Davidson K.R., Donsig A.P. Real Analysis with Real Applications

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8.5 Power Series 233

8.5.1. HADAMARD’S THEOREM.

Given a power series

∞

n=0

, there is R in [0, +∞) ∪ {+∞} so that the series

converges for all x with |x| < R and diverges for all x with |x| > R. Moreover,

the series converges uniformly on each closed interval [a, b] contained in (−R, R).

Finally, if α = lim sup

n→∞

1/n

, then

R =











+∞ if α = 0,

0 if α = +∞,

if α ∈ (0, +∞).

We call R the radius of convergence of the power series.

PROOF. Fixing x ∈ R and applying the Root Test to

∞

n=0

gives

limsup|a

1/n

= |x|lim sup |a

1/n

= |x|α.

So if α = 0, then |x|α < 1 for all choices of x and so the series always converges,

as claimed. If α = +∞, then |x|α > 1 for all x 6= 0, and so the series diverges

for nonzero x, again as claimed. Otherwise, |x|α < 1 if and only if |x| < R and

|x|α > 1 if and only if |x| > R. By the Root Test (Exercise 3.2.4), it follows that

we have convergence and divergence on the required intervals.

It remains only to show uniform convergence on each interval [a, b] contained

in (−R, R). There is some c < R so that [a, b] ⊂ [−c, c] for some c < R. Observe

that, for x ∈ [−c, c], |a

| ≤ |a

. As c < R, the previousparagraph showsthat

∞

n=0

converges. By the M-test, it follows that

∞

n=0

converges uniformly

on [−c, c] and hence on [a, b]. ¥

By Exercise 3.3.F, if lim

n→∞

n+1

is deﬁned, then lim

n→∞

1/n

= lim

n→∞

n+1

Thus, we can often use ratios instead of roots to compute radii of convergence. See

Exercise 8.5.C.

8.5.2. EXAMPLE. The previous theorem contains no information about what

happens if |x| = R. The answer, like the Ratio and Root Tests for series of num-

bers, is that the series may converge or diverge at these points. We consider three

series,

∞

n=1

∞

n=1

∞

n=1

to illustrate this. For the ﬁrst series, the limit ratio of successive coefﬁcients is

lim

n→∞

n+1

(n + 1)

= lim

n→∞

2(n + 1)

Therefore, the radius of convergence is 2. Similarly, the second series also has

radius of convergence 2.

234 Limits of Functions

Consider the ﬁrst series at x = ±2. We have

∞

n=1

1/n

and

∞

n=1

(−1)

both of which converge. For the second series at x = ±2, we have

∞

n=1

1/n and

∞

n=1

(−1)

/n, which diverge and converge, respectively. The ﬁrst series has interval

of convergence [−2, 2] while the second has [−2, 2).

The third series differs crucially from the other two. To write this series in

the form

∞

n=1

, we cannot deﬁne a

to be 1/(2

n), but rather a

2k+1

= 0 and

= 2

−k

/k for k ≥ 0. Using this formula for a

, only the even terms matter:

limsup

n→∞

1/n

= lim

k→∞

1/2k

√

lim

k→∞

1/2k

√

So the series has radius of convergence

√

2. At x = ±

√

2, this series is

∞

n=1

which diverges. Therefore, the interval of convergence is (−

√

2).

It seems natural to differentiate and integrate a power series term-by-term. That

is, the derivative and indeﬁnite integral of f (x) =

∞

n=0

should be the sum of

the terms na

n−1

and a

n+1

/(n+1), respectively. The badly behaved examples

of previous sections show that this intuition cannot be trusted for arbitrary series of

functions. However, we will prove that this is true for power series.

This pair of properties make power series particularly useful. For example, the

following theorem implies that if a power series has radius of convergence R > 0,

then it is inﬁnitely differentiable on (−R, R).

8.5.3. TERM-BY-TERM OPERATIONS ON SERIES.

If f (x) =

∞

n=0

has radius of convergence R > 0, then

∞

n=1

n−1

has

radius of convergence R, f is differentiable on (−R, R) and, for x ∈ (−R, R),

(x) =

∞

n=1

n−1

Further,

∞

n=0

n + 1

n+1

has radius of convergence R and, for x ∈ (−R, R),

f(t) dt =

∞

n=0

n + 1

n+1

8.5 Power Series 235

PROOF. Observe that

∞

n=0

n−1

and

∞

n=0

have the same radius of conver-

gence. We have

limsup

n→∞

|na

1/n

= lim

n→∞

1/n

limsup

n→∞

1/n

Thus

∞

n=0

n−1

has radius of convergence R. As the partial sums

n=0

n−1

converge uniformly on each interval [−a, a] ⊂ (−R, R), we can apply Corol-

lary 8.3.2 (with c = 0) to show that f is differentiable and f

(x) =

∞

n=0

n−1

Similarly,

∞

n=0

n + 1

n+1

and

∞

n=0

n + 1

have the same radius of conver-

gence and

limsup

n→∞

n + 1

1/n

= lim

n→∞

(n + 1)

1/n

limsup

n→∞

1/n

∞

n=0

n + 1

n+1

has radius of convergence R. If x ∈ (−R, R), then since

n=0

converges uniformly to f (t) on the interval [0, x], Theorem 8.3.1 implies

that the sequence of integrals

(x) =

k=0

dt =

k=0

k + 1

k+1

converges uniformly to F (x) =

f(t) dt on each interval [−a, a] ⊂ (−R, R).

Thus,

∞

n=0

n + 1

n+1

f(t) dt

as required. ¥

8.5.4. EXAMPLE. We return to the series f (x) =

n≥0

of Example 8.4.3. It

was shown using the M-test that this series has inﬁnite radius of convergence, and

converges uniformly on [−A, A] for all ﬁnite A. Use term-by-term differentiation

to compute

(x) =

∞

n=1

n−1

(n − 1)!

∞

k=0

= f (x).

The differential equation f

(x) = f (x) may be rewritten as

1 =

(x)

f(x)

= (log f )

(x).

236 Limits of Functions

Integrating from 0 to t, we obtain

t =

1dx =

(logf)

(x) dx = log f (t) − logf(0).

It is evident that f (0) = 1 and therefore logf(t) = t, whence f (t) = e

8.5.5. EXAMPLE. Consider the power series

n≥1

. Since lim

n→∞

(n+1)

= 1,

the Ratio Test tells us that the radius of convergence is 1. When |x| = 1, the terms

do not tend to 0, and thus the series diverges. So the function f(x) =

n≥1

deﬁned for x ∈ (−1, 1).

Deﬁne the function g(x) =

n≥0

. This also has radius of convergence 1.

Since it is a geometric series, it is easily evaluated as g(x) =

1 − x

for |x| < 1.

Apply Theorem 8.5.3 to obtain

(1 − x)

= g

(x) =

n≥1

n−1

This series has the same radius of convergence, 1, as does

(1 − x)

= xg

(x) =

n≥1

A second application of Theorem 8.5.3 yields

n≥1

n−1

(x)

(1 − x)

1 + x

(1 − x)

Therefore,

f(x) =

n≥1

x(1 + x)

(1 − x)

In particular,

n≥1

= f (

) = 6.

8.5.6. EXAMPLE. In this example, we obtain the Binomial Theorem for frac-

tional powers. That is, we derive the power series expansion of (1+x)

for α ∈ R.

If g(x) = (1 + x)

, then g

(x) = α(1 + x)

α−1

and so g satisﬁes the differential

equation (DE)

(1 + x)g

(x) = αg(x), g(0) = 1.

Suppose there is a power series f (x) =

∞

n=0

that satisﬁes this DE. Then we

have

(1 + x)

∞

n=1

n−1

= α

∞

n=0

8.5 Power Series 237

Collecting terms, we have

∞

n=0

+ (n + 1)a

n+1

∞

n=0

αa

and so na

+ (n + 1)a

n+1

= αa

giving a

n+1

α − n

n + 1

. As a

= f(0) = 1,

we have a

= α, a

α(α − 1)

, a

α(α − 1)(α − 2)

, and so on. In general,

we obtain the fractional binomial coefﬁcients

α(α − 1) ···(α − n + 2)(α − n + 1)

It remains to show that this series has a positive radius of convergence and that it

actually converges to (1 + x)

If α is a nonnegative integer, then the a

are eventually zero, and so the series

reduces to the usual Binomial Theorem. In this case, the radius of convergence is

inﬁnite. Otherwise a

6= 0 for all n, and we can apply the Ratio Test to obtain

lim

n→∞

n+1

= lim

n→∞

α − n

n + 1

= 1.

Hence the series has radius of convergence 1.

To show that f(x) = (1+x)

, consider the ratio f(x)/(1+x)

. Differentiating

the ratio with respect to x gives

(1 + x)

(x) − α(1 + x)

α−1

f(x)

(1 + x)

2α

and since we have shown (1 + x)f

(x) = αf (x), it follows that the derivative is

zero. However, setting x = 0 in f(x)/(1 + x)

gives 1/1 = 1 and so the ratio is

constantly equal to 1, showing that f (x) = (1 + x)

Thus, for |x| < 1 and any real α,

(1 + x)

∞

n=0

Exercises for Section 8.5

A. Determine the interval of convergence of the following power series:

(a)

∞

n=0

(b)

∞

n=1

(−1)

(c)

∞

n=0

(d)

∞

n=0

√

n x

(e)

∞

n=0

(−1)

(2n)!

(f)

∞

n=0

(g)

∞

n=1

(h)

∞

n=0

(n!)

(2n)!

(i)

∞

n=0

238 Limits of Functions

B. Find a power series

∞

n=0

that does not have the same interval of convergence as

∞

n=0

n−1

C. Suppose that lim

n→∞

n+1

= L exists. Find the radius of convergence of the power

series

∞

n=1

. (See Exercise 3.3.F.)

D. Using the method of Example 8.5.6, show that if f(x) =

∞

n=0

satisﬁes the DE

(x) = f(x) and f(0) = 1, then f(x) =

∞

n=0

/n!.

E. Repeat the previous exercise with the conditions f

(x) = −f(x), f is anodd function,

and f(0) = 0.

F. Prove that if inﬁnitely many of the a

are nonzero integers, then the radius of conver-

gence of

∞

n=1

is at most 1.

G. (a) Compute f(x) =

∞

n=1

(b) Compute

∞

n=1

. Justify your method.

H. (a) Compute f(x) =

∞

n=0

(n + 1)x

(b) Compute

∞

n=0

. Justify your method.

I. (a) Compute g(x) =

∞

n=0

+ n)x

(b) Compute

∞

n=0

+ n

. Justify your method.

J. Using the binomial series for (1 − x)

−1/2

and the formula

π/2

sin

t dt =

1 · 3 ·5 ···(2n − 1)

2 · 4···(2n)

show that, for κ ∈ (0, 1), the integral

π/2

(1 − κ

sin

−1/2

dt equals

1 +

1 · 3

2 · 4

1 · 3 ·5

2 · 4 ·6

+ ···

K. Recall the Fibonacci sequence deﬁned recursively by F (0) = F (1) = 1 and

F (n + 2) = F (n) + F (n + 1) for n ≥ 0. Set f(x) =

∞

n=0

F (n)x

(a) Show that F (n) ≤ 2

and hence deduce a positive lower bound for the radius of

convergence of this power series.

8.6 Compactness and Subsets of C(K) 239

(b) Compute lim

n→∞

F (n + 1)

F (n)

and hence ﬁnd the radius of convergence R.

HINT: Let r

= F (n + 1)/F (n). Show by induction that

n+1

− r

= −

− r

n−1

and that r

n+1

≥ 2. Hence deduce that the limit R exists. Show that R satisﬁes a

quadratic equation.

)f(x) for |x| < R, and justify your steps. Hence compute

f(x).

(d) Show that f(x) =

∞

n=0

(x + x

)

, and that this converges for |x| < R.

8.6. Compactness and Subsets of C(K)

We sawin Chapter 4 that compactness is a very powerful property. This showed

up particularly in Chapter 5 in the proofs of the Extreme Value Theorem and of

uniform continuity for a continuous function on a compact set. In this section, we

characterize the subsets of C(K) that are themselves compact. We will need this

characterization to prove Peano’s Theorem (Theorem 12.8.1), which shows that a

wide range of differential equations have solutions.

We restrict our attention to sets of functions in C(K) when K is a compact

subset of R

. As usual, a subset F of C(K) is called compact if every sequence

) of functions in F has a subsequence (f

) that converges uniformly to a func-

tion f in F. The Heine–Borel Theorem showed that a subset of R

is compact if

and only if it is closed and bounded. However, R

is a ﬁnite-dimensional space;

and this is a critical fact. C(K) is inﬁnite dimensional, so some of those arguments

are invalid—as is the conclusion.

The arguments of Lemma 4.4.3 are still valid. If F is not closed, there is a

sequence (f

) in F that has a uniform limit f = lim

n→∞

that is not in F. Every

subsequence (f

) also has limit f , which is not in F. So F is not compact.

Likewise, if F is unbounded, it contains a sequence (f

) such that kf

k > n

for n ≥ 1. Any subsequence (f

) satisﬁes lim

n→∞

k = ∞. Consequently, it

cannot converge uniformly to any function.

However, this is not the whole story. There are other ways in which a subset of

C(K) can fail to be compact, as we now show.

8.6.1. EXAMPLE. Look again at Example 8.1.5. We will show that the set

F = {f

(x) = x

: n ≥ 1} is closed and bounded but not compact. The functions

on [0, 1] are all bounded by 1. Suppose that (f

) is any subsequence of (f

By Example 8.1.5,

lim

i→∞

(x) = lim

n→∞

(x) =

(

0 for 0 ≤ x < 1

1 for x = 1.

However, this limit

{1}

is not continuous, and the convergence is not uniform.

Thus no subsequence converges. It follows that the only limit points of F are the

240 Limits of Functions

points in F themselves. In particular, F contains all of its limit points, and therefore

it is closed. On the other hand, F is not compact because the sequence (f

) has no

convergent subsequence.

8.6.2. EXAMPLE. Consider a sequence (g

) of continuous functions on K such

that g

converges uniformly to a function g. Then the set

G = {g

: n ≥ 1} ∪ {g}

is compact. Indeed, suppose that (f

) is a sequence in G. Either some element of G

is repeated inﬁnitely often, or inﬁnitely many g

’s are represented in this sequence.

In the ﬁrst case, there is a constant subsequence that evidently converges in G.

Otherwise, there is a subsequence (f

) such that f

= g

and lim

i→∞

= ∞. In

this case, the subsequence converges uniformly to g.

Now consider a point a in K and an ε > 0. Since g is continuous, there is an

> 0 such that

kg(x) − g(a)k <

whenever kx − ak < r

Since g

converges uniformly to g, there is an integer N so that

kg − g

∞

for all n ≥ N.

Combining these estimates, we can show that for n ≥ N and kx − ak < r

(x) − g

(a)k ≤ kg

(x) − g(x)k + kg(x) − g(a)k + kg(a) − g

(a)k

≤

= ε.

Now we can modify this to obtain a statement for all functions in G. Each g

for n < N is continuous. So there are positive real numbers r

> 0 such that

(x) − g

(a)k < ε whenever kx − ak < r

Set r = min{r

: 0 ≤ n ≤ N}. We have shown that

kf(x) − f(a)k < ε whenever f ∈ G and kx − ak < r.

This example suggests a new variant of continuity in which a whole family of

functions satisfy the same inequalities.

8.6.3. DEFINITION. A family of functions F mapping a set S ⊂ R

into R

is equicontinuous at a point a ∈ S if for every ε > 0, there is an r > 0 such that

kf(x) − f(a)k < ε whenever kx − ak < r and f ∈ F.

The family F is equicontinuous on a set S if it is equicontinuous at every point in

S. The family F is uniformly equicontinuous on S if for each ε > 0, there is an

r > 0 such that

kf(x) − f(y)k < ε whenever kx − yk < r, x, y ∈ S and f ∈ F.

8.6 Compactness and Subsets of C(K) 241

Reconsider the previous two examples. In the second example, we established

that G is equicontinuous. However, in the ﬁrst example, the set F = {x

: n ≥ 1}

is not equicontinuous if x = 1. Indeed, take ε = 1/10 and let 0 < r < 1 be

an arbitrary positive number. Take x = 1 − r/2. Since lim

n→∞

= 0, there is an

integer N sufﬁciently large that

|1 − x

| > .5 for all n ≥ N.

Hence this r does not work in the deﬁnition of equicontinuity as |1 − x| < r and

|1 − x

| > .5. Since r is arbitrary, there is no choice of r that will work, and so F

is not equicontinuous at 1.

8.6.4. LEMMA. Let K be a compact subset of R

. A compact subset F of

C(K, R

) is equicontinuous.

PROOF. Suppose to the contrary that for a certain point a in K and ε > 0, the

deﬁnition of equicontinuity is not satisﬁed for F. This means that for each choice

of r = 1/n, there is a function f

∈ F and a point x

∈ K such that

− ak <

and kf

) − f

(a)k ≥ ε.

It is evident that no subsequence of (f

) can be equicontinuous either.

Now if F were compact, there would be a subsequence (f

) that converges

uniformly to some function f . By Example 8.6.2, this subsequence would be

equicontinuous. This contradiction shows that F must be equicontinuous. ¥

8.6.5. PROPOSITION. If F is an equicontinuous family of functions on a com-

pact set, then it is uniformly equicontinuous.

PROOF. This is a modiﬁcation of Theorem 5.5.9. If the result is false, there is an

ε > 0 for which the deﬁnition of equicontinuity fails. This means that for each

r = 1/n, there are points x

and y

in K and a function f

in F such that

− y

k <

and kf

) − f

)k ≥ ε.

Since K is compact, the sequence (x

) has a convergent subsequence with

lim

i→∞

= a. Hence

lim

i→∞

= lim

i→∞

+ lim

i→∞

− x

= a + 0 = a.

By the equicontinuity of F at a, there is an r > 0 so that

kf(x) − f(a)k <

for all f ∈ F, kx − ak < r.

There is an integer I so large that

− ak < r and ky

− ak < r for all i ≥ I.

242 Limits of Functions

Therefore

) − f

)k ≤ kf

) − f

(a)k + kf

(a) − f

= ε.

This is a contradiction to the hypothesis that uniform equicontinuity fails; and

thus it must hold. ¥

We need a new property, total boundedness, which follows from compactness.

Conversely, a closed totally bounded set is compact, and we will prove this, as

part of the Borel–Lebesgue Theorem (Theorem 9.2.3), in the next chapter. For

convenience, we give a simple proof of the direction we need valid in R

8.6.6. DEFINITION. A subset S of K is called an ε-net of K if

K ⊂

[

a∈S

(a).

A set K is totally bounded if it has a ﬁnite ε-net for every ε > 0.

8.6.7. LEMMA. Let K be a bounded subset of R

. Then K is totally bounded.

PROOF. Fixε > 0. Choose N so that ε > 1/N. Since K is bounded, it is contained

in a large m-cube

C = {x ∈ R

: |x

| ≤ L, 1 ≤ i ≤ m}.

The ﬁnite set of points

A = {a ∈ R

: a

, k

∈ Z, |k

| ≤ 2LN }

forms an ε-net for C. For each a in A, either B

ε/2

(a) ∩K is empty or it contains a

point x

. Let B = {x

: a ∈ A}.

We will show that B is an ε-net for K. Indeed, if x is any point in K, there is

a point a ∈ A such that kx − ak < ε/2. Since B

ε/2

(a) ∩ K is nonempty, x

deﬁned and belongs to this intersection. Therefore,

kx − x

k ≤ kx − ak + ka − x

k <

= ε.

8.6.8. COROLLARY. Let K be a bounded subset of R

. Then K contains a

sequence {x

: i ≥ 1} that is dense in K. Moreover, for any ε > 0, there is an

integer N so that {x

: 1 ≤ i ≤ N} forms an ε-net for K.

PROOF. Let B

be a ﬁnite

-net for K for each integer k ≥ 1. Form a se-

quence {x

: i ≥ 1} by listing all of the elements of each B

in turn as the points

k−1

, . . . , x

. If x ∈ K, then for each k there is an element x

correspond-

ing to a point in B

such that kx − x

k < 1/k. Thus x = lim

k→∞

. As x is an

arbitrary point in K, the sequence is dense. By construction, if ε > 1/k, the set of

points {x

: 1 ≤ i ≤ N

} forms an ε-net for K. ¥