2Series of functions

IB Analysis II


2.2 Power series
A particularly interesting kind of series is a power series. We have already met
these in IA Analysis I and proved some results about them. However, our results
were pointwise results, discussing how
P
c
n
(
x−a
)
n
behaves at a particular point
x
. Here we will quickly look into how the power series behaves as a function of
x. In particular, we want to know whether it converges absolutely uniformly.
Theorem. Let
∞
P
n=0
c
n
(
x −a
)
n
be a real power series. Then there exists a unique
number R ∈ [0, +∞] (called the radius of convergence) such that
(i) If |x −a| < R, then
P
c
n
(x − a)
n
converges absolutely.
(ii) If |x −a| > R, then
P
c
n
(x − a)
n
diverges.
(iii)
If
R >
0 and 0
< r < R
, then
P
c
n
(
x −a
)
n
converges absolutely uniformly
on [a − r, a + r].
We say that the sum converges locally absolutely uniformly inside the circle
of convergence, i.e. for every point
y ∈
(
a − R, a
+
R
), there is some open
interval around y on which the sum converges absolutely uniformly.
These results hold for complex power series as well, but for concreteness we will
just do it for real series.
Note that the first two statements are things we already know from IA
Analysis I, and we are not going to prove them.
Proof. See IA Analysis I for (i) and (ii).
For (iii), note that from (i), taking
x
=
a − r
, we know that
P
|c
n
|r
n
is
convergent. But we know that if x ∈ [a − r, a + r], then
|c
n
(x − a)
n
| ≤ |c
n
|r
n
.
So the result follows from the Weierstrass M-test by taking M
n
= |c
n
|r
n
.
Note that uniform convergence need not hold on the entire interval of con-
vergence.
Example. Consider
P
x
n
. This converges for
x ∈
(
−
1
,
1), but uniform conver-
gence fails on (−1, 1) since the tail
n
X
j=m
x
j
= x
n
n−m
X
j=0
x
j
≥
x
m
1 − x
.
This is not uniformly small since we can make this large by picking
x
close to 1.
Theorem (Termwise differentiation of power series). Suppose
P
c
n
(
x − n
)
n
is
a real power series with radius of convergence R > 0. Then
(i) The “derived series”
∞
X
n=1
nc
n
(x − a)
n−1
has radius of convergence R.
(ii)
The function defined by
f
(
x
) =
P
c
n
(
x − a
)
n
,
x ∈
(
a − R, a
+
R
) is
differentiable with derivative
f
′
(
x
) =
P
nc
n
(
x − a
)
n−1
within the (open)
circle of convergence.
Proof.
(i) Let R
1
be the radius of convergence of the derived series. We know
|c
n
(x − a)
n
| = |c
n
||x − a|
n−1
|x − a| ≤ |nc
n
(x − a)
n−1
||x − a|.
Hence if the derived series
P
nc
n
(
x − a
)
n−1
converges absolutely for some
x, then so does
P
c
n
(x − a)
n
. So R
1
≤ R.
Suppose that the inequality is strict, i.e.
R
1
< R
, then there are
r
1
, r
such that
R
1
< r
1
< r < R
, where
P
n|c
n
|r
n−1
1
diverges while
P
|c
n
|r
n
converges. But this cannot be true since
n|c
n
|r
n−1
1
≤ |c
n
|r
n
for sufficiently
large n. So we must have R
1
= R.
(ii)
Let
f
n
(
x
) =
n
P
j=0
c
j
(
x − a
)
j
. Then
f
′
n
(
x
) =
n
P
j=1
jc
j
(
x − a
)
j−1
. We want
to use the result that the derivative of limit is limit of derivative. This
requires that
f
n
converges at a point, and that
f
′
n
converges uniformly.
The first is obviously true, and we know that
f
′
n
converges uniformly on
[
a−r, a
+
r
] for any
r < R
. So for each
x
0
, there is some interval containing
x
0
on which f
′
n
is convergent. So on this interval, we know that
f(x) = lim
n→∞
f
n
(x)
is differentiable with
f
′
(x) = lim
n→∞
f
′
n
(x) =
∞
X
j=1
jc
j
(x − a)
j
.
In particular,
f
′
(x
0
) =
∞
X
j=1
jc
j
(x
0
− a)
j
.
Since this is true for all x
0
, the result follows.