math 63: winter 2021 lecture 24

Math 63: Winter 2021Lecture 24

Dana P. Williams

Dartmouth College

Wednesday, March 3, 2021

Dana P. Williams Math 63: Winter 2021 Lecture 24

Getting Started

1 We should be recording.

2 The final will be administered in a manner similar to theprelim and midterm exams. My plan is to release it on thefirst day of the final period, Saturday, March 13, at 8am EST,and require it to be completed by Monday, March 15 at 10pmEST. There will be what I hope is a comfortable window tocomplete the exam. More details later.

3 Time for some questions!


Series of Functions

Remark

Let (fn) be a sequence of real-valued functions on a metric spaceE . We say that the series

∑∞n=1 fn converges pointwise if the series∑∞

n=1 fn(x) is convergent for all x ∈ E .

Then we can let f (x) =∑∞

n=1 fn(x) and say that∑∞

n=1 fnconverges pointwise to f on E . If we let sn =

∑nk=1 fn then∑∞

n=1 fn converges pointwise to f on E exactly when (sn)converges pointwise to f on E .

We say that∑∞

n=1 fn converges absolutely on E is∑∞

n=1 fn(x) isabsolutely convergent for all x ∈ E .

Naturally, we say that∑∞

n=1 fn converges uniformly to f on E if(sn) converges uniformly to f on E .


Example

Example (What are Midterms Good for Anyway?)

Let fn : (−1, 1)→ R be given by fn(x) = xn and consider the series∑∞n=0 x

n on (−1, 1). Then

sn(x) = 1 + x + · · ·+ xn−1 =1− xn

1− x.

As we saw on the midterm,

∞∑n=0

xn =1

1− x

pointwise on (−1, 1).

Clearly the series is absolutely convergent on (−1, 1).

However the convergence is not uniform on (−1, 1).

But it is uniform on [−a, a] provided 0 < a < 1.


Low Hanging Fruit

Proposition

Let∑∞

n=1 fn be a series of real-valued functions on a metric spaceE . Then

∑∞n=1 fn converges uniformly on E if and only if for all

ε > 0 there is a N ∈ N such that m > n ≥ N implies

|fn+1(x) + · · ·+ fm(x)| < ε for all x ∈ E . (∗)

Sketch of the Proof.

The series converges uniformly if and only if the sequence of partialsums (sn) converges uniformly. This happens if and only if (sn) isuniformly Cauchy. This translates immediately into (∗).


Keep Picking

Corollary (Weierstrass M-Test)

Suppose that∑∞

n=1 fn is a series of real valued functions on ametric space E such that |fn(x)| ≤ Mn for all x ∈ E . If∑∞

n=1Mn <∞, then∑∞

n=1 fn converges uniformly and absolutelyon E .


Absolute convergence is immediate by the Comparison Test. Foruniform convergence we appeal to the proposition. Given ε > 0,there is a N such that m > n ≥ N implies Mn+1 + · · ·+ Mm < ε,etc.


Without Proof

Proposition

If∑∞

n=1 fn is a uniformly convergent series of continuousreal-valued functions on a metric space E , then its sum iscontinuous on E .

Proposition

If∑∞

n=1 fn is a uniformly convergent series of of Riemann integrablefunctions on [a, b], then the sum f is Riemann integrable and∫ b

af (x) dx =

∞∑n=1

∫ b

afn(x) dx .


Another

Proposition

Let (fn) be a sequence of continuously differentiable real-valuedfunctions on an open interval U ⊂ R. Suppose that

∑∞n=1 f

′n

converges uniformly to g on U. If∑∞

n=1 fn(a) converges for somea ∈ U, then

∑∞n=1 fn converges to a continuously differentiable

function f on U such that f ′ = g


Break Time

Time for a break and questions.


Power Series

Definition

Suppose that (cn) is sequence of real numbers and a ∈ R, then theseries

∞∑n=0

cn(x − a)n = c0 + c1(x − a) + c2(x − a)2 + · · ·

is called a power series about a.

Remark

A power series is very specific example of a series of functions on Rwhere fn(x) = cn(x − a)n. We will see power series have somegeneral properties which are special to power series and do notapply to more general settings.


Three Cases To Keep in Mind

Example

The ratio test is especially useful when applied pointwise to apower series.

1 The series∑∞

n=0xn

n! converges absolutely for all x ∈ R.

2 The series∑∞

n=0 xn converges absolutely if |x | < 1 and

diverges if |x | ≥ 1.

3 The series∑∞

n=0(n!)xn converges if x = 0 and diverges for allother x .

Remark

As a consequence of item (1), we see that

limn→∞

Mn

n!= 0

for all M ∈ R.


General Case

Theorem

Suppose that∑∞

n=0 cn(x − a)n is a power series. Then exactly onethe following three cases applies.

1 The series converges absolutely for all x ∈ R.

2 There is a R > 0 such that the series converges absolutely if|x − a| < R and diverges if |x − a| > R.

3 The series converges only if x = a.

Furthermore, if R1 is such that 0 < R1 < R in case (2) or for anyR1 > 0 in case (1), the convergence is uniform on{ x : |x − a| ≤ R1 }.


Proof

Proof.

Assume that case (3) does not apply and that the series convergesfor some x0 6= a. Let 0 < b < |x0 − a|. I claim that the seriesconverges uniformly and absolutely on { x : |x − a| ≤ b }. Since∑∞

n=0 cn(x0 − a)n converges, limn cn(x0 − a)n = 0. Hence there isa M such that |cn(x0 − a)n| ≤ M for all n. Thus if |x − a| ≤ b,

|cn(x − a)n| = |cn(x0 − a)n| ·∣∣∣ x − a

x0 − a

∣∣∣n ≤ M ·∣∣∣ b

x0 − a

∣∣∣n.Since |b/(x0 − a)| < 1,

∑∞n=0M

∣∣b/(x0 − a)∣∣n is a convergent

geometric series. Thus by the Weierstrass M-Test,∑∞n=0 cn(x − a)n converges uniformly and absolutely on

{ x : |x − a| ≤ b }.


Proof

Proof Continued.

Now let S = { y :∑∞

n=0 cn(y − a)n converges }. By assumption,x0 ∈ S . If S is not bounded, then for all R1 > 0 there is a y ∈ Ssuch that R1 < |y − a|. Then we just saw that

∑∞n=0 cn(x − a)n

converges uniformly and absolutely on { x : |x − a| ≤ R1 }. SinceR1 is arbitrary, this implies we are in case (1).

Is S is bounded, let R = l. u. b.{ |y − a| : y ∈ S }. ThenR ≥ |x0 − a| > 0 and the series diverges if |x − a| > R. If0 < R1 < R, then there is a y ∈ S such that |y − a| > R1. Then∑∞

n=0 cn(x − a)n converges uniformly and absolutely on{ x : |x − a| ≤ R1 }. Since 0 < R1 < R was arbitrary, this showsthat case (2) holds.


Radius of Convergence

Remark

In case (2), we call R the radius of convergence. In case (1), wesay that the radius of convergence is ∞ and in case (3) we say it is0. It is common practice to speak of the radius of convergence of apower series as R ∈ [0,∞] ignoring that issue that ∞ is not a realnumber.

If R ∈ (0,∞), then the series is guaranteed to converge on(a− R, a + R) and diverge off [a− R, a + R], but all bets are off atthe endpoints a± R. For all p ≥ 0, the series

∞∑n=1

xn

np

has radius of convergence R = 1. But if p = 0, the seriesconverges only on (−1, 1). If p = 1, then the series converges on[−1, 1) using the Alternating Series Test. If p = 2, and you believethat

∑n

1n2

converges, then the series converges on [−1, 1] by theComparison Test.


Break Time

Time for a break and some questions.


Term-By-Term Operations

Lemma

Suppose that∑∞

n=0 cn(x − a)n has radius of convergenceR ∈ [0,∞]. Then the series

∞∑n=1

ncn(x − a)n−1 and∞∑n=0

cnn + 1

(x − a)n+1

both have radius of convergence R.

Remark

These are just the series we obtain from∑∞

n=0 cn(x − a)n bydifferentiating term-by-term in the first case and integratingterm-by-term in the second case.


Proof

Proof.

We first show that if∑∞

n=0 cn(x − a)n converges at x0 6= a, thenthe other two series converge for all |x − a| < |x0 − a|. But asbefore, there is a M such that |cn(x0 − a)n| ≤ M for all n. Thus

|ncn(x − a)n−1| =n|cn(x0 − a)n||x0 − a|

·∣∣∣ x − a

x0 − a

∣∣∣n−1≤ nM

|x0 − a|·∣∣∣ x − a

x0 − a

∣∣∣n−1Since the series

∑∞n=1

nM|x0−a| ·

∣∣∣ x−ax0−a

∣∣∣n−1 converges by the Ratio

Test,∑∞

n=1 ncn(x − a)n−1 converges by the Comparison Test.


Proof

Proof Continued.

But you can check that∣∣∣ cnn + 1

(x − a)n+1∣∣∣ ≤ M|x0 − a|

n + 1·∣∣∣ x − a

x0 − a

∣∣∣n+1.

Again, we can use the Ratio Test and the comparison test to showthat

∑∞n=0

cnn+1(x − a)n+1 converges.

This shows the two new series have radi of convergence at least aslarge as the original series. However, we can integrate the seriesobtained by differentiating each term to see that its radius ofconvergence is no larger that the original. Similarly, we candifferentiate the series obtained by integrating the original seriesterm by term to see that its radius of convergence is that same asthe original.


Term-By-Term Differentiating and Integration

Theorem

Suppose that∑∞

n=0 cn(x − a)n has radius of convergenceR ∈ (0,∞] and we define

f (x) =∞∑n=0

cn(x − a)n for all x ∈ (a− R, a + R).

Then f is differentiable on (a− R, a + R) and for allx ∈ (a− R, a + R)

f ′(x) =∞∑n=1

ncn(x−a)n−1 and

∫ x

af (t) dt =

∞∑n=0

cnn + 1

(x−a)n+1.

Remark

It should be clear that if R =∞, then (a− R, a + R) is meant tobe interpreted as (−∞,∞) = R.


Proof

Proof.

Note that if 0 < R1 < R, then the convergence is uniform on[a− R1, a + R1]. Thus if sn(x) =

∑nk=0 ck(x − a)k , then (sn) and

(s ′n) converge uniformly on [a− R1, r + R1] to continuous functionsf and g . But our silly theorem implies f ′ = g on (a− R1, a + R1)and this holds for all R1 < R. This proves the first assertion. Butour theorem on term-by-term integration of a uniformly convergentsequence implies for any x ∈ [a− R1, a + R1] we have∫ x

af (t) dt = lim

n

∫ x

asn(t) dt.

This gives us the second assertion.


Analytic Functions

Corollary

Suppose that f (x) =∑∞

n=0 cn(x − a)n has radius of convergenceR > 0. (That is, R ∈ (0,∞].) Then f has derivatives of all orderson (a− R, a + R) and for all n ≥ 0

cn =f (n)(a)

n!.

In particular, the power series expansion for f about a is unique.


You can use induction to prove that for all k ≥ 0, f (k) exists and

f (k)(x) =∞∑n=0

(n + k)(n + k − 1) · · · (n + 1)an+k(x − a)n.

Then f (k)(a) = k!ck .


Enough

1 That is enough for today.


math 63: winter 2021 lecture 24

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