Download Chapter 12 Infinite series, improper integrals, and Taylor series

Chapter 12
Infinite series, improper integrals, and
Taylor series
12.1
Introduction to series
In studying calculus, we have explored a variety of functions. Among the most basic are polynomials,
i.e. functions such as
p(x) = x5 + 2x2 + 3x + 2.
These functions have features that make them particularly easy to handle: it is elementary to
compute the derivatives and integrals of polynomials. Further, the value of such a function can
be computed by simple multiplications and additions, i.e. by basic operations easily handled by
an ordinary calculator. This is not the case for, say, trigonometric functions. For example, to
find the decimal value of sin(2.5) we require a scientific calculator. The same difficulty occurs for
exponentials and logarithms. For this reason, being able to approximate a function by a polynomial
is an attractive proposition. This idea forms the core topic of this chapter.
We have discussed linear approximations in differential calculus, Here we extend such elementary approximation methods. Our goal is to increase the accuracy of the linear approximation by
including higher order terms (quadratic, cubic, etc), i.e. to find a polynomial that approximates
the given function.
We first review the idea of series introduced in Chapter 1.
12.2
The geometric series
Recall the geometric series seen earlier this term:
2
3
n
Sn = 1 + r + r + r + . . . + r =
n
X
rk
k=0
v.2005.1 - January 5, 2009
1
Math 103 Notes
Chapter 12
We have seen a trick that helps to sum the series, namely multiplying both sides by r. This leads
to
Sn = 1+ r + r 2 + . . . + r n
rSn =
r + r 2 + . . . + r n + r n+1
Subtracting the second line from the first and cancelling terms leads to
(1 − r)Sn = 1 − r n+1 .
Solving for the sum Sn leads to the observation made previously, namely that
The sum of a finite geometric series,
2
n
Sn = 1 + r + r + · · · + r =
n
X
rk ,
k=0
is
Sn =
1 − r n+1
.
1−r
(12.1)
We would like to investigate how this “sum” behaves when more and more terms of the series
are included. Our goal is to find a way of attaching a meaning to the expression
Sn =
∞
X
rk ,
k=0
when the series becomes an infinite series. We will use the following definition:
12.2.1
The infinite geometric series
Definition
An infinite series that has a finite sum is said to be convergent. Otherwise it is divergent.
Definition
Suppose that S is an (infinite) series whose terms are ak . Then the partial sums, Sn , of this series
are
n
X
Sn =
ak .
k=0
We say that the sum of the infinite series is S, and write
S=
∞
X
ak ,
k=0
v.2005.1 - January 5, 2009
2
Math 103 Notes
Chapter 12
provided that
S = lim
n→∞
n
X
ak .
k=0
That is, we consider the infinite series as the limit of the partial sums as the number of terms n is
increased. In this case we also say that the infinite series converges to S.
We will see that only under certain circumstances will infinite series have a finite sum, and we
will be interested in exploring two questions:
1. Under what circumstances does an infinite series have a finite sum.
2. What value does the partial sum approach as more and more terms are included.
In the case of a geometric series, the sum of the series, 12.1 depends on the number of terms in
the series, n via r n+1 . Whenever r > 1, or r < −1, this term will get bigger as n increases, whereas,
for 0 < r < 1, this term decreases with n. We can say that
lim r n+1 = 0 provided |r| < 1.
n→∞
These observations are illustrated by two specific examples below. This leads to the following
conclusion:
The sum of an infinite geometric series,
S = 1 + r + r2 + · · · + rk + · · · =
∞
X
rk ,
k=0
exists provided |r| < 1 and is
S=
1
.
1−r
(12.2)
Examples of convergent and divergent geometric series are discussed below.
12.2.2
Example: A geometric series that converges.
Consider the geometric series with r = 21 , i.e.
2 3
n X
n k
1
1
1
1
1
+
+ ...+
=
Sn = 1 + +
2
2
2
2
2
k=0
Then
Sn =
v.2005.1 - January 5, 2009
1 − (1/2)n+1
1 − (1/2)
3
Math 103 Notes
Chapter 12
We observe that as n increases, i.e. as we retain more and more terms, we obtain
1
1 − (1/2)n+1
=
= 2.
n→∞ 1 − (1/2)
1 − (1/2)
lim Sn = lim
n→∞
In this case, we write
∞ n
X
1
n=0
2
=1+
1
1
+ ( )2 + . . . = 2
2
2
and we say that “the (infinite) series converges to 2”.
12.2.3
Example: A geometric series that diverges
In contrast, we now investigate the case that r = 2: then the series consists of terms
2
3
n
Sn = 1 + 2 + 2 + 2 + . . . + 2 =
n
X
2k
k=0
1 − 2n+1
= 2n+1 − 1
1−2
We observe that as n grows larger, the sum continues to grow indefinitely. In this case, we say that
the sum does not converge, or, equivalently, that the sum diverges.
Sn =
It is important to remember that an infinite series, i.e. a sum with infinitely many terms added
up, can exhibit either one of these two very different behaviours. It may converge in some cases, as
the first example shows, or diverge (fail to converge) in other cases. We will see examples of each of
these trends again. It is essential to be able to distinguish the two. Divergent series (or series that
diverge under certain conditions) must be handled with particular care, for otherwise, we may find
contradictions or seemingly reasonable calculations that have meaningless results.
12.3
Trading a function for an infinite series
The expression found above holds true for any number r whose magnitude is less than 1. Suppose
we think of this number as a quantity we wish to vary - let us call it x to remind us that we consider
it a variable. Then rewritten in terms of x, the expression we found is
1
= 1 + x + x2 + x3 + . . .
1−x
when |x| < 1
We wish to understand this in a new way: on the left we have a function, namely
f (x) =
1
.
1−x
On the right we have what looks like an “infinite polynomial”, i.e. a sum of terms that are all powers
of x. This type of expansion is called a power series, since it consists of a sum of powers of x. We
v.2005.1 - January 5, 2009
4
Math 103 Notes
Chapter 12
will soon see that this power series is a particular example of a Taylor series, an expansion that
can be helpful in approximating more complicated functions such as trigonometric or exponential
functions.
One way to think of this new connection between the function on the left and the series on the
right is that it defines a whole set of polynomials, each one formed by keeping one more term in
the (infinite) expansion, and each one a better and better approximation of the given function. For
example, we could list the successive approximations as
T0 (x)
T1 (x)
T2 (x)
Tn (x)
=
=
=
=
1
1+x
1 + x + x2
1 + x + x2 + . . . + xn
The first polynomial is just the value of the function at one point (at the point x = 0, it turns out).
The second is the linear approximation, arrived at by using information about the derivative. The
subsequent polynomials retain more and more details of the function, as we will shortly see. We so
arrive at the Taylor polynomial of degree n, shown at the end of this list. As long as |x| < 1, the
Taylor polynomials provide a better and better approximation for the function f (x) = 1/(1 − x) as
n gets larger.
But we must take care, since if |x| > 1, the expansion will fail to converge - as we have already
established previously. For example, the value x = 1 leads to
1 + 1 + 1 + ...,
which does not converge. As another example, if we let x = −1, the sum looks like this:
1 + (−1) + 1 + (−1) + . . .
If we add the first two terms, we get zero. If we add the third term, we get 1. Adding the fourth
term gives zero. As we continue term by term, the sum keeps flipping back and forth between 0
and 1. In this case, we also say that the sum does not converge.
Thus this Taylor series is valid only so long as |x| < 1. The restriction |x| < 1 is required to
ensure that the infinite sum will converge to a finite result. We say that the radius of convergence
of this series is 1.
We have seen two ways in which a series can diverge: first, the sum may become unbounded,
i.e. infinitely large. Second, the sum could jump from value to value without approaching any fixed
value. In both cases, the series is divergent and will have little practical significance.
12.4
Improper integrals
We will see that there is a close connection between certain infinite series and improper integrals,
i.e. integrals over an infinite domain. Recall the following definition:
v.2005.1 - January 5, 2009
5
Math 103 Notes
Chapter 12
Definition
An improper integral is an integral performed over an infinite domain, e.g.
Z ∞
f (x) dx.
a
The value of such an integral is understood to be the limit
Z
∞
f (x) dx = lim
b→∞
a
Z
b
f (x) dx.
a
i.e. we evaluate an improper integral by first computing a definite integral over a finite domain
a ≤ x ≤ b, and then taking a limit as the endpoint b moves off to larger and larger values.
Definition
When the limit shown above exists, we say that the improper integral converges. Other wise we
say that the improper integral diverges.
With this in mind, we compute a number of classic integrals:
12.4.1
Example: A divergent integral
Z b
1
1
I=
dx = lim
dx
b→∞
x
a x
1
b
I = lim ln(x) = lim (ln(b) − ln(1))
b→∞
b→∞
Z
∞
1
I = lim ln(b) = ∞
b→∞
The fact that we get an infinite value for this integral follows from the observation that ln(b)
increases without bound as b increases. Thus the area under the curve f (x) = 1/x over the infinite
interval 1 ≤ x ≤ ∞ is not finite. We say that the improper integral of 1/x diverges (or does not
converge).
12.4.2
Example: A convergent integral
Now consider the related integral
I=
Then
Z
∞
1
1
dx = lim
b→∞
x2
Z
b
x−2 dx.
1
b
1
I = lim (−x ) = − lim
− 1 = 1.
b→∞
b→∞
b
−1
1
Thus, this integral converges and we see that its value is 1.
v.2005.1 - January 5, 2009
6
Math 103 Notes
12.4.3
Chapter 12
When does the integral of 1/xp converge?
More generally, consider
I=
By a similar calculation, we find that
Z
∞
1
1
dx.
xp
b
x1−p I = lim
b→∞ (1 − p) 1
1
I = lim
b1−p − 1
b→∞ 1 − p
Thus this integral converges provided that the term b1−p does not “blow up” as b increases. This
can only be true so long as p > 1. In this case, we have
I=
1
.
p−1
To summarize our result,
Z
1
12.4.4
∞
(
1
dx
xp
converges if
diverges if
p>1
p≤1
Comparisons
The integrals discussed above can be used to make comparisons that help us to identify when
other improper integrals converge or diverge. The following important result establishes how these
comparisons work:
Suppose we are given two functions, f (x) and g(x), both continuous on some infinite interval
[a, ∞). Suppose, moreover, that, at all points on this interval, the first function is smaller than the
second, i.e.
0 ≤ f (x) ≤ g(x).
Then the following conclusions can be made:
1.
Z
∞
f (x) dx ≤
a
Z
∞
g(x) dx. (This means that the area under f (x) is smaller than the area
a
under g(x).)
Z ∞
Z
2. If
g(x) dx converges, then
a
smaller one)
Z ∞
Z
3. If
f (x) dx diverges, then
a
larger one.)
v.2005.1 - January 5, 2009
∞
f (x) dx converges. (If the larger area is finite, so is the
a
∞
g(x) dx diverges. (If the smaller area is infinite, so is the
a
7
Math 103 Notes
Chapter 12
Example
We can determine that the integral
Z
converges by noting that for all x > 0
∞
1
x
dx
1 + x3
x
x
1
≤ 3 = 2.
3
1+x
x
x
Thus
Z
1
∞
x
dx ≤
1 + x3
Z
1
∞
1
dx
x2
Since the larger integral on the right is known to converge, so does the smaller integral on the left.
12.5
Infinite series
In this section we present a few classical series which are important landmarks and which carry
special significance. These will be useful in our appreciation of the behaviour of infinite series.
Geometric series
We have already seen that
The geometric series converges when |r| < 1 and its sum is then
S=
∞
X
rn =
n=0
1
.
1−r
If r ≥ 1 the geometric series diverges.
The harmonic series
The harmonic series is a sum of terms of the form 1/k where k = 1, 2, . . . . At first appearance,
this series might seem to have the desired qualities of a convergent series, simply because the
successive terms being added are getting smaller and smaller, but this is a misleading and deceptive
observation!
The harmonic series
∞
X
1
k=1
v.2005.1 - January 5, 2009
k
=1+
1 1 1
+ + + ...
2 3 4
diverges
8
Math 103 Notes
Chapter 12
1.0
<==== the function y=1/x
The harmonic series
0.0
0.0
11.0
Figure 12.1: The harmonic series is a sum that corresponds to the area under the staircase shown
above.
It can be shown that the harmonic series does not converge. by comparing it to an integral
of the function
1
y = f (x) = .
x
In Figure 12.1 we show a graph of this function superimposed on a bar graph. We will compare the
areas of the steps in the bar graph with the area under the graph of this function. For the area of
the steps, we note that the base width of each step is 1, and the heights form the sequence
1 1 1
{1, , , , . . . }
2 3 4
Thus the area of (infinitely many) of these steps is
∞
X1
1
1
1 1 1
1
.
A = 1 · 1+ 1· + 1 · + 1 · + ··· = 1 + + + + ... =
2
3
4
2 3 4
k
k=1
Thus the area of the bar graph is a geometric representation of precisely the series we are investigating, the harmonic series.
On the other hand, the area under the graph of the function y = f (x) = 1/x is given by the
improper integral
Z ∞
1
dx
x
1
We have seen previously that this integral diverges!
v.2005.1 - January 5, 2009
9
Math 103 Notes
Chapter 12
¿From Figure 12.1 we see that the areas under the function, Af and under the bar graph, Ab ,
satisfy
Af < Ab .
Thus, since the smaller of the two is infinite, so is the larger. We have established, using this
comparison, that the the sum of the harmonic series cannot be finite, so that this series diverges.
Other comparisons: The “p” series
More generally, we can compare series of the form
∞
X
1
kp
k=1
to the integral
Z
∞
1
1
dx
xp
in precisely the same way. This leads to the conclusion that
The “p” series,
∞
X
1
kp
k=1
(
converges if
diverges if
p>1
p≤1
For example, the series
∞
X
1
1
1 1
=1+ + +
+ ...
2
k
4 9 16
k=1
converges. Notice, however, that the comparison does not give us a value to which the sum converges.
It merely indicates that the series does converge.
Alternating series
We can modify the harmonic series to produce the alternating harmonic series in which the sign
alternates from term to term.
∞
X
k=1
(−1)k+1
1 1 1
1
= 1 − + − + ...
k
2 3 4
The alternating harmonic series converges due to the alternation of signs. (Consecutive terms
nearly cancel out). We will see later how to show that this series converges, remarkably to the value
ln(2). However, the convergence is “slow”: it takes many terms added together to get close to the
v.2005.1 - January 5, 2009
10
Math 103 Notes
Chapter 12
“final value”. The speed of convergence is important for practical reasons: if we want to devise a
calculation that will help us to find an approximate decimal expansion of some number (such as
ln(2)), we would certainly prefer to use a series that converges quickly, to reduce the number of
calculations that have to be made.
More generally, for alternating series with terms that are decreasing to zero, we have convergence,
as summarized below.
Suppose that successive terms ak and ak+1 have opposite signs. Then,
provided the terms are getting smaller so that ak → 0, the alternating
series
∞
X
ak
k=0
converges.
12.5.1
Example
In this example, we will consider the series
∞
X
1
1 1
1
= 1+1+ + +
+ ...
k!
2 6 24
k=0
We can see that this series converges by noticing:
1
1 1 1
1
1 1
1
1
= · · · . . . ≤ 1 · · · . . . = k−1 =
k!
1 2 3
k
2 2
2
2
k−1
1
2
if k ≥ 2. In other words, the series eventually becomes smaller than a geometric series which
converges.
∞
X
1
1 1
1
= 1+1+ + +
+ ...
k!
2 6 24
k=0
< 1+1+
1+1+
1 1 1
+ + + ...
2 4 8
1
1
1
1 1 1
+ + + . . . = 1 + (1 + + ( )2 + ( )3 + . . .)
2 4 8
2
2
2
∞ k
X
1
= 1+
=1+2=3
2
k=0
Note that the series converges, and, moreover, converges to a number smaller than 3.
v.2005.1 - January 5, 2009
11
Math 103 Notes
12.6
Chapter 12
Finding Taylor series
Now that we understand a little bit about when series converge, we will turn our attention to finding
the series representations of some elementary functions.
Using one series to find another
Once we have identified the expansion of some function in terms of an infinite “Taylor polynomial”,
we can build up many other expansions of related functions. In this section, we repeatedly use
our geometric summation formula, together with substitution and/or some other operation to find
interesting new conclusions.
All examples are based on the fact (discussed earlier) that
∞
X
1
= 1 + u + u2 + u3 + . . . =
uk
1−u
k=0
when |u| < 1.
12.6.1
Example: Simple substitution
We make the substitution u = −t, then |u| < 1 means that | − t| < 1, so the above formula for the
sum of a geometric series implies that:
1
= 1 + (−t) + (−t)2 + (−t)3 + . . .
1 − (−t)
1
= 1 − t + t2 − t3 + t4 + . . .
1+t
This means we have produced the series for the function 1/(1 + t).
12.6.2
Example: Term by term integration
We will use the results of Example 12.6.1, but we follow our substitution by integration. On the
left, we integrate the function f (t) = 1/(1 + t) (to arrive at a ln type integral) and on the right we
integrate the power terms of the expansion. We are permitted to integrate the power series term
by term provided that the series converges.
Z
0
x
1
dt =
1+t
Z
x
(1 − t + t2 − t3 + t4 − . . .) dt
0
∞
X
x2 x3 x4
xk
ln(1 + x) = x −
+
−
+ ... =
(−1)k+1
2
3
4
k
k=1
v.2005.1 - January 5, 2009
12
Math 103 Notes
Chapter 12
This procedure has allowed us to find a series representation for a new function, ln(1 + x).
ln(1 + x) = x −
x2 x3 x4
+
−
+ ...
2
3
4
When is this series expected to converge? We obtained this by starting with the geometric series
which only converged when |u| < 1. We made the substitution u = −t, and then integrated for
0 < t < x. To assure convergence, the value of t is not permitted to leave the interval |t| < 1 so we
certainly cannot expect the series for ln(1 + x) to converge when |x| > 1 . Indeed, for x = −1, we
have ln(1 + x) = ln(0) which is undefined. We also get a right hand side of
−(1 +
1 1 1
+ + + . . .)
2 3 4
which is the divergent harmonic series preceded by a negative sign. Thus, we must avoid x = −1,
since the expansion will not converge there, and neither is the function defined.
An expansion for ln(2)
Strictly speaking, our analysis does not predict what happens if we substitute x = 1 into the
expansion of the function found in Example 2, because this value of x is outside of the permitted
range −1 < x < 1 in which the Taylor series can be guaranteed to converge. It takes some deeper
mathematics (Abel’s theorem) to prove that the result of this substitution actually makes sense,
and converges, i.e. that
1 1 1
ln(2) = 1 − + − + . . .
2 3 4
We state without proof here that the alternating harmonic series converges to ln(2).
12.6.3
Example
Suppose we make the substitution u = −t2 into the geometric series formula, and recall that we
need |u| < 1 for convergence. Then
1
= 1 + (−t2 ) + (−t2 )2 + (−t2 )3 + . . .
1 − (−t2 )
∞
X
1
2
4
6
8
(−1)n t2n
= 1 − t + t −t + t + ... =
1 + t2
k=0
This series will converge provided |t| < 1. Now integrate both sides, and recall that the antiderivative
of the function 1/(1 + t2 ) is arctan(t). Then
Z x
Z x
1
dt =
(1 − t2 + t4 − t6 + t8 + . . .) dt
2
1
+
t
0
0
∞
X
x(2k−1)
x3 x5 x7
+
−
+ ... =
(−1)k+1
arctan(x) = x −
3
5
7
(2k − 1)
k=1
v.2005.1 - January 5, 2009
13
Math 103 Notes
Chapter 12
An expansion for π
For a particular application of this expansion, consider plugging in x = 1. Then
arctan(1) = 1 −
1 1 1
+ − + ...
3 5 7
But arctan(1) = π/4. Thus we have found a way of computing the irrational number π, namely
!
∞
X
1 1 1
1
k+1
π = 4 1 − + − + ... = 4
.
(−1)
3 5 7
(2k − 1)
k=1
While it is true that this series converges, the convergence is slow. (This can be seen by adding up
the first 100 or 1000 terms of this series with a spreadsheet.) This means that it is not practical
to use such a series as an approximation for π. (There are other series that converge to π very
rapidly.)
12.6.4
Finding the coefficients directly from the function
As indicated in the introduction, approximating functions by the sum of power functions (i.e. by
a polynomial or a series) is helpful in practical calculations. For one thing, we can evaluate a
polynomial (or finite series) more readily than functions such as arctan or ln. In the preceding
discussions, we found some series representations of functions fortuitously. Here we set out to
identify the connection by starting with the function of interest.
Suppose we have a function which we want to write as
2
3
f (x) = a0 + a1 x + a2 x + a3 x + . . . =
∞
X
ak xk .
k=0
We can use information about the function to directly determine the coefficients. To determine a0 ,
let x = 0 and note that
f (0) = a0 + a1 0 + a2 02 + a3 03 + . . . = a0
We conclude that
a0 = f (0).
By differentiating both sides we find the following:
f ′ (x) = a1 + 2a2 x + 3a3 x2 + . . . + kak xk−1 + . . .
f ′′ (x) = 2a2 + 2 · 3a3 x + . . . + (k − 1)kak xk−2 + . . .
f ′′′ (x) = 2 · 3a3 + . . . + (k − 2)(k − 1)kak xk−3 + . . .
f (k) (x) = 1 · 2 · 3 · 4 . . . kak + . . .
v.2005.1 - January 5, 2009
14
Math 103 Notes
Chapter 12
Here we have used the notation f (k) (x) to denote the k’th derivative of the function. Now evaluate
each of the above derivatives at x = 0. Then
f ′ (0) = a1 ,
⇒ a1 = f ′ (0)
f ′′ (0) = 2a2 ,
⇒ a2 =
f ′′ (0)
2
f ′′′ (0) = 2 · 3a3 ,
⇒ a3 =
f ′′′ (0)
2·3
f (k) (0) = k!ak ,
⇒ ak =
f (k) (0)
k!
This gives us a recipe for calculating all coefficients ak . This means that if we can compute all the
derivatives of the function f (x), then we know the coefficients of the Taylor series as well. Because
we have evaluated all the coefficients by the substitution x = 0, we say that the resulting power
series is the Taylor series of the function about x = 0.
12.6.5
Example: A Taylor series for ex
Consider the function f (x) = ex . All the derivatives of this function are equal to ex . In particular,
f (k) (x) = ex
⇒
f (k) (0) = 1.
So that the coefficients of the Taylor series are
ak =
1
f (k) (0)
=
k!
k!
Therefore the Taylor series for ex about x = 0 is
∞
a0 + a1 x + a2 x2 + a3 x3 + . . . + ak xk + . . . = 1 + x +
X xk
xk
x2 x3 x4
+
+
+ ...+
+ ... =
2
6
24
k!
k!
k=0
This is a very interesting series. We state here without proof that this series converges for all values
of x. Further, the function defined by the series is in fact equal to ex that is,
∞
ex = 1 + x +
X xk
x2 x3
+
+ ... =
2
6
k!
k=0
The implication is that the function ex is completely determined (for all x values) by its behaviour
(i.e. derivatives of all orders) at x = 0. In other words, the value of the function at x = 1, 000, 000
is determined by the behaviour of the function around x = 0. This means that ex is a very special
function with superior “predictable features”. If a function f (x) agrees with its Taylor polynomial
on a region (−a, a), as was the case here, we say that f is analytic on this region. It is known that
ex is analytic for all x.
v.2005.1 - January 5, 2009
15
Math 103 Notes
Chapter 12
Remark: How the exponential function grows
We can use the results of this example to establish the fact that the exponential function grows
“faster” than any power function xn . That is the same as saying that the ratio of ex to xn (for any
power n) increases with x.
We can see this fact directly from the results of our Taylor series for ex and the following simple
“division”:
2
3
n
n+1
x
1 + x + x2! + x3! + . . . xn! + (n+1)!
+ ...
ex
=
xn
xn
1
1
x
1
+ ...
= n + n−1 + . . . + +
x
x
n! (n + 1)!
x
>
(n + 1)!
We see that the last term becomes arbitrarily large as x → ∞, implying that
ex
=∞
x→∞ xn
lim
This shows that the exponential function ex grows faster than any power function, and hence also
faster than any (finite) polynomial.
12.6.6
Example: Other (related) exponential functions
We can also easily obtain a Taylor series expansion for functions related to ex , without assembling
the derivatives. We start with the result that
∞
X uk
u2 u3
e =1+u+
+
+ ... =
2
6
k!
k=0
u
Then the substitution u = −x leads to the result
∞
e−x = 1 − x +
X (−x)k
x2 x3
−
+ ... =
.
2
6
k!
k=0
The substitution u = x2 leads to
∞
x2
e
X (x2 )k
(x2 )2 (x2 )3
+
+ ... =
=1+x +
2
6
k!
k=0
2
In other words,
∞
x2
e
v.2005.1 - January 5, 2009
X x2k
x4 x6
+
+ ... =
=1+x +
2
6
k!
k=0
2
16
Math 103 Notes
Chapter 12
2.0
T1
T3
sin(x)
T2
T4
-2.0
0.0
7.0
Figure 12.2: An approximation of the function y = sin(x) by successive Taylor polynomials,
T1 , T2 , T3 , T4 . The higher Taylor polynomials do a better job of approximating the function on
a larger interval about x = 0.
12.6.7
Example: Taylor series of simple trigonometric functions
In this example we determine the Taylor series for the function f (x) = sin x.
The derivatives of y = sin(x) are
f (x) = sin x
f ′ (x) = cos x
f ′′ (x) = − sin x
f ′′′ (x) = − cos x
f (4) (x) = sin x
and then the cycle repeats. This means that f (0) = 0, f ′ (0) = 1, f ′′ (0) = 0, f ′′′ (0) = −1 and then
the pattern repeats. In other words,
1
1
a0 = 0, a1 = 1, a2 = 0, a3 = − , a4 = 0, a5 = , . . .
3!
5!
Now it is a fact that sin x is analytic for all x so that
∞
sin x = x −
v.2005.1 - January 5, 2009
X
x2n+1
x3 x5 x7
+
−
+ ... =
(−1)n
3!
5!
7!
(2n + 1)!
n=0
17
Math 103 Notes
Chapter 12
Consider the sequence of polynomials
T1 (x) = x
x3
3!
x3 x5
T3 (x) = x −
+
3!
5!
3
x5 x7
x
+
−
T4 (x) = x −
3!
5!
7!
T2 (x) = x −
Then these polynomials provide a better and better approximation to the function sin(x) close
to x = 0. The first of these is just a linear (or tangent line) approximation that we had studied
long ago. The second improves this with a quadratic approximation, etc. Figure 12.2 illustrates
how the first few Taylor polynomials approximate the function sin(x) near x = 0.
To determine the Taylor series for cos x, we can simply differentiate the series for sin x.
d
sin x
dx
x3 x5 x7
d
(x −
+
−
+ . . .)
=
dx
3!
5!
7!
3x2 5x4 7x6
+
−
+ ...
= 1−
3!
5!
7!
cos x =
We find that
∞
2n
X
x2 x4 x6
n x
+
−
+ ... =
(−1)
cos(x) = 1 −
2
4!
6!
(2n)!
n=0
12.7
Application of Taylor series
In this section we illustrate some of the applications of Taylor series to problems that may be
difficult to solve using other conventional methods
12.7.1
Evaluating definite integrals
Some functions do not have an antiderivative that can be expressed in terms of other simple functions. Integrating these functions can be a problem, as we cannot use the convenience of evaluating
that antiderivative at endpoints, as the Fundamental Theorem of Calculus specifies. In some cases,
we can approximate the value of the definite integral using a Taylor series, as the following example
illustrates.
v.2005.1 - January 5, 2009
18
Math 103 Notes
Chapter 12
Example 1
Evaluate the definite integral
Z
1
sin(x2 ) dx
0
2
A simple substitution (e.g. u = x ) will not work here, and we cannot find an antiderivative. Here
is how we might approach the problem using Taylor series: We know that the series expansion for
sin(t) is
t3 t5 t7
sin t = t − + − + . . .
3! 5! 7!
2
Substituting t = x , we have
sin(x2 ) = x2 −
x6 x10 x14
+
−
+ ...
3!
5!
7!
In spite of the fact that we cannot antidifferentiate the function, we can antidifferentiate the Taylor
series, just as we would a polynomial:
Z
1
1
x6 x10 x14
+
−
+ . . .) dx
3!
5!
7!
sin(x ) dx =
Z
=
=
1
1
1
1
−
+
−
+ ...
3 7 · 3! 11 · 5! 15 · 7!
2
0
(x2 −
0
1
x3
x7
x11
x15
−
+
−
+ . . . 3
7 · 3! 11 · 5! 15 · 7!
0
This is an alternating series so we know that it converges. If we add up the first four terms, the
pattern becomes clear: the series converges to 0.31026.
12.7.2
Solving differential equations
Some differential equations cannot be solved in terms of familiar functions (just as some functions
do not have antiderivatives which can be expressed in terms of familiar functions). However, Taylor
series can come to the rescue again. Here we will present two examples of this idea.
Example
We are already familiar with the differential equation and initial condition that describes unlimited
exponential growth.
dy
= y,
dx
y(0) = 1.
v.2005.1 - January 5, 2009
19
Math 103 Notes
Chapter 12
Indeed, from previous work, we know that the solution of this differential equation and initial
condition is y(x) = ex , but we will pretend that we do not know this fact in illustrating the
usefulness of Taylor series. In some cases, where separation of variables does not work, this option
would have great practical value.
Let us express the “unknown” solution to the differential equation as
y = a0 + a1 x + a2 x2 + a3 x3 + a4 x4 + . . .
Our task is to determine values for the coefficients ai
Since this function satisfies the condition y(0) = 1, we must have y(0) = a0 = 1.
Differentiating this power series leads to
dy
= a1 + 2a2 x + 3a3 x2 + 4a4 x3 + . . .
dx
But according to the differential equation,
match, i.e.
dy
dx
= y, Thus, it must be true that the two Taylor series
a0 + a1 x + a2 x2 + a3 x3 + a4 x4 + . . . = a1 + 2a2 x + 3a3 x2 + 4a4 x3 + . . .
This can only happen if the coefficients of like terms are the same, i.e. if the constant terms on
either side of the equation are equal, if the terms of the form Cx2 on either side are equal, and so
on for all powers of x. Equating coefficients, we obtain:
a0 = a1 = 1,
a1 = 2a2 ,
⇒ a1 = 1
⇒ a2 = a21 =
a2 = 3a3 ,
⇒ a3 =
a2
3
=
1
2·3
a3 = 4a4 ,
⇒ a4 =
a3
4
=
1
2·3·4
an-1 = nan ,
⇒ an =
an-1
n
This means that
y =1+x+
1
2
=
1
1·2·3...n
=
1
n!
x2 x3
xn
+
+ ...+
+ . . . = ex
2!
3!
n!
which, as we have seen, is the expansion for the exponential function that is the real solution we
have been expecting. In the example here shown, we would hardly need to use series to arrive at
the right conclusion, but in the next example, we would not find it as easy to discover the solution
by other techniques discussed previously.
Example 2: Airy’s equation
Airy’s equation arises in the study of optics, and (with initial conditions) is as follows:
v.2005.1 - January 5, 2009
20
Math 103 Notes
Chapter 12
y ′′ = xy
y(0) = 1
y ′ (0) = 0
As before, we will write the solution as a series:
y = a0 + a1 x + a2 x2 + a3 x3 + a4 x4 + a5 x5 + . . .
Using the information from the initial conditions, we get y(0) = a0 = 1 and y ′(0) = a1 = 0. Now
we can write down the derivatives:
y ′ = a1 + 2a2 x + 3a3 x2 + 4a4 x3 + 5a5 x4 + . . .
y ′′ = 2a2 + 2 · 3x + 3 · 4x2 + 4 · 5x3 + . . .
The equation then gives
y ′′ = xy
2a2 + 2 · 3a3 x + 3 · 4a4 x2 + 4 · 5a5 x3 + . . . = x(a0 + a1 x + a2 x2 + a3 x3 + . . .)
2a2 + 2 · 3a3 x + 3 · 4a4 x2 + 4 · 5a5 x3 + . . . = a0 x + a1 x2 + a2 x3 + a3 x4 + . . .
Again, we can equate the coefficients of x, and use a0 = 1 and a1 = 0, to obtain
2a2 = 0
2 · 3a3 = a0
3 · 4a4 = a1
4 · 5a5 = a2
5 · 6a6 = a3
⇒ a2
⇒ a3
⇒ a4
⇒ a5
⇒ a6
=0
1
= 2·3
=0
=0
1
= 2·3·5·6
This gives us the first few terms of the solution:
x3
x6
+
+ ...
2·3 2·3·5·6
If we continue in this way, we can write down many terms of the series.
y =1+
v.2005.1 - January 5, 2009
21