NOTES
THEOREM 4.2 Suppose that lim
1. One reason that it is important to distinguish absolute from con- con-ditional convergence of a series is the rearrangement of series,
7.6 POWER SERIES
We now want to expand our discussion of series to the case where the terms of the series are functions of the variable x. Pay close attention to what we are about to introduce, for this is the culmination of all your hard work in sections 7.1 through 7.5. The primary reason for studying series is that we can use them to represent functions. This opens up all kinds of possibilities for us, from approximating the values of transcendental functions to calculating derivatives and integrals of such functions, to studying differential equations.
As well, defining functions as convergent series produces an explosion of new functions available to us. In fact, many functions of great significance in applications (for instance, Bessel functions) are defined as a series. We take the first few steps in this section.
As a start, consider the series
∞ k=0
(x− 2)k= 1 + (x − 2) + (x − 2)2+ (x − 2)3+ · · · .
Notice that for each fixed x, this is a geometric series with r = (x − 2). Recall that this says that the series will converge whenever|r| = |x − 2| < 1 and will diverge whenever |r| =
|x − 2| ≥ 1. Further, for each x with |x − 2| < 1 (i.e., 1 < x < 3), the series converges to a
1− r = 1
1− (x − 2) = 1 3− x. That is, for each x in the interval (1, 3), we have
∞ k=0
(x− 2)k= 1 3− x.
3 2 1
1 1 2
3 y f (x)
y P1(x) y P2(x)
y P3(x) y
x
FIGURE 7.38
y= 1
3− x and the first three partial sums of∞
k=0(x− 2)k.
For all other values of x, the series diverges. In Figure 7.38, we show a graph of f (x)= 1
3− x, along with the first three partial sums Pn, of this series, where Pn(x)=
n k=0
(x− 2)k= 1 + (x − 2) + (x − 2)2+ · · · + (x − 2)n,
on the interval [1, 3]. Notice that as n gets larger, Pn(x) appears to get closer to f (x), for any given x in the interval (1, 3). Further, as n gets larger, Pn(x) tends to stay close to f (x) for a larger range of x-values.
Make certain that you understand what we’ve observed here: we have taken a series and noticed that it is equivalent to (i.e., it converges to) a known function on a certain interval.
You might ask why anyone would care if you could do that. Certainly, f (x)= 1 3− x is a perfectly good function and anything you’d want to do with it will most definitely be easier using the algebraic expression 1
3− x than using the equivalent series representation,
∞
k=0(x− 2)k. However, imagine what benefits you might find if you could take a given function (say, one that you don’t know a whole lot about) and find an equivalent series representation. This is precisely what we are going to do in section 7.7. For instance, we will be able to show that for all x,
ex=∞
k=0
xk
k! = 1 + x +x2 2! +x3
3! + x4
4! + · · · . (6.1)
So, who cares? Well, suppose you wanted to calculate e1.234567. How would you do that? Of course, you’d use a calculator. But, haven’t you ever wondered how your calculator does it? The problem is that exis not an algebraic function. That is, we can’t compute its values by using algebraic operations (i.e., addition, subtraction, multiplication, division and nth roots). Over the next few sections, we begin to explore this question. For the moment, let us say this: if we have the series representation (6.1) for ex, then for any given x, we can compute an approximation to ex, simply by summing the first few terms of the equivalent series. This is easy to do, since the partial sums of the series are simply polynomials.
In general, any series of the form
∞ k=0
bk(x− c)k = b0+ b1(x− c) + b2(x− c)2+ b3(x− c)3+ · · ·
Power series
NOTES
According to the expansion indicated to the right, the first term of the power series is b0, even when x= c, although the summation notation might suggest that the first term is b0(0)0, which is not defined.
is called a power series in powers of (x− c). We refer to the constants bk, k = 0, 1, 2,. . . as the coefficients of the series. The first question is: for what values of x does the series converge? Saying this another way, the power series∞
k=0bk(x− c)kdefines a function of x.
Its domain is the set of all x for which the series converges. The primary tool for investigating the convergence or divergence of a power series is the Ratio Test. Notice again that the partial sums of a power series are all polynomials (the simplest functions around).
EXAMPLE 6.1 Determining Where a Power Series Converges Determine the values of x for which the power series∞
k=0
k
3k+1xkconverges.
Solution Using the Ratio Test, we have convergence if
klim→∞
ak+1
ak
= limk→∞
(k+ 1)xk+1 3k+2
3k+1 kxk
= lim
k→∞
(k+ 1)|x|
3k = |x|
3 lim
k→∞
k+ 1 k
Since xk+1= xk· x1 and 3k+2= 3k+1· 31.
= |x|
3 < 1,
for |x| < 3 or −3 < x < 3. So, the series converges absolutely for −3 < x < 3 and diverges for|x| > 3 (i.e., for x > 3 or x < −3). Since the Ratio Test gives no conclusion for the endpoints x= ±3, we must test these separately.
For x = 3, we have the series
∞ k=0
k
3k+1xk=∞
k=0
k
3k+13k=∞
k=0
k 3. Since
klim→∞
k
3 = ∞ = 0,
the series diverges by the kth-term test for divergence. The series diverges when x = −3, for the same reason. Thus, the power series converges for all x in the interval (−3, 3) and diverges for all x outside this interval.
Observe that example 6.1 has something in common with the introductory example.
In both cases, the series have the form ∞
k=0bk(x− c)k and there is an interval of the form (c− r, c + r) on which the series converges and outside of which the series diverges. (In the case of example 6.1, notice that c= 0.) This interval on which a power series converges is called the interval of convergence. The constant r is called the radius of convergence (i.e., r is half the length of the interval of convergence). It turns out that there is such an interval for every power series. We have the following result.
THEOREM 6.1
Given any power series,∞
k=0bk(x− c)k, there are exactly three possibilities:
(i) The series converges for all x ∈ (−∞, ∞) and the radius of convergence is r = ∞;
(ii) The series converges only for x = c (and diverges for all other values of x) and the radius of convergence is r= 0; or
(iii) The series converges for x∈ (c − r, c + r) and diverges for x < c − r and for x> c + r, for some number r with 0 < r < ∞.
The proof of the theorem can be found in Appendix F.
EXAMPLE 6.2 Finding the Interval and Radius of Convergence Determine the radius and interval of convergence for the power series
∞ k=0
10k
k!(x− 1)k. Solution From the Ratio Test, we have
klim→∞
ak+1
ak
= limk→∞
10k+1(x− 1)k+1 (k+ 1)!
k!
10k(x− 1)k
= 10|x − 1| lim
k→∞
k!
(k+ 1)k!
Since (x− 1)k+1= (x − 1)k(x− 1)1 and (k+ 1)! = (k + 1)k!
= 10|x − 1| lim
k→∞
1
k+ 1 = 0 < 1,
for all x. This says that the series converges absolutely for all x. Thus, the interval of convergence for this series is (−∞, ∞) and the radius of convergence is r = ∞.
The interval of convergence for a power series can be a closed interval, an open interval or a half-open interval, as in example 6.3.
EXAMPLE 6.3 A Half-Open Interval of Convergence Determine the radius and interval of convergence for the power series∞
k=1
xk k4k. Solution From the Ratio Test, we have
klim→∞
ak+1
ak
= limk→∞
xk+1 (k+ 1)4k+1
k4k xk
= |x|
4 lim
k→∞
k
k+ 1 = |x|
4 < 1.
So, we are guaranteed absolute convergence for|x| < 4 and divergence for |x| > 4. It remains only to test the endpoints of the interval: x = ±4. For x = 4, we have
∞ k=1
xk k4k =∞
k=1
4k k4k =∞
k=1
1 k,
which you will recognize as the harmonic series, which diverges. For x= −4, we have
∞ k=1
xk k4k =∞
k=1
(−4)k k4k =∞
k=1
(−1)k k ,
which is the alternating harmonic series, which we know converges (see example 4.2).
So, in this case, the interval of convergence is the half-open interval [−4, 4) and the radius of convergence is r= 4.
Notice that (as stated in Theorem 6.1) every power series,∞
k=0bk(x− c)kconverges at least for x= c, since for x = c, we have the trivial case
∞ k=0
bk(x− c)k= b0+∞
k=1
bk(c− c)k= b0+∞
k=1
bk0k = b0+ 0 = b0.
EXAMPLE 6.4 A Power Series That Converges at Only One Point Determine the radius of convergence for the power series∞
k=0k!(x− 5)k. Solution From the Ratio Test, we have
klim→∞
ak+1
ak
= limk→∞
(k+ 1)!(x − 5)k+1 k!(x− 5)k
= lim
k→∞
(k+ 1)k!|x − 5|
k!
= lim
k→∞[(k+ 1)|x − 5|]
=
0, if x= 5
∞, if x = 5.
Thus, this power series converges only for x= 5 and so, its radius of convergence is r= 0.
Suppose that the power series∞
k=0bk(x− c)khas radius of convergence r > 0. Then the series converges absolutely for all x in the interval (c− r, c + r) and might converge at one or both of the endpoints, x = c − r and x = c + r. Notice that since the series converges for each x ∈ (c − r, c + r), it defines a function f on the interval (c − r, c + r),
f (x)=∞
k=0
bk(x− c)k = b0+ b1(x− c) + b2(x− c)2+ b3(x− c)3+ · · · . It turns out that such a function is continuous and differentiable, although the proof is beyond the level of this course. In fact, we differentiate exactly the way you might expect,
f(x)= d
d x f (x)= d
d x[b0+ b1(x− c) + b2(x− c)2+ b3(x− c)3+ · · ·]
= b1+ 2b2(x− c) + 3b3(x− c)2+· · · =∞
k=1
bkk(x− c)k−1,
Differentiating a power series
where the radius of convergence of the resulting series is also r . Since we find the deriva-tive by differentiating each term in the series, we call this term-by-term differentiation.
Likewise, we can integrate a convergent power series term-by-term:
f (x) d x =
∞
k=0
bk(x− c)kd x =∞
k=0
bk
(x− c)kd x
=∞
k=0
bk
(x− c)k+1 k+ 1 + K,
Integrating a power series
where the radius of convergence of the resulting series is again r and where K is a constant of integration. The proof of these two results can be found in a text on advanced calculus.
It’s important to recognize that these two results are not obvious. They are not simply an application of the rule that a derivative or integral of a sum is simply the sum of the derivatives or integrals, respectively, since a series is not a sum, but rather, a limit of a sum.
(What’s the difference, anyway?) Further, these results are true for power series, but are not true for series in general.
We summarize the term-by-term differentiation and integration of power series in the following.
Suppose that
f (x)=∞
k=0
bk(x− c)k, where the radius of convergence is r > 0. Then,
f(x)=∞
k=0
kbk(x− c)k−1 and
f (x) d x =∞
k=0
bk
k+ 1(x− c)k+1+ K, where both of these series also have radius of convergence r .
EXAMPLE 6.5 A Convergent Series Whose Series of Derivatives Diverges Find the interval of convergence of the series∞
k=1
sin (k3x)
k2 and show that the series of derivatives does not converge for any x.
Solution Notice that
sin (k3x) k2
≤ 1
k2, for all x, since|sin (k3x)| ≤ 1. Further, ∞
k=1
1
k2 is a convergent p-series ( p= 2 > 1) and so, it follows from the Comparison Test that∞
k=0
sin (k3x)
k2 converges absolutely, for all x. On the other hand, the series of derivatives (found by differentiating the series term-by-term) is
∞ k=1
d d x
sin (k3x) k2
=∞
k=1
k3cos (k3x)
k2 =∞
k=1
[k cos (k3x)],
which diverges for all x, by the kth-term test for divergence, since the terms do not tend to zero as k→ ∞, for any x.
Keep in mind that∞
k=1
sin (k3x)
k2 is not a power series. (Why not?) The result of exam-ple 6.5 (a convergent series whose series of derivatives diverges) cannot occur with any power series with radius of convergence r> 0.
In example 6.6, we find that once we have a convergent power series representation for a given function, we can use this to obtain power series representations for any number of other functions, by differentiating and integrating the series term by term.
EXAMPLE 6.6 Differentiating and Integrating a Power Series Use the power series∞
k=0(−1)kxkto find power series representations of 1
(1+ x)2, 1 1+ x2 and tan−1x.
Solution Notice that ∞
k=0(−1)kxk =∞
k=0(−x)kis a geometric series with ratio r = −x.
This series converges, then, whenever|r| = |−x| = |x| < 1, to a
1− r = 1
1− (−x) = 1 1+ x. That is, for−1 < x < 1,
1
1+ x =∞
k=0
(−1)kxk. (6.2)
Differentiating both sides of (6.2), we get
−1
(1+ x)2 =∞
k=0
(−1)kkxk−1, for −1 < x < 1.
Multiplying both sides by−1 gives us a new power series representation:
1
(1+ x)2 =∞
k=0
(−1)k+1kxk−1,
valid for−1 < x < 1. Notice that we can also obtain a new power series from (6.2) by substitution. For instance, if we replace x with x2, we get
1
1+ x2 =∞
k=0
(−1)k(x2)k=∞
k=0
(−1)kx2k, (6.3)
valid for−1 < x2< 1 (which is equivalent to having x2< 1 or −1 < x < 1).
Integrating both sides of (6.3) gives us
1
1+ x2d x=∞
k=0
(−1)k
x2kd x=∞
k=0
(−1)kx2k+1
2k+ 1 + c. (6.4)
You should recognize the integral on the left-hand side of (6.4) as tan−1x. That is, tan−1x=∞
k=0
(−1)kx2k+1
2k+ 1 + c, for −1 < x < 1. (6.5) Taking x = 0 gives us
tan−10=∞
k=0
(−1)k02k+1
2k+ 1 + c = c,
so that c= tan−10= 0. Equation (6.5) now gives us a power series representation for tan−1x, namely:
tan−1x=∞
k=0
(−1)kx2k+1
2k+ 1 = x −1 3x3+1
5x5−1
7x7+ · · · , for −1 < x < 1.
Notice that working as in example 6.6, we can produce power series representations of any number of functions. In section 7.7, we present a systematic method for producing power series representations for a wide range of functions.
EXERCISES 7.6
WRITING EXERCISES
1. Power series have the form ∞
k=0
ak(x− c)k. Explain why the farther x is from c, the larger the terms of the series are and the less likely the series is to converge. Describe how this general trend relates to the radius of convergence.
2. Applying the Ratio Test to∞
k=0
ak(x− c)krequires you to eval-uate lim
k→∞
ak+1
ak
(x− c)
. For x = c, this limit equals 0 and the series converges. As x increases or decreases,|x − c| in-creases. If the series has a finite radius of convergence r> 0, what is the value of the limit when |x − c| = r? Explain how the limit changes when|x − c| < r and |x − c| > r and
how this determines the convergence or divergence of the series.
3. As shown in example 6.2,∞
k=0
10k
k! (x− 1)kconverges for all x.
If x= 1001, the value of (x − 1)k= 1000kgets very large very fast, as k increases. Explain why, for the series to converge, the value of k! must get large faster than 1000k. To illustrate how fast the factorial grows, compute 50!, 100! and 200! (if your calculator can handle these).
4. In a power series representation of√
x+ 1 about c = 0, ex-plain why the radius of convergence cannot be greater than 1.
(Think about the domain of√ x+ 1.)
In exercises 1–10, find a power series representation of f (x) about c 0 (refer to example 6.6). Also, determine the radius and interval of convergence and graph f (x) together with the partial sums
3 k0
akxkand
6 k0
akxk.
1. f (x)= 2
1− x 2. f (x)= 3
x− 1 3. f (x)= 3
1+ x2 4. f (x)= 2
1− x2 5. f (x)= 2x
1− x3 6. f (x)= 3x
1+ x2 7. f (x)= 4
1+ 4x 8. f (x)= 3
1− 4x 9. f (x)= 2
4+ x 10. f (x)= 3
6− x
In exercises 11–16, determine the interval of convergence and the function to which the given power series converges.
11.
∞ k=0
(x+ 2)k 12.
∞ k=0
(x− 3)k
13.
∞ k=0
(2x− 1)k 14.
∞ k=0
(3x+ 1)k
15.
∞ k=0
(−1)k x
2
k
16.
∞ k=0
3 x
4
k
In exercises 17–34, determine the radius and interval of convergence.
17.
∞ k=0
2k
k!(x− 2)k 18.
∞ k=0
3k k!xk
19.
∞ k=0
k
4kxk 20.
∞ k=0
k 2kxk 21.
∞ k=1
(−1)k
k3k (x− 1)k 22.
∞ k=1
(−1)k+1 k4k (x+ 2)k 23.
∞ k=0
k!(x+ 1)k 24.
∞ k=0
k!(x− 2)k
25.
∞ k=1
1
k(x− 1)k 26.
∞ k=0
k(x− 2)k
27.
∞ k=0
k2(x− 3)k 28.
∞ k=1
1 k2(x+ 2)k 29.
∞ k=0
k!
(2k)!xk 30.
∞ k=0
(k!)2 (2k)!xk 31.
∞ k=1
2k
k2(x+ 2)k 32.
∞ k=0
k2 k!(x+ 1)k 33.
∞ k=1
4k
√kxk 34.
∞ k=1
(−1)kxk
√k
In exercises 35–42, find a power series representation and ra-dius of convergence by integrating or differentiating one of the series from exercises 1–10.
35. f (x)= 3 tan−1x 36. f (x)= 2 ln (1 − x) 37. f (x)= 2x
(1− x2)2 38. f (x)= 3 (x− 1)2 39. f (x)= ln (1 + x2) 40. f (x)= ln (1 + 4x) 41. f (x)= 1
(1+ 4x)2 42. f (x)= 2
(4+ x)2 In exercises 43–46, find the interval of convergence of the (non-power) series and the corresponding series of derivatives.
43.
∞ k=1
cos (k3x)
k2 44.
∞ k=1
cos (x/k) k
45.
∞ k=0
ekx 46.
∞ k=0
e−2kx
47. For any constants a and b> 0, determine the interval and ra-dius of convergence of∞
k=0
(x− a)k bk . 48. Prove that if ∞
k=0
akxk has radius of convergence r , with 0<
r< ∞, then∞
k=0akx2khas radius of convergence√ r .
49. If∞
k=0
akxkhas radius of convergence r , with 0< r < ∞, de-termine the radius of convergence of ∞
k=0
ak(x− c)k for any constant c.
50. If∞
k=0
akxkhas radius of convergence r , with 0< r < ∞, deter-mine the radius of convergence of∞
k=0
ak
x b
k
for any constant b = 0.
51. Show that f (x)= x+ 1 (1− x)2 =
2x 1−x+ 1
1− x has the power se-ries representation f (x)= 1 + 3x + 5x2+ 7x3+ 9x4+ · · · . Find the radius of convergence. Set x= 1
1000and discuss the interesting decimal representation of1,001,000
998,001 . 52. Use long division to show that 1
1− x = 1 + x + x2+ x3+ · · · .
53. Even great mathematicians can make mistakes. Leonhard Euler