Chapter 3. Sequences and Series of Functions
Pointwise Convergence
3.1Definition: LetA⊆R,letf:A→R,andforeachintegern≥pletfn :A→R.
We say that the sequence of functions (fn)n≥p converges pointwise to f on A, and we
write fn → f pointwise on A, when lim fn(x) = f(x) for all x ∈ A, that is when for all n→∞
x ∈ A and for all ε > 0 there exists m ≥ p such that for all integers n we have n ≥ m =⇒ | f n ( x ) − f ( x ) | < ε .
3.2 Note: By the Cauchy Criterion for convergence, the sequence (fn)n≥p converges pointwise to some function f(x) on A if and only if for all x ∈ A and for all ε > 0 there exists m ≥ p such that for all integers k, l we have
k , l ≥ m =⇒ f k ( x ) − f l ( x ) < ε .
3.3 Example: Find an example of a sequence of functions (fn)n≥1 and a function f with
fn → f pointwise on [0, 1] such that each fn is continuous but f is not.
0 if x ̸= 1 Solution: Let fn(x) = xn. Then lim fn(x) = 1 if x = 1.
n→∞
3.4 Example: Find an example of a sequence of functions (fn)n≥1 and a function f with fn → f pointwise on [0, 1] such that each fn is differentiable and f is differentiable, but
lim fn′ ̸=f′. n→∞
Solution: Let fn(x) = 1 tan−1(nx). Then lim fn(x) = 0, and fn′(x) = 1
n n→∞ 1+(nx)2
0 if x ̸= 0 lim fn′(x)= 1ifx=0.
so
n→∞
3.5 Example: Find an example of a sequence of functions (fn)n≥1 and a function f with
fn → f pointwise on [0, 1] such that each fn is integrable but f is not. Solution: WehaveQ∩[0,1]={a1,a2,a3,···}where
(an)n≥1 = 0, 1, 0, 1, 2, 0, 1, 2, 3, 0,···, 4,···. 1122233334 4
as an exercise, you can check that an = k where l = −3+√9−8n and k = n− l2+l. l22
0 if x ∈/ {a1,a2,···,an} 0 if x ∈/ Q Forx∈[0,1],letfn(x)= 1ifx∈{a,a,···,a}.Thenlimfn(x)= 1ifx∈Q.
12 n n→∞
3.6 Example: Find an example of a sequence of functions (fn)n≥1 and a function f with fn → f pointwise on [0, 1] such that each fn is integrable and f is integrable but
Solution: Let f1(x) =
1 1
lim fn(x)dx ̸= f(x)dx.
n→∞ 0 0
48x−11−xif 1 ≤x≤1,
2 2 For n ≥ 1 let fn(x) = nf1(nx).
0 otherwise . 1
Then each fn is continuous with fn(x) dx = 1, and lim fn(x) = 0 for all x.
0
n→∞
1
Uniform Convergence
3.7Definition: LetA⊆R,letf:A→R,andforeachintegern≥pletfn :A→R. We say that the sequence of functions (fn)n≥p converges uniformly to f on A, and we write fn → f uniformly on A, when for all ε > 0 there exists m ∈ Z≥p such that for all x∈Aandforallintegersn∈Z≥p wehave
n ≥ m =⇒ f n ( x ) − f ( x ) < ε .
3.8 Theorem: (Cauchy Criterion for Uniform Convergence of Sequences of Functions) Let (fn)n≥p be a sequence of functions on A ⊆ R. Then (fn) converges uniformly (to some functionf)onAifandonlyifforallε>0thereexistsm∈Z≥p suchthatforallx∈A and for all integers k, l ∈ Z≥p we have
k , l ≥ m =⇒ f k ( x ) − f l ( x ) < ε .
Proof: Suppose that (fn) converges uniformly to f on A. Let ε > 0. Choose m so that
forallx∈Awehaven≥m=⇒fn(x)−f(x)< ε. Thenfork,l≥mwehave
fk(x)−f(x)<ε andfl(x)−f(x)<ε andso 22
2
fk(x)−fl(x)≤fk(x)−f(x)+fl(x)−f(x)< ε + ε =ε. 22
Conversely, suppose that (fn) satisfies the Cauchy Criterion for uniform convergence, that is for all ε > 0 there exists m such that for all x ∈ A and all integers n, l we have
n, l ≥ m =⇒ fn(x) − fl(x) < ε .
For each fixed x ∈ A, (fn(x)) is a Cauchy sequence, so (fn(x)) converges, and we can
define f(x) by
We know that fn → f pointwise on A, but we must show that fn → f uniformly on A.
Letε>0. Choosemsothatforallx∈Aandforallintegersn,lwehave n,l≥m=⇒fn(x)−fl(x)< ε .
2
Let x ∈ A. Since lim fl(x) = f(x), we can choose l ≥ m so that fl(x)−f(x) < ε . Then l→∞ 2
for n ≥ m we have
fn(x)−f(x)≤fn(x)−fl(x)+fl(x)−f(x)< ε + ε =ε.
22
3.9 Theorem: (Uniform Convergence, Limits and Continuity) Suppose that fn → f uniformly on A. Let x be a limit point of A. If lim fn(y) exists for each n, then
y→x
lim lim fn(y) = lim lim fn(y) .
y→x n→∞ n→∞ y→x In particular, if each fn is continuous in A, then so is f.
Proof: Suppose that lim fn(y) exists for all n. Let bn = lim fn(y). We must show that y→x y→x
lim f(y) = lim bn. We claim first that (bn) converges. Let ε > 0. Choose m so that y→x n→∞
k,l≥m=⇒fk(y)−fl(y)< ε forally∈A. Letk,l≥m. Choosey∈Asothat 3
fk(y)−bk< ε andfl(y)−bl< ε. Thenwehave 33
bk −bl|≤|bk −fk(y)+fk(y)−fl(y)+fl(y)−bl< ε + ε + ε =ε. 333
By the Cauchy Criterion for sequences, (bn) converges, as claimed. 2
f(x)= lim fn(x). n→∞
Now,letb= lim bn. Wemustshowthat limf(x)=b. Letε>0. Choosemsothat n→∞ y→x
whenn≥mwehavefn(y)−f(y)< ε forally∈Aandwehave|bn−b|< ε. Letn≥m. 33
Since lim fn(y) = bn we can choose δ > 0 so that 0 < |y − x| < δ =⇒ fn(y) − bn < ε . y→x 3
Then when 0 < |y − x| < δ we have f(y)−b≤f(y)−fn(y)+fn(y)−bn+bn −b< ε + ε + ε =ε.
Thus lim f(x) = b, as required. y→x
In particular, if x ∈ A and each fn is continuous at x then we have
lim f(y) = lim lim fn(y) = lim lim fn(y) = lim fn(x) = f(x)
y→x y→x n→∞ n→∞ y→x n→∞ so f is continuous at x.
3.10 Theorem: (Uniform Convergence and Integration) Suppose that fn → f uniformly x
on [a,b]. If each fn is integrable on [a,b] then so is f. In this case, if gn(x) = x
fn(t)dt
and g(x) =
f (t) dt, then gn → g uniformly on [a, b]. In particular, we have b b
lim fn(x)dx= lim fn(x)dx. n→∞ a a n→∞
a
Proof: Suppose that each fn is integrable on [a, b]. We claim that f is integrable on [a, b]. Letε>0. ChooseN sothatn≥N =⇒fn(x)−f(x)< ε forallx∈[a,b]. Fix
333
a
4(b−a)
n ≥ N. Choose a partition X of [a,b] so that U(fn,X)−L(fn,X) < ε. Note that since
εε2ε fn(x)−f(x)<4(b−a) wehaveMi(f)
and so
U(f, X) − L(f, X) = Mi(f) − mi(f)∆ix < Mi(fn) − mi(fn) +
nn i=1 i=1
ε ∆ix 2(b−a)
=U(fn,X)−L(fn,X)+ε <ε +ε =ε. 222
Thus f is integrable on [a, b].
x x
Now define gn(x) = fn(t)dt and g(x) = f(t)dt. We claim that gn → g aaε
uniformly on [a,b]. Let ε > 0. Choose N so that n ≥ N =⇒ fn(t)−f(t) < 2(b−a) for all t∈I. Letn≥N. Letx∈[a,b]. Thenwehave
x x x gn(x) − g(x) = fn(t) dt − f(t) dt = fn(t) − f(t) dt
aaa x x
≤ fn(t)−f(t)dt≤ ε
aa Thus gn → g uniformly on [a, b], as required.
In particular, we have lim gn(b) = g(b), that is n→∞
dt=
ε (x−a)≤ε <ε. 2(b−a) 2
b b
lim fn(x)dx= lim fn(x)dx.
n→∞ a a n→∞
3
2(b−a)
3.11 Theorem: (Uniform Convergence and Differentiation) Let (fn) be a sequence of functions on [a, b]. Suppose that each fn is differentiable on [a, b], fn′ converges uniformly on [a, b], and fn (c) converges for some c ∈ [a, b]. Then (fn ) converges uniformly on [a, b],
lim fn(x) is differentiable, and n→∞
d limfn(x)=lim dfn(x). dx n→∞ n→∞ dx
Proof: We claim that (fn) converges uniformly on [a, b]. Let ε > 0. Choose N so that when n, m ≥ N we have fn′(t)−fm′(t) < ε for all t ∈ [a, b] and we have fn(c)−fm(c) < ε .
2(b−a) 2 Let n,m ≥ N. Let x ∈ [a,b]. By the Mean Value Theorem applied to the function
fn(x) − fm(x), we can choose t between c and x so that
fn(x) − fm(x) − fn(c) + fm(c) = fn′(t) − fm′(t)(x − c) .
Then we have
fn(x) − fm(x) ≤ fn(x) − fm(x) − fn(c) + fm(c) + fn(c) − fm(c)
2
Let f(x) = lim fn(x). We claim that f is differentiable with f′(x) = lim fn′(x) for n→∞ n→∞
= fn′(t) − fm′(t)|x − c| + fn(c) − fm(c) < ε (b−a)+ε =ε.
2(b−a) Thus (fn ) converges uniformly on [a, b].
all x ∈ [a,b]. Fix x ∈ [a,b]. Note that
f′(x) = lim fn′(x) ⇐⇒ lim f(y) − f(x) = lim lim fn(y) − fn(x)
n→∞ y→x y − x n→∞ y→x y − x
⇐⇒ lim lim fn(y) − fn(x) = lim lim fn(y) − fn(x)
y→x n→∞ y − x n→∞ y→x y − x so it suffices to show that gn converges uniformly on [a, b] \ {x}, where
gn(y) = fn(y) − fn(x) . y−x
Let ε > 0. Choose N so that n,m ≥ N =⇒ fn′(t)−fm′(t) < ε for all t ∈ [a,b]. Let n,m ≥ N. Let y ∈ [a,b] \ {x}. By the Mean Value Theorem, we can choose t between x and y so that
fn(y) − fm(y) − fn(x) + fm(x) = fn′(t) − fm′(t)(y − x) .
fn(y)−fm(y)−fn(x)+fm(x) ′ ′
Then
gn(y)−gm(y)= y−x =fn (t)−fm (t)<ε.
Thus gn converges uniformly on [a, b] \ {x}, as required.
4
Series of Functions
3.12 Definition: Let (fn)n≥p be a sequence of functions on A ⊆ R. The series of
l
functions fn(x) is defined to be the sequence Sl(x) where Sl(x) = fn(x). The
n≥p n=p
function Sl(x) is called the lth partial sum of the series. We say the series fn(x) n≥p
converges pointwise (or uniformly) on A when the sequence {Sl} converges, pointwise (or uniformly) on A. In this case, the sum of the series of functions is defined to be the
function
∞
f(x)=fn(x)= limSl(x).
l→∞
3.13 Theorem: (Cauchy Criterion for the Uniform Convergence of a Series of Functions)
The series fn(x) converges uniformly (to some function f) on A if and only if for every n≥p
ε > 0 there exists N ≥ p such that for all x ∈ A and for all k, l ≥ p we have
then
n=p
n=k+1
Proof: This follows immediately from the analogous theorem for sequences of functions.
3.14 Theorem: (Uniform Convergence, Limits and Continuity) Suppose that fn(x) n≥p
converges uniformly on A. Let x be a limit point of A. If lim fn(y) exists for all n ≥ p, y→x
l l > k ≥ N = ⇒
f n ( x ) < ε .
∞∞
lim fn(y) = lim fn(y) . y→x y→x
n=p n=p
∞
In particular, if each fn(x) is continuous on A then so is fn(x). n=p
Proof: This follows immediately from the analogous theorem for sequences of functions. 3.15 Theorem: (Uniform Convergence and Integration) Suppose that fn(x) converges
n≥p ∞
uniformly on [a, b]. If each fn (x) is integrable on [a, b], then so is fn (x). In this case, n=p
if we define gn(x) =
x x∞
fn(t)dt and g(x) = fn(t)dt, then gn(x) converges
a
a n=p n≥p
uniformly to g(x) on A. In particular, we have b∞ ∞b
fn(x)dx = fn(x)dx. a n=p n=p a
Proof: This follows immediately from the analogous theorem for sequences of functions.
5
3.16 Theorem: (Uniform Convergence and Differentiation) Suppose that each fn(x) is differentiable on [a, b], fn′(x) converges uniformly on [a, b], and fn(c) converges for
n≥p n≥p some c ∈ [a, b]. Then fn (x) converges uniformly on [a, b] and
n≥p
dx n=p n=p dx
Proof: This follows immediately from the analogous theorem for sequences of functions.
3.17 Theorem: (The Weierstrass M-Test) Suppose that fn is bounded with |fn(x)| ≤ Mn for all n ≥ p and x ∈ A, and Mn converges. Then fn(x) converges uniformly on A.
n≥p n≥p
l
Proof: Letε>0. ChooseN sothatl>k≥N =⇒ Mn <ε. Letl>k≥N. Let
d∞ ∞d
fn(x) = fn(x) .
x ∈ A. Then
n=k+1 ll l
fn(x)≤ fn(x)≤ Mn <ε.
n=k+1 n=k+1 n=k+1
3.18 Example: Find a sequence of functions fn(x)
on R, such that fn(x) converges uniformly on R, but the sum f(x) = fn(x) is
∞ n≥0 n=0
nowhere differentiable.
Solution: Let f (x) = 1 sin2(8nx). Since |f (x)| ≤ 1 and 1 converges, f (x)
n2n n2n2n n n≥0
n≥0
, each of which is differentiable
∞
converges uniformly on R. Let f (x) = fn (x). We claim that f (x) is nowhere differen- n=0
tiable.Letx∈R.Foreachn,letm,a andb besuchthata = mπ,b =(m+1)π and n n n 2·8n n 2·8n
x∈[a ,b ). Notethatoneoff (a )andf (b )isequalto 1 andtheotherisequalto0 nnnnnnn
12
so we have fn(bn)−fn(an) = 2n . Note also that for k > n we have fk(an) = fk(bn) = 0.
Also, for all k we have f (x) = 1 sin2(8kx), f ′(x) = 4k sin(2·8kx), and f ′(x) ≤ 4k, so k2k k k
by the Mean Value Theorem,
fk(bn) − fk(an) ≤ 4k|bn − an| .
Finally, note that if f′(x) did exist, then we would have f′(x) = lim f(bn) − f(an), but n→∞ bn − an
f(b)−f(a) ∞ f(b)−f(a) n f(b)−f(a) n n k n k n k n k n
==
bn − an
k=0 bn − an k=0 bn − an fn(bn) − fn(an) n−1 fk(bn) − fk(an)
≥
− bn − an k=0 bn − an
n−1
ππ3π33
1
≥ 2n −4k=2·4n −4n−1=2−14n+1→∞asn→∞
2·8n
k=0
6
Power Series
3.19 Definition: A power series centred at a is a series of the form an(x − a)n n≥0
for some real numbers an, where we use the convention that (x − a)0 = 1.
3.20 Example: The geometric series xn is a power series centred at 0. It converges n≥0
when |x| < 1 and for all such x the sum of the series is ∞ 1
xn = 1 − x . n=0
3.21 Lemma: (Abel’s Formula) Let {an} and {bn} be sequences. Then we have l l−1p l
Proof: We have
l−1 p an
p=m k=m
anbn+ an (bp−1−bp)= an bl. n=m p=m n=m n=m
(bp+1 − bp) = am(bm+1 − bm) + (am + am+1)(bm+2 − bm+1)
+ (am + am+1 + am+2)(bm+3 − bm+2)
+···+(am +am+1 +am+2 +···+al−1)(bl −bl−1)
= −ambm − am+1bm+1 − · · · − al−1bl−1 +(am +am+1 +···al−1)bl −albl +albl
ll
= an bl−anbn.
n=m n=m
3.22 Definition: Let (an) be a sequence in R. We define limsupan = lim sn where n→∞ n→∞
sn = sup{ak | k ≥ n} (with lim sup an = ∞ when (an) is not bounded above). n→∞
3.23 Theorem: (The Interval and Radius of Convergence) Let an(x − a)n be a power n≥0
1
series and let R = limsup n |an| ∈ [0,∞]. Then the set of x ∈ R for which the power
n→∞
series converges is an interval I centred at a of radius R. Indeed
(1)if|x−a|>Rthen lim an(x−a)̸=0sonn(x−a)n diverges,
n→∞
(2) if |x − a| < R then an(x − a)n converges absolutely,
n≥0
(3) if 0 < r < R then an(x−a)n converges uniformly in [a−r,a+r], and
n≥0
(4) (Abel’s Theorem) if an(x − a)n converges when x = a + R then the convergence is
n≥0
uniform on [a, a + R], and similarly if an(x − a)n converges when x = a − R then the n≥0
convergence is uniform on [a − R, a].
7
n≥0
Proof: To prove part (1), suppose that |x − a| > R. Then
limsupn |an(x−a)n|=|x−a|limsupn |an|>R· 1 =1,
n→∞ n→∞ R
and so lim an(x − a)n ̸= 0 and an(x − a)n diverges, by the Root Test.
n→∞
To prove part (2), suppose that |x − a| < R. Then
limsupn |an(x−a)n|=|x−a|limsupn |an|
anR <ε.
Then by Abel’s Formula and using telescoping we have
l
n=m
l n
n x−an
an(x−a) =
n=m
anR R
l
=anRR− anRR−R
l−1 p
n x−al n x−ap+1 x−ap
n=m
l
p=m n=m
l−1 p
nx−al
n=m p=m n=m
n x−ap
≤ anR R + anR R − R
x−ap+1 <εx−al +εx−am −x−al=εx−am <ε.
RRRR
3.24 Definition: The number R in the above theorem is called the radius of conver- gence of the power series, and the interval I is called the interval of convergence of the power series.
3.25 Example: Find the interval of convergence of the power series (3−2x)n
√n (−2)n
.
3
(3 − 2x)n
n≥1 Solution: First note that this is in fact a power series, since √n =
3n x − 2 ,
(3 − 2x)n
n (−2)n
√n
for n ≥ 1 and a = 2 .
√n = (3−2x)n
cn(x − a) , where c0 = 0, cn = an+1 (3−2x)n+1
√
so lim = |3−2x|. By the Ratio Test,
and so
Now, let an =
√n ann+1(3−2x)n n+1
n≥1
n≥0
√n n =
an+1 n→∞ an
n
|3 − 2x|, an converges when |3−2x| < 1 and
. Then = √
diverges when |3 − 2x| > 1. Equivalently, it converges when x ∈ (1, 2) and diverges when x∈/[1,2]. Whenx=1so(3−2x)=1,wehavea = 1 ,whichdiverges(itsa
n √n
p-series), and when x = 2 so (3−2x) = −1, we have a = (−1)n which converges by
n √n
the Alternating Series Test. Thus the interval of convergence is I = (1, 2 ].
8
Operations on Power Series
3.26 Theorem: (Continuity of Power Series) Suppose that the power series an(x−a)n
∞
converges in an interval I. Then the sum f(x) = an(x − a)n is continuous in I. n=0
Proof: This follows from uniform convergence of an(x − a)n in closed subintervals of I.
3.27 Theorem: (Addition and Subtraction of Power Series) Suppose that the power series an(x − a)n and bn(x − a)n both converge in the interval I. Then (an + bn)(x − a)n and(an−bn)(x−a)n bothconvergeinI,andforallx∈Iwehave
∞ ∞ ∞
an(x−a)n ± bn(x−a)n = (an ±bn)(x−a)n .
n=0 n=0 n=0 Proof: This follows from Linearity.
3.28 Theorem: (Multiplication of Power Series) Suppose the power series an(x − a)n n
and bn(x − a)n both converge in an open interval I with a ∈ I. Let cn = akbn−k.
Thencn(x−a)n convergesinI andforallx∈I wehave
∞ ∞∞
cn(x−a)n = an(x−a)n bn(x−a)n . n=0 n=0 n=0
k=0
Proof: This follows from the Multiplication of Series Theorem, since the power series converge absolutely in I.
3.29 Theorem: (Division of Power Series) Suppose that an(x − a)n and bn(x − a)n both converge in an open interval I with a ∈ I, and that b0 ̸= 0. Define cn by
c0=a0 ,andforn>0,cn=an −bnc0 −bn−1c1 −···−b1cn−1 . b0 b0b0b0b0
Then there is an open interval J with a ∈ J such that cn(x − a)n converges in J and for all x ∈ J,
∞
an(x−a)n cn(x−a)n = n=0 .
n=0 bn(x − a)n n=0
Proof: Choose r > 0 so that a + r ∈ I. Note that |anrn| and |bnrn| both converges. n n anrn
∞
Since |anr | → 0 and |bnr | → 0 and b0 ̸= 0, we can choose M so that M ≥
and
∞
bnrn a0 a1 b1c0 M ≥ for all n. Note that |c0| = ≤ M and since c1 = +
b0 we have
b0 b0 b0b0 a1r b1r 2
|c1r|≤ + |c0|≤M+M =M(1+M). b0 b0
Suppose, inductively, that |ck rk | ≤ M (1 + M )k for all k < n. Then since an =bnc0 +bn−1c1 +···+b1cn−1 +b0cn,
9
we have
n n n−1
n a r b r bn−1r b r n−1 |cnr |≤ n + n |c0|+ |c1r|+···+ 1 |cn−1r |
b0 b0 b0 b0
≤ M + M 2 + M 2 (1 + M ) + M 2 (1 + M )2 + M 2 (1 + M )3 + · · · + M 2 (1 + M )n−1
2 (1 + M )n − 1 n =M+M M =M(1+M) .
Buinduction,wehave|cnrn|≤M(1+M)n foralln≥0. LetJ1 =a− Letx∈J1 so|x−a|< r . Thenforallnwehave
1+M
r ,a+ 1+M
r . 1+M
n n 1 x−an x−an |cn(x−a) |=|cnr |· · ≤M (1 + M)n r/(1 + M) r/(1 + M)
and so |cn(x − a)n| converges by the Comparison Test. n
Note that from the definition of cn we have an = ckbn−k, and so by multiplying k=0
power series, we have
∞ ∞ ∞ cn(x−a)n bn(x−a)n = an(x−a)n
n=0 n=0
for all x ∈ I ∩J1. Finally note that f(x) = bn(x−a)n is continuous in I and we have
n=0
f(0) = b0 ̸= 0, and so there is an interval J ⊂ I ∩ J1 with a ∈ J such that f(x) ̸= 0 in J.
∞
3.30 Theorem: (Composition of Power Series) Let f(x) = an(x − a)n in an open
n=0
interval I with a ∈ I, and let g(y) = bm(y−b)m in an open interval J with b ∈ J
m=0
andwitha0 ∈J. LetKbeanopenintervalwitha∈Ksuchthatf(K)⊂J. For
each m ≥ 0, let cn,m be the coefficients, found by multiplying power series, such that
∞ n=0
cn,m(x−a)n = bn
∞ n=0
an(x−a)n −b
m
for all x ∈ K,
∞ n cn,m (x − a)
converges and
n≥0
m=0
∞
∞
∞ ∞ n cn,m (x−a) =g f(x) .
m=0
Proof: This follows from Fubini’s Theorem for Series since
n=0
∞∞∞∞∞ gf(x)= b f(x)−bm = b a (x−a)n−b = c (x−a)n .
m mn n,m m=0 m=0 n=0 m=0 n=0
10
n=0
. Then cn,m converges for all m ≥ 0, and m≥0
3.31 Theorem: (Integration of Power Series) Suppose that an(x − a)n converges in ∞
the interval I. Then for all x ∈ I, the sum f(x) = an(x − a)n is integrable on [a, x] (or n=0
[x, a]) and
(x − a)n+1 .
3.32 Theorem: (Differentiation of Power Series) Suppose that an(x − a)n converges
∞
in the open interval I. Then the sum f(x) = an(x − a)n is differentiable in I and n=0
∞
f ′ (x) = n an (x − a)n−1 .
n=1
Proof: We claim that the radius of convergence of an(x − a)n is equal to the radius of convergence of nan(x − a)n−1. Let R be the radius of convergence of an(x − a)n and let S be the radius of convergence of nan(x − a)n−1. Fix x ∈ (a − R,a + R) so |x−a| < R and an(x−a)n converges. Choose r,s with |x−a| < r < s < R. Since
(r/s)n n
lim =0,wecanchooseN sothatn≥N =⇒ r < 1. Thenforn≥N we
x∞ ∞x an(t − a)n dt =
a n=0 n=0 a
∞ an an(t − a)n dt =
Proof: This follows from uniform convergence.
n=0n+1
n→∞ (1/n) s n
have
nan (x − a)n = n r n x−a n an sn ≤ 1 · 1 · |an sn | . sr
Since |ansn| converges, nan(x − a)n converges by the Comparison Test, and so nan(x − a)n−1 converges by Linearity. Thus R ≤ S.
Now fix x ∈ (a−S,a+s) so that |x−a| < S and nan(x−a)n−1 converges. Then nan(x−a)n converges by Linearity, and an(x−a)n ≤ nan(x−a)n so an(x−a)n converges by Comparison. Thus S ≤ R and so R = S as claimed.
The theorem now follows from the uniform convergence of nan (x − a)n−1 . ∞
3.33 Example: We have 1 = (−1)nxn for |x| < 1. By Integration of Power Series,
lnx= ∞ (−1)nxn+1 = ∞ (−1)nxn for|x|<1. Inparticular,wecantakex= 1 toget n+1n 2
1+x
n=0
n=0 n=1 ∞∞∞
ln3 = (−1)n andwecantakex=−1 togetln1 = −1 ,thatisln2= 1 .
2 n·2n 2 2 n=1
n·2n n·2n n=1
n=1
Let us also argue that we can also take x = 1. Note that the series ∞ (−1)n+1 xn
n n=1
diverges when x = −1 (by the Integral Test) and converges when x = 1 (by the Alternating Series Test), so the interval of convergence is (−1, 1]. Thus the sum f (x) = ∞ (−1)n+1 xn
n n=1
isdefinedfor−1
m(t1 − a) ≤ f(l)(t1) − f(l)(a) ≤ M(t1 − a)
1m(t2 −a)2 ≤f(l−1)(t2)−f(l)(a)(t2 −a)≤ 1M(tt −a)2
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for all t2 > a in I. Integrating with respect to t2 from a to t3 gives
1m(t3−a)3 ≤f(l−2)(t3)−f(l−2)(a)−1f(l)(a)(t3−a)3 ≤ 1M(t3−a)3 3! 23!
for all t3 > a in I. Repeating this procedure eventually gives
1 m(tl+1 − a)l+1 ≤ f(tl+1) − Tl(tl+1) ≤ 1 M(tl+1 − a)l+1
(l+1)! (l+1)!
for all tl+1 > a in I. In particular 1 m(x−a)l+1 ≤ f(x)−Tl(x) ≤
1 M(x−a)l+1, (l+1)!
(l+1)!
m≤f(x)−T(x) (l+1)! ≤M.
so
By the Intermediate Value Theorem, there is a number c ∈ [a, x] such that
l (x−a)l+1
f(l+1)(c) = f(x) − Tl(x) (l + 1)! (x − a)l+1
.
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3.40 Theorem: The functions ex , sin x and (1 + x)p are all exactly equal to the sum of their Taylor series centered at 0 in the interval of convergence.
Proof: First let f(x) = ex and let x ∈ R. By Taylor’s Theorem, f(x) − Tl(x) = ecxl+1 (l + 1)!
for some c between 0 and x, and so
e|x||x|l+1 (l+1)!
f(x) − Tl(x) ≤
converges by the Ratio Test, we have lim
Since
gence Test, so lim f(x) − Tl(x) = 0, and so f(x) = lim Tl(x) = T(x).
e|x| |x|l+1 (l + 1)! .
e|x||x|l+1 l→∞ (l+1)!
= 0 by the Diver-
l→∞ l→∞ Nowletf(x)=sinxandletx∈R. ByTaylor’sTheorem,f(x)−T(x)= f(l+1)(c)xl+1
(l + 1)!
for some c between 0 and x. Since f(l+1)(x) is one of the functions ± sin x or ± cos x, we
have f(l+1)(c) ≤ 1 for all c and so
|x|l+1
f(x)−T(x)≤ (l+1)!.
Since
and so we have and f(x) = T(x) as above.
|x|l+1
|x|l+1 l→∞ (l+1)!
converges by the Ratio Test, lim
Finally, let f(x) = (1 + x)p. The Taylor series centered at 0 is
(l+1)!
= 0 by the Divergence Test,
T (x) = 1 + px + p(p−1) x2 + p(p−1)(p−2) x3 + p(p−1)(p−2)(p−3) x4 + · · · 2! 3! 4!
and it converges for |x| < 1. Differentiating the power series gives
T ′ (x) = p + p(p−1) x + p(p−1)(p−2) x2 + p(p−1)(p−2)(p−3) x3 + · · ·
and so
1! 2! 3!
(1+x)T′(x)=p+p+p(p−1)x+p(p−1) +p(p−1)(p−2)x2 1! 1! 2!
+ p(p−1)(p−2) − p(p−1)(p−2)(p−3) x3 + · · · 2! 3!
= p + p·p x + p·p(p−1) x2 + p·p(p−1)(p−2) x3 + · · · 1! 2! 3!
= pT(x).
Thus we have (1+x)T′(x) = pT(x) with T(0) = 1. This DE is linear since we can write it
1+x
as T′(x)− p T(x) = 0. An integrating factor is λ = e − p dx = e−pln(1+x) = (1+x)−p
1+x
and the solution is T (x) = (1+x)−p
0 dx = b(1+x)p for some constant b. Since T (0) = 1 wehaveb=1andsoT(x)=(1+x)p =f(x).
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