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CHAPTER 6 Sequences and series of functions

创建日期：2024-05-25 16:14:53

所属分类：数学mathematics

标签： CHAPTER 6

Sequences and series of functions

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CHAPTER 6

Sequences and series of functions
1. The Power of Power Series
We know that
∞∑
n=0
xn =
1
1− x if |x| < 1.
If we differentiate both sides, on the right hand side we get 1
(1−x)2 . If we assume that
d
dx
( ∞∑
n=0
xn
)
=
∞∑
n=0
d
dx
(xn),
then we get
∞∑
n=1
nxn−1 =
1
(1− x)2 .
Multiplying by x and setting x = 1/2 gives
∞∑
n=1
nxn =
x
(1− x)2 =⇒
∞∑
n=1
n
2n
=
(1/2)
(1− 1/2)2 = 2.
Is this valid? Is the derivative of the infinite sum the infinite sum of the derivatives?
Another, potentially more striking example is the following. Take
∞∑
n=0
xn =
1
1− x if |x| < 1.
and set x = −y2. Then we find
1
1 + y2
= 1− y2 + y4 − y6 + · · ·
The left hand side is the derivative of tan−1(y), so antidifferentiating both sides we
find
tan−1(y) = y − 1
3
y3 +
1
5
y5 − 1
7
y7 + · · ·
Since tan(pi
4
) = 1⇐⇒ pi
4
= tan−1(1), we get
pi
4
= 1− 1
3
+
1
5
− 1
7
+ · · · =⇒ pi = 4− 4
3
+
4
5
− 4
7
+ · · ·
1
2 6. SEQUENCES AND SERIES OF FUNCTIONS
yet, plugging x = 1 or y = 1 to most of the above equations, the result is undefined.
So is this really pi?
In this chapter, we will discuss when the order of two limiting processes can be
reversed. We will talk about power series, and about approximating functions with
polynomials.
2. Convergence of a sequence of functions
Definition 6.2.1. For each n ∈ N, let fn(x) be a function defined on a set D ⊂ R.
We say the sequence {fn(x)}∞n=1 converges pointwise on D to a function f(x) if for
every x ∈ D, we have lim
n→∞
fn(x) = f(x). In other words, for every > 0 and for
every x ∈ D, there exists an N ∈ N (that can depend on and x) such that if n > N ,
|fn(x)− f(x)| < .
(1) What does fn(x) =
x
n
converge to on R?
It converges to 0. For all x ∈ R we have limn→∞ xn = x · limn→∞ 1n =
x · 0 = 0.
(2) What does fn(x) =
1
1+xn
converge to on [0, 1]?
If x < 1, we have limn→∞ xn = 0.
But 1n = 1 and so limn→∞ xn =
{
0 if x < 1
1 if x = 1
. This gives
lim
n→∞
1
1 + xn
=
{
1 if x < 1
1/2 if x = 1
.
Notice that in this case, fn(x) is continuous for all n, but limn→∞ fn(x) is
not continuous.
(3) For a set A, define χA(x) =
{
1 if x ∈ A
0 if x 6∈ A .
This is called the characteristic function of A. What does fn(x) =
χ[n,n+1](x) =
{
1 if x ∈ [n, n+ 1]
0 if x otherwise
converge to on (0, 1)?
0. Choose N ∈ N with N > x. Then for n ≥ N , fn(x) = 0 since fn(x) is
only nonzero for inputs ≥ n ≥ N > x. In this case, we have ∫∞−∞ fn(x) dx = 1
for all x, but
∫∞
−∞ limn→∞ fn(x) dx = 0.
2. CONVERGENCE OF A SEQUENCE OF FUNCTIONS 3
(4) What does fn(x) = x
n converge to on [0, 1]?
As before, we have limn→∞ xn = 0 if x < 1 and limn→∞ xn = 1. Thus,
the limit is
f(x) =
{
0 if x ∈ [0, 1)
1 if x = 1.
So pointwise converges is not sufficient to conclude the limit function is differen-
tiable or even continuous. Also, we see that the sequence {fn(x)}∞n=1 may converge
to zero and yet {∫ fn(x)dx}∞n=1 does not converge to zero, which is strange. We need
a stronger notion of convergence: uniform convergence.
Definition 6.2.2. For each n ∈ N, let fn(x) be a function defined on a set
D ⊂ R. We say the sequence {fn(x)}∞n=1 converges uniformly on D to a function
f(x) if for every > 0, there exists an N ∈ N such that for all x ∈ D, and n > N ,
|fn(x)− f(x)| < .
What’s the difference between uniform and pointwise convergence? Which is
stronger? Does one type of convergence imply the other? How do you prove a
sequence of functions does not converge uniformly?
• In pointwise convergence, the N chosen depends on the x. In uniform conver-
gence, it doesn’t.
• Uniform convergence is stronger, and uniform convergence implies pointwise
convergence.
• Show that there exists some 0 > 0 so that for all N ∈ N, ∃n ≥ N and some
xn ∈ D so that |fn(xn)− f(xn)| ≥ 0.
Take a look at figures 6.4 and 6.5 on page 179 in the textbook.
(1) Does fn(x) =
1
n(1+x2)
converge uniformly on R?
Yes. Fix > 0 and choose N ∈ N so that 1
N
< . Then
|fn(x)− 0| = 1
n(1 + x2)
≤ 1
n
≤ 1
N
< .
(2) Does fn(x) =
x2+nx
n
converge uniformly on R?
First, we have that limn→∞ x
2+nx
n
= limn→∞ x+ limn→∞ x
2
n
= x+ 0.
Let 0 = 1. For any n ∈ N we can choose some x ∈ R so that
|fn(x)− x| =
∣∣∣∣x2n
∣∣∣∣ ≥ 1,
for example x =
√
n.
(3) Does fn(x) = x
n converge uniformly on [0, 1]?
4 6. SEQUENCES AND SERIES OF FUNCTIONS
No. Let 0 = 1/2. Suppose that N ∈ N and n ≥ N . Let x = 121/n . Then
|xn − 0| =
∣∣∣∣12 − 0
∣∣∣∣ ≥ .
We have an analogue of the Cauchy criterion for uniform convergence.
Theorem 6.2.3. A sequence of functions {fn(x)}∞n=1 converges uniformly on A if
and only if for every > 0, there exists N ∈ N such that for all m,n ≥ N and all
x ∈ A, we have |fn(x)− fm(x)| < .
Proof: Homework problem.
Theorem 6.2.4. If {fn(x)}∞n=1 is a sequence of functions which converges uni-
formly to f(x) on a domain D, and each fn(x) is continuous on D, then f(x) is
continuous on D.
Fix c ∈ D and > 0. Since fn → f uniformly, there is some N ∈ N so that
n ≥ N =⇒ |fn(x)− f(x)| < /3 for all x ∈ D.
We’ll use that
|f(x)− f(c)| ≤ |f(x)− fN(x)|+ |fN(x)− fN(c)|+ |fN(c)− f(c)|.
Since fN is continuous, ∃δ > 0 so that if |x− c| < δ, |fN(x)− fN(c)| < /3.
We have that |f(x) − fN(x)| < /3 and |fN(c) − f(c)| < /3 by the uniform
convergence also.
Note: The place we use the uniform convergence is in handling |f(x) − fN(x)|.
We can’t control x in this inequality and so if there were values of x close to c for
which |f(x)− fN(x)| were large, the conclusion might not be true.
2. CONVERGENCE OF A SEQUENCE OF FUNCTIONS 5
Is the converse true? If fn(x) is continuous and f(x) is continuous and fn → f
pointwise, is the convergence uniform?
A1: No. Above we showed that if fn(x) =
x2+nx
n
, then fn → f where f(x) = x.
A2: If we add an additional condition, then yes.
Theorem 6.2.5. Assume that fn → f pointwise on a compact set K and assume
that for each x ∈ K the sequence (fn(x)) is increasing. If each fn(x) is continuous
and f(x) is continuous, then the convergence is uniform.
Proof: Homework.
6 6. SEQUENCES AND SERIES OF FUNCTIONS
3. Uniform convergence and differentiation
Theorem 6.3.1. If {fn(x)}∞n=1 is a sequence of functions which converges point-
wise to f(x) on [a, b], and each fn(x) is differentiable on [a, b], and {f ′n(x)}∞n=1 con-
verges uniformly to a function g(x) on [a, b], then f(x) is differentiable and g(x) =
f ′(x).
Proof: Our goal is the following. Fix a c ∈ [a, b]. Then we need to show that for
all > 0, we need to show that ∃δ > 0 so that if |x− c| < δ, then∣∣∣∣f(x)− f(c)x− c − g(c)
∣∣∣∣ < .
The strategy is to use the triangle inequality with∣∣∣∣f(x)− f(c)x− c − g(x)
∣∣∣∣ ≤ ∣∣∣∣f(x)− f(c)x− c − fn(x)− fn(c)x− c
∣∣∣∣+∣∣∣∣fn(x)− fn(c)x− c − f ′n(c)
∣∣∣∣+|f ′n(c)−g(c)|.
We handle the third term using that f ′n converges pointwise to g. We handle the
second term using that fn is differentiable. The first term is the trickiest to handle.
Our hypothesis is that f ′n → g uniformly. So we relate the first term to derivatives
by using the MVT and use the Cauchy criterion for uniform convergence.
Fix > 0.
Last term: Since f ′n(c) converges pointwise to g(c), ∃N1 so that for n ≥ N1,
|f ′n(c)− g(c)| < /3.
Middle term: Since limx→c
fn(x)−fn(c)
x−c = f
′
n(c), ∃δ > 0 so that if |x− c| < δ,∣∣∣∣fn(x)− fn(c)x− c − f ′n(c)
∣∣∣∣ < /3.
First term: Since fn → f pointwise, we can rewrite∣∣∣∣f(x)− f(c)x− c − fn(x)− fn(c)x− c
∣∣∣∣ = ∣∣∣∣ limm→∞ fm(x)− fm(c)x− c − fn(x)− fn(c)x− c
∣∣∣∣ .
Now, let h(x) = fm(x)−fn(x). The MVT says that there is some α (that depends
on m) between x and c so that h(x)−h(c)
x−c = h
′(αm). Rewriting this gives
fm(x)− fn(x)− fm(c) + fn(c)
x− c = f
′
m(αm)− f ′n(αm)
and so the first term becomes |limm→∞ f ′m(αm)− fn(αm)|. Since {f ′n} converges uni-
formly, the Cauchy criterion for uniform convergence gives that there is some N2 so
that for m,n ≥ N2, we have |f ′m(x)− f ′n(x)| < /3 for all x ∈ [a, b]. This implies that∣∣∣∣fm(x)− fm(c)x− c − fn(x)− fn(c)x− c
∣∣∣∣ < /3.
4. SERIES OF FUNCTIONS 7
Thus
fn(x)− fn(c)
x− c − /3 <
fm(x)− fm(c)
x− c <

3
+
fn(x)− fn(c)
x− c .
The order limit theorem now gives that
fn(x)− fn(c)
x− c − /3 ≤
f(x)− f(c)
x− c ≤

3
+
fn(x)− fn(c)
x− c
and this shows that for n ≥ max{N1, N2} we have that the second term is ≤ /3.
In total, for |x− c| < δ, we have
∣∣∣f(x)−f(c)x−c − g(c)∣∣∣ < .
The previous theorem can be relaxed a little. In fact, you can prove the following
theorem.
Theorem 6.3.2. Suppose {fn(x)}∞n=1 is a sequence of differentiable functions de-
fined on [a, b], and {f ′n(x)}∞n=1 converges uniformly to a function g(x) on [a, b]. If
there exists one point x0 ∈ [a, b] such that {fn(x0)}∞n=1 converges, then the sequence
{fn(x)}∞n=1 converges uniformly to a differentiable function f(x) and f ′(x) = g(x).
Proof. Exercise.
Q: Why do we need to assume that there is some x0 ∈ [a, b] so that (fn(x0)
converges?
A: Knowing f ′n(x) doesn’t uniquely determine f . For example, if fn(x) = x + n,
then f ′n(x) = 1, which converges uniformly to g(x) = 1. However, there is no function
f(x) with f ′(x) = 1 so that fn(x)→ f(x) pointwise.
4. Series of functions
Now that we understand sequences of functions, it is time to study series of func-
tions.
Definition 6.4.1. The formal series
∞∑
n=1
fn(x) = f1(x) + f2(x) + . . .
is said to converge pointwise to a function f(x) on a domain D if for each x ∈ D,
the sequence of partial sums
SN(x) = f1(x) + f2(x) + · · ·+ fN(x)
converges to f(x) for each x ∈ D. The series is said to converge uniformly to f(x)
if lim
N→∞
SN(x) = f(x) uniformly.
8 6. SEQUENCES AND SERIES OF FUNCTIONS
Theorem 6.4.2. Suppose {fn(x)}∞n=1 is a sequence of continuous functions on D,
and assume
∞∑
n=1
fn(x) = f(x) uniformly on D. Then f(x) is continuous.
Idea: Let Sn(x) = f1(x) + f2(x) + · · ·+ fn(x). Then each Sn(x) is continuous and
Sn(x)→ f(x) uniformly. By Theorem 2.4, f(x) is continuous.
Theorem 6.4.3. Suppose {fn(x)}∞n=1 is a sequence of differentiable functions de-
fined on [a, b], and assume
∞∑
n=1
f ′n(x) = g(x) uniformly on [a, b]. If there exists one
point x0 ∈ [a, b] such that
∞∑
n=1
fn(x0) = f(x0), then the series
∞∑
n=1
fn(x) converges
uniformly to a differentiable function f(x) and f ′(x) = g(x) on [a, b].
Idea: This follows from Theorem 3.2 above.
4. SERIES OF FUNCTIONS 9
The same theorems we discussed in chapter 2 for series can be applied to series
of functions. Therefore, re-writing them in the context of function series we have the
two theorems.
Theorem 6.4.4 (Cauchy criteria for uniform convergence of series). The series
∞∑
n=1
fn(x) = g(x) uniformly on D if and only if for all x ∈ D and every > 0 there
exists an N ∈ N such that for all n > m > N
|fm+1(x) + · · ·+ fn(x)| < .
Idea: This is a version for series of Theorem 2.3.
Theorem 6.4.5 (Comparison Test, also known as the Weierstrass M-Test). Sup-
pose {fn(x)}∞n=1 is a sequence of functions on D. For each n = 1, 2, . . . and for
all x ∈ D, suppose |fn(x)| ≤ an. Suppose that
∑∞
n=1 an converges. Then
∞∑
n=1
fn(x)
converges uniformly on D.
Proof: Since
∑
an converges, the sequence sn = a1 +a2 + · · ·+an is Cauchy. This
implies that for all > 0, ∃N ∈ N such that n > m > N implies that
am+1 + am+2 + · · ·+ an < .
However,
|fm+1(x) + · · ·+ fn(x)| ≤ am+1 + am+2 + · · ·+ an <
and so the previous theorem implies the convergence is uniform.
Example 6.4.6. Show f(x) =
∞∑
n=1
xn
n2
is continuous on [−1, 1].
Let fn(x) =
xn
n2
. Then |fn(x)| ≤ 1n2 for x ∈ [−1, 1]. and since
∑∞
n=1
1
n2
converges,
the Weierstrass M -test applies and gives that
∑
fn(x) converges uniformly. The
uniform limit of continuous functions is continuous.
10 6. SEQUENCES AND SERIES OF FUNCTIONS
Example 6.4.7. Show f(x) =
∞∑
n=1
xn
n2
is differentiable on (−1, 1).
This is similar but a little bit more involved. Let fn(x) =
xn
n2
. We have f ′n(x) =
nxn−1
n2
= x
n−1
n
. Let a be any positive real number < 1. We’ll show that
∑
f ′n(x)
converges uniformly on [−a, a].
We have |f ′n(x)| ≤ a
n−1
n
and
∑∞
n=1
an−1
n
converges, since a
n−1
n
≤ an−1 and∑∞n=1 an−1
is a convergent geometric series.
Finally, we need to know that there is some x0 ∈ [−a, a] so that
∑
fn(x0) = h(x0).
This is true for x0 = 0. So Theorem 4.3 implies that f(x) is differentiable and that
f ′(x) =
∑∞
n=1
xn−1
n
for all x ∈ [−a, a]. But since a < 1 was arbitrary, we have that
these statements are true in
⋃
a<1[−a, a] = (−1, 1).
4. SERIES OF FUNCTIONS 11
Example 6.4.8. Show f(x) =
∞∑
n=1
sin(nx)
n3
is differentiable on R. Is it twice
differentiable?
This starts like the last argument. We let fn(x) =
sin(nx)
n3
. Then f ′n(x) =
cos(nx)
n2
.
We have |f ′n(x)| ≤ 1n2 and
∑
1
n2
converges. Since
∑
fn(0) = 0 = f(0), Theorem 4.3
implies f(x) is differentiable and
f ′(x) =
∞∑
n=1
cos(nx)
n2
.
Now, the plot thickens. We have f ′′n(x) =
− sin(nx)
n
, and so |f ′′n(x)| ≤ 1n and
∑
1
n
diverges. So the Weierstrass M -test doesn’t work.
Is it the case that the convergence is still uniform, but the M -test just isn’t strong
enough to prove it? No, the convergence is not uniform.
Since sin(x)
x
→ 1, there is some δ > 0 so that for 0 < x < δ, sin(x)
x
≥ 1/2.
Let sn(x) = − sin(x)− sin(2x)2 −· · ·− sin(nx)n be the partial sums of
∑
f ′′n(x). Choose
N ∈ N and consider |s2N(δ/2N)− sN(δ/2N)|. This is
2N∑
n=N+1
sin(n(δ/2N))
n
≥
2N∑
n=N+1
1
2
· nδ
2N
n
=
2N∑
n=N+1
δ
4N
≥ δ
2
.
So we have infinitely many cases of |sm(x)−sn(x)| ≥ δ/2, and by the Cauchy criterion,
the convergence is not uniform.
Q: Does
∑
f ′′n(x) converge? Is f(x) twice differentiable?
A: Yes, the series
∑
f ′′n(x) does converge. No, f(x) is not twice differentiable. In
particular, f ′′(0) does not exist.
12 6. SEQUENCES AND SERIES OF FUNCTIONS
5. Power Series
Some functions can be expressed as a power series; series of the form
f(x) =
∞∑
n=0
anx
n = a0 + a1x+ a2x
2 + . . . .
Clearly all power series converge when x = 0. Where else must they converge? We
will show a power series converges on {0}, R, or a bounded interval (−R,R), [−R,R),
(−R,R], or [−R,R]. The maximum value R is call the radius of convergence.
(1) Find the values of x for which
∞∑
n=1
n!xn converges.
The ratio test was Exercise 2.7.9 and we skipped it. The series converges
if x = 0 and diverges otherwise.
Assume x 6= 0. Let an = |n!xn| so ln(an) = ln(n!) + n ln(|x|). We have
n! ≥ n · (n− 1) · · · · 1.
At least half the terms in this product are ≥ n/2 and so n! ≥ (n

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MAS8383 ECO 512 STAT 8310 CMN3102 MA577 CMPT 225 COMP0045 Methods Equations MATH 113 Math 3283W ICS-33 EN4302/ENT603 BIA B350F MA318 MTHM506 MECH 5315M ECONOMICS MAS8382 COM 2004 PX3142 XJEL 3662 STAT 231 MGT B399F PX4131 COM2108 30AM MSCI521 ECONM1015 ACFI304 GR5241 MATH10101 ECON 2070 COMS31900 EFIM20011 CSE 250B ACSE-5 STAT4207 SOFT2412 ELEC 331 AY2020 MSIN0209 FIT2094 CS 241L ST404 CS 411 3D CA3 INF4002 COMP3506 SIT718 CS 188 CS 170 MATH 138 CSE 251A CMT118 DATA5207 CMPT307 CMPT459 T1 PGEE11108 COSC 3P32 ME 163 MATH96007 7QQMM203 EE6227 BA 574 STA 4234 ECO206Y1Y ECO2008 MT2300 MATH 11158 CS 527 MACM 316 CSE474 2DI70 ISE 525 COMP3411 EMAT30007 7CCSMBDT MFIN704 MFIN704W21 MfIB ECE 438 CS 436 F36 CS5344 FE 621A FE621 TBA 612 ECM3151 ECMM422 MF1 MSIN0107 ESSMENT CS 5854 MFIT 5008 MIPS R4000 ENG5274 BU52018 ENG2079 BIOA02 STA2570 F0 INF553 STAT 440 STAT3008 COMP9417 COMP6451 6COM1042 BU52006 ST303 HW-1 CISC-235 CS 510 MATH3161 ECN 140 ECE391 Comp 3613 CVEN 9625 PRG455 UESTION 1 EARN 5 FE 825 IERG 5130 STAT 8178 EE161 ADS2 G54FUZ CS326 113B ECN 226 QBUS6860 ELEC3202 COMP9334 MGT11B MET CS 777 STA238 00099F Python SMM306 CMT206 EC116 SP MAST90084 CS480/680 BCPM0091 CMPT 300 ECON2209 FY20 STAT 432 CVEN9625 MATH1712 CMP9771M CMP3750M CIS 325 AI506 MAT344H1S MGT7180 PSTAT 160B CSYS5010 1D STAT 453/558 ME152B STA304H1F CSCI-128 ME428 CO3002 FINE 7640 COMP3234 ECSE 223 EE497 – 3D MATH5905 ECS648U ENG5009 MATH10242 EC665 A CPI221 ISOM5220 MACE40402 MGT B456F COMP9315 AFM 113 STAT 778 CS 564 MA231 CS 230 CS 124 COMP5511 CSC317 ITEC1010 ELEC207 INFT 3019 7CCMFM06 EE 161 5G ECLT5810 MATH 7349 CS526 O2 COMP90054 ELE00063M ACCT5034 ARC1015 CSCI 40 CSCI E40 FIN 524B CS 462 389COM CMT309 MFIN703 COMP4250 STA304 STA304H1S STA 1003 FIN 448 BUFN 732 CS520 CVEN3701 ECMT6006 PHAS0100 CSIT314 MDIA 5028 CSCI 2134 MAT 451 CMPT 459 STAT 457 INFO20003 COMP4141 IELE8066 F540 ECM9 CSE 6321 FIN 430 MATH3911 BFC5916 COMP1011 EEE 6209 STATS 240 CW3 COMP206 COMP1003 4331W ISE 530 CMPT 295 MATH 239 QF5203 INFR08010 IT114105 FIT2093 ECON3124 ECON1195 MATH 337 STAT 3006 COMP202 COMP3130 MATH3609 COMP 250 CS6140 IEEE ELE00041I FIN40090 CIV 5899 MATH3066 IGNMENT MATH 2009 STAT3405 HIST 101 MATH 3281 CEGE0042 ICT532 LAB 7 LAB 5 MCO455 CSIT214 Q4 IFB104 A1 ECO220Y5Y DA5020 CMPSC473 FIN9007 DSCI553 MSCI534 COMP3308 CSE-112 CMPT 310 EE104 F 2C P20 ALY2010 BUSM060 STA258 ECO00030M R BIOSTAT 274 COS6012-B FIT3155 CMPUT 340 EC421 INFS4205 ECOM135 NATS 1740 COMP809 MBE 6005 STA465 CSSE2310 CIVL2700 DATA4700 CMPSC 473 ISE 529 CS5044 MAST30020 COMP2207 ENGG1001 COMP0046 FIT3165 STAT GR5241 MECH0007 vATHK1001 ANALYTIC ATHK1001 MATH352001 HW3 S1 BMAN 24662 COMP 636 Stat-461 STAT 302 STAT 2132 STAR534 ENG4137 CMP3752M CS3103 BISM3222 STA 135 ME155A IIMT3601 ECON3371 ECON 3371 HW4 FINA 5840 DT211C CTEC3902 MAT005 MATH2148 S261F Econ 332 MATH3734 MIS41270 7CCMFM13 SYST 664 CIS434 CENG10014 MT4111 MT4527 ECMM424 CMT219 Math 136 MATH 3120 COMP10002 MAST20009 MATH0058 BEEM117 ISYE 6420 MGMTMFE 405 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