Trevor Hyde
Math 116
Even and Odd Power Series
When confronted with a power series like
2
c0 + c1 (x − a) + c2 (x − a) + . . . =
∞
X
cn (x − a)n ,
n=0
we learned that the radius of convergence R may be computed via a limit
cn .
R = lim n→∞ cn+1 Implicit in this definition and in all the examples we have seen is that this limit makes sense
and exists. In this note, I want to consider a common phenomenon where this limit does not
make sense, and how we can circumvent the difficulty.
Let’s start with an example. Say we are given the following power series and asked to find
its radius of convergence.
1 + 31 (x − 1)2 + 91 (x − 1)4 +
1
(x
27
− 1)6 + . . . .
Notice that the center is 1 and that we only have even exponents of (x − 1). If we follow
the rules for defining the sequence of coefficients, then cn is the coefficient of (x − 1)n ; so we
need to fill in the missing terms:
1 + 0(x − 1) + 31 (x − 1)2 + 0(x − 1)3 + 19 (x − 1)4 + 0(x − 1)5 +
1
(x
27
− 1)6 + . . . .
Our formula for cn is a bit complicated to write down—it depends on whether n is even or
odd.
(
1
if n is even, and
cn = 3(n/2)
0
if n is odd.
Compute the first few examples of cn using this formula to make sure you understand how
I came up with it.
We run into a problem when we try to compute the radius R: whenever n is even, cn+1 = 0,
cn
cn
so the fraction cn+1
does not make sense. On the other hand, when n is odd we have cn+1
= 0.
cn
So, half the time cn+1 does not make sense, and the other half of the time it’s zero. What
do we do? Do we give up? Not yet.
Let’s try changing our variables. Introduce a new variable y and let y = (x − 1)2 . Now we
may rewrite our original power series conveniently in terms of y.
(1)
1+
1
y
3
+
1 2
y
9
+
1 3
y
27
+ ... =
∞
X
yn
n=0
3n
.
If we had started with this series, we would have no trouble computing the radius of convergence. Our sequence of coefficients is much easier to write down now, it’s just cn = 31n . No
2
more zeros! Then we compute
Math 116
n+1
cn = lim 3
R = lim = lim 3 = 3.
n→∞ cn+1 n→∞ 3n
n→∞
Remember what this means: it means that the power series (1) converges when |y| < 3. But
y was a variable we introduced for convenience, so we should switch back to our original
variable x, which gives us
3 > |y| = |(x − 1)2 | = |x − 1|2 .
Here is the crucial step: we must take a square root! This leaves us with the inequality
√
|x − 1| < 3.
√
So our radius of convergence is 3. We had to take a 1/2 power because our original series
had all even exponents.
Next, I want to give one more example which I hope will help with this week’s team homework. Consider the series
∞
X
(x − 2)5n+1
1
1
= 13 (x − 2)6 + 9·4
(x − 2)11 + 27·9
(x − 2)16 + . . .
n n2
3
n=1
Our situation is similar but slightly different; we are still skipping many exponents at regular
intervals but its no longer every other exponent, instead we are skipping exponents 5 at a
time. We begin again by introducing a new variable y but this time we set y = (x − 2)5 . We
can’t quite write the entire power series in terms of y in a nice way, but we can get close.
Check this out:
∞
∞
∞
X
X
(x − 2)5n+1 X (x − 2)(x − 2)5n
yn
=
=
(x
−
2)
.
3n n2
3n n2
3n n2
n=1
n=1
n=1
There is an extra factor of (x − 2) floating around, but its not mixed up with the index n
so I can treat it like a scalar and pull it out of the series—that’s what I do in the last step
above. Now we focus on the power series in y,
∞
X
yn
,
3n n2
n=1
and compute its radius of convergence in terms of y. The sequence of coefficients is cn =
so we have
2
n+1
2
cn 3
(n
+
1)
n
+
2n
+
1
= lim
R = lim = lim 3
= 3.
n→∞ cn+1 n→∞
n→∞
3n n2
n2
1
,
3n n2
Finally, we switch back to the original variable to compute the radius of convergence in terms
of x:
3 > |y| = |(x − 2)5 | = |x − 2|5 .
This time, we need to take 1/5 powers of both sides to get
|x − 2| < 31/5 ,
hence the radius of convergence in terms of x is 31/5 .
Math 116
page 3
There is a pattern linking the size of the gaps between exponents
(2 in the first example, 5
cn in the second) and the exponent we need to raise limn→∞ cn+1 to (1/2 in the first example,
1/5 in the second.) Once you practice a few problems, you may get the hang of this and no
longer need to follow through these extra steps. The book talks about series with only even
or odd exponents, the common feature being that in both cases the gap between exponents
is 2. On the team homework, there is a problem where the gap between exponents is 3.