measure. Then f dµ = inf ψ dµ ψ≥f ZE ZE where ψ is a simple function.

Theorem Let f be bounded and Riemann integrable on [a, b]. Then f is measurable and b f(x) dx = f dµ Za Z[a,b]

Wm C Bauldry ([email protected]) MAT 5930. Analysis for Teachers. Summer, 2010 128 / 139 Integral Properties

Theorem Let f and g be bounded and measurable functions on a set E of ﬁnite measure, and let c R. Then ∈ 1 cf dµ = c f dµ ZE ZE 2 f + g dµ = f dµ + g dµ ZE ZE ZE 3 if f g a.e., then f dµ g dµ ≤ ≤ ZE ZE 4 if m f(x) M a.e. on E, then m µ(E) f dµ M µ(E) ≤ ≤ · ≤ ≤ · ZE 5 if E E = E with E E = , then f dµ + f dµ = f dµ 1 ∪ 2 1 ∩ 2 ∅ ZE1 ZE2 ZE 6 f dµ f dµ ≤ | | ZE ZE

Wm C Bauldry ([email protected]) MAT 5930. Analysis for Teachers. Summer, 2010 129 / 139

Homework

1 Problems: 3.2 pg. 166; # 21,§ 24, 25, 26, 45.

2 Read 4.2 §

3 Study Lebesgue’s 1926 address to the “Societ´ e´ Mathematique”´ in Copenhagen. See the appendix of Chae’s Lebesgue Integration for an English translation.

4 Explore Lebesgue’s criterion for Riemann integrability:

f R([a, b]) µ ( x [a, b] f is discontinuous at x ) = 0 ∈ ⇔ { ∈ | }

Wm C Bauldry ([email protected]) MAT 5930. Analysis for Teachers. Summer, 2010 130 / 139 TI and Numerical Integration ( 4.2+) § Deﬁnition (TI Floating Point Number (decimal value)) A decimal value is coded internally as a 10 byte object: sign biased mantissa exponent (1 bit) (15 bits) (64 bits: 14 × 4 bit BCD #’s, 0, 0)

s (exp 16384)B x = ( 1) mantissa 10 − − · BCD · (TI math functions restrict exponent range to [−999, +999])

Examples π: 0 16384 3 1415926 535897 00 ˆimplied decimal point 10: 1 16385 1 0000000 000000 00 −

Wm C Bauldry ([email protected]) MAT 5930. Analysis for Teachers. Summer, 2010 131 / 139

Gaussian Quadrature

The ‘calculus-course techniques’ for numerical integration—left sum, trapezoid rule, Simpson’s rule—all use equally spaced partitions. Can we do better by adjusting the points? 1 n n The basic formula is f(x) dx f(xk) ∆xk = f(xk) wk 1 ≈ Z− kX=1 kX=1 Gauss treated both xk and wk as unknowns and looked for a set that would integrate 2n 1 degree polynomials exactly over [ 1, 1]. Let’s take n =− 3, so 2n 1 = 5 and f(x) = 1, x, x2,− . . . , x5, successively. Then −

w1 + w2 + w3 = 2 x1w1 + x2w2 + x3w3 = 0 n 1  2 2 2  n n x1w1 + x2w2 + x3w3 = 2/3 xk wk = x dx =  3 3 3  1 ⇒ x1w1 + x2w2 + x3w3 = 0  k=1 Z−  4 4 4  X x1w1 + x2w2 + x3w3 = 2/5 x5w + x5w + x5w = 0  1 1 2 2 3 3      giving x = 3/5, x = 0, x = 3/5 and w = 5/9, w = 8/9, w = 5/9. 1 − 2 3 1 2 3 p p Wm C Bauldry ([email protected]) MAT 5930. Analysis for Teachers. Summer, 2010 132 / 139 Gaussian Quadrature Examples

Examples

1 [n = 3, 5th degree.] Set p(x) = x5 2x4 x3 + 7x2 + 8x + 1. Then 1 − −88 p(x) dx = 1 15 and Z− 88 GQ = p 3/5 (5/9) + p(0) (8/9) + p 3/5 (5/9) = − · · · 15 p RelativeError = 0 p

2 [n = 3, 6th degree.] Set q(x) = 3x6 2x5 + 7x4 4x2 + 7x 8. Then 1 − 1576 − − q(x) dx = 1 − 105 and Z− 1136 GQ = q 3/5 (5/9) + q(0) (8/9) + q 3/5 (5/9) = − · · · − 75 RelativeErrorp = ( 1576/105 1136/75)p/( 1576/105) 2% − − − − ≈

Wm C Bauldry ([email protected]) MAT 5930. Analysis for Teachers. Summer, 2010 133 / 139

Gaussian Quadrature Error

Advantages: Simple weighted sum formula Minimal number of function evaluations Published tables list nodes and weights for many n Disadvantages: Nodes don’t take into account a function’s properties Transformations for domains other than [ 1, 1] New nodes and weights discard old computations−

Theorem (Error Estimate for Gaussian Quadrature) 2n (2n) If f C [ 1, 1] with M2n = maxx∈[a,b] f (x) , then the error ∈ − n | | 1 22n+1[n!]4 f(x) dx wkf(xk) 3 M2n −1 − ≤ (2n + 1) [(2n)!] Z k=1 X

Wm C Bauldry ([email protected]) MAT 5930. Analysis for Teachers. Summer, 2010 134 / 139 C.F. Gauss, A.S. Kronrod, and TI

Aleksandr Kronrod (1964) discovered a way to extend Gaussian quadrature by adding nodes to an existing set rather than using all new nodes. The error analysis is relatively difﬁcult, but provides a good heuristic. Deﬁnition (Gauss-Kronrod 7-15 Quadrature)

Gauss-Kronrod 7-15 quadrature (GK15) is a combination of a 7 node Gaussian (GQ7) with 8 new nodes and weights that gives exact integrals for polynomials up to degree 29. A “conservative, realistic” error estimate is

1 15 1.5 f(x) dx f(xk)wk 2000√2 GQ7 GK15 1 − ≤ − Z− k=1 X

GQ7 nodes weights GK15 new nodes weights ±0.94910791234276 0.12948496616887 ±0.99145537112081 0.02293532201053 ±0.74153118559939 0.27970539148928 ±0.86486442335977 0.10479001032225 ±0.40584515137740 0.38183005050512 ±0.58608723546769 0.16900472663927 0.0 0.41795918367347 ±0.20778495500790 0.20443294007530

Explore Gauss-Kronrod 7-15 quadrature with a Java applet. Consider f(x) = x sin(x2) + |x|.

Wm C Bauldry ([email protected]) MAT 5930. Analysis for Teachers. Summer, 2010 135 / 139

The Points

GK15 points on f

Wm C Bauldry ([email protected]) MAT 5930. Analysis for Teachers. Summer, 2010 136 / 139

GQ7 points GK15 points Example

1 2 Compute e−x dx. (This function has no elementary antiderivative.) Z−1 TI-84: TI-89: 1.4936482656246 (but to 6-digits) TI-Nspire CAS: 1.4936482656246 (but to 6-digits) Maple 14: 1.493648266; using GK15 gives 1.493648264 PocketCAS: 1.49364826562

Wm C Bauldry ([email protected]) MAT 5930. Analysis for Teachers. Summer, 2010 137 / 139

TI’s Quadrature

From the TI Guidebook: TI-84: fnInt( (function integral) returns the numerical integral (Gauss-Kronrod method) of expression with respect to variable, given lower limit, upper limit, and a tolerance (if not speciﬁed, the default is 1E 5). fnInt( is valid only for − real numbers. [TI-84: p. 61; TI-83: p. 71]

TI-898: nInt( returns an approximation of (expression1, var, lower, upper). This approximation is a weighted average R of some sample values of the integrand in the interval lower < var < upper.[TI-89: p. 472; TI-Nspire CAS: p. 76] (Uses adaptive G-K with tolerance = “ 6 signiﬁcant digits.”)

8Also for TI-92, Voyage 200, Nspire, and Nspire CAS

Wm C Bauldry ([email protected]) MAT 5930. Analysis for Teachers. Summer, 2010 138 / 139 Homework

1 Problems: 4.2 pg. 214; # 13,§ 14, 15, 24, 59.

2 Read everything

3 Investigate numerical quadrature.

4 Explore David Bressoud’s texts A Radical Approach to Real Analysis and A Radical Approach to Lebesgue’s Theory of Integration. Two great books.

Wm C Bauldry ([email protected]) MAT 5930. Analysis for Teachers. Summer, 2010 139 / 139