Download Introduction to, or Review of, Series The absolute value of a complex

Introduction to, or Review of, Series
The absolute value of a complex number is that number’s distance from 0. We can
calculate it using the Pythagorean Theorem. For instance,
√
√
√
|3 + 7i| = 32 + 72 = 9 + 49 = 58.
The absolute value is always a positive (real) number or 0.
The distance between two complex numbers is the absolute value of their difference.
For instance, the distance between 3 + 7i and 5 − 2i is
√
√
|(3 + 7i) − (5 − 2i)| = | − 2 + 9i| = 22 + 92 = 85.
A sequence of complex numbers s0 , s1 , s2 , s3 , . . . is said to converge to a limit
L if L is a complex number such that limn→∞ |sn − L| = 0. Then we may also write
limn→∞ sn = L. The idea is that the terms of the sequence are getting closer to L as we
go further out in the sequence.
However, we do not require that the sn ’s are always getting strictly closer to L; we
merely require that they get closer to L “in the long run.” For instance, the sequence
1
2+ ,
2
2+
100
,
22
1
,
23
2+
2+
100
,
24
2+
1
,
25
2+
100
,
26
...
(alternating between 2 + 21n and 2 + 100
) is sometimes getting closer to 2 and sometimes
2n
getting farther away. But it gets close to 2 in the long run, so it does converge to 2.
An infinite series is an expression of the form
∞
X
pk = p0 + p1 + p2 + · · ·
k=0
where p0 , p1 , p2 , . . . are complex numbers. That series might or might not represent a
complex number (see below); it might just be a string of symbols. The partial sums of
that series are the complex numbers
s0 = p0 ,
s1 = p0 + p1 ,
s2 = p0 + p1 + p2 ,
s3 = p0 + p1 + p2 + p3 ,
...
P
and so on. In other words, sn = nk=0 pk .
P
Some series converge, and some do not. We say that the series ∞
k=0 pk converges if
there is some complex number L such that
lim (p0 + p1 + p2 + · · · + pn ) = L;
n→∞
that is, some number L such that limn→∞ sn = L. When this is the case, we say that the
P∞
P
series ∞
k=0 pk = L. (If that doesn’t happen, we say
k=0 pk converges to L; we write
the series diverges.)
Here are some examples:
The series 1 + 12 + 41 + 18 + · · · converges to 2. Indeed, its partial sums are
1
1 + 12
1 + 12 + 41
1 + 12 + 41 +
1
8
=
1
= 1.5
= 1.75
= 1.875
···
and those numbers get closer and closer to 2.
The series 1 + i + i2 + i3 + i4 + · · · does not converge to anything. Indeed, its partial
sums are
1
=
1
1+i
= 1+i
1 + i + i2
= 1+i−1
=
i
2
3
1+i+i +i
= 1+i−1−i
=
0
1 + i + i2 + i3 + i4
= 1+i−1−i+1
=
1
1 + i + i2 + i3 + i4 + i5 = 1 + i − 1 − i + 1 + i = 1 + i
and so on. The partial sums form the repeating sequence
1,
1 + i,
i,
0,
1,
1 + i,
i,
...
There does not exist a single number L that this sequence gets close to.
The series 1 + 12 + 31 + 14 + 15 + · · · does not converge to anything. To prove that,
we won’t calculate its partial sums; we will just give a lower bound for them. (Only the
certification, not the method of discovery, is shown here.) We use the fact that, whenever
n is a positive integer and n < x < n + 1, then n1 > x1 . Consequently,
Z n+1
Z n+1
1
1
1
=
dx >
dx = [ln x]n+1
= ln(n + 1) − ln n.
n
n
n
x
n
n
So we can compute
1
1 + 12
1 + 12 +
1
3
>
[ln 2 − ln 1]
= ln 2
>
[ln 2 − ln 1] + [ln 3 − ln 2]
= ln 3
> [ln 2 − ln 1] + [ln 3 − ln 2] + [ln 4 − ln 3] = ln 4
and so on. In general, the partial sums sn = 1 + 12 + 13 + · · · + n1 satisfy sn > ln(n + 1), so
we have limn→∞ sn = ∞. Thus there does not exist a complex number L
2
The series 112 + 212 + 312 + 412 + · · · converges to π6 , but that’s hard to prove, so we won’t.
Power Series
A power series is an expression of the form
f (z) =
∞
X
k=0
ck (z − z0 )k = c0 + c1 (z − z0 ) + c2 (z − z0 )2 + c3 (z − z0 )3 + · · ·
where c0 , c1 , c2 , . . . and z0 are some given complex numbers, and z is a variable. In the
simplest cases, we have z0 = 0, and so the formula takes the slightly simpler form
f (z) =
∞
X
ck z k = c0 + c1 z + c2 z 2 + c3 z 3 + · · · .
k=0
But we will discuss the more general case where z0 need not be zero.
If we plug in some particular complex number for z, then the expression
f (z) =
∞
X
ck (z − z0 )k = c0 + c1 (z − z0 ) + c2 (z − z0 )2 + c3 (z − z0 )3 + · · ·
k=0
becomes just an infinite series, of the sort discussed earlier in this document. If you want
to use our earlier notation, it would be
p0 = c0 ,
p1 = c1 (z − z0 ),
p2 = c2 (z − z0 )2 ,
...
So for each different value of the complex number z, we get a different infinite series. The
series may converge for some values of z, and diverge for other values of z. Here are a few
examples: It can be shown that
2
3
4
5
• The series 1 + z + z2 + z6 + z4! + z5! + · · · converges when z is any complex number.
• The series 1 + z + z 2 + z 3 + z 4 + · · · converges when z is any complex number with
|z| < 1, and diverges when z is any complex number with |z| ≥ 1.
2
3
4
n
z
• The series z + z4 + z9 + 16
+ · · · + zn2 + · · · converges when z is any complex number
with |z| ≤ 1, and diverges whenever |z| > 1.
1
• The series 1 + 12 (z − 3) + 14 (z − 3)2 + 81 (z − 3)3 + 16
(z − 3)4 + · · · converges whenever
z is a complex number with |z − 3| < 2, and diverges for any other z.
• The series 1 + z + 2z 2 + 6z 3 + 24z 4 + 5!z 5 + 6!z 6 + · · · converges when z = 0, and
diverges for any other choice of z.
Radius of Convergence
It turns out that the values of z where the series converges form a disk in the complex
plane. More precisely, we have this theorem (which is proved in some other math courses,
but we won’t prove it):
Theorem on the shape of the region of convergence. Let c0 , c1 , c2 , c3 , . . .
and z0 be any given complex numbers. Then those numbers uniquely determine a number R, which may be 0 or a positive number or +∞, and which
has these two properties:
• Whenever z is a complex number with |z − z0 | < R, then the series
P∞
n
n=0 cn (z − z0 ) converges. (The set {z : |z − z0 | < R} is called the disk
of convergence; it is the inside of a circle. It is the whole complex plane
if R = ∞.)
• Whenever z is a complex number with |z − z0 | > R, then the series
P∞
n
n=0 cn (z − z0 ) diverges. (The set {z : |z − z0 | > R} is the outside of
that circle.)
The number R is called the radius of convergence of the power series.
Here are some further remarks about R:
The set {z : |z − z0 | = R} is called the circle of convergence. Note that the
theorem does not tell us whether we get convergence or divergence at points on the circle
of convergence. That is a more complicated question, which cannot be answered so easily.
Almost anything can happen:
• For some power series, we get convergence at every point on the circle.
• For some power series, we get divergence at every point on the circle.
• For some power series, we get convergence at some points on the circle, and divergence at other points on the circle.
Our textbook talks about the interval of convergence, but that’s because they’re only
considering real numbers for z. The real line is a straight line in the complex plane; its
intersection with a disk is an interval.
The radius R is actually determined by the complex numbers c0 , c1 , c2 , . . ., but does
not depend on the number z0 . If we change our choice of z0 , that moves the center of the
circle, but it does not change the radius of the circle.
There is actually a formula for R, but it is complicated. Here it is:
1
R
q
=
lim sup
n→∞
n
|cn |.
Here’s what that means: Consider the sequence of positive numbers
q
|c1 |,
|c2 |,
q
q
3
4
|c3 |,
|c4 |,
...
In general, this sequence might not converge. However, some of its subsequences do
converge. (A subsequence is what we get if we skip some of the terms in the sequence.)
Consider the limits of the convergent subsequences. The largest of those limits is called
the limsup of the sequence. It turns out to be equal to the reciprocal of the radius of
convergence.
That complicated formula always works, but sometimes it is hard to use. There are
simpler formulas that sometimes work. Here is the main one that we will use:
The ratio formula. For some power series, |cn /cn+1 | converges to a limit.
Whenever that limit exists, it is equal to the radius R. (When that limit
doesn’t exist, then there is still a radius R, but then R may be harder to find.)
(How can you have two different
qformulas for the same radius R? Well, here is what’s
going on: the number 1/ lim sup n |cn | exists for every power series, and the number
lim |cn /cn+1 | exists for some power series. But when both those numbers exist, they are
equal. That’s proved in some other courses.)
Some students get confused between cn /cn+1 and cn+1 /cn . To see where the confusion comes from, and how to avoid it, consider these conditions (all of which are stated
correctly):
¯
¯
¯ c ¯
¯ n ¯
R = lim ¯
¯
n→∞ ¯ c
¯
n+1
convergence if
convergence if
¯
¯
¯ c ¯
¯ n ¯
¯
|z − z0 | < n→∞
lim ¯¯
cn+1 ¯
¯
¯
¯ cn+1 ¯
¯
¯<1
lim |z − z0 | ¯
n→∞
cn ¯
Our textbook uses a condition like that last equation, but you see that now the fraction
has been turned upside down, because it has been moved to the other side of the inequality.
One easy way to remember which way the fraction goes is by looking at a simple example,
such as
z z2 z3 z4
1 + + 2 + 3 + 4 + ···.
5 5
5
5
In some cases, even if limn→∞ |cn /cn+1 | doesn’t exist, we can still use the ratio formula
indirectly. For instance, the power series
f (z) = 1 + 3z 2 + 9z 4 + 27z 6 + 81z 8 + · · ·
has what radius of convergence? Here the coefficients are 30 , 0, 31 , 0, 32 , 0, 33 , 0, . . .. So
|cn /cn+1 | alternates between 0 and ∞, and does not converge to a limit. To figure this
one out, substitute w = z 2 , and look at the related function
g(w) = 1 + 3w + 9w2 + 27w3 + 81w4 + · · ·
For this series, we have cn = 3n , so cn /cn+1 = 1/3, and the radius is 1/3. That means
the series for g(w) converges whenever |w| < 13 . But the series for g(w) is the same as the
series for f (z), if we substitute w = z 2 . So the series for f (z) converges whenever |z 2 | < 31
— that is, whenever |z| < √13 . So the radius of convergence for f (z) is √13 .
It can be proved (though not in this course) that the radius of convergence is also
equal to the distance from z0 to the nearest “bad” point of the function. For instance, it
can be shown that
1
1
1
1
1 + (z − 3) + (z − 3)2 + (z − 3)3 + (z − 3)4 + · · ·
2
4
8
16
2
converges to 5−z
at points near z = 3. The circle is centered at z = 3. The only place
2
where 5−z behaves badly is at z = 5. The distance from 3 to 5 is 2. So the radius of
convergence is 2.
Power Series Coefficients
Power series can be integrated or differentiated term by term. If we start from
∞
X
f (z) =
ck (z − z0 )k = c0 + c1 (z − z0 ) + c2 (z − z0 )2 + c3 (z − z0 )3 + · · ·
k=0
and differentiate zero or more times, and then plug in z = z0 , we get these results:
0
c0 = f (z0 ),
c1 = f (z0 ),
1
c2 = f 00 (z0 ),
2
...
f (n) (z0 )
cn =
n!
Exercises
A. Find the radius of convergence of these power series:
1.
∞
X
(z − 1)n
n=0
2.
3.
∞
X
nz n
2
n=0 (n + 1)
∞
X
n2 z n
n=0
4.
(−3)n
2n
∞
X
2n (z+i)n
n=0
5.
∞ µ
X
n=0
n!
¶
n+
1
zn
n+1
6. 1 + 14 z 2 +
1 4
z
16
+
1 6
z
64
+ ··· +
1 2k
z
22k
+ ···
B. Find the first three coefficients (c0 , c1 , c2 ) for these power series.
1. cos(z) = c0 + c1 z + c2 z 2 + c3 z 3 + · · ·
2. sin(z) = c0 + c1 z + c2 z 2 + c3 z 3 + · · ·
3.
1
z
= c0 + c1 (z − 2) + c2 (z − 2)2 + c3 (z − 2)3 + · · ·
4. ln z = c0 + c1 (z − 1) + c2 (z − 1)2 + c3 (z − 1)3 + · · ·
5. ln(1 + z) = c0 + c1 z + c2 z 2 + c3 z 3 + · · ·
6. arctan(z) = c0 + c1 z + c2 z 2 + c3 z 3 + · · ·
7. z 3 + 3z + 7 = c0 + c1 z + c2 z 2 + c3 z 3 + · · ·
...