MATH 1D, WEEK 5 UNIFORM CONVERGENCE AND

POWER SERIES

INSTRUCTOR: PADRAIC BARTLETT

Abstract. These are the lecture notes from week 5 of Ma1d, the Caltech
mathematics course on sequences and series.

1. Midterm information
Midterm average: 79%.
Good things: People, on the whole, did rather well! In particular, I was im-
pressed with the style people displayed in their proofs; a number of students
really stepped up their mathematical game for this test.
However, there were a few issues: in particular, people didnt seem very
comfortable with some of the basic definitions. There were more than a
few tests that confused series and sequences, and a plurality that forgot
how to deal with conditionally convergent series altogether; given that the
test was open-book and open-notes, this was a little worrisome.
That said, as a whole, the class exceeded my expectations. Weve covered
a huge amount of material through the first four weeks of this course; a
retention rate of about 80%, considering how fast weve been moving, is
wonderful. Good job!

2. Uniform Convergence
So: last class, we discussed what it might mean for a sequence of functions
to converge towards this goal, we came up with two possible definitions of
convergence, which are reprinted below for your convenience:
Definition 2.1. For a sequence of functions {fn }, from some set A to R, we say
that
lim fn = f pointwise
n
if for every x in A, we have that
lim fn (x) = f (x).
n

Definition 2.2. For a sequence of functions {fn }, from some set A to R, we say
that
lim fn = f uniformly
n
if for every  > 0, there is some N N such that whenever n > N , we have
|fn (x) = f (x)| < 
for every x in A.
1
2 INSTRUCTOR: PADRAIC BARTLETT

Both definitions have their own advantages and disadvantages; as we showed in

the last lecture, pointwise convergence is an easier condition to check and study,
while uniform convergence has the advantage of conserving a number of qualities
that we like in functions. Specifically, we showed that uniform convegence pre-
served the properties of continuity and the integral, in certain well-defined ways.
Given the results of our last lecture, then, a natural question to ask here is
whether uniform convergence preserves other properties of functions: namely, whether
uniform convergence preserves the concept of differentiability. To answer this ques-
tion, we first consider the following pair of examples:

Example 2.3. Suppose that

1
fn (x) = sin(n2 x).
n
Does the sequence {fn } uniformly converge? If so, where does it converge to? Also,
what happens to the derivatives of these functions at 0?

Proof. We first graph the fn s, to better illustrate whats going on here:

So: from the picture above, it seems very likely that these functions are uniformly
converging to zero.
To prove this, simply take any  greater than zero, and let N > 1 . Then, for
any n > N , we have that
 
1 1  1 1
|fn (x) 0| =  sin(n x) = sin(n2 x) <
2 
 
 < ;
n n n N
so the sequence fn uniformly converges to 0, as we claimed.
 MATH 1D, WEEK 5 UNIFORM CONVERGENCE AND POWER SERIES 3

The derivatives at zero, however, are far less well-behaved: in fact, we have that

lim (fn )0 (0) = lim n cos(n2 x)

n n 0
= lim n cos(0)
n
= lim n
n
= ,

whereas the derivative of 0 at 0 is merely 0. So the derivatives here were not

convserved i.e. the limit of the derivative of the functions was not equal to the
derivative of the limit! 

Example 2.4. Suppose that

r
1
fn (x) = x2 +
n2
Does the sequence {fn } uniformly converge? If so, where does it converge to? Also,
what happens to the derivatives of these functions at 0?

Proof. Again, we graph the fn s to better show whats going on here:

So: the graphs above are just those of the upper branch of the hyperbola y 2 x2 =
1
n2 ;as such, we can visually see that these graphs will converge uniformly to the
graph of their asymptotes, |x|.
4 INSTRUCTOR: PADRAIC BARTLETT

To prove this, simply notice that for any x, n,

2 21 2 1 x2
x x + 2 x + 2 +2 2
r n n rn
1 1
x2 x2 + 2 x2 +
n n2
r
1 1
|x| x2 + 2 |x| + ;
n n

1
so, if we let N >  we will again have that
 
 1  1
|fn (x) |x|| |x| + |x| = < 
n n

and thus that these functions uniformly converge to |x|.

However, again, we have that the derivatives are not conserved at zero! I.e. we
explicitly have that

0 2x 
(fn ) (0) = q  = 0,

1
2 x2 +  n2 0

but |x| doesnt have a derivative at zero! So, again, the limit of the derivatives is
not the derivative of the limits. 

These initial results might seem somewhat discouraging for our hopes of pre-
serving any notion of the derivative. However, it bears noting that both of these
sequences failed us in the same way specifically, in both cases, we had the limit
of the derivatives of the fn s converging to something discontinuous and otherwise
ill-behaved. In fact, it turns out that this is the only way in which differentiabil-
ity can not be conserved i.e. if we force the derivatives of the fn s to converge
to something continuous, then the derivatives of the fn s in fact converge to the
derivative of their limit! The following theorem simply states this:

Theorem 2.5. If the limit limn fn = f uniformly, and the limit limn fn0 con-
verges uniformly to some continuous function, then f is differentiable and limn fn0 (x) =
f 0 (x).

Proof. So; because the function limn fn0 converges uniformly, we have that
Z x Z x
lim fn0 (t)dt = lim fn0 (t)dt
a n n a
= lim fn (x) fn (a)
n
= f (x) f (a).

Then, because limn fn0 (x) is continuous, we can use the fundamental theorem
to conclude that limn fn0 (x) = f 0 (x). 
 MATH 1D, WEEK 5 UNIFORM CONVERGENCE AND POWER SERIES 5

3. Power Series
Just like with sequences of numbers, it turns out that a large part of why we
study sequences of functions is to better understand series of functions. Explicitly,
we make the following definition:
DefinitionP3.1. For a sequence of functions fn from some set A to R, we say that
the sum n=0 fnn converges
PN o uniformly to some function f : A R if and only
if the sequence n=0 fn converges uniformly to f .
N =1
So: we have a quick triple of theorems, the proofs of which are immediate from
our proofs of the same results for sequences (and are thus omitted:)
P
Theorem 3.2. Suppose that the sum n=0 fn converges uniformly to some func-
tion f : [a, b] R. Then,
If all of the fn s are continuous on [a, b], then so is f .
Rb P R b
If all of the fn s are integrable on [a, b],then a f = n=1 a fn .
P
If
Pthe sum n=1 fn0 converges to a continuous function on [a, b], then f 0 =
0
n=1 fn .

So: the above theorems are amazingly powerful, but getting to use them, at
the moment, is something of a hassle; while we know a number of things about a
series once weve established that its uniformly convergent, we really dont have
any tools beyond the definitions to show that a sequence is uniformly convergent.
The theorem below, thankfully, will change that for us:
Theorem 3.3. (Weierstrass M-test:) Suppose that {fn } is a sequence of functions
from some set A to R, and {Mn } is a sequence of numbers such that |fn (x)|
Mn , x A, with the following properties:
|f
Pn (x)| Mn , for every x in A, and
n=1 Mn converges absolutely.
P
Then the sum n=1 fn converges uniformly.
Proof. First, note that for every x A,
X
X
|fn (x)| Mn <
n=1 n=1
P
and thus (by the comparison test) we have that P n=1 |fn (x)| converges absolutely,
for every x A. Denote the function that n=1 fn (x) converges pointwise to by
f (x).
Thus, we have that for every N N,

 
X  X 
|f (x) f1 (x) f2 (x) . . . fN (x)|  fn (x)
 
 
n=1 n=N +1

X
|fn (x)|
n=N +1
X
Mn .
n=N +1
6 INSTRUCTOR: PADRAIC BARTLETT

P P
Because n=1 Mn converges absolutely, we know that this quantity n=N +1 Mn
goes to 0 as N goes to infinity; so, for any  > 0 we can pick a N such that for
every n > N ,

X
|f (x) f1 (x) f2 (x) . . . fn (x)| < .
n=1

As this is P
precisely the condition for uniform convergence, weve thus shown that

the series n=1 fn (x) converges uniformly. 

We will use this theorem mostly to study when certain series of functions, called
power series, converge. To illustrate how this goes, we first define what a power
series is, and work several example problems discussing their convergence:
Definition 3.4. A power series around the point a is a series of the form

X
f (x) = an (x a)n ,
n=0

where the an are a collection of real numbers. Typically, we will be concerned with
power series around the point 0, which take the form

X
f (x) = an xn .
n=0

Example 3.5. Suppose that

X
f (x) = xn .
n=0

Where does this series converge? How does it converge?

Proof. First, note that if |x| 1, we have that limn xn 6= 0, and thus this series
cannot possibly converge. So it suffices to consider the case when x lies in the
interval (1, 1); in this situation, we have that
  
X   1 
n
|f (x)| =  x  =  <


n=0
 1 x

by using our summation identity for the geometric series. P

This, furthermore, proves that on any interval [a, a] (1, 1), the sum n=0 xn
1
converges uniformly to 1x , as on any such interval we have
|xn | an .
P
Because Pn=1 an converges, we can then apply the Weierstrass M-test to show that

the sum n=0 xn must converge uniformly on [a, a]. 

Example 3.6. Suppose that

X (1)n n
f (x) = x
n=1
n

Where does this series converge? How does it converge?

 MATH 1D, WEEK 5 UNIFORM CONVERGENCE AND POWER SERIES 7

n
Proof. First, note that if |x| > 1, we have that limn (1) n x
n
6= 0, because
n
x >> n as n grows large. Consequently, in those cases, f cannot possibly converge;
so it suffices to consider the case when x lies in the interval [1, 1].
P n
If x = 1, we have that f (x) = n=0 (1) n , which we have shown several times
in the past to be convergent (via Leibnizs theorem.)
P n P
If x = 1, we have that f (x) = n=0 (1) n (1)n = n=1 n1 , which we know
to be the harmonic series and is thus divergent.
Finally, if x (1, 1), we have that
 
 X (1)n  X 1 n X n X |x|
n
|f (x)| =  x  |x| |x| = |x| |x|n = <


n=1
n 
n=1
n n=1 n=0
1 |x|
by again using our summation identity for the geometric series; so this converges
for all x (1, 1).
So this series converges for all xin(1, 1]. As well, by the same argument as
before, on any interval [a, a] (1, 1], we have
 (1)n n 
 
n
 n x a
 
P n
and also
Pthatn the sum n=1 a converges; so, again by the Weierstrass M-test, the
sum n=0 x converges uniformly on [a, a]. 
Example 3.7. Suppose that

X 1 n
f (x) = x
n=1
n!
Where does this series converge? How does it converge?
Proof. So: simply note that by the ratio test, we have that
xn+1 /(n + 1)! x
lim n
= lim = 0,
n x /n! n n + 1

and thus that this series converges for every x in R. As well, just as in the two
proceeding examples, we have that this converges uniformly on every single subset
[a, a] R, by combining the observations that
 n
 x  an
 
 n!  n!
P n
and that the sum n=1 a converges (via the ratio test), along with the Weierstrass
M-test. 
Example 3.8. Suppose that

X
f (x) = n! xn
n=1
Where does this series converge? How does it converge?
Proof. So: again note that by the ratio test, we have that
xn+1 (n + 1)!
lim = lim x(n + 1) = ,
n xn n! n
for any x 6= 0. So this series only converges for x = 0. 
8 INSTRUCTOR: PADRAIC BARTLETT

By glancing at the examples above, a few patterns might leap out at you; (1)
that all of the power series above were convergent in symmetric intervals around the
origin (up to the endpoints), and (2) that if a power series converges on a set, then
it uniformly converges on any closed interval lying strictly inside of that set. At the
least, thats what we saw in the examples above; we had power series converging on
the sets (1, 1), (1, 1], R, and 0, and in every case these series uniformly converged
on every closed interval [a, a] lying within those sets.
It turns out this is actually true in general, for any power series! The following
theorem demonstrates this result, along with another result which will hopefully
illustrate just why we do all of this work:
P
Theorem 3.9. Let f (x) = n=0 anP xn , and suppose that x0 is a number in R such

that f (x0 ) converges; i.e. such that n=0 an xn0 converges.
Pick any a such that 0 < a < x0 . Then, we have that
P
on [a, a], n=0 an xnPconverges uniformly,

on [a, a], the series n=1 an nxn1 converges uniformly as well, and

f 0 (x) = n=1 an nxn1 on all of [a, a].
P

Basically, our theorem claims the following: if a power series converges at one
point x0 , it converges uniformly on every interval [a, a] with a < x0 , and fur-
thermore we can calculate its derivative by simply differentiating the power series
itself termwise. This second statement is perhaps the most remarkable and useful
result of this theorem, as it essentially makes the basic operations of calculus on any
function given by a power series completely trivial! because it lets us simply
perform derivations on the functions power series, which is just a giant polynomial
(and, consequently, is something we can handle with ease.)
P
Proof. So: because n=1 an xn0 converges, we have that the terms an xn0 go to zero;
thus, they must be bounded! Pick a bound M such that |an xn0 < M , for all n.
Then, if x [a, a], we have that

|an xn | = |an | |xn |

|an | |an |
 n
a 
= |an | |xn0 |  n 
x0
 n
a
M  
x0

 
Because a < x0 , we know that  xa0  < 1; thus, the series
 

 n 
 a n

X a X
M   =M
   
n=0
x 0 n=0
 x0 

is geometric,
 n and thus converges! So, by the Weierstrass M-test, taking Mn =
P
M  xa0  , we know that n=1 an xn converges uniformly on [a, a].
 
 MATH 1D, WEEK 5 UNIFORM CONVERGENCE AND POWER SERIES 9

P
To see that the series n=1 an nxn1 converges, we perform mostly the same
trick: again, for any x [a, a], we have that

|nan xn1 | = n|an | |xn1 |

n|an | |an1 |
n|an | |an |

|a|
 n
n|an | |xn0 |  xan 
 
0
=
|a|
 n
n M  xa0 
 
.
|a|

 
Thus, because  xa0  < 1, the sum
 

 n
n M  a   n
X x0 M X a
= n  
n=1
|a| |a| n=1 x0

converges via the ratio test (as we have shown several times). Applying the Weier-
strass M-test, we can then conclude that this series converges uniformly as well! Fi-
nally, because uniform convergence preserves the derivative, wePcan see at last that
this means that f 0 (x) is actually given by this power series, n=1 an nxn1 . 

To illustrate a little bit of the power inherent in the above theorem, here are a
pair of examples:

Example 3.10. Using only the observations, from our earlier work with Taylor
series, that

x3 x5 x7 x9
sin(x) = x + + ...,
3! 5! 7! 9!
x x2 x3 x4
ex = 1 + + + + + ...,
1 2! 3! 4!

calculate the derivatives and antiderivatives of these functions.

Proof. So: by the above theorem, we know that taking derivatives or antiderivatives
of these functions can be done just on their power series. Explicitly, then, we have
10 INSTRUCTOR: PADRAIC BARTLETT

that
x3 x5 x7 x9
 
d d
sin(x) = x + + ...
dx dx 3! 5! 7! 9!
3x2 5x4 7x6 9x8
=1 + + ...
3! 5! 7! 9!
x2 x4 x6 x8
=1 + + ...
2
2! 4!
2
6!
4
8!
6
x8

d d x x x
2 sin(x) = 1 + + ...
dx dx 2! 4! 6! 8!
2x1 4x3 6x5 8x7
= + + ...
2! 4! 6! 8!
x3 x5 x7
= x + + ...
3! 5! 7!
= sin(x).

Thus, weve shown using no properties of sin other than its power series that
its second derivative is just sin(x), and consequently that its fourth derivative is
just sin(x) again. As well, weve found the power series for its derivative; so if we
knew sin(x)s power series, but not cos(x)s, this method just told us what it was!
As for ex : because
x x2 x3 x4
 
d x d
e = 1+ + + + + ...
dx dx 1 2! 3! 4!
1 2x1 3x2 4x3
= + + + + ...
1 2! 3! 4!
x x2 x3
=1+ + + + ...
1 2! 3!
= ex ,

we can see that taking derivatives or antiderivatives of ex wont change it! So,
again, from just its power series, weve been able to derive properties about ex
which pretty much completely define it as a function.