程序代写代做代考 C Chapter 3. Sequences and Series of Functions

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 􏰀48􏰂x−1􏰃􏰂1−x􏰃if 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 and􏰈􏰈fl(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 and􏰈􏰈fl(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≥mwehave􏰈􏰈fn(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)mi(fn)−4(b−a),
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 aa􏰈􏰈ε 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􏰈􏰈 􏰈aa􏰈􏰈a􏰈 􏰓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 􏰈􏰈l􏰈􏰈l 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=􏰂2−1􏰃4n+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−1􏰏p 􏰐 􏰏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 􏰏l􏰐l = 􏰒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)̸=0so􏰒nn(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
limsup􏰘n |an(x−a)n|=|x−a|limsup􏰘n |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 limsup􏰘n |an(x−a)n|=|x−a|limsup􏰘n |an|0. ChooseN sothatl>m>N =⇒ 􏰈􏰈 􏰈n=m
anR 􏰈􏰈<ε. 􏰈 Then by Abel’s Formula and using telescoping we have 􏰈􏰈 l 􏰈􏰒 􏰈 􏰈n=m 􏰈􏰈 􏰈􏰈 l n􏰈 􏰈􏰒 􏰈􏰈 n 􏰂x−a􏰃n􏰈 an(x−a) 􏰈=􏰈 􏰈 􏰈n=m anR R 􏰈 􏰈 􏰈􏰈􏰏 l 􏰊􏰈􏰈 =􏰈anRR− anRR−R􏰈 l−1 􏰏 p 􏰐􏰉 􏰈 􏰒 n 􏰂x−a􏰃l 􏰒 􏰒 n 􏰂x−a􏰃p+1 􏰂x−a􏰃p 􏰈 􏰐 􏰈 n=m 􏰈􏰈 l 􏰈􏰈 p=m n=m l−1 􏰈􏰈 p 􏰈􏰈􏰉 􏰈 􏰊 􏰈􏰒 n􏰈􏰂x−a􏰃l 􏰈n=m 􏰈 p=m 􏰈n=m 􏰈 􏰒 􏰈􏰒 n􏰈 􏰂x−a􏰃p ≤􏰈 anR􏰈 R + 􏰈 anR􏰈 R − R 􏰂x−a􏰃p+1 <ε􏰂x−a􏰃l +ε􏰉􏰂x−a􏰃m −􏰂x−a􏰃l􏰊=ε􏰂x−a􏰃m <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 = 3􏰃n 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 􏰈an􏰈􏰈n+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,wehave􏰁a =􏰁 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.
Then􏰁cn(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−a􏰈n 􏰈x−a􏰈n |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 ∞∞􏰌∞􏰍∞􏰌∞􏰍 g􏰂f(x)􏰃= 􏰒b 􏰂f(x)−b􏰃m = 􏰒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 a (the case that x < a is similar). Since f(l+1) is differentiable and hence continuous, by the Extreme Value Theorem it attains its maximum and minimum values, say M and m. Sincem≤f(l+1)(t)≤M forallt∈I,wehave 􏰓 t1 􏰓 t1 􏰓 t1 mdt ≤ f(l+1)(t)dt ≤ M dt aaa that is for all t1 > a in I. Integrating each term with respect to t1 from a to t2, we get
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
22
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
.
13

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). 14