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
2
)n/2
. Thus
ln(n!) ≥ n
2
ln(n/2).
So
ln(an) ≥ n
2
ln(n/2) + n ln(|x|) = n
(
1
2
ln(n/2) + ln(|x|)
)
.
For n large enough, the factor in parentheses is ≥ 1 and this implies that
ln(an) ≥ n and so an ≥ en.
In order to converge, we need an → 0.
5. POWER SERIES 13
(2) Find the values of x for which
∞∑
n=1
xn
n!
converges.
We will show the series converges for all x ∈ R. This is clear if x = 0.
Assume x 6= 0. Now, let an =
∣∣xn
n!
∣∣. We have
ln(an) = n ln(|x|)− ln(n!) ≤ n ln(|x|)− n
2
ln(n/2) = n
(
ln(|x|)− 1
2
ln(n/2)
)
.
For n large enough, ln(an) ≤ −n. This implies that an ≤ e−n and so
∑
an
converges by the comparison test.
14 6. SEQUENCES AND SERIES OF FUNCTIONS
Theorem 6.5.1. If
∞∑
n=0
anx
n converges at x0 > 0, then it converges on (−x0, x0)
as well.
Proof: Since
∑
anx
n
0 converges, we have limn→∞ anx
n
0 → 0 and so the sequence
(anx
n
0 ) converges and is hence bounded. Let C be a constant so that |anxn0 | ≤ C for
all n. If y ∈ (−x0, x0), then
|anyn| ≤ |anxn0 | ·
∣∣∣∣ yx0
∣∣∣∣n ≤ C ∣∣∣∣ yx0
∣∣∣∣n .
We have that
∑
C
∣∣∣ yx0 ∣∣∣n is a convergent geometric series since |y/x0| < 1. Thus,∑ |anyn| converges.
Note: The same argument works if x0 < 0. This theorem implies the state-
ment earlier that if
∑
anx
n is a power series, the interval of convergence is either
{0},R, [−R,R], [−R,R), (−R,R) or (−R,R].
5. POWER SERIES 15
Theorem 6.5.2. If
∞∑
n=0
anx
n converges absolutely at x0 > 0, then it converges
uniformly on [−x0, x0] as well.
Let An = |an|xn0 . By assumption
∑∞
n=0An converges. We have |anxn| ≤ An for
x ∈ [−x0, x0] and so
∑
anx
n converges uniformly by the Weierstrass M -test.
Question: Is it true that if a power series converges conditionally at x0 > 0, then
it must converge on [−x0, x0] as well? Must it converge uniformly on [0, x0]?
A1: No. If the series converges conditionally at x0, the radius of convergence is
equal to x0, but the interval of convergence might be either [−x0, x0] or (−x0, x0].
A2: Yes, the series must converge uniformly on [0, x0]. This is a strengthening
of the above theorem is a good bit harder to prove. (See Lemma 6.5.3 and Theorem
6.5.4 in the book.)
16 6. SEQUENCES AND SERIES OF FUNCTIONS
Example 6.5.3. We have
∑∞
n=0
(−1)n
2n+1
= pi
4
.
We are returning to the example from the beginning of the chapter. Let f(x) =∑∞
n=0
(−1)nx2n+1
2n+1
and fn(x) =
(−1)nx2n+1
2n+1
. Since
∑ (−1)n
2n+1
converges by the alternating
series test, the theorem at the bottom of the previous page implies that the series for
f(x) converges uniformly for all x ∈ [0, 1]. In particular, f(x) is continuous on [0, 1].
In particular, f(1) = limx→1− f(x).
Fix a number 0 < c < 1. For x ∈ [−c, c], we have that f ′n(x) = (−1)nx2n and so
|f ′n(x)| ≤ c2n. Since
∑
c2n converges (it’s a geometric series), by the M -test,
∑
f ′n(x)
converges uniformly. For x = 0,
∑
fn(0) = f(0) and by Theorem 4.3, we have that
f(x) is differentiable on [−c, c], and
f ′(x) =
∞∑
n=0
f ′n(x) =
∞∑
n=0
(−1)nx2n = 1
1− (−x2) =
1
1 + x2
.
Since f(0) = 0 and f ′(x) = 1
1+x2
, it follows that f(x) = tan−1(x) on [−c, c]. Since
these are true for all x ∈ [−c, c] for all c < 1, we have f(x) = tan−1(x) for x ∈ (−1, 1).
Thus, by continuity of f and tan−1(x), we have
∞∑
n=0
(−1)n
2n+ 1
= f(1) = lim
x→1−1
f(x) = lim
x→1−
tan−1(x) = tan−1(1) =
pi
4
.
6. TAYLOR SERIES 17
Corollary 6.5.4. If a power series converges on a set A ⊆ R, then it converges
uniformly on any compact set K ⊆ A.
Corollary 6.5.5. If a power series converges on (−R,R), then the differentiated
series
∞∑
n=1
nanx
n−1 converges uniformly on compact subsets of (−R,R) as well.
Note, this does not say anything about the convergence at ±R.
Corollary 6.5.6. If a power series f(x) =
∞∑
n=0
anx
n converges on (−R,R), then
f(x) is continuously differentiable, and its derivatives are given by the term-by-term
differentiation of the series.
6. Taylor series
Theorem 6.6.1 (Taylor’s Formula). Let f(x) =
∞∑
n=0
anx
n converge on some nonempty
interval. Then an =
f (n)(0)
n!
.
Proof: If the power series has radius of convergence R, term-by-term differentia-
tion is allowed, and the radius of convergence does not change. This means that
f (n)(x) = n!an + (n+ 1)(n)(n− 1) · · · 2 · an+1x+ · · · .
In particular, setting x = 0 gives f (n)(0) = n!an and so an =
f (n)(0)
n!
.
18 6. SEQUENCES AND SERIES OF FUNCTIONS
Q: Suppose that f(x) is an infinitely differentiable function, we define the numbers
an by an =
f (n)(0)
n!
and
∑∞
n=0 anx
n converges for some x 6= 0. Does it follow that
∞∑
n=0
anx
n = f(x)
for all x ∈ (−R,R)?
A: No! Let
f(x) =
{
e−1/x
2
if x 6= 0
0 if x = 0.
We have f(0) = 0. Also,
f ′(0) = lim
x→0
f(x)− f(0)
x− 0 = limx→0
e−1/x
2
x
= lim
x→0
1/x
e1/x2
.
We apply L’Hopital’s rule in the ∞/∞ case. We get
f ′(0) = lim
x→0
−1/x2
e1/x2 · (−2/x3)
= lim
x→0
x
2e1/x2
= 0.
Exercise: Prove that f (n)(0) = 0 for all n ≥ 1.
This means that the graph of f(x) is very flat near x = 0.
So in this case, an = 0 for all n and this means that the Taylor series
∞∑
n=0
anx
n = 0
for all x ∈ R. However, f(x) > 0 if x 6= 0. So ∑∞n=0 anxn 6= f(x).
Note: This cannot happen if f(x) were a differentiable function of a complex
variable. A differentiable function of a complex variable is always equal to the sum
of its Taylor series in some neighborhood.
7. WEIERSTRASS APPROXIMATION THEOREM 19
7. Weierstrass Approximation Theorem
Theorem 6.7.1 (Weierstrass approximation theorem). Let f : [0, 1] → R be
continuous. For all > 0, there exists a polynomial p(x) so that |f(x)− p(x)| < for
all x ∈ [0, 1].
Commentary: If {fn(x)} is a sequence of polynomials and fn(x)→ f(x) uniformly,
Theorem 2.4 implies that f(x) is continuous.
This theorem says the converse is true. Every continuous function on [0, 1] is the
uniform limit of polynomials!
(The theorem is also true on [a, b], not just on [0, 1], but the proof is simpler on
[0, 1].)
Overview: Let n be a positive integer. For 0 ≤ k ≤ n, define
bk,n(x) =
(
n
k
)
xk(1− x)n−k.
This is called a Bernstein polynomial.
This is a polynomial that is “largeish” near k/n but is small away from k/n.
If f is a continuous function on [0, 1] and n is a positive integer, the function
fn(x) =
n∑
k=0
f
(
k
n
)
bk,n(x)
is a reasonable approximation to f(x). We’ll prove that fn(x)→ f(x) uniformly.
Lemma 7.2:
• We have ∑nk=0 bk,n(x) = ∑nk=0 (nk)xk(1− x)n−k = 1.
• We have ∑nk=0 knbk,n(x) = x.
• We have ∑nk=0 (x− kn)2 bk,n(x) = x(1−x)n .
20 6. SEQUENCES AND SERIES OF FUNCTIONS
Proof that
∑n
k=0 bk,n(x) = 1.
We have
n∑
k=0
bk,n(x) =
n∑
k=0
(
n
k
)
xk(1− x)n−k
which is exactly the binomial theorem expansion of (x+ (1− x))n = 1n = 1.
Proof that
∑n
k=0
k
n
bk,n(x) = x.
We have that k
n
(
n
k
)
= k
n
· n!
k!(n−k)! =
(n−1)!
(k−1)!(n−k)! =
(n−1)!
(k−1)!((n−1)−(k−1))! =
(
n−1
k−1
)
. Thus,
n∑
k=0
k
n
bk,n(x) =
n∑
k=1
k
n
(
n
k
)
xk(1− x)n−k
=
n∑
k=1
(
n− 1
k − 1
)
xk(1− x)n−k
= x
n∑
k=1
(
n− 1
k − 1
)
xk−1(1− x)(n−1)−(k−1).
Now, we set ` = k − 1. As k varies from 1 to n, ` varies from 0 to n− 1. This gives
x
n−1∑
`=0
(
n− 1
`
)
x`(1− x)(n−1)−` = x · 1
since
∑n−1
`=0 b`,n−1(x) = 1 by above.
7. WEIERSTRASS APPROXIMATION THEOREM 21
Proof that
∑n
k=0
(
x− k
n
)2
bk,n(x) =
x(1−x)
n
.
We have
(
x− k
n
)2
= x2 − 2xk
n
+ k
2
n2
. We already know how to sum
∑
x2bk,n and∑
2xk
n
bk,n, so let’s focus on
∑
k2
n2
bk,n. We will use again that
k
n
(
n
k
)
=
(
n−1
k−1
)
. We get
n∑
k=0
k2
n2
bk,n(x) =
n∑
k=1
k
n
(
n− 1
k − 1
)
xk(1− x)n−k
=
n−1∑
`=0
`+ 1
n
(
n− 1
`
)
x`+1(1− x)n−1−`
=
(n− 1)x
n
n−1∑
`=0
`
n− 1
(
n− 1
`
)
x`(1− x)n−1−` + x
n
n−1∑
`=0
(
n− 1
`
)
x`(1− x)n−1−`
=
(n− 1)x2
n
+
x
n
.
Now,
n∑
k=0
(
x− k
n
)2
bk,n(x) =
n∑
k=0
x2bk,n(x)−
n∑
k=0
2xk
n
bk,n(x) +
n∑
k=0
k2
n2
bk,n(x)
= x2 − 2x2 + (n− 1)x
2
n
+
x
n
= − 1
n
x2 +
x
n
=
x− x2
n
=
x(1− x)
n
,
as desired.
22 6. SEQUENCES AND SERIES OF FUNCTIONS
Proof of the Weierstrass Approximation Theorem:
Suppose that f : [0, 1]→ R is continuous. Fix > 0.
Since [0, 1] is compact, f is uniformly continuous on [0, 1]. Thus there exists some
δ > 0 so that if a, b ∈ [0, 1] with |a − b| < δ, then |f(a) − f(b)| < /2. Also, |f(x)|
is continuous on [0, 1] and so there is a real number M so that |f(x)| ≤ M for all
x ∈ [0, 1]. Choose n ∈ N so that n > 2M
δ2
.
For x ∈ [0, 1] we have
fn(x)− f(x) =
(
n∑
k=0
f(k/n)bk,n(x)
)
− f(x)
=
(
n∑
k=0
f(k/n)bk,n(x)
)
−
(
n∑
k=0
f(x)bk,n(x)
)
=
n∑
k=0
(f(k/n)− f(x))bk,n(x),
and so |fn(x) − f(x)| ≤
∑n
k=0 |f(k/n) − f(x)|bk,n(x). We break the sum on k into
two pieces: the k’s where |k/n− x| < δ and the k’s where |k/n− x| ≥ δ.
First piece: Here, |k/n− x| < δ and so |f(k/n)− f(x)| < /2. This means that
n∑
k=0
|f(k/n)− f(x)|bk,n(x) <
n∑
k=0
(/2)bk,n(x) = /2.
Second piece: If |k/n− x| ≥ δ, then (x−k/n)2
δ2
≥ 1. So∑
k
|k/n−x|≥δ
|f(k/n)− f(x)|bk,n(x) ≤
n∑
k=0
(x− k/n)2
δ2
|f(k/n)− f(x)|bk,n(x).
Since |f(t)| ≤M for all t ∈ [0, 1] we have |f(k/n)− f(x)| ≤ 2M . So the second piece
is bounded by
2M
δ2
n∑
k=0
(
x− k
n
)2
bk,n(x) =
2M
δ2
· x(1− x)
n
and x(1− x) ≤ 1/2 for all x ∈ [0, 1] so we get that the second term is bounded by
M
δ2n
<
M
δ2
· δ
2
2M
=

2
.
In total |fn(x)− f(x)| < for all x ∈ [0, 1] and this proves the theorem.