Lecture 18: Power Series and Analytic Functions (11.8-11.10)
Rahul Krishna
1
Introduction to Power Series
We have already seen an example of a power series. There is the prototypical example
1
1
x
=
1
X
xk
k=0
which we showed is valid for all values of x with jxj < 1: What does this mean, in terms of the partial sums?
Well, let’s plot these partial sums together with the function in question, 1 1 x : [Draw on board.]
So s0 = 1: s1 = 1 + x: s2 = 1 + x + x2 : s3 = 1 + x + x2 + x3 : :::
As is visible from the picture, for each x 2 ( 1; 1) ; the partial sums s0 ; s1 ; ::: progressively better
approximate 1 1 x as long as we remain in the interval ( 1; 1) :
So what is this kind of phenomenon saying? Well 1 1 x is not a polynomial, but we are saying that, in the
interval ( 1; 1) ; we can approximate it quite will by polynomials.
Well, it takes a bit of thought to realize that in this canonical example, there are really two things going
on:
P1
1. The function 1 1 x = k=0 xk can literally be realized as an in…nite sum for jxj < 1:
2. The function 1 1 x is, for x close to zero, well-approximated by the polynomials s0 = 1; s1 = 1 + x; s2 =
1 + x + x2 ; :::
These two phenomenon, which in this example seem to blend together, should really be thought of
as distinct. The …rst is summarized by saying that f (x) = 1 1 x is an analytic function. The second is
summarized by saying that f (x) = 1 1 x is well-approximated by its Taylor polynomials.
But before we dig into these interesting ideas, let’s make a preliminary de…nition.
De…nition 1 A power series centered around a point a 2 R is an in…nite series of the form
1
X
cn (x
n
a)
n=0
P1
The sum of the series de…nes a function f (x) = n=0 cn (x
the domain of all x for which the series converges.
n
a) , in the domain where it is de…ned, i.e. in
Note that the example above was a = 0 and cn = 1 for all n:
In some sense, a power series is completely analagous to a polynomial, only with (possibly) in…nite degree.
So how can we …nd the domain of de…nition of a power series? We should use the Ratio test (or Root
n
test, if you prefer). We see, letting an = cn (x a)
lim
n!1
so if limn!1
cn+1
cn
n+1
cn+1 (x
an+1
= lim
n!1
an
cn (x
a)
n
a)
= lim
n!1
cn+1
jx
cn
= ; then we have that the series converges absolutely if
jx
jx
aj < 1
1
aj <
1
aj
and will diverge if jx
aj >
1
1
: Call R =
= limn!1
cn
cn+1
: As usual, we adopt the convention that if
= 1; then we write = 1 and R = 1 = 0: Similarly, if = 0; then R = 1: We have
limn!1 cn+1
cn
e¤ectively shown the following (modulo some important technical details)
P1
n
Theorem 1 For power series n=0 cn (x a) ; there are only three possibilities:
(i). The series converges only when x = a: [ = 1; R = 0]
(ii). The series converges for all x: [ = 0; R = 1]
(iii). The series converges for jx aj < R and diverges for jx aj > R: [0 < R < 1]
At the endpoints, we have to check for convergence, usually using the alternating series test.
Example 1 For what values of x does
1
X
(x
k
3)
k
k=1
converge? Use ratio test.
n+1
=1
n
= lim
n!1
P1
so R = 1= = 1: If x = 4, we get k=1 k1 which is divergent; if x = 2; we get the alternating harmonic
series, which is convergent. So the domain is [2; 4):
Let’s do another example:
Example 2 Find the domain of the Bessel function
J0 (x) =
1
k
X
( 1) x2k
2
k=0
22k (k!)
We use the ratio test. Note that here, since we’re not given the power series in a nice form where the power
of x is k; we have to go back to the ratio test on the full thing. This gives
k+1
( 1)
lim
2
x2k+2
22k (k!)
2
k
22k+2 ((k + 1)!) ( 1) x2k
k!1
x2
=
2
4 (k + 1)
=0
for all x 2 R. So by the Ratio test, the series converges for all x 2 R. So the domain is R.
One more, for good measure:
Example 3 Find the radius of convergence for the power series
1
X
1+
k=0
1
k
k
k
(x + 3)
Well, we use the Root Test here. We …nd
lim
k!1
so
1
1+
k
k
!1=k
= lim 1 +
k!1
1
=1
k
= 1 and R = 1: Note that here it is very hard to say anything about convergence at the endpoints!
Good, so time for a de…nition:
2
De…nition 2 A function f (x) is said to be analytic around a point a if f (x) can be expanded in a power
series around a
1
X
n
f (x) =
cn (x a)
n=0
which has a positive radius of convergence, i.e. limn!1
cn
cn+1
6= 0:
So, almost tautologically, a convergent power series de…nes an analytic function.
Here are some properties of analytic functions. We won’t prove these, because the proofs are a little
messy and not terribly enlightening if we insist on trying to do them by hand. [An aside: the proofs are
quite easy if we knew some complex analysis, but that would take us too far a…eld.]
P
P
n
n
Sums, Products of Analytic Functions Suppose that f (x) = cn (x a) and g (x) = dn (x a)
are both analytic in (a R; a + R) : Then f (x) + g (x) and f (x) g (x) are both analytic as well, and in the
same region (a R; a + R) : In fact, the power series found are
X
n
f (x) + g (x) =
(cn + dn ) (x a)
f (x) g (x) =
1 X
X
(cl dk ) (x
n
a)
n=0 n=l+k
= c0 d0 + (c1 d0 + c0 d1 ) (x
a) + (c2 d0 + c1 d1 + c0 d2 ) (x
2
a) + :::
P
n
Quotients
by Non-Vanishing Analytic Functions Suppose that f (x) =
cn (x a) and g (x) =
P
n
dn (x a) are both analytic in (a R; a + R) : Suppose further than g (x) 6= 0 for all x 2 (a R; a + R) : Then
f (x) =g (x) is also analytic in (a R; a + R) :
P1
n
Derivatives and Antiderivatives If f (x) = n=0 cn (x a) is an analyticPfunction in (a R; a + R) ;
n
then so is f 0 (x) ; and it is given, within the radius of convergence of the series
cn (x a) , by
f 0 (x) =
1
X
ncn (x
n 1
a)
n=0
Similarly, any antiderivative of f (x) is also analytic in (a
F (x) = C +
R; a + R) : They are given by
1
X
cn 1
(x
n
n=1
n
a)
These are called term-by-term di¤erentiation and integration–we simply di¤erentiate and integrate termwise.
1.1
Some Exercises on Basic Power Series
Here are some nice exercises from the book that we’re going to do together. I think they will help clarify
some things.
pp 746
37. Suppose f is de…ned by
f (x) = 1 + 2x + x2 + 2x3 + x4 + :::
That is, its coe¢ cients are c2n = 1 and c2n+1 = 2 for all n
series and what is a formula for f (x)?
3
0: What is the interval of convergence for this
Well,
1
1
and
x
1
x2
1
so
= 1 + x + x2 + x3 + :::
= 1 + x2 + x4 + x5 + :::
x
1
and so
x2
= x + x3 + x5 + :::
1 + 2x
1
x
=
= 1 + 2x + x3 + 2x4 + :::
+
1 x2
1 x 1 x2
since the intervals of convergence of both of these series is ( 1; 1) ; we conclude that this formula is true for
all x in ( 1; 1) :
P
P
41. Suppose the series
cn xn has a radius of convergence 2 and the series
dn xn has a radius of
convergence 3: Then the radius of convergence of
X
(cn + dn ) xn
P
P
is 2: This
for jxj <P2; the series
cn xn and
dn xn converge absolutely; on the other hand, for
Pis because
n
n
jxj > 2;
cn x diverges, so
(cn + dn ) x should as well.
P
42. Suppose that the radius of convergence of the power series
cn xn is R: What is the radius of
convergence of the power series
X
cn x2n
Well, we compute: we want
lim
n!1
cn+1 x2n+2
2 cn+1
= lim jxj
n!1
cn x2n
cn
1
= x2
<1
R
p
so we want jxj < R:
Examples:
Ex1: Find a power series expansion for f (x) = ln (1 + x) : We have
1
1
and substituting t =
x; we …nd
t
=
1
X
tk
k=0
1
X
1
n
=
( 1) xn
1 + x n=0
and integrating termwise to get ln (1 + x) gives
ln (1 + x) =
1
X
n+1
( 1)
n=1
Ex2: Find a power series expansion for f (x) = tan
d
tan
dx
1
1
(x) =
4
xn
n
x: Well:
1
1 + x2
and we can write a power series for
1
1+x2
using the power series
1
1
and subsituting in t =
x2 to …nd
tk
k=0
1
X
1
n
=
( 1) x2n
1 + x2
n=0
so
tan
t
1
X
=
1
x3
x5
+
3
5
(x) = C + x
x7
+ :::
7
evaluating at x = 0;R we …nd C = 0:
1
Ex3: Evaluate 1+x
7 dx as a power series. Well:
1
X
1
n
=
( 1) x7n
1 + x7
n=0
so integrating, we …nd
2
Z
x8
x15
+
8
15
1
=C +x
1 + x7
x22
x29
+
+ :::
22
29
Analytic Functions and Taylor Series
The main point here is the following: suppose that f (x) is an analytic function, i.e. for jx
f (x) = c0 + c1 (x
a) + c2 (x
2
a) + :::
We know:
f (a) = c0
can we determins the other coe¢ cients c1 ; c2 ; :::? Well,
f 0 (x) = c1 + 2c2 (x
a) + 3c3 (x
2
a) + :::
so
f 0 (a) = c1
similarly
f 00 (x) = 2c2 + (3) (2) c3 (x
a) + (4) (3) c4 (x
2
a) + :::
so
f 00 (a) = 2c2
and
f 000 (x) = (3) (2) c3 + (4) (3) (2) (x
a) + (5) (4) (3) (x
so
f 000 (a) = 3!c3
We can clearly see the pattern, and can prove by induction that
f (n) (a) = n!cn
for all n: This gives us the following:
5
2
a) + :::
aj < R;
Theorem 2 Let f (x) be an analytic function around a; i.e. let
f (x) =
1
X
cn (x
n
a)
n=0
for jx
aj < R: Then the coe¢ cients cn are given by the formula
cn =
so we have
f (x) =
f (n) (a)
n!
1
X
f (n) (a)
(x
n!
n=0
n
a)
This is called the Taylor Series expansion of the analytic function f:
If we only look at the case a = 0, i.e. power series centered at the origin, we have:
f (x) =
2.1
1
X
f (n) (0) n
x
n!
n=0
Important Analytic Functions
Let’s see what happens if we apply this to ex : This gives, since
dn x
(e ) jx=0 = 1
dxn
for all n; so we write the power series as
1
X
xn
n!
n=0
P1 n
So if ex is analytic at 0; then it is given by the series n=0 xn! : Moreover, we can check that the radius of
convergence for this power series is 1; so if it analytic at 0; then for all x 2 R,
ex =
1
X
xn
n!
n=0
We’ll show tomorrow, when we cover Taylor’s inequality, that in fact ex is analytic at 0:
Then we can do a similar argument to see, if f (x) = sin x; that
8
if n is even
< 0
1
if n = 1; 5; 9; 13; :::
f (n) (0) =
:
1 if n = 3; 7; 11; 15; :::
so if sin x is analytic at 0; then (check radius of convergence using n-th root test) it is equal to:
sin x = x
x3
x5
+
3!
5!
x7
+ :::
7!
We can do a similar computation for g (x) = cos x to see
8
if n is odd
< 0
1
if n = 0; 4; 8; 12; :::
g (n) (0) =
:
1 if n = 2; 6; 12; 16; :::
then
cos x = 1
x2
x4
+
2!
4!
6
x6
+ :::
6!