6Complex power series
IA Analysis I
6.2 Differentiating power series
We shall show that inside the circle of convergence, the derivative of
P
∞
n=0
a
n
z
is
given by the obvious formula
P
∞
n=1
na
n
z
n−1
.
We first prove some (seemingly arbitrary and random) lemmas to build up
the proof of the above statement. They are done so that the final proof will not
be full of tedious algebra.
Lemma. Let a and b be complex numbers. Then
b
n
−a
n
−n(b −a)a
n−1
= (b −a)
2
(b
n−2
+ 2ab
n−3
+ 3a
2
b
n−4
+ ···+ (n −1)a
n−2
).
Proof. If b = a, we are done. Otherwise,
b
n
− a
n
b − a
= b
n−1
+ ab
n−2
+ a
2
b
n−3
+ ··· + a
n−1
.
Differentiate both sides with respect to a. Then
−na
n−1
(b − a) + b
n
− a
n
(b − a)
2
= b
n−2
+ 2ab
n−3
+ ··· + (n − 1)a
n−2
.
Rearranging gives the result.
Alternatively, we can do
b
n
− a
n
= (b − a)(b
n−1
+ ab
n−2
+ ··· + a
n−1
).
Subtract n(b − a)a
n−1
to obtain
(b − a)[b
n−1
− a
n−1
+ a(b
n−2
− a
n−2
) + a
2
(b
n−3
− a
n−3
) + ···]
and simplify.
This implies that
(z + h)
n
− z
n
− nhz
n−1
= h
2
((z + h)
n−2
+ 2z(z + h)
n−3
+ ··· + (n − 1)z
n−2
),
which is actually the form we need.
Lemma. Let
a
n
z
n
have radius of convergence
R
, and let
|z| < R
. Then
P
na
n
z
n−1
converges (absolutely).
Proof.
Pick
r
such that
|z| < r < R
. Then
P
|a
n
|r
n
converges, so the terms
|a
n
|r
n
are bounded above by, say, C. Now
X
n|a
n
z
n−1
| =
X
n|a
n
|r
n−1
|z|
r
n−1
≤
C
r
X
n
|z|
r
n−1
The series
P
n
|z|
r
n−1
converges, by the ratio test. So
P
n|a
n
z
n−1
|
converges,
by the comparison test.
Corollary. Under the same conditions,
∞
X
n=2
n
2
a
n
z
n−2
converges absolutely.
Proof. Apply Lemma above again and divide by 2.
Theorem. Let
P
a
n
z
n
be a power series with radius of convergence
R
. For
|z| < R, let
f(z) =
∞
X
n=0
a
n
z
n
and g(z) =
∞
X
n=1
na
n
z
n−1
.
Then f is differentiable with derivative g.
Proof. We want f (z + h) − f (z) − hg(z) to be o(h). We have
f(z + h) − f(z) − hg(z) =
∞
X
n=2
a
n
((z + h)
n
− z
n
− hnz
n
).
We started summing from
n
= 2 since the
n
= 0 and
n
= 1 terms are 0. Using
our first lemma, we are left with
h
2
∞
X
n=2
a
n
(z + h)
n−2
+ 2z(z + h)
n−3
+ ··· + (n − 1)z
n−2
We want the huge infinite series to be bounded, and then the whole thing is a
bounded thing times h
2
, which is definitely o(h).
Pick
r
such that
|z| < r < R
. If
h
is small enough that
|z
+
h| ≤ r
, then the
last infinite series is bounded above (in modulus) by
∞
X
n=2
|a
n
|(r
n−2
+ 2r
n−2
+ ··· + (n − 1)r
n−2
) =
∞
X
n=2
|a
n
|
n
2
r
n−2
,
which is bounded. So done.
In IB Analysis II, we will prove the same result using the idea of uniform
convergence, which gives a much nicer proof.
Example. The derivative of
e
z
= 1 + z +
z
2
2!
+
z
3
3!
+ ···
is
1 + z +
z
2
2!
+ ··· = e
z
.
So we have another proof that of this fact.
Similarly, the derivatives of sin z and cos z work out as cos z and −sin z.