MAT 5610, Analysis I

MAT 5610: 0

Introduction Review Derivative Integral Riemann-Stieltjes Integral Sequences & Series

MAT 5610, Analysis I

Wm C Bauldry

[email protected]

Spring, 2011

0 Today

http://www.mathsci.appstate.edu/~wmcb/

[email protected]

MAT 5610: 0


MAT 5610: 1


Analysis I

MAT 5610. Analysis I/(3).S. A rigorous treatment of topology of thereal numbers, continuity, differentiation, and sequences and series offunctions. Prerequisite: MAT 3220 (Introduction to Real Analysis I) orpermission of the instructor.

Our goal is a rigorous development of differential and integral calculusthrough sequences and series of functions. Chapters 5→ 8 of ourtext, Witold Kosmala’s A Friendly Introduction to Analysis, 2nd ed.Optional Supplement: David Bressoud’s A Radical Approach to RealAnalysis, 2nd ed.

Grading:

Projects / Presentations ≈ 100 pt.Homework & Proofs ≈ 100 pt.Midterm Exam ≈ 100 pt.Final Exam ≈ 100 pt.Total ≈ 400 pt.

http://www.amazon.com/gp/product/0130457965/ref=pd_sim_b_1/002-5116236-5854423?ie=UTF8

http://www.amazon.com/Radical-Approach-Real-Analysis-Classroom/dp/0883857472/

http://www.amazon.com/Radical-Approach-Real-Analysis-Classroom/dp/0883857472/

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Semester Projects

• Individual ProjectGlossary: Build a glossary of the terms we use in analysis.

Start with basic items such as ‘open set’.

• Class ProjectsBibliography: Generate an annotated list of references for

• real analysis and advanced calculus,• calculus and teaching calculus. (sample)

Concept Map: Create a concept map of analysis.Look at the Derivative Map for a sample. There isfree software at the Institute for Human andMachine Cognition (IHMC) site.

http://mathsci2.appstate.edu/~wmcb/Class/5610/SampleAnnBib.pdf

http://en.wikipedia.org/wiki/Concept_map

http://mathsci2.appstate.edu/~wmcb/Class/1110/QuizTest/CurrentFiles/DerivativeCmap.html

http://cmap.ihmc.us/

http://cmap.ihmc.us/

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Reprise

DefinitionsOpen ball: B(a, ρ) = { |x− a| < ρ} (also called neighborhood)

Punctured ball: B′(a, ρ) = {0 < |x− a| < ρ} (deleted neighborhood)Open set: O is open⇐⇒ (∀x ∈ O) (∃δx > 0) [B(x, δx) ⊆ O]

Accum. point: a is an acc. pt. of S ⇐⇒ (∀ρ > 0) [B′(a, ρ) ∩ S 6= ∅ ]

Closed set: F is closed⇐⇒ F ⊇ {all accumulation pts of F}

Definition (Limit)Let φ :D → R and a be an accumulation point of D. Then the limit ofφ as x goes to a is L iff for every ε > 0 there is a δ > 0 such thatwhenever x ∈ B′(a, δ) ∩D, then f(x) ∈ B(L, ε). In symbols:

limx→a

φ(x) = L⇐⇒ (∀ε > 0) (∃δ > 0)[φ (B′(a, δ) ∩D) ⊆ B(L, ε)

]

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Re2prise

Definition (Continuity)Let f :D → R and a be a point of D. Then f is continuous at a iff forevery ε > 0 there is a δa > 0 such that whenever x ∈ B(a, δa) ∩D,then f(x) ∈ B(f(a), ε). In symbols:

f is cont at a⇐⇒ (∀ε > 0) (∃δa > 0)[f (B(a, δa) ∩D) ⊆ B(f(a), ε)

]

Definition (Uniform Continuity)Let f :D → R and E ⊆ D. Then f is uniformly cont. on E iff for everyε > 0 there is a δ > 0 such that whenever x1, x2 ∈ D with|x1 − x2| < δ, then |f(x1)− f(x2)| < ε. In symbols:

f is u.c. on E ⇔ (∀ε > 0) (∃δ > 0)[|x1 − x2|<δ ⇒ |f(x1)− f(x2)|<ε

]

The Four Types of Discontinuity:Simple: removable or jump. Essential: infinite or oscillating.

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Borzellino’s Example

f(x) =�x4 − x2

�3/2

What is limx→0

f(x)? Is f continuous at a = 0? 1

1Preview of a coming attraction: Is f differentiable at a = 0?

http://www.calpoly.edu/~jborzell/

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Borzellino’s Second Example

f(x) =�

x sin(1/x)

What is limx→0

f(x)? Is f continuous at a = 0? 1

1Preview of a coming attraction: Is f differentiable at a = 0?

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Re3prise

PropositionsLet I be a closed, bounded interval. Let f be cont on I.• Then f is bounded on I.• Then f attains maximum and minimum values on I.• Then f has Bolzano’s Intermediate Value Property on I.• If f is nonconstant, then the range of f is also an interval.

Propositions• If f is cont on a closed, bounded set, then f is uniformly cont.• If f is unif cont on (a, b), then f(a+) and f(b−) are finite.• If f : [a,∞)→ R is cont and f(∞) is finite, then f is unif cont.

http://www-gap.dcs.st-and.ac.uk/~history/Biographies/Bolzano.html

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Re4prise

Definition (Open Cover)An open cover of a set E is a family of open sets A = {Aδ | δ ∈ ∆} s.t.

• each Aδ is an open set, and

• E ⊆⋃δ∈∆ Aδ.

Definition (Compact)A set K is compact iff every open cover of K has a finite subcover; that is,K ⊆

⋃nj=1 Aδj whenever K ⊆

⋃δ∈∆ Aδ.

Theorem (Heine-Borel Theorem)A set K ⊆ R is compact iff K is closed and bounded.

PropositionIf f is continuous on a compact set K, then it is uniformly continuous on K.

http://www-gap.dcs.st-and.ac.uk/~history/Biographies/Heine.html

http://www-gap.dcs.st-and.ac.uk/~history/Biographies/Borel.html

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Re5prise

TheoremThe continuous image of a compact set is compact.

Proof.Suppose that f is continuous on the compact set K and that f(K)has an open cover O = {Oδ | δ ∈ ∆}. Whence f(K) ⊆ ⋃δ∈∆Oδ.

• Thus K ⊆ f−1 (f(K)) ⊆ f−1(⋃

δ∈∆Oδ)

=⋃δ∈∆ f−1(Oδ).

(i) Since f is continuous, each f−1(Oδ) is open, and(ii) since K is compact, there is a finite subcover s.t.K ⊆ ⋃nj=1 f

−1(Oδj ).

• Thence f(K) ⊆ f(⋃n

j=1 f−1(Oδj )

)⊆ ⋃nj=1Oδj ; i.e., f(K) is

contained in a finite subcover of O.

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The Derivative

The “differential triangle” from Lectiones Opticæ et Geometricæ,Isaac Barrow, 1669. (Lect X, App, p 80(291). Fig. 115 to 121, p 386)

“. . . illustrates the way in which Barrow, Hudde and de Sluzewere working on the lines suggested by Fermat . . . ”

http://books.google.com/books?id=GWg_AAAAcAAJ&printsec=frontcover&source=gbs_ge_summary_r

http://www-gap.dcs.st-and.ac.uk/~history/Biographies/Barrow.html

http://www-gap.dcs.st-and.ac.uk/~history/Biographies/Hudde.html

http://www-gap.dcs.st-and.ac.uk/~history/Biographies/Sluze.html

http://www-gap.dcs.st-and.ac.uk/~history/Biographies/Fermat.html

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Definitions

Definition (Elementary Calculus Version)The derivative of f at x = a is

f ′(a) = limx→a

f(x)− f(a)

x− aprovided the limit exists (and is finite).

Definition (Analysis Version1)The derivative of f at x = a, an acc pt of f ’s domain in the domain, isf ′(a) iff for any ε > 0 there is a δa > 0 s.t. whenever x ∈ B′(a, δa),then ∣∣∣∣

f(x)− f(a)

x− a − f ′(a)

∣∣∣∣ < ε.

Newton: y Leibniz:dy

dxArbogast / Euler: Dxy Lagrange: y′

1Cauchy, 1823. He used ε for ‘l’erreur’ and δ for ‘la distance’. [Bolzano (1817) but unknown andWeierstrass (c.1840)]. See Grabiner, “Who Gave You the Epsilon? Cauchy and the Origins ofRigorous Calculus,” MAA Monthly 90 (3), p 185–194. (Abel on Cauchy)

http://www-gap.dcs.st-and.ac.uk/~history/Biographies/Newton.html

http://www-gap.dcs.st-and.ac.uk/~history/Biographies/Leibniz.html

http://www-gap.dcs.st-and.ac.uk/~history/Biographies/Arbogast.html

http://www-gap.dcs.st-and.ac.uk/~history/Biographies/Euler.html

http://www-gap.dcs.st-and.ac.uk/~history/Biographies/Lagrange.html

http://www-gap.dcs.st-and.ac.uk/~history/Biographies/Cauchy.html

http://www-gap.dcs.st-and.ac.uk/~history/Biographies/Abel.html

http://mathsci2.appstate.edu/~wmcb/Class/5610/Vignettes/AbelOnCauchy.html

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I. Differentiable?

Example (Good)Set f(x) = xn.Then

f ′(x) = limx→a

xn − an

x− a= limx→a

(xn−1 + xn−2a+ xn−3a2 + · · ·+ an−1) = nan−1.

Example (Bad)

Set f(x) =√|x| =

{√x x ≥ 0√−x x < 0

.

Then f ′(x) =

1/(2√x ) x > 0

? x = 0

−1/(2√−x ) x < 0

. What about f ′(0)?

limx→0−

f ′(x) = −∞ and limx→0+ f ′(x) =∞ Is there a tan line at 0?

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II. Differentiable?

Example

Set h(x) =

{2− x x ≥ 019x

2 − x− 2 x < 0.

Then h′(x) =

−1 x > 0

? x = 029x− 1 x < 0

. What about h′(0)?

limx→0−

h′(x) = −1 = limx→0+

h′(x)

h(x) =

�2 − x x ≥ 019x2 − x − 2 x < 0

MAT 5610: 14


Contrasting Definitions

Compare:Calculus Analysis

Limit ∀ε > 0 ∃δ > 0f(B′(a, δ) ∩D) ⊆ B(L, ε)

add ‘a is an accumulationpt. of D’

Continuity limx→a

f(x) = f(a) add ‘or a is an isolated pt.’

Derivative f ′(a) = limx→a

f(x)− f(a)

x− a essentially the same

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What’s in a Definition

SubtletiesMany authors1 use:

The function y = f(x) has a tangent line at the point(x0, f(x0)) if and only if f is differentiable at x0.

• What about f(x) = 3√x at x = 0?

Others add:The curve has a vertical tangent at (x0, y0) if and only if theslope approaches ±∞ as x→ x0 from either side.

• What about g(x) =

{3√x+ 1 x ≥ 0

3√x x < 0

at x = 0?

1Eg., Calculus, Stewart, 6th ed, Brooks/Cole, 2008, p 113.

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Differentiability Implies . . .

TheoremIf f is differentiable at x = a, then f is continuous at a.

Proof. (outline).

limx→a

[f(x)− f(a)] = limx→a

[(f(x)− f(a)) · x− a

x− a

]

= limx→a

f(x)− f(a)

x− a · limx→a

x− a

= f ′(a) · 0 = 0

PropositionLet f :D → R and a ∈ D where a is an acc pt of D. Then f ′(a) existsiff qn = f(xn)−f(a)

xn−a always converges to the same real number z forany sequence xn → a with xn 6= a for all n. Then z = f ′(a).

MAT 5610: 17


Algebra of Derivatives

Theorem (Algebra of Differentiability)Suppose f, g :D → R are differentiable at a and c ∈ R. Then

1. [c · f ]′(a) = c · [f ′(a)]

2. [f ± g]′(a) = [f ′(a)]± [g′(a)]

3. [f · g]′(a) = [f ′(a) · g(a)] + [f(a) · g′(a)]

4. [f/g]′(a) =[f ′(a) · g(a)]− [f(a) · g′(a)]

g2(a)provided g(a) 6= 0

CorollarySuppose fi :D → R is differentiable at a for i = 1..n. Then

1.

[n∑i=1

fi

]′(a) =

n∑i=1

[f ′i(a)]

2.

[n∏i=1

fi

]′(a) =

(n∏i=1

fi(a)

)n∑i=1

[f ′i(a)]

fi(a)(when fi(a) 6= 0, else . . . )

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Legis Catena

Theorem (The Chain Rule)Suppose f is continuous on [a, b], f ′(x) exists at some point x ∈ (a, b), g isdefined on an interval I containing the range of f, and g is differentiable atthe point f(x). Then [g ◦ f ]′(x) = g′(f(x)) f ′(x)

Proof.• Set y = f(x).

f(t)− f(x) = (t− x)[f ′(x) + u(t)]

g(s)− g(y) = (s− y)[g′(y) + v(s)]

where t ∈ [a, b], s ∈ I, and u→ 0 as t→ x and v → 0 as s→ y.

• Set s = f(t).g(f(t))− g(f(x)) = (t− x) · [f ′(x) + u(t)] · [g′(y) + v(s)]∗

g(f(t))− g(f(x))

t− x = [g′(y) + v(s)] · [f ′(x) + u(t)] when t 6= x

• Let t→ x. Then f cont⇒ s→ y. And the result obtains.

∗Needs f(B′(x, δ) ∩ dom(f)

)∩ dom(g) 6= ∅ !

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Malum Exemplum

ExampleLet f(x) = x4 − x2 and g(u) = u3/2. Let x = 0. What is (g ◦ f)′(0)?

Then f ′(x) = 4x3 − 2x and g′(u) = 3u1/2/2. So (g ◦ f)′(0) = 0?

No! Graph it! (Explain the problem.)

Exercise

Set Φ(x) =

{[x3 sin(1/x)

]3/2x 6= 0

0 x = 0.

1. Find Φ′(0), if it exists.2. Is Φ continuous at 0?3. Find limx→0 Φ(x), if it exists.

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Extremus Cuspis

TheoremSuppose f : (a, b)→ R is differentiable at x = c. If c is a relativeextremum of f , then f ′(c) = 0.

Proof.Wolog c is a rel max of f . There is a δ > 0 such that f(x) ≤ f(c) on(c− δ, c+ δ). Whence, for |h| < δ,

f(c+ h)− f(c)

h=

{negative h > 0

positive h < 0

Thus f ′(c) ≤ 0 and f ′(c) ≥ 0. So f ′(c) = 0.

Relation to calculus course’s First and Second Derivative Tests?

MAT 5610: 21


Vado Domus

Theorem (Inverse Function Theorem)Suppose f is differentiable on the interval I and f ′ exists and isnonzero on I. Then

1. f is one-to-one on I2. f−1 is continuous on I3.(f−1

)′(y) = 1/f ′(x) or, with x = f−1(y), write x′ = 1/y′.

Proof.1. f ′ is nonzero on the interior of I, so f ′ is either always pos or

always neg. Hence f has no extrema on I. Thence f ismonotone. Whence f is one-to-one on I. (And f−1 exists on I.)

2. f is diffy on I ⇒ f is cont on I ⇒ f−1 is cont on I

3. f−1(yn)−f−1(y0)yn−y0 = xn−x0

f(xn)−f(x0) → 1f ′(x0) where xn → x0

MAT 5610: 22


Horrendus Exempoator

Example (Weierstrass (1872) )

Define f(x) =∞∑n=0

an cos(bnπx) where a ∈ (0, 1), b > 0 is an odd

integer, and ab > 1 + 3π/2.

1. f is continuous everywhere2. f is differentiable nowhere

W (x) =∞�

n=0

cos(7nπx)

4n

FactoidThe set of discontinuities of f is an Fσ set (a countable union of closed sets).What about the set of f ’s nondifferentiable points?

MAT 5610: 23


Outerlude

A theorem today relates . . .

MAT 5610: 24


Most Valuable Theorems, 1691

Theorem (Rolle’s Theorem (1691))If f is (i) continuous on [a, b], (ii) differentiable on (a, b), and (iii)f(a) = f(b), then there is a point c ∈ (a, b) where f ′(c) = 0.

Proof.Wolog f is nonconstant. Assume f ′(x) 6= 0 on (a, b).

1. Since f is continuous on a compact interval, it has max and minvalues, say M 6= m.

2. If f ′(x) 6= 0 on (a, b), there are no internal extreme vals. So themax and min must be the endpoints: M = f(a) = f(b) = m BC

http://www-gap.dcs.st-and.ac.uk/~history/Biographies/Rolle.html

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Theorem (Lagrange’s Mean Value Theorem (1797))If f is (i) continuous on [a, b] and (ii) differentiable on (a, b), then there

is a point c ∈ (a, b) where f ′(c) =f(b)− f(a)

b− a .

CorollaryIf f is (i) continuous on [a, b] and (ii) differentiable on (a, b), then thereis a point θ ∈ (0, 1) with f(x+ h)− f(x) = h f ′(x+ θh) wherex, x+ h ∈ [a, b].

Proof.

Apply Rolle’s thm to γ(x) = f(x)− f(b)− f(a)

b− a · x.

Simple Eg: | sin(x1)− sin(x2)| ≤ ?

http://www-gap.dcs.st-and.ac.uk/~history/Biographies/Lagrange.html

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Theorem (Cauchy’s Mean Value Theorem (1821))If f and g are (i) continuous on [a, b] and (ii) differentiable on (a, b),then there is a point c ∈ (a, b) where

f ′(c)g′(c)

=f(b)− f(a)

g(b)− g(a)(g(a) 6= g(b))

or, when g(b) = g(a), then g′(c) · [f(b)− f(a)] = 0.

Proof.Apply Rolle’s thm to

κ(x) = f(x) · [g(b)− g(a)]− g(x) · [f(b)− f(a)].


MAT 5610: 27


Most Valuable Theorems

Let f(x) = cos(x2/5) and

g(x) = sin(x/3) on [0, 2π].

Set R(t) = [g(t), f(t)].

Example: Cauchy’s Mean Value Theorem

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Problem Children, I

Counterexamples1. The Heaviside function H(x) is not the derivative of any fcn.

2. f(x) =

{x2 sin(1/x) x 6= 0

0 x = 0has a discontinuous derivative

with f ′(0) = 0

3. g(x) =

{x4 (2 + sin(1/x)) x 6= 0

0 x = 0has abs min at 0, but g′(x)

oscillates in every neighborhood of 0.

4. h(x) =

{x+ 2x2 sin(1/x) x 6= 0

0 x = 0. f ′(0) = 1 but f is not

monotonic in any neighborhood of 0.

5. j(x) = eix fails LMVT

MAT 5610: 29


Problem Children, II

Exercise (Prin of Math Analysis, Rudin. No. 13, Pg. 115.)13. Suppose a and c are real numbers, c > 0, and f is defined on

[0, 1] by

f(x) =

{xa sin(x−c) (if x 6= 0)

0 (if x = 0)

Prove the following statements:(a) f is continuous iff a > 0.

(b) f ′(0) exists iff a > 1.

(c) f ′ is bounded iff a ≥ 1 + c.

(d) f ′ is continuous iff a > 1 + c.

(e) f ′′(0) exists iff a > 2 + c.

(f ) f ′′ is bounded iff a ≥ 2 + 2c.

(g) f ′′ is continuous iff a > 2 + 2c.

⇒ Homework link (due Friday, Jan 28)

http://en.wikipedia.org/wiki/Walter_Rudin

http://mathsci2.appstate.edu/~wmcb/Class/5610/DerivQuestion.pdf

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Darboux’s Feat

Theorem (Darboux’s Intermediate Value Theorem (1875))Let I be an open interval, and let f :I → R be a differentiablefunction. Suppose [a, b] ⊆ I. For any m between f ′(a) and f ′(b),there exists a value c ∈ [a, b] such that f ′(c) = m.

Proof.Wolog f ′(a) ≤ m ≤ f ′(b). Define

fa(t) =

{f(a)−f(t)

a−t t 6= a

f ′(a) t = aand fb(t) =

{f(t)−f(b)

t−b t 6= a

f ′(b) t = b

Note: fa and fb are continuous and fa(b) = fb(a). So m is eitherbetween fa(a) and fa(b) or fb(a) and fb(b). Wolog, the former.Hence the IVT applied to fa gives an s ∈ (a, b] s.t.m = fa(s) = [f(s)− f(a)]/(s− a). Now, LMVT applied to f givesc ∈ [a, s] where [f(s)− f(a)]/(s− a) = f ′(c); i.e., m = f ′(c).

http://www-gap.dcs.st-and.ac.uk/~history/Biographies/Darboux.html

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Uniformity

Definition (Uniformly Differentiable)Let f be defined on an open interval I. Then f is uniformly differ-entiable iff f is diffy on I, and for every ε > 0 there is a δ > 0 s.t.

∣∣∣∣f(x1)− f(x2)

x1 − x2

∣∣∣∣ < ε

whenever 0 < |x1 − x2| < δ and xi ∈ I.

PropositionIf f is uniformly differentiable on I, then f ′ is continuous on I.

Example• L(t) = 1/t is uniformly diffy on every interval (a,∞) where a > 0.

• But L(t) = 1/t is not uniformly diffy on the interval (0,∞).

MAT 5610: 32


Higher

Definition (Higher Derivatives)For n ≥ 2, define

dn

dxnf(x) =

d

dx

(dn−1

dxn−1f(x)

).

Exercises

1. Computedn

dxn(f ◦ g)(x) for n = 1..4.

2. Determine all derivatives of Vn(x) = |x|n.

3. Discuss derivatives of R(x) =

{x x ≥ 0

0 x < 0wrt Darboux’s thm.

MAT 5610: 33


In Search of Roots

DefinitionA root of multiplicity m ∈ N is a zero of f s.t. m is the least integer forwhich f can be written as

f(x) = (x− r)m q(x)

where limx→r q(x) 6= 0.

TheoremIf f is m times continuously differentiable, r is a root of multiplicity miff 0 = f(r) = f ′(r) = · · · = f (m−1)(r), but 0 6= f (m)(r).

Proof.⇒ Differentiate f(x) = (x− r)mq(x) and let x→ r.

⇐ Apply LMVT successively to [f(x)− f(r)]/(x− r).

MAT 5610: 34


Taylor Made Theorems

Definition (Taylor Polynomial)If f has sufficiently many derivatives at x = a, the Taylor polynomial ofdegree n (or order n) is

pn(x) =

n∑

k=0

f (k)(a)

k!(x− a)k

where f (0)(a) = f(a).

Theorem (Taylor’s Theorem (1715∗))Suppose f has n+ 1 derivatives on a neighborhood of a. Thenf(x) = pn(x) +Rn(x) where

Rn(x) =f (n+1)(c)

(n+ 1)!(x− a)n+1

for some c between x and a.

∗Actually, first discovered by Gregory in 1671, ∼ 14 years before Taylor was born!

http://www-gap.dcs.st-and.ac.uk/~history/Biographies/Gregory.html

MAT 5610: 35


Taylor Remainders

Forms of the Remainder∗

Lagrange (1797) Rn(x) =1

(n+ 1)!f (n+1)(c) (x− a)n+1

Cauchy (1821) Rn(x) =1

n!f (n+1)(c) (x− a)(x− c)n

Integral Form Rn(x) =1

n!

∫ x

a

f (n+1)(t) (x− t)n dt

Uniform Estimate Rn(x) = max∣∣f (n+1)(x)

∣∣ · rn+1

(n+ 1)!on B(a, r)

∗See, e.g., Whitaker and Watson, A Course of Modern Analysis, Cambridge, 1927.

MAT 5610: 36


The Proof Sewed Up

Proof.∗

1. The FToC⇒ f(x) = f(a) +

∫ x−a

0

f ′(x− t) dt.

2. Integrate by parts with u = f ′(x− t) and dv = dt:

f(x) = f(a) + f ′(a)(x− a) +

∫ x−a

0

f ′′(x− t) · t dt

3. Repeat the process choosing u = f (k)(x− t) and dv = tk−1/(k − 1)!arriving at

f(x) = f(a)+f ′(a)(x−a)+f ′′(a)

2!(x−a)2+· · ·+f (n)(a)

n!(x−a)n+Rn(a, x)

where

Rn(a, x) =1

n!

∫ x−a

0

f (n+1)(x− t) · tn dt

∗Why can’t this proof be used in our text?

MAT 5610: 37


The Second Proof

Proof.Let pn(x) be the Taylor polynomial for f centered at a.

1. Define M by f(b) = pn(b) +M(b− a)n+1.

2. Set φ(x) = f(x)− pn(x)−M(x− a)n+1.

Then φ(k)(a) = f (k)(a)− p(k)n (a) = 0 for k = 0..n by the defn of pn.

3. Now φ(a) = 0 = φ(b). So Rolle’s thm gives x1 ∈ (a, b) s.t. φ′(x1) = 0

4. Whence φ′(a) = 0 = φ′(x1). So Rolle’s thm gives x2 ∈ (a, x1) s.t.φ′′(x2) = 0.

5. Continue the process generating xi+1 ∈ (a, xi) s.t. φ(i+1)(xi+1) = 0.

6. After n+ 1 steps, we have

0 = φ(n+1)(xn+1) = f (n+1)(xn+1)−M (n+ 1)!;

i.e., M = f (n+1)(xn+1)/(n+ 1)! proving the thm.

MAT 5610: 38


Taylor’s Proof

http://mathsci2.appstate.edu/~wmcb/Class/5610/Vignettes/Taylor1.jpg

http://mathsci2.appstate.edu/~wmcb/Class/5610/Vignettes/Taylor2.jpg

MAT 5610: 39


A Sample

ExampleFind the Maclaurin polynomial p3 with error bounds on [−1, 1] for the

sine integral Si(x) =

∫ x

0

sin(t)

tdt.

n f (n)(x) 1n!f

(n)(0)xn

0 Si(x) 0

1 sin(x)x x

2 cos(x)x − sin(x)

x2 0

3 − sin(x)x − 2 cos(x)

x2 + 2 sin(x)x3 − 1

18 x3

p3(x) = x− x3

18with |R3| ≤ max |Si(4)(x)| · 24

4!≤ 1

5

24

4!=

2

15

MAT 5610: 40


Si and p3

-5 -2.5 0 2.5 5

-2.5

2.5Si(x) =

� x

0

sin(t)

tdt

p3(x) = x − x3

18

MAT 5610: 41


Si and p3 Error

-1 -0.75 -0.5 -0.25 0 0.25 0.5 0.75 1

5!10-4

0.001

0.0015

Terr(x) = |Si(x) − p3(x)|

maxx∈[−1,1]

Terr(x) = Terr(1) ≈ 1.6386260× 10−3

MAT 5610: 42


Outerlude

XKCD

http://xkcd.com/854/

MAT 5610: 43


Try It!

ExerciseLet f(x) = tan(sin(x)) defined for all x ∈ R. Note f ∈ C∞.

1. Find an n ∈ N so that max |Rn(x)| ≤ 1 on [−π, π].

-π -0.5π 0 0.5π π

-2

-1

1

2f(x) = tan(sin(x))

To compute the nth Taylor polynomial in Maple, defineT:=n->convert(taylor(f(x),x,n+1),polynom)

MAT 5610: 44


Try It! Computed

In Maple 15:

> f := x -> tan(sin(x)):> d := n -> (D@@n)(f);

n→ D(n)(f)

> c := n -> d(n)(0)/n!;

n→ d(n)(0)

n!

> N := 20:> <<$i=1..N>

|<d(i))(0)$i=1..N>

|evalf[5](<c(i)$i=1..N>)>

n f (n)(0) f (n)(0)/n!1 1 12 0 03 1 0.166674 0 05 -3 -0.0250006 0 07 -107 -0.0212308 0 09 -1095 -0.0030175

10 0 011 41897 0.001049612 0 013 3027637 0.0004862114 0 015 34528445 0.00002640416 0 017 -11832720271 -0.00003326718 0 019 -1190157296815 -0.000009783820 0 0

MAT 5610: 45


Try It! Graphed

f(x) = tan(sin(x)) and T37(x).

MAT 5610: 46


A Differentiable Inequality

Theorem (Jensen’s Inequality, I (1906))Let f be twice continuously differentiable on (a, b) with f ′′(x) ≥ 0.Suppose xi ∈ (a, b) for i = 1..n. Then

f

(1

n

n∑

k=1

xk

)≤ 1

n

n∑

k=1

f(xk)

Proof.1. Let x∗ = 1

n

∑nk=1 xk. Define L(x) = f(x∗) + f ′(x∗) · (x− x∗).

2. L is linear gives f(x∗) = L(x∗) = 1n

∑nk=1 L(xk).

3. Since f is concave up (convex), then L(x) ≤ f(x) on (a, b).Whence

f

(1

n

n∑

k=1

xk

)=

1

n

n∑

k=1

L(xk) ≤ 1

n

n∑

k=1

f(xk).

http://www-groups.dcs.st-and.ac.uk/~history/Biographies/Jensen.html

MAT 5610: 47


A Mean Application

Theorem (Jensen’s Inequality (1906))If f is convex on (a, b), xk ∈ (a, b), and µk > 0 with

∑nk=1 µi = 1, then

f

(n∑k=1

µkxk

)≤

n∑k=1

µkf(xk)

Example (Arithmetic-Geometric Mean Inequality)1. Set f(x) = − ln(x) for x > 0. Then f ′′(x) > 0. Let xk > 0.

2. By Jensen’s inequality,

− ln

(1

n

n∑k=1

xk

)≤ − 1

n

n∑k=1

ln(xk) = − 1

nln

(n∏k=1

xk

)

3. Whence

ln

n

√√√√ n∏k=1

xk

≤ ln

(1

n

n∑k=1

xk

)⇒

A-GMI

n√x1x2 · · ·xn ≤

x1 + x2 + · · ·+ xnn

MAT 5610: 48


Guillaume de l’Hopital

Actually:M. Guillaume-Francois-Antoine Marquis de l’Hopital, Marquis deSainte-Mesme, Comte d’Entremont et Seigneur d’Ouques-la-Chaise.

TheoremSuppose that f and g are

• continuous on [a, b],

• differentiable on (a, b),

• limx→a+ f(x) = 0 = limx→a+ g(x), and

• g′(x) 6= 0 on the interval (a, a+ ε).

If limx→a+

f ′(x)

g′(x)= L, then lim

x→a+

f(x)

g(x)= L

Proof.If xn → a, then [f(xn)− f(a)] / [g(xn)− g(a)] = f ′(cn) / g′(cn) & cn → a.

http://www-groups.dcs.st-and.ac.uk/~history/Biographies/De_L'Hopital.html

MAT 5610: 49


Example

Example

Determine limn→∞

(√n2 + n+ 1− n

).

1.√n2 + n+ 1− n =

(√n2 + n+ 1

)2 − n2

√n2 + n+ 1 + n

=n+ 1√

n2 + n+ 1 + n

2.n+ 1√

n2 + n+ 1 + n=

1 + (1/n)√1 + (1/n) + (1/n2) + 1

→ 1√1 + 1

=1

2

3. Whence limn→∞

(√n2 + n+ 1− n

)=

1

2

MAT 5610: 50


Diwrnod o Meddwl

Aha! If true, prove; if false, fix, then prove.1. Set y = (4x− 1)x, then y′ = x (4x− 1)x−1(4).

2. f(x) = x3 + x−√

2 has exactly one real root.

3. If f is differentiable at a, then f ′(a) is unique.

4. If f ′ is bounded, then f is bounded.

5. If f is bounded, then f ′ is bounded.

6. One point can be both an inflection pt. and a rel. maximum.

7. a is an inflection point iff f ′′(a) = 0.

8. If x ≥ 0, then√

1 + x < 1 + 12x.

9. If f is continuous and one-to-one, then f is differentiable.

10. If f and g are diffy and f(x)g(x) = 1, then f ′(x)f(x) + g′(x)

g(x) = 0.

http://mathsci2.appstate.edu/~wmcb/Class/5610/MonotoneContNondiffy.pdf

MAT 5610: 51


Diwrnod o Meddwl, II

Aha! If true, prove; if false, fix, then prove.

11. If limn→∞

n

[f

(a+

1

n

)− f(a)

]is finite, then f ′(a) exists.

12. If f is differentiable and uniformly continuous on (a, b), then f ′ isbounded on (a, b).

13. If f is continuously differentiable on an interval I, then f isuniformly continuous on I.

14. If a is a triple root of a three-times differentiable fcn f, then fmust have an inflection pt at a.

15. f(x) = e−x − 12x

2 + x− 1 has a triple root at 0.

16. f ′ may have more roots than f .

17. Let f ∈ C1(R) and f(0) = 0. If f ′(x) > 0 always, then f(x) > 0for x > 0.

MAT 5610: 52


Bounded Variation

Definition (Partition)Let [a, b] be a compact interval. A partition of [a, b] is a set of n+ 1points P = {a = x0 < x1 < · · · < xn = b}. Set ‖P‖ = n.

Definition (Bounded Variation — Jordan(1881))Suppose f is defined on the compact interval [a, b]. The variation of fon [a, b] is

Vf (a, b) = supP∈P

‖P‖∑

k=1

|f(xk)− f(xk−1)|

where P is the collection of all partitions of [a, b]. Then f is ofbounded variation on [a, b] iff Vf (a, b) is finite.

¿How does variation relate to arclength (“rectification,” Neile (1657))?

http://www.youtube.com/watch?v=GNnua9IlsTg

http://www-gap.dcs.st-and.ac.uk/~history/Biographies/Jordan.html

MAT 5610: 53


Sampling Variation

Examples• f(x) = sin(x) has bounded variation on any compact interval.

• g(x) =

{sin(1/x) x 6= 0

0 x = 0is not of bounded variation on any

compact interval containing 0.

• h(x) =

{x2 sin(1/x) x 6= 0

0 x = 0is of bounded variation on any

compact interval.

• j(x) =

{x2 sin(1/x2) x 6= 0

0 x = 0is not of bounded variation on any

compact interval containing 0 but j is differentiable at 0.

• A montone function?

MAT 5610: 54


Variations on a Theorem

TheoremIf f and g are BV on [a, b], then f ± g and f · g are BV . If g isbounded away from 0, then f/g is of BV .

TheoremSuppose f is continuous on [a, b]. If f has a bounded derivative on(a, b), then f is BV on [a, b].

Proposition1. f is BV on [a, b] iff f can be written as the difference of two

increasing functions. (Jordan (1893))2. A BV fcn on [a, b] has only jump discontinuities.3. The discont of a BV fcn on [a, b] are countable. (A. Froda (1929))4. A BV fcn on [a, b] has a derivative a.e. (Lebesgue (1904)∗)

∗See Botsko

http://en.wikipedia.org/wiki/Froda's_theorem

http://www-gap.dcs.st-and.ac.uk/~history/Biographies/Lebesgue.html

http://mathsci2.appstate.edu/~wmcb/Class/5610/LebesgueDiffThm_Botsko.pdf

MAT 5610: 55


Interlude: Planimeter to ENIAC

Amler’s Planimeter, 1854

Thomson Differential Analyzer, 1876

ENIAC: Electronic NumericalIntegrator And Computer, 1946

http://en.wikipedia.org/wiki/Planimeter

http://en.wikipedia.org/wiki/James_Thomson_(engineer)

MAT 5610: 56


The Integral

Classical Greek ProblemsTrisect an Angle: divide an angle into three equal parts. (Wantzel, 1837)

Square the Circle: create a square having the same area as a given circle.(Lindemann, 1882)

Duplicate the Cube: construct a second cube with double the volume of agiven cube. (Descartes, 1637)

Quadrature of the Lune: Hippocrates of Chios’ construction (c 440 BC) isone of the “Great Theorems.” (See Journey Through Genius)

Universal Etymological English Dictionary, Bailey (1731)QUADRATURE [quadratura, L] the making of a thing fquare, or the finding afquare equal to the area of any figure given.

QUADRATURE of Curves [in the higher Geometry] is the meafuring of theirarea, or the finding of a rectilinear fpace, equal to a curvilinear fpace.

http://mathdl.maa.org/mathDL/46/?pa=content&sa=viewDocument&nodeId=1203&bodyId=1531

http://www.amazon.com/Journey-through-Genius-Theorems-Mathematics/dp/014014739X

http://www.google.com/books?id=o-gIAAAAQAAJ&printsec=frontcover&dq=quadrature#v=onepage&q=quadrature&f=false

MAT 5610: 57


The Integrators

Cavalieri, 1630’s Newton, 1670’s Leibniz, 1670’s Cauchy, 1820’s

Riemann, 1840’s Darboux, 1870’s Stieltjes, 1880’s Lebesgue, 1900’s

http://www-gap.dcs.st-and.ac.uk/~history/Biographies/Cavalieri.html

http://www-gap.dcs.st-and.ac.uk/~history/Biographies/Newton.html

http://www-gap.dcs.st-and.ac.uk/~history/Biographies/Leibniz.html


http://www-gap.dcs.st-and.ac.uk/~history/Biographies/Riemann.html

http://www-gap.dcs.st-and.ac.uk/~history/Biographies/Darboux.html

http://www-gap.dcs.st-and.ac.uk/~history/Biographies/Stieltjes.html

http://www-gap.dcs.st-and.ac.uk/~history/Biographies/Lebesgue.html

MAT 5610: 58


Cauchy’s Integral

Definition (Cauchy’s Integral (1823))Suppose f is continuous on [a, b]. For the partition P = {x0 < x1 < · · · < xn}(x0 = a, xn = b) set ∆xk = (xk − xk−1). Define the Cauchy sum to be

C(P, f) =n∑k=1

f(xk−1)∆xk

If there is a number I s.t. for any ε > 0 there is a δ > 0 where |C(f, P )− I|<εwhenever max

P∆x < δ, then f is Cauchy integrable. Write I =

∫ baf(x) dx.

Theorem (Cauchy (1823))If f is continuous on [a, b], then f is Cauchy integrable.

Oops!The function f(x) = 1 for x 6= 1 & f(1) = 2 is not Cauchy integrable on [0, 2].

http://www.mathsci2.appstate.edu/~wmcb/Class/5610/Vignettes/CauchyDefinesIntegral.pdf

MAT 5610: 59


Riemann’s Integral

Definition (Partition)Let [a, b] be a compact interval. A partition of [a, b] is a set of n+ 1 pointsP = {a = x0 < x1 < · · · < xn = b}. Set |P | = maxk=1..n ∆xk where∆xk = xk − xk−1. P is refinement of Q iff Q ⊆ P.

Definition (Riemann Sum)

The Riemann sum of a bounded function on [a, b] is R(P, f) =

n∑k=1

f(ck) ∆xk

where the “tags” ck ∈ [xk, xk−1].

Definition (Darboux Sums)The upper- & lower-Darboux sums of a bounded function on [a, b] are

U(P, f) =n∑k=1

Mk ∆xk and L(P, f) =

n∑k=1

mk ∆xk

respectively, with Mk= supx∈Ik

f(x) and mk= infx∈Ik

f(x) for Ik=[xk−1, xk].

MAT 5610: 60


Order Lemmas

LemmaSuppose m ≤ f(x) ≤M on [a, b], and let P be a partition of [a, b].Then

m(b− a) ≤ L(P, f) ≤ R(P, f) ≤ U(P, f) ≤M(b− a)

for any set {ck} of tags with P .

LemmaSuppose f is bounded on [a, b]. Let P and Q be partitions of [a, b].Then

1. If Q is a refinement of P, then

L(P, f) ≤ L(Q, f) ≤ U(Q, f) ≤ U(P, f).

2. L(P, f) ≤ U(Q, f) for any partitions P and Q.

MAT 5610: 61


Order Lemmas

Proof. (Second Order Lemma).1. Suppose Q = P ∪ {x∗} with x∗ ∈ (xi−1, xi). Since

infx∈[xi−1,xi]

f(x)

︸︷︷︸mi

≤ infx∈[xi−1,x∗]

f(x)

︸︷︷︸mi,1

or infx∈[x∗,xi]

f(x)

︸︷︷︸mi,2

,

we have

mi∆xi = mi(x∗−xi−1)+mi(xi−x∗) ≤ mi,1(x∗−xi−1)+mi,2(x∗−xi)

Whence L(P, f) ≤ L(Q, f).

If Q = P ∪ {x∗k | k = 1..m}, repeat the argument m times.

2. By 1, P ∪Q refines both P and Q, whence L(P ) ≤ L(P ∪Q) andU(P ∪Q) ≤ U(Q).

MAT 5610: 62


Upper and Lower Riemann-Darboux Integrals

DefinitionFor a bounded function f on [a, b], define∗∫ b

a

f(x) dx = supPL(P, f) and

∫ b

a

f(x) dx = infPU(P, f).

PropositionIf f is bounded on [a, b] and P is any partition of [a, b], then

L(P, f) ≤∫ b

a

f(x) dx ≤∫ b

a

f(x) dx ≤ U(P, f).

Definition (Riemann/Darboux Integrable)A bounded function f is Riemann- or Riemann-Darboux integrable iff∫ b

a

f(x) dx =

∫ b

a

f(x) dx = A.

We write f ∈ R[a, b] and A =∫ baf(x) dx. Sample

∗Why don’t we use limits?

http://www.mathsci2.appstate.edu/~wmcb/Class/5610/RD_Int_Eg.pdf

MAT 5610: 63


Short Readings

Exercise (Read and Ponder)• Go to our “AsULearn class website.” Read the selections from:

1. A Garden of Integrals by Frank E. Burk

2. Lebesgue’s Theory of Integration: Its Origins and Development byThomas Hawkins

• Read §6.2 from David Bressoud’s A Radical Approach to RealAnalysis

• Investigate the theorem:Proposition: A bounded function f on [a, b] is Cauchy

integrable iff f is continuous.

How did Cauchy deal with simple discontinuities?

http://asulearn.appstate.edu/course/view.php?id=40444

MAT 5610: 64


“Bad Integral! No Biscuit!”

Example (Bad Integral)

Determine∫ 1

−1

χQ(x) dx where χQ(x) =

{1 x ∈ Q0 x /∈ Q

.

For any P , we have L(P, χQ) = 0 and U(P, χQ) = 2. Thence∫ 1

−1

χQ(x) dx = 0 < 2 =

∫ 1

−1

χQ(x) dx,

whereupon the integral fails to exist.

Example (Good Integral)

Determine∫ 2

0

f(x) dx where f(x) =

{1 x 6= 1

2 x = 1.

The only “problem point” is at x = 1 where f is not continuous. Refine anypartition containing x = 1 as a tag by making ∆xk of that interval less thanε/8. The total contribution to any R-D sum from those terms is less than ε/2.Whence 2 ≤ L(P, f) ≤ U(P, f) ≤ 2 + ε, and so the integral equals 2.

MAT 5610: 65


The Function’s View

TheoremLet f be bounded on [a, b]. Then f ∈ R[a, b] iff for every ε > 0 there isa partition P s.t. U(P, f)− L(P, f) < ε.

Proof.(⇒) Let A =

∫ baf and ε > 0. Then

1. ∃PU s.t. U(PU ) < A+ ε/2 (Why?) and2. ∃PL s.t. L(PL) > A− ε/2. (Why?)

Set P = PL ∪ PU . Whence

A− ε/2 < L(PL) ≤ L(P ) ≤ U(P ) ≤ U(PU ) < A+ ε/2

(⇐) Let ε > 0. Suppose ∃P with U(P )− L(P ) < ε. ThenU(P ) < L(P ) + ε. Whence

L(P ) ≤∫ b

a

f ≤∫ b

a

f ≤ U(P ) < L(P ) + ε

MAT 5610: 66


Who Lives in R?

Theorem{f | f is monotone on [a, b]} ⊂ R[a, b]

Proof.Wolog f is mono increasing. Let ε > 0 and set Pn = {xk | k = 0..n}where xk = a · (n− k)/n+ b · (k/n). Then ∆xk = (b− a)/n. We have

U(Pn)−L(Pn) =

n∑

k=1

[f(xk)−f(xk−1)] ·∆xk = [f(b)−f(a)] · b−an

=K

n

Hence, U(P )− L(P ) < ε for large enough n. Thus f ∈ R[a, b].

Example

Heaviside(x) =

{1 x ≥ 0

0 x < 0∈ R[a, b] for any a < b.

MAT 5610: 67


Who Else Lives in R?

TheoremC[a, b] ⊂ R[a, b]: All continuous functions are integrable.

Proof.Let ε > 0.

1. Since f is continuous on a compact set, it is uniformly continuous.Whence ∃δ > 0 s.t. |f(t1)− f(t2)| < ε/(b− a) whenever |t1 − t2| < δ.

2. Choose P s.t. |P | < δ.

3. f cont on [a, b] says ∃tk, sk s.t. Mk = f(tk) and mk = f(sk) for each k.

4. Hence U(P )− L(P ) ≤n∑k=1

|f(tk)− f(sk)|∆xk

5. Since |P | < δ, thence U(P )− L(P ) < ε/(b− a) ·n∑k=1

∆xk = ε.

I.e., there is a partition P s.t. U(P )− L(P ) < ε

MAT 5610: 68


The Crazy Neighbors

Example

D Bernoulli and Goldbach soughtto extend n! to real values(1720’s). Euler’s solution (1729)is the Gamma function∗ definedby

Γ(x) =

∫ ∞

0

tx−1e−tdt

for x 6= 0,−1,−2, . . . .

Γ(x) =

� ∞

0

tx−1e−tdt

y = 1/Γ(x)

For x > 0, we see that Γ(x) is continuous. Whence Γ(x) is Riemannintegrable on any compact interval [a, b] for 0 < a < b <∞.

(Curious factoid:∫ x

1Γ(t) dt < Γ(x) for x ≥ 4.)

∗See the Handbook of Mathematical Functions, M Abramowitz & I Stegun, pg 255.(A result of Roosevelt’s WPA!) See also the DLMF Chap 5.

http://www-gap.dcs.st-and.ac.uk/~history/Biographies/Bernoulli_Daniel.html

http://www-gap.dcs.st-and.ac.uk/~history/Biographies/Goldbach.html


http://people.math.sfu.ca/~cbm/aands/frameindex.htm

http://people.math.sfu.ca/~cbm/aands/page_255.htm

http://www.digitalhistory.uh.edu/database/article_display.cfm?HHID=473

http://dlmf.nist.gov/

http://dlmf.nist.gov/5

MAT 5610: 69


A Degenerate Function

Example (Burk’s Bad Badger)

Set f(x) =

{2x sin(1/x2)− 2x−1 cos(1/x2) x 6= 0

0 x = 0.∫ 1/

√π

0

f(t)dt =?

Cauchy: No! f is not continuous at 0!Riemann: No! f is unbounded at 0!

Lebesgue: No! |f | can’t be integrated, so neither can f !Newton/Leibniz: Dudes, whatev. Since d

dx

[x2 sin(1/x2)

]=f(x), it’s 0.

MAT 5610: 70


Props to Riemann

Theorem (Riemann Integration is Linear)Suppose f , g ∈ R[a, b] and c ∈ R. Then

1. f ± g ∈ R[a, b], and∫ ba

(f ± g) =∫ baf ±

∫ bag.

2. cf ∈ R[a, b], and∫ ba

(cf) = c∫ baf.

Proof. (Key steps).

• ∃Pf , Pg s.t.∫ baf ≤ U(Pf ) < ε+

∫ baf and

∫ bag ≤ U(Pg) < ε+

∫ bag

• U(Pf ∪ Pg, f + g) ≤ U(Pf , f) + U(Pg, g)

∴∫ ba

(f + g) ≤ 2ε+∫ baf +

∫ bag

An analogous argument gives∫ ba(f + g) ≥

∫ baf +

∫ bag.

Whence∫ ba

(f + g) ≤∫ baf +

∫ bag ≤

∫ ba(f + g).

Hence∫ ba

(f + g) =∫ baf +

∫ bag

MAT 5610: 71


More Props to Riemann

TheoremSuppose f , g ∈ R[a, b]. If f(x) ≤ g(x) on [a, b], then

∫ baf ≤

∫ bag.

TheoremSuppose f is bounded on [a, b], and c ∈ [a, b]. Then (i) f ∈ R[a, b]⇐⇒(f ∈ R[a, c] and f ∈ R[c, b]) and (ii)

∫ baf =

∫ caf +

∫ bcf .

TheoremIf (i) f ∈ R[a, b], (ii) g ∈ R[c, d], and (iii) f([a, b]) ⊆ [c, d], then f ◦ g ∈ R[a, b].

CorollaryIf f, g ∈ R[a, b], and n ∈ N, then (i) fn ∈ R[a, b], and (ii) fg ∈ R[a, b].

Proof.X (i) g(x) = xn and (ii) fg = (f + g)2/4− (f − g)2/4

MAT 5610: 72


Outerlude 2/17/11 9:10 PMPiled Higher and Deeper

Page 2 of 2http://www.phdcomics.com/comics/archive_print.php?comicid=1421

Piled Higher and Deeper by Jorge Cham www.phdcomics.com

title: "Guide to T.A. Office Hours" - originally published 2/16/2011

MAT 5610: 73


MVTs, I

Theorem (Bonnet’s Mean Value Theorem (1849))Suppose f ∈ C[a, b] and g ∈ R[a, b] with g(x) ≥ 0. There is a c ∈ (a, b) s.t.∫ b

a

f(x)g(x) dx = f(c) ·∫ b

a

g(x) dx

Proof.1. f ∈ C[a, b] =⇒ f is bounded: m ≤ f(x) ≤M2. g ≥ 0 =⇒ mg(x) ≤ f(x)g(x) ≤Mg(x)

=⇒ m∫ bag ≤

∫ bafg ≤M

∫ bag

3. Wolog∫ bag 6= 0. Then set A =

∫ bafg /

∫ bag.

4. Since m ≤ A ≤M and f ∈ C[a, b], there is a c ∈ (a, b) s.t. f(c) = A.

Corollary (Cauchy’s ‘Water in the Bucket’ Thm (1821))If f ∈ C[a, b], then there is a c ∈ (a, b) s.t.

∫ baf = f(c)(b− a).

http://www-gap.dcs.st-and.ac.uk/~history/Biographies/Bonnet.html

MAT 5610: 74


MVTs, II

Theorem (The Second Mean Value Thm for Integrals)If f is monotonic on [a, b], then there is a c ∈ (a, b) s.t.∫ b

a

f(x) dx = f(a)(c− a) + f(b)(b− c)

Proof.• Define g(t) = f(a)(t− a) + f(b)(b− t) = µ t+ β.

• Since f is monotone, g(a) ≤∫ baf ≤ g(b)

• g has the IVP (Why?) which finishes the proof.

Theorem (Generalized Second MVT for Integrals)If f is monotonic on [a, b] and g ∈ C[a, b], then there is a c ∈ (a, b) s.t.∫ b

a

f(x)g(x) dx = f(a)

∫ c

a

g(x) dx+ f(b)

∫ b

c

g(x) dx

http://www.mathsci2.appstate.edu/~wmcb/Class/5610/SMVT.mov

MAT 5610: 75


Integral Inequalities

Theorem (Holder’s Inequality∗)If f, g ∈ R[a, b] and p, q ∈ [1,∞] are such that 1

p+ 1

q= 1, then∫ b

a

|fg| ≤[∫ b

a

|f |p]1/p

·[∫ b

a

|g|q]1/q

.

Corollary (Cauchy-Bunyakovsky-Schwarz Inequality)If f and g ∈ R[a, b], then[∫ b

a

|fg|]2

≤∫ b

a

|f |2 ·∫ b

a

|g|2.

Theorem (Minkowski’s Inequality)If f, g ∈ R[a, b] and p ≥ 1, then[∫ b

a

|f + g|p]1/p

≤[∫ b

a

|f |p]1/p

+

[∫ b

a

|g|p]1/p

∗See handout.

http://www-gap.dcs.st-and.ac.uk/~history/Biographies/Holder.html

http://www-gap.dcs.st-and.ac.uk/~history/Biographies/Bunyakovsky.html

http://www-gap.dcs.st-and.ac.uk/~history/Biographies/Schwarz.html

http://www-gap.dcs.st-and.ac.uk/~history/Biographies/Minkowski.html

http://mathsci2.appstate.edu/~wmcb/Class/5610/IneqFToC.pdf

MAT 5610: 76


FToC

TheoremSuppose f is differentiable on [a, b] and f ′ ∈ R[a, b]. Then

∫ b

a

f ′(x) dx = f(x)∣∣∣b

a= f(b)− f(a)

Proof.1. Let P be a partition with tags ck chosen via the MVT s.t.

f ′(ck)(xk − xk−1) = f(xk)− f(xk−1).

2. With mk = inf f and Mk = sup f on [xk − xk−1], we seeL(P, f) ≤ R(P, f) ≤ U(P, f) =⇒ L(P, f) ≤ f(b)−f(a) ≤ U(P, f)

3. Since P is arbitrary and f ′ ∈ R[a, b], the result follows.

What about:∫ 1

0e−x

2

dx?∫ +1

−1U(x) dx?

∫ +1

−1

[x2 sin(1/x2)

]′dx?

MAT 5610: 77


Results of FToC

Theorem (Integration by Parts)Suppose f and g are differentiable and f ′, g′ ∈ R[a, b]. Then

∫ b

a

f(x)g′(x) dx = f(b)g(b)− f(a)g(a)−∫ b

a

f ′(x)g(x) dx

Proof.1. f, g diffy⇒ f, g cont⇒ f, g ∈ R ⇒ f ′g, f g′ ∈ R ⇒ (fg)′ ∈ R2. Integrate the equation (fg)′ = f ′g + f g′ to finish.

Theorem (Indefinite Integral Theorem)Let f ∈ R[a, b]. Define F (x) =

∫ xaf(t) dt. Then

1. F is uniformly continuous on [a, b], and2. if f is cont at c, then F is diffy at c and F ′(c) = f(c).

MAT 5610: 78


IIT’s Proof

Proof. (Indefinite Integral Theorem).1. Let ε > 0. f ∈ R ⇒ f bounded, say by M . Choose δ < ε/M.

Let |x1 − x2| < δ. Then

|F (x1)− F (x2)| =∣∣∣∣∫ x1

a

f(t) dt−∫ x2

a

f(t) dt

∣∣∣∣ =

∣∣∣∣∫ x1

x2

f(t) dt

∣∣∣∣

Whence |F (x1)− F (x2)| ≤M · |x1 − x2| < ε.

2. Let f ∈ C[a, b] and ε > 0. Then ∃δ > 0 s.t. if |x− c| < δ, then|f(x)− f(c)| < ε. Then

∣∣∣∣F (x)− F (c)

x− c − f(c)

∣∣∣∣ =1

|x− c|

∣∣∣∣∫ x

c

f(t) dt− f(c)(x− c)∣∣∣∣

≤ 1

|x− c|

∫ x

c

|f(t)− f(c)| dt < 1

|x− c|

∫ x

c

ε dt < ε

MAT 5610: 79


Fun Example

Example (O.M.G.)

Determine F (x) =

∫ x

0

sin(2t) dt.

• F1(x) = − 12 cos(2x) (check with differentiation)

• sin(2t) = 2 sin(t) cos(t). Let u = sin(t) and du = cos(t) dt. Then

F2(x) = sin2(x)

• sin(2t) = 2 cos(t) sin(t). Let u = cos(t) and du = − sin(t) dt. Then

F3(x) = − cos2(x)

Which is the correct F (x)? F (x) = F2(x), = F1(x) + 12 , = F3(x) + 1.

MAT 5610: 80


Part II

Outerlude

MAT 5610: 81


Riemann-Stieltjes Integration

Thomas Jan Stieltjes (1856–1894 ) modeled the mo-ments of thin rods with nonuniform mass distribu-tions. The nth moment of f about the point c is

µn =

∫ ∞

−∞(x− c)nf(x) dx

Stieltjes considered the distributions that came from nonuniform den-sities or from discrete masses placed along the x-axis, problems thatRiemann’s integral didn’t handle. He published his generalization ofRiemann’s integral in 1894, the year he died.

Stieltjes problem, now called the “Stieltjes Moment Problem,” is still thesubject for much research. (Also see the “Hamburger Moment Problem”.)

http://www-history.mcs.st-and.ac.uk/Biographies/Stieltjes.html

http://en.wikipedia.org/wiki/Stieltjes_moment_problem

http://scholar.google.com/scholar?q=stieltjes+moment+problem&hl=en&as_sdt=0&as_vis=1&oi=scholart

http://en.wikipedia.org/wiki/Hamburger_moment_problem

MAT 5610: 82


Riemann-Stieltjes Integration

Definition (Upper and Lower Riemann-Stieltjes Integrals)Let f be bounded and α be monotonically increasing on [a, b]. For apartition P, define the upper- and lower Riemann-Stieltjes sums by

U(P, f, α) =

n∑

k=1

Mk∆αk and L(P, f, α) =

n∑

k=1

mk∆αk

where ∆αi = α(xi)− α(xi−1). Define the upper- and lowerRiemann-Stieltjes integrals as

∫ b

a

f dα = infPU(P, f, α) and

∫ b

a

f dα = supPL(P, f, α)

When α(x) = x, the Riemann-Stieltjes integral is just a Riemann integral.

MAT 5610: 83


Riemann-Stieltjes Integrability

DefinitionIf∫fdα=

∫fdα, thenf is Riemann-Stieltjes integrable; write f ∈ R(α).

TheoremA function f is Riemann-Stieltjes integrable on [a, b] if and only if forevery ε > 0 there is a partition P of [a, b] such that

U(P, f, α)− L(P, f, α) < ε

Proof.Essentially the same as for Riemann-Darboux sums.

MAT 5610: 84


Continuity & Riemann-Stieltjes Integrals

TheoremIf f is continuous on [a, b], then f ∈ R(α) on [a, b].

Proof.1. Let ε > 0. f is uniformly cont, so ∃δ > 0 s.t. if |x1 − x2| < δ, then|f(x1)− f(x2)| < ε/(α(b)− α(a)).

2. If |P | < δ, then on [xk−1, xk], we have Mk −mk<ε/(α(b)− α(a)).So

U(P, f, α)− L(P, f, α) =

n∑

k=1

(Mk −mk)∆αk

≤n∑

k=1

ε

α(b)− α(a)∆αk

=ε

α(b)− α(a)

n∑

k=1

∆αk = ε

MAT 5610: 85


Monotonicity & Riemann-Stieltjes Integrals

TheoremIf f is monotonic on [a, b] and α is continuous, then f ∈ R(α) on [a, b].

Proof.1. Let ε > 0 and n ∈ N. Choose∗ P s.t. ∆αk = (α(b)− α(a))/n.

2. Wolog f mono incr. Then Mk = f(xk) and mk = f(xk−1).3. Whence

U(P, f, α)− L(P, f, α) =α(b)− α(a)

n

n∑

k=1

[f(xk)− f(xk−1)]

=α(b)− α(a)

n· [f(b)− f(a)] < ε.

∗Requires α to be continuous. Remember, α is still monotone increasing.

MAT 5610: 86


Pointing to Continuity

TheoremIf f is bounded and has finitely many discontinuities on [a, b], and α iscontinuous at each discontinuity of f, then f ∈ R(α).

Proof.1. Let ε > 0, set M = sup |f(x)|, and let E = {discontinuities of f}.2. Since E is finite and α is cont on E, cover E with finitely many disjoint

intervals (uj , vj) s.t.∑

(α(vj)− α(uj)) < ε.

3. The set A = [a, b]−⋃

(uj , vj) is compact, so f is unif cont on A. Then∃δ > 0 s.t. if |x1 − x2| < δ, then |f(x1)− f(x2)| < ε.

4. Form P s.t.: • uj , vj ∈ P , • no point of (uj , vj) is in P , and • If xi−1 6=any uj , then ∆xi = xi − xi−1 < δ.

5. Then Mi −mi < 2M , and Mi −mi ≤ ε unless xi−1 = uj .

Whence U(P, f, α)− L(P, f, α) ≤ [α(b)− α(a)] ε︸︷︷︸xi 6=vj terms

+ 2Mε︸︷︷︸xi=vj terms

MAT 5610: 87


Exercises

Exercises

1. Compute∫ 1

−1

5(x2 + 1) d(x3 + 1). ( HINT: 16 )

2. Define Uj(x) =

1 x > 0

uj x = 0

0 x < 0

for j = 1, 2, and 3

2.a Let u1 = 0.Prove: f ∈ R(U1) iff f(0+) = f(0) and then

∫f dU1 = f(0).

2.b Let u2 = 1.State and prove a similar result for U2.

2.c Let u3 = 1/2.Prove: f ∈ R(U3) iff f is continuous at 0.

2.d Prove:If f is continuous at 0, then

∫f dU1 =

∫f dU2 =

∫f dU3 = f(0).

MAT 5610: 88


Exercise Hints

Exercises (Hints)1. Use equipartitions. Split U and L into decr and incr parts for f .

(A ‘coming attraction’ theorem makes it very easy.)

2. Let P be any partition containing the point x = 0.

2.1 The only interval with ∆αk 6= 0 is [0, xk]. All but the intervalcontaining 0 have ∆αk = 0. Then consider lim

xk→0

(Mk −mk)

2.2 Repeat with [xk, 0].Let P be any partition not containing the point x = 0.

2.3 Then 0 ∈ (xk∗−1, xk∗), so U(P )− L(P ) = (Mk∗ −mk∗)∆αk∗ .Consider what happens as both xk∗−1, xk∗ → 0.

MAT 5610: 89


Outerlude, II

MAT 5610: 90


Alpha’s Bucket

PropositionIf f is bounded on [a, b] and f ∈ R(α), then there exists m and M ∈ Rsuch that

m[α(b)− α(a)] ≤∫ b

a

f(x) dα(x) ≤M [α(b)− α(a)]

This result matches Cauchy’s Water in the bucket theorem.

Example

Consider∫ 1

−1

5(x2 + 1) d(x3 + 1).

Since ∆α = α(1)− α(−1) = 2 and 5 ≤ f(x) ≤ 10, we see

5 · 2 = 10 ≤∫ 1

−1

5(x2 + 1) d(x3 + 1) ≤ 10 · 2 = 20.

MAT 5610: 91


Composition

TheoremIf f is bounded, f ∈ R(α), and φ is cont on range(f), then φ◦f ∈ R(α).

Proof.1. Let ε > 0. Suppose m ≤ f(x) ≤M. Then ∃δ > 0 s.t. δ < ε and if|s− t| < δ, then |φ(s)− φ(t)| < ε. Let M∗i ,m∗i be the “M & m”s for φ◦f

2. ∃P s.t. U(P, f, α)− L(P, f, α) < δ2

3. Divide P by i ∈ A if Mi −mi < δ and i ∈ B otherwise

4. For i ∈ A, we have M∗i −m∗i <ε⇒∑i∈A

(M∗i −m∗i )∆αi<ε[α(b)−α(a)]

5. For i ∈ B, M∗i −m∗i ≤ 2 sup |φ|.

6. Now∑i∈B

δ∆αi ≤(3.)

∑i∈B

(Mi−mi)∆αi <(2.)

δ2 ⇒∑i∈B

∆αi < δ

7. Whence U(P, φ◦f, α)− L(P, φ◦f, α) =

[∑i∈A

+∑i∈B

](M∗i −m∗i )∆αi

≤ ε[α(b)−α(a)] + [2 sup |φ|] δ < const · ε

MAT 5610: 92


Exercises, II

Exercises1. Evaluate

1.1∫ π/2

0

cos(x) d sin(x)

1.2∫ 2

0

x d[x]

1.3∫ π

0

x d cos(x) (Surely, cos is decreasing! Yes, but don’t call me ...)

2. Prove: If g is monotone increasing on [a, b], then∫ b

a

dg exists

and is equal to g(b)− g(a).

3. Suppose that f and α are both discontinuous at x = 0. What can

you say about∫ 1

−1

f dα?

MAT 5610: 93


Outerlude, III

MAT 5610: 94


RSI: Algebra

Theorem (RS Integration is Bilinear)

Let f and g ∈ R(α) on [a, b] and c ∈ R. Then

1.∫ b

a

(cf) dα = c

∫ b

a

f dα and∫ b

a

f d(cα) = c

∫ b

a

f dα

2.∫ b

a

(f + g) dα=

∫ b

a

f dα+

∫ b

a

g dα and∫ b

a

f d(α+ β)=

∫ b

a

f dα+

∫ b

a

f dβ

3. f · g ∈ R(α) (NB: f and f · g ∈ R(α) 6⇒ g ∈ R. Eg? )

Theorem (RS Integration is Ordered)If f and g ∈ R(α) on [a, b] and f(x) ≤ g(x), then

1.∫ b

a

f dα ≤∫ b

a

g dα

2.∣∣∣∣∫ b

a

f dα

∣∣∣∣ ≤ ∫ b

a

|f | dα

MAT 5610: 95


Making a Massive Point

DefinitionDefine the shifted Heaviside function to be

Ux0(x) =

{0 x ≤ x0

1 x > x0

(Note: Ux0(x) = U(x− x0).)

TheoremIf f is bounded on [a, b] and continuous at x0 ∈ (a, b), then

∫ b

a

f dUx0 = f(x0)

MAT 5610: 96


Wow

TheoremLet

∑n

cn be a convergent sequence of positive terms and {xn} be a

sequence of distinct points in (a, b). Set α(x) =

∞∑k=1

ck Uxk (x). If f is

continuous on [a, b]. Then ∫ b

a

f dα =∞∑k=1

ckf(xk)

ExampleLet ck = 1/k2 and xk = (k − 1)/k. Set α =

∑n cn Uxn . Then∫ 1

0

1

1 + xdα =

∞∑k=1

1

k2

1

1 + (k − 1)/k=

∞∑k=1

1

k(2k − 1)= ln(4)

(Note:∫ 1

0

1/(1 + x) dx = ln(2).)

MAT 5610: 97


Wow2

TheoremIf f is bounded and α′ ∈ R(α), then f ∈ R(α) iff fα′ ∈ R and∫ b

a

f dα =

∫ b

a

f · α′ dx

Example

Compute∫ 1

−1

5(x2 + 1) d(x3 + 1).

∫ 1

−1

5(x2 + 1) d(x3 + 1)=

∫ 1

−1

5(x2 + 1)(3x2) dx=15

∫ 1

−1

x4 + x2 dx = 16

For a ‘reasonable’ α, separate∫fdα into an integral and a sum as∫ b

a

fdα =

∫ b

a

fα′dx+∑n

Jα(xn)f(xn)

where Jα(xn) = α(xn+)− α(xn−) for the ‘mass points’ xn of α.

http://mathsci2.appstate.edu/~wmcb/Class/5610/Discontinuities.pdf

MAT 5610: 98


RSI: Last Word

Theorem (Holder’s Inequality)Let f and g be in R(α) and let p, q > 0 be such that 1/p+ 1/q = 1.Then ∣∣∣∣∣

∫ b

a

fg dα

∣∣∣∣∣ ≤[∫ b

a

|f |pdα]1/p [∫ b

a

|g|qdα]1/q

For p = 2, this is the Cauchy-Bunyakovski-Schwarz inequality.

Theorem (Minkowski’s Inequality)Let p > 1 and let fp and gp be in R(α). Then

[∫ b

a

|f + g|p dα]1/p

≤[∫ b

a

|f |pdα]1/p

+

[∫ b

a

|g|pdα]1/p

MAT 5610: 99


Exercises, III

Exercises1. Compute

1.1∫ 7/2

0

x d[x]

1.2∫ 2

0

x d[x]

2. Set α(x) = x+ bxc. Find the value of∫ 3

0

exdα.

3. Prove the “RS Integration is Bilinear theorem” (#86)

4. State and prove analogs of the First and Second Mean ValueTheorems for Integration for Riemann-Stieltjes Integrals.

5. “Write a homework problem suitable for this section and solve it.”

MAT 5610: 100


Interlude

MAT 5610: 101


The Gamma Function

Γ(x) =

� ∞

0

tx−1e−tdt

y = 1/Γ(x)

MAT 5610: 102


Who? What? Where? When?

Leonhard Euler at 22 (in letters to Goldbach in 1729−30) briefly described the gamma function. Then Eu-ler published the details in De progressionibus tran-scendentibus seu quarum termini generales alge-braice dari nequeunt.† Euler merged an interpola-tion problem (a generating function for the triangularnumbers) with integration. Euler discovered that[(

2

1

)n1

n+ 1

] [(3

2

)n2

n+ 2

] [(4

3

)n3

n+ 3

]· · · = n!

and used interpolation via integration to extend thefunction to noninteger values. He called it C(x); laterLegendre named it Γ(x).

†On transcendental progressions whose general term cannot be expressedalgebraically


http://www-gap.dcs.st-and.ac.uk/~history/Biographies/Legendre.html

http://www.ams.org/notices/200911/rtx091101440p.pdf

MAT 5610: 103


Defining Gamma

Definition (Euler’s Gamma Integral (1730))

If x is not a negative integer, define Γ(x) =

∫ ∞0

tx−1e−tdt.

Examples

1. Γ(1) =

∫ ∞0

e−tdt. So Γ(1) = 1.

2. Γ(2) =

∫ ∞0

t e−tdt. So Γ(2) = − (t+ 1)e−t∣∣∣∞0

= 1.

3. Γ(x+ 1) =

∫ ∞0

tx e−tdt. So

Γ(x+ 1) = txe−t∣∣∣∣∞0

+ x

∫ ∞0

tx−1e−tdt = xΓ(x)

4. Γ(n+ 1) = nΓ(n) = n(n− 1)Γ(n− 1) = n(n− 1)(n− 2)Γ(n− 2)= n(n− 1)(n− 2)(n− 3)Γ(n− 3) = · · · = n!

MAT 5610: 104


Gamma’s Rays

Theorem (Weierstrass’s Product Formula)

Γ(x) =e−γx

x

∞∏

n=1

(1 +

x

n

)−1

ex/n

where γ ≈ 0.5772156649 is the Euler-Mascheroni constant.

Theorem (Euler’s Reflection Formula)

Γ(x) · Γ(1− x) =π

sin(πx)

Corollary

Γ

(1

2

)= (−0.5)! =

√π and Γ

(3

2

)= 0.5 ! =

√π

2

http://www.enotes.com/effects-gamma

http://mathworld.wolfram.com/Euler-MascheroniConstant.html

MAT 5610: 105


Gamma’s Mirror

Proof (Euler’s Reflection Formula).

1.1

Γ(x)= x eγx

∞∏

n=1

(1 +

x

n

)e−x/n

2.1

Γ(x)· 1

Γ(−x)= −x2eγxe−γx

∞∏

n=1

(1− x2

n2

)e−x/nex/n

= −x2∞∏

n=1

(1− x2

n2

)

3. Since Γ(1− x) = (−x)Γ(−x), then

1

Γ(x)· 1

Γ(−x)= x

∞∏

n=1

(1− x2

n2

)

4. Euler had previously shown the product on the right equal tosin(πx)/π (in his solution of the Basel Problem.)

MAT 5610: 106


Almost Gamma

Theorem (Stirling’s Approximation)

Γ(x+ 1) ≈(xe

)x√2πx

Corollary

n! ≈ (n/e)n√

2πn

Theorem (Ramanujan’s Double Inequality)

[8x3 + 4x2 + x+

1

100

]1/6

<Γ(x+ 1)

(x/e)x√π<

[8x3 + 4x2 + x+

1

30

]1/6

http://www-groups.dcs.st-and.ac.uk/~history/Biographies/Ramanujan.html

MAT 5610: 107


Project

Exercises

1. Show√

2

2Γ

(1

4

)Γ

(3

4

)= π.

2. Verify the identities√

3

2

3∏k=1

Γ

(k

3

)= π and

27

8

6∏k=1

Γ

(k

3

)= π2

Project (Spring Break Project for the Intrepid Student)1. Problem #11426, Am. Math. Monthly, Vol.116, April, 2009:

Γ(

114

)Γ(

914

)Γ(

1114

)Γ(

1314

)Γ(

514

)Γ(

314

) = 2

2. For some rational r(n) = p/q,n∏

k=1gcd(k,n)=1

Γ

(k

n

)=p

q· π

MAT 5610: 108


Interlude: An Application of Integration

MAT 5610: 109


Fourier’s Problem

J B J Fourier had an interesting life. He was arrestedat times as a revolutionary, others as a counter-revolutionary. He taught at the Ecole Polytechnique,then served as Napolean’s scientific adviser in theinvasion/occupation of Egypt. He retuned to Franceand the Ecole in 1801. Then Napolean appointedhim as Prefect of the Department of Isere. (Fourierwas not happy, but couldn’t refuse.) He presentedOn the Propagation of Heat in Solid Bodies in 1807.His theory was controversial. However, in 1822, Fourier, was awardeda prize for the second version of this work. (In it, he introduced thenotation

∫ ba

.) Fourier, who wasn’t overly modest, wrote in his preface

The theory of heat will hereafter form one of the mostimportant branches of general physics.

http://www-gap.dcs.st-and.ac.uk/~history/Biographies/Fourier.html

MAT 5610: 110


Defining Fourier’s SeriesThe main tool Fourier introduced was trigonometric series.

Theorem (Fourier Series of f )

f(x) =a0

2+

∞∑k=1

ak cos(kx) + bk sin(kx) −→Dirichlet

f(x+) + f(x−)

2

where

ak =1

π

∫ π

−πf(x) cos(kx) dx and bk =

1

π

∫ π

−πf(x) sin(kx) dx

The definition depends on the following identities. Suppose m,n ∈ N, then

1.∫ π

−πcos(mx) cos(nx) dx =

0 m 6= n

π m = n 6= 0

2π m = n = 0

2.∫ π

−πcos(mx) sin(nx) dx = 0

3.∫ π

−πsin(mx) sin(nx) dx =

0 m 6= n

π m = n 6= 0

0 m = n = 0

MAT 5610: 111


A Square Example

-4 -3 -2 -1 0 1 2 3 4

-1.2

-0.8

-0.4

0.4

0.8

1.2

-4 -3 -2 -1 0 1 2 3 4

-1.2

-0.8

-0.4

0.4

0.8

1.2

n = 10 n = 20

-4 -3 -2 -1 0 1 2 3 4

-1.2

-0.8

-0.4

0.4

0.8

1.2

-4 -3 -2 -1 0 1 2 3 4

-1.2

-0.8

-0.4

0.4

0.8

1.2

n = 30 n = 40

f(x) = sign(sin(2x)) =⇒ SN (x) =4

π

b(N+2)/4c∑k=1

sin(2(2k − 1)x)

2k − 1

MAT 5610: 112


Selected Fourier Series

Function on [−π, π] Fourier Series

f(x) = x 2

∞∑k=1

(−1)k+1 sin(kx)

k

f(x) = |x| π

2− 4

π

∞∑k=1

cos((2k − 1)x)

(2k − 1)2

f(x) = x2 π2

3+ 4

∞∑k=1

(−1)kcos(kx)

k2

f(x) = sin2(x)1

2− 1

2cos(2x)

f(x) = | sin(x)| π

2− 4

π

∞∑k=1

cos(2kx)

4k2 − 1

f(x) =

2π − 4|x|

π2|x| < π/2

0 otherwise

1

2π+

8

π3

∞∑k=1

1− cos(kπ/2)

k2cos(kx)

MAT 5610: 113


Selected Graphs

S10(f) for ...

-4 -3 -2 -1 0 1 2 3 4

-2.5

2.5

-4 -3 -2 -1 0 1 2 3 4

0.25

0.5

0.75

f(x) = |x| f(x) = triangle pulse

-4 -3 -2 -1 0 1 2 3 4

2.5

5

7.5

10

-4 -3 -2 -1 0 1 2 3 4

-2.5

2.5

f(x) = x2 f(x) = x

MAT 5610: 114


Fourier Spectrum

Definition (Spectrum of a Fourier Series)The spectrum φ of a series is the amplitude of the harmonics. Thus

φ(k) =√a2k + b2k.

(Also called line spectrum, power spectrum, etc.)

The spectrum indicates the amount of “energy” or “power” of a givenharmonic in the Fourier series. (Think of an audio signal.)

Example

Suppose that f(x) = 2

∞∑

k=1

(−1)k+1 sin(kx)

k. Then φ(k) = 2/k.

The line spectrum is a plot of φ v k.

MAT 5610: 115


Line Spectrum

ExampleConsider the square wave f(x) = sign(sin(2x)). Then

ak = 0 and bk =4

π·{

12k−1 k = 2 mod 4

0 otherwise

Whereupon f ’s line spectrum is

MAT 5610: 116


The Application

A line spectrum lets us compare instruments or assists with creatingelectronically generated music. Other applications include measuringflow and pressure in arteries, measuring biological morphology, etc.

Brown’s‡ spectrum analysis of the opening chord in the Beatle’s “AHard Day’s Night,” solved the ‘what instruments were played’ mystery.‡“Mathematics, Physics and A Hard Day’s Night,” Jason Brown, 2004.

http://mathsci2.appstate.edu/~wmcb/Class/5610/HardDaysNightOpeningChord.m4v

http://en.wikipedia.org/wiki/A_Hard_Day's_Night_(song)#Opening_chord

http://www.mscs.dal.ca/~brown/n-oct04-harddayjib.pdf

MAT 5610: 117


Outerlude

MAT 5610: 118


Seriesly Fun

Definition• A series is a sequence of partial sums with the nth term given bySn =

∑nk=p ak for some fixed p ∈ Z

• A sequence {bn}n≥p can be written as a series Sn =∑nk=p ak by

setting an = bn − bn−1 and bp−1 = 0.

Examples

1.∞∑

k=1

1

k2=π2

6

2.∞∑

k=0

1

k!= e

3.∞∑k=1

1

k(k + 1)=1=

∞∑k=2

1

k(k − 1)

4.∞∑

k=1

(−1)k+1

k= ln(2)

5.∞∑

k=1

(−1)k+1 sin(k)

k=

arctan

(sin(1)

1 + cos(1)

)

MAT 5610: 119


Harmonious Series

RECALL:

DefinitionA series

∑∞k=p ak converges to A ∈ R iff for every ε > 0 there is an

n∗ ∈ N s.t. whenever n ≥ n∗, then |Sn −A| < ε.

Proposition• If

∑ak converges, then ak → 0.

• If ak 6→ 0, then∑ak diverges.

Example (Nicole Oresme (c.1360))The harmonic series

∑1/k diverges.

Take segments that are greater than 1/2 to show S2n ≥ 1 + n/2.

http://www-gap.dcs.st-and.ac.uk/~history/Biographies/Oresme.html

MAT 5610: 120


A Critical Series

PropositionThe geometric series

∑a rk converges iff the constant |r| < 1, then∞∑

k=0

a rk =a

1− r .

Proof.1. If |r| ≥ 1, then a rk 6→ 0. Whence

∑a rk diverges.

2. Let |r| < 1 and n > 1. Then Sn − rSn = a− a rn+1. Hence

Sn =a(1− rn+1)

1− r

3. Since |r| < 1, then rn → 0. Thence Sn converges.

What about∑

sin(k)k?

MAT 5610: 121


A Geometric Sample

ExampleWrite 0.1011 as a rational number.

1. 0.1011 = 1011 · 10−4 + 1011 ·(10−4

)2+ 1011 ·

(10−4

)3+ · · ·

2. =

∞∑

k=1

1011 ·(10−4

)k

3. = 1011

(−1 +

∞∑

k=0

(10−4

)k)

4. = 1011

(−1 +

1

1− 10−4

)

5. = 1011

(−1 +

10000

9999

)= 1011

(1

9999

)=

337

3333

MAT 5610: 122


A Maple Sample

Maple Summing

How does Maple find that∞∑

k=1

(−1)k+1 sin(k)

k= arctan

(sin(1)

1 + cos(1)

)?

1. First, enter> infolevel[sum]:=5:

to raise the level of “user information” printed.

2. Now enter the summation

>∞∑

k=1

(−1)k+1

ksin(k)

and wait for the results.∗

∗Plan B.

http://mathsci2.appstate.edu/~wmcb/Class/5610/MapleSumEg.pdf

MAT 5610: 123


Cauchy and Friends

Theorem (Cauchy Criterion for Series)∑ak converges iff for each ε > 0 there is an n∗ ∈ N s.t. whenever

n ≥ m ≥ n∗, then

|Sn − Sm−1| = |am + am+1 + am+2 + · · ·+ an| < ε

Proof.X (Series are sequences; cf. Cauchy’s Criterion for Sequences.)

Corollary1. If ak = bk for k ≥ k∗, and

∑bk converges, then

∑ak converges.

2. If∑ |ak| converges, then

∑ak converges.

3. If∑ak diverges, then

∑ |ak| diverges

MAT 5610: 124


Convergence Tests, I

Theorem (The Integral Test)If f : [1,∞)→ R is continuous, decreasing, and nonnegative, then

∞∑

k=1

f(k) converges iff∫ ∞

1

f(x) dx converges

Proof.Let ak = f(k). Since f is decr and cont, ak+1 ≤ f(x) ≤ ak for k ≥ 1.Whence

ak+1 ≤∫ k+1

k

f(x) dx ≤ ak

ThereforeSk+1 − a1 ≤

∫ k+1

1

f(x) dx ≤ Sk

The series bounds the integral, the integral bounds the series. Sinceall terms are nonneg, Bolzano-Weierstrass Thm gives the result.

MAT 5610: 125


Integral Test Figure

∞∑k=2

ak ≤∫ ∞

1

f(x) dx

∫ ∞1

f(x) dx ≤∞∑k=1

ak

So∫ ∞

2

f(x) dx ≤∞∑

k=2

ak ≤∫ ∞

1

f(x) dx ≤∞∑

k=1

ak

MAT 5610: 126


Convergence Tests, II

Theorem (The Comparison Test)Suppose

∑ak and

∑bk are two series with bk ≥ 0 for all k.

1. If |ak| ≤Mbk eventually (for some M ∈ R+) and∑bk converges,

then∑ |ak| and

∑ak both converge.

2. If mbk ≤ |ak| eventually (for some m ∈ R+) and∑bk diverges,

then∑ |ak| diverges.∗

Proof.1. Wolog |ak| ≤Mbk for all k. Since

∑bk converges, so does∑

Mbk, say to S. Then

0 ≤N∑

k=1

|ak| ≤N∑

k=1

Mbk ≤∞∑

k=1

Mbk = S

Whence∑ |ak| is a bounded, increasing series, thus convergent.

∗ ∑ ak may or may not diverge; e.g., ak = (−1)k+1/k and bk = 1/k.

MAT 5610: 127


A Useful Series Set

Proposition (p-Series∗)

The real series ζ(p) =

∞∑k=1

1

kpconverges iff p > 1.

Proof.X (Apply the integral test.)

Examples

1.∞∑k=1

1/k =∞

2.∞∑k=1

1/k2 = π2/6

3.∞∑k=1

1/k3 = ζ(3)

4.∞∑k=1

1/k4 = π4/90

5.∞∑k=1

1/k5 = ζ(5)

6.∞∑k=1

1/k6 = π6/945

∗See the Riemann Zeta function. Also $

http://mathworld.wolfram.com/RiemannZetaFunction.html

http://www.claymath.org/millennium/Riemann_Hypothesis/

MAT 5610: 128


Convergence Tests, III

Theorem (The Limit Comparison Test)Suppose

∑ak and

∑bk are two positive-term series.

1. If limn→∞

anbn

is finite and∑bn converges, then

∑an converges.

2. If limn→∞

bnan

is finite and∑bn diverges, then

∑an diverges.

Proof.Set L = lim an/bn.

1. If L = 0, then ∃M ∈ R+ s.t. an ≤Mbn. Now compare.

2. If L > 0, then 12L bn ≤ an ≤ 3

2L bn eventually. Again compare.

Corollary

Suppose∑ak &

∑bk are positive-term series, and lim

n→∞

anbn

= L with

0 < L <∞. Then∑ak &

∑bk converge or diverge together.

MAT 5610: 129


Fun Time

Exercises

1.∞∑k=1

(−1)kk

(k + 1) ek

2.∞∑n=1

102n

(2n− 1)!

3.∞∑j=1

3j

j3

4.∞∑i=1

(−1)i 23i

32i

5.∞∑m=1

(√

5− 1)m

m2 + 1

6.∞∑k=1

k!

kk

7.∞∑n=1

(−1)n−1

n2 + 1

8.∞∑j=2

(−1)j

j ln(j)

9.∞∑i=1

(−1)nn

n2 + 1

10.∞∑m=1

(−1)mm3

(m2 + 1)4/3

11. 13

+ 1·43·6 + 1·4·7

3·6·9 + · · ·

12. 29

+ 2·59·12

+ 2·5·89·12·15

+

13.∞∑k=1

k + 2

(k+1)√k+3

Challenge:∞∑k=1

(a+ kb)rk =a− (a− b)r

(1− r)2; |r| < 1

(cacdcadcac2c2cddcd)

http://mathsci2.appstate.edu/~wmcb/Class/5610/SeriesFunTime1/SeriesFunTime.html

MAT 5610: 130


Convergence Tests, IV

Theorem (Ratio Comparison Test)Let

∑an &

∑bn be positive term series. If an+1/an ≤ bn+1/bn eventually,

and∑bn converges, then

∑an converges.

Theorem (D’Alembert’s Ratio Test)Let

∑an be a series with positive terms. If there is a constant α ∈ [0, 1) s.t.

an+1/an ≤ α eventually, then∑an converges. If an+1/an ≥ 1 eventually,

then∑an diverges.

Corollary (Cauchy’s Ratio Test)Let

∑an be a series with positive terms. Set r = limn→∞

an+1

an. Then:

• If r < 1, the series converges.

• If r > 1, the series diverges.

• If r = 1, the test fails.

MAT 5610: 131


Convergence Proofs

Proof (Ratio Comparison Test).1. an/bn is decreasing eventually, thus bounded. Thus an ≤Mbn.2. Since

∑bn converges, then

∑an converges by comparison.

Proof (D’Alembert’s Ratio Test).1. α < 1 gives

∑αn converges.

2. Then an+1/an ≤ αn+1/αn = α implies that∑an converges.

Proof (Cauchy’s Ratio Test).X

MAT 5610: 132


Convergence Tests, V

Theorem (Root Test)Let

∑an be a series with positive terms. If ∃α ∈ [0, 1) s.t. n

√an ≤ α

eventually, then∑an converges. If n

√an > 1 eventually, then

∑an diverges.

Theorem (Cauchy’s Root Test)Let

∑an be a series with positive terms. Set ρ = lim

n→∞n√an. Then:

• If ρ < 1, the series converges.

• If ρ > 1, the series diverges.

• If ρ = 1, the test fails.

Proof.1. n√an ≤ ρ < 1 implies an ≤ ρn < 1.

2. Since ρ < 1, then∑ρn converges. By comparison,

∑an converges.

MAT 5610: 133


Convergence Tests, VI

Theorem (Raabe’s Test∗)

Let∑an be a series with lim

n→∞

∣∣∣an+1

an

∣∣∣ = 1. Set ρ = limn→∞

n[1−∣∣∣an+1

an

∣∣∣].

• If ρ < 1, the series diverges.• If ρ > 1, the series converges absolutely.• If ρ = 1, the test fails.

Example

Consider the series S = 1− 1

2−∞∑

k=2

1 · 3 · · · (2n− 3)

2n n!

1.∣∣∣∣an+1

an

∣∣∣∣ =2n− 1

2n+ 2

2. n[1−∣∣∣∣an+1

an

∣∣∣∣]

=3n

2n+ 2→ 3

2

∗Raabe’s Test (1832) is a special case of Kummer’s Test (1835)

http://mathsci2.appstate.edu/~wmcb/Class/5610/RaabeKummer.pdf

http://mathworld.wolfram.com/KummersTest.html

MAT 5610: 134


Challenge Prize Problem

Ready. Set. Go. . .What conditions are needed on α, β, and γ (in addition to γ 6= 0, −1,−2, . . . ) so that the hypergeometric function

2F1(α, β; γ; z) =

∞∑

k=0

(α)k(β)k(γ)k

· zk

k!

=

∞∑

k=0

α(α+ 1) · · · (α+ k − 1) · β(β + 1) · · · (β + k − 1)

γ(γ + 1) · · · (γ + k − 1)· z

k

k!

converges at z = 1.

(See Gauss’s Hypergeometric Theorem.)

http://mathworld.wolfram.com/HypergeometricFunction.html

http://mathworld.wolfram.com/GausssHypergeometricTheorem.html

MAT 5610: 135


Converge? Absolutely!

Definition• A series of the form

∑(−1)kak or

∑(−1)k+1ak with ak ≥ 0 is an

alternating series.• If

∑ |ak| converges, then∑ak converges absolutely.

• If∑ |ak| diverges but

∑ak converges, then

∑ak converges

conditionally.

Theorem (Leibniz’s Alternating Series Test)Let an ≥ 0 for all n. If 1. an is eventually decreasing, and 2. an → 0,then

∑(−1)k+1ak converges. (N.B. |S − Sn| < an+1.)

Examples

1.∑

1k =∞;

∑ (−1)k+1

k = ln(2) 2.∑

1k2 = π2

6 ;∑ (−1)k+1

k2 = π2

12

Fun Time

MAT 5610: 136


Rearranging the Terms

Theorem• If

∑ak converges to S absolutely and

∑bk is a rearrangement of

∑ak,

then∑bk converges to S absolutely.

• If∑ak converges to S conditionally and α ∈ [−∞,+∞], then there is a

rearrangement∑bk that converges to α.

• If∑ak converges to S conditionally and α, β ∈ R, then there is a

rearrangement∑bk that oscillates between α and β.

Example∞∑k=1

(−1)k+1

k= ln(2), conditionally convergent. Choose α ∈ R.

1. Add positive terms until 1 + 1/3 + · · ·+ 1/(2n1 − 1) first exceeds α

2. Add negative terms until 1 + · · ·+ 1/(2n1 − 1)− 1/2− · · · − 1/(2n2) firstis less than α.

3. Rinse and repeat.

MAT 5610: 137


Outerlude

MAT 5610: 138


Functioning Sequences & Series

Definition (Pointwise Convergence)A sequence of fcns {fn} converges pointwise to f on D ⊆ R iff foreach xo ∈ D, the sequence fn(x0)→ f(x0). In symbols:

fn→f ⇐⇒ (∀ε>0)(∀x0∈D)(∃n∗∈N)(∀n>n∗)[|fn(x0)− f(x0)| < ε

]

Examples

1. fn(x) = xn → f(x) =

{0 |x| < 1

1 x = 1on D = (−1, 1]. ((−1)n div.)

2. gn(x) = 1n cos(nx)→ 0 on D = R

3. hn(x) = sin(nx)→ 0 on D = {kπ | k ∈ Z} and div otherwise.

4. jn(x) = 2π arctan(nx)→ signum(x) on D = R

MAT 5610: 139


Problems

ExamplesContinuity: Both xn and 2

π arctan(nx) are continuous fcns that havediscontinuous limits

limn→∞

limx→a

fn(x) 6= limx→a

limn→∞

fn(x)

Differentiability: fn(x) = 1n sin(nx)→ 0, but f ′n(x) = cos(nx) diverges

for all x 6= 0

d

dxlimn→∞

fn(x) 6= limn→∞

d

dxfn(x)

Integrability: nxn → 0 on [0, 1), but∫ 1

0nxn dx→ 1

∫ b

a

limn→∞

fn(x) dx 6= limn→∞

∫ b

a

fn(x) dx

MAT 5610: 140


Interlude: L2

Definition (L2-norm)For f , define the mean-square or L2-norm on [a, b] to be

‖f‖2 =

[∫ b

a

|f(x)|2 dx]1/2

Definition (Mean-Square Convergence)A seq {fn} converges to f in the mean on [a, b] iff lim

n→∞‖fn − f‖2 = 0.

Examples

1. Let fn(x)= xn on D=[0, 1]. Since ‖xn‖2 =√∫ 1

0x2ndx =1/

√2n+1→0,

thence xn → 0 in the mean on [0, 1].

2. Theorem. Riesz-Fischer (1907).The Fourier series of an L2 fcn f converges to f in the mean.

‡ Fourier Squares

MAT 5610: 141


Uniform Convergence

Definition (Uniform Convergence)A sequence of fcns {fn} converges uniformly to f on D ⊆ R iff forevery ε > 0 there is an n∗ = n∗(ε) ∈ N such that for any x ∈ D, ifn > n∗, then |fn(x)− f(x)| < ε.

Examples

1. fn(x) =xn

1 + nxn→ 0 uniformly on [0, 1]

2. gn(x) = 1√n

cos(nx)→ 0 uniformly on R

3. hn(x) =1

n(x2 + 1)→ 0 uniformly on R (but

1

nx2 + 1doesn’t)

4. jn(x) = xn → f(x) =

{0 |x| < 1

1 x = 1pw, not uniformly, on [0, 1].

MAT 5610: 142


Two Graphs

Uniform Convergence Nonuniform Convergence

MAT 5610: 143


How To...

Proposition

fn(x) =x

nx+ 1converges uniformly to f(x) = 0 on [0,∞).

Proof.1. If x = 0, then fn(0) ≡ 0. Hence |fn(0)− 0| < ε for any ε > 0.

2. If x > 0, then |fn(x)− 0| =∣∣∣∣ x

nx+ 1

∣∣∣∣ < ∣∣∣ xnx ∣∣∣ =1

n< ε for n suffic large.

Propositiongn(x)→0 pw, not unif on [0, 1], where gn(x) = 4n2x for 0 ≤ x < (2n)−1, andgn(x) = 4n(1−nx) for (2n)−1≤x≤n−1, and 0 otherwise.

Proof.Consider gn(1/(2n)) = 2n 6< ε for n sufficiently large.

MAT 5610: 144


Outerlude

MAT 5610: 145


Props

QueryWe know that a function continuous on a compact set is uniformlycontinuous. Does a sequence converging on a compact set convergeuniformly?

Theorem (Cauchy Criterion for Sequences)A sequence of fcns {fn} converges uniformly on D ⊆ R iff for eachε > 0 there is an n∗ s.t. for all x ∈ D, if n,m ≥ n∗, then|fn(x)− fm(x)| < ε.

Proof.1. Build an f by considering fn(x) for each x ∈ D2. 4 shows fn → f uniformly

MAT 5610: 146


Interlude: L∞

Definition (L∞-norm)For f , define the sup- or L∞-norm on [a, b] to be

‖f‖∞ = supx∈[a,b]

|f(x)|

��x�1 = 1��x�2 = 1 ��x�∞ = 1

TheoremA sequence fn converges to f in the sup norm on [a, b]; i.e., ‖fn − f‖∞ → 0iff fn converges to f uniformly on [a, b].

Proof.Observe that |fn(x)− f(x)| < ε for all x ∈ [a, b] iff ‖fn − f‖∞ < ε.

MAT 5610: 147


Interlude: L∞ Examples

Example

Consider fn(x) =x

1 + n2x2. (fn → 0 pw on [−1, 1])

L∞: fn → 0 in L∞[−1, 1].

• f ′n(x) =1− n2x2

(1 + n2x2)2 =⇒ f ′n(x) = 0 at x = ±1/n. Thus

−1

2n≤ fn(x) ≤ 1

2n

• So ‖fn − 0‖∞ = 12n→ 0 on [−1, 1] (fn → 0 unif on [−1, 1])

L2: fn → 0 in L2[−1, 1].

•√∫ 1

−1|fn(x)|2dx =

√arctan(n)

n3− 1

n2(n2 + 1)→ 0.

• Thus ‖fn − 0‖2 → 0 on [−1, 1]

MAT 5610: 148


Interlude II: Extra for Experts

Theorem (Dini’s Theorem (1878))Suppose {fn(x)} is a sequence of functions converging to f onD ⊆ R. If

1. each fn is continuous on D,2. f is continuous on D,3. fn → f monotonically, and4. D is compact (on R: compact = closed and bounded),

then fn converges to f uniformly on D.

1. fn(x) = 1 on 0 < x < 1/n and 0 otherwise converges to 0 onD = [0, 1] (fn is not continuous)

2. xn → U(x− 1) on [0, 1] (f = U is not continuous)3. “Marching triangles”→ 0 on [0, 1] (convergence is not monotonic)4. xn → 0 on [0, 1) ([0, 1) is not compact)

http://www-gap.dcs.st-and.ac.uk/~history/Biographies/Dini.html

http://en.wikipedia.org/wiki/Dini%27s_theorem

MAT 5610: 149


Props to UniCon

TheoremIf a sequence of continuous functions {fn} converges uniformly on D,the limit function f is continuous on D.

Proof.The proof hinges on the inequalities

|f(x)− f(x0)| ≤ |f(x)−fn1(x)|︸︷︷︸n1>n∗

+ |fn1(x)−fn1(x0)|︸︷︷︸|x−x0|<δ

+ |fn1(x0)−f(x0)|︸︷︷︸n1>n∗

<ε

3+ε

3+ε

3= ε

for some n1 > n∗ and all x ∈ B(x0, δ).

Converse? No — the “marching triangles”.

MAT 5610: 150


More Props to UniCon

TheoremIf a sequence of continuous functions {fn} converges uniformly to fon D = [a, b], then

limn→∞

[∫ b

a

fn(x) dx

]=

∫ b

a

[limn→∞

fn(x)]dx.

I.e., if Fn =∫ bafn and F =

∫ baf , then Fn → F.

Proof.The proof hinges on the inequalities∣∣∣∣∣

∫ b

a

fn(x) dx−∫ b

a

f(x) dx

∣∣∣∣∣ =

∣∣∣∣∣

∫ b

a

[fn(x)−f(x)] dx

∣∣∣∣∣

≤∫ b

a

|fn(x)−f(x)| dx <∫ b

a

ε dx = O(ε)

MAT 5610: 151


Yet Even More Props to UniCon

TheoremSuppose {fn} is a sequence of continuously∗ differentiable functionsconverging pointwise to f on [a, b]. If the sequence of derivatives {f ′n}converges uniformly on [a, b], then f is continuously differentiable and• fn → f uniformly on [a, b], and• f ′n → f ′ uniformly on [a, b].

Proof.The proof is based on three ideas:

1.∫ xaf ′n(t) dt = fn(x)− fn(a)

2. f(x)−f(a)= limn→∞

[fn(x)−fn(a)] = limn→∞

∫ xaf ′n(t) =

∫ xa

[limn→∞

f ′n(t)]

3. f ′(x) = ddx

∫ xaf ′(t) dt

∗‘Continuously’ can be relaxed. ‘Pointwise’ can be relaxed to a single point.

MAT 5610: 152


A Sample Case

Exercises

1.{

n√

sin(x)}

converges uniformly on [ε, 12π] for ε > 0, but pw on (0, 1

2π].

2. Determine limn→∞

∫ a

−ae−nx

2

dx where a > 0.

3. Set Sn(x) = nxe−nx2

. What does Sn converge to? Show

limn→∞

∫ a

0

Sn(x) dx 6=∫ a

0

limn→∞

Sn(x) dx

4. Let fn(x) =x

1− nx2for |x| ≤ 1.

4.1 What is f(x) = limn→∞

fn(x)?

4.2 Is the convergence uniform?

4.3 Is f ′(x) = limn→∞

f ′n(x)? Explain.

MAT 5610: 153


Series

Definition

Function Series: The sequence of partial sums Sn(x) =n∑k=1

fk(x)

Series Conv PW: The series converges pointwise on D if {Sn} convergespointwise on D

Series Conv Unif: The series converges uniformly on D if {Sn} convergesuniformly on D

Series Conv Abs: The series converges absolutely on D if {Sn} convergesabsolutely on D

Series Representation: A function f has a series representation on Diff ∃Sn → f on D

TheoremIf a series

∑k fk(x) of continuous functions converges uniformly to S(x),

then S is continuous.

MAT 5610: 154


Samples

Exercise

1. S(x) =

∞∑

k=1

x2

(1 + x2)k

• S(0) = 0

• 0 <1

(1 + x2)< 1 for x 6= 0

• Set a = x2 and r = 1/(1 + x2), then use∞∑k=0

ark = a/(1− r).

Whence for x 6= 0,

S(x) =∞∑k=1

x2

(1 + x2)k=

x2

1− 11+x2

− x2 = 1

• Thus S(x) =

{1 x 6= 0

0 x = 0.

• Since S is not continuous, but each fn is, the convergence ispointwise and absolutely.

MAT 5610: 155


Convergence Modes

Pointwise convergence

Absoluteconvergence

Uniformconvergence

Cesàroconvergence

Cesaro Convergence: Let Sn(x) =n∑k=0

fn(x) be a series. Then

SnCesaro−→ S(x) iff

1

n+ 1(S0 + S1 + · · ·+ Sn)→ S

E.g.,∞∑n=0

(−1)nCesaro−→ 1

2

MAT 5610: 156


Two Big Theorems

Theorem (Cauchy Criterion)The series S(x) =

∑fn(x) converges uniformly on D iff for any ε > 0

there is an n∗ ∈ N s.t. whenever m,n ≥ n∗, then for all x ∈ D,

|Sn(x)− Sm(x)| =∣∣∣∣∣

n∑

k=m+1

fn(x)

∣∣∣∣∣ < ε

Theorem (Weierstrass M -Test)Suppose that |fn(x)| ≤Mn for all x ∈ D and for each n. If

∑Mn

converges, then∑fn(x) converges uniformly and absolutely on D.

Example

What can be said about∞∑

k=1

(−1)k+1 sin(k4x)

k2?

MAT 5610: 157


The Example

-5 -4 -3 -2 -1 0 1 2 3 4 5

-1.2

-0.8

-0.4

0.4

0.8

1.2

-5 -4 -3 -2 -1 0 1 2 3 4 5

-1.2

-0.8

-0.4

0.4

0.8

1.2

n = 25

-5 -4 -3 -2 -1 0 1 2 3 4 5

-16

-8

8

16

-5 -4 -3 -2 -1 0 1 2 3 4 5

-16

-8

8

16

d

dx

∞∑

k=1

(−1)k+1 sin(k4x)

k26=∞∑

k=1

(−1)k+1 d

dx

sin(k4x)

k2

MAT 5610: 158


Integrating Series

Theorem

If Sn(x) =n∑fn(x) is integrable on [a, b] for each n ∈ N and Sn → S

uniformly, then ∫ b

a

[n∑

k=0

fn(x)

]dx =

n∑

k=0

[∫ b

a

fn(x) dx

]

Example

1

1 + t2=

∞∑

k=0

(−1)kt2k on |t| < 1 and uniformly on |t| ≤ ρ < 1

Whence∫ x

0

1

1 + t2dt =

∞∑

k=0

(−1)k∫ x

0

t2kdt

Wherefore arctan(x) =

∞∑

k=0

(−1)kx2k+1

2k + 1for |x| ≤ ρ < 1

MAT 5610: 159


Arctan? < 1?

arctan(x) =

∞∑

k=0

(−1)kx2k+1

2k + 1for |x| ≤ ρ < 1︸︷︷︸

Why?

x

y

zf(z) = |arctan(z)|

f(z) = arctan(z) has poles at z = ±i

MAT 5610: 160


Differentiating Series

Theorem

Suppose Sn(x) =n∑fn(x) is a series of continuously differentiable

functions converging pointwise to S(x) on [a, b]. If the series of

derivativesn∑f ′n(x) converges uniformly on [a, b], then

• S(x) is continuously differentiable,•∑fn(x)→ S(x) uniformly on [a, b], and

•∑f ′n(x)→ S′(x) uniformly on [a, b].

Example ∑xn =

1

1− x =⇒∑

nxn−1 =1

(1− x)2=⇒∑

n(n− 1)xn−2 =2

(1− x)3=⇒

∑n(n− 1)(n− 2)xn−3 =

3!

(1− x)4

MAT 5610: 161


Fun Day

ProblemI. Where does

∞∑

n=1

n(x− 1)n

2n(3n− 1)

converge?

II. Let a > 0. Investigate

d

dx

∞∑

n=1

xn

n2an?=

∞∑

n=1

d

dx

xn

n2an

III. Set fn(x) = nxe−nx2

andf(x) =

∑n fn(x) on [0, 1].

Investigate∫ 1

0

f(x) dx?=

∞∑

n=0

∫ 1

0

fn(x) dx

IV. Bessel’s function is

Jp(x) =

∞∑

n=0

(−1)n(x/2)2n+p

n! (n+ p)!.

Show that y = Jp(x) solves

x2y′′ + xy′ + (x2 − p2)y = 0.

MAT 5610: 162


Fun Day, II

ProblemV. Is∞∑

n=1

(−1)(n+1)/2 · 1− (−1)n

2

Cesaro-convergent?

VI. Is∞∑

n=1

(−1)n · n− 1

n+ 1

Cesaro-convergent?

VII. Calculate1

f(x)when

f(x) =

∞∑

n=0

xn.

VIII. Calculate1

g(x)when

g(x) =

∞∑

n=0

xn

n!.

MAT 5610: 163


Very Powerful Series

Definition (Power Series)

• A power series is a series of the form∞∑

k=0

ak(x− a)k.

• The radius of convergence is the largest value R ∈ [0,∞] s.t. thepower series converges for all |x− a| < R.

TheoremSuppose

∑k ak(x− a)k converges for x∗ s.t. |x∗ − a| = r0. Then∑

k ak(x− a)k converges absolutely for all x s.t. |x− a| < r0.

Proof.Based on the inequalities |(x− a)/r0| < 1 and

∣∣ak(x− a)k∣∣ ≤

∣∣akrk0∣∣ ·∣∣∣∣x− ar0

∣∣∣∣k

≤M ·∣∣∣∣x− ar0

∣∣∣∣k

MAT 5610: 164


More Power

CorollaryIf∑k ak(x− a)k diverges for x∗, then

∑k ak(x− a)k diverges for all x

s.t. |x− a| ≥ |x∗ − a|.

TheoremLet R be the radius of conv of

∑k ak(x− a)k. Then

∑k ak(x− a)k

1. converges absolutely on |x− a| < R,2. converges uniformly on |x− a| ≤ ρ < R, and3. diverges on |x− a| > R.

• If∑k ak(R)k conv, then

∑k ak(x− a)k conv uniformly on

−R < −ρ ≤ |x− a| ≤ R.

• If∑k ak(−R)k conv, then

∑k ak(x− a)k conv uniformly on

−R ≤ |x− a| ≤ ρ < R.

MAT 5610: 165


More Power

Examples (“Try not. Do, or do not. There is no try.”)

1. sin(x) =

∞∑

k=0

(−1)kx2k+1

(2k + 1)!

2. f(x) =

∞∑

k=0

k!xk

3. ln(1 + x) =

∞∑

k=1

(−1)k+1xk

k

4. ln

[√1 + x

1− x

]=

∞∑

k=1

x2k−1

2k − 1

5. g(x) =

∞∑

k=1

ln(k + 1)√k + 1

(x− 5)k

6.xex

(ex − x)2=

∞∑

k=1

k xke−kx

7.1√

1 + x= 1− 1

2x+

1 · 32 · 4 x

2 − 1 · 3 · 52 · 4 · 6 x

3 +1 · 3 · 5 · 72 · 4 · 6 · 8 x

4 − . . .

8. ln

[x− 3

x− 4

]=

1

x− 3+

1

2(x− 3)2+

1

3(x− 3)3+

1

4(x− 3)4+ . . .

MAT 5610: 166


More Power Solutions, I

Examples

7.1√

1 + x= 1− 1

2x+

1 · 32 · 4 x

2 − 1 · 3 · 52 · 4 · 6 x

3 +1 · 3 · 5 · 72 · 4 · 6 · 8 x

4 − . . .

a.1 · 3 · 52 · 4 · 6 =

1 · 2 · 3 · 4 · 5 · 6(2 · 4 · 6)2

=6!

26 · (3!)2=

6!

43 · (3!)2(n = 3)

b.1 · 3 · 5 · 72 · 4 · 6 · 8 =

8!

28 · (4!)2=

8!

44 · (4!)2(n = 4)

c. an =(−1)n(2n)!

4n(n!)2=

(−1)n

4n·

(2n

n

)

d. rn(x) =

∣∣∣∣ (2(n+ 1))!xn+1

4n+1((n+ 1)!)2· 4n(n!)2

(2n)!xn

∣∣∣∣ =2n+ 1

2n+ 2|x| −→

n→∞|x|

Whence the RoC is 1; i.e., the series converges on |x| < 1.

Testing the endpoints shows the series converges at x = 1 anddiverges at x = −1. (The recursive form of an is easier.)

MAT 5610: 167


More Power Solutions, II

Examples

8. ln

[x− 3

x− 4

]=

1

x− 3+

1

2(x− 3)2+

1

3(x− 3)3+

1

4(x− 3)4+ . . .

a. Let the series be S. Substitute y = 1/(x− 3). Then

S = y +1

2y2 +

1

3y3 +

1

4y4 + . . .

b. Thus an = 1/n.

c. So rn =

∣∣∣∣yn+1/(n+ 1)

yn/n

∣∣∣∣ =n

n+ 1|y| −→

n→∞|y|

d. S converges for |y| < 1. That is, for1

|x− 3| < 1. Or |x− 3| > 1.

Testing the endpoints shows S(4) diverges and S(2) converges.

MAT 5610: 168


Cauchy–Hadamard Theorem

Theorem (Cauchy (1821)–Hadamard (1888) Theorem)

The radius of convergence R of the power series∞∑

k=0

ak(x− a)k is

given by 1

R= lim sup

n→∞n√|an|

Proof.Wolog a = 0. Since

lim sup n√|anxn| = |x| lim sup n

√|an| =

|x|R

the result follows from Cauchy’s root test.

http://www-history.mcs.st-andrews.ac.uk/Biographies/Hadamard.html

MAT 5610: 169


Differentiate and Integrate

Theorem

If the power series∞∑

k=0

ak(x− a)k has RoC R ∈ (0,∞], the power

series∞∑

k=1

k ak(x− a)k−1 and∞∑

k=0

akk + 1

(x− a)k+1

have the same radius of convergence R.

Proof.L TTRAA E!

NB: Differentiating may lose convergence at the endpoints; e.g.,∑

1n x

n

MAT 5610: 170


Unicity

Theorem (Unicity of Power Series Representation)Suppose f(x) =

∑n an(x− a)n and f(x) =

∑n bn(x− a)n on some

interval |x− a| < R where R > 0. Then an = bn for all n ∈ N.

Proof.Setting x = a shows that a0 = b0. Differentiate both series, then setx = a again. Now we have a1 = b1. (Do it again!) Induction showsan = bn for all n. Actually, we can show an = f (n)(a)/n!.

Examplesin(x) = x− 1

6x3 + 1

120x5 − 1

5040x7 + 1

362880x9 + . . .

sin(x) = −(x− π) + 16 (x− π)

3 − 1120 (x− π)

5+ 1

5040 (x− π)7 − . . .

a0 = b0 =∑∞k=1

(−1)k+1

(2k−1)! π2k−1, a1 = b1 =

∑∞k=0

(−1)k

(2k)! π2k, . . .

MAT 5610: 171


Unicity: Caveat Emptor

-5 -4 -3 -2 -1 0 1 2 3 4 5

-3

-2

-1

1

2

3

T11(x − 1)

T11(x)

f(x) = tan(sin(x))

MAT 5610: 172


Unicity: Caveat Emptor, II

f(x) = tan(sin(x))

T11,0(x) = x+ 0.1667x3 − 0.0250x5 − 0.0212x7 − 0.0030x9 + 0.0011x11

converging on (−1.87, 1.87)

T11,1(x) = 1.119 + 1.217(x− 1)− 0.2122(x− 1)2 − 0.7861(x− 1)3

− 0.6239(x− 1)4 + 0.0625(x− 1)5 + 0.4961(x− 1)6

+ 0.3716(x− 1)7 − 0.0054(x− 1)8 − 0.3161(x− 1)9

− 0.2243(x− 1)10 + 0.0042(x− 1)11

= −0.0910 + 2.045x− 5.126x2 + 15.50x3 − 31.53x4

+ 44.06x5 − 44.75x6 + 30.28x7 − 13.97x8 + 4.241x9

− 0.6887x10 + 0.0435x11

converging on (−0.17, 2.17)

MAT 5610: 173


Taylor Series

Theorem (Taylor Series∗)Suppose f(x) =

∑∞k=0 ak(x− a)k converges with RoC R > 0. Then

• f ∈ C∞(|x− a| < R)

• f (n)(a) = ann! for n = 1, 2, . . .

Example (Bad example! No Biscuit! — Cauchy (1823))

Consider f(x) = e−1/x2 with f(0) = 0.

Since f (n)(0) = 0 for all n, we have the Maclaurin series for f is 0.

Example (Good example! Biscuit!)Consider f(x) = sin(x2).

Since sin(x)=

∞∑k=1

(−1)k+1 x2k−1

(2k − 1)!, then f(x)=

∞∑k=1

(−1)k+1 x2(2k−1)

(2k − 1)!.

‡Taylor (1712). Earlier: Gregory (1671), Leibniz (1670s), Newton (1691), J Bernoulli(1694), de Moivre (1708). Remainder formulas: Cauchy (1822), Lagrange (1772)

MAT 5610: 174


Taylor Series Convergence

Taylor ConvergenceIF

1. f (n)(a) has a “reasonable” form, and2. one of the remainders

Lagrange’s form Rn(x) = f(n+1)(c)(n+1)! (x− a)n+1

Cauchy’s form Rn(x) = f(n+1)(c)n! (x− c)n(x− a)

Integral form Rn(x) = 1n!

∫ xaf (n+1)(t)(x− t)n dt

etc., is “reasonably” easy to find the limit of as n→∞,THEN we can determine a Taylor series and its radius ofconvergence.

MAT 5610: 175


Weierstrass’s Two Stone Caps

Theorem (Weierstrass (1872))Let

f(x) =

∞∑

n=1

bn cos(anx).

Then1. f converges uniformly for b < 1, and2. f is nowhere differentiable when ab > 1 + 3

2 π.

Theorem (Weierstrass (1885))Let f ∈ C([a, b]). For every ε > 0 there is a polynomial pε(x) s.t.

|f(x)− pε(x)| < ε

for all x ∈ [a, b]. This is also written as ‖f − pε‖∞ < ε.

A continuous function f may not have a derivative anywhere, butis still within ε of an infinitely differentiable function everywhere.

MAT 5610: 176


Interlude: Algebra for Analysis

Definition (The Field of Rational Functions on R)The field of rational functions on R is

R(x) =

{p(x)

q(x)

∣∣∣∣ p, q ∈ R[x] and q 6≡ 0

}.

with the usual addition and multiplication.

Definition (Derivation)An operator D on R(x) is a derivation if and only if for any c ∈ R andany α and β ∈ R(x), we have

1. D(x) = 1,2. D(c α) = cD(α) for any constant c,3. D(α+ β) = D(α) +D(β),4. D(α · β) = D(α) · β + α ·D(β) (the Leibniz rule).

MAT 5610: 177


Properties of D

TheoremLet c ∈ R and n ∈ N. Then

1. D(1) = 0

2. D(c) = 0

3. D(xn) = nxn−1

4. D(x−n) = −nx−n−1

Proof.1. Apply the Leibniz rule to 1 = 1 · 12. Use linearity on c = c · 13. Apply the Leibniz rule to xk+1 = xk · x and induct4. Apply the Leibniz rule to xn · x−n = 1

MAT 5610: 178


Chains

Definition (The Chain Rule)For α and β ∈ R(x), define

D(α ◦ β) = (D(α) ◦ β) ·D(β)

TheoremLet p, q ∈ N and α, β ∈ R(x) with β 6≡ 0. Then

1. D(x1/q) = (1/q)x1/q−1

2. D(xp/q) = (p/q)xp/q−1

3. D (α/β) = [D(α) · β − α ·D(β)]/β2

Proof.1. Apply the chain rule to (x1/q)q = x

2. Modified ditto

3. Look at (α/β) · β = α

MAT 5610: 179


Two Extensions to R(x)

Definition• Define the function L by the relation: for any nonzero α ∈ R(x),

D(L ◦ α) =D(α)

α

• Define the function E by the relation: for any nonzero α ∈ R(x),

D(E ◦ α) = E(α) ·D(α)

Exercise1. Determine D(L ◦ α) when α(x) = x.2. Determine D(E ◦ β) when β(x) = x.

DefinitionFor any γ ∈ R(x)[L,E], define

∫γ = Γ iff D(Γ) = γ.

MAT 5610: 180


J-Lio’s Result

Theorem (Liouville’s Principle)If a function f has an elementary integral, then the integral has theform ∫

f = F0 +

n∑

k=1

ckL(Fk)

where the ck ∈ R and the Fk ∈ R(x). The function L is the logarithmdefined above.

Problem

• Does∫xx + xx ln(x)

?=

∫xx +

∫xx ln(x)? No!

Read Risch’s two papers:• “The Problem of Integration in Finite Terms,” Robert H. Risch, Trans

AMS, Vol. 139 (May, 1969), pp. 167–189.• “Algebraic Properties of the Elementary Functions of Analysis,” Robert

H. Risch, Am J Math, Vol. 101, No. 4 (Aug., 1979), pp. 743–759.

http://www-gap.dcs.st-and.ac.uk/~history/Biographies/Liouville.html

http://en.wikipedia.org/wiki/Risch_algorithm

MAT 5610: 181


Bad Function! No Biscuit!

TheoremThere is no elementary antiderivative for f(x) = ex

2

.

Proof.

Suppose F =

∫ex

2

. We know D(ex2

) = ex2

2x. Applying Liouville’s principle

implies F = α(x) ex2

with α ∈ R(x). Thus F ′ = α′ex2

+ α 2x ex2

= ex2

. Soα′ + 2xα = 1. Comparing degrees shows α can’t be a poly, and so must be arational function. Let α = p(x)/q(x). Then q must have a root (possiblycomplex), so write α = h(x)/(x− z0)m where h(z0) 6= 0 is a rational function.Substitute into the DE α′ + 2xα = 1:[

h′(x)

(x− z0)m−m h(x)

(x− z0)m+1

]+ 2x

h(x)

(x− z0)m= 1

Taking the limit of both sides as x→ z0 yields the contradiction∞ = 1.Whence F cannot exist in R(x, L,E).

MAT 5610: 182


The End

MAT 5610, Analysis I

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